keirp/tiny_math · Datasets at Hugging Face

url stringlengths 17 172	text stringlengths 44 1.14M	metadata stringlengths 820 832
http://mathhelpforum.com/statistics/167501-probability-check.html	# Thread: 1. ## Probability check I came across this task, and I could not solve it furthermore. Can anyone check my equation and help me solve it? -The probability to succeed in a random test is 0.8. It's known that if N people attending to the test the probability that all of them will succeed is 0.4096. A- Find N. I was only able to do this equation (Bernouli's table) $(N/K)(0.8)^k(0.2)$ $^n^-K=0.4096$ Can anyone help me solving furthermore with this equation please. that's ^n-k 2. Originally Posted by Mathematicsfan I came across this task, and I could not solve it furthermore. Can anyone check my equation and help me solve it? -The probability to succeed in a random test is 0.8. It's known that if N people attending to the test the probability that all of them will succeed is 0.4096. A- Find N. I was only able to do this equation (Bernouli's table) $(N/K)(0.8)^k(0.2)$ $^n^-K=0.4096$ Can anyone help me solving furthermore with this equation please. that's ^n-k $(0.8)^N=0.4096$ Can you solve for N? 3. No, I don't know I tried desperately. can you guide me ? 4. Originally Posted by mathematicsfan no, i don't know i tried desperately. Can you guide me ? $(0.8)^N=0.4096$ $(0.8)^N=(0.8)^4$ Find N. 5. How did you get 4? cause that's the answer . and where did the 0.2 ^n-k go ? and N\K ? 6. Originally Posted by Mathematicsfan How did you get 4? cause that's the answer . and where did the 0.2 ^n-k go ? and N\K ? This is a very simple question which does not require the use of Bernoulli's tables. You're probably overthinking it. 7. then i don't know how to solve it. can you help me with what to do next ? 8. Originally Posted by Mathematicsfan then i don't know how to solve it. can you help me with what to do next ? I solved it for you in post #4. N=4. 9. lolllllllllllll I got it lollll this is so easy wow.w I just had to play with the Powers / lol thanks a lot #### Search Tags View Tag Cloud Copyright © 2005-2013 Math Help Forum. All rights reserved.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9689220190048218, "perplexity_flag": "middle"}
http://nrich.maths.org/public/leg.php?code=-38&cl=2&cldcmpid=6745	# Search by Topic #### Resources tagged with Permutations similar to Weekly Problem 45 - 2009: Filter by: Content type: Stage: Challenge level: ##### Other tags that relate to Weekly Problem 45 - 2009 Place value. Mathematical reasoning & proof. Working systematically. Addition & subtraction. Algebra - generally. Number - generally. Factors and multiples. Combinations. Properties of numbers. Permutations. ### There are 13 results Broad Topics > Decision Mathematics and Combinatorics > Permutations ### Six Times Five ##### Stage: 3 Challenge Level: How many six digit numbers are there which DO NOT contain a 5? ### Even Up ##### Stage: 3 Challenge Level: Consider all of the five digit numbers which we can form using only the digits 2, 4, 6 and 8. If these numbers are arranged in ascending order, what is the 512th number? ### And So on and So On ##### Stage: 3 Challenge Level: If you wrote all the possible four digit numbers made by using each of the digits 2, 4, 5, 7 once, what would they add up to? ### Flagging ##### Stage: 3 Challenge Level: How many tricolour flags are possible with 5 available colours such that two adjacent stripes must NOT be the same colour. What about 256 colours? ### Permute It ##### Stage: 3 Challenge Level: Take the numbers 1, 2, 3, 4 and 5 and imagine them written down in every possible order to give 5 digit numbers. Find the sum of the resulting numbers. ### Power Crazy ##### Stage: 3 Challenge Level: What can you say about the values of n that make $7^n + 3^n$ a multiple of 10? Are there other pairs of integers between 1 and 10 which have similar properties? ### Painting Cubes ##### Stage: 3 Challenge Level: Imagine you have six different colours of paint. You paint a cube using a different colour for each of the six faces. How many different cubes can be painted using the same set of six colours? ### Bell Ringing ##### Stage: 3 Challenge Level: Suppose you are a bellringer. Can you find the changes so that, starting and ending with a round, all the 24 possible permutations are rung once each and only once? ### Ding Dong Bell ##### Stage: 3, 4 and 5 The reader is invited to investigate changes (or permutations) in the ringing of church bells, illustrated by braid diagrams showing the order in which the bells are rung. ### Card Shuffle ##### Stage: 3 and 4 This article for students and teachers tries to think about how long would it take someone to create every possible shuffle of a pack of cards, with surprising results. ### Shuffle Shriek ##### Stage: 3 Challenge Level: Can you find all the 4-ball shuffles? ### Master Minding ##### Stage: 3 Challenge Level: Your partner chooses two beads and places them side by side behind a screen. What is the minimum number of guesses you would need to be sure of guessing the two beads and their positions? ### Euromaths ##### Stage: 3 Challenge Level: How many ways can you write the word EUROMATHS by starting at the top left hand corner and taking the next letter by stepping one step down or one step to the right in a 5x5 array? The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8663181066513062, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/52524/is-it-part-of-special-relativity-that-mass-possessing-energy-is-more-dense/52526	# Is it part of special relativity that mass possessing energy is more dense? I was reading http://www.edge.org/3rd_culture/hillis/hillis_p2.html and it says that a charged battery weighs more than a dead one or a rotating object weighs more than a stationary one (i.e. mass containing energy weighs more than mass that doesn't). Is this true in special relativity? Because the energy would be confined to the area of the original mass, wouldn't it become more dense if it is true? Also, if the mass we are speaking of does become more gravity-sensitive on energy gain, does it actually gain physical mass (protons, electrons and neutrons)?? - ## 2 Answers Is this true in special relativity? Well, it is a consequence of special relativity. So yes, I guess. Because the energy would be confined to the area of the original mass, wouldn't it become more dense if it is true? It is not true in the general that the volume of the mass remains the same. There are shortening effects as well. For example in the linear case: if you just consider a linear mass with rest mass $m_0$ and rest length $l_0$, then its linear mass density is $\mu_0=m_0/l_0$. And if the mass is now moving along its length at velocity $v$, then its new mass is, $m = \gamma m_0$ and new length is $l = l_0/\gamma$, where $\gamma=\frac{1}{\sqrt{1-v^2/c^2}}$ is the Lorentz factor. Which makes its new mass density: $\mu = \gamma^2 m_0/l_0$. So the mass density has increased by a quadratic factor. Since energy is simply $mc^2$, the energy density has also increased quadratically. But for a linear mass but with some other shape (say in the shape of a U), it gets more complicated. Say its moving along the $x$-directions, then its length along $x$ will decrease but its length along $y$ and $z$ will stay the same. So the total length might change in some complicated way. Other issues in higher dimensions is that the Lorentz factor for the mass will use the total velocity while the Lorentz factor for length along each direction will use the velocity along that direction. I have no idea how to do the calculation for a rotating body. A rotating sphere would be a really interesting calculation to do. Also, if the mass we are speaking of does become more gravity-sensitive on energy gain, does it actually gain physical mass (protons, electrons and neutrons)? It doesn't actually gain more protons, electrons, neutrons. What happens is that the fundamental particles themselves get 'heavier'. Are you wondering whether energy gains might contribute to gravitational mass but not to inertial mass (I guess that's what you are referring to as physical mass)? As far as we know, those two things are the same, but we don't know why yet. - The reason a rotating object weighs more than a non-rotating one is that weight is a quantity that indicates how strongly gravity is coupling to (interacting with) the object in question. When the object rotates, its energy increases, and since what determines the strength of the gravitational interaction is energy and momentum of the object, not just rest energy (mass), the weight of the object increases. This might be confusing at first because Newton's law of gravitation indicates that the strength of the gravitational interaction between two objects goes like the product of their masses and says nothing about energy, but this law is only approximately true, and general relativity tells us otherwise. The "mass we are speaking of" as you say, does not change. The term "mass" in modern physics is used to describe what people often call "rest mass." The number of atoms in a spinning object does not change. This is true both in special, and in general relativity. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9540835022926331, "perplexity_flag": "head"}
http://www.haskell.org/haskellwiki/index.php?title=User:Michiexile/MATH198/Lecture_1&diff=46142&oldid=30053	# User:Michiexile/MATH198/Lecture 1 ### From HaskellWiki (Difference between revisions) \| \| \| \| \| \|----------------------------------------\|----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------\|-----------------------------------------------------------------\|----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------\| \| \| \| Current revision (02:50, 22 June 2012) (edit) (undo)m ( Typo.) \| \| \| (10 intermediate revisions not shown.) \| \| \| \| \| Line 1: \| \| Line 1: \| \| \| - \| IMPORTANT NOTE: THESE NOTES ARE STILL UNDER DEVELOPMENT. PLEASE WAIT UNTIL AFTER THE FIRST LECTURE WITH HANDING ANYTHING IN, OR TAKING THE NOTES AS READY TO READ. \| + \| This page now includes additional information based on the notes taken in class. Hopefully this will make the notes reasonably complete for everybody. \| \| \| \| \| \| \| - \| ==Welcome, administrativia== \| + \| \| \| \| \| + \| ==Welcome, administratrivia== \| \| \| \| \| \| \| \| I'm Mikael Vejdemo-Johansson. I can be reached in my office 383-BB, especially during my office hours; or by email to mik@math.stanford.edu. \| \| I'm Mikael Vejdemo-Johansson. I can be reached in my office 383-BB, especially during my office hours; or by email to mik@math.stanford.edu. \| \| Line 32: \| \| Line 33: \| \| \| \| * Matrices \| \| * Matrices \| \| \| * Homomorphisms \| \| * Homomorphisms \| \| \| \| + \| \| \| \| \| + \| ===Good references=== \| \| \| \| + \| \| \| \| \| + \| On reserve in the mathematics/CS library are: \| \| \| \| + \| * ''Mac Lane'': '''Categories for the working mathematician''' \| \| \| \| + \| :This book is technical, written for a mathematical audience, and puts in more work than is strictly necessary in many of the definitions. When Awodey and Mac Lane deviate, we will give Awodey priority. \| \| \| \| + \| * ''Awodey'': '''Category Theory''' \| \| \| \| + \| :This book is also available as an ebook, accessible from the Stanford campus network. The coursework webpage has links to the ebook under Materials. \| \| \| \| + \| \| \| \| \| + \| ===Monoids=== \| \| \| \| + \| \| \| \| \| + \| In order to settle notation and ensure everybody's seen a definition before: \| \| \| \| + \| \| \| \| \| + \| \| \| \| \| + \| A ''semigroup'' is a monoid without the requirement for an identity element. \| \| \| \| + \| \| \| \| \| + \| A function <math>f:M\to N</math> is a monoid homomorphism if the following conditions hold: \| \| \| \| + \| * <math>f(\emptyset) = \emptyset</math> \| \| \| \| + \| * <math>f(mm') = f(m)f(m')</math> \| \| \| \| + \| \| \| \| \| + \| '''Examples''' \| \| \| \| + \| * Any group is a monoid. Thus, specifically, the integers with addition is a monoid. \| \| \| \| + \| * The natural numbers, with addition. \| \| \| \| + \| * Strings <math>L^</math> in an alphabet <math>L</math> is a monoid with string concatenation forming the operation and the empty string the identity. \| \| \| \| + \| Non-empty strings form a semigroup. \| \| \| \| + \| \| \| \| \| + \| \| \| \| \| + \| Awodey: p. 10. \| \| \| \| + \| \| \| \| \| + \| ===Partially ordered set=== \| \| \| \| + \| \| \| \| \| + \| '''Definition''' A ''partially ordered set'', or a ''partial order'', or a ''poset'' is a set <math>P</math> equipped with a binary relation <math>\leq</math> which is: \| \| \| \| + \| * Reflexive: <math>x\leq x</math> for all <math>x\in P</math> \| \| \| \| + \| * Anti-symmetric: <math>x\leq y</math> and <math>y\leq x</math> implies <math>x=y</math> for all <math>x,y\in P</math>. \| \| \| \| + \| * Transitive: <math>x\leq y</math> and <math>y\leq z</math> implies <math>x\leq z</math> for all <math>x,y,z\in P</math>. \| \| \| \| + \| \| \| \| \| + \| If <math>x\leq y</math> or <math>y\leq x</math>, we call <math>x,y</math> ''comparable''. Otherwise we call them ''incomparable''. A poset where all elements are mutually comparable is called a ''totally ordered set'' or a ''total order''. \| \| \| \| + \| \| \| \| \| + \| If we drop the requirement for anti-symmetry, we get a ''pre-ordered set'' or a ''pre-order''. \| \| \| \| + \| \| \| \| \| + \| If we have several posets, we may indicate which poset we're comparing ''in'' byindicating the poset as a subscript to the relation symbol. \| \| \| \| + \| \| \| \| \| + \| A ''monotonic'' map of posets is a function <math>f:P\to Q</math> such that <math>x\leq_P y</math> implies <math>f(x)\leq_Q f(y)</math>. \| \| \| \| + \| \| \| \| \| + \| '''Examples''' \| \| \| \| + \| * The reals, natural numbers, integers all are posets with the usual comparison relation. A poset in which all elements are comparable. \| \| \| \| + \| * The natural numbers, excluding 0, form a poset with <math>a\leq b</math> if <math>a\|b</math>. \| \| \| \| + \| * Any family of subsets of a given set form a poset with the order given by inclusion. \| \| \| \| + \| \| \| \| \| + \| \| \| \| \| + \| Awodey: p. 6. Preorders are defined on page 8-9. \| \| \| \| \| \| \| \| ==Category== \| \| ==Category== \| \| \| \| + \| \| \| \| \| + \| Awodey has a slightly different exposition. Relevant pages in Awodey for this lecture are: sections 1.1-1.4 (except Def. 1.2), 1.6-1.8. \| \| \| \| \| \| \| \| ===Graphs=== \| \| ===Graphs=== \| \| Line 44: \| \| Line 101: \| \| \| \| \| \| \| \| \| We shall not, in general, require either of the collections to be a set, but will happily accept larger collections; dealing with set-theoretical paradoxes as and when we have to. A graph where both nodes and arrows are sets shall be called ''small''. A graph where either is a class shall be called ''large''. \| \| We shall not, in general, require either of the collections to be a set, but will happily accept larger collections; dealing with set-theoretical paradoxes as and when we have to. A graph where both nodes and arrows are sets shall be called ''small''. A graph where either is a class shall be called ''large''. \| \| - \| \| + \| 0 \| \| \| If both <math>G_0</math> and <math>G_1</math> are finite, the graph is called ''finite'' too. \| \| If both <math>G_0</math> and <math>G_1</math> are finite, the graph is called ''finite'' too. \| \| \| \| \| \| \| Line 74: \| \| Line 131: \| \| \| \| ===Categories=== \| \| ===Categories=== \| \| \| \| \| \| \| - \| We now are ready to define a category. A ''category'' is a graph <math>C</math> equipped with an associative composition operation <math>\circ:G_2\to G_1</math>, and an identity element for composition <math>1_x</math> for each node <math>x</math> of the graph. \| + \| We now are ready to define a category. A ''category'' is a graph <math>G</math> equipped with an associative composition operation <math>\circ:G_2\to G_1</math>, and an identity element for composition <math>1_x</math> for each node <math>x</math> of the graph. \| \| \| \| \| \| \| \| Note that <math>G_2</math> can be viewed as a subset of <math>G_1\times G_1</math>, the set of all pairs of arrows. It is intentional that we define the composition operator on only a subset of the set of all pairs of arrows - the composable pairs. Whenever you'd want to compose two arrows that don't line up to a path, you'll get nonsense, and so any statement about the composition operator has an implicit "whenever defined" attached to it. \| \| Note that <math>G_2</math> can be viewed as a subset of <math>G_1\times G_1</math>, the set of all pairs of arrows. It is intentional that we define the composition operator on only a subset of the set of all pairs of arrows - the composable pairs. Whenever you'd want to compose two arrows that don't line up to a path, you'll get nonsense, and so any statement about the composition operator has an implicit "whenever defined" attached to it. \| \| Line 109: \| \| Line 166: \| \| \| \| \| \| \| \| \| A category with objects a collection of sets and morphisms a selection from all possible set-valued functions such that the identity morphism for each object is a morphism, and composition in the category is just composition of functions is called ''concrete''. Concrete categories form a very rich source of examples, though far from all categories are concrete. \| \| A category with objects a collection of sets and morphisms a selection from all possible set-valued functions such that the identity morphism for each object is a morphism, and composition in the category is just composition of functions is called ''concrete''. Concrete categories form a very rich source of examples, though far from all categories are concrete. \| \| \| \| + \| \| \| \| \| + \| Again, the Wikipedia page on Category (mathematics) [[http://en.wikipedia.org/wiki/Category_%28mathematics%29]] is a good starting point for many things we will be looking at throughout this course. \| \| \| \| \| \| \| \| ===New Categories from old=== \| \| ===New Categories from old=== \| \| Line 137: \| \| Line 196: \| \| \| \| # We can also form the category with objects <math>\langle A,B\rangle</math> for every pair of objects <math>A\in C, B\in D</math>. A morphism in <math>Hom(\langle A,B\rangle,\langle A',B'\rangle)</math> is simply a pair <math>\langle f:A\to A',g:B\to B'\rangle</math>. Composition is defined componentwise. This category is the categorical correspondent to the cartesian ''product'', and we denot it by <math>C\times D</math>. \| \| # We can also form the category with objects <math>\langle A,B\rangle</math> for every pair of objects <math>A\in C, B\in D</math>. A morphism in <math>Hom(\langle A,B\rangle,\langle A',B'\rangle)</math> is simply a pair <math>\langle f:A\to A',g:B\to B'\rangle</math>. Composition is defined componentwise. This category is the categorical correspondent to the cartesian ''product'', and we denot it by <math>C\times D</math>. \| \| \| \| \| \| \| - \| In these three constructions - the dual, the product and the coproduct - he arrows in the categories are formal constructions, not functions; even if the original category was given by functions, the result is no longer given by a function. \| + \| In these three constructions - the dual, the product and the coproduct - the arrows in the categories are formal constructions, not functions; even if the original category was given by functions, the result is no longer given by a function. \| \| \| \| \| \| \| \| Given a category <math>C</math> and an object <math>A</math> of that category, we can form the ''slice category'' <math>C/A</math>. Objects in the slice category are arrows <math>B\to A</math> for some object <math>B</math> in <math>C</math>, and an arrow <math>\phi:f\to g</math> is an arrow <math>s(f)\to s(g)</math> such that <math>f=g\circ\phi</math>. Composites of arrows are just the composites in the base category. \| \| Given a category <math>C</math> and an object <math>A</math> of that category, we can form the ''slice category'' <math>C/A</math>. Objects in the slice category are arrows <math>B\to A</math> for some object <math>B</math> in <math>C</math>, and an arrow <math>\phi:f\to g</math> is an arrow <math>s(f)\to s(g)</math> such that <math>f=g\circ\phi</math>. Composites of arrows are just the composites in the base category. \| \| Line 143: \| \| Line 202: \| \| \| \| Notice that the same arrow <math>\phi</math> in the base category <math>C</math> represents potentially many different arrows in <math>C/A</math>: it represents one arrow for each choice of source and target compatible with it. \| \| Notice that the same arrow <math>\phi</math> in the base category <math>C</math> represents potentially many different arrows in <math>C/A</math>: it represents one arrow for each choice of source and target compatible with it. \| \| \| \| \| \| \| - \| There is a dual notion: the ''coslice category'', where the objects are paired with maps <math>A\to B</math>. \| + \| There is a dual notion: the ''coslice category'' <math>A\backslash C</math>, where the objects are paired with maps <math>A\to B</math>. \| \| \| \| \| \| \| - \| Slice categories can be used, among other things, to specify the idea of parametrization. The slice category <math>(C\downarrow A)</math> gives a sense to the idea of ''objects from <math>C</math> labeled by elements of <math>A</math>''. \| + \| Slice categories can be used, among other things, to specify the idea of parametrization. The slice category <math>C/A</math> gives a sense to the idea of ''objects from <math>C</math> labeled by elements of <math>A</math>''. \| \| \| \| \| \| \| \| We get this characterization by interpreting the arrow representing an object as representing its source and a ''type function''. Hence, in a way, the <hask>Typeable</hask> type class in Haskell builds a slice category on an appropriate subcategory of the category of datatypes. \| \| We get this characterization by interpreting the arrow representing an object as representing its source and a ''type function''. Hence, in a way, the <hask>Typeable</hask> type class in Haskell builds a slice category on an appropriate subcategory of the category of datatypes. \| \| Line 151: \| \| Line 210: \| \| \| \| Alternatively, we can phrase the importance of the arrow in a slice categories of, say, Set, by looking at preimages of the slice functions. That way, an object <math>f:B\to A</math> gives us a family of (disjoint) subsets of <math>B</math> ''indexed'' by the elements of <math>A</math>. \| \| Alternatively, we can phrase the importance of the arrow in a slice categories of, say, Set, by looking at preimages of the slice functions. That way, an object <math>f:B\to A</math> gives us a family of (disjoint) subsets of <math>B</math> ''indexed'' by the elements of <math>A</math>. \| \| \| \| \| \| \| - \| Finally, any graph yields a category by just filling in the arrows that are missing. The result is called the ''free category generated by the graph'', and is a concept we will return to in some depth. Free objects have a strict categorical definition, and they serve to give a model of thought for the things they are free objects for. Thus, categories are essentially graphs, possibly with restrictions or relations imposed; and monoids are essentially strings in some alphabet, with restrictions or relatinos. \| + \| Finally, any graph yields a category by just filling in the arrows that are missing. The result is called the ''free category generated by the graph'', and is a concept we will return to in some depth. Free objects have a strict categorical definition, and they serve to give a model of thought for the things they are free objects for. Thus, categories are essentially graphs, possibly with restrictions or relations imposed; and monoids are essentially strings in some alphabet, with restrictions or relations. \| \| \| \| \| \| \| \| ===Examples=== \| \| ===Examples=== \| \| Line 168: \| \| Line 227: \| \| \| \| Sets for objects. Arrows are pairs <math>(S'\subseteq S,f:S'\to T)\in PFn(S,T)</math>. \| \| Sets for objects. Arrows are pairs <math>(S'\subseteq S,f:S'\to T)\in PFn(S,T)</math>. \| \| \| <math>PFn(A,B)</math> is a partially ordered set. <math>(S_f,f)\leq(S_g,g)</math> precisely if <math>S_f\subseteq S_g</math> and <math>f=g\|_{S_f}</math>. \| \| <math>PFn(A,B)</math> is a partially ordered set. <math>(S_f,f)\leq(S_g,g)</math> precisely if <math>S_f\subseteq S_g</math> and <math>f=g\|_{S_f}</math>. \| \| \| \| + \| ** The exposition at Wikipedia uses the construction here: [[http://en.wikipedia.org/wiki/Partial_function]]. \| \| \| * There is an alternative way to define a category of partial functions: For objects, we take sets, and for morphisms <math>S\to T</math>, we take subsets <math>F\subseteq S\times T</math> such that each element in <math>S</math> occurs in at most one pair in the subset. Composition is by an interpretation of these subsets corresponding to the previous description. We'll call this category <math>PFn'</math>. \| \| * There is an alternative way to define a category of partial functions: For objects, we take sets, and for morphisms <math>S\to T</math>, we take subsets <math>F\subseteq S\times T</math> such that each element in <math>S</math> occurs in at most one pair in the subset. Composition is by an interpretation of these subsets corresponding to the previous description. We'll call this category <math>PFn'</math>. \| \| \| * Every partial order is a category. Each hom-set has at most one element. \| \| * Every partial order is a category. Each hom-set has at most one element. \| \| \| Objects are the elements of the poset. Arrows are unique, with <math>A\to B</math> precisely if <math>A\leq B</math>. \| \| Objects are the elements of the poset. Arrows are unique, with <math>A\to B</math> precisely if <math>A\leq B</math>. \| \| - \| * Every monoid is a category. Only one object. \| + \| * Every monoid is a category. Only one object. The elements of the monoid correspond to the endo-arrows of the one object. \| \| - \| *** Kleene closure. Free monoids. \| + \| \| \| \| * The category of Sets and injective functions. \| \| * The category of Sets and injective functions. \| \| \| \| + \| ** Objects are sets. Morphisms are injective functions between the sets. \| \| \| * The category of Sets and surjective functions. \| \| * The category of Sets and surjective functions. \| \| \| \| + \| ** Objects are sets. Morphisms are surjective functions between the sets. \| \| \| * The category of <math>k</math>-vector spaces and linear maps. \| \| * The category of <math>k</math>-vector spaces and linear maps. \| \| \| * The category with objects the natural numbers and <math>Hom(m,n)</math> the set of <math>m\times n</math>-matrices. \| \| * The category with objects the natural numbers and <math>Hom(m,n)</math> the set of <math>m\times n</math>-matrices. \| \| \| \| + \| ** Composition is given by matrix multiplication. \| \| \| * The category of Data Types with Computable Functions. \| \| * The category of Data Types with Computable Functions. \| \| \| Our ideal programming language has: \| \| Our ideal programming language has: \| \| Line 189: \| \| Line 251: \| \| \| \| This allows us to model a functional programming language with a category. \| \| This allows us to model a functional programming language with a category. \| \| \| * The category with objects logical propositions and arrows proofs. \| \| * The category with objects logical propositions and arrows proofs. \| \| - \| \| + \| * The category Rel has objects finite sets and morphisms <math>A\to B</math> being subsets of <math>A\times B</math>. Composition is by <math>(a,c)\in g\circ f</math> if there is some <math>b\in B</math> such that <math>(a,b)\in f, (b,c)\in g</math>. Identity morphism is the diagonal <math>(a,a): a\in A</math>. \| \| \| \| \| \| \| \| \| \| \| \| Line 197: \| \| Line 259: \| \| \| \| ===Homework=== \| \| ===Homework=== \| \| \| \| \| \| \| - \| For a passing mark, a written, acceptable solution to at least 2 of the 5 questions should be given no later than midnight before the next lecture. \| + \| For a passing mark, a written, acceptable solution to at least 3 of the 6 questions should be given no later than midnight before the next lecture. \| \| \| \| \| \| \| \| For each lecture, there will be a few exercises marked with the symbol . These will be more difficult than the other exercises given, will require significant time and independent study, and will aim to complement the course with material not covered in lectures, but nevertheless interesting for the general philosophy of the lecture course. \| \| For each lecture, there will be a few exercises marked with the symbol . These will be more difficult than the other exercises given, will require significant time and independent study, and will aim to complement the course with material not covered in lectures, but nevertheless interesting for the general philosophy of the lecture course. \| \| \| \| \| \| \| - \| # Prove the general associative law: that for any path, and any bracketing of that path, the same composition may be found. \| + \| # Prove the general associative law: that for any path, and any bracketing of that path, the same composition results. \| \| \| \| + \| # Which of the following form categories? Proof and disproof for each: \| \| \| \| + \| #* Objects are finite sets, morphisms are functions such that <math>\|f^{-1}(b)\|\leq 2</math> for all morphisms f, objects B and elements b. \| \| \| \| + \| #* Objects are finite sets, morphisms are functions such that <math>\|f^{-1}(b)\|\geq 2</math> for all morphisms f, objects B and elements b. \| \| \| \| + \| #* Objects are finite sets, morphisms are functions such that <math>\|f^{-1}(b)\|<\infty</math> for all morphisms f, objects B and elements b. \| \| \| \| + \| :Recall that <math>f^{-1}(b)=\{a\in A: f(a)=b\}</math>. \| \| \| # Suppose <math>u:A\to A</math> in some category <math>C</math>. \| \| # Suppose <math>u:A\to A</math> in some category <math>C</math>. \| \| \| ## If <math>g\circ u=g</math> for all <math>g:A\to B</math> in the category, then <math>u=1_A</math>. \| \| ## If <math>g\circ u=g</math> for all <math>g:A\to B</math> in the category, then <math>u=1_A</math>. \| \| \| ## If <math>u\circ h=h</math> for all <math>h:B\to A</math> in the category, then <math>u=1_A</math>. \| \| ## If <math>u\circ h=h</math> for all <math>h:B\to A</math> in the category, then <math>u=1_A</math>. \| \| - \| ## These two results characterize the objects in a category by the properties of their corresponding identity arrows completely. \| + \| ## These two results characterize the objects in a category by the properties of their corresponding identity arrows completely. Specifically, there is a way to rephrase the definition of a category such that everything is stated in terms of arrows. \| \| \| # For as many of the examples given as you can, prove that they really do form a category. Passing mark is at least 60% of the given examples. \| \| # For as many of the examples given as you can, prove that they really do form a category. Passing mark is at least 60% of the given examples. \| \| \| #* Which of the categories are subcategories of which other categories? Which of these are wide? Which are full? \| \| #* Which of the categories are subcategories of which other categories? Which of these are wide? Which are full? \| \| Line 211: \| \| Line 278: \| \| \| \| ## For which sets is the free monoid on that set commutative. \| \| ## For which sets is the free monoid on that set commutative. \| \| \| ## Prove that for any category <math>C</math>, the set <math>Hom(A,A)</math> is a monoid under composition for every object <math>A</math>. \| \| ## Prove that for any category <math>C</math>, the set <math>Hom(A,A)</math> is a monoid under composition for every object <math>A</math>. \| \| \| \| + \| :For details on the construction of a free monoid, see the Wikipedia pages on the Free Monoid [[http://en.wikipedia.org/wiki/Free_monoid]] and on the Kleene star [[http://en.wikipedia.org/wiki/Kleene_star]]. \| \| \| # * Read up on <math>\omega</math>-complete partial orders. Suppose <math>S</math> is some set and <math>\mathfrak P</math> is the set of partial functions <math>S\to S</math> - in other words, an element of <math>\mathfrak P</math> is some pair <math>(S_0,f:S_0\to S)</math> with <math>S_0\subseteq S</math>. We give this set a poset structure by <math>(S_0,f)\leq(S_1,g)</math> precisely if <math>S_0\subseteq S_1</math> and <math>f(s)=g(s)\forall s\in S_0</math>. \| \| # * Read up on <math>\omega</math>-complete partial orders. Suppose <math>S</math> is some set and <math>\mathfrak P</math> is the set of partial functions <math>S\to S</math> - in other words, an element of <math>\mathfrak P</math> is some pair <math>(S_0,f:S_0\to S)</math> with <math>S_0\subseteq S</math>. We give this set a poset structure by <math>(S_0,f)\leq(S_1,g)</math> precisely if <math>S_0\subseteq S_1</math> and <math>f(s)=g(s)\forall s\in S_0</math>. \| \| \| #* Show that <math>\mathfrak P</math> is a strict <math>\omega</math>-CPO. \| \| #* Show that <math>\mathfrak P</math> is a strict <math>\omega</math>-CPO. \| \| Line 216: \| \| Line 284: \| \| \| \| ## <math>k(0) = 1</math> \| \| ## <math>k(0) = 1</math> \| \| \| ## <math>k(n)</math> is defined only if <math>h(n-1)</math> is defined, and then by <math>k(n)=nh(n-1)</math>. \| \| ## <math>k(n)</math> is defined only if <math>h(n-1)</math> is defined, and then by <math>k(n)=nh(n-1)</math>. \| \| - \| :Describe <math>\phi(n\mapsto n^2)</math> and <math>\phi(n\mapsto n^3)</math>. Show that <math>\phi</math> is ''continuous''. Find a fixpoint <math>(S_0,f)</math> of <math>\phi</math> such that any other fixpoint of the same function is less than this one. \| + \| :Describe <math>\phi(n\mapsto n^2)</math> and <math>\phi(n\mapsto n^3)</math>. Show that <math>\phi</math> is ''continuous''. Find a fixpoint <math>(S_0,f)</math> of <math>\phi</math> such that any other fixpoint of the same function is larger than this one. \| \| \| :Find a continuous endofunction on some <math>\omega</math>-CPO that has the fibonacci function <math>F(0)=0, F(1)=1, F(n)=F(n-1)+F(n-2)</math> as the least fixed point. \| \| :Find a continuous endofunction on some <math>\omega</math>-CPO that has the fibonacci function <math>F(0)=0, F(1)=1, F(n)=F(n-1)+F(n-2)</math> as the least fixed point. \| ## Current revision This page now includes additional information based on the notes taken in class. Hopefully this will make the notes reasonably complete for everybody. ## 1 Welcome, administratrivia I'm Mikael Vejdemo-Johansson. I can be reached in my office 383-BB, especially during my office hours; or by email to mik@math.stanford.edu. I encourage, strongly, student interactions. I will be out of town September 24 - 29. I will monitor forum and email closely, and recommend electronic ways of getting in touch with me during this week. I will be back again in time for the office hours on the 30th. ## 2 Introduction ### 2.1 Why this course? An introduction to Haskell will usually come with pointers toward Category Theory as a useful tool, though not with much more than the mention of the subject. This course is intended to fill that gap, and provide an introduction to Category Theory that ties into Haskell and functional programming as a source of examples and applications. ### 2.2 What will we cover? The definition of categories, special objects and morphisms, functors, natural transformation, (co-)limits and special cases of these, adjunctions, freeness and presentations as categorical constructs, monads and Kleisli arrows, recursion with categorical constructs. Maybe, just maybe, if we have enough time, we'll finish with looking at the definition of a topos, and how this encodes logic internal to a category. Applications to fuzzy sets. ### 2.3 What do we require? Our examples will be drawn from discrete mathematics, logic, Haskell programming and linear algebra. I expect the following concepts to be at least vaguely familiar to anyone taking this course: • Sets • Functions • Permutations • Groups • Partially ordered sets • Vector spaces • Linear maps • Matrices • Homomorphisms ### 2.4 Good references On reserve in the mathematics/CS library are: • Mac Lane: Categories for the working mathematician This book is technical, written for a mathematical audience, and puts in more work than is strictly necessary in many of the definitions. When Awodey and Mac Lane deviate, we will give Awodey priority. • Awodey: Category Theory This book is also available as an ebook, accessible from the Stanford campus network. The coursework webpage has links to the ebook under Materials. ### 2.5 Monoids In order to settle notation and ensure everybody's seen a definition before: Definition A monoid is a set M equipped with a binary associative operation * (in Haskell: mappend ) and an identity element $\emptyset$ (in Haskell: mempty ). A semigroup is a monoid without the requirement for an identity element. A function $f:M\to N$ is a monoid homomorphism if the following conditions hold: • $f(\emptyset) = \emptyset$ • f(m * m') = f(m) * f(m') Examples • Any group is a monoid. Thus, specifically, the integers with addition is a monoid. • The natural numbers, with addition. • Strings L * in an alphabet L is a monoid with string concatenation forming the operation and the empty string the identity. • Non-empty strings form a semigroup. Awodey: p. 10. ### 2.6 Partially ordered set Definition A partially ordered set, or a partial order, or a poset is a set P equipped with a binary relation $\leq$ which is: • Reflexive: $x\leq x$ for all $x\in P$ • Anti-symmetric: $x\leq y$ and $y\leq x$ implies x = y for all $x,y\in P$. • Transitive: $x\leq y$ and $y\leq z$ implies $x\leq z$ for all $x,y,z\in P$. If $x\leq y$ or $y\leq x$, we call x,y comparable. Otherwise we call them incomparable. A poset where all elements are mutually comparable is called a totally ordered set or a total order. If we drop the requirement for anti-symmetry, we get a pre-ordered set or a pre-order. If we have several posets, we may indicate which poset we're comparing in byindicating the poset as a subscript to the relation symbol. A monotonic map of posets is a function $f:P\to Q$ such that $x\leq_P y$ implies $f(x)\leq_Q f(y)$. Examples • The reals, natural numbers, integers all are posets with the usual comparison relation. A poset in which all elements are comparable. • The natural numbers, excluding 0, form a poset with $a\leq b$ if a \| b. • Any family of subsets of a given set form a poset with the order given by inclusion. Awodey: p. 6. Preorders are defined on page 8-9. ## 3 Category Awodey has a slightly different exposition. Relevant pages in Awodey for this lecture are: sections 1.1-1.4 (except Def. 1.2), 1.6-1.8. ### 3.1 Graphs We recall the definition of a (directed) graph. A graph G is a collection of edges (arrows) and vertices (nodes). Each edge is assigned a source node and a target node. $source \to target$ Given a graph G, we denote the collection of nodes by G0 and the collection of arrows by G1. These two collections are connected, and the graph given its structure, by two functions: the source function $s:G_1\to G_0$ and the target function $t:G_1\to G_0$. We shall not, in general, require either of the collections to be a set, but will happily accept larger collections; dealing with set-theoretical paradoxes as and when we have to. A graph where both nodes and arrows are sets shall be called small. A graph where either is a class shall be called large. 0 If both G0 and G1 are finite, the graph is called finite too. The empty graph has $G_0 = G_1 = \emptyset$. A discrete graph has $G_1=\emptyset$. A complete graph has $G_1 = \{ (v,w) \| v,w\in G_0\}$. A simple graph has at most one arrow between each pair of nodes. Any relation on a set can be interpreted as a simple graph. • Show some examples. A homomorphism $f:G\to H$ of graphs is a pair of functions $f_0:G_0\to H_0$ and $f_1:G_1\to H_1$ such that sources map to sources and targets map to targets, or in other words: • s(f1(e)) = f0(s(e)) • t(f1(e)) = f0(t(e)) By a path in a graph G from the node x to the node y of length k, we mean a sequence of edges $(f_1,f_2,\dots,f_k)$ such that: • s(f1) = x • t(fk) = y • s(fi) = t(fi − 1) for all other i. Paths with start and end point identical are called closed. For any node x, there is a unique closed path () starting and ending in x of length 0. For any edge f, there is a unique path from s(f) to t(f) of length 1: (f). We denote by Gk the set of paths in G of length k. ### 3.2 Categories We now are ready to define a category. A category is a graph G equipped with an associative composition operation $\circ:G_2\to G_1$, and an identity element for composition 1x for each node x of the graph. Note that G2 can be viewed as a subset of $G_1\times G_1$, the set of all pairs of arrows. It is intentional that we define the composition operator on only a subset of the set of all pairs of arrows - the composable pairs. Whenever you'd want to compose two arrows that don't line up to a path, you'll get nonsense, and so any statement about the composition operator has an implicit "whenever defined" attached to it. The definition is not quite done yet - this composition operator, and the identity arrows both have a few rules to fulfill, and before I state these rules, there are some notation we need to cover. #### 3.2.1 Backwards! If we have a path given by the arrows (f,g) in G2, we expect $f:A\to B$ and $g:B\to C$ to compose to something that goes $A\to C$. The origin of all these ideas lies in geometry and algebra, and so the abstract arrows in a category are supposed to behave like functions under function composition, even though we don't say it explicitly. Now, we are used to writing function application as f(x) - and possibly, from Haskell, as f x . This way, the composition of two functions would read g(f(x)). On the other hand, the way we write our paths, we'd read f then g. This juxtaposition makes one of the two ways we write things seem backwards. We can resolve it either by making our paths in the category go backwards, or by reversing how we write function application. In the latter case, we'd write x.f, say, for the application of f to x, and then write x.f.g for the composition. It all ends up looking a lot like Reverse Polish Notation, and has its strengths, but feels unnatural to most. It does, however, have the benefit that we can write out function composition as $(f,g) \mapsto f.g$ and have everything still make sense in all notations. In the former case, which is the most common in the field, we accept that paths as we read along the arrows and compositions look backwards, and so, if $f:A\to B$ and $g:B\to C$, we write $g\circ f:A\to C$, remembering that elements are introduced from the right, and the functions have to consume the elements in the right order. The existence of the identity map can be captured in a function language as well: it is the existence of a function $u:G_0\to G_1$. Now for the remaining rules for composition. Whenever defined, we expect associativity - so that $h\circ(g\circ f)=(h\circ g)\circ f$. Furthermore, we expect: 1. Composition respects sources and targets, so that: • $s(g\circ f) = s(f)$ • $t(g\circ f) = t(g)$ 2. s(u(x)) = t(u(x)) = x In a category, arrows are also called morphisms, and nodes are also called objects. This ties in with the algebraic roots of the field. We denote by HomC(A,B), or if C is obvious from context, just Hom(A,B), the set of all arrows from A to B. This is the hom-set or set of morphisms, and may also be denoted C(A,B). If a category is large or small or finite as a graph, it is called a large/small/finite category. A category with objects a collection of sets and morphisms a selection from all possible set-valued functions such that the identity morphism for each object is a morphism, and composition in the category is just composition of functions is called concrete. Concrete categories form a very rich source of examples, though far from all categories are concrete. Again, the Wikipedia page on Category (mathematics) [[3]] is a good starting point for many things we will be looking at throughout this course. ### 3.3 New Categories from old As with most other algebraic objects, one essential part of our tool box is to take known objects and form new examples from them. This allows us generate a wealth of examples from the ones that shape our intuition. Typical things to do here would be to talk about subobjects, products and coproducts, sometimes obvious variations on the structure, and what a typical object looks like. Remember from linear algebra how subspaces, cartesian products (which for finite-dimensional vectorspaces covers both products and coproducts) and dual spaces show up early, as well as the theorems giving dimension as a complete descriptor of a vectorspace. We'll go through the same sequence here; with some significant small variations. A category D is a subcategory of the category C if: • $D_0\subseteq C_0$ • $D_1\subseteq C_1$ • D1 contains 1X for all $X\in D_0$ • sources and targets of all the arrows in D1 are all in D0 • the composition in D is the restriction of the composition in C. Written this way, it does look somewhat obnoxious. It does become easier though, with the realization - studied closer in homework exercise 2 - that the really important part of a category is the collection of arrows. Thus, a subcategory is a subcollection of the collection of arrows - with identities for all objects present, and with at least all objects that the existing arrows imply. A subcategory $D\subseteq C$ is full if D(A,B) = C(A,B) for all objects A,B of D. In other words, a full subcategory is completely determined by the selection of objects in the subcategory. A subcategory $D\subseteq C$ is wide if the collection of objects is the same in both categories. Hence, a wide subcategory picks out a subcollection of the morphisms. The dual of a category is to a large extent inspired by vector space duals. In the dual C * of a category C, we have the same objects, and the morphisms are given by the equality C * (A,B) = C(B,A) - every morphism from C is present, but it goes in the wrong direction. Dualizing has a tendency to add the prefix co- when it happens, so for instance coproducts are the dual notion to products. We'll return to this construction many times in the course. Given two categories C,D, we can combine them in several ways: 1. We can form the category that has as objects the disjoint union of all the objects of C and D, and that sets $Hom(A,B)=\emptyset$ whenever A,B come from different original categories. If A,B come from the same original category, we simply take over the homset from that category. This yields a categorical coproduct, and we denote the result by C + D. Composition is inherited from the original categories. 2. We can also form the category with objects $\langle A,B\rangle$ for every pair of objects $A\in C, B\in D$. A morphism in $Hom(\langle A,B\rangle,\langle A',B'\rangle)$ is simply a pair $\langle f:A\to A',g:B\to B'\rangle$. Composition is defined componentwise. This category is the categorical correspondent to the cartesian product, and we denot it by $C\times D$. In these three constructions - the dual, the product and the coproduct - the arrows in the categories are formal constructions, not functions; even if the original category was given by functions, the result is no longer given by a function. Given a category C and an object A of that category, we can form the slice category C / A. Objects in the slice category are arrows $B\to A$ for some object B in C, and an arrow $\phi:f\to g$ is an arrow $s(f)\to s(g)$ such that $f=g\circ\phi$. Composites of arrows are just the composites in the base category. Notice that the same arrow φ in the base category C represents potentially many different arrows in C / A: it represents one arrow for each choice of source and target compatible with it. There is a dual notion: the coslice category $A\backslash C$, where the objects are paired with maps $A\to B$. Slice categories can be used, among other things, to specify the idea of parametrization. The slice category C / A gives a sense to the idea of objects from C labeled by elements of A. We get this characterization by interpreting the arrow representing an object as representing its source and a type function. Hence, in a way, the Typeable type class in Haskell builds a slice category on an appropriate subcategory of the category of datatypes. Alternatively, we can phrase the importance of the arrow in a slice categories of, say, Set, by looking at preimages of the slice functions. That way, an object $f:B\to A$ gives us a family of (disjoint) subsets of B indexed by the elements of A. Finally, any graph yields a category by just filling in the arrows that are missing. The result is called the free category generated by the graph, and is a concept we will return to in some depth. Free objects have a strict categorical definition, and they serve to give a model of thought for the things they are free objects for. Thus, categories are essentially graphs, possibly with restrictions or relations imposed; and monoids are essentially strings in some alphabet, with restrictions or relations. ### 3.4 Examples • The empty category. • No objects, no morphisms. • The one object/one arrow category 1. • A single object and its identity arrow. • The categories 2 and 1 + 1. • Two objects, A,B with identity arrows and a unique arrow $A\to B$. • The category Set of sets. • Sets for objects, functions for arrows. • The catgeory FSet of finite sets. • Finite sets for objects, functions for arrows. • The category PFn of sets and partial functions. • Sets for objects. Arrows are pairs $(S'\subseteq S,f:S'\to T)\in PFn(S,T)$. • PFn(A,B) is a partially ordered set. $(S_f,f)\leq(S_g,g)$ precisely if $S_f\subseteq S_g$ and $f=g\|_{S_f}$. • The exposition at Wikipedia uses the construction here: [[4]]. • There is an alternative way to define a category of partial functions: For objects, we take sets, and for morphisms $S\to T$, we take subsets $F\subseteq S\times T$ such that each element in S occurs in at most one pair in the subset. Composition is by an interpretation of these subsets corresponding to the previous description. We'll call this category PFn'. • Every partial order is a category. Each hom-set has at most one element. • Objects are the elements of the poset. Arrows are unique, with $A\to B$ precisely if $A\leq B$. • Every monoid is a category. Only one object. The elements of the monoid correspond to the endo-arrows of the one object. • The category of Sets and injective functions. • Objects are sets. Morphisms are injective functions between the sets. • The category of Sets and surjective functions. • Objects are sets. Morphisms are surjective functions between the sets. • The category of k-vector spaces and linear maps. • The category with objects the natural numbers and Hom(m,n) the set of $m\times n$-matrices. • Composition is given by matrix multiplication. • The category of Data Types with Computable Functions. • Our ideal programming language has: • Primitive data types. • Constants of each primitive type. • Operations, given as functions between types. • Constructors, producing elements from data types, and producing derived data types and operations. • We will assume that the language is equipped with • A do-nothing operation for each data type. Haskell has id . • An empty type 1, with the property that each type has exactly one function to this type. Haskell has () . We will use this to define the constants of type t as functions $1\to t$. Thus, constants end up being 0-ary functions. • A composition constructor, taking an operator $f:A\to B$ and another operator $g:B\to C$ and producing an operator $g\circ f:A\to C$. Haskell has (.) . • This allows us to model a functional programming language with a category. • The category with objects logical propositions and arrows proofs. • The category Rel has objects finite sets and morphisms $A\to B$ being subsets of $A\times B$. Composition is by $(a,c)\in g\circ f$ if there is some $b\in B$ such that $(a,b)\in f, (b,c)\in g$. Identity morphism is the diagonal $(a,a): a\in A$. ### 3.5 Homework For a passing mark, a written, acceptable solution to at least 3 of the 6 questions should be given no later than midnight before the next lecture. For each lecture, there will be a few exercises marked with the symbol . These will be more difficult than the other exercises given, will require significant time and independent study, and will aim to complement the course with material not covered in lectures, but nevertheless interesting for the general philosophy of the lecture course. 1. Prove the general associative law: that for any path, and any bracketing of that path, the same composition results. 2. Which of the following form categories? Proof and disproof for each: • Objects are finite sets, morphisms are functions such that $\|f^{-1}(b)\|\leq 2$ for all morphisms f, objects B and elements b. • Objects are finite sets, morphisms are functions such that $\|f^{-1}(b)\|\geq 2$ for all morphisms f, objects B and elements b. • Objects are finite sets, morphisms are functions such that $\|f^{-1}(b)\|<\infty$ for all morphisms f, objects B and elements b. Recall that $f^{-1}(b)=\{a\in A: f(a)=b\}$. 1. Suppose $u:A\to A$ in some category C. 1. If $g\circ u=g$ for all $g:A\to B$ in the category, then u = 1A. 2. If $u\circ h=h$ for all $h:B\to A$ in the category, then u = 1A. 3. These two results characterize the objects in a category by the properties of their corresponding identity arrows completely. Specifically, there is a way to rephrase the definition of a category such that everything is stated in terms of arrows. 2. For as many of the examples given as you can, prove that they really do form a category. Passing mark is at least 60% of the given examples. • Which of the categories are subcategories of which other categories? Which of these are wide? Which are full? 3. For this question, all parts are required: 1. For which sets is the free monoid on that set commutative. 2. Prove that for any category C, the set Hom(A,A) is a monoid under composition for every object A. For details on the construction of a free monoid, see the Wikipedia pages on the Free Monoid [[5]] and on the Kleene star [[6]]. 1. Read up on ω-complete partial orders. Suppose S is some set and $\mathfrak P$ is the set of partial functions $S\to S$ - in other words, an element of $\mathfrak P$ is some pair $(S_0,f:S_0\to S)$ with $S_0\subseteq S$. We give this set a poset structure by $(S_0,f)\leq(S_1,g)$ precisely if $S_0\subseteq S_1$ and $f(s)=g(s)\forall s\in S_0$. • Show that $\mathfrak P$ is a strict ω-CPO. • An element x of S is a fixpoint of $f:S\to S$ if f(x) = x. Let $\mathfrak N$ be the ω-CPO of partially defined functions on the natural numbers. We define a function $\phi:\mathfrak N\to\mathfrak N$ by sending some $h:\mathbb N\to\mathbb N$ to a function k defined by 1. k(0) = 1 2. k(n) is defined only if h(n − 1) is defined, and then by k(n) = n * h(n − 1). Describe $\phi(n\mapsto n^2)$ and $\phi(n\mapsto n^3)$. Show that φ is continuous. Find a fixpoint (S0,f) of φ such that any other fixpoint of the same function is larger than this one. Find a continuous endofunction on some ω-CPO that has the fibonacci function F(0) = 0,F(1) = 1,F(n) = F(n − 1) + F(n − 2) as the least fixed point. Implement a Haskell function that finds fixed points in an ω-CPO. Implement the two fixed points above as Haskell functions - using the ω-CPO fixed point approach in the implementation. It may well be worth looking at Data.Map to provide a Haskell context for a partial function for this part of the task.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 104, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.869941234588623, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/131921/very-interesting-sequence-problem	# Very Interesting Sequence Problem I need help primarily with the finding a solution, I already came up with an answer. We define an Eden sequence to be a subset of the set $\{1,2,3,4,\ldots ,N\}$. The Eden sequence has three conditions. 1. each of its terms is an element of the set of consecutive integers $\{1,2,3,4,\ldots ,N\}$, 2. the sequence is increasing, and 3. the terms in odd numbered positions are odd and the terms in even numbered positions are even. We then define a function $e(N)$ such that $e(N)$ denotes the number of Eden sequences of the set $\{1,2,3,4,\ldots ,N\}$. If we are given that $e(17)=4180$ and $q(20)=17710$, how would we find $e(18)$ and $e(19)$ using a mathematical approach? I am pretty sure the answers are that $e(18) = 6764$ and $e(19) = 10945$. Thanks for your help in advance! - ## 3 Answers Ok, either you or I are off by one, but here goes. The fact that the sequence is increasing but the parity stays the same means that the missing numbers go in consecutive blocks of two. So the number of Eden sequences is the same as the number of ways of tiling the numbers from 1-n with dominoes of one size one or two. These are well known to be the Fibonacci numbers. It appears that the sequence itself is not counted as the numbers you quote are one greater than the corresponding Fibonacci number. As will has observed, this does not count the sequences where an odd number of numbers are left out of the tail of the Eden sequence. These are eqivalent to the tilings that have a 2-domino extending one unit past the end. There are $F(n-1)$ of these (the sequences of length $n-1$ with a 2 domino slapped on the end). So there are $F(n)+F(n-1)=F(n+1)$ possible sequences, and it appears that the empty sequence is the missing one. - Hi thanks for the response! my calculation was based on trial and error it has no legit proof behind it. I tested it for small numbers and say this: q(1)=1, q(2)=q(1)+1=2,q(3)=q(2)+2=4,q(4)=q(3)+3=7,q(5)=q(4)+5=12,q(6)=q(5)+8=20. So obviously there is some relation to fibonacci numbers, but if you go along in this pattern my answer comes up and so does the conditions given in the problem. How can we show that q(n)=q(n-1)+F_n . Where F_n is the nth fibonacci number. Once again thanks for the help! – ahuang Apr 15 '12 at 3:53 Also can you please explain why the missing numbers go in consecutive blocks of two, and the domino idea. Thanks! – ahuang Apr 15 '12 at 3:59 @alan If you have an odd number followed by a larger even number there are an even number of integers skipped, Likewise for an even number followed by an odd number. – deinst Apr 15 '12 at 4:21 I believe the quoted numbers are one less than the corresponding Fibonacci numbers, most likely because the empty sequence is not counted. (More about this in my answer.) Nice observation about the dominoes, by the way. – Will Orrick Apr 15 '12 at 15:51 An observation related to domino tilings: use "." and "=" to represent dominoes of lengths 1 and 2. The tilings we get for the first few $N$ are {.} ($N=1$), {.., =} ($N=2$), {...,=.,.=} ($N=3$), {....,=..,.=.,..=,==} ($N=4$). These correspond to the Eden sequences {{1}}, ($N=1$), {{1,2},{}} ($N=2$), {{1,2,3},{3},{1}} ($N=3$), {{1,2,3,4},{3,4},{1,4},{1,2},{}} ($N=4$). Observe that (1) for $N$ even, only even-length Eden sequences correspond to domino tilings, while for $N$ odd, only odd-length Eden sequences correspond to domino tilings; – Will Orrick Apr 15 '12 at 16:33 show 2 more comments The answers for q(18) and q(19) are correct. The formula that can be established is that q(N)=q(n-1) + q(N-2) + 1, N>2. I found that examining q(N) separately for N odd and for N even helped. Also, for each N up to 6 I looked separately at the sequences that ended in an even number and those that ended in an odd number; this led me to the above equation. Using the formula along with the given values for q(17) and q(20) it was relatively easy to calculate q(18) and then q(19). I could go into more detail but if I can establish the above formula using this approach I'm sure you'll have no problem. - I tried to do just as you said but I still am not able to come up with the formula I dont know what it is that im doing wrong. Can you please go in a little bit more detail. Once again thank you very much! – ahuang Apr 15 '12 at 4:22 I was thinking similarly to the above answer using o(N) and e(N) but I did get q(18) = 6764 so I'll quickly outline my approach. Starting with N odd, we have q(N) = o(N) + e(N). Now for q(N+1), we can add (N+1) to the end of each odd sequence to form o(N) more sequences giving us q(N+1) = 2o(N) + e(N). For q(N+2), with (N+2) being odd we can add it to the end of the e(N) sequences and the o(N) sequences we just made even, plus have (N+2) as a single digit sequence, giving us q(N+2) = 3o(N) + 2 *e(N) + 1 = q(N+1) + q(N) + 1, which can be rewritten as q(N) = q(N-1) + q(N-2) + 1. – Brian Apr 15 '12 at 5:51 The same thing can be done starting with N even, yielding the same formula. I hope my approach isn't too muddled for you. – Brian Apr 15 '12 at 5:55 I think the discrepancy between our answers comes down to whether the empty sequence is allowed or not. I did not check, although I should have, whether the original poster's numbers for $q(17)$ and $q(20)$ included the empty sequence. Since in fact they do not, your recurrences are the correct ones. I will add a note about this to my answer. – Will Orrick Apr 15 '12 at 15:38 Denote $\{1,2,\ldots,N\}$ by $[N]$. Let $e(N)$ and $o(N)$ be the numbers of Eden sequences from $[N]$ of even length and of odd length. So $q(N)=e(N)+o(N)$. If $N$ is even, then $e(N)=e(N-1)+o(N-1)$ since an even-length sequence from $[N-1]$ is also an even-length sequence from $[N]$, while one may add $N$ to an odd-length sequence from $[N-1]$ to get an even-length sequence from $[N]$. Furthermore, $o(N)=o(N-1)$ since an odd-length sequence from $[N]$ cannot have $N$ as its last element. The $N$ odd case can be handled similarly. Using these recurrences, one may express $e(N)$ and $o(N)$, and hence $q(N)$, for $N>17$, in terms of $e(17)$ and $o(17)$. Since $q(17)=e(17)+o(17)$, one may use the known values of $q(17)$ and $q(20)$ to solve for $e(17)$ and $o(17)$. One may then compute $q(18)$ and $q(19)$. I find the same value of $q(19)$ that you found, but a different value of $q(18)$. Edit: The analysis above assumes that the empty sequence counts as one of the even sequences. This is not the assumption that was made in the original poster's question since $q(17)=4180$ and $q(20)=17710$ are the correct numbers when the empty sequence is omitted. We can modify the analysis above to account for this by leaving the definitions of $e(N)$ and $o(N)$ as is (that is, $e(N)$ includes the empty sequence), but redefining $q(N)$ as $q(N)=e(N)-1+o(N)$ (so that $q(N)$ does not include the empty sequence). With this modification our numbers are in perfect agreement. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 67, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9429876804351807, "perplexity_flag": "head"}
http://mathhelpforum.com/number-theory/212816-rearrangements-infinite-series.html	1Thanks • 1 Post By johng # Thread: 1. ## Rearrangements of an infinite Series If ak>=0 for all k in N, and Σak= prove that Σa'k= for any rearrangement Σa'k of Σak I think I understand this proof, I'm just having trouble making it rigorous. I was going to go the route that since ak is positive, then its sum is monotone increasing. In order to be convergent it would need to be bounded, so I can assume it's unbounded since it diverges. Then any rearrangement of the series is also going to be unbounded and then it will diverge. Is that correct thinking? 2. ## Re: Rearrangements of an infinite Series For a formal proof, I think you have to explicitly use the definition of series rearrangement. Here's my version: 3. ## Re: Rearrangements of an infinite Series to help decipher what johng has done: we know we can find a partial sum for the original series that is greater than any given M. this is a finite sum of terms of the original series. now in the "re-arranged" series, these terms may be scattered apart, but we can surely find a partial sum of the re-arranged series that contains ALL of them (even if we have to go out much further than we did in the original series), first we find "how far out (the largest index in the re-arragement)" we went, and include all those terms in our partial sum for the re-arranged series. this includes every term in our original partial sum, plus perhaps a lot more, but certainly no less. so this partial sum is at least the same as the partial sum of the original series (but probably a lot bigger). in any case, its bigger than M, and M can be as large as we like. EDIT: a note about the formalism involved. with infinite things, we can't really explicitly write them out. so, for example, to say a quantity (our series, in this instance) is infinite, we have to say something like: for every (finite) real number M, there exists a partial sum Sn, with Sn > M. this lets us say informally: $\sum_{k = 0}^{\infty} a_k = \infty$ but it must be kept in mind that the "=" sign here is just a SHORTHAND for the formal definition above (which doesn't even mention infinity, because infinity is NOT a number). re-arranging "infinite" things can have some counter-intuitive properties (it is possible with alternating series to get finite AND infinite "re-arrangements", which justifies the cautious approach taken here), so we want to take "behavior at infinity" back to the realm of the finite whenever possible, or else the reasoning can get a little "soft".	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9545174241065979, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/tagged/q-analogs	# Tagged Questions Use this tag for questions pretaining to q-analogs of functions, for example q-Binomials, $q$-derivatives, the q-theta function, the q-Pochammer symbol, etc. 1answer 61 views ### What does the $q$-Catalan Numbers count? I had completed a paper describing the $q$-Catalan numbers, which is the $q$-analog of the Catalan numbers. The $n$-th Catalan numbers can be represented by: $$C_n=\frac{1}{n+1}{2n \choose n}$$ and ... 1answer 54 views ### Combinatorial identity. Using echelon matrices. Determine the exponents $e_i$ s.t. the following identity is correct. $$\sum\limits_{i=0}^k q^{e_i} {\binom mi}_q {\binom{n}{k-i}}_q = {\binom{n+m}{k}}_q$$ Note: When $q=1$ the equation reduces to a ... 3answers 101 views ### How to prove it? (one of the Rogers-Ramanujan identities) Prove the following identity (one of the Rogers-Ramanujan identities) on formal power series by interpreting each side as a generating function for partitions: ... 1answer 182 views ### Which families of groups have interesting formulas for the number of elements of given order? Suppose that $G$ is a group and that $n$ is a positive integer diving the order of $G$. Let $f_n(G)$ be the number of elements satisfying $x^n = 1$ in $G$. According to a theorem of Frobenius, then we ... 0answers 60 views ### A general Combinatorics problem (Coefficients of the q factorial) I was solving a combinatorics problem when I encountered difficulties. The problem was: $x_1 \in \{0,1\}$ $x_2 \in \{0,1,2\}$ . . $x_{n-1}\in\{0,1,2..,n-1\}$ We have to find the number of ways ... 4answers 337 views ### Proving q-binomial identities I was wondering if anyone could show me how to prove q-binomial identities? I do not have a single example in my notes, and I can't seem to find any online. For example, consider: \${a + 1 + b \brack ... 0answers 133 views ### Different notions of q-numbers It seems that most of the literature dealing with q-analogs defines q-numbers according to $$[n]_q\equiv \frac{q^n-1}{q-1}.$$ Even Mathematica uses this definition: with the built-in function QGamma ... 1answer 128 views ### Reciprocity Law of the Gaussian (or $q$-Binomial) Coefficient It is a standard exercise in combinatorics to show that the binomial coefficient satisfies the reciprocity law $\binom{-n}{k} = (-1)^k \binom{n+k-1}{k}$ for $n, k \geqslant 0$, which is the multiset ... 0answers 53 views ### Necessary and sufficient condition for $f(q^n)$ to be in $\mathbb{Z}[q,q^{-1}]$ when $f\in\mathbb{Q}(q)[x]$? In this question, user bgins shows that for each $k$ there is a unique polynomial $P_k(x)$ of degree $k$ whose coefficients are in $\mathbb{Q}(q)$, the field of rational functions, such that ... 2answers 217 views ### q-Analogue of the formula $x^n=\sum_k\left\{n\atop k\right\}(x)_k$. The stirling numbers of the second kind satisfy the formula $x^n=\sum_k\left\{n\atop k\right\}(x)_k$, where $(x)_k$ is the falling factorial. Consider the $q$-analog recursive definition of the ... 2answers 188 views ### Recurrence for $q$-analog for the Stirling numbers? I read in some papers that the Stirling numbers (of the second kind) have a natural $q$-analog $S_q(n,k)$, which satisfy the recurrence $$S_q(n,k)=(k)_qS_q(n-1,k)+q^{k-1}S_q(n-1,k-1)$$ with the ... 1answer 150 views ### Inferring Jacobi Triple product from $q$-binomial theorem? Quite a while ago I asked a question about deducing the Jacobi triple product from the $q$-binomial theorem. In fact, from the $q$-binomial theorem ... 1answer 148 views ### Is the following product of $q$-binomial coefficients a polynomial in $q$? $$\frac{\binom{n}{j}_q\binom{n+1}{j}_q \cdots\binom{n+k-1}{j}_q}{\binom{j}{j}_q\binom{j+1}{j}_q\cdots\binom{j+k-1}{j}_q}$$ where $n,j,k$ are non-negative integers. 1answer 219 views ### Intermediate step in deducing Jacobi's triple product identity. An intermediate step deduces Jacobi's triple product identity by taking the $q$-binomial theorem $$\prod_{i=1}^{m-1}(1+xq^i)=\sum_{j=0}^m\binom{m}{j}_q q^{\binom{j}{2}}x^j$$ and deducing ... 2answers 243 views ### Representing the $q$-binomial coefficient as a polynomial with coefficients in $\mathbb{Q}(q)$? Trying a bit of combinatorics this winter break, and I don't understand a certain claim. The claim is that for each $k$ there is a unique polynomial $P_k(x)$ of degree $k$ whose coefficients are ... 1answer 119 views ### Primitive roots as roots of a $q$-multinomial. If $n$ is divisible by $m$, why is it the case that the $m$th primitive roots of unity are also roots of $\binom{n}{k}_q$ if and only if $m$ does not divide $k$? I'm viewing $\binom{n}{k}_q$ as a ... 1answer 152 views ### Deriving Cauchy's identity from the $q$-binomial theorem? Cauchy's identity states that $$\prod_{i\geq 0}\frac{1-axq^i}{1-xq^i}=\sum_{n\geq 0}\frac{(1-a)(1-aq)\cdots(1-aq^{n-1})}{(1-q)(1-q^2)\cdots(1-q^n)}x^n.$$ Is it possible to somehow derive this ... 1answer 130 views ### What's the reasoning for this recurrence on $q$-multinomial coefficients? I'm familiar with the recurrence for binomial coefficients based on Pascal's triangle. However, in general, there is the recurrence for $q$-multinomial coefficients given by ... 1answer 114 views ### Closed form for $\sum_{m \geq 1} (-1)^m q^{m(m+1)/2 + m \Delta}$? Is there a useful closed form for the following series ($\|\Delta\|$ is a small integer)? $$f(q,\Delta) =\sum_{m=1}^{\infty} (-1)^m q^{m(m+1)/2 + m \Delta}$$ It is a large-$n$ approximation of the ... 4answers 447 views ### Intriguing polynomials coming from a combinatorial physics problem For real $0<q<1$, integer $n >0$ and integer $k\ge 0$, define $$[k, n]_q \equiv -\sum_{m=1}^{n} q^{m(k+1)} (q^{-n}; q)_m = -\sum_{m=1}^{n} q^{m(k+1)} \prod_{l=0}^{m-1} (1-q^{l-n})$$ ... 1answer 105 views ### Interpolation of a sequence of polynomials (viewed in terms of q-analogue powers) I'm trying to find closed-form expressions for a sequence of coefficients, such that the index of the coefficient occurs as number such that I can later interpolate to fractional indexes as well. ... 1answer 245 views ### Is there a gamma-like function for the q-factorial? I'm looking at quantum calculus and just trying to understand what is going with this subject. Looking at the q-factorial made me wonder if this function could take all real or even complex numbers in ... 2answers 239 views ### What is Eulerian? I encountered an interesting function which is called "Eulerian" by the Wolfram's MathWorld: $$\phi(q)=\prod_{k=1}^{\infty} (1-q^{k})$$ It is interesting because it seems that roots of any ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 65, "mathjax_display_tex": 10, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9183622002601624, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?t=611510	Physics Forums Why electric field is zero inside a conductor? 1.Why electric field is zero inside a conductor? Can anyone explain it with Gauss's law and without Gauss's law? 2.Is it possible to create electric field inside a conductor? If so,how and in which condition it is created? PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus 1. In electrostatics the electric field is 0 inside a perfect conductor because otherwise there would be moving charges. The charges must redistribute themselves to make a net electric field inside the conductor 0. 2. It is possible to have an electric field inside a conductor. You can put one there by e.g. using a battery. The trick is that this prompts charges to move inside the conductor and therefore a current to develop. This is no longer electrostatics. Well,I got this.But how you will explain this with Gauss's law?Explain why electric field is zero inside a conductor by the help of Gauss's law. Now if i ask,why charges are always distributed on the surface?Why there is no charge inside the conductor?I prepared like this... Consider a gaussian surface inside the conductor A (see the figure attached).As electric field is zero inside the conductor,Gauss's law requires that there could be no net charge inside the surface.Now imagine shrinking the surface like a collapsing balloon until it encloses a region so small that we may consider it as a point.Then the charge at that point must be zero.We can do this anywhere inside the conductor,so there could be no excess charge an any point within a solid conductor.Any excess charge must reside on the conductor surface. Attached Thumbnails Why electric field is zero inside a conductor? Gauss' Law can not be used since as stated that is for an electrostatic condition. In other words it is not used when you have a moving charge. If you have a charge within an electric field that puts a force on that charge. If it is free to move it will move due to that force. In your illustration each charge would create its own electric field that other charges will react to. All of the charges push against all of the other charges and they end up on the outer boundary of the conductor where they cannot move away from each other any further. Ohm's law (differential form) $\textbf{j}=\sigma\textbf{E}$ If E is not zero, then a current will very quickly form in a conductor. If the conductor is surrounded by insulator, then charge can't escape from the conductor, so you can't have net current escaping from the conductor. Therefore, the flux of E over the surface of the conductor is 0. There's no law that says E is zero inside the conductor, but it goes to 0 very quickly in steady state or slowly varying conditions. Internal E field will generate internal currents, which will quickly dissipate electrical energy into heat via internal resistance. Tags electric field, electricity, gauss's law Thread Tools \| \| \| \| \|---------------------------------------------------------------------\|-------------------------------\|---------\| \| Similar Threads for: Why electric field is zero inside a conductor? \| \| \| \| Thread \| Forum \| Replies \| \| \| Classical Physics \| 12 \| \| \| Introductory Physics Homework \| 3 \| \| \| Classical Physics \| 37 \| \| \| Introductory Physics Homework \| 1 \| \| \| Introductory Physics Homework \| 1 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9228706955909729, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/75096/list	## Return to Answer 2 reversed inclusion for $H^1$ and $L^4$ I am not going to try to find the most general conditions under which lower semi-continuity holds but for that I suggest the standard reference for all of this is "Direct methods in the Calculus of Variations" by Bernard Dacorogna which covers all of this in full detail. I will give a brief outline of the answer however. In order for $\int f(x,u,du)dx$ to be lower semi-continuous with respect to weak convergence one does not in general need any sort of convexity in the $u$ variable and what is more important generally is 1) Coercivity (see below) of the functional $f$ in the $du$ variable. 2) The right embedding along with continuity of $f$ in the $u$ variable with respect to the topology of strong convergence in this space. Example: A good example is $E(u) = \int_0^1 \|\nabla u \|^2 + (1-u^2)^2$ with $u = 0$ on $\partial \Omega$ in $\mathbb{R}^d$. Observe that when $d \leq 4$ one has $L^4 H^1(\Omega) \subset \subset H^1(\Omega)$ L^4(\Omega)$and consequently the non-convex term depending on$u$is lower order. Therefore one can pass to the limit in any minimizing sequence even though$(1-u^2)^2$is very non convex (but it is continuous with respect to strong convergence). Notice however that for the energy$\bar E(u) = \int_0^1 \|\nabla u\|^2 + u^4$in$\mathbb{R}^5$with$u$prescribed on$\partial \Omega\$, the second term is not lower order but here we may use convexity to conclude lower semi-continuity of the second term. The function $p \mapsto f(x,z,p)$ is coercive if there exists some constant $C > 0$ so that $f(x,z,p) \geq C\|p\|^q$ for some range of $q$. It then depends on what spaces one is working in but the goal is to use an embedding theorem such as $L^q \subset\subset W^{1,p}$ for $1 \leq q < p^$ and to conclude that for a minimizing sequence $u_n$, there is in fact a strongly convergent subsequence in $L^q$ for some $q$. Then one will generally expect some sort of continuity of $f$ in the $u$ variable. There are many more interesting examples but in almost all cases in practice the goal is to show that one has strong convergence of the $u_n$s in your minimizing sequence in some $L^p$ space. The book by Braides focuses mostly on asymptotics of functionals which depend on some large (or small) parameter and I'm not sure how much he talks about the assumptions needed in the direct method. 1 I am not going to try to find the most general conditions under which lower semi-continuity holds but for that I suggest the standard reference for all of this is "Direct methods in the Calculus of Variations" by Bernard Dacorogna which covers all of this in full detail. I will give a brief outline of the answer however. In order for $\int f(x,u,du)dx$ to be lower semi-continuous with respect to weak convergence one does not in general need any sort of convexity in the $u$ variable and what is more important generally is 1) Coercivity (see below) of the functional $f$ in the $du$ variable. 2) The right embedding along with continuity of $f$ in the $u$ variable with respect to the topology of strong convergence in this space. Example: A good example is $E(u) = \int_0^1 \|\nabla u \|^2 + (1-u^2)^2$ with $u = 0$ on $\partial \Omega$ in $\mathbb{R}^d$. Observe that when $d \leq 4$ one has $L^4 \subset \subset H^1(\Omega)$ and consequently the non-convex term depending on $u$ is lower order. Therefore one can pass to the limit in any minimizing sequence even though $(1-u^2)^2$ is very non convex (but it is continuous with respect to strong convergence). Notice however that for the energy $\bar E(u) = \int_0^1 \|\nabla u\|^2 + u^4$ in $\mathbb{R}^5$ with $u$ prescribed on $\partial \Omega$, the second term is not lower order but here we may use convexity to conclude lower semi-continuity of the second term. The function $p \mapsto f(x,z,p)$ is coercive if there exists some constant $C > 0$ so that $f(x,z,p) \geq C\|p\|^q$ for some range of $q$. It then depends on what spaces one is working in but the goal is to use an embedding theorem such as $L^q \subset\subset W^{1,p}$ for $1 \leq q < p^$ and to conclude that for a minimizing sequence $u_n$, there is in fact a strongly convergent subsequence in $L^q$ for some $q$. Then one will generally expect some sort of continuity of $f$ in the $u$ variable. There are many more interesting examples but in almost all cases in practice the goal is to show that one has strong convergence of the $u_n$s in your minimizing sequence in some $L^p$ space. The book by Braides focuses mostly on asymptotics of functionals which depend on some large (or small) parameter and I'm not sure how much he talks about the assumptions needed in the direct method.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 64, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9544402956962585, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/tagged/work?sort=faq&pagesize=30	# Tagged Questions The work tag has no wiki summary. 7answers 3k views ### Why does holding something up cost energy while no work is being done? I read the definition of work as $$W ~=~ \vec{F} \cdot \vec{d}$$ $$\text{ Work = (Force) $\cdot$ (Distance)}.$$ If a book is there on the table, no work is done as no distance is covered. If I ... 4answers 337 views ### Violation of Newton's Second Law (?) Here the big circle denotes the circular path of a stone(small circle on path) tied to a string from the centre of the circular path . This is COMPLETELY HORIZONTAL At an instant the velocity in ... 2answers 555 views ### Why is there a $\frac 1 2$ in $\frac 1 2 mv^2$? For elastic collisions of n particles, we know that momentum in the three orthogonal directions are independently conserved:$$\frac{d}{dt}\sum\limits_i^n m_iv_{ij} =0,\quad j=1,2,3$$ From this, it ... 1answer 304 views ### A Question about Virtual Work related to Newton's Third Law In describing D'Alembert's principle, the lecture note I was provided with states that the total force $\mathbb F_l$ acting on a particle can be taken as, $$\mathbb F_l=F_l+\sum_mf_{ml}+C_l,$$ ... 3answers 534 views ### Is the normal force a conservative force? Most of the time the normal force doesn't do any work because it's perpendicular to the direction of motion but if it does do work, would it be conservative or non-conservative? For example, consider ... 2answers 241 views ### Intuitively Understanding Work and Energy It is easy to understand the concepts of momentum and impulse. The formula $mv$ is simple, and easy to reason about. It has an obvious symmetry to it. The same cannot be said for kinetic energy, ... 6answers 324 views ### Electrostatic Potential Energy Derivation How is the boxed step , physically as well as mathematically justified and correct ? Source:Wiki http://en.wikipedia.org/wiki/Electric_potential_energy As work done = $- \Delta U$. for Conservative ... 3answers 170 views ### Why does it require such little energy to create the fastest thing in the universe? I have noticed when I turn on the light switch in my house light comes from the bulb. How is this light created?(process occurring in the bulb) and why is this small amount of electricity enough to ... 4answers 601 views ### Work done by the Magnetic Force The magnetic part of the Lorentz force acts perpendicular to the charge's velocity, and consequently does zero work on it. Can we extrapolate this statement to say that such a nature of the force ... 1answer 997 views ### How much work is needed to compress a certain volume of gas? I want to know the formula (and what does the symbols stand for) for how much work is needed to compress a certain volume of gas?	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9234683513641357, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/97100/jordan-algebra-and-quaternionic-projective-space/97101	## Jordan algebra and quaternionic projective space ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) How we can define quaternionic projective space and metric on it using Jordan algebra? Thank you in advance! - ## 2 Answers I am not much educated in this subject so please take my answer with a grain of salt. The appropriate Jordan algebra is $J_n(\mathbb{H}) := \{ A \in M(n,\mathbb{H})\,\|\, \overline{A}^T=A \}$ with multiplication defined as $A\circ B = (AB+BA)/{2}$. The metric is defined by $\mathrm{Tr}(A^2)$. The quaternionic projective space can be then defined as the space of elements of $J_n(\mathbb{H})$ of rank one and trace one (think of these matrices as projectors to one-dimensional subspaces) or alternatively as the (real) projectivization of rank one matrices. I am little bit more familiar with the case of octonionic projective plane, so let me explain that case. There the relevant algebra is $J_3(\mathbb{O})$ and the plane $\mathbb{OP}^2$ and its metric is defined in the same way[1] as in the quaternionic case. Now there is a well defined cubic form on $J_3(\mathbb{O})$ which is basically the determinant. The group that preserves this cubic form is $E_6$. In fact one defines some sort of cross product (I think it is called Jordan cross product.) out of the cubic form which then gives the incidence relation of the projective geometry of $\mathbb{OP}^2$. If I remember correctly, the product $A\times B$ is defined by the relation $(A\times B,C) = (A,B,C)$, where on the left hand side the brackets denote the polarization of the quadratic form while on the right hand side the brackets denote the polarization of the cubic form. The group $E_6$ is then the group of collineations of $\mathbb{OP}^2$ The group that preserves the determinant and the quadratic form (the norm) is the group $F_4$. This sheds some light on the fact that $F_4$ is in fact the isometry group of $\mathbb{OP}^2$. It can be proven that $F_4$ is in fact the automorphism group of the Jordan algebra $J_3(\mathbb{O})$! In the quaternionic case one finds that $E_6$ is replaced by $Sl(n,\mathbb{H})$ and $F_4$ by $Sp(n)$. The projective plane $\mathbb{OP}^2$ was much studied by Freudenthal and others but I do not know of any reference where the quaternionic case is treated via Jordan algebras. [1] Of course one needs to be a little careful with the definition of rank because of the nonassociativity of $\mathbb{O}$. Either one uses the fact that any element of $J_3(\mathbb{O})$ can be mapped by the action of $F_4$ to a diagonal matrix and then one defines the rank by the number of nonzero elements in this diagonalization. Or one can experiment a little bit and discover that it is possible to define determinants of 2x2 minors in such a way that their vanishing is equivalent to matrix of being of rank one according to first definition. - 1 How does the structure of the Jordan algebra comes into play ? Is the space of rank 1 closed with respect to Jordan's product ? I would be very thankful for further comments... For the long time I keep some papers on Jordan algs nearby me but always does not see key how to open them :) – Alexander Chervov May 16 2012 at 10:28 @robot Still I am not clear about the role of Jordan algebra structure. If you will omit this word and just write matrices A^t=A , norm Tr(A^2)... nothing changes.. Any way thank you for yours answer, it useful (at least for me). – Alexander Chervov May 16 2012 at 12:00 I do not know of a nice direct way to connect Jordan algebra structure and the geometry of the projective space. In the octonionic case, the incidence relation is given by the cross product which can be expressed in terms of Jordan product and traces. (See arxiv.org/abs/0902.0431) – robot May 16 2012 at 12:05 You are right, the conditions $\overline{A}^t=A$ and being of rank one and trace one are sufficient to define the projective space and its metric. The Jordan multiplication lurks in the background, because - at least in the octonionic case - the group of isometries is the group of automorphisms of the Jordan algebra. – robot May 16 2012 at 12:08 Is the condition $\overline{A}^{T}=A$ here because these matrices preserve $H$-inner product? Is there any connection between $Sp(n)$ and and $J_{n}(H)$? – Mirjana May 17 2012 at 9:33 show 3 more comments ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. (Perhaps my answer will be too general for your purposes.) The description of the projective space that you mention, is only one special case of a very general construction (due to O. Loos) of a projective space from a Jordan pair, which is a pair of vector spaces $(V^+, V^-)$ equipped with certain quadratic operators. The notion of a Jordan pair generalizes the notion of a Jordan algebra, in the sense that any Jordan algebra $J$ can be made into a Jordan pair (but the converse is not true). If $V = (V^+, V^-)$ is such a Jordan pair, then the projective space $X(V)$ of $V$ is the quotient of $V^+ \times V^-$ by projective equivalence, defined as $$(x, y) \sim (x', y') \iff (x, y - y') \text{ is quasi-invertible, and } x' = x^{y-y'},$$ where $x^{y-y'}$ denotes the quasi-inverse of the pair $(x, y-y')$. In the case where $V$ is the Jordan pair arising from the Jordan algebra of hermitian matrices over any skew field $D$ (of which $D = \mathbb{H}$ is just a particular case), the projective space $X(V)$ is the classical projective space coordinatized by the skew field $D$. - what is "projective space" ? Some nice manifold ? Symmetric space ? homogeneous space ? – Alexander Chervov May 16 2012 at 12:29 For general fields, it's just an algebraic object, but for instance if $k = \mathbb{C}$, then one obtains the compact hermitian symmetric spaces. – Tom De Medts May 16 2012 at 12:33 @Tom thank you ! Do all compact hermitian spaces can be obtained in this way ? – Alexander Chervov May 16 2012 at 13:00 I guess so (but I'm not very familiar with symmetric spaces though). You might be interested in the lecture notes "Jordan pairs and bounded symmetric domains" by O. Loos, available on the website homepage.uibk.ac.at/~c70202/jordan/archive/irvine/…. – Tom De Medts May 16 2012 at 13:10	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 45, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9127271175384521, "perplexity_flag": "head"}
http://mathoverflow.net/questions/91939?sort=votes	## Higher computability : Constructive ordinal and $\Delta^1_1$ predicates ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Everything I know on this subject comes from Sacks book : "Higher recursion theory" Let $\mathcal{O^Y}$ be the set of codes for ordinals constructive in $Y$. We should have the result that $A \subseteq \omega \times 2^\omega$ is $\Delta^1_1$ iff $\exists a \in \mathcal{O}\ \ \exists e \in \omega$ such that $A(n, Y) \leftrightarrow \varphi_e^{H_a^Y}(n) \downarrow$ where $H_a^Y$ is the $\|a\|$-th iteration of the turing jump of $Y$. Two things are now in contradiction in my mind : The first one : $X$ is $\Delta^1_1(Y)$ iff $\exists a \in \mathcal{O^Y}\ \ \exists e \in \omega$ such that $n \in X \leftrightarrow \varphi_e^{H_a^Y}(n) \downarrow$. Potentially we can have $\|a\| \geq \omega_1^{ck}$ if $\omega_1^{Y} \geq \omega_1^{ck}$ The second one : $X$ is $\Delta^1_1(Y)$ iff then there exists a $\Delta^1_1$ predicate $A \subseteq \omega \times 2^\omega$ and $\exists a \in \mathcal{O}\ \ \exists e \in \omega$ such that $n \in X \leftrightarrow A(n, Y) \leftrightarrow \varphi_e^{H_a^Y}(n) \downarrow$ This time, the code $a$ for the constructive ordinal is always smaller than $\omega_1^{ck}$. Can anyone see where I made a mistake ? Thanks in advance - ## 1 Answer I finaly got an answer from another forum. The answer is simple, I was assuming that if $A \subseteq \omega$ is $\Delta^1_1(Y)$ it means that there is a $\Pi^1_1$ predicate $F \subseteq \omega \times 2^\omega$ and a $\Sigma^1_1$ predicate $E \subseteq \omega \times 2^\omega$ such that $\forall n\ \ A(n) \leftrightarrow F(n, Y) \leftrightarrow E(n, Y)$ and $\forall X\ \ \forall n\ \ F(n, X) \leftrightarrow E(n, X)$. But it does just mean that $\forall n\ \ F(n, Y) \leftrightarrow E(n, Y)$ which is indeed very different. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 33, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9488143920898438, "perplexity_flag": "head"}
http://climateaudit.org/2007/05/04/more-on-hegerl-et-al-2006-non-confidence-intervals/?like=1&source=post_flair&_wpnonce=134bd2d71b	by Steve McIntyre ## More on Hegerl et al 2006 Non-Confidence Intervals There are a few other blogs that from time to time do detailed analyses of what people are doing, not dissimilar in format to what I do. Last year, in Apr 2006 shortly after publication, I observed here that the upper and lower confidence intervals of Hegerl et al crossed over. In Feb 2007, Tapio Schneider published a Comment in Nature observing that the confidence intervals in Hegerl et al were wrong. Hegerl published a Reply and replaced the Supplementary Information with new data (I kept the old version in case anyone wants a comparison.) James Annan recently discussed the matter , linking to my graph, acknowledging it in a business-like way. About the new Supplementary Information, he said: There is now a file giving the reconstruction back to 1500 with new confidence intervals, which no longer vanish or swap over. This new data doesn’t match the description of their method, or the results they plotted in their Fig 1 (which is almost surely a smoothed version of the original supplementary data). He went on to say: Hegerl et al used a regression to estimate past temperatures anomalies as a function of proxy data, and estimated the uncertainty in reconstructed temperature as being entirely due to the uncertainty in the regression coefficient. The problem with this manifests itself most clearly when the tree ring anomaly is zero, as in this event the uncertainty in the reconstructed temperature is also zero! Maybe UC (or Jean S who we haven’t heard from for a while) can comment on this. This comment still doesn’t seem right to me as I can’t think of why the uncertainty would be zero merely because all of the uncertainty was allocated to the regression coefficient. I still can’t get a foothold on what they’re doing here; Annan said that Tapio Schneider had been unsuccessful in getting Hegerl to document what they did in any of the calculations. I’ll write but I’m not optimistic about my chances. I’m up to about 20 emails with Crowley trying to find out how they got their Mongolia and Urals series, without any success. Eli Rabett observed that Huang et al appeared to have done the same thing in a borehole study. In the caption, Huang refer to Bayesian methods being used, so maybe there’s a clue for someone. Whatever these folks are doing, it’s not a totally isolated incident. Who knows – one day,we might even find out how the MBH99 confidence intervals were calculated – presently one of the 21st Century Hilbert Problems in climate science. References: Hegerl JClim 2006 here Hegerl Nature 2006 here Hegerl SI here ### Like this: Like Loading... This entry was written by , posted on May 4, 2007 at 11:06 AM, filed under Hegerl 2006. Bookmark the permalink. Follow any comments here with the RSS feed for this post. Post a comment or leave a trackback: Trackback URL. ### 25 Comments 1. John Hekman Posted May 4, 2007 at 11:32 AM \| Permalink \| Reply if y = a + bx + e, and you want to know the variance of the estimated value, y^, that variance is not zero when you are comparing the actual time series with the fitted values and one of the fitted values is equal to the actual value. If that is what he is saying here, he’s nuts. You just have to look at the ANOVA. http://en.wikipedia.org/wiki/ANOVA 2. Posted May 4, 2007 at 4:18 PM \| Permalink \| Reply Confidence intervals, regardless of method, cannot vanish or cross over. If one does end up with such confidence intervals, it is a sure sign that something is seriously wrong in the way calculations have been done. If it happened to me, I would withdraw to an isolated room, not get out until I figured out my error and I would be ashamed to tell anyone about whatever stupid mistake I made. However, this paper got published. Sad. Here is what I mean: The width of the confidence interval for mean response in a simple OLS regression with one independent variable depends on the standard error of the mean response (which, in turn, depends on the x-value at which mean response is being evaluated). We have, from the Intro Stats textbook I use in the undergrad stats class I teach, $\mathrm{SE}_{\hat{\mu}} = s\sqrt{\frac{1}{n} + \frac{(x^-\bar{x})^2}{\Sum (x_i-\bar{x})^2}}$ If $x^ = \bar{x}$, this reduces to $s\sqrt{\frac{1}{n}}$ which cannot be zero (unless the regression line is a perfect fit) because s is the square root of MSE. Now, confidence intervals for mean response will be narrower than prediction confidence intervals (for the same confidence) but they cannot vanish and the lower and upper bounds cannot cross. I am baffled. Sinan 3. 2dogs Posted May 4, 2007 at 4:21 PM \| Permalink \| Reply It looks like the uncertainty is all assigned to the regression co-effecient of the anomoly (i.e. to b in y = bx + e); consequently, the upper and lower bounds flip over when the anomoly passes zero. 4. Posted May 4, 2007 at 4:22 PM \| Permalink \| Reply Correction to #2: In the formula for the standard error of mean response, there is a sum $\Sigma$ sign missing from the denominator of the second term in the square root. Also, I mention the undergrad intro stats textbook to underline the fact that this is basic, elementary stuff. Vanishing confidence intervals or crossing upper and lower bounds is an indication that the author of the article ought not be trusted to balance his own checkbook. Sinan 5. Steve McIntyre Posted May 4, 2007 at 10:07 PM \| Permalink \| Reply I sent the following inquiry to Gabi Hegerl: I have not received the requested information on the Mongolian and Urals series from Tom. I realize that you have other concerns but this has been going on far too long. Also can you provide a replicable description of how you calculated your confidence intervals. I realize that the original SI required correction, but in the wake of this there is no formal description of your procedures or statistical references. If it would save you time to merely provide source code for the results, that would be fine with me. 6. Steve McIntyre Posted May 5, 2007 at 7:46 AM \| Permalink \| Reply #5. Gabi Hegerl replied as follows: Tom, can you send Steve some more information, please? Steve, Tom produced the individual timeseries, but not with fortran code but with other software, so I am not sure there is a source code. The correction in the SI was minimal and it would have been totally sufficient to change order 3 words in the caption, but I thought it was more helpful to instead link the reconstructions with full uncertainty] range. The description in the SI is very complete on that, particularly together with my J Climate paper. It should make it completely reproducable. Gabi 7. Nicholas Posted May 5, 2007 at 8:42 AM \| Permalink \| Reply Did you plot that diagram yourself? If so, is the R code easily available somewhere? I’m asking because they look an awful lot like they are merely scaled versions of each other. It would be interesting to check if that’s true. 8. Steve McIntyre Posted May 5, 2007 at 9:12 AM \| Permalink \| Reply I’ll try to post up the code today. It might be an interesting project for statistically-interested people to decode their uncertainty methodology. I’ll post up as much relevant data as I have. 9. Henry Posted May 5, 2007 at 5:45 PM \| Permalink \| Reply It looks to me as if they originally used a statistical recipe in the wrong context. Trying to imagine a proper context: suppose you tried to regress national GDP against national population using data for 100 countries. $y_i=\alpha +\beta x_i +\varepsilon_i$ would be a bad model since it is obvious that a country with a tiny population would have a tiny GDP. $y_i=\beta x_i +\varepsilon_i$ would be almost as bad as the absolute values of errors would clearly be related to population. $y_i=\beta \varepsilon_i x_i$ starts to make more sense. But that model only makes sense if all the values are positive and if errors tend to increase with the independent variable. That doesn’t apply here, so it was the wrong method to apply. It probably stemmed from a desire to have everything at zero in the base period for calculating anomalies. And the graph should have been spotted by the peer reviewers. 10. Vince Causey Posted May 6, 2007 at 12:06 PM \| Permalink \| Reply I have heard it said before, that a lot of these results that involve statistical techniques lack the input of suitably qualified statisticians. Am I right in thinking that elementary statistical errors are being made due to lack of competence? If this is the case, then the quality of a lot of this kind of research must be suspect. 11. Posted May 6, 2007 at 1:04 PM \| Permalink \| Reply Very briefly: something odd in the manuscript figure 2 as well, cross overs.. BTW, ‘negative bias of the variance of the reconstruction’ in supplementary refers to vS04. I believe that this negative bias is a result of ‘natural calibration’ assumption, i.e. optimal solution in the case where P and T are obtained from joint normal distribution. I’ll return to this topic later ( and if I’m wrong I just disappear ) Recommended reading: Confidence and Conflict in Multivariate Calibration (PJ Brown, R Sundberg, Journal of the Royal Statistical Society, Ser B, Vol 49, No 1, 1987) Multivariate Calibration – Direct and Indirect Regression Methodology (R Sundberg) Those papers contain some ideas how to deal with uncertainties in this context (filtering theory is another way, but it doesn’t seem to interest climate people ). 12. Tim Posted May 8, 2007 at 7:22 PM \| Permalink \| Reply Here is how one can get “confidence” intervals of zero width in temperature reconstructions: 1) Estimate a regression model T – [T] = b(p – [p] + eta) + eps () where T-[T] are temperature anomalies (mean temperature [T]), p-[p] are proxy anomalies, and eta and eps are error terms. 2) “Reconstruct” temperature anomalies by taking expected values in the regression model () and plugging in estimated regression coefficients b’ (and estimated mean values [T] and [p]). -> The “reconstructed” temperature anomalies T-[T] are zero for any value b’ of the estimated regression coefficient if the proxy anomaly p-[p] is zero. 3) Vary the estimated regression coefficient b’ within estimated confidence limits and “reconstruct” temperature anomalies as in (2) for each value of b’. -> The “reconstructed” temperature anomalies will still be zero for all values of b’ whenever p-[p] is zero. 4) If one infers confidence intervals from the reconstructed temperature anomalies for the different b’ for each year, they will have zero width whenever p-[p] is zero. Of course, it makes no sense to estimate confidence intervals in this way. This seems to be what Hegerl et al. were doing and what led to their manifestly wrong confidence intervals for the reconstructed temperatures. Schneider’s point in the Nature comment, however, seems to be more general. Hegerl et al. base their inference about climate sensitivity on the residual difference between the “reconstructed” temperature anomalies and temperature anomalies simulated with an EBM. Schneider points out that What should enter the calculation of the likelihoods of the temperature-anomaly time series T-[T] is the estimated variance of the residuals r, not just the sample variance proportional to their sum of squares, sum(r^2). While the error terms eta and eps do not enter the expected value of the difference between EBM temperature anomalies and reconstructed temperature anomalies, they do enter the variance of the residual difference. The variance of the estimated regression coefficients contributes to the residual variance, but so do the variances of eta and eps. This is elementary regression analysis: The variances of eta and eps are generally greater (by a factor of order sample size) than the variance of b’. 13. Posted May 9, 2007 at 6:59 AM \| Permalink \| Reply #12 Clarifying posts are appreciated, thks (and posters should not be worried about making mistakes, if you make an explicit mistake, gavin drops by and corrects it, so no worries ). It is 1959 in climate science in statistics sense, Williams just published Regression Analysis. Next year will be interesting, as Kalman will publish his A New Approach to Linear Filtering and Prediction Problems. 14. Steve McIntyre Posted May 9, 2007 at 7:06 AM \| Permalink \| Reply #12. That makes total sense as an explanation. That’s unbelievable. What a joke. I’ll make a separate post on it in a few days. 15. bender Posted May 15, 2007 at 10:56 PM \| Permalink \| Reply Any updates on this? It’s hard to imagine an error of this magnitude being made by an author and then not being caught by co-authors, internal reviewers, peer reviewers, co-editors, editor. The crossing over of the curves is an obvious sign something’s gone wrong. An undergraduate could tell you that. 16. bender Posted May 15, 2007 at 11:11 PM \| Permalink \| Reply Steve M, look at the original confidence interval in the Nature 2006 paper, the region in gray. Why is the confidence interval so thick in those problem areas where the upper and lower bounds cross over? The answer, which you get in the form of a delay in graphics redrawing when you zoom in close, is that the authors have heavily thickened the lines of the gray graphic object. (If they were thin, the pinch would be obvious.) Now why would they do that? To hide the fact that the confidence region pinches as the bounds cross over? Anyways, this is probably why reviewers never caught the error. But it would be nice to know to what degree this deception was intentional. • bender Posted Jul 9, 2010 at 9:03 AM \| Permalink \| Reply “Hide the crossover”. 17. bender Posted May 15, 2007 at 11:39 PM \| Permalink \| Reply The new SI at least answers john lichtenstein’s #5 from the previous thread: Am I the only one perplexed with the confidence interval around 1400, 1650, and 1750? The new SI (which starts from 1505) shows the CI’s around 1650 and 1750 to be very wide. Still, as Annan argues, the CI’s are wrong. In order to get the uncertainty on the inferred value, the uncertainty in the observed value (the proxy value on which the temp reconstruction is based) has to propagate through the uncertainty in the regression coefficients. It makes no sense to assume all the uncertainty comes from the standard errors in the estimated regression coefficients. Some of it must come from sampling error from the estimation of the means. It is very important to know exactly what these folks are doing. 18. Posted May 16, 2007 at 1:46 AM \| Permalink \| Reply Some notes / questions: TLS uses first principal component, minimizes perpendicular distances from data points to the line ( a new line to my calibration-line-plot ) ? In Hegerl J. Climate paper Note that if the uncertainties in the paleo reconstruction are much larger than in instrumental data, an alternative is the use of inverse regression, neglecting error in instrumental data (Coehlo et al., 2004). I think this is an misunderstanding. IMO Coehlo explains CCE and ICE quite well, and Hegerl et al didn’t read carefully. ICE requires prior distribution for temperatures, and another viewpoint of ICE was just developed in http://www.climateaudit.org/?p=1515#comment-108922 and following comments. If I were a reviewer, I’d stop reading that manuscript right there, page 7. ( I don’t have the final version, was anything corrected? ). I’m not surprised that their CIs went wrong.. 19. bender Posted May 16, 2007 at 7:41 AM \| Permalink \| Reply Suppose: inferred Temperature = A * Proxy measure + B where A, B are regression parameters, each with standard errors, and Proxy P is subject to sampling error. Then the error in the inference T is the quadrature sum of the errors in AP and B, and the error in AP here is the straight sum of the relative errors in A and P. Is that not the correct way to compute the confidence level for a quantity inferred from an error-prone calibration? That’s what I learned in high school physics, anyways. 20. Steve McIntyre Posted May 16, 2007 at 7:48 AM \| Permalink \| Reply Gabi Hegerl has sent me the following email last week in response to my request for further particulars on her method: if you tell me which part you find hard to understand, I can send you an algorithm. The records are processed by a number of programs. Tom;s teaching is over for the semester so I think you’ll get more detail out of him soon. Just so that I meet the requirements of other readers, UC and bender, can you summarize any questions? BTW I got a copy of the Hegerl J CLim article as published posted up here Ther references ot an SI have been removed and there is no SI at the J Climate website. 21. Earle Williams Posted May 16, 2007 at 1:43 PM \| Permalink \| Reply Re #20 Steve M, The URL for the Hegerl paper is a bit off. It looks to be http://data.climateaudit.org/pdf/Hegerll07.jclim.pdf 22. Steve McIntyre Posted May 21, 2007 at 10:17 PM \| Permalink \| Reply James Annan writes: Zero-width confidence intervals are not necessarily wrong. They are rather unconventional, perhaps, but that doesn’t make them incorrect. 23. Willis Eschenbach Posted May 22, 2007 at 2:55 PM \| Permalink \| Reply Zero-width confidence intervals are not necessarily wrong. They are rather unconventional, perhaps, but that doesn’t make them incorrect. Man, that’s just plain and fancy footwork. I can’t offhand think of any physical measurement that has a confidence interval of zero. Examples, anyone? w. 24. bender Posted May 22, 2007 at 3:10 PM \| Permalink \| Reply Re #23 That remark baffled me, too. (I didn’t say anything because I’ve been commenting too much lately.) • ### Tip Jar (The Tip Jar is working again, via a temporary location) • ### NOTICE Click on the "Reply" link to respond to a comment. Frequent visitors will want the CA Assistant. Sort/hide comments; improved reply box, etc. • ### Blog Stats • 10,052,573 hits since 2010-Sep-12 • ### Recent Comments %d bloggers like this:	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 7, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9246657490730286, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/75425-reduction-formula.html	# Thread: 1. ## reduction formula hey there, first time poster, long time struggler...so i need some help with a problem. it says to find the reduction formula for the integral (e^x)(sin x)^n in terms of the integral (e^x)(sin x)^n-2. i know its integration by parts but i have no idea what to do for my substitutions because (sin x)^n on its own is difficult enough so how do I incorporate the e^x? cheers! 2. Originally Posted by jackm7 hey there, first time poster, long time struggler...so i need some help with a problem. it says to find the reduction formula for the integral (e^x)(sin x)^n in terms of the integral (e^x)(sin x)^n-2. i know its integration by parts but i have no idea what to do for my substitutions because (sin x)^n on its own is difficult enough so how do I incorporate the e^x? cheers! Integration by parts twice. For the first use $u = \sin^n x,\; dv = e^x\,dx$ so $du = n \sin ^{n-1}x \cos x\, dx \;\;v = e^x$ and $e^x \sin^n x - n \int e^x \sin ^{n-1}x \cos x\,dx$ For the second by parts $u = \sin ^{n-1}x \cos x,\; dv = e^x\,dx$ and use some identites - see how that goes.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9163363575935364, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/1706/does-set-mathbbr-include-zero/1740	# Does set $\mathbb{R}^+$ include zero? I've been trying to find answer to this question for some time but in every document I've found so far it's taken for granted that reader know what $\mathbf ℝ^+$ is. - 19 It depends on the choice of the person using the notation: sometimes it does, sometimes it doesn't. It is just a variant of the situation with $\mathbb N$, which half the world (the mistaken half!) considers to include zero. – Mariano Suárez-Alvarez♦ Aug 6 '10 at 9:05 8 It is just as Mariano says. In this case, the "correct" convention is that it should not include zero (after all, zero is not positive), but you certainly can't count on this: about half the time, the author means to include $0$. – Pete L. Clark Aug 6 '10 at 9:17 7 You will often find $\mathbf R^+$ for the positive reals, and $\mathbf R^+_0$ for the positive reals and the zero. – zar Aug 6 '10 at 10:20 2 Let @zar post it as an answer and we are done. :) – Pratik Deoghare Aug 6 '10 at 13:38 5 I tend to use $\mathbb{R}_{> 0}$ or $\mathbb{R}_{\geq 0}$ and avoid that notation altogether. – Andrea Ferretti Aug 6 '10 at 14:22 show 8 more comments ## 5 Answers It depends on the choice of the person using the notation: sometimes it does, sometimes it doesn't. It is just a variant of the situation with $\mathbb N$, which half the world (the mistaken half!) considers to include zero. - You will often find $\mathbf R^+$ for the positive reals, and $\mathbf R^+_0$ for the positive reals and the zero. - As a rule of thumb most mathematicians of the anglo saxon school consider that positive numbers (be it $\mathbb{N}$ or $\mathbb{R}^{+}$) do not include while the latin (French, Italian) and russian schools make a difference between positive and strictly positive and between negative and strictly negative. This means by the way that $0$ is the intersection of positive and negative numbers. One needs to know upfront the convention. - Mh, here in Italy when we say "positive" we mean "$>0$", not "$\ge 0$". – zar Aug 6 '10 at 17:10 3 Hmm. As above, for me $\mathbb{R}^+ = (0,\infty)$, but $\mathbb{N}$ includes zero. Hence I wouldn't call the latter the set of positive integers: for that I use $\mathbb{Z}^+$. Seems logical... – Pete L. Clark Aug 6 '10 at 17:35 As well in France the anglo saxon school is taking over, but if you read Italian mathematicians of the beginning of the 20 th century you will realise that like Bourbaki slightly later in France positive included zero – marwalix Aug 6 '10 at 19:14 2 Early editions of Bourbaki indeed defined zero to be both positive and negative, but by the 1930s even Bourbaki changed their mind. As a general rule, $\mathbb{N}$ excludes zero if and only if you are a number theorist. – JeffE Aug 23 '10 at 19:54 1 In my first semester of graduate school (in the U.S.), my algebra professor told us on the first day that $\mathbb{N}$ includes zero and $\mathbb{Z}_+$ does not. That afternoon, my analysis professor told us the exact opposite. How tedious! I prefer to say "positive" or "nonnegative" rather than use these symbols, but have also come to appreciate $\mathbb{Z}_{\geq 0}$ and $\mathbb{Z}_{>0}$ where brevity is desired. – Jonas Meyer Aug 27 '10 at 23:06 show 1 more comment I write, e.g., $\mathbb R_{>0}$, $\mathbb R_{\geq0}$, $\mathbb N_{>0}$. - I met (in IBDP programme, UK and Poland) the followin notation: $$\mathbb{R}^{+} = \{ x \| x \in \mathbb{R} \land x > 0 \}$$ $$\mathbb{R}^{+} \cup {0} = \{ x \| x \in \mathbb{R} \land x \geq 0 \}$$ With the explanation that $\mathbb{R}^{+}$ denotes set of positive reals and $0$ is neither positive nor negative. $\mathbb{N}$ is possibly a sligthly different case and it usually differ from branch of mathematics to branch of mathematics. I belive that is usually includes $0$ but I belive theory of numbers is easier without it. It can be easilly extended in such was to have $\mathbb{N}^+ = \mathbb{Z}^+$ denoting positive integers/naturals. Of course as noted before it is mainly a question of notation. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 32, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9524615406990051, "perplexity_flag": "middle"}
http://mathhelpforum.com/algebra/45224-how-factorise.html	# Thread: 1. ## how to factorise this Trying to jog my memory here, how would I go about factorising this, as I'm trying to find its limit as x approaches 3 from the left. \[ \frac{{x^2 - x + 12}}{{x + 3}} \] 2. fyi, the numerator will not factor, but why factor at all? there is no discontinuity at x = 3 ... just plug in 3 for x and calculate the limit straight up. 3. sorry should be as x approaches -3 (not 3). How would I go about finding the limit then in this case? 4. Originally Posted by Craka sorry should be as x approaches -3 (not 3). How would I go about finding the limit then in this case? It's already been pointed out that the numerator does not factorise. So you can't cancel a common factor of x + 3. The limit approaches -oo as x --> -3 from the left. If you don't like this answer, I suggest you go back to the original question and check whether you made a mistake in copying it. If you haven't made a mistake in the equation posted, then I'm afraid you're stuck with the answer I've given. 5. Sorry to clarify, the question ask to find the value for the for the following expression. $<br /> \mathop {\lim }\limits_{x \to ( - 3)} \frac{{x^2 - x + 12}}{{x + 3}}<br />$ The answer given is -7. 6. Originally Posted by Craka Sorry to clarify, the question ask to find the value for the for the following expression. $<br /> \mathop {\lim }\limits_{x \to ( - 3)} \frac{{x^2 - x + 12}}{{x + 3}}<br />$ The answer given is -7. Did you check the question like I asked? To get that answer the numerator must be $x^2 - x {\color{red}-} 12$ NOT $x^2 - x {\color{red}+} 12$ as you've posted several times now. $x^2 - x - 12 = (x - 4)(x + 3)$. 7. Yes I have checked the question several times. That is the question being asked. From what you are saying it appears there is a mistake in the question being asked on the paper. 8. Originally Posted by Craka Yes I have checked the question several times. That is the question being asked. From what you are saying it appears there is a mistake in the question being asked on the paper. Looks like it. 9. Apply the L'HOSPITAL RULE differentiate with respect rule u will get =lim x-->-3 (2x-1) =2(-3)-1 =-6-1 =-7 10. Originally Posted by Craka Trying to jog my memory here, how would I go about factorising this, as I'm trying to find its limit as x approaches 3 from the left. \[ \frac{{x^2 - x + 12}}{{x + 3}} \] Originally Posted by Craka sorry should be as x approaches -3 (not 3). How would I go about finding the limit then in this case? Originally Posted by tutor Apply the L'HOSPITAL RULE differentiate with respect rule u will get =lim x-->-3 (2x-1) =2(-3)-1 =-6-1 =-7 NO!! The limit does NOT have an indeterminant form 0/0. Using l'Hopital's Rule here is totally wrong. This is the problem with l'Hopital's Rule - it's a lazy approach that is easily abused. (And in a pre-algebra forum I doubt that l'Hopital's Rule has ever even been heard of).	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.967134416103363, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?p=4065657	Physics Forums Page 2 of 4 < 1 2 3 4 > ## The problem with infinity. Quote by bahamagreen Thanks for mentioning those, I had never run across the Ant On A Rubber Rope before. Might Hilbert's Hotel have a flaw in the premise that relates to problems with infinity? If an infinite number of rooms are each occupied by a guest, where does the new guest come from? Some think that an infinite collection must necessarily contain all instances... It was Georg Cantor that demonstrated that not only was actual infinities perfectly logical but that it necessarily entailed orders of infinity, called aleph numbers $\aleph$ or cardinality (a powerset). Cardinality is basically the size of an infinite set. This was controversial in Cantor's day since the only infinity acceptable prior to that was unbounded sets, or potential infinities. You can look up his work, including his diagonal argument, for a more quantitative treatment. I'll just give a more intuitive description. If you have a finite interval, i.e., distance between two points, then logically you can divide it into an infinite set of infinitesimal points. To answer the above question one way, it cannot be said that the infinite set of points between 0 and 1 contain all possible numbers. There is not only more than one infinite set, there are an infinite set of infinite sets. There is no flaw in Hilbert's Hotel. There is also countably infinite sets and uncountable sets. There is also open sets and closed sets, which differ only in whether they contain the infinitesimal boundary points or not. There are also dense sets, which play a role in many quantum foundation arguments, including EPR and the scalability of quantum computing. The one thing you cannot do is make a priori generalized statements, like infinity must contain the entirety of the whole Universe to be infinite. Any finite subset of the Universe also contains an infinite set of infinitesimal points. Neither can you, for the same reason, make the claim that two infinite sets must be the same size. You must restrict your statements about infinity to those statements that can be mathematically demonstrated to be consistent, and avoid the intuitively implied and inconsistent properties Zeno's arguments depended on. Quote by my_wan To answer the above question one way, it cannot be said that the infinite set of points between 0 and 1 contain all possible numbers. There is not only more than one infinite set, there are an infinite set of infinite sets. There is no flaw in Hilbert's Hotel. I'm not thinking that the infinite set of points between 0 and 1 contains all possible numbers, only that it contains all possible numbers between 0 and 1. It seems to me by definition, the set of points between 0 and 1 must include every point between 0 and 1. Are you suggesting otherwise? I'm thinking that any arbitrary number I specify between 0 and 1 must already be included in the set of points between 0 and 1; so I don't see how any possible number between 0 and 1 is not already a member of the set of points between 0 and 1. The "new guest" coming to Hilbert's Hotel's is like a point between 0 and 1 that is not a member of the set of points between 0 and 1... I see this as a flaw in the premise. If the 0 to 1 range is problematic, we can do the same with the set of natural numbers... I'm thinking that the set of natural numbers must include any and all arbitrary natural numbers that I may specify... this seems clear by definition. If each occupied room is mapped to a natural number, an infinite number of rooms means all the natural numbers are mapped, as are their corresponding guests... the "new guest" would need to represent an unmapped natural number, but there are none, by definition. Maybe I'm missing something...? Quote by bahamagreen The "new guest" coming to Hilbert's Hotel's is like a point between 0 and 1 that is not a member of the set of points between 0 and 1... I see this as a flaw in the premise. How does it remotely resemble this? EDIT: Upon closer inspection, the paradox is simply mapping natural numbers to new natural numbers. The set ##\mathbb{Z}\cup\left[1,\infty\right)## has the same cardinality as the set ##\mathbb{Z}\cup\left[2,\infty\right)##. This should solve the paradox quite easily. Quote by Whovian Actually, in most cases, infinite means unbounded, so there would be no such edge. or finite without boundaries, like a torus. Quote by bahamagreen I'm not thinking that the infinite set of points between 0 and 1 contains all possible numbers, only that it contains all possible numbers between 0 and 1. It seems to me by definition, the set of points between 0 and 1 must include every point between 0 and 1. Are you suggesting otherwise? In the original post I addressed it was implied that maybe if there was an infinite number of hotel rooms, then these rooms being occupied implied infinite guest such that there could be no new guest. If there are an infinite set of point between the two points, [0,1], and each of these correspond to a hotel room occupied by a point, then the original suggestion to work around the hotel hotel paradox implies that this infinity of points, [0,1], contains all points that might occupy the infinity of hotel rooms. Hence I made the suggestion in order to provide proof by contradiction that the hotel paradox was not flawed. I'm thinking that any arbitrary number I specify between 0 and 1 must already be included in the set of points between 0 and 1; so I don't see how any possible number between 0 and 1 is not already a member of the set of points between 0 and 1. The "new guest" coming to Hilbert's Hotel's is like a point between 0 and 1 that is not a member of the set of points between 0 and 1... I see this as a flaw in the premise. If the points between 0 and 1 are an infinite set of occupied hotel rooms, and yet "new guest" are still available from members that are not member of the set of points between 0 and 1, why is this a special case? The original suggestion was that an infinite number of guest implied no more guest exist, but here you add a special case to say there are more guest available from sets other that [0,1]. If the 0 to 1 range is problematic, we can do the same with the set of natural numbers... Precisely. The infinity problem is just as big in the interval [0,1] as it is in the interval [0,∞]. I'm thinking that the set of natural numbers must include any and all arbitrary natural numbers that I may specify... this seems clear by definition. If each occupied room is mapped to a natural number, an infinite number of rooms means all the natural numbers are mapped, as are their corresponding guests... the "new guest" would need to represent an unmapped natural number, but there are none, by definition. Maybe I'm missing something...? Only problem is that I can pull new guest from the infinite set of real number which you didn't included here. Note that the numbers are merely name tags on the guest, and it make no difference which ones you label with which numbers. I can relabel an infinite number of guest labeled with even numbers with odd numbers, and visa versa, and the count remains the same. I can also relabel all natural numbers as real numbers simply by multiplying their name tags with an irrational number and assigning them that number. It changes nothing about the total number of guest. Cantor showed that all these infinities existed, but we should not lose sight of the fact that they are mathematical infinities. Mathematically, what does it mean to say that something exists? If a mathematician can write down a set of non-contradictory axioms, and set down rules for deducing mathematically true statements from them, then those statements can be said to ‘exist’. This existence requires only logical self-consistency. Physical existence is completely unnecessary. If there can be a profound difference between physical and mathematical “existence” then it seems reasonable to identify a similar difference between physical and mathematical “truth”. Cantor’s infinities were all mathematical infinities, as are the rooms and guests in Hilbert’s Hotel. They may bear no relation to any possible physical infinity, which would include an infinite universe. The actual, ancient, fear of infinity was not removed; it was just that Cantor provided the world with a “label” that could be attached to infinity, which reads: “this is a mathematical infinity – it doesn’t bite”. my_wan, Your responses support my thinking that the Hilbert Hotel premise is flawed. A "complete" mapping to the natural numbers would be to include all of them by taking the natural numbers in order... 1,2,3... any other mapping scheme like 2,4,6... is an obvious mechanism to skip some numbers, yet there would be objection to a scheme that skipped points between 0 and 1. Claiming that the new guest could be from the set of real numbers when the set of guests is represented by the natural numbers misses the whole point. If the points between 0 and 1 represent the rooms, it is the guests that are mapped to the natural numbers. The paradox is based on the assumption that all guests, including any possible new guests, are all of the same class of thing - natural numbers, and violating that by suggesting a new guest could be a real number is like solving the question of the origin of the new guest by finding a chair and checking that chair into the hotel as a new guest... no, the new guest has to be a person like all the other guests. Quote by bahamagreen Claiming that the new guest could be from the set of real numbers when the set of guests is represented by the natural numbers misses the whole point. It doesn't miss the point any more than saying that the set of all hotel customers must consist of all possible hotel customers, and that is the only way you can claim there is nobody remaining to request a room in the hotel. If the real numbers represent the set of all present hotel customers, and the natural numbers represent the set of all people that might request a room, then there are an infinity of people that may request a room even after an infinite number of people have already filled the hotel. To assume otherwise is effectively an attempt to impose a boundary condition on an unbounded variable. Quote by Endervhar Cantor showed that all these infinities existed, but we should not lose sight of the fact that they are mathematical infinities. Very true, and so far the justification for "actual infinities" is fairly slim. However, attempting to avoid them has a number of problems. Avoiding actual infinities is just as problematic and paradoxical as accepting them. Modern mathematics didn't select the axioms simply to avoid contradictions with infinities, they where selected to avoid mathematical contradiction as a result of attempting to avoid them. In the physical sciences the scalability of quantum computers actually depends on these mathematical properties associated with mathematical infinities. This comes from the fact that a Hilbert space must be "complete", i.e., is a complete metric space. The very thing that allows the calculus of limits, and from which we derive our mathematical justifications for infinities. Quote by StJohnRiver There are physicists who insist that the universe is finite and has a distinct geometry. So what'd be the problem if the universe were infinite? Quote by my_wan It doesn't miss the point any more than saying that the set of all hotel customers must consist of all possible hotel customers, and that is the only way you can claim there is nobody remaining to request a room in the hotel. If the real numbers represent the set of all present hotel customers, and the natural numbers represent the set of all people that might request a room, then there are an infinity of people that may request a room even after an infinite number of people have already filled the hotel. To assume otherwise is effectively an attempt to impose a boundary condition on an unbounded variable. The set of natural numbers must include all possible natural numbers. The set of points between 0 and 1 must include all possible points between o and 1. The set of all hotel customers must comprise all possible hotel customers. You are defining: The reals is the set of present hotel customers. The naturals is the set of people that might request a room (not presently hotel customers). I'm suggesting that is a flaw because you are defining some persons as both a non-customer and a customer - because the naturals and reals share some members in common (all naturals are members of the reals, some reals are members of the naturals). But maybe I'm still missing something? Quote by bahamagreen The set of all hotel customers must comprise all possible hotel customers. This can't be justified. The set of all actual hotel customers anywhere in the world does not comprise all possible hotel customers. If that was necessarily true then in order for hotels to have any customers they must have every person on earth as a customer. Conversely, by this logic, since I am not a customer, either hotels have no customers or I am not a potential customer. Neither of which is true. Quote by MY Wan This can't be justified. The set of all actual hotel customers anywhere in the world does not comprise all possible hotel customers. As ObsessiveMathsFreak pointed out, this "depends on what you mean by infinite." In terms of mathematical infinities your assertion is undoubtedly true, but it is a mathematical "truth" and has no significance in reality. You cannot have an infinite number of rooms, an infinite number of people, or an infinite number of anything. Earlier, someone suggested that infinite and boundless might be synonymous; this is not so. Anything that is infinite is boundless, but not everything that is boundless is infinite. Quote by nikkoo Actually, infinite means unbounded Sorry, Nikkoo, I missed your quote when I was looking for it. You will probably want to take issue with my previous post. :) Quote by Endervhar As ObsessiveMathsFreak pointed out, this "depends on what you mean by infinite." In terms of mathematical infinities your assertion is undoubtedly true, but it is a mathematical "truth" and has no significance in reality. I have already gone over how it is relevant to the physical sciences, by way of Hilbert space. You can call Hilbert space a mathematical fiction, or slide rule of sorts, used to calculate. But a rejection of mathematically defined infinities has very real empirical consequences. The inability to scale quantum computers being a major one. Others involves issues surrounding Bell's theorem, and other no-go theorems. You cannot escape the issues mathematicians have worked around with a "mathematical fiction" clause. Finite mathematics is inconsistent without mathematical infinities. How do you propose the reinstate consistency if you reimpose a finiteness condition on the physical world? You can't simply close your eyes and pretend it has no physical consequences, whether those consequences ultimately justify actual infinities or not. Quote by Pianoasis Infinity is a number that cannot be divided, cannot be measured, and cannot be contained. This infinite universe obviously does not exist due to the fact that all pieces of space are made of this ultimately small unit. Every quantity can be described by this unit, thus making the concept of infinity null. That's an absurd assertion. The number line is infinite, but we can still use integers to measure and divide segments of the line. Quote by my_wan This can't be justified. The set of all actual hotel customers anywhere in the world does not comprise all possible hotel customers. If that was necessarily true then in order for hotels to have any customers they must have every person on earth as a customer. Conversely, by this logic, since I am not a customer, either hotels have no customers or I am not a potential customer. Neither of which is true. How about addressing my previous post #28 first? Page 2 of 4 < 1 2 3 4 > Thread Tools \| \| \| \| \|-------------------------------------------------\|----------------------------\|---------\| \| Similar Threads for: The problem with infinity. \| \| \| \| Thread \| Forum \| Replies \| \| \| Calculus & Beyond Homework \| 6 \| \| \| Calculus & Beyond Homework \| 8 \| \| \| Cosmology \| 28 \| \| \| General Math \| 13 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9594243168830872, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/84231/how-do-i-set-up-the-following-problem-to-arrive-at-the-answer?answertab=votes	# How do I set up the following problem to arrive at the answer? A warehouse has 10 unlabelled rows of pallets. Each row of pallets contains thousands of cell phones destined for different countries. Each 100 gram cell phone is exactly the same except for those in the row destined for Japan, which have a “special” 2 gram chip encased within each phone to make sure it works on the Japanese networks. How can the warehouse manager make sure the right phones go to Japan in the quickest time possible? All the warehouse manager has at his disposal is a digital balance. - If this is homework, please tag it as 'homework'. Regardless, please share with us what you have attempted so far. What is the maximum capacity of the digital scale? We don't want to break the manager's only scale, do we? – jthetzel Nov 21 '11 at 1:29 1 How does statistics enter into this problem? It seems to be similar to one of several math puzzles involving minimum number of steps to identify the odd piece. If so, this question may be off-topic here and perhaps on topic on math.SE. – varty Nov 21 '11 at 4:57 1 Do the phones intended for Japan weigh $102$ grams instead of $100$ grams? I don't get this from the problem statement, e.g. the $2$ gram chip in phones intended for the Japanese market might be replacing a $1.5$ gram chip, which would make @PeterFlom's solution not work.... Please edit your problem statement to state the weight of the special phones explicitly. – Dilip Sarwate Nov 21 '11 at 15:55 ## 1 Answer This is an old old puzzle that can be found in many books. Solution: Weigh 1 cell phone from row 1, 2 from row 2 and so on, all at the same time. Divide the weight by 100. Divide the remainder by 2. That's the row number with the Japanese phones. - You probably want to subtract $5500$, not divide by $100$. A total of $55$ phones are being weighed, and so the total weight is $5500 + 2n$ where the $n$ phones from pallet $n$ are contributing an extra $2n$ grams of weight. – Dilip Sarwate Nov 21 '11 at 19:53	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 11, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9435410499572754, "perplexity_flag": "middle"}
http://stats.stackexchange.com/questions/22940/bayesian-prior-corresponding-to-penalized-regression-coefficients	# Bayesian prior corresponding to penalized regression coefficients I'm working on a Bayesian Regression problem where I would like to estimate the beta coefficients subject to this constraint (penalty): $\sum\|\beta_i\|<C$ or similarly $\sum \beta_i^2<C$ Which is basically a Lasso or L2 Penalty. Now, if I understand correctly, we constrain the coefficients through the prior in Bayesian analysis. Therefore my question is what would an appropriate prior be for the Betas? I should note, that for my case, the betas are restricted to be positive, they cannot be negative. - ## 4 Answers L2 penalty penalizes the sum of squared betas but not via a constraint such as $< C$. The L1 penalty is the lasso. For the Bayesian lasso see the 2008 JASA paper by Trevor Park and George Cassella. - Thanks, I wasn't sure what modifications I need however if I want to restrict the prior so only positive values are possible. – Glen Feb 16 '12 at 4:46 @Frank: Do you mind explaining your first comment? For each value of the regularization parameter $\lambda$, there is, indeed a constant $C$ corresponding to the constrained optimization problem of the OP. And vice versa. (Just think about the Lagrangian formulation.) – cardinal Feb 16 '12 at 15:13 My understanding of $L_2$ penalty is that large values of sum of squared $\beta$s are penalized against but there is no cutoff on how large this sum can be. – Frank Harrell Feb 17 '12 at 3:22 @FrankHarrell - this is true, but characterising the maximum $\beta$ is 1-to-1 between $C$ and $\lambda$. For any $C$ which leads to a maximum $\hat{\beta}_C$ there is a corresponding $\lambda(C)$ which leads to the same maximum $\hat{\beta}_{\lambda(C)}=\hat{\beta}_{C}$ – probabilityislogic May 6 '12 at 0:12 @prob: The correspondence is not quite bijective, unless restricted to subintervals of the positive half-line. See my previous comment to this effect above. – cardinal May 6 '12 at 0:16 For the lasso penalty this corresponds to a double exponential prior - so long as you are taking the posterior mode as your estimate. If you constraint the betas to be positive then you have an exponential prior. The parameter of the exponential distribution $\lambda$ has a correspondence with your $C$ in that you can choose a value of each such that the same naximum is achieved. The bayesian prior is the langrangian form (on log scale) of the constraint. For the ridge penalty constraning the parameter to be positive is just a truncated normal distribution. - The $L_2$ constraint on the coefficients is Tikhonov regularization. It turns out that for the case where the prior is multivariate normal and the model is linear, the posterior is also multivariate normal. It turns out that the mean of the posterior distribution occurs at a point in parameter space that can also be obtained by Tikhonov regularization and the relationship between the Tikhonov regularization parameter and the equivalent multivariate normal prior is pretty simple. This is textbook material that can be found in Tarantola's textbook among other places. When the model being fit is nonlinear, or there are additional constraints on the parameters (such as your nonnegativity constraint), or the prior isn't MVN, it all becomes much more complicated and this simple equivalence breaks down. - If you want your $\beta$s to be non-negative and sum to a given value then it seems a scaled Dirichlet prior would make sense. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 18, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9094456434249878, "perplexity_flag": "head"}
http://mathoverflow.net/questions/61829?sort=newest	## Best introduction to probability spaces, convergence, spectral analysis ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I'm not sure if this stuff all falls under what most would just term "probability", but I'm researching applied macroeconomics and need to get a handle on the following concepts ASAP: • probability spaces and sigma algebras • Borel sets • convergence • stationarity/ergodicity • martingales • laws of large numbers • spectral analysis The text I'm using right now, Fabio Canova's Methods for Applied Macroeconomic Research, touches on all these things briefly in the space of about twenty-five pages, but it's pretty impenetrable. Does anyone know anything off-hand that presents these topics in a more practical, easier-to-digest way? This is a little bit better, but does anyone have any other suggestions? - As a mathematician, I really like the book by Durrett. But if you are not familiar with sigma algebras... maybe you should begin with a book on integration (for example the book by Rudin "Real and complex analysis"). Note that "ASAP" will probably require some time (a couple of month ?). Note also that you may get more answers on a forum devoted to teaching. – camomille Apr 15 2011 at 15:38 Jacod and Protter's book is a reasonably friendly introduction, and doesn't require any background beyond multivariable calculus. As camomille said, you're looking at a couple of months of work to take all this stuff on board. – Simon Lyons Apr 15 2011 at 16:00 Try "Measure, integral and probability", too. It doesn't cover all the topics you mentioned, but you should be able to work through it quite quickly, and it will establish a good base to build on. – Simon Lyons Apr 16 2011 at 14:51 ## 11 Answers There is no royal road to probability. The closest is W. Feller's book, which has many (but not all) of the topics you mention, but I strongly advise reading (at least parts of) it first. Otherwise, you will go through life hopelessly confused. - After looking at most of the other books listed it seems like Feller's book is the easiest for the non-mathematician to pick up. Thanks for the recommendation. – jefflovejapan Apr 16 2011 at 14:09 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. I just started up a blog and have put a link to my own alternative introduction to advanced probability theory. It could be useful as an introduction to Probability With Martingales, and might help you see the route ahead, as it were. http://fermatslastspreadsheet.wordpress.com/2011/11/30/introduction-to-probability-theory/ - Any thoughts on how Cinlar compares to Klenke, "Probability Theory A Comprehensive Course"? They both appear to be excellent recent alternatives to the likes of Billingsley and Chung. Also what would be good or required prep for Cinlar? Capinski and Kopp (seems to have a lot of overlap with Cinlar and Klenke)? Royden and Fitzpatrick (certain chapters)? - The book "Probability and Stochastics" by Professor Erhan Cinlar is among the best introductory (actually it touches deeper things too) book on probability theory, assuming a little background on integration in general measure spaces. It starts with basic things (probability spaces, convergence, conditioning), and ends with some deeper waters like Levy-Ito for Levy processes, Hunt processes. Professor Cinlar is an expert in the theory of Markov processes (e.g. he co-authored papers that characterize Markov semimartingales), among many other things. The book uses very precise language to describe things, in a very streamlined way. The author guides the reader through more difficult materials very easily, without making the reader feel the difficulty. In other words, pedagogically (in addition to the broad spectrum of things and the depth of the treatment) this book is among the best IMHO. I would recommend this book to anyone with an interest in probability theory and stochastic processes. In terms of coverage of the OP's list, the book covers at least the following: probability spaces and sigma algebras, Borel sets, convergence, martingales, laws of large numbers - Many of the really good introduction-type books have already been mentioned. As a current grad student I encounter many of them on an almost daily basis and would suggest the following: • Billingsley - "Probability and measure", although I would skip the first part about the dyadic intervals. • Durrett - "Probability: Theory and examples". I used the 3rd version when I was taught from this book and then it did not have that much measure theory in the, sense that it was confined to the appendix. As I understand it this is not the case for the 4th edition and I really love the way Durrett presents the material so this is a good starting point. • Shiryayev - "Probability". Great book from one of the current masters. It starts with an intuitive discussion about probability theory and then moves on to develop the mathematical theory needed (sample spaces and so on) in order to "do real probability". I haven't read the entire thing so that's why I can't put this as my #1 choice, I seems as if it has the potential though. • Williams - "Probability with martingales". Very nicely written account of measure theoretic probability. It is quite concise though and depending on your mathematical maturity it could perhaps be a bit difficult to follow completely. One possibility is to keep the author's "Weighing the odds" at your side to get the elementary theory as a complement to the more advanced book. • Feller - There is a reason why it's considered a classic. Another, perhaps somewhat odd and unconventional, choice might also be Tomas Björk - "Arbitrage theory in continuous time". Now it's not an introduction to probability but depending on what type of economics you are interested this could be a nice reference book. It has an appendix devoted to measure theory and probability and Prof. Björk being one of the best teachers I've had in terms of providing you with an intuitive feeling for the subject at hand, I'm sure that such sections would work very well for someone who is just trying to get a basic understanding of the subject. - OK ... "probability spaces" or "probability theory"?? Most of the other responses tell you texts for probability theory, where probability spaces are mentioned only in passing if at all. This would include the last 4 of your topics. On the other hand, "probability spaces" would be a branch of "measure and integration" theory. - That's an important distinction, but it certainly looks from the list of topics like the OP is interested in learning about probability theory, and most books on probability theory do devote a chapter to probability spaces before those recede to the background. – Mark Meckes Apr 17 2011 at 23:54 Given your starting point, specific interests, and desire to get to the bottom line fast, I recommend starting with Probability with Martingales by David Williams, which is yet another first-year graduate text, but more concise than the others. - I agree with tipanverella that you're probably looking for textbooks designed for first-year graduate students in this field. I think you'd be hard-pressed to find one book that covers all the topics listed, but there's nothing wrong with reading parts of several books to get you up to speed. For everything except ergodic theory and spectral analysis, I highly recommend Probability Theory and Elements of Measure Theory by Heinz Bauer. This book has a fantastic first chapter and really develops $\sigma$-algebras from scratch (most books I've seen just throw the definitions at you). Throughout the development of measure theory, connections are made to probability and part II of the book is all probability theory. There's a whole chapter on the Law of Large Numbers and another on Martingales. By the way, I would NOT read Rudin for this. Rudin's great if you already know the material and want a reference with the shortest, most elegant proofs possible (with many details skipped), but for actually learning the material and for connecting it to probability theory his book is not the greatest. Williams' Probability with Martingales really emphasizes exercises, so if you want examples with all the details worked out they will not be as prevalent. But the examples and exercises he gives are really great if you work through them yourself. I don't know much about the other books mentioned. As for Ergodic Theory, I really like An Outline to Ergodic Theory by Steven Kalikow and Randall McCutcheon. If you click that link you can check out the table of contents and the introduction, which includes great examples to get you started as well as statements of all the big theorems you will need. I know this wasn't part of your question, but you may also want to add entropy to your list of things to learn. This book would help with that as well. Another thing you may find yourself needing is the Borel-Cantelli Lemma (found in Bauer's book). Good Luck! - Those are probability topics, so I would go with what graduate students in probability are trained with. These past few years it seems that the standard is Rick Durrett's "Probability: Theory and Examples" (4th edition, Cambridge U. Press, 2010). MOst of what you want are in the first 5 chapters. Also a version of the book is provided in PDF format by the author, http://www.math.duke.edu/~rtd/PTE/PTE4_Jan2010.pdf. You could also get Shiryaev's "Probability" from Springer-Verlag or Stroock's "Probability Theory, an analytic view. Hope that helps. Cheers, Tipan - These don't have anything on spectral analysis, but cover the rest pretty well: Infinite Dimensional Analysis: A Hitchhiker's Guide by Aliprantis and Border, and Stochastic Limit Theory by Davidson. The first book was written for economic theorists, the second one for econometricians. So they specialize quite nicely to what you seem to be working on. However, they are both rather advanced. A great introduction to measure theoretic probability is Probability with Martingales by Williams. It's quite chatty and fun, but does still require some mathematical sophistication. A book that I think is a bit dry but that proceeds in small and easy steps with all the details included is A Probability Path by Resnick. - 1 +1 for mentioning Williams' Probability with martingales, aka the blue book. – Didier Piau Apr 16 2011 at 5:56 I looked at Probability with Martingales, and I really enjoy the style, but it's a little out of my reach. I'm going to check out Hitchhiker's Guide on Monday. Davidson and Resnick aren't available at my library, but I put requests in for them. Thanks for the advice. – jefflovejapan Apr 16 2011 at 14:11 Hitchhiker's guide is considerably more technical than Williams. It's very complete, but there's a danger of information overload there. – Simon Lyons Apr 16 2011 at 14:49 The first two books are both great pure math texts, and its odd that they both come out of economics. – arsmath Jun 30 2011 at 12:39 I think Probability and Measure by Patrick Billingsley is what you're looking for. It contains all of the topics you're asking about (except maybe spectral theory), and I feel it does a good job introducing the world of probability to the reader without assuming you already understand it, which Durrett's book tends to do. - It looks good, but the starting point's too advanced for me. Thanks for the advice. – jefflovejapan Apr 16 2011 at 14:19	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9563212990760803, "perplexity_flag": "middle"}
http://mathhelpforum.com/geometry/1637-geometry-help-needed-please.html	# Thread: 1. ## geometry help needed, please Why is line segment ab = line segment ab? what is the name of the reason/postulate/whatever it is called? Also, if you know that you have 2 triangles and you are trying to prove they are congruent and you know that 2 sides are equal and 1 angle is equal, which reason/law/postulate do you cite? Thank you! 2. Originally Posted by e-mom Why is line segment ab = line segment ab? what is the name of the reason/postulate/whatever it is called? Also, if you know that you have 2 triangles and you are trying to prove they are congruent and you know that 2 sides are equal and 1 angle is equal, which reason/law/postulate do you cite? Thank you! $AB=AB$ Called the Reflexsive Postulate. When two triangles have the same sides and and included angle they are congruent. This is called S.A.S. which means SIDE ANGLE SIDE. With ANGLE in between SIDE and SIDE, which was required. Thank you! 4. ## geometry help needed, again If you have 2 line segments, which form the 2 sides of a traingle, and have been given the fact that corresponding segments of the sides are equal, what rule could you cite to prove that the entire length of the 2 sides are the same. I am eventually trying to prove that the triangle is isosceles. A diagram is attached. Thanks! Attached Files • math help document.doc (19.0 KB, 59 views) 5. The way how I understand the problem there is no way to prove what you are saying because it is wrong. I understand it as that you know the small one is isoseles and and to show the big one is also isoseles but there is not enough infromation. Perhaps you wanted to say that line down the middle is a bisector? 6. ## thanks Thank you for the info that my reasoning wasn't going to prove it that way. I'll look for another way. Thank you! #### Search Tags View Tag Cloud Copyright © 2005-2013 Math Help Forum. All rights reserved.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9681377410888672, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/52075/solution-of-plateau-problem-for-a-simple-smooth-closed-curve-on-a-riemannian-man/52168	## Solution of Plateau Problem for a simple, smooth closed curve on a Riemannian Manifold (Kahler) gives a surface that can be parametrized by a closed disk? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Hi, Perhaps it's a stupid question, in that case i'll delete it. Let M be a compact orientable smooth (Kahler if changes things) manifold of dimension $dim_{\mathbb{R}}(M)=2n$ with $n\geq1$, let $\gamma$ be a simple smooth closed curve that lies in a (holomorphic) coordinate chart and that can be taken as small as necessary (one can choose $\gamma$ such $diam(\gamma)<\epsilon$ with $\epsilon>0$). If i want to solve Plateau problem for a $\gamma$ so small such that the solution is contained in the coordinate chart, do i get something that can be parametrized by closed disc? Thank you in advance. Edit: I'll try to clarify my question, this is what i wanted to know. Let $U$ be a sufficiently small geodesically convex set of a manifold $M$ and $\gamma$ a smooth simple closed curve lying in $U$ (no other assumptions on $\gamma$). 1) Can $\gamma$ be the boundary of an embedded closed disk? 2) If $\gamma$ can be the boundary of a closed disk, then can it be the boundary of a minimal (as a surface, not only among the disks that it bounds) embedded closed disk? I anticipate that i couldn't see works of Douglas so i don't know if the answer to my question is there. My suspect was that the answer could be yes for dimension 2 (i think about jordan curve theorem), Professor Thusrston example of the knotted curve suggests me that in dimension 3 i need additional assumptions on the curve not only on the linking number. But what happens for dimension $n\geq4$? - I think this is a good question. I haven't worked out the details myself, but I am pretty sure that you can prove this on any Riemannian manifold using standard tools in differential geometry. I'm hoping that a real expert will provide a reference, more explicit details, or a counterexample. The idea is to use the inverse function theorem applied to the appropriate functional to demonstrate the existence of a minimal surface parameterized by the closed disk. Then use a comparison argument to show existence and minimality. – Deane Yang Jan 14 2011 at 16:53 Just to echo what has been said, you should be more precise in your question. Namely, given your $\gamma$ there is a solution that is an (immersed) minimal disk spanning $\gamma$ namely a Douglas-Rado solution (see Bill's great answer for some details). This is because the $\gamma$ you describe is spanned by some disk and hence by direct methods in the calculus of variations is spanned by a minimal disk (of course easier said then proved). However, there may be many other minimal surfaces (or weaker concepts) spanning the curve so it is not correct to speak of THE solution. – Rbega Jan 15 2011 at 18:01 I edited the question, i hope it is more clear now – Italo Jan 16 2011 at 0:39 Bill Thurston's answer (which is quite good) shows that my comment is not right. At best the approach outlined by me might work if the initial curve is "sufficiently flat", whatever that means. – Deane Yang Jan 16 2011 at 3:24 ## 1 Answer I've been waiting for someone with more expertise than me to answer, but since they haven't (so far) I'll say something. There are different versions of Plateau-like problems; I'm not sure if there's a specific single one that's generally accepted as "the Plataeu problem". One can ask for a mapped-in disk with minimal area having the given boundary, a mapped-in surface of minimal area, a current of minimal area, an integral current of minimal area, an embedded "minimal surface" meaning that it's just a critical point for area among embedded (or mapped in if you prefer) surfaces, a minimal disk ... A lot is known about these different questions, and the answers aren't the same. First: even for a curve in Euclidean space, there might not be an embedded minimal disk. The easiest examples are for a knotted curve in R^3. However, there are also unknotted curves in R^3 that do not bound a disk in their convex hull, so they do not bound any embedded minimal disk. The minimum area of a disk bounding an unknotted curve grows exponentially in an appropriate measure of the complexity of the curve; the genus of a minimum area surface also grows exponentially. (Fred Almgren and I once wrote a paper about this). The same examples can be transported to any higher dimension, e.g. taking the product with another manifold. Second: Jesse Douglas showed how to find mapped-in minimal disks in great generality. This will work locally within any manifold of dimension. The basic technique is that for any parametrization of the curve by the boundary of a disk, first find an energy-minimizing map of the disk that extends this parametrization --- a harmonic map. Now consider the energy as a function of the parametrization (which is a kind of Teichmuller space). The critical points of the harmonic energy with respect to the Teichmuller space are minimal surfaces: the basic insight is that critical points of the harmonic maps within the Teichmuller space are when the harmonic map is conformal. When it is not conformal, you can change the conformal structure on the disk (which is equivalent to giving a reparametrization, by the uniformization theorem) and reduce energy. Third: minimal surfaces of any type (inclding currents) whose boundary is inside a convex set are always contained in the convex set. Here convex means that geodesic arcs between nearby points on the boundary are contained within the set. Even weaker conditions are sufficient, but this is good enough for your purposes --- a metric ball of small radius in any Riemannian manifold is convex in this sense. Fourth: there always exist minimizing objects of some sort, but they might not be things that you are happy with: one thing that happens is that $k$ times a curve might bound a surface of area less than $k$ times that for the original curve (in dimensions > 3). A limit of $1/k$ times the minimum area for $k$ times the curve might be a diffuse current, spread out in an entire region. I think this can happen even locally in a Kahler manifold, but I'm not sure. Even without using fractional weights, the minimizing object would in general be an integral current that is like a surface but with singularities. I don't know the classification for the 2-d case, but i'm sure the experts do. On the other hand, if you take a nearly round circle in a small coordinate chart, there should be an embedded minimal disk --- basically because nearly flat minimal surfaces are stable, so changing the metric a little bit gives you a new minimal surface by implicit function theorem type arguments (in the space of surfaces, which in this case can be described as graphs of functions). In R^3, there's a theorem that any curve whose total curvature is less than 4$\pi$ bounds an embedded minimal disk; I suspect there are known estimates likke this in Riemannian manifolds of higher dimension, using curvature assumptions about some convex set contining the curve, but I don't actually know. Is there an expert who can correct or extend what I've said? - Thank you very much for the detailed answer! I'll wait a little more before accepting it because i want to see if you or someone has anything else to say. – Italo Jan 16 2011 at 0:45 Not an expert, but I'll add one bit to (part 4 of) this excellent answer: Given a boundary, you can always find an area-minimizing rectifiable current with the given boundary. Almgren's big regularity paper says that the interior of this minimizer is an embedded smooth manifold away from a Hausdorff codimension two singular set. In the case where the boundary is 1-dimensional, Sheldon Chang further showed that the singular points are isolated classical branch points. – Dan Lee Jan 24 2011 at 22:08	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 23, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9383137226104736, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/122296-help-particle-motion-derivatives-etc-print.html	# Help with particle motion, derivatives, etc. Printable View • January 3rd 2010, 12:04 PM macroecon Help with particle motion, derivatives, etc. I have never seen any examples of these problems, and they showed up on a review sheet for the final exam in May. If you can help me through these or explain to me what I should do for each step it would help me a lot. Thanks. 1. A particle moves along a line so that at any time (t) its position is given by x(t)=πt+cos(πt). btw, π is pi. a) Find the velocity at time t. b) Find the acceleration at time t. c) What are all the values of t, 0≤t≤3, for which the particle is at rest? d) What is the maximum velocity over the interval 0≤t≤3? 2. Let f be the function defined by f(x)=(x^2 +1)e^-x for all x such that -4≤x≤4. a) For what value of x does f reach its absolute maximum? Justify. b) Find the x-coordinates of all points of inflection of f. Justify. 3. Let p and q be real numbers and let f be the function defined by: f(x)={1+3p(x-2)+(x-2)^2 for x≤2 f(x)={qx+p for x>2 a) Find the value of q, in terms of p, for which f is continuous at x=2. b) Find the values of p and q for which f is differentiable at x=2. c) If p and q have the values determined in part b, is f" a continuous function? Justify. • January 3rd 2010, 01:09 PM ANDS! The position, velocity and acceleration functions are all related in that: Velocity is the derivative of the position function, and acceleration is the derivative of the velocity function, and the second derivative of the position function. Therefore if the position function is defined as $x(t)=\pi t+cos(\pi t)$ (taking $\pi$ as a constant): A. $x'(t)=v(t)=\pi-\pi sin(\pi t)$ B. $x''(t)=v'(t)=a(t)=-\pi^2 cos(\pi t)$ C. For this problem, understand that for an object to be at rest, its velocity must be equal to 0. Therefore solve the equation given in A for 0: $v(t)=\pi-\pi sin(\pi t) \Longrightarrow 0=\pi-\pi sin(\pi t) \Longrightarrow \pi sin(\pi t)=\pi \Longrightarrow sin (\pi t)=1$ For what value(s) of t, between 0 and 3 will make the above equation 1? D. You have your velocity function - now you are being asked for what values of t will you have maximum velocity. In others words, optimize the velocity function. • January 3rd 2010, 01:21 PM ANDS! For your second problem, you are doing much of the same, so I will not go into detail in that, simply note that to find the maximum (or minimum of a fucntion) you need to find the derivatve and set it to zero. If you function has multiply zeros, then test (in the original function) for the critical point that has the largest value. For the inflection point, the method is the same, except you are take the second derivative, and finding the critical points. Once your critical points have been established, test points around them to see if your values of f(x) are changing sign. If so, then the critical point is an inflection point. Review your notes on this if you are having trouble but it is a simple exercise. For your third problem, remember that a function is continuous at a point if and only if the limit as x approaches a from either direction (the left or the right) is the same, and f(a) actually exists. Assuming it does exist: First equation: $f(2)=1+3p(2-)+(2-2)^2 \Longrightarrow f(2)=1$ Second equation: $f(2)=2q+p$ Equating the two equations: $1=2q+p \Longrightarrow \frac{1-p}{2}=q$ I will leave parts B and C to you. If you want a check on your answer post work here. • January 3rd 2010, 07:00 PM macroecon We aren't supposed to use a calculator on any of these. So for par c of number 1, what work am I supposed to show to justify my answers of t=1/2 and 2 1/2? Thanks. • January 3rd 2010, 07:10 PM pickslides $v(t) = 0$ $0= \pi-\pi\sin(\pi t)$ $\pi\sin(\pi t)= \pi$ $\sin(\pi t)= 1$ Form the grpah of the sine function $\pi t= \frac{\pi}{2}$ $t= \frac{1}{2}$ As the period of the fucntion is 2 the next solution will be at $t= \frac{5}{2} = 2.5$ • January 3rd 2010, 08:08 PM macroecon And on the second one, I get stuck at f'(x)=(-e^-x)((x-1)^2) and I'm not sure exactly how to progress from there. Any help? • January 3rd 2010, 08:46 PM ANDS! The great thing about optimization problems, is that when it comes time to set our differentiated function equal to zero, ANYTHING in the denominator dies a swift death at the hands of Mr. Multiplication-By-Zero: $f'(x)=\frac{(x-1)^2}{-e^x} \Longrightarrow 0=\frac{(x-1)^2}{-e^x} \Longrightarrow 0=(x-1)^2$ From here it should be simple to see what your solution is. Perhaps you weren't seeing the next step because you were writing $-e^-x$ instead of $\frac{1}{-e^x}$ All times are GMT -8. The time now is 01:20 PM.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 18, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9258739352226257, "perplexity_flag": "middle"}
http://stats.stackexchange.com/questions/1430/example-of-a-2nd-order-stationary-but-not-strictly-stationary-process	# Example of a 2nd order stationary, but not strictly stationary process Does anybody have a nice example of a stochastic process that is 2nd-order stationary, but is not strictly stationary? - I would like to add a stochastic processes tag, but don't have the reputation, perhaps somebody could edit this for me? – Robby McKilliam Aug 9 '10 at 6:51 retag done ! – robin girard Aug 9 '10 at 6:59 ## 1 Answer Take any process $(X_t)_t$ with independent components that has a constant first and second moment and put a varying third moment. It is second order stationnary because $E[ X_t X_{t+h} ]=0$ and it is not strictly stationnary because $P( X_t \geq x_t, X_{t+1} \geq x_{t+1})$ depends upon $t$ - Perhaps I am a little confused, or we are using different definitions? Am I correct in thinking that a process is 2nd order stationary if the joint marginal cdfs $F_{X(t),X(t+τ)}$ are equal for all $\tau$? Similarly for a process to be 1st-order stationary the marginal cdfs $F_{X(t)}$ need to be the same for every $t$. So all the moments of the $X(t)$ must be equal. 2nd-order stationary implies 1st-order stationary, correct? – Robby McKilliam Aug 9 '10 at 7:44 Extending this, a process is $N$th order stationary if for every $t_1, t_2, \dots, t_N$ the marginal cdfs $F_{X(t_1 + \tau), X(t_2 + \tau)\dots,X(t_N+\tau)}$ are the same for all $\tau$. Strictly stationary is Nth order stationary for all N. – Robby McKilliam Aug 9 '10 at 7:45 – robin girard Aug 9 '10 at 8:05 – Robby McKilliam Aug 9 '10 at 8:24 @robby OK... I didn't know this "second order stationnarity" I think you should not say second order stationnarity in the question and give the definintion instead. For more clarity you should ask another question. do you have a paper that refers to this second order stationnarity ? – robin girard Aug 9 '10 at 8:32 show 1 more comment	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9149402379989624, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/121260/isometric-embeddings-of-cayley-graphs-in-nice-spaces/121286	isometric embeddings of Cayley graphs in “nice” spaces Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) This is from a physicist I know and as may be expected, I am threading my way between poorly defined and poorly translated. What groups have Cayley graphs (w.r.t. a fixed finite generating set, and metrized so that each edge has length 1) that can be isometrically embedded in some reasonably nice space? Let us start with either f.d. Euclidean space, or f.d. hyperbolic space as the list of reasonably nice spaces. This fails to mention whether the embedding should be a quasi-isometry, but let us bring in that question only if it seems to be looming in importance. Yes, Cartesian products of finitely many copies of $\mathbf Z$ did occur to me, as well as (in dimension 2) triangle groups, but the length restrictions will severely restrict which ones work. Helpful contributions will be either a few examples, or a few families, or pointers to what references might be relevant and anything else that might be relevant. - 1 If you are talking about the geodesic metric on the Cayley graph, there will never be even a locally isometric embedding in a Riemannian manifold. – Lee Mosher Feb 9 at 1:34 The question is too vague, so I do not think you will get any answer. One only can say something related. First of all isometric should mean quasiisometric (otherwise the question has no sense). Say if you are interested in Euclidean space then the group has polynomial growth --- it is sufficient, but that is for an other question. Asymptotic dimension is yet an other thing which should forbid embeddings in Hyperbolic spaces as well as Euclidean spaces. I am sure it can be continued forever... – Anton Petrunin Feb 9 at 2:31 For embedding groups into Euclidean spaces, try googling "hilbert compression exponents". But then again, this is very far from isometry, since in this theory you can get, at best, only Lipschitz embeddings. The requirement of embeddings being isometries, rather than some weaker classes of maps, seems very stringent. Is that really what you want? – Marcin Kotowski Feb 9 at 2:38 Okay, so the word metric does not survive so well in the ambient target space. My apologies. Now that I am brought to a bit more sense in this, I would have thought the taxi metric more useful than the max metric suggested below, but I don't trust myself at the moment. I will discuss all this with my friend. – Matt Brin Feb 9 at 3:56 1 What Lee said is 99.9% correct, since cycles embed isometrically into $S^1$ and the doubly infinite path embeds isometrically into $\mathbb{R}$. But these are the only cases. Indeed, a 3-pointed star in a graph (with 3 vertices at distance 1 of a central vertex) cannot be realized in a Riemannian manifold, using standard properties of geodesics. So a regular graph isometrically embeddable in a Riemannian manifold, necessarily has degree 2. – Alain Valette Feb 9 at 9:42 4 Answers You can get a general answer avoiding the discussion on the choice of the generating set or of the metric on the larger space: which finitely generated (f.g.) groups have a bilipschitz embedding into a Euclidean space? The answer is: virtually abelian f.g. groups (i.e. having a f.g. abelian group of finite index). Indeed such groups admit proper cocompact actions on $\mathbf{R}^n$ for some $n$, giving rise to a quasi-isometric embedding which can easily be deformed to a bilipschitz embedding (alternatively, if you like to stay with an action, you can add some extra dimensions to make the action [whose kernel is finite] faithful but you lose cocompactness). Conversely, if a group $\Gamma$ admits a bilipschitz embedding (coarse embedding --aka uniform embedding-- would be enough) into a Euclidean space, by a growth argument, the group has polynomial growth. So it is virtually nilpotent, by Gromov's polynomial growth Theorem. Then a result of general result of Pauls (2001) shows that a virtually nilpotent f.g. group admits a bilipschitz embedding into a CAT(0) space (or a uniformly convex Banach space). However, I guess that this follows, in this special case (Euclidean target) from Pansu's 1989 Annals paper about metric differentiability. The argument consists of observing that a bilipschitz map $\Gamma\to\mathbf{R}^n$ induces a bilipschitz embedding from the asymptotic cone (which is a simply connected nilpotent group $G$ with a Carnot-Caratheodory metric, by an older result of Pansu) into $\mathbf{R}^n$, to show that generically it is metrically differentiable and the differential is generically an injective homomorphism; in particular $G$ is abelian, and this holds iff the original discrete group is virtually abelian. To go back to the original question: if you stick to isometric embeddings, what you get as a corollary is that if $\Gamma$ is a group with a finite generating set $S$, and $E$ is a finite-dimensional Banach space, then if the Cayley graph of $\Gamma$ w.r.t. $S$ embeds isometrically into $E$ then $\Gamma$ is virtually abelian (which is an extremely restrictive condition). The converse, however, depends on the specific choice of $S$ and the norm on $E$, as pointed out in other answers. - Even more, Scott Pauls proved in his thesis that a nilpotent group QI embeds in a CAT(0) space if and only if the group is virtually abelian. – Misha Feb 9 at 14:44 @Misha: it's the same: you can trivially change a QI (from a f.g. group) into a bilipschitz map (into a geodesic space not reduced to a point) by a little perturbation. – Yves Cornulier Feb 9 at 19:27 You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. I'd like to suggest to study the embeddings of Cayley graphs into $\mathbf Z^n$ with the max metrics $d$: $$d(x \ y) := \max_{k=1...n}\ \|x_k-y_k\|$$ My short note, on (the degree of) the universality of the $\mathbf R^n$ space with the max distance function (metrics), published in AMS Notices (in the late 1970s I'd think), may motivate the above approach. Let me stress that a finite metric space with integer distances is isometrically embeddable in $\mathbf R^n\quad\Leftrightarrow\quad$ it is embeddable into $\mathbf Z^n$, both spaces considered with the max metrics. The max metrics makes embeddings easy. - Could you please give a more precise reference to your short note? It is impossible for me to guess either title of the paper or its author from the information on this page. – Martin Feb 9 at 3:41 I am sorry. My name is Włodzimierz Holsztyński. The title was something like "$\mathbf R^n$ as a universal metric space". (It's not easy for me to access a mathematical library, and I don't have my reprints at this time, and some of them I don't have at all). I'll post my name in my profile (thought that it was easily available to the participants of OM). – Wlodzimierz Holsztynski Feb 9 at 5:44 The taxi metrics is inferior to the max metrics in the context of your question. Indeed, you need ease of embeddings. The max product preserves the injectivity. – Wlodzimierz Holsztynski Feb 9 at 7:30 Hm, your question admits more than one interpretation. I was concerned with the finite metric space which has the group as its set of points (there would be no other points), and the distance function induced by the Cayley graph. I was not concerned with the edges as parts of the space to be embedded. – Wlodzimierz Holsztynski Feb 9 at 7:38 Thank you. I found the article: W. Holsztyński, $\mathbf R^n$ as a universal metric space, Notices AMS 25 (3) (1978) A- 367. – Martin Feb 9 at 10:27 In my first answer above I considered the embeddings of the group only (according to the Cayley graph). The requirement to embed isometrically the whole graph (an infinite metric space) in a sense does not add any complication at all when you embed the whole graph into $\mathbf R^n$ with the max metrics, while you may insist on embedding the vertices into $\mathbf Z^n\subseteq\mathbf R^n$ (if you want to). Indeed, once vertices are embedded, the embedding can be extended to an isometric embedding of the whole graph (injectivity); furthermore, there is exactly one canonical (the simplest) isometric extension. While considering other distance functions in the Euclidean spaces can be of some interest, it's clear to me that max is special, is "the right one"--"the best". This is also what the theory of categories strongly suggests. - The Cayley graph of arbitrary cyclic group, for any $1$-element generating set, can be isometrically embedded in $R^{\lceil\frac n2\rceil}$ with the distance function given by maximum (while the group itself gets embedded in $Z^3$). On the other hand it is not difficult to show that the cyclic group $Z/5$, with a 1-generator presentation, cannot be isometrically embedded in $R^3$ with the max distance. - How do you that, say, for the cyclic group of order 3 ? – Alain Valette Feb 17 at 9:01 $Z/3$ is embeddable in $Z^^2$ with the max metric: $$\\{(0\ 0)\ \ (0\ 1)\ \ (1\ \ 0)\\}$$ while mapping onto the unit vectors of $R^3$ is nicer. – Wlodzimierz Holsztynski Feb 17 at 9:34 A typo correction: that "Z^^2" above is really $Z^2$. – Wlodzimierz Holsztynski Feb 17 at 9:36	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 37, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9221734404563904, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/279003/full-hessian-spherical-coordinates	# full hessian, spherical coordinates The question itself is pretty simple. I am running into confusion. Seems like there is a typo in the book. I wanna check myself. Maybe I am doing something wrong. Suppose we have the function (which I term full hessian): $$w=\sum_{i,j}f_if_{ij}f_j$$ with $$i,j=x,y,z$$ where $f_{ij}$ and $f_i$ means partial derivatives with respect to $i$ and $j$, and $i$, respectively. How do I write $w$ for some function $\rho(r)$ expressed in spherical coordinates? Note that $\rho$ is independent of $\phi$ and $\theta$. Let's take $\rho(r)=e^{-ar}$ as an example. Thanks in advance:) - ## 1 Answer You are free to call it full Hessian, but the established name for $w$ is $\infty$-Laplacian. Here is a nice expository article, though being 10 years old, it's a little out of date in the fast-developing subject. Recall that for any vector $v$ the expression $\sum_{i,j} v_i f_{ij} v_j$ is the directional second derivative of $f$, that is, $\frac{d^2}{dt^2}f(x+tv)$ evaluated at $t=0$. When we plug in $v=\nabla f(x)$, the result is the second derivative along the gradient. For your function $f(x)=\rho(\|x\|)$, the gradient is $\rho\,'(\|x\|)\frac{x}{\|x\|}$; that is, a radial vector with magnitude $\|\rho\,'\|$. The second derivative along this vector is $\rho\,''(\rho\,')^2$. This is the $\infty$-Laplacian of a spherically symmetric function. In particular, we observe that spherically symmetric $\infty$-harmonic functions are precisely the cones $f(x)=a\|x\|+b$. Comparison with cones is an effective method of gaining control over $\infty$-harmonic functions, as seen in the aforementioned article by Aronsson, Crandall and Juutinen. - 1 thnx for your nice answer!:) It was really helpful for me. – molkee Jan 26 at 22:47	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 27, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9418821930885315, "perplexity_flag": "head"}
http://mathoverflow.net/questions/38825/what-kind-of-lagrangians-can-we-have/39185	## What kind of Lagrangians can we have? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) In any physics book I've read the Lagrangian is introuced as as a functional whose critical points govern the dynamics of the system. It is then usually shown that a finite collection of non-interacting particles has a Lagrangian $\frac{1}{2}(m_1\dot{x}_1^2 + \cdots + m_n \dot{x}_n^2)$. It is then generally argued that $L=T-U$. I feel like something is missing here. What exactly are the physical hypotheses that go into this? Can we have other forms of the Lagrangian? How do we know those are "right"? Do we always have to compare them to the form of the equations we derived previously? For example, the Lagrangian formalism seems to be justified usually in so far as it 'works' for a finite collection of particles. Then you can solve any dynamics problem involving a collection of particles. I have been vague so let me try to be more precise in my question. Is the principle of least action an experimental hypothesis? Is it always true that $L=T-U$? When we don't know what the Lagrangian is, do we have to just guess and hope it is compatible with the dynamical equations we had already? Or can we perhaps start with the ansatz of a Lagrangian in some cases? I hope this is sufficiently precise. - 1 It seems to me you have to guess which forms of energies come into play in your physical system, i.e. guess the Lagrangian/Hamiltonian, from which you can derive equations of motion (through the principle of least action), which allows you to test whether particular guesses are suitable or not. – Sam Derbyshire Sep 15 2010 at 14:54 3 @Dorian This is a question that I have often thought of or asked but with no good reply. I haven't even seen a book which explain this point clearly. One can frame the question in various parts. Q1. Given a function on the phase space when is the function a valid Lagrangian? Q2: Given a "physical system" when does it have a Lagrangian? Q3: When is a Lagrangian system also Hamiltonian? (issues about dissipative forces). – Anirbit Sep 15 2010 at 14:56 1 Also I would like to know how does the theory of "Lagrangian submanifolds" (as understood in symplectic geometry) connect to the Physicist's intuitions. Thanks for motivating this topic. I really hope enlightening answers come up. – Anirbit Sep 15 2010 at 14:57 @Anirbit, you may find an answer to your final question in Arnol'ds book on mechanics. I think Q2 is a bit too vague to be answered, as it's basically asking "when does a physical model admit a variational formulation at one level or another". – jc Nov 5 2010 at 17:11 ## 12 Answers This question has a very wide scope, judging by the comments which have appeared already. I will try to narrow the scope a little in my answer. Classical mechanics is all about how a system evolves in time. The system itself is defined by specifying a configuration space, which we take to be a smooth manifold $M$. (The coordinates in the configuration space are usually known as "generalized coordinates" in the physics literature, to distinguish them from the standard coordinates in $\mathbb{R}^n$.) A lagrangian is then simply a differentiable function $L: TM \to \mathbb{R}$ on the tangent bundle. Relative to local coordinates $(x,v)$ for $TM$ (where $x$ --- the "positions" --- are local coordinates on $M$ and $v$ --- the "velocities" --- are the fibre coordinates), the lagrangian $L$ determines the equations of motion via the principle of least action, which results in the Euler-Lagrange equations: $$\frac{d}{dt} \frac{\partial L}{\partial v} = \frac{\partial L}{\partial x}$$ evaluated on curves $(x,\dot x)$. In addition one would like the fibre derivative $\frac{\partial L}{\partial v}$ to be nondegenerate, so that we get a fibrewise isomorphism $TM \to T^M$. Then $p = \frac{\partial L}{\partial v}$ are the canonical momenta and nondegeneracy means that we can invert this relation and write $v$ as a function of $p$. The hamiltonian $H : T^M \to \mathbb{R}$ is defined as the Legendre transform of the lagrangian $$H(x,p) = p v - L$$ where $v$ is given as a function of $p$. The cotangent bundle $T^*M$ has a natural symplectic form $\omega$ and the hamiltonian vector field associated to $H$ defines a flow which evolves the state $(x,p)$ in the same way as do the Euler-Lagrange equations. In general there is no canonical way to choose $L$ (or $H$): it ultimately depends on the physical system one is trying to model. Of course, we have had so much experience, that there are many heuristics one can use to guess the appropriate lagrangian. In doing so, symmetries play an important rôle and nowhere is this more evident than in the business of "model building" in particle physics. Here we are dealing with field theories, whence the configuration space consists of sections of certain homogeneous vector (or more general fibre) bundles on the spacetime and there are a number of symmetries (both Lie and discrete) that one might want our lagrangian to have. More interesting perhaps are the theories for which no lagrangians are known, or at least no lagrangians which are manifestly invariant under the desired symmetry. There are many examples of such theories, some very interesting ones in six dimensions which are intimately linked to the geometric Langlands program as described in this paper of Witten's, for istance. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Let me comment on your second question: When we don't know what the Lagrangian is, do we have to just guess and hope it is compatible with the dynamical equations we had already? If you want the variational derivative of your Lagrangian to yield your equations of motion (or a system which is equivalent to your equations of motion) you should solve the inverse problem of calculus of variations (googling will land you plenty of references; for the "mechanical" case which is apparently your primary interest, you can start here; see also this book by Ian Anderson and Gerard Thompson). In general, there are certain necessary conditions for your equations of motion to satisfy in order that the Lagrangian exists at all. If these are satisfied, finding the Lagrangian $L$ essentially boils down to using an appropriate homotopy operator to reconstruct $L$ from its variational derivative, see e.g. Ch. 5 of the book Applications of Lie groups to Differential Equations by Peter Olver. As a historical remark, it is interesting to note that one of the first Fileds medalists, Jesse Douglas, has made significant contributions to this field of research. Sorry for being rather sketchy, maybe I'll expand this answer a bit later. - As a theoretical physicist who shifted to pure mathematics, I think to answer this question and as a clarification to previous posts, we should not forget the historical side of the evolution and origins of the terms involved in physics theory modeling. The origin of the principle of stationary ('least' most of the time) action comes as a variational formulation using a functional $S[q(t)]:=\int_{t_1}^{t_2}L(q^i (t),\dot{q}^i (t);t)dt$ for the Lagrange equations of the real motion of a system with generalized coordinates $q^i (t)$ (obtained this way they are called Euler-Lagrange equations): $$\delta S[q_{real}(t)]=0\Rightarrow \frac{d}{dt}\frac{\partial L}{\partial \dot{q}^i}-\frac{\partial L}{\partial q^i}=0$$ The original Lagrange equations of motion were obtained as a reformulation of Newton's second law of mechanics for the case of generalized coordinates (the remaining degrees of freedom after introducing constraints) as an easier to handle version of D'Alembert's principle. D'Alembert wrote the 'virtal work' principle as a way to state a general equation for statics, which in a way amounts to postulate that constraint forces do not exert work (as they should be internal and the weak 3rd law applies). Now given Newton's second law for a system of particles, D'Alembert reformulates constrained dynamics as statics introducing the inertial force (you can see details at wikipedia) $$\sum_i (\vec{F}_{i}^{ext}-\frac{d\vec{p}_i}{dt})\cdot\delta \vec{r_i}=0$$ Lagrange wanted to work only with generalized coordinates which changing variables in D'Alembert's principle led him to the original Lagrange's equations of motion in terms of a kinetic generalised energy $T$ and generalized forces $Q_i$; furthermore he absorbed those conservative forces which derived from a potential $Q_i^c=-\partial U/\partial q_i$ (or more generally those which even depended on speed or time but had the required form of the left-hand side) along with the kinetic energy, into a function $L=T-U$ now called the Lagrangian $$\frac{d}{dt}\frac{\partial T}{\partial \dot{q}^i}-\frac{\partial T}{\partial q^i}=Q_i\Rightarrow \frac{d}{dt}\frac{\partial L}{\partial \dot{q}^i}-\frac{\partial L}{\partial q^i}=Q_i^{no-c.}$$ That is the origin of the form of the lagrangian as $L=T-U$. Now, the non-conservative forces $Q_i^{no-c.}$ appear because we are considering general 'open systems' which interchange energy with their enviroment, other systems as sources of energy, etc. To check this form in detail, first consdier that a closed system is always of the form $L=T-U$ because of a constructive reasoning as follows. For a free particle, at least locally, because of the homogeneity of space and time, and isotropy of space, we must have $L(q_i,\dot{q}_i;t)=T(v^2)$, that is a scalar which only depends on the particle and its speed (but no direction); since a free particle must still be free in another inertial system by def. the relation must be linear $L=a\cdot v^2$ (this is because any other Lagrangian giving the same equations of motion has to be of the form $L'=A\cdot L+\frac{d}{dt}\Omega(q;t)$ for which an inertial transformation $\vec{v}'=\vec{v}-\vec{\epsilon}$ must imply at first order $-2\vec{v}\cdot\vec{\epsilon}\partial L(v^2)/\partial (v^2)\propto d\Omega(q;t)/dt\Leftrightarrow \partial L/\partial (v^2)=0$); now the constant $a$ characterizes the particle, and must reduce to the free Newtonian second law ($p_i=m\cdot v_i =0$) therefore $a=m/2$. For a system of free particles without interactions the same applies summing over the individual kinetic energies $L_{free}=T=\frac{1}{2}\sum_i m_i v_i^2=\frac{1}{2}\sum_{i,j} A_{ij}\dot{q}^i\dot{q}^j$ (when $T$ is a cuadratic form the system is called 'natural' which happens when $\partial \vec{r}_k(q^i(t),\dot{q}^i(t);t)/\partial t = 0$ for all particles $k$, common for the natural case of constraints not dependent on time). To add interactions between particles in a closed system, their degrees of freedom must be coupled, which just means that the relative energies are not lost to the exterior, and this is satisfactorily implemented by a potential function $U=U(q_i,\dot{q}_i)$. By this method we obtain a general framework for mechanics using Lagrange's equations: postulating different potentials we obtain different models of interactions which shall be contrasted with experiment to constrain the form of $U$ (this may be done to obtain just effective phenomenological models, nevertheless all fundamental interactions of physics seem to fit this framework very well). Finally, when part B of the system A+B is fixed (like considered external) its motion can be treated like already 'solved', which opens the previous closed system, separating some of the degrees of freedom $q_B^i(t)$; therefore the Lagrangian decouples $L_{open}=T_A+T_B(\dot{q}^i_B(t))-U(q_A,q_B(t),\dot{q}_A,\dot{q}_B(t))$, in this case $T_B=d\Omega(q_B,t)/dt$ for some $\Omega$, so we can neglect that part since does not contribute to the equations of motion and therefore $L_{open}=T_A-U_{open}(q_A,\dot{q}_A,t)$. This way $U$ creates the effect of non-conservative forces which are just those that add or subtract energy from the system without taking into account where this energy goes. Therefore we can consider the Lagrange equation (without non-conservative forces) as the general more fundamental equation given individual interactions between the particles of the Universe, at least in principle. This is why the form $L=T-U$ is the usual approach in theoretical physics to deal with dynamics, since the very concept of instantaneous force at a distance à-la Newton is anti-natural (above all after special-general relativity) and the energetic local approach is not only mathematically better but philosophically more satisfactory. The Hamiltonian approach can be taken also as a starting point, or as a Legendre transformation of Lagrange dynamics, but that way one makes a non-relativistic break down of the coordinates $x^i, t$ which is useful for non-relativistic quantum mechanics. Now in general, physics is a theory of fields so we want to add these to the framework. This is easily done by considering them as continuous systems of degrees of freedom, providing generalizations of kinetic and potential energies of those. Furthermore one wishes for a (special or general) relativistic invariant theory which forces the $Ldt$ to be a Lorentz scalar in order to get Lorentz-covariant ($SO(3,1)$) equations of motion (i.e. covariant as tensor equations in a pseudo-Riemannian manifold). This way the free case reasoned above leads us to $ds=Ldt$ (where $ds$ is the relativistic space-time interval or 'proper time' measured by the particle) which indeed reduces to $ds\approx \frac{1}{2}mv^2$ for speeds $v\ll c$, and therefore can be taken as a relativistic generalization of the framework. Nevertheless point-particles are idealizations and both the classical and quantum theories require a field-theoretic treatment of matter (in the classical case, particles appear as small density lumps, and in the quantum case particle behaviour arises for discrete-like energy levels of the quantum field states). In the wikipedia article about Lagrangians different examples can be seen from point particles to fields, their kinetic terms and so on. In any case, the traditional form $L=T-U$ has its roots in the very nature of mechanics and any modern field theory is created by building up possible $T-U$ functionals from the field's degrees of freedom, i.e. the fields $\phi_a(x^\mu)$ and their "velocities" $\partial \phi_a/\partial x^\nu$ (for relativistic invariance reasons time-like speed $\frac{\partial}{\partial t}$ is not enough and the whole $\partial_\nu$ must be used). Hence one constructs kinetic terms like (Einstein's summation convenction) $T_\phi=\frac{1}{2}\partial_\mu\phi_a\partial^\mu\phi_a$ for scalar fields or $T_A=F_{\mu\nu}F^{\mu\nu}$ for vector fields ($F_{\mu\nu}:=\partial_\mu A_\nu-\partial_\nu A_\mu$ is in this case the tensor constructed from the field used to define its kinetic energy because we want to build invariant 'speed-like' $\partial_\mu$ scalars that are as well Gauge invariant which is another physical symmetry typically required besides Lorentz; furthermore, requiring gauge invariance for different Lie groups, tyipcally $SU(N)$, one forces the automatic appearance of fields and couplings between the matter fields responsible for the interactions, which is just working with connections and curvatures of fiber bundles). Since any term which makes the field equation of motion non-homogeneous is considered a source of perturbation, or force, one sees the corresponding couplings in the Lagrangian like a potential energy, giving as always $\mathcal{L}=T_\phi+T_A-U(A_\mu,\phi)$. With this, the coupled physical equations of motion are deduced for the principle of stationary action, where now the integration must be over the whole space and an interval of time (i.e. $\mathcal{L}$ is a density), the field Euler-Lagrange equations of motion: $$\partial_\mu\frac{\partial\mathcal{L}}{\partial (\partial_\mu\phi)}-\frac{\partial\mathcal{L}}{\partial\phi}=0$$ Finally these equations give the solutions of motion, that is the real field configurations at any point in space-time, which contribute the most in a quantum-theoretic framework, where one weighs complex transition amplitudes by Feynman operational methods using the action for each possible field-configuration ('motion'): $e^{\frac{i}{\hbar}S[\phi,A_\mu]}$. As remarked in other comment, the semi-classical approximation $\hbar\rightarrow 0$ makes the classical solution of Euler-Lagrange the predominant one, hence deducing the principle of stationary action. (Also, in the non-relativistic hamiltonian quantum-mechanics Ehrenfest's theorem provides a classical Newton-like equation of motion for the average degrees of freedom) Indeed as science makes progress we must see previous theories and models as limiting cases of new more precise theories, but in the beginning any physical theory must be constructed by physical insight, intuition and permanent comparison with experiment. Even though Feynman titled his thesis that way, he does not develop his approach to quantum theory starting from the action-principle, on the contrary he works with standard (hamiltonian) quantum mechanics and obtains a novel method for computing quantum probability amplitudes in which the classical action appears. You could start with quantum mechanics as an axiomatic system, develop Feynman's approach to quantum propagators of motion and deduce that classical approximate solutions of motion must obey, at first order of quantum corrections, the classical Euler-Lagrange equations of motion. At the same time, you need the input from the experimentally succsessful classical mechanics to get to quantum mechanics... and so forth. In the end at any moment of time in the history of science we are getting better and better mathematical structures that model reality to different degrees of precision; the important point is that a new one must contain an old one as an approximation in some regime, and explain even more. This way, theoretical physics tends to get better structures to encompass more phenomena from Nature in a simpler and simpler manner as it explains more effective laws with less theories (currently almost all the phenomena observed is explained by general relativity and quantum field theory, and unifying both is the neverending quest for the holy grail of physics). - 1 This is for sure the most comprehensive and enlightening answer! I really enjoyed reading this response and it should have more up-votes. – Samuel Reid Mar 23 2012 at 6:37 As far as the priniciples are concerned that enter the choice of a Lagrangian it is important to distinguish between classical systems and quantum systems. In the case of classical particle systems, or classical field theories, the primary input has historically been the dynamics of the system, as encoded in the equations of motion. The choice of Lagrangian, or rather the class of equivalent Lagrangians, then is determined by the constraint that the variational principle needs to reproduce this known dynamics. In the case of quantum field theories the strategy is somewhat different, and Feynman's approach to the computation of physical observables via the path integral puts more emphasis on the Lagrangian as the fundamental quantity. Here a powerful guide are symmetries that constrain the possible forms of the Lagrangian. Generally, symmetries are not enough to determine its form, and more experimental constraints are necessary to specify the theory. The final justification of any given Lagrangian is that its physical consequences are consistent with the observed phenomena. - I believe the standard answer to your question is that if one starts from Newton's Laws of Motion for a system of particles, then the second law says essentially $F = Ma$, giving the equations of motion as a second order ODE that is completely determined if we know the forces between particles. Then, and this is the important point that is often un-appreciated, Newton's Third Law (action and reaction are equal and opposite) IMPLIES that the forces are derivable from a potential function $U$. [The latter statement is WRONG ! See the comments and correction below.] It then follows by easy formal manipulation that $F=Ma$ is equivalent to the Euler Lagrange Equations for the Lagrangian $L= K - U$ where the kinetic energy $K$ is given by the formula you gave above for the Lagrangian without interactions. If you are interested in further details, I have a recent book (published by AMS and co-authored with my son Robert) called "Differential Equations, Mechanics, and Computation" in which I go to great lengths to give a conceptual approach to exactly the sort of question you are asking. There is a companion website at http://ode-math.com/ at which you can download more than half the material in the book as pdf files. - 1 I don't understand how Newton's third law can imply that forces are conservative. It seems to me that the consideration of a two-particle system provides a counterexample: Let the force on the first particle be an arbitrary function of the two positions and just let the force on the second be the precise opposite. Did you have further hypotheses in mind for this claim? – Harald Hanche-Olsen Sep 15 2010 at 16:19 4 Thank you Harald. Yes, you are correct (and your example is excellent); Newton's Third Law does NOT imply conservation of energy but rather conservation of linear momentum. To get conservation of energy, or that the force is a gradient of a potential, you must add to Newton's Laws a "no free lunch" axiom: The work done (line integral of the force) around a closed loop is zero. Sorry---another example of "If you say something complicated too early, you may have too eat your words for breakfast". :-) – Dick Palais Sep 15 2010 at 17:14 "Is the principle of least action an experimental hypothesis?" A physicist views classical mechanics as a semiclassical limit of quantum mechanics, valid in the limit that $\hbar \rightarrow 0$. In the path integral approach to quantum mechanics one studies integrals over an infinite dimensional space of paths (yes,yes I know these are not completely rigorous) of the form $\int {\cal D}x e^{i S[x]/\hbar}$ with $S= \int {\cal L} dt$ where ${\cal L}$ is the Lagrangian of the system. The principle of least action can then be viewed as a stationary phase approximation to the path integral, valid in the semiclassical limit. Thus rather than a hypothesis I would say the principle of least action is derived as part of the process of taking the classical limit of quantum mechanics. - 1 But is this not actually circular? I thought the path integral was arrived at departing from the principle of least action. In fact, Feynman's PhD thesis was titled "The principle of least action in quantum mechanics." (I confess it's been a while since I looked at it, though.) – José Figueroa-O'Farrill Nov 13 2010 at 14:37 I don't think it is circular, except perhaps in some historical sense. You can't derive QM from classical mechanics, but you can definitely take a classical limit of QM. – Jeff Harvey Dec 9 2010 at 16:27 Here are two nice, natural, examples of Lagrangians not of the form $T-U$ which occur naturally in physics. For a relativistic particle of charge e, mass m, travelling under the influence of an electromagnetic field F, with one-form potential A (so $F = dA$), the Lagrangian is $L = mds - eA$ where $ds$ is the proper time, or element of arc-length in Minkowski space (or more generally, the space-time within which said particle is travelling). [References: Feynman, vol. II, p. 19-7; Folland `Quantum Field Theory: A tourist's guide, ch. 2, p. 29. Most intermediate level physics texts] This can be parsed so as to have the form $L(q, \dot q)$ in two different ways. Way 1: choose an arbitrary parameter $u$ for the time and parameterize the space-time path with $u$. Then the $\dot q$ is $dq/du$. Way 2: choose to parameterize with time, which in one of the space-time coordinates. The non-relativistic limit of this $L$ is the $L$ of Alex R.s answer. (This variational principle is particularly interesting for photons, where $ds = 0$.) The oldest variational principle in mechanics (excluding Fermat's for optics) is that attributed to Maupertuis. He says that solutions to Newton's equations at fixed energy E, extremize $L = \sqrt{(E-U)T}$, using your terminology above. This Lagrangian represents a metric conformal to the Euclidean metric, as represented by $T$. (The geodesic time of these extremals is not Newtonian time.) I agree with Moduli that symmetry is one of the key ingredients that goes into the choice of a Lagrangian. In the symmetry light, it is worth noting that the two above Lagrangians are homogeneous of degree 1 in `velocities', so invariant under reparameterization of paths. - The most valuable discovery of the 20th century physics is symmetry. When you try to write down a Lagrangian for a system, usually there are two "principles": (1) the symmetries this system has and (2) experience. So no, L is not always T-U. For example, in General Relativity the Lagrangian (or rather Lagrangian density) is R, the scalar curvature of the spacetime manifold. As another example, in WZW models, the Lagrangian has two parts: a "kinetic" term plus a Wess–Zumino term. It is in quantum field theory that all kinds of Lagrangians pop up. What is more interesting is that the best you can do is to write down an “effective Lagrangian” at a certain energy scale, which may (and usually does) get quantum corrections at higher energy scales. - 2 "For example, in General Relativity the Lagrangian (or rather Lagrangian density) is R, the scalar curvature of the spacetime manifold." My knowledge of general relativity is admittedly limited but I don't see the contradiction. General relativity is a theory of statics (so there is no kinetic term) and the scalar curvature is a potential energy of intrinsic bending. It's analogous to how the energy of a static membrane is the integral of $\|\nabla u\|^2$ where u is the scalar displacement. – Per Vognsen Sep 18 2010 at 5:44 Good point. But energy in GR has never been well defined, so it is tricky to talk about "energy" in GR. You can say R is like a potential energy by some sort of analogies, but personally I do not think this is a good argument. – Moduli Sep 18 2010 at 15:33 1 The statement above that GR is a theory of statics is wrong. Static spacetimes are very special solutions to the Einstein equations (see Wald 6.1). Generic solutions are not static. If you expand the metric about flat spacetime you will find that the perturbations, which describe gravitational waves, have kinetic terms. etc. etc. – Jeff Harvey Nov 5 2010 at 16:49 It seems wildly arrogant to say `symmetry' is a 20th century discovery. I seem to recall a certain Galileo. Symmetry is central to Newton's formulation of the N-body equations and all the great mechanicians -Euler, Lagrange, Hamilton - used symmetry all the time. – Richard Montgomery Mar 22 2012 at 23:13 Is $L=T-V$ ? Depends what you mean by "$V$". Try finding $V$ for a system with nonconservative forces. There's a sleight of hand here of course, since when we talk about $V$ we really mean $V=V(q)$, a function of the coordinates $q$. This doesn't mean we can't find a Lagrangian. Take electromagnetism for example, say a charged particle moving through an electromagnetic field. It's lagrangian can we written as: $L=\frac{1}{2}m\dot{\mathbf{q}}^2-e\phi(\mathbf{q},t)+e\dot{\mathbf{q}}\cdot \mathbf{A}(\mathbf{q},t)$ where $\phi$ and $\mathbf{A}$ are the scalar and vector potentials respectively. How did we get this Lagrangian? One way is working backwards from the Lorenz force $\mathbf{F}=e\left(E+\frac{1}{c}\mathbf{v}\times\mathbf{B}\right)$, which gives us the equations of motion, and then we go back to our course in electromagnetism to remember that electromagnetism comes from scalar and vector potentials. If you are handed the equations of motion of a system (EOM), then there can be multiple Lagrangians that give rise to these EOM. In fact, at least locally, when two Lagrangians $L_1,L_2$ give rise to the same EOM, then $L_1-L_2=d\Phi/dt$, where $\Phi$ is a function on the position coordinates $\mathbb{Q}$. The point is, if you already have equations of motion, then finding a Lagrangian becomes a derivative matching game. - The Lagrangians in physics usually take the form $L=m\dot{x}^2/2+...$ (or $\mathcal{L}=(\partial \phi)^2/2+...$ in field theory), i.e. the velocity dependence is quadratic. An important consequence of this form can be seen in the path integral formulation of quantum mechanics. The amplitude as a path integral in the phase space is $\int \mathcal{D}x \ \mathcal{D}p \ e^{i\int(p \ dx - H \ dt)}$ where the Hamlitonian $H$ is a function of $p, x$ and $t$. The thing is, this amplitude is in general NOT equal to the form we usually use $\int \mathcal{D}x \ e^{i\int L \ dt}$. The two forms are equal when $H$ depends on $p$ as $p^2$, i.e. when $L$ depends on $\dot{x}$ as $\dot{x}^2$. (the derivation can be found in Polchinski, Peskin etc., basically $e^{ip^2}$ will give a constant by Gaussian integration) I am not sure if this is the reason that physical Lagrangian must take the form $L=m\dot{x}^2/2+...$, but at least I feel this shows some deep connection between the Lagrangian formulation and Hamiltonian formulation in quantum theory. - Here are two nice, natural, examples of Lagrangians not of the form $T-U$ which occur naturally in physics. For a relativistic particle of charge e, mass m, travelling under the influence of an electromagnetic field F, with one-form potential A (so $F = dA$), the Lagrangian is $L = mds - eA$ where $ds$ is the proper time, or element of arc-length in Minkowski space (or more generally, the space-time within which said particle is travelling). [References: Feynman, vol. II, p. 19-7; Folland `Quantum Field Theory: A tourist's guide, ch. 2, p. 29. Most intermediate level physics texts] This can be parsed so as to have the form $L(q, \dot q)$ in two different ways. Way 1: choose an arbitrary parameter $u$ for the time and parameterize the space-time path with $u$. Then the $\dot q$ is $dq/du$. Way 2: choose to parameterize with time, which in one of the space-time coordinates. The non-relativistic limit of this $L$ is the $L$ of Alex R.s answer. (This variational principle is particularly interesting for photons, where $ds = 0$.) The oldest variational principle in mechanics (excluding Fermat's for optics) is that attributed to Maupertuis. He says that solutions to Newton's equations at fixed energy E, extremize $L = \sqrt{(E-U)T}$, using your terminology above. This Lagrangian represents a metric conformal to the Euclidean metric, as represented by $T$. (The geodesic time of these extremals is not Newtonian time.) I agree with Moduli that symmetry is one of the key ingredients that goes into the choice of a Lagrangian. In the symmetry light, it is worth noting that the two above Lagrangians are homogeneous of degree 1 in `velocities', so invariant under reparameterization of paths. - 1 Somehow you posted this answer twice. – arsmath Dec 9 2010 at 12:26 In general, I would say that you have to check that a variational principle implies some physical law (usually in the form of a differential equation). In classical mechanics, in particular, Hamilton's principle of least (or stationary) action implies Newton's second law of motion (in Cartesian coordinates). But then you can use Hamilton's principle in any system of coordinates because the Euler-Lagrange equation is valid in any system of coordinates. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 148, "mathjax_display_tex": 6, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9398801326751709, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/52527/can-low-gravity-planets-sustain-a-breathable-atmosphere	# Can low-gravity planets sustain a breathable atmosphere? If astronauts could deliver a large quantity of breathable air to somewhere with lower gravity, such as Earth's moon, would the air form an atmosphere, or would it float away and disappear? Is there a minimum amount of gravity necessary to trap a breathable atmosphere on a planet? - ## 3 Answers Gravity is a major factor in planets retaining atmospheres over the eons. But there are other factors that must be taken into consideration to consider the volatility of an atmosphere. Solar wind is the main factor of erosion on any atmosphere. But a healthy magnetic field can deflect most of the solar radiation and decrease the erosion. It has been a matter of debate recently if exo-moons of jovian planets in habitable zones of their host stars would be able to sustain atmospheres: such moons are most likely tidally-locked, so their magnetic fields are not expected to be high, but their host planets will likely have strong radiation belts. But is not clear at the moment if the radiation belts will protect or erode the atmosphere. Saturn has a benign level of radiation, so we have Titan, which has an atmosphere that is thicker than earth's - 2 I'm not so sure that Solar wind is always the dominant factor. There is also the possibility that thermal fluctuations in the molecules' speeds can take them above the escape velocity. This would be the case on Earth if our atmosphere was made of $\text{H}_2$, IIRC. – Nathaniel Jan 30 at 14:05 good point, @Nathaniel – lurscher Jan 30 at 19:22 I guess the devil is in the details. For example, if the celestial body in question is far from its star, so its temperature is very low, it is easier to retain low-temperature air around the body. On the other hand, very cold air is not breathable anyway. There is another way though. If the astronauts can bring so much air to the body, why don't they arrange a membrane around the body to keep the air? Furthermore, they don't need the membrane around the entire body, they can arrange it just over some limited area where they want to live. On the other hand, they would need to protect such a membrane from meteorites... So I guess there is a lot they can do and a lot of factors that could make their life miserable:-) - The speed of oxygen at room temperature (293k) is 1720km per hour so if the escape velocity of the moon or planet is greater than that then at least you will have oxygen. If you want some nitrogen in the mix then you will have to google it's speed like I did for oxygen;-) - 2 The speed you give is (I think) the RMS speed, but there's a distribution of speeds around it. Some molecules will have higher speeds, and may escape, and if the temperature is maintained some other molecules will gain energy to fill those high velocity states in the distribution... leading to a gradual evaporation of the atmosphere. – Kyle Jan 30 at 14:30	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9504743218421936, "perplexity_flag": "middle"}
http://mathhelpforum.com/pre-calculus/96095-how-find-maximum-minimum-points.html	# Thread: 1. ## How to find Maximum and Minimum Points? 1. $y=(1-x^2)(4-x^2)$ 2. $y=3x^4-16x^3+18x^2$ Answer: 1. $(-\sqrt{\frac{5}{2}}, -2) (\sqrt{\frac{5}{2}}, -2)$ 2. $(1, 5) (3, -27)$ 2. The condition is : dy/dx = 0 but i get three values for each question, not two 3. ## Differentiation The only way I know to solve this question is to employ some calculus and differentiate. Differentiate to find $\frac{dy}{dx}$, then set $\frac{dy}{dx}=0$, and solve the resulting equation. Eg. No.2 $\frac{dy}{dx}=12x^3-48x^2+36x$ $=12x(x^2-4x+3)$ $\frac{dy}{dx}=12x(x^2-4x+3)=0$ $=(12x)(x-1)(x-3)=0$ $x=0,1,3$ Place these results into the original equation When $x=1$ $y=3(1)^4-16(1)^3+18(1)^2=5$ When $x=3$ $y=3(3)^4-16(3)^3+18(3)^2=-27$ When $x=0,y=0$ $(0,0)$	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 17, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.848802924156189, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-algebra/174018-orthogonal-matrices-help.html	# Thread: 1. ## orthogonal matrices help using the definition of an orthogonal matrix prove : P is an orthogonal matrix if and only if P^t is an orthogonal matrix. I know P is orthogonal if we can multiply P by P^t and get the identity matrix but dont know where to go from there!! any help appreciated!! 2. Originally Posted by eleahy using the definition of an orthogonal matrix prove : P is an orthogonal matrix if and only if P^t is an orthogonal matrix. I know P is orthogonal if we can multiply P by P^t and get the identity matrix but dont know where to go from there!! any help appreciated!! Just use the fact that $\left(P^{\top}\right)^{-1}=\left(P^{-1}\right)^{-1}=P$ and so $\left(P^{\top}\right)^{-1}=P=\left(P^{\top}\right)^{\top}$. 3. still not exactly sure but have stuff to work with ill figure it thanks so much	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9092798829078674, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/23729/what-is-the-condition-for-a-field-to-make-the-degree-of-its-algebraic-closure-ov	# What is the condition for a field to make the degree of its algebraic closure over it infinite? As we all know, the algebraic closure often has an infinite degree. Also, this shows the necessary and sufficient condition for a Galois extension to be a finite extension of fields. However, we may want to characterize the cases of the case where the extension is not Galois, which is actually my question. And this question is related to this question. - 1 Geez. The second link is to Pete Clark's notes which are 73 pages long; I'm supposed to go hunting through it to see what specifically you are refering to? Perhaps you might consider quoting it instead of sending the reader on a scavenger hunt through a 73 page long document... – Arturo Magidin Feb 25 '11 at 16:22 @Arturo Magidin: Sorry, according to the last link, it is in section 12.5, thanks in any case. – awllower Feb 25 '11 at 16:25 5 As far as I can see, all the questions asked here are already answered in $\S 12.4$ of my notes. – Pete L. Clark Feb 25 '11 at 19:31 I am very sorry. – awllower Feb 26 '11 at 2:55 ## 2 Answers Let me summarize the results of $\S 12.4$ of my notes. (This involves making explicit some things which were left as "exercises", but I'm okay with that.) Let $K$ be any field, let $K^{\operatorname{sep}}$ be any separable closure and let $\overline{K}$ be any algebraic closure containing $K^{\operatorname{sep}}$. Then: 1) Suppose first that $K = K^{\operatorname{sep}}$, i.e., $K$ is separably closed. Then either: 1a) $K$ is algebraically closed, or 1b) It isn't, i.e., $K$ has positive characteristic $p$ and there exists $a \in K$ such that the polynomial $t^p - a$ is irreducible. Then $t^{p^n}-a$ is irreducible for all $n$, so $[\overline{K}:K]$ is infinite. (For this, see Lemma 32 in $\S 6.1$.) 2) Suppose that $K$ is not separably closed, so $G = \operatorname{Aut}(K^{\operatorname{sep}}/K)$ is nontrivial. Then by the Artin-Schreier theorem, either 2a) $\# G = 2$, in which case $K$ is formally real and $K(\sqrt{-1})$ is algebraically closed, or 2b) $\# G > 2$, in which case $G$ is infinite, which implies $[\overline{K}:K]$ is infinite. Note that some of the exercises outline further extensions of the Artin-Schreier Theorem, i.e., if $\overline{K}/K$ is "small" in other ways then $K$ is either real closed or algebraically closed. - Thanks very much, @Pete L. Clark. – awllower Feb 26 '11 at 3:25 I suddenly found that I forgot to accept it as an answer, hope you won't mind, @Pete L. Clark... – awllower Mar 12 '11 at 3:20 The Artin-Schreier theorem asserts that, if the field $k$ is not algebraically closed and $[\bar{k} : k]$ is finite, then in fact it equals $2$ and $\bar{k} = k[i]$ and $k$ is real closed. So I guess the condition you're looking for is: if and only if $k$ is not algebraically closed and also not real closed. I am answering the title question, not the body question. The body question is not equivalent to the title question; the answer to it is that the algebraic closure is Galois if and only if it's separable, hence if and only if $k$ is perfect. Perhaps you have only seen the definition of Galois extension for finite extensions. This is not the definition which works in maximal generality. The definition which does work in maximal generality is the following: Definition: An algebraic extension $k \to L$ is Galois if the fixed field of $\text{Aut}_k(L)$ is $k$. The separable closure $k_s$ of $k$ is always Galois (in fact it is the maximal Galois extension of $k$), and agrees with the algebraic closure if and only if $k$ is perfect. - Thanks very much. – awllower Feb 25 '11 at 16:33	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 41, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9470527768135071, "perplexity_flag": "head"}
http://mathoverflow.net/questions/20381/crystalline-cohomology-of-abelian-varieties/20385	Crystalline cohomology of abelian varieties Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I am trying to learn a little bit about crystalline cohomology (I am interested in applications to ordinariness). Whenever I try to read anything about it, I quickly encounter divided power structures, period rings and the de Rham-Witt complex. Before looking into these things, it would be nice to have an idea of what the cohomology that you construct at the end looks like. The l-adic cohomology of abelian varieties has a simple description in terms of the Tate module. My question is: is there something similar for crystalline cohomology of abelian varieties? More precisely, let $X$ be an abelian scheme over $\mathbb{Z}_p$. Is there a concrete description of $H^1(X_0/\mathbb{Z}_p)$? (or just `$H^1(X_0/\mathbb{Z}_p) \otimes_{\mathbb{Z}_p} \mathbb{Q}_p$`?) I think that this should consist of three things: a $\mathbb{Z}_p$-module $M$, a filtration on `$M \otimes_{\mathbb{Z}_p} \mathbb{Q}_p$` (which in the case of an abelian variety has only one term which is neither 0 nor everything) and a Frobenius-linear morphism $M \to M$. I believe that the answer has something to do with Dieudonné modules, but I don't know what they are either. - 3 The $\ell$-adic Tate mod. of an ab. var. $A$ over field $k$ of char. distinct from is not the etale cohom. (with $\mathbf{Z}_ {\ell}$ coefficients) of the $A$ in general (unless $k = k_s$), but rather of $A_{k_s}$ (with such coefficients). The etale coh. of $A$ itself is rather more complicated! Just as you probably learned about Tate modules and $\ell$-adic Galois representations before etale cohomology, you should learn about Dieudonne modules and their relation with $p$-divisible groups before learning crystalline cohomology. – BCnrd Apr 5 2010 at 12:25 Indeed, I meant the l-adic cohomology of the abelian variety over $k_s$. – Martin Orr Apr 6 2010 at 21:40 1 Answer To add a bit more to Brian's comment: the crystalline cohomology of an abelian variety (over a finite field of characteristic p, say) is canonically isomorphic to the Dieudonné module of the p-divisible group of the abelian variety (which is a finite free module over the Witt vectors of the field with a semi-linear Frobenius). If you start with an abelian scheme over the Witt vectors of this field then the crystalline cohomology of the special fibre is canonically isomorphic to the algebraic de Rham cohomology of the thing upstairs, hence receives a Hodge filtration also. A good starting place to understand Dieudonné modules is Demazure's 'Lectures on p-divisible groups', which appears in Springer LNM. In particular, he gives a nice description of the analogy with Tate modules (and the relation between the various Frobenii that appear). For a general picture of crystalline cohomology, and the various structures that can be placed on it, I would look at Illusie's survey in the Motives volumes (this is a little out of date now, but gives a good description of the basic theory). -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9087642431259155, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/112217/velocity-radius-and-angular-speed-finding-units	Velocity, Radius and Angular Speed - Finding Units I have been trying to figure this one out for two days, and I really have no idea what to do. To begin with, I need to know what units will be produced in the following situations: • Velocity's units when angular speed and radius are rpm and feet, respectively. (I think it is feet/minute) • Anuglar speed's units when radius and velocity are foot and feet per minute, respectively. (It stays as feet/minute?) • Radius' units when velocity and angular speed are mph and rpm, respectively. (I am guessing it will come out as miles, according to what my teacher tried to communicate) Those are just a few issues, but there are more. I believe that, if I am correct/find why I am incorrect, I will be able to apply the concept to the other problems. The second problem I am having is that I have a problem where: Velocity is 50 feet/second, Angular Speed is 100 revolutions/second and I need to find the radius in miles. According to what I know, velocity is angular speed times theta, so I just have to divide by angular speed. However, I do not know where to go when there are different units. I know it is a lot of text, but I appreciate anyone who can help me; my teacher said she is preoccupied and cannot answer any more of my questions today, so I'm stuck. - 2 Answers I always think of this sort of stuff in terms of forcing the the result to have the units you want and/or the only units that can possibly make sense. In the first case, you are given $\frac{rotations}{minutes}$ and $feet=\frac{feet}{1}$ and you want to end up with velocity, which has the general form $\frac{distance}{time}$. In this case, the pertinent units for distance and time are feet and minutes, respectively. The key, though, is the following: one rotation equals $2\pi$ times the radius (in feet). That is, $$\frac{rotations}{minutes}=2\pi\cdot radius\cdot\frac{feet}{minutes}.$$ But remember that in the end we just want an expression for velocity; that is, $\frac{feet}{minutes}=$ something. That's simple, though. Just divide both sides of the above equation by $2\pi\cdot radius$. The second and third ones are very similar, but angular velocity has units $\frac{rotations}{minutes}$. The last thing you asked is a specific instance of this sort of dimensional analysis, except that you have to convert everything to miles in the end. Alternatively, you can think of it as follows: if the velocity is 50 feet/second and the angular speed is 100 revolutions/second, that means in one second, the apparatus is making 100 revolutions - and furthermore, that these 100 revolutions = 50 feet. From this you get that one revolution is 1/2 a foot. That means that the circumference of the circle drawn out by the motion of the apparatus is 1/2 a foot, meaning that the radius of the circle is $\frac{1}{2}\cdot\frac{1}{2\pi}$ feet. To turn this into miles, just divide by 5280. Hope that helps. - Yay! Much obliged. – nmagerko Feb 22 '12 at 23:52 Angular speed has to be some measure of angle divided by some measure of time. In your second bullet point it will probably be revolutions per minute (rpm). You presumably have some formula saying velocity is related the product of radius and angular speed. Here you probably have $$50 \text{ feet/ second} = r \times 2\pi \text{ radians/revolution }\times 100 \text{ revolutions/ second}$$ which gives a radius of $r = 0.25 / \pi \text{ feet}$, not very much. If you must convert it into miles, then use $1 \text{ mile} = 5280 \text{ feet}$, though that will give you an even smaller number. - That is why I don't understand why we do problems like this! – nmagerko Feb 23 '12 at 0:42 1 It does get slightly easier using SI units. – Henry Feb 23 '12 at 0:54	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 10, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.962917685508728, "perplexity_flag": "head"}
http://mathoverflow.net/questions/75964?sort=votes	## Extending a vector bundle to a torsion free sheaf ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let's say that $X$ is an integral scheme of finite type over a field and $Y \subset X$ is a closed subscheme. Given a vector bundle $E$ on $Y$, is $E$ the restriction to $Y$ of a vector bundle on a neighborhood $U$ of $Y$ in $X$? - Isn't this exactly the Leff(X,Y) condition? This is true if $Y$ is a complete intersection of dimension at least 2 in a smooth projective variety $X$. Given $E$ on $Y$, you take $\hat{E}$ on the completion $\hat{X}$ of $X$ along $Y$, and Leff(X,Y) allows you to lift it to a bundle $E'$ in a neighborhood of $Y$ in $X$. – Parsa Apr 8 2012 at 17:14 What is Leff(X,Y)? – Ian Shipman Apr 8 2012 at 17:24 Ok, I found it. Even if the pair (X,Y) does not satisfy Leff, is there a way to understand when E on Y extends to a bundle on the formal completion? Some kind of obstruction perhaps to extending to various infinitesimal thickenings of Y... – Ian Shipman Apr 8 2012 at 17:29 ## 1 Answer No, this is not even true for line bundles. For an example, let $X = \mathbb{P}^2$ and $Y$ a smooth curve in $X$ of genus $>0$ (over an algebraically closed field). Since $X$ is smooth, any line bundle on an open set $U$ extends to a line bundle on $X$ so the map $\operatorname{Pic}(X) \to \operatorname{Pic}(U)$ is surjective. Since $\operatorname{Pic}(X) \cong \mathbb{Z}$, it follows that the image of $\operatorname{Pic}(U)$ in $\operatorname{Pic}(Y)$ is of rank $1$ and is independent of $U \supset Y$. Since $\operatorname{Pic}(Y)$ is not even finitely generated we see that there exist (many) line bundles on $Y$ which do not extend to any open $U \supset Y$. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 36, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9326929450035095, "perplexity_flag": "head"}
http://unapologetic.wordpress.com/2009/09/01/row-echelon-form/?like=1&source=post_flair&_wpnonce=d91e0e6d9a	# The Unapologetic Mathematician ## Row Echelon Form For now, I only want to focus on elementary row operations. That is, transformations of matrices that can be effected by multiplying by elementary matrices on the left, not on the right. Row echelon form is a (non-unique) simplification of the form of a matrix that can be reached by elementary row operations. A matrix is in row echelon form if all the empty (all-zero) rows are below all the nonzero rows, and the leftmost nonzero entry in a given row is to the right of the leftmost nonzero entry in the row above it. For example, an upper-triangular matrix is in row echelon form. We put a matrix into row echelon form by a method called “Gaussian elimination”, which always moves a matrix closer and closer to row echelon form. First, we look at the first column. If it’s empty we leave it alone and move on the next column. If there are any nonzero entries, there must be only one of them at the top of the column, or the matrix can’t be in row echelon form. So, if this isn’t the case then we pick one of the nonzero entries and call it our “pivot”. Swap its whole row up to the top row, and then use shears to eliminate all the other nonzero entries in the column before moving on to the second column. Before we continue, I’ll point out that the pivot is now the leading entry on a nonzero row at the top of the matrix. We’ll regard it as fixed, so whenever I say “all rows” I mean all the rows that we haven’t locked down yet. Now that we’ve moved on to the next column, we again look for any nonzero entries. If there are none, we can move on. But if there are, we choose one to be a pivot, swap it to the top (below the previously locked rows) and use shears to remove all the other nonzero entries in the column (other than the locked rows). We continue like this, choosing pivots where there are nonzero entries to deal with, moving them to the top, and eliminating all the other nonzero entries below them, until we’ve locked down all the rows or run out of columns. Let’s look at this method applied to the matrix $\displaystyle\begin{pmatrix}2&1&-1&8\\-3&-1&2&-11\\-2&1&2&-3\end{pmatrix}$ in the first column we have the nonzero value $2$ in the first row, so we may as well leave it where it is. We add $\frac{3}{2}$ times the first row to the second to eliminate the $-3$ there, and add the first row to the third to eliminate the $-2$ there. At this point our matrix looks like $\displaystyle\begin{pmatrix}2&1&-1&8\\{0}&\frac{1}{2}&\frac{1}{2}&1\\{0}&2&1&5\end{pmatrix}$ Now in the second row we have $\frac{1}{2}$ at the top (since the very top row is finished). We leave it where it is and add $-4$ times the second row to the third to eliminate the $2$ there. Now the matrix looks like $\displaystyle\begin{pmatrix}2&1&-1&8\\{0}&\frac{1}{2}&\frac{1}{2}&1\\{0}&0&-1&1\end{pmatrix}$ now there are no empty rows, so “all” of them are below all the nonzero rows. Further, within the nonzero rows the leading entries move from left to right as we move down the rows. The matrix is now in row echelon form. Now let’s try this one: $\displaystyle\begin{pmatrix}2&1&-1&8\\-4&-2&2&-11\\4&4&-3&6\\-2&-3&2&-3\end{pmatrix}$ Again, we’ll start with using the upper-left $2$ as our first pivot. We use shears to eliminate the rest of the entries in the first column and get $\displaystyle\begin{pmatrix}2&1&-1&8\\{0}&0&0&5\\{0}&2&-1&-10\\{0}&-2&1&5\end{pmatrix}$ Now there are nonzero entries in the second column, but none of them are at the top. So we swap the second and third rows to bring a pivot to the top $\displaystyle\begin{pmatrix}2&1&-1&8\\{0}&2&-1&-10\\{0}&0&0&5\\{0}&-2&1&5\end{pmatrix}$ and eliminate using a shear $\displaystyle\begin{pmatrix}2&1&-1&8\\{0}&2&-1&-10\\{0}&0&0&5\\{0}&0&0&-5\end{pmatrix}$ the third column has no nonzero entries (other than the two locked-in entries at the top), so we skip it. In the fourth column we use $5$ as a pivot and eliminate, getting $\displaystyle\begin{pmatrix}2&1&-1&8\\{0}&2&-1&-10\\{0}&0&0&5\\{0}&0&0&0\end{pmatrix}$ Now the empty row is on the bottom, and the leading entries of the other rows move from left to right as we move down the rows. Thus, the matrix is now in row echelon form. ### Like this: Posted by John Armstrong \| Algebra, Linear Algebra ## 2 Comments » 1. [...] Equations with Gaussian Elimination The row echelon form of a matrix isn’t unique, but it’s still useful for solving systems of linear [...] Pingback by \| September 2, 2009 \| Reply 2. [...] Row Echelon Form Let’s take row echelon form and push it a little further into a form that is unique: reduced row echelon [...] Pingback by \| September 3, 2009 \| Reply « Previous \| Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 17, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8943586349487305, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/132062/asymptotic-notation	# Asymptotic notation [THIS IS HOMEWORK so please, don't just provide solutions, help me reasoning] I'm dealing with the following problem: Let $f(n)$ be the worst case time complexity of an algorithm A. Assume that there exist $c > 0$ and $n_{0}$, such that for each $n > n_{0}$ there exists an input of size $n$ for which A runs in exactly $c\cdot{g(n)}$ steps. a. Is it necessarily true that $f(n) = \Omega(g(n))$? b. Assume $f(n) = O(g(n))$. Is it necessarily true that $f(n) = \Theta(g(n))$? For a. I think it must be true since the $\Omega$ notation has on it all the elements that are either $f(n) = c\cdot g(n)$ or $f(n) > c\cdot g(n)$ so following the assumption leads to $f(n) = c\cdot g(n)$ being always true, therefore making the whole sentence true. (hope I'm right). For b. I assume $f(n) = O(g(n))$ and $f(n)=c\cdot{g(n)}$ ($\forall{n>n_{0}}$... of course) then I concluded that since (a) is true then $f(n) = \Theta(g(n))$. any help will be appreciated, thanks. - Part b. seems a little strange. If you're already given that $f(n) = c\,g(n)$ when $n$ is large enough why do you need to assume that $f(n) = O(g(n))$? Is that exactly how the question was written? – Antonio Vargas Apr 15 '12 at 13:15 @Antonio yes by 100% – JanosAudron Apr 15 '12 at 13:27 1 In that case then I think your reasoning is correct. Everything is easy if you know that $f(n) = c \,g(n)$ eventually. – Antonio Vargas Apr 15 '12 at 14:07	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 21, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9411389231681824, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/160577/population-change-differential-equations?answertab=active	# Population change differential equations Lake has 400 fish, estimated carrying capacity is 10,000 for that year. Fish tripled in first year. a) Assuming size of the fish population satisfies the logisitic equation find an expression for the size of population after t years. b) How long will it take for population for increase to 5000? I do not know which logisitical equation to use or how to put the numbers in it, this is not like the practice problems. - 5 Am away from your book. The DE will have shape $\frac{dy}{dt}=ky(10000-y)$. – André Nicolas Jun 20 '12 at 0:18 ## 1 Answer As Andre commented above, the logistic equation has the form $$\frac{dy}{dt}=ry(10000-y),$$ and your initial condition $y(0)=400$. This is a separable equation, whose solution is given by $$y(t)=\frac{1000 e^{10000rt}}{24+e^{10000 rt}}.$$ You also have the condition that $y(1)=3\cdot y(0)=1200$, from which it is possible to estimate $r$: $$r\approx 0.0001185.$$ Now you can find the moment $t$ when $y(t)=5000$ (it is $2.6805$). -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 7, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9436778426170349, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/102107/combinatorial-proof-to-the-identity-sum-k-0n-binomxkk-binomxn1n	Combinatorial proof to the identity $\sum_{k=0}^n\binom{x+k}{k}=\binom{x+n+1}{n}$? [duplicate] Possible Duplicate: Combinatorial proof for two identities The identity $$\sum_{k=0}^n\binom{x+k}{k}=\binom{x+n+1}{n}$$ can be verified by a straightforward induction, but is there some nice combinatorial argument which proves it? Here $x$ is a nonnegative integer, by the way. I've always found combinatorial arguments make it easier to remember such identities and intuitively grasp them so I would appreciate seeing one. Thank you. - Sorry, it was hard to tell out of the questions with similar titles if one was a duplicate, since the identities weren't in the title. I tried to include it in my title. – Threepio Jan 24 '12 at 21:35 marked as duplicate by Arturo Magidin, Byron Schmuland, Aryabhata, Fabian, Grumpy ParsnipJan 24 '12 at 22:02 This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question. 1 Answer Rewrite it as $$\sum_{k=0}^n\binom{x+k}x=\binom{x+n+1}{x+1}\;.$$ The righthand side clearly counts the subsets of $[x+n+1]$ of size $x+1$. The lefthand side breaks that count down according to the largest element of the subset. That is, if the largest element is $x+k+1$, the other $x$ elements must be chosen from the set $[x+k]$, and this can be done in $\binom{x+k}k$ ways. (Here I use $[n]$ for $\{1,\dots,n\}$.) -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 10, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9427751302719116, "perplexity_flag": "head"}
http://cs.stackexchange.com/questions/9831/context-free-grammar-for-language-l	# Context Free Grammar for language L Can someone help with this: $L=\{a^ib^j \mid i,j \ge 1 \text{ and } i \ne j \text{ and } i<2j\}$ I'm trying to write a grammar for this language? I tried this: $S \to S_1 \mid S_2 \\ S_1 \to aXb \\ X \to aXb \mid aaXb \mid aab \\ S_2 \to aYb \\ Y \to aYb \mid Yb \mid b \\$ What do you think? - 1 Seems correct. Is there a problem? – Karolis Juodelė Feb 16 at 11:46 – Paresh Feb 16 at 13:37 ## 1 Answer The solution in the question is correct. The constraint $i\ne j$ is the one that gives us trouble. To get around it we have to split into two cases: (i) $i<j$, and (ii) $i>j$ (but still $i<2j$) thus $S\to S_{(i)} \mid S_{(ii)}$ as the first production splits into this two cases. Now, divide and conquer: case (i) is very simple, since we only need $i<j$, $S_{(i)} \to aS_{(i)}b \mid B$ $B \to Bb \mid b$ This is your $S_2$. For case (ii), we need to $j< i < 2j$, so for every single $a$, the variable $C$ will generate exactly one $b$, and at a certain point we switch to the variable $D$ that will generate two $a$'s for each $b$. $S_{(ii)} \to aCb$ $C \to aCb \mid D$ $D \to aaDb \mid aab$ This is your $S_1$. - To get $i < j$ you need $S_{(i)} \rightarrow a S_{(i)} b \mid B b$, $B \rightarrow B b \mid \epsilon$. The $B$ alternative adds at least one $b$ here. – vonbrand Mar 20 at 13:29 yes, you're right. – Ran G. Mar 20 at 16:19	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 27, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8735233545303345, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/55383?sort=oldest	## Impact of discrepancy between Kunen’s and Jech’s definition of iterated forcing on full support iterations ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) One of the first things we usually learn when we study iterated forcing is that we can force over a model of ZFC + GCH to make the continuum function ($\lambda \mapsto 2^{\lambda}$) restricted to some set $S$ of regular cardinals follow any nondecreasing pattern satisfying König's Theorem ($\text{cof}(2^{\lambda}) > \lambda$). This is accomplished by an Easton product of adding the desired number of Cohen subsets to each cardinal in $S$. For example, assuming that the GCH holds in $V$, we can force $2^{\aleph_n} = \aleph_{n+2}$ for all Natural numbers $n$ with the full support $\omega$-iteration where we force with the ground model poset adding $\aleph_{n+2}$ many Cohen subsets of $\aleph_n$ at every stage $n$. However, if we instead force with the posets adding $\aleph_{n+2}$ Cohen subsets of $\aleph_n$ from the successive extensions, then the iteration will preserve the GCH. Such a phenomenon is easily explained away by the fact that after forcing to add Cohen subsets of $\aleph_n$, we have necessarily changed the definition of the forcing adding Cohen subsets of $\aleph_{n+1}$ in the extension. Now consider a full support $\omega$-iteration $\mathbb{P}$ where at every stage $n$, we force to add a Cohen real. Since the poset adding a Cohen real is the set of finite partial binary sequences with domain $\omega$, all transitive models of ZFC agree on its definition. Therefore, the aforementioned concern disappears in this context. Because the forcing to add a Cohen real satisfies the countable chain condition (c.c.c.), it is proper. Then since proper forcing is closed under countable support iterations using Jech's definition of a forcing iteration, $\mathbb{P}$ will be proper and hence $\omega_1$-preserving. However, using Kunen's definition, the iteration can collapse $\omega_1$. For example, from Chapter VIII, Exercise (E4): Let $\mathbb{P}_{\omega}$ be defined by countable supports, and let each $\pi_n$ be $(\text{Fn}(\omega, 2)){}^{\check{}}$). Show that $\mathbb{P}_{\omega}$ collapses $\omega_1$. Kunen then notes that this problem goes away when full names are used. On the other hand, Jech seems to circumvent the problem altogether by defining $\mathbb{P}_n * \dot{\mathbb{Q}}_n$ to be the set of pairs $\langle p, \dot{q}\rangle \in \mathbb{P} \times \text{dom}(\dot{\mathbb{Q}}_n)$ such that all conditions from $\mathbb{P}_n$ force $\dot{q}$ to be in $\dot{\mathbb{Q}}_n$ rather than just requiring that the condition $p$ forces this as Kunen does. What I would like to see then is an elaboration of the explanation of why these forcing iterations are different including the intuition behind these differences. Specifically, my question is as follows: Why does the full support $\omega$-iteration forcing with $(\text{Fn}(\omega, 2)){}^{\check{}}$ (forcing to add a Cohen real) at every stage collapse $\omega_1$ under Kunen's definition but not under Jech's? For example, how is it that every binary $\omega$-sequence from the ground model is coded in the generic filter using Kunen's definition but not Jech's? As a side request related to how this question came up, can you provide an example of a full support $\omega$-iteration that must have size at least $\aleph_2$, but each individual stage of forcing is necessarily proper and of size at most $\aleph_1$? - For the side request, please also assume CH. – Jason Feb 14 2011 at 6:11 1 Jech is hiding the use of full names in his construction. The assertion that every element of a PO forces a statement is exactly the statement that 1 forces that statement, which when viewed from the confines of Kunens construction is exactly his definition of full-name. There is a nice intuition and kinda picture that I imagine when thinking about both not all that sure I can describe it adequately though. – Michael Blackmon Feb 14 2011 at 11:21 ## 1 Answer I hope I can answer this question in a reasonable way. The natural way to define the iteration $\mathbb P\dot{\mathbb Q}$ is to consider all pairs $(p,\dot q)$ with $p\in\mathbb P$ and $p\Vdash\dot q\in\dot{\mathbb Q}$. Unfortunately this is a proper class. (The definition in Jech's book also gives a proper class.) The problem in Kunen's definition of the iteration $\mathbb P\dot{\mathbb Q}$ is not so much that he defines $(p,\dot q)\in\mathbb P\dot{\mathbb Q}$ if $p$ forces $\dot q$ to be in $\dot{\mathbb Q}$, but that he only considers names $\dot q$ in the domain of $\dot{\mathbb Q}$. He does this to make sure that the iteration $\mathbb P\dot{\mathbb Q}$ is a set rather than a proper class. But there are other ways to make sure that the iteration is a set, for example by allowing arbitrary names for $\dot q$ that are in some sufficiently large $H_\chi$ (sets with transitive closure of size $<\chi$). "Sufficiently large" depends on $\mathbb P$ and $\dot{\mathbb Q}$. Now Jech only considers pairs $(p,\dot q)$ such that $\dot q$ is forced by $1_{\mathbb P}$ to be in $\dot{\mathbb Q}$. But this doesn't really make a difference since by the existential completeness lemma, if $p\Vdash\dot q\in\dot{\mathbb Q}$, then there is a name $\dot r$ which is forced by $1_{\mathbb P}$ to be a condition in $\dot{\mathbb Q}$ and such that $p\Vdash\dot q=\dot r$. Now in the case of a two step iteration, the definitions by Jech and Kunen are equivalent, and both are equivalent to the one that I suggested (arbitrary names in the second coordinate, but cutting off somewhere). The right way to see this is to first show that if you cut off Jech's iteration at a sufficiently large $H_\chi$, then you get a dense subset of Jech's iteration. This cut off version of Jech's iteration is dense in my version (with the same sufficiently large $H_\chi$). Now Kunen's iteration is equivalent to mine since given $(p,\dot q)$ in my iteration, there are $s\leq p$ and $\dot t$ in the domain of $\dot Q$ such that $s\Vdash\dot q=\dot t$. We have $(s,\dot t)\leq(p,\dot q)$, showing that Kunen's iteration is dense in mine. Things start to fall apart if we go to longer iterations, and iterations with infinite support. My argument above still shows that Kunen's iteration of arbitrary length, with finite supports, is equivalent to my version for long iterations. This is because given a finitely supported condition in the long iteration in my sense, say with support of length $n$, you can decrease the first $n-1$ coordinates to force the last coordinate to be something given by a name in the domain of the respective name of a partial order. Then you decrease the first $n-2$ coordinates to take care of the $n-1$-th coordinate and so on. This argument clearly does not go through with countable supports. Hence the countable support iteration in Kunen's sense is not necessarily equivalent to my version or Jech's version. It is equivalent, however, if all the iterands are given by full names. Now, what is the problem in terms of properness? The proof of properness of countable support iterations of proper forcing (and I am referring to the proof in Goldstern's "Tools for your forcing construction") makes frequent use of the existential completeness lemma to cook up names for conditions. Now in the Kunen version of the iteration it may happen that these names cannot be used since they may not be in the domain of the respective name for a partial order. And in fact, as the exercise shows, even the countable support iteration of Cohen forcing of length $\omega$ in Kunen's sense depends on which names you choose for the iterands: Full names yield a proper forcing (which would be equivalent to the iteration in my sense), but taking canonical names yields a forcing notion that collapses $\omega_1$. I hope this helps. - I still want to think through some stuff, but this is very helpful. Thank you very much. – Jason Mar 11 2011 at 20:06	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 80, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9434811472892761, "perplexity_flag": "head"}
http://cms.math.ca/10.4153/CJM-2012-042-9	Canadian Mathematical Society www.cms.math.ca \| \| \| \| \| \|----------\|----\|-----------\|----\| \| \| \| \| \| \| \| \| Site map \| \| \| CMS store \| \| location: Publications → journals → CJM Abstract view # Cumulants of the $q$-semicircular Law, Tutte Polynomials, and Heaps Read article [PDF: 242KB] Published:2012-11-13 • Matthieu Josuat-Vergès, CNRS and Institut Gaspard Monge, Université Paris-Est Marne-la-Vallée, 5 Boulevard Descartes, Champs-sur-Marne, 77454 Marne-la-Vallée cedex 2, France Features coming soon: Citations (via CrossRef) Tools: Search Google Scholar: Format: LaTeX MathJax PDF ## Abstract The $q$-semicircular distribution is a probability law that interpolates between the Gaussian law and the semicircular law. There is a combinatorial interpretation of its moments in terms of matchings where $q$ follows the number of crossings, whereas for the free cumulants one has to restrict the enumeration to connected matchings. The purpose of this article is to describe combinatorial properties of the classical cumulants. We show that like the free cumulants, they are obtained by an enumeration of connected matchings, the weight being now an evaluation of the Tutte polynomial of a so-called crossing graph. The case $q=0$ of these cumulants was studied by Lassalle using symmetric functions and hypergeometric series. We show that the underlying combinatorics is explained through the theory of heaps, which is Viennot's geometric interpretation of the Cartier-Foata monoid. This method also gives a general formula for the cumulants in terms of free cumulants. Keywords: moments, cumulants, matchings, Tutte polynomials, heaps MSC Classifications: 05A18 - Partitions of sets 05C31 - Graph polynomials 46L54 - Free probability and free operator algebras	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.7995302081108093, "perplexity_flag": "middle"}
http://www.all-science-fair-projects.com/science_fair_projects_encyclopedia/Moment_(physics)	# All Science Fair Projects ## Science Fair Project Encyclopedia for Schools! Search Browse Forum Coach Links Editor Help Tell-a-Friend Encyclopedia Dictionary # Science Fair Project Encyclopedia For information on any area of science that interests you, enter a keyword (eg. scientific method, molecule, cloud, carbohydrate etc.). Or else, you can start by choosing any of the categories below. # Moment (physics) See also moment (mathematics) for a more abstract concept of moments that evolved from this concept of physics. In physics, the moment is a quantity that represents the magnitude of force applied to a rotational system at a distance from the axis of rotation. The concept of the moment arm, this characteristic distance, is key to the operation of the lever, pulley, gear, and most other simple machines capable of generating mechanical advantage. Contents ## Overview In general, the moment M of a vector B is $\mathbf{M_A} = \mathbf{r} \times \mathbf{B} \,$ where r is the position where quantity B is applied. If r is a vector relative to point A, then the moment is the "moment M with respect to the axis that goes through the point A", or simply "moment M around A". If A is the origin, one often omits A and says simply moment. ## Parallel axis theorem Since the moment is dependent on the given axis, the moment expression possess a common property when the observation axis is changed. If MA is the moment around A, then the moment around the axis that goes through a point B is $\mathbf{M_B} = \mathbf{M_A} + \mathbf{R} \times \mathbf{B} \,$ where R is the vector from point B to point A. This expression is usually referred to as the parallel axis theorem. For cases when the moment is the sum of individual "submoments", such as in rigid body dynamics where each particle of the body contribute to a moment, the axis change is the sum of a macroscopic and microscopic quantity, $\mathbf{M_B} = \mathbf{R} \times \mathbf{B} + \sum_{i=0}{\mathbf{r_i} \times \mathbf{b_i}} \,$ where $\mathbf{B} = \sum_{i=0}{\mathbf{b_i}} \,$ or alternatively, $\mathbf{M_B} = \mathbf{R} \times \mathbf{B} + \mathbf{M_A} \,$ ## Related quantities Some notable physical quantities arise from the application of moments: • Angular momentum (L = Iω), which is typically the cause of rotational motion of a body. • Moment of inertia (I = mω×r), which is analogous to mass in discussions of rotational motion. • Torque (τ = rF), which is a force applied on a position of a body. When no net torque is applied, angular momentum is conserved. ## History The principle of moments is derived from Archimedes' discovery of the operating principle of the lever. In the lever one applies a force, in his day most often human muscle, to an arm, a beam of some sort. Archimedes noted that the amount of force applied to the object, the moment of force, is defined as M = rF, where F is the applied force, and r is the distance from the applied force to object. 03-10-2013 05:06:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 5, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8755354881286621, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/294443/would-it-also-be-useful-to-include-an-ordered-pair-function-in-first-order-logic	# Would it also be useful to include an ordered pair function in first order logic? Typically, first-order logic is assumed to include an equality relation $=$, even though this is "non-logical," together with some postulates about equality. Would it also be useful to include an ordered pair function $(,)?$ One could assume that $(x,y)=(x',y')$ precisely when $x=x$ and $y=y'$. Or perhaps it would be best to add denumerably many such functions, $()$, $(,)$, $(,,)$, etc. The upshot of doing (either) of these is that relations can now all be unary. This would prettify a lot of notation. For instance, the axiom schema of replacement would look much neater. Thoughts, anyone? EDIT: Replacement would read as follows. $$\forall w \forall A[(\forall x \in A \exists ! y \phi(x,y,w,A)) \Rightarrow \exists B\forall y(y \in B \Leftrightarrow \exists x \in A \phi(x,y,w,A))]$$ - 1 I am very confused about how the Axiom Schema of Replacement would "look much neater" with the introduction of pairing/tripling/$n$-tupling operations in the syntax itself. Please express this Schema in your new syntax so I/we can judge. – Arthur Fischer Feb 4 at 12:14 So now every function has to be defined on tuples of every arity? Not to mention on tuples whose elements are themselves tuples? This doesn't sound very neat to me. – Chris Eagle Feb 4 at 12:31 @ArthurFischer Done. – user18921 Feb 4 at 20:48 ## 4 Answers The main issue with this that I can see would be that it wouldn't allow for any finite models, because if a model has $n$ distinct elements $a_i$, then it also has to have $n^2$ distinct elements of the form $(a_i, a_j)$. I think this on its own is reason enough to avoid it for first order logic in general. Having said that, there are several individual theories where it is convenient to have a pairing operator. Set theory and number theory both have conservative extensions with pairing operators (that is, there are already "implicit" pairing operators in the theory). Because of this, in these theories someone will often use brackets $(,)$ in their formulas as if there is a pairing operator. Like you said, this does make the notation much nicer. - The fact that you mention Replacement leads me to believe that you're especially interested in first-order theories which are meant to capture some sort of set theory. In a set theory, we have a way of identifying ordered pairs with elements of the universe - the standard choice in ZFC is to code $(a,b)$ by the element $\{\{a\},\{a,b\}\}$. Now we have a definable predicate Pair($x$) which determines whether some set $x$ codes a pair (it says "This set has two elements. One of these has one element, $a$. The other has at most two elements, one of which is $a$.") and definable functions First($x$) and Second($x$) which return $a$ and $b$. The theory then proves $\forall a \forall b \exists x\, \text{Pair}(x) \land (\text{First}(x) = a) \land (\text{Second}(x) = b)$. By adding the ordered pair function, you're adding a Skolem function for the sentence above. This eliminates the existential quantifier, so you now have $\forall a \forall b \text{Pair}((a,b)) \land (\text{First}((a,b)) = a) \land (\text{Second}((a,b)) = b)$. This is a totally standard thing to do in first-order logic - it simplifies syntax without changing the strength of the theory or its class of models (this is what the user aws means by saying that set theory has a "conservative" extension with a pairing operator). It's analogous to adding the inverse symbol ($^{-1}$) to the theory of groups. You can read more details on the wikipedia link. But set theory is somehow special in that there is a canonical way to code a pair of elements as a single element. That is, there is a definable injection $M^2 \rightarrow M$ for any model $M$. In general first-order logic, you don't have this kind of structure, and adding a pairing function to a general theory could really change the theory. However, there is a "safe" way to add pairing (and n-tuple) functions to an arbitrary first-order theory, which I'll describe now. We move from single-sorted logic to many-sorted logic, with one sort for each natural number. The $n^{th}$ sort represents the cartesian power $M^n$. Now for each $n$, we can also add an $n$-ary function $f_n$ which takes the $n$ elements $x_1,\dots,x_n$ to the element in the $n^{th}$ sort representing the $n$-tuple ($x_1,\dots,x_n$). We also add axioms stating that $f_n$ is a bijection for each $n$. One can check that this is construction gives a "conservative extension" in some sense. The advantage of this situation is that we can now view all definable functions and relations as unary, as you wanted. The expense is that we're in a many-sorted context, which you have to get used to. Model theorists often go further, adding new sorts not just for all $M^n$, but for all quotients of $M^n$ by definable equivalence relations - the resulting theory is called $T^{eq}$, and it is a nice place to work for a variety of reasons. - In fact, there are lots of pairing functions in set theory – one benefit of using an explicit pairing function is to get rid of the "theorems" that depend on the precise implementation of pairing. – Zhen Lin Feb 4 at 19:14 2 @ZhenLin Yes, that's a good point - there's another approach, which I didn't talk about, where you add a pairing function to the language and an axiom saying that it's injective, but you don't specify the form of output. This is attractive in that you don't get "meaningless theorems" like $\{a,b\} \in (a,b)$. On the other hand, it might be unattractive to people who want set theory to be "as close to complete as possible" - it makes lots of very simple sentences independent! Of course, this might support the point of view that this desire for completeness is misguided... – Alex Kruckman Feb 4 at 19:38 A comment on an incidental (but interesting) feature of the question, rather than a direct answer. According to the OP: Typically, first-order logic is assumed to include an equality relation =, even though this is "non-logical", ... But why suppose identity is non-logical? We need a demarcation criterion for distinguishing logical from non-logical operators. Let's consider two: A. There's the idea going back to Gentzen that a logical operator is one that is fully defined by introduction rules and (harmonious) elimination rules. This idea is explored in e.g. a famous paper by Ian Hacking 'What is Logic?' (1979) reprinted in e.g. Dov Gabbay (ed) What is a Logical System?. B. There's the idea going back to Tarski that a logical operator is one that exhibits a certain invariance under arbitrary permutations of the domain of objects. This idea is explored in e.g. Tarski's 'What are Logical Notions?' History and Philosophy of Logic (1986). (Transcript of a 1966 talk.) Both ideas A and B, for a start, make identity a logical operator. - I was implicitly defining non-logical as follows: A logical symbol can be viewed as acting on formulae and returning a formula, whereas a non-logical symbol can be viewed as acting on acting on expressions and returning a formula. However, the possible definitions of "logical operator" you mention are probably better definitions than what I had in mind. – user18921 Feb 4 at 20:58 Pairing does exist in first-order logic. It is implicit in untyped logic, as pairing is managed through the way we organize formulae. e.g. in the theory of real closed fields (arithmetic of real numbers), we can't have a variable $P$ that represents a point of the Euclidean plane $\mathbb{R}^2$, but we can systematically use two variables $x$ and $y$ always quantified in pairs and used together in relations and functions. In untyped logic, we can introduce pairing at a purely syntactic level as well. We can define a new language that allows variables that range over tuples of objects, and provide a purely syntactic translation from this new language into the old language where all variables are merely objects. There is also typed logic, which generally have explicit product types. e.g. taking again the theory of real closed fields, we can let $R$ denote the type comprising numbers, and there are product types such as $R \times R$ comprising pairs of numbers. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 51, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9543578028678894, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/237246/find-orthonormal-basis-for-image-plane-given-camera-matrix	# Find Orthonormal Basis for Image Plane Given Camera Matrix Given a projection matrix $P = [M \| p_4]$, ($M \in 3 \times 3$, $p_4 \in 3 \times 1$), the principal axis (the vector that passes through the center of projection and is perpendicular to the image plane) $v$ is $$v = det(M)m_3$$ where $m_3$ is the third row of $M$. Is this correct? Now I want to find two more vectors, say $v_1$ and $v_2$, that form an orthonormal basis with $v$ and span the image plane. I know that $v_1$ and $v_2$ are not unique so I want them to be equal to the $x$- and $y$-axes once the principal axis coincides with the $z$-axis (of some world coordinate system). If the principal axis differs from the $z$-axis due to some transformation I want that same transformation to give me $v_1$ and $v_2$ if applied to the $x$- and $y$-axes. Hope you get what I mean by that! Thanks in advance! - Your requirement still does not force a unique representation: you can rotate about the $z$ axis while maintaining alignment of that axis with your principal axis. As a consequence, “that same transformation” isn't well defined. Basically you want two more vectors orthogonal to $m_3$, right? Simply compute the cross product between that axis and any vector, and the result will be orthogonal. If may be zero, so take the cross product with the three unit vectors and choose the result with maximal length as $v_1$. You can then compute $v_2=m_3\times v_1$. – MvG Nov 14 '12 at 17:54 Hmmmmm i see :( ... what if I know the extrinsic parameters of my camera? Doesn't that give me that transformation? – Yasmin Nov 15 '12 at 9:29	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 23, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9151535630226135, "perplexity_flag": "head"}
http://mathoverflow.net/questions/42189/number-of-edges-in-low-complexity-graphs	## Number of edges in low complexity graphs ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Is the number of edges in the 3-connected graph with the minimum number of spanning trees always $\lceil {\frac{3}{2}}n\rceil$? - 1 What are some examples of such 3c-min-span graphs? Presumably you know some classes that achieve the min... – Joseph O'Rourke Oct 14 2010 at 19:40 It seems there are no classes which achieve the minimum always. An example of a class which achieve a low number of spanning trees are Prism Graphs, but they do not always have the minimum number of spanning trees. – utdiscant Oct 14 2010 at 20:50 For what values of n do you know this to be true? – jc Oct 15 2010 at 0:18 ## 1 Answer Are the minimally 3-connected graphs (edge removal pushes you into a 2-connected graph) the same as the class of 3-connected graphs with the minimal number of spanning trees? There is a nice classification of the minimal 3-connected graphs (or maximal not-3-connected graphs, I forget which direction he does it in) in Jonsson's book "Simplicial Complexes of Graphs". I think there's also a classification of minimal 3-connected graphs inside a paper by David Fisher, Kathryn Fraughnaugh, and Larry Langley which could help ["3-connected graphs of minimal size".] At the very least, they note that all 3-connected graphs on n vertices have at least 3/2(n-2) edges, and produce graphs achieving this bound. All of your graphs with the minimal number of spanning trees must be one of these minimal 3-connected graphs (as if your hypothetical "minimal spanning tree" graph wasn't also a minimal 3-connected graph, removing edges to a minimal 3-connected graph would reduce the number of spanning trees.) - It is true that the 3-connected graph(s) with the minimum number of spanning trees must be found among the 3-connected graphs with the minimal number of edges, but this is not enough to draw a conclusion, since it is easy to find examples showing that minimal 3-connected graphs have more than ceil(3n/2) edges. For instance the Wheel Graph on 6 vertices is a minimal 3-connected graph, but has 10 edges. – utdiscant Oct 15 2010 at 1:57 Also, I can't seem to find the paper ["3-connected graphs of minimal size".] which you are referring too, do you have a link? – utdiscant Oct 15 2010 at 2:01 The actual title appears to be "P_3 - connected graphs of minimal size" in Ars Combinatorica. I found this by putting one of the author's names into my favorite search engine. This led me to the personal website of the author, which included a CV, which included a list of papers. (MathSciNet would have been even quicker.) – JBL Oct 15 2010 at 2:31 Yeah, that's a good point. Since the complex of not 3-connected graphs is homotopic to a wedge of spheres of the same dimension, I was under the impression that all of the minimal 3-connected graphs were of the same size - not the case. The critical graphs left over after an appropriately chosen discrete Morse matching are though... Do you have these graphs classified for small n? – Gwyn Whieldon Oct 15 2010 at 5:00	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9455713033676147, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/15098/linear-acceleration-vs-angular-acceleration-equation?answertab=oldest	# Linear acceleration vs angular acceleration equation I'm learning about angular velocity, momentum, etc. and how all the equations are parallel to linear equations such as velocity or momentum. However, I'm having trouble comparing angular acceleration to linear acceleration. Looking at each equation, they are not as similar as some of the other equations are: • Anglular acceleration = velocity squared / radius • Linear acceleration = force/ mass I would think angular acceleration would take torque into consideration. How is Vsquared similar in relation to force, and how is radius's relation to Vsquared match the relationship between mass and force? I suppose the root of this misunderstanding is how I'm thinking of angular acceleration, which is only an vector representing an axis's direction, and having a magnitude equal to the number of radians rotated per second. I also am confused on what exactly 'V' (tangential velocity) represents and how it's used. Is it a vector whose magnitude is equal to the number of radians any point on a polygon should rotate? What is the explanation? - ## 3 Answers You made a mistake in assuming that the angular acceleration ($\alpha$) is equal to $v^2/r$ which actually is the centripetal acceleration. In simple words, angular acceleration is the rate of change of angular velocity, which further is the rate of change of the angle $\theta$. This is very similar to how the linear acceleration is defined. $$a=\frac{d^2x}{dt^2} \rightarrow \alpha=\frac{d^2\theta}{dt^2}$$ Like the linear acceleration is $F/m$, the angular acceleration is indeed $\tau/I$, $\tau$ being the torque and I being moment of inertia (equivalent to mass). I also am confused on what exactly 'V' (tangential velocity) represents and how it's used. Is it a vector who's magnitude is equal to the number of radians any point on a polygon should rotate? The tangential velocity in case of a body moving with constant speed in a circle is same as its ordinary speed. The name comes from the fact that this speed is along the tangent to the circle (the path of motion for the body). Its magnitude is equal to the rate at which it moves along the circle. Geometrically you can show that $v = r\omega$. - $a_c = \frac{v^2}{r}$ isn't angular acceleration. It's the magnitude of the linear acceleration towards the centre of an object following a circular path at constant angular velocity. Angular acceleration is the derivative of angular velocity, and the analogue of Newton's second law is that angular acceleration equals torque divided by moment of inertia. - Always start with the units. They'll tell you a lot about the equations, and allow you to fix consistency errors. Incidentally, this is why I prefer Leibniz's notation over Newton's for derivatives, the units are immediately determined by examining the derivative, e.g. $dx/dt$ has units of distance over time assuming the usual definition of $x$ and $t$. In this case, the angle, $\theta$, is the equivalent of distance traveled in linear kinematics, and it has units of radians (${rad}$). (Radians, being unit less, are to some extent a place holder, but place holders can be very useful, so keep them in mind.) So, then the rate of change of angle with respect to time, $\omega$, has the units of ${rad}/s$. Angular acceleration, $\alpha$, will then have units of ${rad}/s^2$. With those in mind, you can immediately tell that $a_c = \frac{v^2}{r}$ is not an angular acceleration, but a linear acceleration, as described by Peter. Similarly, angular acceleration is not directly related to force, but to torque, $\tau = I \alpha$, where $I$ is the moment of inertia. (From a mathematical perspective, the moment of inertia is the second moment of the mass distribution where the center of mass is the first moment.) Torque has the units ${kg}\ m^2/s^2$, where the radians were dropped. Note, it has units of energy, or $(Force)(distance)$, and $\tau = r \times F$. On any single parameter curve in $\mathbb{R}^n$, $n\geq2$, the derivative with respect to that parameter always lies tangent to the curve. The derivative is literally showing us how the position is going to change. From a physics perspective, you can think of this as attaching the velocity and acceleration vectors to the moving object, itself, as in drawing a free body diagram. To be concrete, for uniform circular motion, the position is $$r(t) = R( \cos(t) \hat{i} + \sin(t) \hat{j} )$$ where $R$ is the radius of the circle, $\hat{i}$ and $\hat{j}$ are the unit vectors in the $x$ and $y$ directions, respectively, and the velocity is $$v(t) = R( -\sin(t) \hat{i} + \cos(t) \hat{j} ).$$ Note, that the velocity is perpendicular to the position which is a property of circular motion. From this, you should be able to mathematically demonstrate that the acceleration is perpendicular to the velocity and anti-parallel to the position. I'll leave you with the problem of understanding why this makes sense physically, also. - +1 Nice point on examining the units – Griffin Dec 13 '11 at 6:38	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 30, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9479528069496155, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/59846/list	## Return to Answer 1 [made Community Wiki] Well, there are of course many examples but perhaps the Dirac $\delta$-function is the most striking one. With quite some violence to the math of usual calculus, Dirac's idea turned out to be exremely useful: the "function" everywhere zero and very high at $0$ so that the integral is $1$. Applications are ranging from Fourier analysis, (linear) PDE, quantum mechanics, and many more. Now of course we know how to make things rigorous in the framework of Schwartz's test function and distribution theory. But the intuition is clearly from physics. Similar and also in this framework is the notion of oscillatory integrals...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9420386552810669, "perplexity_flag": "middle"}
http://bmet.wikia.com/wiki/Optics	# Optics Talk0 1,606pages on this wiki ## About Optics is the study of the behavior and properties of light, including its interactions with matter and the construction of optical instruments that use or Photodetector it. Optics usually describes the behavior of visible, ultraviolet, and infrared light. Because light is an electromagnetic wave, other forms of electromagnetic radiation such as X-rays, microwaves, and radio waves exhibit similar properties.[1] Optics is regarded within theoretical physics as a subfield of electromagnetism, and the vast majority of optical phenomena can be accounted for using the classical electromagnetic description of light. Complete electromagnetic descriptions of light are often difficult to apply in practice, however, so practical optics is usually done using simplified models. The most common of these, geometric optics, treats light as a collection of Ray (optics)s that travel in straight lines and bend when they pass through or reflect from surfaces. Physical optics is a more comprehensive model of light, which includes wave effects such as diffraction and interference that cannot be accounted for in geometric optics. Not all optical phenomena can be treated using classical electromagnetism. Some phenomena depend on the fact that light has both wave-particle duality. Explanation of these effects requires quantum mechanics. In quantum mechanics, light is treated as a collection of particles called photons. Quantum optics deals with the application of quantum mechanics to optical systems. Optical science is relevant to and studied in many related disciplines including astronomy, electrical engineering, photography, and medicine (particularly ophthalmology and optometry). Practical applications of optics are found in a variety of technologies and everyday objects, including mirrors, lenses, optical telescope, microscopes, lasers, and fiber optics. ## Early history of optics The earliest known lenses were made from polished crystal, often quartz, and have been dated as early as 700 BC for Assyrian lenses such as the Layard / Nimrud lens. There are many similar lenses from ancient Egypt, Greece and Babylon. The ancient Romans and Greeks filled glass spheres with water to make lenses. However, glass lenses were not thought of until the Middle Ages. Some lenses fixed in ancient Egyptian statues are much older than those mentioned above. There is some doubt as to whether or not they qualify as lenses, but they are undoubtedly glass and served at least ornamental purposes. The statues appear to be anatomically correct schematic eyes. In ancient India, the philosophical schools of Samkhya and Vaisheshika, from around the 6th–5th century BC, developed theories on light. According to the Samkhya school, light is one of the five fundamental "subtle" elements (tanmatra) out of which emerge the gross elements. In contrast, the Vaisheshika school gives an atomic theory of the physical world on the non-atomic ground of ether, space and time. (See Indian atomism.) The basic atoms are those of earth (prthivı), water (apas), fire (tejas), and air (vayu), that should not be confused with the ordinary meaning of these terms. These atoms are taken to form binary molecules that combine further to form larger molecules. Motion is defined in terms of the movement of the physical atoms. Light rays are taken to be a stream of high velocity of tejas (fire) atoms. The particles of light can exhibit different characteristics depending on the speed and the arrangements of the tejas atoms. Around the first century BC, the Vishnu Purana refers to sunlight as the "the seven rays of the sun". In the fifth century BC, Empedocles postulated that everything was composed of four elements; fire, air, earth and water. He believed that Aphrodite made the human eye out of the four elements and that she lit the fire in the eye which shone out from the eye making sight possible. If this were true, then one could see during the night just as well as during the day, so Empedocles postulated an interaction between rays from the eyes and rays from a source such as the sun. In his Optics Greek mathematician Euclid observed that "things seen under a greater angle appear greater, and those under a lesser angle less, while those under equal angles appear equal". In the 36 propositions that follow, Euclid relates the apparent size of an object to its distance from the eye and investigates the apparent shapes of cylinders and cones when viewed from different angles. Pappus believed these results to be important in astronomy and included Euclid's Optics, along with his Phaenomena, in the Little Astronomy, a compendium of smaller works to be studied before the Syntaxis (Almagest) of Ptolemy. In 55 BC, Lucretius, a Roman who carried on the ideas of earlier Greek atomists, wrote: "The light and heat of the sun; these are composed of minute atoms which, when they are shoved off, lose no time in shooting right across the interspace of air in the direction imparted by the shove." —Lucretius , On the nature of the Universe Despite being similar to later particle theories of light, Lucretius's views were not generally accepted and light was still theorized as emanating from the eye. In his Catoptrica, Hero of Alexandria showed by a geometrical method that the actual path taken by a ray of light reflected from a plane mirror is shorter than any other reflected path that might be drawn between the source and point of observation. In a twelfth-century translation assigned to Roman mathematician Claudius Ptolemy, a study of refraction, including atmospheric refraction, was described. It was suggested that the angle of refraction is proportional to the angle of incidence. Later in 499, Aryabhata, who proposed a heliocentric solar system of gravitation in his Aryabhatiya, wrote that the planets and the Moon do not have their own light but reflect the light of the Sun. The Indian Buddhists, such as Dignāga in the 5th century and Dharmakirti in the 7th century, developed a type of atomism that is a philosophy about reality being composed of atomic entities that are momentary flashes of light or energy. They viewed light as being an atomic entity equivalent to energy, similar to the modern concept of photons, though they also viewed all matter as being composed of these light/energy particles. ## Classical optics In pre-quantum-mechanical optics, light is an electromagnetic wave composed of oscillating electric and magnetic fields. These fields continually generate each other, as the wave propagates through space and oscillates in time. The frequency of a light wave is determined by the period of the oscillations. The frequency does not normally change as the wave travels through different materials ("media"), but the speed of the wave depends on the medium. The speed, frequency, and wavelength of a wave are related by the formula $v=\lambda\,f ,$ where v is the speed, λ is the wavelength and f is the frequency. Because the frequency is fixed, a change in the wave's speed produces a change in its wavelength. The speed of light in a medium is typically characterized by the index of refraction, n, which is the ratio of the speed of light in vacuum, c, to the speed in the medium: $n=c/v.$ The speed of light in vacuum is a constant, which is exactly 299,792,458 meters per second. Thus, a light ray with a wavelength of λ in a vacuum will have a wavelength of λ / n in a material with index of refraction n. The amplitude of the light wave is related to the intensity of the light, which is related to the energy stored in the wave's electric and magnetic fields. Traditional optics is divided into two main branches: geometrical optics and physical optics. ### Geometrical optics Main article: Geometrical optics As light wave travels through space, it oscillates in amplitude. In this image, each maximum amplitude crest is marked with a plane to illustrate the wavefront. The ray is the arrow perpendicular to these parallel surfaces. Geometrical optics, or ray optics, describes light propagation in terms of "rays". The "ray" in geometric optics is an abstraction, or "instrument", that can be used to predict the path of light. A light ray is a ray that is perpendicular to the light's wavefronts (and therefore collinear with the wave vector). Light rays bend at the interface between two dissimilar media and may be curved in a medium in which the refractive index changes. Geometrical optics provides rules for propagating these rays through an optical system, which indicates how the actual wavefront will propagate. This is a significant simplification of optics that fails to account for optical effects such as diffraction and polarization. It is a good approximation, however, when the wavelength is very small compared with the size of structures with which the light interacts. Geometric optics can be used to describe the geometrical aspects of imaging, including optical aberrations. A slightly more rigorous definition of a light ray follows from Fermat's principle which states that the path taken between two points by a ray of light is the path that can be traversed in the least time. #### Approximations Geometrical optics is often simplified by making the paraxial approximation, or "small angle approximation." The mathematical behavior then becomes linear, allowing optical components and systems to be described by simple matrices. This leads to the techniques of Gaussian optics and paraxial ray tracing, which are used to find basic properties of optical systems, such as approximate image and object positions and magnifications. #### Reflections Main article: Reflection (physics) Reflections can be divided into two types: specular reflection and diffuse reflection. Specular reflection describes glossy surfaces such as mirrors, which reflect light in a simple, predictable way. This allows for production of reflected images that can be associated with an actual (real) or extrapolated (virtual) location in space. Diffuse reflection describes matte surfaces, such as paper or rock. The reflections from these surfaces can only be described statistically, with the exact distribution of the reflected light depending on the microscopic structure of the surface. Many diffuse reflectors are described or can be approximated by Lambert's cosine law, which describes surfaces that have equal luminance when viewed from any angle. In specular reflection, the direction of the reflected ray is determined by the angle the incident ray makes with the surface normal, a line perpendicular to the surface at the point where the ray hits. The incident and reflected rays lie in a single plane, and the angle between the reflected ray and the surface normal is the same as that between the incident ray and the normal. This is known as the Law of Reflection. For flat mirrors, the law of reflection implies that images of objects are upright and the same distance behind the mirror as the objects are in front of the mirror. The image size is the same as the object size. (The magnification of a flat mirror is unity.) The law also implies that mirror images are parity inverted, which we perceive as a left-right inversion. Images formed from reflection in two (or any even number of) mirrors are not parity inverted. Corner reflectors retro-reflect light, producing reflected rays that travel back in the direction from which the incident rays came. Mirrors with curved surfaces can be modeled by ray-tracing and using the law of reflection at each point on the surface. For mirrors with parabolic surfaces, parallel rays incident on the mirror produce reflected rays that converge at a common focus. Other curved surfaces may also focus light, but with aberrations due to the diverging shape causing the focus to be smeared out in space. In particular, spherical mirrors exhibit spherical aberration. Curved mirrors can form images with magnification greater than or less than one, and the magnification can be negative, indicating that the image is inverted. An upright image formed by reflection in a mirror is always virtual, while an inverted image is real and can be projected onto a screen. #### Refractions Main article: Refraction Refraction occurs when light travels through an area of space that has a changing index of refraction; this principle allows for lenses and the focusing of light. The simplest case of refraction occurs when there is an interface between a uniform medium with index of refraction n1 and another medium with index of refraction n2. In such situations, Snell's Law describes the resulting deflection of the light ray: $n_1\sin\theta_1 = n_2\sin\theta_2\$ where θ1 and θ2 are the angles between the normal (to the interface) and the incident and refracted waves, respectively. This phenomenon is also associated with a changing speed of light as seen from the definition of index of refraction provided above which implies: $v_1\sin\theta_2\ = v_2\sin\theta_1$ where v1 and v2 are the wave velocities through the respective media. Various consequences of Snell's Law include the fact that for light rays traveling from a material with a high index of refraction to a material with a low index of refraction, it is possible for the interaction with the interface to result in zero transmission. This phenomenon is called total internal reflection and allows for fiber optics technology. As light signals travel down a fiber optic cable, it undergoes total internal reflection allowing for essentially no light lost over the length of the cable. It is also possible to produce polarized light rays using a combination of reflection and refraction: When a refracted ray and the reflected ray form a right angle, the reflected ray has the property of "plane polarization". The angle of incidence required for such a scenario is known as Brewster's angle. Snell's Law can be used to predict the deflection of light rays as they pass through "linear media" as long as the indexes of refraction and the geometry of the media are known. For example, the propagation of light through a prism results in the light ray being deflected depending on the shape and orientation of the prism. Additionally, since different frequencies of light have slightly different indexes of refraction in most materials, refraction can be used to produce dispersion spectra that appear as rainbows. The discovery of this phenomenon when passing light through a prism is famously attributed to Isaac Newton. Some media have an index of refraction which varies gradually with position and, thus, light rays curve through the medium rather than travel in straight lines. This effect is what is responsible for mirages seen on hot days where the changing index of refraction of the air causes the light rays to bend creating the appearance of specular reflections in the distance (as if on the surface of a pool of water). Material that has a varying index of refraction is called a gradient-index (GRIN) material and has many useful properties used in modern optical scanning technologies including photocopiers and scanners. The phenomenon is studied in the field of gradient-index optics. A ray tracing diagram for a converging lens. A device which produces converging or diverging light rays due to refraction is known as a lens. Thin lenses produce focal points on either side that can be modeled using the lensmaker's equation. In general, two types of lenses exist: convex lenses, which cause parallel light rays to converge, and concave lenses, which cause parallel light rays to diverge. The detailed prediction of how images are produced by these lenses can be made using ray-tracing similar to curved mirrors. Similarly to curved mirrors, thin lenses follow a simple equation that determines the location of the images given a particular focal length (f) and object distance (S1): $\frac{1}{S_1} + \frac{1}{S_2} = \frac{1}{f}$ where $S_2$ is the distance associated with the image and is considered by convention to be negative if on the same side of the lens as the object and positive if on the opposite side of the lens. The focal length f is considered negative for concave lenses. Incoming parallel rays are focused by a convex lens into an inverted real image one focal length from the lens, on the far side of the lens. Rays from an object at finite distance are focused further from the lens than the focal distance; the closer the object is to the lens, the further the image is from the lens. With convex lenses, incoming parallel rays diverge after going through the lens, in such a way that they seem to have originated at an upright virtual image one focal length from the lens, on the same side of the lens that the parallel rays are approaching on. Rays from an object at finite distance are associated with a virtual image that is closer to the lens than the focal length, and on the same side of the lens as the object. The closer the object is to the lens, the closer the virtual image is to the lens. Likewise, the magnification of a lens is given by $M = - \frac{S_2}{S_1} = \frac{f}{f - S_1}$ where S2 is the distance associated with the image and is considered by convention to be negative if on the same side of the lens as the object and positive if on the opposite side of the lens.[39] The focal length f is considered negative for concave lenses. Incoming parallel rays are focused by a convex lens into an inverted real image one focal length from the lens, on the far side of the lens. Rays from an object at finite distance are focused further from the lens than the focal distance; the closer the object is to the lens, the further the image is from the lens. With convex lenses, incoming parallel rays diverge after going through the lens, in such a way that they seem to have originated at an upright virtual image one focal length from the lens, on the same side of the lens that the parallel rays are approaching on. Rays from an object at finite distance are associated with a virtual image that is closer to the lens than the focal length, and on the same side of the lens as the object. The closer the object is to the lens, the closer the virtual image is to the lens. Lenses suffer from aberrations that distort images and focal points. These are due to both to geometrical imperfections and due to the changing index of refraction for different wavelengths of light (chromatic aberration). ### Physical optics Main article: Physical optics hysical optics or wave optics builds on Huygens's principle, which states that every point on an advancing wavefront is the center of a new disturbance. When combined with the superposition principle, this explains how optical phenomena are manifested when there are multiple sources or obstructions that are spaced at distances similar to the wavelength of the light.[40] Complex models based on physical optics can account for the propagation of any wavefront through an optical system, including predicting the wavelength, amplitude, and phase of the wave. Additionally, all of the results from geometrical optics can be recovered using the techniques of Fourier optics which apply many of the same mathematical and analytical techniques used in acoustic engineering and signal processing. Using numerical modeling on a computer, optical scientists can simulate the propagation of light and account for most diffraction, interference, and polarization effects. Such simulations typically still rely on approximations, however, so this is not a full electromagnetic wave theory model of the propagation of light. Such a full model is computationally demanding and is normally only used to solve small-scale problems that require extraordinary accuracy. Gaussian beam propagation is a simple paraxial physical optics model for the propagation of coherent radiation such as laser beams. This technique partially accounts for diffraction, allowing accurate calculations of the rate at which a laser beam expands with distance, and the minimum size to which the beam can be focused. Gaussian beam propagation thus bridges the gap between geometric and physical optics. #### Superposition and interference Main article: Superposition principle In the absence of nonlinear effects, the superposition principle can be used to predict the shape In the absence of nonlinear effects, the superposition principle can be used to predict the shape of interacting waveforms through the simple addition of the disturbances. This interaction of waves to produce a resulting pattern is generally termed "interference" and can result in a variety of outcomes. If two waves of the same wavelength and frequency are in phase, both the wave crests and wave troughs align. This results in constructive interference and an increase in the amplitude of the wave, which for light is associated with a brightening of the waveform in that location. Alternatively, if the two waves of the same wavelength and frequency are out of phase, then the wave crests will align with wave troughs and vice-versa. This results in destructive interference and a decrease in the amplitude of the wave, which for light is associated with a dimming of the waveform at that location. See below for an illustration of this effect Since Huygens's principle states that every point of a wavefront is associated with the production of a new disturbance, it is possible for a wavefront to interfere with itself constructively or destructively at different locations producing bright and dark fringes in regular and predictable patterns. Interferometry is the science of measuring these patterns, usually as a means of making precise determinations of distances or angular resolutions. The Michelson interferometer was a famous instrument which used interference effects to accurately measure the speed of light. The appearance of thin films and coatings is directly affected by interference effects. Antireflective coatings use destructive interference to reduce the reflectivity of the surfaces they coat, and can be used to minimize glare and unwanted reflections. The simplest case is a single layer with thickness one-fourth the wavelength of incident light. The reflected wave from the top of the film and the reflected wave from the film/material interface are then exactly 180° out of phase, causing destructive interference. The waves are only exactly out of phase for one wavelength, which would typically be chosen to be near the center of the visible spectrum, around 550 nm. More complex designs using multiple layers can achieve low reflectivity over a broad band, or extremely low reflectivity at a single wavelength. Constructive interference in thin films can create strong reflection of light in a range of wavelengths, which can be narrow or broad depending on the design of the coating. These films are used to make dielectric mirrors, interference filters, heat reflectors, and filters for color separation in color television cameras. This interference effect is also what causes the colorful rainbow patterns seen in oil slicks. #### Diffraction and optical resolution Diffraction is the process by which light interference is most commonly observed. The effect was first described in 1665 by Francesco Maria Grimaldi, who also coined the term from the Latin diffringere, 'to break into pieces'. Later that century, Robert Hooke and Isaac Newton also described phenomena now known to be diffraction in Newton's rings while James Gregory recorded his observations of diffraction patterns from bird feathers. The first physical optics model of diffraction that relied on Huygens' Principle was developed in 1803 by Thomas Young in his accounts of the interference patterns of two closely spaced slits. Young showed that his results could only be explained if the two slits acted as two unique sources of waves rather than corpuscles. In 1815 and 1818, Augustin-Jean Fresnel firmly established the mathematics of how wave interference can account for diffraction. The simplest physical models of diffraction use equations that describe the angular separation of light and dark fringes due to light of a particular wavelength (λ). In general, the equation takes the form $m \lambda = d \sin \theta$ where d is the separation between two wavefront sources (in the case of Young's experiments, it was two slits), θ is the angular separation between the central fringe and the mth order fringe, where the central maximum is m = 0. This equation is modified slightly to take into account a variety of situations such as diffraction through a single gap, diffraction through multiple slits, or diffraction through a diffraction grating that contains a large number of slits at equal spacing. More complicated models of diffraction require working with the mathematics of Fresnel or Fraunhofer diffraction. X-ray diffraction makes use of the fact that atoms in a crystal have regular spacing at distances that are on the order of one angstrom. To see diffraction patterns, x-rays with similar wavelengths to that spacing are passed through the crystal. Since crystals are three-dimensional objects rather than two-dimensional gratings, the associated diffraction pattern varies in two directions according to Bragg reflection, with the associated bright spots occurring in unique patterns and d being twice the spacing between atoms. Diffraction effects limit the ability for an optical detector to optically resolve separate light sources. In general, light that is passing through an aperture will experience diffraction and the best images that can be created (as described in diffraction-limited optics) appear as a central spot with surrounding bright rings, separated by dark nulls; this pattern is known as an Airy pattern, and the central bright lobe as an Airy disk. The size of such a disk is given by $\sin \theta = 1.22 \frac{\lambda}{D}$ where θ is the angular resolution, λ is the wavelength of the light, and D is the diameter of the lens aperture. If the angular separation of the two points is significantly less than the Airy disk angular radius, then the two points cannot be resolved in the image, but if their angular separation is much greater than this, distinct images of the two points are formed and they can therefore be resolved. Rayleigh defined the somewhat arbitrary "Rayleigh criterion" that two points whose angular separation is equal to the Airy disk radius (measured to first null, that is, to the first place where no light is seen) can be considered to be resolved. It can be seen that the greater the diameter of the lens or its aperture, the finer the resolution. Interferometry, with its ability to mimic extremely large baseline apertures, allows for the greatest angular resolution possible. For astronomical imaging, the atmosphere prevents optimal resolution from being achieved in the visible spectrum due to the atmospheric scattering and dispersion which cause stars to twinkle. Astronomers refer to this effect as the quality of astronomical seeing. Techniques known as adaptive optics have been utilized to eliminate the atmospheric disruption of images and achieve results that approach the diffraction limit. #### Dispersion and scattering Main article: Dispersion (optics) Refractive processes take place in the physical optics limit, where the wavelength of light is similar to other distances, as a kind of scattering. The simplest type of scattering is Thomson scattering which occurs when electromagnetic waves are deflected by single particles. In the limit of Thompson scattering, in which the wavelike nature of light is evident, light is dispersed independent of the frequency, in contrast to Compton scattering which is frequency-dependent and strictly a quantum mechanical process, involving the nature of light as particles. In a statistical sense, elastic scattering of light by numerous particles much smaller than the wavelength of the light is a process known as Rayleigh scattering while the similar process for scattering by particles that are similar or larger in wavelength is known as Mie scattering with the Tyndall effect being a commonly observed result. A small proportion of light scattering from atoms or molecules may undergo Raman scattering, wherein the frequency changes due to excitation of the atoms and molecules. Brillouin scattering occurs when the frequency of light changes due to local changes with time and movements of a dense material Dispersion occurs when different frequencies of light have different phase velocities, due either to material properties (material dispersion) or to the geometry of an optical waveguide (waveguide dispersion). The most familiar form of dispersion is a decrease in index of refraction with increasing wavelength, which is seen in most transparent materials. This is called "normal dispersion". It occurs in all dielectric materials, in wavelength ranges where the material does not absorb light. In wavelength ranges where a medium has significant absorption, the index of refraction can increase with wavelength. This is called "anomalous dispersion". Dispersion occurs when different frequencies of light have different phase velocities, due either to material properties (material dispersion) or to the geometry of an optical waveguide (waveguide dispersion). The most familiar form of dispersion is a decrease in index of refraction with increasing wavelength, which is seen in most transparent materials. This is called "normal dispersion". It occurs in all dielectric materials, in wavelength ranges where the material does not absorb light. In wavelength ranges where a medium has significant absorption, the index of refraction can increase with wavelength. This is called "anomalous dispersion". Material dispersion is often characterized by the Abbe number, which gives a simple measure of dispersion based on the index of refraction at three specific wavelengths. Waveguide dispersion is dependent on the propagation constant. Both kinds of dispersion cause changes in the group characteristics of the wave, the features of the wave packet that change with the same frequency as the amplitude of the electromagnetic wave. "Group velocity dispersion" manifests as a spreading-out of the signal "envelope" of the radiation and can be quantified with a group dispersion delay parameter: $D = \frac{1}{v_g^2} \frac{dv_g}{d\lambda}$ where $v_g$ is the group velocity.[2] For a uniform medium, the group velocity is $v_g = c \left( n - \lambda \frac{dn}{d\lambda} \right)^{-1}$ where n is the index of refraction and c is the speed of light in a vacuum.[3] This gives a simpler form for the dispersion delay parameter: $D = - \frac{\lambda}{c} \, \frac{d^2 n}{d \lambda^2}.$ If D is less than zero, the medium is said to have positive dispersion or normal dispersion. If D is greater than zero, the medium has negative dispersion. If a light pulse is propagated through a normally dispersive medium, the result is the higher frequency components slow down more than the lower frequency components. The pulse therefore becomes positively chirped, or up-chirped, increasing in frequency with time. This causes the spectrum coming out of a prism to appear with red light the least refracted and blue/violet light the most refracted. Conversely, if a pulse travels through an anomalously (negatively) dispersive medium, high frequency components travel faster than the lower ones, and the pulse becomes negatively chirped, or down-chirped, decreasing in frequency with time. The result of group velocity dispersion, whether negative or positive, is ultimately temporal spreading of the pulse. This makes dispersion management extremely important in optical communications systems based on optical fibers, since if dispersion is too high, a group of pulses representing information will each spread in time and merge together, making it impossible to extract the signal. #### Polarization Main article: Polarization Polarization is a general property of waves that describes the orientation of their oscillations. For transverse waves such as many electromagnetic waves, it describes the orientation of the oscillations in the plane perpendicular to the wave's direction of travel. The oscillations may be oriented in a single direction (linear polarization), or the oscillation direction may rotate as the wave travels (circular or elliptical polarization). Circularly polarized waves can rotate rightward or leftward in the direction of travel, and which of those two rotations is present in a wave is called the wave's chirality. The typical way to consider polarization is to keep track of the orientation of the electric field vector as the electromagnetic wave propagates. The electric field vector of a plane wave may be arbitrarily divided into two perpendicular components labeled x and y (with z indicating the direction of travel). The shape traced out in the x-y plane by the electric field vector is a Lissajous figure that describes the polarization state. The following figures show some examples of the evolution of the electric field vector (blue), with time (the vertical axes), at a particular point in space, along with its x and y components (red/left and green/right), and the path traced by the vector in the plane (purple): The same evolution would occur when looking at the electric field at a particular time while evolving the point in space, along the direction opposite to propagation. In all other cases, where the two components either do not have the same amplitudes and/or their phase difference is neither zero nor a multiple of 90°, the polarization is called elliptical polarization because the electric vector traces out an ellipse in the plane (the polarization ellipse). This is shown in the above figure on the right. Detailed mathematics of polarization is done using Jones calculus and is characterized by the Stokes parameters. Media that have different indexes of refraction for different polarization modes are called birefringent. Well known manifestations of this effect appear in optical wave plates/retarders (linear modes) and in Faraday rotation/optical rotation (circular modes). If the path length in the birefringent medium is sufficient, plane waves will exit the material with a significantly different propagation direction, due to refraction. For example, this is the case with macroscopic crystals of calcite, which present the viewer with two offset, orthogonally polarized images of whatever is viewed through them. It was this effect that provided the first discovery of polarization, by Erasmus Bartholinus in 1669. In addition, the phase shift, and thus the change in polarization state, is usually frequency dependent, which, in combination with dichroism, often gives rise to bright colors and rainbow-like effects. In mineralogy, such properties, known as pleochroism, are frequently exploited for the purpose of identifying minerals using polarization microscopes. Additionally, many plastics that are not normally birefringent will become so when subject to mechanical stress, a phenomenon which is the basis of photoelasticity. ## Modern optics Main article: Optical physics Modern optics encompasses the areas of optical science and engineering that became popular in the 20th century. These areas of optical science typically relate to the electromagnetic or quantum properties of light but do include other topics. A major subfield of modern optics, quantum optics, deals with specifically quantum mechanical properties of light. Quantum optics is not just theoretical; some modern devices, such as lasers, have principles of operation that depend on quantum mechanics. Light detectors, such as photomultipliers and channeltrons, respond to individual photons. Electronic image sensors, such as CCDs, exhibit shot noise corresponding to the statistics of individual photon events. Light-emitting diodes and photovoltaic cells, too, cannot be understood without quantum mechanics. In the study of these devices, quantum optics often overlaps with quantum electronics. Specialty areas of optics research include the study of how light interacts with specific materials as in crystal optics and metamaterials. Other research focuses on the phenomenology of electromagnetic waves as in singular optics, non-imaging optics, non-linear optics, statistical optics, and radiometry. Additionally, computer engineers have taken an interest in integrated optics, machine vision, and photonic computing as possible components of the "next generation" of computers. Today, the pure science of optics is called optical science or optical physics to distinguish it from applied optical sciences, which are referred to as optical engineering. Prominent subfields of optical engineering include illumination engineering, photonics, and optoelectronics with practical applications like lens design, fabrication and testing of optical components, and image processing. Some of these fields overlap, with nebulous boundaries between the subjects terms that mean slightly different things in different parts of the world and in different areas of industry. A professional community of researchers in nonlinear optics has developed in the last several decades due to advances in laser technology. ## Applications Optics is part of everyday life. The ubiquity of visual systems in biology indicate the central role optics plays as the science of one of the five senses. Many people benefit from eyeglasses or contact lenses, and optics are integral to the functioning of many consumer goods including cameras. Rainbows and mirages are examples of optical phenomena. Optical communication provides the backbone for both the Internet and modern telephony. ### Human eye Main article: Human eye The human eye functions by focusing light onto an array of photoreceptor cells called the retina, which covers the back of the eye. The focusing is accomplished by a series of transparent media. Light entering the eye passes first through the cornea, which provides much of the eye's optical power. The light then continues through the fluid just behind the cornea—the anterior chamber, then passes through the pupil. The light then passes through the lens, which focuses the light further and allows adjustment of focus. The light then passes through the main body of fluid in the eye—the vitreous humor, and reaches the retina. The cells in the retina cover the back of the eye, except for where the optic nerve exits; this results in a blind spot. There are two types of photoreceptor cells, rods and cones, which are sensitive to different aspects of light. Rod cells are sensitive to the intensity of light over a wide frequency range, thus are responsible for black-and-white vision. Rod cells are not present on the fovea, the area of the retina responsible for central vision, and are not as responsive as cone cells to spatial and temporal changes in light. There are, however, twenty times more rod cells than cone cells in the retina because the rod cells are present across a wider area. Because of their wider distribution, rods are responsible for peripheral vision. In contrast, cone cells are less sensitive to the overall intensity of light, but come in three varieties that are sensitive to different frequency-ranges and thus are used in the perception of color and photopic vision. Cone cells are highly concentrated in the fovea and have a high visual acuity meaning that they are better at spatial resolution than rod cells. Since cone cells are not as sensitive to dim light as rod cells, most night vision is limited to rod cells. Likewise, since cone cells are in the fovea, central vision (including the vision needed to do most reading, fine detail work such as sewing, or careful examination of objects) is done by cone cells. Ciliary muscles around the lens allow the eye's focus to be adjusted. This process is known as accommodation. The near point and far point define the nearest and farthest distances from the eye at which an object can be brought into sharp focus. For a person with normal vision, the far point is located at infinity. The near point's location depends on how much the muscles can increase the curvature of the lens, and how inflexible the lens has become with age. Optometrists, ophthalmologists, and opticians usually consider an appropriate near point to be closer than normal reading distance—approximately 25 cm. Defects in vision can be explained using optical principles. As people age, the lens becomes less flexible and the near point recedes from the eye, a condition known as presbyopia. Similarly, people suffering from hyperopia cannot decrease the focal length of their lens enough to allow for nearby objects to be imaged on their retina. Conversely, people who cannot increase the focal length of their lens enough to allow for distant objects to be imaged on the retina suffer from myopia and have a far point that is considerably closer than infinity. A condition known as astigmatism results when the cornea is not spherical but instead is more curved in one direction. This causes horizontally extended objects to be focused on different parts of the retina than vertically extended objects, and results in distorted images. All of these conditions can be corrected using corrective lenses. For presbyopia and hyperopia, a converging lens provides the extra curvature necessary to bring the near point closer to the eye while for myopia a diverging lens provides the curvature necessary to send the far point to infinity. Astigmatism is corrected with a cylindrical surface lens that curves more strongly in one direction than in another, compensating for the non-uniformity of the cornea. The optical power of corrective lenses is measured in diopters, a value equal to the reciprocal of the focal length measured in meters; with a positive focal length corresponding to a converging lens and a negative focal length corresponding to a diverging lens. For lenses that correct for astigmatism as well, three numbers are given: one for the spherical power, one for the cylindrical power, and one for the angle of orientation of the astigmatism #### Visual effects Main article: Optical illusions Optical illusions (also called visual illusions) are characterized by visually perceived images that differ from objective reality. The information gathered by the eye is processed in the brain to give a percept that differs from the object being imaged. Optical illusions can be the result of a variety of phenomena including physical effects that create images that are different from the objects that make them, the physiological effects on the eyes and brain of excessive stimulation (e.g. brightness, tilt, color, movement), and cognitive illusions where the eye and brain make unconscious inferences. Cognitive illusions include some which result from the unconscious misapplication of certain optical principles. For example, the Ames room, Hering, Müller-Lyer, Orbison, Ponzo, Sander, and Wundt illusions all rely on the suggestion of the appearance of distance by using converging and diverging lines, in the same way that parallel light rays (or indeed any set of parallel lines) appear to converge at a vanishing point at infinity in two-dimensionally rendered images with artistic perspective. This suggestion is also responsible for the famous moon illusion where the moon, despite having essentially the same angular size, appears much larger near the horizon than it does at zenith. This illusion so confounded Ptolemy that he incorrectly attributed it to atmospheric refraction when he described it in his treatise, Optics Another type of optical illusion exploits broken patterns to trick the mind into perceiving symmetries or asymmetries that are not present. Examples include the café wall, Ehrenstein, Fraser spiral, Poggendorff, and Zöllner illusions. Related, but not strictly illusions, are patterns that occur due to the superimposition of periodic structures. For example transparent tissues with a grid structure produce shapes known as moiré patterns, while the superimposition of periodic transparent patterns comprising parallel opaque lines or curves produces line moiré patterns #### Optical instruments Main article: Optical instruments Single lenses have a variety of applications including photographic lenses, corrective lenses, and magnifying glasses while single mirrors are used in parabolic reflectors and rear-view mirrors. Combining a number of mirrors, prisms, and lenses produces compound optical instruments which have practical uses. For example, a periscope is simply two plane mirrors aligned to allow for viewing around obstructions. The most famous compound optical instruments in science are the microscope and the telescope which were both invented by the Dutch in the late 16th century. Microscopes were first developed with just two lenses: an objective lens and an eyepiece. The objective lens is essentially a magnifying glass and was designed with a very small focal length while the eyepiece generally has a longer focal length. This has the effect of producing magnified images of close objects. Generally, an additional source of illumination is used since magnified images are dimmer due to the conservation of energy and the spreading of light rays over a larger surface area. Modern microscopes, known as compound microscopes have many lenses in them (typically four) to optimize the functionality and enhance image stability. A slightly different variety of microscope, the comparison microscope, looks at side-by-side images to produce a stereoscopic binocular view that appears three dimensional when used by humans. The first telescopes, called refracting telescopes were also developed with a single objective and eyepiece lens. In contrast to the microscope, the objective lens of the telescope was designed with a large focal length to avoid optical aberrations. The objective focuses an image of a distant object at its focal point which is adjusted to be at the focal point of an eyepiece of a much smaller focal length. The main goal of a telescope is not necessarily magnification, but rather collection of light which is determined by the physical size of the objective lens. Thus, telescopes are normally indicated by the diameters of their objectives rather than by the magnification which can be changed by switching eyepieces. Because the magnification of a telescope is equal to the focal length of the objective divided by the focal length of the eyepiece, smaller focal-length eyepieces cause greater magnification. Since crafting large lenses is much more difficult than crafting large mirrors, most modern telescopes are reflecting telescopes, that is, telescopes that use a primary mirror rather than an objective lens. The same general optical considerations apply to reflecting telescopes that applied to refracting telescopes, namely, the larger the primary mirror, the more light collected, and the magnification is still equal to the focal length of the primary mirror divided by the focal length of the eyepiece. Professional telescopes generally do not have eyepieces and instead place an instrument (often a charge-coupled device) at the focal point instead. ### Atmospheric optics The unique optical properties of the atmosphere cause a wide range of spectacular optical phenomena. The blue color of the sky is a direct result of Rayleigh scattering which redirects higher frequency (blue) sunlight back into the field of view of the observer. Because blue light is scattered more easily than red light, the sun takes on a reddish hue when it is observed through a thick atmosphere, as during a sunrise or sunset. Additional particulate matter in the sky can scatter different colors at different angles creating colorful glowing skies at dusk and dawn. Scattering off of ice crystals and other particles in the atmosphere are responsible for halo (optical phenomenon), afterglows, Corona (meteorology), rays of sunlight, and sun dogs. The variation in these kinds of phenomena is due to different particle sizes and geometries.[4] Mirages are another sort of optical phenomena due to variations in the refraction of light through the atmosphere. Other dramatic optical phenomena associated with this include the Novaya Zemlya effect where the sun appears to rise earlier than predicted with a distorted shape. A spectacular form of refraction occurs with a temperature inversion called the Fata Morgana (mirage) where objects on the horizon or even beyond the horizon, such as islands, cliffs, ships or icebergs, appear elongated and elevated, like "fairy tale castles".[5] Rainbows are the result of a combination of optical effects: total internal reflection and dispersion of light in raindrops. A single reflection off the backs of an array of raindrops produces a coherent rainbow with an angular size on the sky that ranges from 40 to 42 degrees with red on the outside. Double rainbows are produced by two internal reflections with angular size of 50.5 to 54 degrees with violet on the outside. Because rainbows must be seen with the sun 180 degrees away from the center of the rainbow, rainbows are more prominent the closer the sun is to the horizon. ## References 1. ↑ 2. ↑ Rajiv Ramaswami and Kumar N. Sivarajan, Optical Networks: A Practical Perspective (Academic Press: London 1998). 3. ↑ Brillouin, Léon. Wave Propagation and Group Velocity. Academic Press Inc., New York (1960) 4. ↑ Ahrens,C. Donald, Meteorology Today: an introduction to weather, climate, and the environment, West Publishing Company,1994 5. ↑ • M. Born and E.Wolf. Principles of Optics (7th ed.). Pergamon Press., 1999 • E. Hecht. Optics (4th ed.). Pearson Education. 2001. isbn=0805385665 • R. A. Serway and J. W. Jewett. Physics for Scientists and Engineers (6th ed.). Brooks/Cole. 2004. isbn=0534408427 • P. Tipler. Physics for Scientists and Engineers: Electricity, Magnetism, Light, and Elementary Modern Physics (5th ed.). W. H. Freeman. 2004. isbn=0716708108 • S. G. Lipson. Optical Physics (3rd ed.). Cambridge University Press. 1995. isbn=0521436311 ## Read more • Formulas • The Tyndall effect, also known as Tyndall scattering, is the scattering of light by colloidal... Tyndall effect • Rayleigh scattering (named after the English physicist Lord Rayleigh) is the elastic scattering... Rayleigh scattering # Photos Add a Photo 1,165photos on this wiki • by BiomedGuy 2013-02-26T00:32:14Z Posted in Radiation Safety • by BiomedGuy 2013-02-10T00:57:16Z • by BiomedGuy 2013-02-10T00:55:59Z • by BiomedGuy 2013-02-10T00:54:57Z • by BiomedGuy 2013-02-10T00:53:55Z • by BiomedGuy 2013-02-10T00:53:14Z • by BiomedGuy 2013-02-10T00:51:45Z • by BiomedGuy 2013-02-02T00:52:21Z • by BiomedGuy 2013-01-27T19:09:40Z Posted in Main Page • by BiomedGuy 2013-01-27T19:08:35Z Posted in Main Page • by BiomedGuy 2013-01-27T19:07:48Z Posted in Main Page • See all photos See all photos >	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 13, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.933448076248169, "perplexity_flag": "middle"}
http://mathhelpforum.com/discrete-math/120872-not-understanding.html	Thread: 1. not understanding have to do a couple problems for homework and im not understanding exactly what is going on and why its the answer 27. Let m, n, and d be integers. Show that if d \ m and d \ n, then d\ (m - n). my professor gave us the answer to this but i do not understand why it works answer: m = dq1 n = dq2 m - n = dq1 - dq2 = d(q1 - q2) therefore d \ (m - n) these are the other problems i have to do 28. Let m, n, and d be integers. Show that if d \ m, then d \ mn. 31. Let a, b, and c be integers. Show that if a \ b and b \ c, then a \ c. 33. Give an example of consecutive primes p1 = 2, p2 ..., pn where p1p2... pn + 1 is not prime. any help would be much appreciated 2. Originally Posted by pooshipple have to do a couple problems for homework and im not understanding exactly what is going on and why its the answer 27. Let m, n, and d be integers. Show that if d \ m and d \ n, then d\ (m - n). If $d\mid m,d\mid n$ then $m=dz,n=dz'$ for some $z,z'\in\mathbb{Z}$. Clearly then $m-n=dz-dz'=d\left(z-z'\right)$ and since $z,z'\in\mathbb{Z}\implies z-z'\in\mathbb{Z}$ the conclusion follows. my professor gave us the answer to this but i do not understand why it works answer: m = dq1 n = dq2 m - n = dq1 - dq2 = d(q1 - q2) therefore d \ (m - n) Sorry, I just noticed this. What it means to say that $d\mid n$ is that $d$ divides $n$ or that $\frac{n}{d}\in\mathbb{Z}$. Does that make a little more sense? So if $m-n=d\left(z-z'\right)$ then $\frac{m-n}{d}=\frac{d\left(z-z'\right)}{d}=z-z'\in\mathbb{Z}$ these are the other problems i have to do 28. Let m, n, and d be integers. Show that if d \ m, then d \ mn. $d\mid m\implies m=dz$ so then $mn=dzn$ or $\frac{mn}{d}=zn$ and since the integers are closed under multiplication the conclusion follows. 31. Let a, b, and c be integers. Show that if a \ b and b \ c, then a \ c. These all can be done the same. $a\mid b\implies b=za$ and $b\mid c\implies c=bz'$. So then $c=bz'=\left(az\right)z'$ and the conclusion follows. 33. Give an example of consecutive primes p1 = 2, p2 ..., pn where p1p2... pn + 1 is not prime. What does "consecutive" mean? That $p_2=p_1+1$ or that is merely the next prime in the list of primes? I'm assuming the latter. What do you think?	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 18, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9095651507377625, "perplexity_flag": "middle"}
http://mathhelpforum.com/differential-equations/173762-tricky-equation-first-order-print.html	# A tricky equation of the first order Printable View • March 7th 2011, 10:29 AM getsallad A tricky equation of the first order My teacher gave me a "bonus" assignment the other day, and I've been racking my brain trying to solve it. Here it is: $xy'(ln(\frac{x}{y}) +1) = y(ln(\frac{x}{y}) - 1)$ So far I've figured that I should substitute $ln(\frac{x}{y})$ for $u(x)$, thus giving me $y = \frac{x}{e^{u(x)}}$, but from there I'm kind of lost. Any ideas? • March 7th 2011, 10:35 AM topsquark Quote: Originally Posted by getsallad My teacher gave me a "bonus" assignment the other day, and I've been racking my brain trying to solve it. Here it is: $xy'(ln(\frac{x}{y}) +1) = y(ln(\frac{x}{y}) - 1)$ So far I've figured that I should substitute $ln(\frac{x}{y})$ for $u(x)$, thus giving me $y = \frac{x}{e^{u(x)}}$, but from there I'm kind of lost. Any ideas? I haven't done anything with it, but note that if you divide both sides by x you have an equation in y/x... -Dan • March 7th 2011, 10:41 AM getsallad I'm sorry, my knowledge of differential equations in itself is quite shaky, but how would that help me? If I get what you're saying, is it that I get $y'(ln(\frac{x}{y}) + 1) = \frac{x}{y}(ln(\frac{x}{y}) - 1)$? If that's the case, are you suggesting I could substitute u(x) for $\frac{x}{y}$ instead? • March 7th 2011, 10:53 AM Ackbeet The substitution $u = x/y$ does render the equation separable (the original DE is homogeneous). The problem is, the resulting integral in $u$ is horrendous. I suppose you could say, at that point, that you've "reduced the DE to quadratures", but if your professor is looking for a closed-form solution, that won't do. The substitution $u=\ln(x/y)$ is much better. You have to translate the DE over to the $u$ domain: $u'=\dfrac{y}{x}\cdot\dfrac{y-xy'}{y^{2}}=\dfrac{y-xy'}{xy}.$ Solve this for $y',$ and plug everything into the DE. A few things should cancel, leaving you with a much nicer separable equation. • March 7th 2011, 11:00 AM Ackbeet Very cute problem, by the way! Thanks for posting! • March 7th 2011, 11:05 AM getsallad Thanks for the help. I'm trying it out right now, doing all the steps myself to see if I get it. • March 7th 2011, 11:53 AM topsquark I just did the integral using Akbeet's method as well as using the homogeneous method and I'd rate them to be about the same level of difficulty. For the "connoisseurs" out there, the homogeneous solution includes a nice little integral: $\displaystyle \int \frac{ln(u)~du}{u}$ which I haven't seen done in a long time. (it's not particularly hard to do, just a nice little piece of work.) -Dan • March 7th 2011, 12:04 PM getsallad OK, so I've gotten somewhere. Now I'm stuck on the actual separable differential equation. I have $\frac{2}{u + 1} = xu'$ which leads me to $\frac{2}{u + 1} du = x dx$ but that approach seems to give me a completely different solution from what Wolfram Alpha gives me when I feed it the equation. So I figure I must be doing something wrong, since Wolfram Alpha's solution also seems more in tune with the final solution of the problem. • March 7th 2011, 12:06 PM Ackbeet • March 7th 2011, 01:23 PM getsallad OK, so I've hit one last roadblock and it's mighty frustrating. In the end I get $\frac{u^2}{2} + u = 2ln(x) + c$ and Wolfram Alpha gives the solution $\frac{1}{2}*ln^2(\frac{x}{y}) - ln(\frac{x}{y}) = 2ln(x) + c$ Which is basically the same thing with - instead of +. Should I show you step by step what I am doing? This is driving me nuts. • March 7th 2011, 01:30 PM Ackbeet No, WolframAlpha is not giving you that solution, it's giving you this solution. Notice that the x and y in the WolframAlpha solution are flipped from what you have. Because the left-most logarithm is squared, the change in sign goes unnoticed. That, of course, doesn't happen with the second term. I'm saying your solution is correct. • March 7th 2011, 01:32 PM Ackbeet Incidentally, you could, if you wanted to, use the quadratic formula on log(x/y) and then solve for y. You get a multi-valued "function" then, but it is an explicit formula for y, which is nice. • March 7th 2011, 01:39 PM getsallad Ah, can't believe I missed that. Thanks a million for the help on this one! I'll be sure to include the explicit formula for y just to be on the safe side. You've been most helpful! =) • March 7th 2011, 03:43 PM Ackbeet You're very welcome! Have a good one! All times are GMT -8. The time now is 03:54 PM.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 21, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9576013684272766, "perplexity_flag": "middle"}
http://psychology.wikia.com/wiki/Least_square	Least squares Talk0 31,725pages on this wiki Redirected from Least square Assessment \| Biopsychology \| Comparative \| Cognitive \| Developmental \| Language \| Individual differences \| Personality \| Philosophy \| Social \| Methods \| Statistics \| Clinical \| Educational \| Industrial \| Professional items \| World psychology \| Statistics: Scientific method · Research methods · Experimental design · Undergraduate statistics courses · Statistical tests · Game theory · Decision theory In regression analysis, least squares, also known as ordinary least squares analysis, is a method for linear regression that determines the values of unknown quantities in a statistical model by minimizing the sum of the squared residuals (the difference between the predicted and observed values). This method was first described by Carl Friedrich Gauss around 1794.[1] Today, this method is available in most statistical software packages. The least-squares approach to regression analysis has been shown to be optimal in the sense that it satisfies the Gauss-Markov theorem. A related method is the least mean squares (LMS) method. It occurs when the number of measured data is 1 and the gradient descent method is used to minimize the squared residual. LMS is known to minimize the expectation of the squared residual, with the smallest number of operations per iteration). However, it requires a large number of iterations to converge. Many other types of optimization problems can be expressed in a least squares form, by either minimizing energy or maximizing entropy. History Context The method of least squares grew out of the fields of astronomy and geodesy as scientists and mathematicians sought to provide solutions to the challenges of navigating the Earth's oceans during the Age of Exploration. The accurate description of the behavior of celestial bodies was key to enabling ships to sail in open seas where before sailors had to rely on land sightings to determine the positions of their ships. The method was the culmination of several advances that took place during the course of the eighteenth century[2]: • The combination of different observations taken under the same conditions as opposed to simply trying one's best to observe and record a single observation accurately. This approach was notably used by Tobias Mayer while studying the librations of the moon. • The combination of different observations as being the best estimate of the true value; errors decrease with aggregation rather than increase, perhaps first expressed by Roger Cotes. • The combination of different observations taken under different conditions as notably performed by Roger Joseph Boscovich in his work on the shape of the earth and Pierre-Simon Laplace in his work in explaining the differences in motion of Jupiter and Saturn. • The development of a criterion that can be evaluated to determine when the solution with the minimum error has been achieved, developed by Laplace in his Method of Situation. The method itself Carl Friedrich Gauss is credited with developing the fundamentals of the basis for least-squares analysis in 1795 at the age of eighteen. An early demonstration of the strength of Gauss's method came when it was used to predict the future location of the newly discovered asteroid Ceres. On January 1st, 1801, the Italian astronomer Giuseppe Piazzi discovered Ceres and was able to track its path for 40 days before it was lost in the glare of the sun. Based on this data, it was desired to determine the location of Ceres after it emerged from behind the sun without solving the complicated Kepler's nonlinear equations of planetary motion. The only predictions that successfully allowed Hungarian astronomer Franz Xaver von Zach to relocate Ceres were those performed by the 24-year-old Gauss using least-squares analysis. Gauss did not publish the method until 1809, when it appeared in volume two of his work on celestial mechanics, Theoria Motus Corporum Coelestium in sectionibus conicis solem ambientium. In 1829, Gauss was able to state that the least-squares approach to regression analysis is optimal in the sense that in a linear model where the errors have a mean of zero, are uncorrelated, and have equal variances, the best linear unbiased estimators of the coefficients is the least-squares estimators. This result is known as the Gauss-Markov theorem. The idea of least-squares analysis was also independently formulated by the Frenchman Adrien-Marie Legendre in 1805 and the American Robert Adrain in 1808. Problem statement The objective consists of adjusting a model function to best fit a data set. The chosen model function has adjustable parameters. The data set consist of n points $(y_i,\vec{x}_i)$ with $i = 1, 2,\dots, n$. The model function has the form $y=f(\vec{x},\vec{a})$, where $y$ is the dependent variable, $\vec{x}$ are the independent variables, and $\vec{a}$ are the model adjustable parameters. We wish to find the parameter values such that the model best fits the data according to a defined error criterion. The least sum square method minimizes the sum square error equation $S = \sum_{i=1}^n (y_i - f(\vec{x}_i,\vec{a}))^2$ with respect to the adjustable parameters $\vec{a}$. For an example, the data is height measurements over a surface. We choose to model the data by a plane with parameters for plane mean height, plane tip angle, and plane tilt angle. The model equation is then $y = f ( x_1, x_2 ) = a_1 + a_2 x_1 + a_3 x_2$, the independent variables are $\vec{x}=(x_1,x_2)$, and the adjustable parameters are $\vec{a}=(a_1,a_2,a_3)$. Solving the least squares problem Least square optimization problems can be divided into linear and non-linear problems. The linear problem has a closed form solution. The optimization problem is said to be a linear optimization problem if the first order partial derivatives of S with respect to the parameters $\vec{a}$ results in a set of equations that is linear in the parameter variables. The general, non-linear, unconstrained optimization problem has no closed form solution. In this case recursive methods, such as Newton's method, combined with the gradient descent method, or specialized methods for least squares analysis, such as the Gauss-Newton algorithm or the Levenberg-Marquardt algorithm can be used. Least squares and regression analysis In regression analysis, one replaces the relation $f(x_i)\approx y_i$ by $f(x_i) = y_i + \varepsilon_i,$ where the noise term ε is a random variable with mean zero. Note that we are assuming that the $x$ values are exact, and all the errors are in the $y$ values. Again, we distinguish between linear regression, in which case the function f is linear in the parameters to be determined (e.g., f(x) = ax2 + bx + c), and nonlinear regression. As before, linear regression is much simpler than nonlinear regression. (It is tempting to think that the reason for the name linear regression is that the graph of the function f(x) = ax + b is a line. But fitting a curve like f(x) = ax2 + bx + c when estimating a, b, and c by least squares, is an instance of linear regression because the vector of least-square estimates of a, b, and c is a linear transformation of the vector whose components are f(xi) + εi. Parameter estimates By recognizing that the $y_i = \alpha + \beta x_i + \varepsilon_i$ regression model is a system of linear equations we can express the model using data matrix X, target vector Y and parameter vector $\delta$. The ith row of X and Y will contain the x and y value for the ith data sample. Then the model can be written as $\begin{bmatrix} y_1\\ y_2\\ \vdots\\ y_n \end{bmatrix}= \begin{bmatrix} 1 & x_1\\ 1 & x_2\\ \vdots & \vdots\\ 1 & x_n \end{bmatrix} \begin{bmatrix} \alpha \\ \beta \end{bmatrix} + \begin{bmatrix} \varepsilon_1\\ \varepsilon_2\\ \vdots\\ \varepsilon_n \end{bmatrix}$ which when using pure matrix notation becomes $Y = X \delta + \varepsilon \,$ where ε is normally distributed with expected value 0 (i.e., a column vector of 0s) and variance σ2 In, where In is the n×n identity matrix. The least-squares estimator for $\delta$ is $\widehat{\delta} = (X^T X)^{-1}\; X^T Y \,$ (where XT is the transpose of X) and the sum of squares of residuals is $Y^T (I_n - X (X^T X)^{-1} X^T)\, Y.$ One of the properties of least-squares is that the matrix $X\widehat{\delta}$ is the orthogonal projection of Y onto the column space of X. The fact that the matrix X(XTX)−1XT is a symmetric idempotent matrix is incessantly relied on in proofs of theorems. The linearity of $\widehat{\delta}$ as a function of the vector Y, expressed above by saying $\widehat{\delta} = (X^TX)^{-1}X^TY,\,$ is the reason why this is called "linear" regression. Nonlinear regression uses nonlinear methods of estimation. The matrix In − X (XT X)−1 XT that appears above is a symmetric idempotent matrix of rank n − 2. Here is an example of the use of that fact in the theory of linear regression. The finite-dimensional spectral theorem of linear algebra says that any real symmetric matrix M can be diagonalized by an orthogonal matrix G, i.e., the matrix G′MG is a diagonal matrix. If the matrix M is also idempotent, then the diagonal entries in G′MG must be idempotent numbers. Only two real numbers are idempotent: 0 and 1. So In − X(XTX) -1XT, after diagonalization, has n − 2 1s and two 0s on the diagonal. That is most of the work in showing that the sum of squares of residuals has a chi-square distribution with n−2 degrees of freedom. Regression parameters can also be estimated by Bayesian methods. This has the advantages that • confidence intervals can be produced for parameter estimates without the use of asymptotic approximations, • prior information can be incorporated into the analysis. Suppose that in the linear regression $y = \alpha + \beta x + \varepsilon \,$ we know from domain knowledge that alpha can only take one of the values {−1, +1} but we do not know which. We can build this information into the analysis by choosing a prior for alpha which is a discrete distribution with a probability of 0.5 on −1 and 0.5 on +1. The posterior for alpha will also be a discrete distribution on {−1, +1}, but the probability weights will change to reflect the evidence from the data. In modern computer applications, the actual value of $\beta$ is calculated using the QR decomposition or slightly more robust methods when $X^TX$ is near singular. The code for the MATLAB backslash function, "`\`", is an excellent example of a robust method. Summarizing the data We sum the observations, the squares of the Xs and the products XY to obtain the following quantities. $S_X = x_1 + x_2 + \cdots + x_n \,$ $S_Y = y_1 + y_2 + \cdots + y_n \,$ $S_{XX} = x_1^2 + x_2^2 + \cdots + x_n^2 \,$ $S_{XY} = x_1 y_1 + x_2 y_2 + \cdots + x_n y_n. \,$ Estimating beta (the slope) We use the summary statistics above to calculate $\widehat\beta$, the estimate of β. $\widehat\beta = {n S_{XY} - S_X S_Y \over n S_{XX} - S_X S_X}. \,$ Estimating alpha (the intercept) We use the estimate of β and the other statistics to estimate α by: $\widehat\alpha = {S_Y - \widehat\beta S_X \over n}. \,$ A consequence of this estimate is that the regression line will always pass through the "center" $(\bar{x},\bar{y}) = (S_X/n, S_Y/n)$. Limitations Least squares estimation for linear models is notoriously non-robust to outliers. If the distribution of the outliers is skewed, the estimates can be biased. In the presence of any outliers, the least squares estimates are inefficient and can be extremely slow. When outliers occur in the data, methods of robust regression are more appropriate. References 1. ↑ Bretscher, Otto (1995). Linear Algebra With Applications, 3rd ed., Upper Saddle River NJ: Prentice Hall. 2. ↑ Stigler, Stephen M. (1986). The History of Statistics: The Measurement of Uncertainty Before 1900, Cambridge, MA: Belknap Press of Harvard University Press. Photos Add a Photo 6,465photos on this wiki • by Dr9855 2013-05-14T02:10:22Z • by PARANOiA 12 2013-05-11T19:25:04Z Posted in more... • by Addyrocker 2013-04-04T18:59:14Z • by Psymba 2013-03-24T20:27:47Z Posted in Mike Abrams • by Omaspiter 2013-03-14T09:55:55Z • by Omaspiter 2013-03-14T09:28:22Z • by Bigkellyna 2013-03-14T04:00:48Z Posted in User talk:Bigkellyna • by Preggo 2013-02-15T05:10:37Z • by Preggo 2013-02-15T05:10:17Z • by Preggo 2013-02-15T05:09:48Z • by Preggo 2013-02-15T05:09:35Z • See all photos See all photos >	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 37, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9054163694381714, "perplexity_flag": "head"}
http://mathoverflow.net/questions/43319/do-convex-and-decreasing-functions-preserve-the-semimartingale-property/43389	## Do convex and decreasing functions preserve the semimartingale property? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Some time ago I spent a lot of effort trying to show that the semimartingale property is preserved by certain functions. Specifically, that a convex function of a semimartingale and decreasing function of time is itself a semimartingale. This was needed for a result which I was trying to prove (more details below) and eventually managed to work around this issue, but it was not easy. For twice continuously differentiable functions this is an immediate consequence of Ito's lemma, but this cannot be applied in the general case. After failing at this task, I also spent a considerable amount of time trying to construct a counterexample, also with no success. So, my question is as follows. 1) Let ƒ: ℝ+×ℝ → ℝ be such that ƒ(t,x) is convex in x and continuous and decreasing in t. Then, for any semimartingale X, is ƒ(t,Xt) necessarily a semimartingale? Actually, it can be assumed here that X is both continuous and a martingale which, with some work, would imply the general case. As it turns out, this can be phrased purely as a real-analysis question. 2) Let ƒ: ℝ+×ℝ → ℝ be such that ƒ(t,x) is convex in x and decreasing in t. Can we write ƒ = g − h where g(t,x) and h(t,x) are both convex in x and increasing in t ? Stated like this, maybe someone with a good knowledge of convex functions would be able to answer the question one way or the other. For ƒ(t,x) convex in x and increasing in t then approximation by smooth functions and applying Ito's lemma allows us to express ƒ(t,Xt) as the sum of a stochastic integral and an increasing process $$f(t,X_t)=\int_0^t\frac{\partial f}{\partial x}(s,X_s)\,dX_s+V_t,\qquad()$$ so it is a semimartingale. If, instead, ƒ is decreasing in t, then an affirmative answer to question 2 will reduce it to the easier case where it is increasing in t, also giving a positive answer to the first question. Explaining why question 1 implies 2 is a bit trickier. If 2 was false, then it would be possible to construct martingales X such that the decomposition () holds where the variation of V explodes at some positive time. This problem arose while I was trying to prove the following: is a continuous martingale uniquely determined by its one dimensional marginals? For arbitrary continuous martingales this is false, but is known to be true for diffusions dXt = σ(t,Xt) dW for Brownian motion W and smooth parameter σ. The idea is to back out σ from the Kolmogorov forward equation. This is well-known in finance as the local volatility model. However, I was trying to show rather more than this. All continuous and strong Markov martingales are uniquely determined by the one dimensional marginals. I was able to prove this, and the relation between the marginals and joint distributions of the martingale has many nice properties (I wrote a paper on this, submitted to the arXiv, but not published as I am still working on changes asked for by the referees). The method was to reformulate the Kolmogorov backward equation in terms of the marginals. This does use Ito's lemma, requiring twice differentiability, but can be circumvented with a bit of integration by parts as long as ƒ(t,Xt) is a semimartingale for the kinds of functions mentioned above. The question above arose from trying, and failing, to prove this. Without an answer to this question, the problem becomes much much harder, as many of the techniques from stochastic calculus can no longer be applied (and, approximating by semimartingales didn't seem to help either). The work around was very involved, part of which I published as a standalone paper here and the rest forms most of a paper I submitted to the arXiv here. Adding together those papers, it comes to maybe 50 pages of maths and a lot of effort to work around the question above. - ## 3 Answers I still have no idea what the answer to this question is. However, it is possible to attack the problem in several different ways, and there are various different (but logically equivalent) ways of stating it. I'll post some of these as an answer now, as it seems rather long to fit in the original statement of the question. Maybe one of the reformulations below will help lead to a resolution of the problem. This answer is already very long, and I've been trying to shorten it as much as I can. I can't see any way of giving proper proofs of all the statements below without making it a lot longer. So, I'll only give very few details of he proofs here. I will, however, list each of the equivalent formulations H1-H6 below in the logical order, so that each statement can be proven to be equivalent to the preceding one without too much work. Let's start from statement 2 of the original question, which I will refer to as Hypothesis (H1) for the purposes of this answer. Hypothesis (H1): Let ƒ: ℝ+×ℝ → ℝ be such that ƒ(t,x) is convex in x and decreasing in t. Then, ƒ = g − h where g(t,x) and h(t,x) are both convex in x and increasing in t . The decomposition in (H1) exists if and only if it exists locally. Letting I = [0,1] denote the unit interval, we get the following equivalent statement. Hypothesis (H2): Let ƒ: I2 → ℝ be such that ƒ(t,x) is convex and Lipschitz continuous in x and decreasing in t. Then, ƒ = g − h where g(t,x) and h(t,x) are convex in x and increasing in t. The truth of this statement remains unchanged if it is restricted to functions ƒ which are zero on the three edges I × {0}, I × {1}, {0} × I of the unit square. I'll use D to denote the set of such functions satisfying the conditions of (H2). Then, whenever the decomposition in (H2) exists, it is always possible to choose g, h to be zero on I × {0}, I × {1} and {1} × I. From now on, whenever the decomposition ƒ = g − h in (H2) is referred to, it will be assumed that g, h are chosen to satisfy these conditions. We can strengthen (H2) by also placing a uniform bound on the terms g, h in the decomposition. Here, ‖g‖ denotes the supremum norm and ƒx denotes the partial derivative. Hypothesis (H3): There is a constant K > 0 such that, for all ƒ ∈ D, the decomposition ƒ = g − h as in H2 exists and can be chosen such that ‖g‖,‖h‖ ≤ K‖ƒx‖. Statement (H3) is particularly convenient because of the following: to prove that (H3) holds, it is enough to look at a dense subset of functions in D. Taking limits of the decompositions would then extend to the result to all of D. So, it is enough to concentrate on, say, smooth functions or piecewise linear functions. Next, it is useful to choose the decomposition in (H2) to minimize ‖g‖ and ‖h‖. Lemma 1: Suppose that ƒ ∈ D and that the decomposition ƒ = g − h as in (H2) exists. Then, there is a unique maximal choice for g, h. That is, if ƒ = g1 − h1 is any other such decomposition then g ≥ g1 and h ≥ h1. I'll refer to the decomposition in Lemma 1 as the optimal decomposition. As it not clear that any such decomposition should exist, I'll briefly sketch the argument now. The idea is to discretize time, using a partition of the unit interval 0 = t0 < t1 < … < tr = 1. Denote the convex hull of a function u: I → ℝ by v = H(u), which is the maximum convex function v: I → ℝ bounding u from below, $$\begin{align} v(x) &= \sup\{w(x)\colon w\le u{\rm\ is\ convex}\}\\ &=\inf\{\left((b-x)u(a)+(x-a)u(b)\right)/(b-a)\colon a\le x\le b\}. \end{align}$$ The optimal decomposition in discrete time can be constructed as functions hk: I → ℝ, starting at the final time k = r and working backwards to k = 0, $$h_r(x) = 0,\ h_{k-1}={\rm H}\left(h_k+f(t_k,\cdot)-f(t_{k-1},\cdot)\right).$$ Interpolate this to be piecewise constant in time, defining h(t,x) = hk(x) for tk−1 < t ≤ tk. Then, h(t,x) and g ≡ ƒ + h are convex in x and increasing in time t, restricting to times in the partition. Lemma 2: Suppose that ƒ ∈ D and let 0 = tn,0 < tn,1 <…< tn,rn = 1 be a sequence of partitions of the unit interval. For each n let hn(t,x) be the function corresponding to the partition and piecewise constant in t, as constructed above. We also suppose that the partitions have mesh going to zero and eventually include all times at which ƒ is discontinuous. Then one of the following holds. • ƒ decomposes as in (H2) and hn(t,x) → h(t,x) pointwise on I2, where ƒ = g − h is the optimal decomposition. • ƒ does not have a decomposition as in (H2) and hn(0,x) → -∞ for all 0 < x < 1. The idea is that, if hn has any limit point h, then h(t,x) and g = ƒ + h will be convex in x and increasing in t, giving the decomposition required by (H2). Furthermore, by construction, if ƒ = g' − h' is any other such decomposition, then hn ≥ h' at times in the partition, so h ≥ h'. This shows that the decomposition is optimal and, as the optimal decomposition is unique, all limit limit points of hn are the same, so hn → h. The only alternative is that hn has no limit points, in which case the second statement of the Lemma holds. Using this construction of the optimal decomposition, (H3) can be shown to be equivalent to the following. Hypothesis (H4): There is a constant K > 0 such that, for all smooth functions ƒ,g: I2 → ℝ with ‖ƒ‖, ‖g‖, ‖ƒx‖, ‖gx‖ bounded by 1 and ƒ(t,x), g(t,x) convex in x and respectively decreasing and increasing in t, then, $$\int_0^1\int_0^1 f_{xx}g_t\,dxdt \le K.$$ As this statement is quite different from the preceding ones, I'd better give some explanation now. The idea is to use integration by parts, $$\begin{align} \int_0^1\int_0^1(f_{xx}g_t+g_{xx}f_t)\,dxdt &= \left[\int_0^1(f_xg_t+g_xf_t)\,dt\right]_{x=0}^1-\left[\int_0^1f_xg_x\,dx\right]_{t=0}^1\\ &\le 6(\Vert f_x\Vert \Vert g\Vert + \Vert g_x\Vert \Vert f\Vert). \end{align}$$ If ƒ and g are increasing in time then the terms on the left hand side are both positive, so we get bounds for the integrals of ƒxxgt and gxxƒt individually. Hypothesis (H3) extends this to the case where ƒ is decreasing in time, implying (H4). Conversely, suppose that (H4) holds. Letting f = g − h be the decomposition computed along a partition as described above, we can use the fact that the convex hull v = H(u) of a function u satisfies vxx(u −v) = 0 to get the equality (hk−1)xx(hk − hk−1 + ƒ(tk,.) − ƒ(tk−1,.)) = 0. This leads to the following inequalities, $$\begin{align} \frac12\Vert h\Vert^2&\le\frac12\int_0^1 h_x(0,x)^2\,dx\le\sum_{k=1}^r \int_0^1(h_{k-1})_{xx}(h_{k-1}-h_k)\,dx\\ &=-\sum_{k=1}^r\int_0^1 (h_{k-1})_{xx}(f_k-f_{k-1})\,dx. \end{align}$$ Hypothesis (H4) can be used to bound the final term, showing that h cannot diverge as the mesh of the partition goes to zero, so we get convergence to an optimal decomposition satisfying a bound as in (H3). The hypothesis can now be formulated as a statement about martingales. Hypothesis (H5): There is a constant K > 0 such that, for all ƒ ∈ D and martingales 0 ≤ Xt ≤ 1, ƒ(t,Xt) decomposes as M + V where M is a martingale and V has variation $$\mathbb{E}\left[\int_0^1\,\vert dV\vert\right]\le K\Vert f_x\Vert.$$ The idea is that g(t,x) ≡ E[(Xt − x)+] is convex in x and increasing in t. In the case where ƒ, g are smooth and X is a continuous martingale, Ito's formula can be used to split ƒ(t,Xt) into a martingale term plus the sum of the increasing process ½∫ƒxx(t,X)d[X] and the decreasing process ∫ƒt(t,X)dt, which have expectations ∫∫ƒxxgtdxdt and ∫∫ƒtgxxdxdt respectively. Finally, by adding a randomly occurring jump term to any semimartingale, and with a bit more work, it is possible to reduce this to the martingale case. This gives the statement asked in the original question. Hypothesis (H6): Let ƒ:ℝ+×ℝ → ℝ be such that ƒ(t,x) is convex in x and continuous and decreasing in t. Then, for any semimartingale X, ƒ(t,Xt) is a semimartingale. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. What follows is too long to fit in a comment. Some thoughts on the second problem. Let us consider the problem for $t$ taking values in the compact interval $[0,T]$. The general case perhaps can be approximated by this case. If the $t$ variable were discrete and finite the problem would be: we have a sequence of functions $f_1 \ge f_2 \ge f_3 \cdots \ge f_n$ (let us make the simplifying assumption that $f_i$ are all positive; otherwise, replace $f_i$ with $f_i +C$ where $C$ is $C = -\min_{i,x} f_i(x)$), can one find convex and increasing ${h_i}$ and ${g_i}$ such that $f_i = h_i - g_i$? The answer to this question is obviously yes. For example, $h_i = \sum_{j\le i} f_i$ and $g_i = \sum_{j < i} f_i$ is one possible solution. The problem with this solution when $t$ is continuous is that the $h_i$ and $g_i$ would explode as the discretization of $t$ is refined. Thus, one needs to choose $h$ and $g$ in a way that they increase slowly. How slowly can this increase be? We can start with $h_1 = f_1$ and $g_1 = 0$. $h_2$ and $g_2$ will be of the following form: $$h_2 = h_1 + S_2 = f_1 + S_2,$$ and $$g_2 = g_1 + R_2 = R_2$$ such that: 1) $S_2, R_2 \ge 0$, 2) $h_2 = f_1 + S_2$, $g_2 = R_2$ are convex and 3) $h_2 - g_2 = f_1 + S_2 - R_2 = f_2$. The last of these is equivalent to $R_2 = f_1 - f_2 + S_2$. Thus, what we are looking for is a function $S_2$ satisfying the above conditions. There will be many such $S_2$, the goal is to choose $S_2$ in a minimal way so that $h_i$ and $g_i$ grow slowly.The best would be: $S_2 = 0$, which is indeed a solution when $f_1 - f_i$ is convex. If $f_0 - f_t$ is convex for all $t \in [0,T]$ then this solution directly generalizes to the original continuous time problem (i.e., if $f_0-f_t$ is convex then $h(t,x) = f(0,x)$ and $g(t,x) = f(0,x) - f(t,x)$ is a solution). Now, let us consider the case when $f(t,x)$ is such that $\frac{\partial^2}{\partial x^2}f(t,x)$ is continuous in $(t,x)$ and $x$ also takes values in a compact set $K$. The following type of argument quickly comes to mind. Define $$\tau \doteq \{t: \exists g, h:[0,t]\times K\rightarrow {\mathbb R}, f = h-g, h,g \text{ convex in } x \text{ and increasing in } t \}.$$ $0$ is clearly a member of this set. One can perhaps argue that $\sup \tau$ must be $T$ as follows. If $t_0 = \sup \tau < T$ then one can slightly modify $h(t_0,x)$ and $g(t_0,x)$ to obtain functions $h(t_0+\delta,x)$ and $g(t_0+\delta,x)$ whose difference will be $f(t_0+\delta,x)$ [the slight modification is made possible by the fact that the second derivative of $f$ with respect to $x$ barely changes between $t_0$ and $t_0 + \delta$]. - Thanks for looking at this. I agree that discretizing it is a good first step (it also helps to restrict to a compact interval for x). However, it works out a bit better if you solve for g,h by starting at the last time and working backwards. Doing this, you can show that it converges to the continuous time decomposition if and only if there is a continuous time decomposition. I'm going to post an answer myself, when I have time, describing a few different ways to reformulate this problem. – George Lowther Oct 30 2010 at 21:38 Also, if f is either twice differentiable in x or once in t, then it will have the required decomposition. The problem appears with nondifferentiable functions. You can always approximate by smooth functions but, then, it is possible that the decompositions of these smooth approximations diverge when you take the limit. – George Lowther Oct 30 2010 at 21:42 You are welcome; your questions above and your related results are very interesting, thanks for sharing. What makes the problem difficult seems to be that there are many things that influence how $g$ and $h$ are to evolve in time and there seems to be many choices. The optimization problem in the discrete formulation (the choice of $S_2$) referred to in my answer seems to be nontrivial (I think it can be formulated as a control problem). – has2 Oct 31 2010 at 7:45 Another thing that seem relevant is the following: the region where $x \rightarrow f(x,0) - f(x,t)$ is not convex must be small, because otherwise $f(x,0)$ could not dominate $f(x,t)$. – has2 Oct 31 2010 at 7:50 Hi, Maybe I haven't fully understood your question but I think there is a counter-example (in a simpler context) given in Protter's book on Stochastic Integration in the form of a theorem. It's theorem 71, chapter IV, section 7 (Local Times). It's claimed there that for any $\alpha \in(0,1/2)$ and any continuous local martingale $X_t$, then $Y_t=\|X_t\|^\alpha$ is not a semi-martingale unless it is a null process. The proof is based on local time and Tanaka-Meyer's theorem. Hope it helps, (but if not I would appreciate that you explain where I went wrong). Best regards. - The function $f(x)=\vert x\vert^\alpha$ is not convex for $\alpha < 1$. Convex functions do preserve the semimartingale property (see the Ito-Tanaka formula for example). It is only when you have convex in space and decreasing in time that it gets difficult. – George Lowther Oct 24 2010 at 15:56 Actually, I think what I just called the Ito-Tanaka formula is the same thing as Tanaka-Meyer's theorem that you just mentioned. It is also called the Ito-Tanaka-Meyer formula. – George Lowther Oct 24 2010 at 15:59 Sorry to insit but $f$ is concave though (right ?),so taking $-f$ should do the trick. I must miss something about the convex function definition here. Regards – The Bridge Oct 24 2010 at 16:19 No. f is concave individually on the intervals $[0,\infty)$ and $(-\infty,0]$. It is neither convex nor concave on any open interval about zero. And that is the point of the result you refer to. In any case, a local martingale restricted to being either nonnegative everywhere or nonpositive everywhere must remain at 0 once it hits zero. It can only escape 0 if it can only take both positive and negative values, so the behaviour of f in a neighbourhood of 0 is what is important here. – George Lowther Oct 24 2010 at 16:29 One further point: I don't have Protter's book to hand right now but, if you look at the proof of the result you just quoted, I expect that it is writing the drift term of f(X) as $\int L^x_t f^{\prime\prime}(x)\,dx$ where $L^x_t$ is the local time at x. This explodes if $L^0>0$ in your example but, if f was either convex or concave, $f^{\prime\prime}$ would be locally integrable and you get that f(X) is indeed a semimartingale. – George Lowther Oct 24 2010 at 16:39 show 2 more comments	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 78, "mathjax_display_tex": 10, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9474689364433289, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/189399/expected-number-of-balls-withdrawn-to-get-equal-numbers-of-black-and-white-balls/189413	# expected number of balls withdrawn to get equal numbers of black and white balls There are $n$ black balls and $n$ white balls in a bin. I withdraw the balls one at a time without replacement until I have an equal number of white and black balls. What is the expected number of balls that I have to withdraw? It appears that the answer should be $4^n\left/{2n \choose n}\right.$. So for $n = 3$ it would be: $$4^3\left/{6 \choose 3}\right. = \frac{64}{20} = \frac{16}{5}$$ I have verified the answers for $n = 2$ and $n = 3$, but I am not able to prove the general result. - Of course it is possible that it will never occur. There is a positive probability that you will exhaust say the white balls without ever getting the number of black equal to the number of white. So the expectation only makes sense conditional on being successful. – Michael Chernick Aug 31 '12 at 20:14 4 But after $2n$ drawing, the number balls of each color is bound to be equal. – Sasha Aug 31 '12 at 20:15 Since we have equal numbers of black and white balls, it will always occur. In the worst case, after I have drawn 2n balls, I will have equal numbers of white and black balls. If I have exhausted all the white balls, I would continue by drawing the remaining n black balls so that when I have drawn all 2n balls, I will have equality. In general, I will achieve equality after drawing 2k balls with 1 <= k <= n – keng5 Aug 31 '12 at 20:20 At the start, you have zero balls of each color. The expected number of draws to have equal numbers is zero. – Ross Millikan Aug 31 '12 at 20:33 1 Nice problem, by the way! – Byron Schmuland Aug 31 '12 at 21:02 show 1 more comment ## 3 Answers Your conjecture is true. The following argument is not elegant, but it works! The number of arrangements of $k$ white and $k$ black balls where the first equalization occurs at $2k$ is $2 C_{k-1}$, where $C_{k-1}={1\over k}{2(k-1)\choose k-1}$ is the $k-1$th Catalan number. The number of arrangements of $n$ white and $n$ black balls where the first equalization occurs at $2k$ is therefore $${2\over k}{2(k-1)\choose k-1} {2(n-k)\choose n-k}.\tag 1$$ All ${2n\choose n}$ arrangements are equally likely, so the chance that equalization first occurs at time $2k$ is $$\mathbb{P}(T=2k)= {1\over{2n\choose n}} {2\over k}{2(k-1)\choose k-1} {2(n-k)\choose n-k}.\tag 2$$ The expected time to equalization is therefore $$\mathbb{E}(T)={1\over{2n\choose n}} \sum_{k=1}^n 2k\, {2\over k}\,{2(k-1)\choose k-1} {2(n-k)\choose n-k}.\tag 3$$ Cancelling the $k$'s in (3) and using the known identity $$\sum_{k=1}^n {2(k-1)\choose k-1} {2(n-k)\choose n-k}=4^{n-1}\tag4$$ gives the result. - Sorry: you must have posted while I was writing. – Brian M. Scott Aug 31 '12 at 21:22 No problem. Additional explanations are always welcome. – Byron Schmuland Aug 31 '12 at 21:22 Suppose that the first ball drawn is white and that you stop on draw number $2k$. Then the $2k$-th ball drawn was black, and the $2k-2$ balls drawn in positions $2$ through $2k-1$ form a Dyck word of length $2k-2$. Conversely, any sequence of $2k-2$ white and black balls in which the number of black balls never exceeds the number of white balls can occupy those $2k-2$ positions. Thus, there are $C_{k-1}$ sequences of draws beginning with a white ball that terminate with draw $2k$. If we imagine continuing until the bin is empty, there are $\binom{2n-2k}{n-k}$ ways to complete the draw, for a total of $C_{k-1}\binom{2n-2k}{n-k}$ full draws that start with a white ball and first balance (at equal numbers of white and black balls) on draw $2k$. There is an equal number starting with a black ball, so $$2C_{k-1}\binom{2n-2k}{n-k}=\frac2k\binom{2k-2}{k-1}\binom{2n-2k}{n-k}$$ of the $\binom{2n}n$ possible full draws first balance on draw $2k$. The expected number of draws to the first balanced sample is therefore $$\binom{2n}n^{-1}\sum_{k=1}^n\frac2k\binom{2k-2}{k-1}\binom{2n-2k}{n-k}(2k)=4\binom{2n}n^{-1}\sum_{k=1}^n\binom{2k-2}{k-1}\binom{2n-2k}{n-k}\;,$$ and your conjecture is equivalent to $$4^{n-1}=\sum_{k=1}^n\binom{2k-2}{k-1}\binom{2n-2k}{n-k}=\sum_{k=0}^{n-1}\binom{2k}k\binom{2n-2k-2}{n-k-1}$$ or, after replacing $n-1$ by $n$, to $$4^n=\sum_{k=0}^n\binom{2k}k\binom{2n-2k}{n-k}\;.$$ You can find a proof of this identity in this question together with an outline of a combinatorial proof; this answer gives a full combinatorial proof. - Byron and Brian, Thanks for the proofs. – keng5 Sep 1 '12 at 14:28 I would start by defining $f(n,m), n \ge m$ as the expected number of draws starting from $n$ white balls and $m$ black balls with $n-m$ black balls in hand. Then we have $f(n,n)=1+f(n,n-1)$ $f(n,0)=n$ $f(n,n-1)=\frac n{2n-1}+\frac{n-1}{2n-1}(1+f(n,n-2))$ $f(n,m)=1+\frac m{n+m}f(n,m-1)+\frac n{n+m}f(n-1,m)$ where the cases are checked in this order and seek a solution. No guarantees that this will work. - – leonbloy Sep 1 '12 at 0:06	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 44, "mathjax_display_tex": 9, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9402660727500916, "perplexity_flag": "head"}
http://mathoverflow.net/questions/5761/change-rates-of-the-2nd-chebychevs-function/6064	## change rates of the 2nd Chebychev´s function ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $\psi(x):=\sum_{n\leq x}\Lambda(n)$ denotes the 2nd Chebyshev function, where $\Lambda$ stands for the von-Mangoldt function. Are there any known (and 'nice') estimates for the change rates $\psi(x+h)-\psi(x)$ for general or special $x$ and $h$? Thanks in advance, efq - ## 2 Answers There is the asymptotic estimate $\psi(x+h) - \psi(x) \sim h$ for $x^{7/12 + \epsilon} \leq h \leq x$, valid for any $\epsilon > 0$. This is due to M. N. Huxley, and dates to 1972. I am not aware of any better range for $h$ if you want asymptotic equality. But if you are satisfied with an order of magnitude result, you can have $c_1h \leq \psi(x+h) - \psi(x) \leq c_2h$ with $c_1$ and $c_2$ positive constants and $x^{\theta} \leq h \leq x$ for some $\theta$ slightly larger than $0.5$. I can't give references offhand, but you should be able to find such papers by searching on R. C. Baker, G. Harman, J. Pintz in Mathscinet. - Thanks a lot for you answer! I believe this order-of-magnitude-result may in fact work out for me. I´ll definitely have a look into the reference you mention. – ex falso quodlibet Nov 20 2009 at 19:09 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Also, I think Selberg's result assert that for "almost all x" we have $\psi(x + h) - \psi(x) \sim h$ for $h \asymp (\log x)^2$ this was shown to not be true "pointwise" by Mayer. - That is interesting. Is it known whether this null set is finite? – ex falso quodlibet Nov 20 2009 at 20:49 there is a sequence of x_k -> oo such that psi(x_k + (log x_k)^2) - psi(x_k) > (1+epsilon) (log x_k)^2 ... (I hope this answers your question). – maki Nov 20 2009 at 23:16 yes, of course! ignore my last comment. – ex falso quodlibet Nov 21 2009 at 23:19	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9134442210197449, "perplexity_flag": "middle"}
http://mathhelpforum.com/differential-geometry/122635-fractional-part-integration-print.html	# fractional part and integration Printable View • January 6th 2010, 08:27 AM Shanks fractional part and integration Let $\{x\}$ be the positive fractional part of real number $x$, for example, $\{\pi\}=0.1415926...$ . Prove that the set $S=\{\{nv\}:n\in Z^+\}$ is dense in [0,1] if $v$ is positive irrational number. And for any $f(x)\in C[0,1]$(the class of all real-valued continious function on [0,1]), and any positive irrational number $v$, we have $\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^{N}f(\{nv\} )=\int_0^1f(x)dx$. I know how to do this problem, but i can't write down the formal solution and some details completely. I need a complete and elaborate solution, thanks in advance! • January 6th 2010, 10:47 AM Laurent Quote: Originally Posted by Shanks I know how to do this problem, but i can't write down the formal solution and some details completely. I need a complete and elaborate solution, thanks in advance! You should tell us more precisely which details are lacking in your solution, it would spare some time. • January 7th 2010, 01:38 AM Shanks notice the fact that if $t\in S$, then the fractional part of the multiply of t is also in S. To prove S is dense in [0,1], it is suffice to prove that there is a decreasing subsequence in S such that it converges to 0. Or in another word, 0 is the inf limit of S. This can be easily proved by contradiction! To prove the limit is equal to the integration, we need the definition of Riemann Integral and the uniform continuity of f(x). that is, it is suffice to prove that for any $\delta >0$, there exist N, such that if k > N, then the longest distance of two consecutive( the oder of number, not the index n) among the previous k points is less than $\delta$, Or in another word, the distance of any two consecutive points of the previous k points is less than $\delta$. Although I understand what is actually going on, I can't express by word, can't write down the details here!(Doh) • January 8th 2010, 12:33 AM Laurent Quote: Originally Posted by Shanks To prove the limit is equal to the integration, we need the definition of Riemann Integral and the uniform continuity of f(x). that is, it is suffice to prove that for any $\delta >0$, there exist N, such that if k > N, then the longest distance of two consecutive( the oder of number, not the index n) among the previous k points is less than $\delta$, Or in another word, the distance of any two consecutive points of the previous k points is less than $\delta$. Although I understand what is actually going on, I can't express by word, can't write down the details here!(Doh) You need more than just that: not only need the points of the sequence to be close enough, but they need to be "uniformly distributed" (asymptotically) for the sum to converge to the integral. I give you a magic trick: first prove the limit when $f(x)=e^{2i\pi nx}$ for some non-zero integer $n$ (direct computation), then for trigonometric polynomials, i.e. linear combinations of the previous functions (direct consequence), and finally for general $f$ (using approximation by a trigonometric polynomial). This is the usual way to procede (not very intuitive, I admit), but there may be others. • January 8th 2010, 02:52 AM Shanks Quote: Originally Posted by Laurent You need more than just that: not only need the points of the sequence to be close enough, but they need to be "uniformly distributed" (asymptotically) for the sum to converge to the integral. I give you a magic trick: first prove the limit when $f(x)=e^{2i\pi nx}$ for some non-zero integer $n$ (direct computation), then for trigonometric polynomials, i.e. linear combinations of the previous functions (direct consequence), and finally for general $f$ (using approximation by a trigonometric polynomial). This is the usual way to procede (not very intuitive, I admit), but there may be others. I know that this is the way written in G.polya book. But Here I don't want to using approximation of continious function by trigonometric polynomial. I want to prove it by using the definition of Riemann integral. • January 16th 2010, 03:03 AM Shanks The Problem occurs again in Rudin's analysis book, therefore it is a impressive problem. All times are GMT -8. The time now is 12:49 AM.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 21, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9343944787979126, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/202879/prove-the-existence-of-a-c-infty-bounded-function-with-arbitrary-derivatives	# Prove the existence of a $C^\infty$ bounded function with arbitrary derivatives at a fixed point. (Borel?) I found this problem on an Italian forum and since then I struggled to solve it. The autor there claims it was proposed by Borel. At any rate the problem is as follows (Borel?) For any $\{a_n\}_{n=0}^{\infty}\subseteq \mathbb R,\: x_0\in\mathbb R\,$ and $\,\varepsilon>0$ does there exist a $C^\infty$ function $f\colon\mathbb R\to\mathbb R$ such that $$f^{(n)}(x_0)=a_n,\; \forall n\in\mathbb N\cup\{0\}\tag{1}$$ and moreover $$\left\|f(x)-a_0\right\|<\varepsilon,\;\forall x\in\mathbb R?$$ Clearly even reference about problem are welcomed, but I strongly encourage anybody to think about it because really, when I met it for the first time, I thought: "this is a wonderful problem". I do not have any clue towards the solution. I hope you will have fun with this. Cheers. - – user42754 Sep 26 '12 at 15:56 – user42754 Sep 26 '12 at 15:58 Thanks for the references – uforoboa Sep 27 '12 at 9:53 ## 1 Answer Here is a counterexample. For arbitrary interval $I\subset\mathbb{R}$ denote $M_k(f,I)=\max_{t\in I}\|f^{(k)}(t)\|$, then from exercise 14 Chapter 5 in Rudin's Principles of Mathematical analysis we know that $$M_1^2(f,I)\leq 4M_0(f,I)M_2(f,I)$$ for any interval $I$ and $f\in C^3(I)$ Let's take $a_1>2\sqrt{(\|a_0\|+\varepsilon)(\|a_2\|+\varepsilon)}$ and the rest whatever you want. Assume that there exist a smooth function $f$ with predescribed properties. Since $f\in C^\infty(\mathbb{R})$ then we can find neighbourhood $U$ of $x_0$ where $$M_2(f,U)<\|f^{(2)}(x_0)\|+\varepsilon=\|a_2\|+\varepsilon$$ By construction $\|f(x)-a_0\|<\varepsilon$ for all $x\in\mathbb{R}$, so $$M_0(f,U)<\|a_0\|+\varepsilon$$ Finally $$M_1^2(f,U)>\|f'(x_0)\|^2=a_1^2>4(\|a_0\|+\varepsilon)(\|a_2\|+\varepsilon)>4M_0(f,U)M_2(f,U)$$ Contradiction. - Good! This show that boundedness is restrictive in this case, as we should have $\|a_n\|\leq 4(\|a_n\|+\varepsilon)(\|a_{n+1}\|+\varepsilon)$ for all $n$. – Davide Giraudo Sep 26 '12 at 17:10 @DavideGiraudo, thanks! – Norbert Sep 26 '12 at 17:37 @Norbert I see your point. I was aware of that result you quoted however I never linked it to the solution of this problem. Thank you. – uforoboa Sep 27 '12 at 9:52	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 18, "mathjax_display_tex": 6, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9293622374534607, "perplexity_flag": "head"}
http://stats.stackexchange.com/questions/3438/calculating-percentile-of-normal-distribution	# Calculating percentile of normal distribution http://en.wikipedia.org/wiki/Binomial_proportion_confidence_interval#Agresti-Coull_Interval To get the Agresti-Coull Interval, one needs to calculate a percentile of the normal distribution, called $z$. How do I calculate tha percentile? Is there a ready-made function that does this in Wolfram Mathematica and/or Python/NumPy/SciPy? - 1 The integral expression in the "normal cdf I got exactly from Wiki" is unfortunately off by a factor of $1/\sqrt{\pi}$. There is no known exact formula for the normal cdf or its inverse using a finite number of terms involving standard functions ($\exp, \log, \sin \cos$ etc) but both the normal cdf and its inverse have been studied a lot and approximate formulas for both are programmed into many calculator, spreadsheets, not to mention statistical packages. I am not familiar with R but I would be astounded if it did not have what you are looking for built in already. – Dilip Sarwate Feb 14 '12 at 20:52 @DilipSarwate, it's fixed! I am doing this using inverse tranformation, also "not allowed" to use too much built in. It's for the sake of learning I suppose. – user1061210 Feb 14 '12 at 21:26 1 @Dilip: Not only is there no known exact formula, better yet, it is known that no such formula can exist! – cardinal Feb 14 '12 at 21:30 1 The Box-Muller method generates samples from a joint distribution of independent standard normal random variables. So histograms of the values generated will resemble standard normal distributions. But the Box-Muller method is not a method for computing values of $\Phi(x)$ except incidentally as in "I generated $10^4$ standard normal samples of which $8401$ has value $1$ or less, and so $\Phi(1) \approx 0.8401$, and $\Phi^{-1}(0.8401) \approx 1$. – Dilip Sarwate Feb 14 '12 at 22:29 1 I just chose $8401$ as an example of the kinds of numbers you might expect. $\Phi(1) = 0.8413\ldots$ and so if you generate $10^4$ samples of a standard normal distribution, you should expect close to $8413$ of the $10000$ samples to have value $\leq 1$. You are implementing the Box-Muller method correctly, but are not understanding the results that you are getting and are not relating them to the cdf etc. – Dilip Sarwate Feb 15 '12 at 0:27 show 12 more comments ## 5 Answers For Mathematica, `$VersionNumber > 5` you can use ````Quantile[NormalDistribution[mu,sigma], 100 q] ```` for the q-th percentile. Otherwise, you have to load the appropriate Statistics package first. - (I have version 7.) I have no problem loading the Statistics package. But what's the function in there called? Because I get the impression that this `Quantile` line will do the calculation manually instead of using a formula. – Ram Rachum Oct 9 '10 at 14:13 Evaluate it with symbolic parameters (i.e. don't assign values to `mu`, `sigma`, and `q`); you should get an expression involving the inverse error function. – J. M. Oct 9 '10 at 14:24 John Cook's page, Distributions in Scipy, is a good reference for this type of stuff: ````In [15]: import scipy.stats In [16]: scipy.stats.norm.ppf(0.975) Out[16]: 1.959963984540054 ```` - Thanks for adding a real illustration with ppf. – chl♦ Oct 9 '10 at 16:22 Well, you didn't ask about R, but in R you do it using ?qnorm (It's actually the quantile, not the percentile, or so I believe) ````> qnorm(.5) [1] 0 > qnorm(.95) [1] 1.644854 ```` - 1 Quantile vs. percentile (it's merely a matter of terminology), j.mp/dsYz9z. – chl♦ Oct 9 '10 at 14:22 1 While we are in, in R Wald-adjusted CIs (e.g. Agresti-Coull) are available in the `PropCIs` package. Wilson's method is the default in `Hmisc::binconf` (as suggested by Agresti and Coull). – chl♦ Oct 9 '10 at 14:36 Thanks for the comments chl – Tal Galili Oct 10 '10 at 4:11 In Python, you can use the stats module from the scipy package (look for `cdf()`, as in the following example). (It seems the transcendantal package also includes usual cumulative distributions). - You can use the inverse erf function, which is available in MatLab and Mathematica, for instance. For the normal CDF, starting from $$y=\Phi\left(x\right)=\frac{1}{2}\left[1+\text{erf}\left(\frac{x}{\sqrt{2}}\right)\right]$$ We get $$x=\sqrt{2}\ \text{erf}^{-1}\left(2y-1\right)$$ For the log-normal CDF, starting from $$y=F_{x}(x;\mu,\sigma)=\frac{1}{2}\text{erfc}\left(\frac{-\log x-\mu}{\sigma\sqrt{2}}\right)$$ We get $$-\log \left(x\right)=\mu+\sigma\sqrt{2}\ \text{erfc}^{-1}\left(2y\right)$$ - 1 isn't this more of a comment than an answer? – Macro Feb 15 '12 at 22:18 My idea was that if you have inverses for the erf and erfc functions, then the problem is solved. MatLab, for instance, has such preprogrammed functions. – Jean-Victor Côté Mar 5 '12 at 19:31 @Jean-VictorCôté Please, develop your ideas in your reply. Otherwise, it merely looks like a comment as suggested above. – chl♦ Mar 5 '12 at 22:31 The lognormal calculation doesn't look right. After all, its inverse CDF should be identical to the inverse CDF for the normal apart for the use of $\log(x)$ instead of $x$. – whuber♦ Mar 6 '12 at 19:15 default	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8805760741233826, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/52469/differentiation-and-delta-function	# Differentiation and delta function Need help doing this simple differentiation. Consider 4 d Euclidean(or Minkowskian) spacetime. \begin{equation} \partial_{\mu}\frac{(a-x)_\mu}{(a-x)^4}= ? \end{equation} where $a_\mu$ is a constant vector and the indices are summed over since one really doesn't need to bother about upper and lower indices in flat space. Also $(a-x)^2=(a-x)_\mu(a-x)_\mu$. A simple minded calculation gives me the result $0$. But I think the answer may contain Dirac Delta function like the relation below in 3 dimension \begin{equation} {\bf{\nabla}} . \frac{\hat{r}}{r^2}=4\pi \delta ^3(r) \end{equation} A corollary question. What will be $\partial_{\mu}\frac{1}{(a-x)^2}$ ? Is it $\frac{2(a-x)_\mu}{(a-x)^4}$ or does it also involve the Delta function? - – Qmechanic♦ Jan 29 at 15:31 ## 2 Answers 1. In a $d$-dimensional Euclidean space (with positive definite norm), one has $$\vec{\nabla} \cdot \frac{\vec{r}}{r^d} ~=~{\rm Vol}(S^{d-1})~\delta^d(\vec{r}),$$ cf. the divergence theorem and arguments involving either test functions and integration by part, or $\epsilon$-regularization, similar to methods applied in this Phys.SE answer. Here ${\rm Vol}(S^{d-1})$ is the surface area of the $(d-1)$-dimensional units sphere $S^{d-1}$. 2. By similar arguments one may show that the identity $$\vec{\nabla}(r^{2-d}) ~=~(2-d)\frac{\vec{r}}{r^d} , \qquad d\neq 2,$$ contains no distributional contributions in $d$-dimensional Euclidean space. 3. For the related questions in Minkowski space, one suggestion is to introduce an $\epsilon$-regularization in the Euclidean formulation, and then perform a Wick rotation, and at the end of the calculation, let $\epsilon\to 0^+$. - Thanks Qmechanic. Writing down your equation in the index notation gives \begin{equation} \partial_\mu \frac{x_\mu}{x^2}=2\pi^2\delta^4(x) \end{equation} in 4d which is the result I wanted. Now it looks a bit silly to have asked the question in the first place. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8474989533424377, "perplexity_flag": "head"}
http://mathoverflow.net/questions/tagged/harmonic-analysis	## Tagged Questions 1answer 58 views ### Plancherel formula for non-second-countable (non-unimodular) groups The Plancherel formula for unimodular, second-countable, type 1 groups can be found in A Course in Abstrict Harmonic Analysis by Gerald Folland (theorem 7.44) or here. It states th … 1answer 150 views ### Littlewood-Paley theory and norm estimation In the paper "A Convolution Inequality Concerning Cantor-Lebesgue Measures", the Littlewood-Paley theory is used to estimate the norm of multiplier operator in Lemma 1. It is claim … 0answers 75 views ### Laplacian type operator on compact Lie group Consider the operator $S = \sum_{i,j} X_{ij}^2$ on $L^2(SO(n+1))$, where $X_{ij}$ generates the rotation of the sphere $S^n$ in the $ij$-plane keeping the $(n - 2)$ complementary d … 0answers 187 views ### Fourier coefficients as spectrum Let $\mathbb{T}=[0,1]$ be identified with the circle $\{ e^{2 \pi it} : t \in [0,1] \}$, $\delta_0 \in M(\mathbb{T})$ be the Dirac measure at $0 \in \mathbb{T}$. Suppose \$f \in L^1 … 0answers 74 views ### Character amenability Hello 1)IS ANY relationship between character amenability and weak amenability of Banach algebras?.... If a Banach $A$ is amenable then $A$ is $\phi$-amenable for every \$\phi\i … 1answer 202 views ### For what spaces is the Hardy-Littlewood maximal operator of strong type $(p,p)$ if and only if $p > p_0 > 1$? (This is essentially a continuation of my previous question, here.) Let $(X,d,\mu)$ be a metric measure space, i.e. $\mu$ is a Borel measure on the metric space $(X,d)$. Further a … 1answer 205 views ### Could we interpolate the compactness of compact operators? Classical theorems of Marcinkiewicz and Riesz and their extensions to general Banach spaces by Calderón, Lions, Peetre, et al. allow us to interpolate the continuity of two operato … 0answers 82 views ### Question about a oscillatory integrals on manifold Let $M$ be a compact oriented Riemannian manifold without boundary. Set $f(x)=a(x)+\sqrt{-1}b(x)$ be a complex-valued function on $M$, where $a(x),b(x)$ are real-valued function o … 2answers 544 views ### Is there a “right” proof of Riemann’s Theta Relation? Let $\theta$ denote the usual Jacobi Theta function (with auxiliary parameter $\tau = i$, for simplicity), i.e. \theta(z) = \sum_{n \in \mathbb{Z}} \exp(-\pi (a + n)^2 + 2 \pi … 2answers 408 views ### Image of L^1 under the Fourier Transform The Fourier Transform $\mathcal{F}:L^1(\mathbb{R})\to C_0(\mathbb{R})$ is an injective, bounded linear map that isn't onto. It is known (if I remember correctly) that the range is … 1answer 55 views ### Description of Bessel potential spaces Hi, let $1 < p <\infty$, $0 < \alpha < 1$, and $\mathscr{L}^p_\alpha(R^n)$ be the usual Bessel potential space defined by \mathscr{L}^p_\alpha = (1-\triangle)^{- … 0answers 48 views ### Notation for a functional L2 matrix norm Hi, Let $v(z)$ be a 2x2 matrix depending on a complex variable z, defined on an oriented contour $\Sigma$ in the complex plane. Can anyone tell me the meaning of the notation: … 0answers 137 views ### variant of Haar measure I found a certain trig identity in a discussion on Lie groups \[ \frac{\prod_{i < j} 2 \sin (\mu_i - \mu_j) \prod_{i< j} 2 \sin (\nu_i - \nu_j) }{\prod_{i, j} 2 \cos (\mu_ … 0answers 193 views ### What information about a locally compact group $G$ is encoded in $C_r^\ast(G)$ which is not in $L^1(G)$? Let $G$ be a locally compact group and let $C_r^\ast(G)$ denote its reduced group $C^\ast$-algebra. Many features of a $G$ can be realized from $L^1(G)$ or $C_r^\ast(G)$. For exa … 0answers 452 views ### The Fourier Transform of taking Eigenvalues The purpose of this question is to ask about the Fourier transform of the map which associate to an $n$ by $n$ matrix its $n$ eigenvalues, or some function of the $n$ eigenvalues. … 15 30 50 per page	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 46, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8081178665161133, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/118231/proving-that-11-cannot-be-written-in-the-form-p24pqq2	# Proving that 11 cannot be written in the form $p^2+4pq+q^2$ I have to prove that it is imossible to write 11 in the form $p^2+4pq+q^2$ where $p$ and $q$ are integers. I tried to start a proof by contradiction like this: $$p^2+4pq+q^2 = 11\\ 2(p+q)^2 - (p^2 + q^2) = 11\\ 2(p+q)^2 = 11 + (p^2 + q^2)$$ I'm really not sure where to go from there, though. A pointer in the right direction would be much appreciated! - 12 Try considering the problem mod 4 – Tom Cooney Mar 9 '12 at 14:34 You could simply go through all posibilities: Without loss of generality you can suppose that p >= q. If q = 0 only p is important. 1^2 = 1, 2^2 = 4, 3^2 = 9, 4^2 = 16. If q = 1. 1^2 + 411 + 1^2 = 6, 2^2 + 421 ... if q = 2. 2^2 + 422 + 2^2 = 4 + 16 + 2 != 11. – moose Mar 9 '12 at 18:50 ## 4 Answers In an introductory course in Number Theory, one of the early things one learns is that the sum of two squares cannot be congruent to $-1$ modulo $4$. This can be shown by a short enumeration of cases. Modulo $4$, the middle term $4pq$ is irrelevant, and we are finished. We can also work modulo $3$, directly by a short consideration of cases, made even shorter by the symmetry. Or else we can reluctantly break symmetry and rewrite our expression as $(p+2q)^2-3q^2$, and note that modulo $3$ we need pay no attention to $3q^2$. But the perfect square $(p+2q)^2$ is congruent to $0$ or $1$ modulo $3$, while $11$ is congruent to $-1$. Or else one can note that modulo $3$ we are looking at $p^2-2pq+q^2$, that is, $(p-q)^2$. Remark: The strategy we used is a common one. When we are given a Diophantine equation, and suspect that it has no solutions, it is worthwhile to look at the equation modulo some small numbers. If for some $m$ the equation cannot hold modulo $m$, then it cannot hold in the integers. The simplest such arguments are parity arguments (working modulo $2$). We can think of working modulo numbers greater than $2$ as generalized parity arguments. - For variety, here's a method that doesn't require guessing at which small modulus to try. (however, it does use some more sophisticated mathematics) First, we simplify our quadratic by completing the square $$(p+2q)^2 - 3q^2 = 11$$ Next, we homogenize: $$(pr+2qr)^2 = 11 r^2 + 3 (qr)^2$$ and for clarity, change variable $$z^2 = 11 x^2 + 3 y^2$$ (rational solutions to the original equation would be $q = y/x$ and $p = z/x - 2q$) Now, we invoke a two relevant sledgehammers. This equation obviously a real solution, and the only relevant primes to check are $3$ and $11$. $$(3, 11)_{3} = \left(\frac{11}{3}\right) = -1$$ $$(3, 11)_{11} = \left(\frac{3}{11}\right) = 1$$ (and from this, we can also conclude $(3,11)_2 = -1$) where the left hand side is the Hilbert symbol and the right hand side is the Legendre symbol. Since we got a $-1$, the equation has no rational solutions, and thus it has no integer solutions. Incidentally, this result explains why the other posts were successful when checking modulo $3$ and checking modulo $8$. A lowbrow summary of this method would be ````Complete the square, find a real solution, then try working modulo 8, then try working modulo every prime dividing one of the coefficients. If you found solutions in all cases, then your quadratic has rational solutions. Otherwise, it doesn't. ```` The main thing the sophisticated math tells you is that those are the only primes useful to check, and that if the check passes every test, then there truly are rational solutions. (actually proving none of the rational solutions are integer solutions is generally much trickier) - I'm a number theory noob, but taking all mod 4 keeping in mind what odd or even integers squares can yield mod 4, should solve it?! :) - Looking at $$2(p+q)^2 - (p^2 + q^2),$$ this will be even when $p$ and $q$ have the same parity. But if $p$ and $q$ have different parity then modulo $4$ it is equivalent to $2\times 1 - 1 =1$, while $11 \equiv 3 \mod 4$. So $11$ cannot be expressed in that form. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 41, "mathjax_display_tex": 7, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9226929545402527, "perplexity_flag": "head"}
http://mathoverflow.net/questions/28520/hodge-numbers-of-compactifications/28597	## Hodge numbers of compactifications ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $X$ be a smooth complex quasi-projective variety. We can find good compactification: a smooth proper variety $\bar{X}$ such that ${\bar X} \setminus X$ is a divisor with normal crossing. The variety $\bar{X}$ is then stratified by the singulartities of the divisor. And one can compute the mixed Hodge structure on $H^{\bullet}(X)$ in terms of the pure Hodge structures $H^{\bullet}(S_\alpha)$ of the smooth closed strata using a spectral sequence. Let's say a variety $Y$ is Hodge-Tate if $h^{p,q}(Y) = 0$ for $p\neq q$. If all the closed strata of $\bar{X}$ are Hodge-Tate then $X$ is Hodge-Tate. Question: Let $X$ be a smooth complex quasi-projective variety. Assume $X$ is Hodge-Tate. 1. Can one find a good compactification $\bar{X}$ with Hodge-Tate strata? 2. Are all good compactifications of $X$ of this type? (Edit: Answer is no, see Torsten's elementary example). - ## 2 Answers I asked the question to Claire Voisin and she immediatly gave me the following counter example: Consider Fermat's cubic of dimension 3 $X$. There are 5 cones on elliptic curves $E_i$ inside $X$ and one can verify that $Y=X\setminus U_iE_i$ is Hodge-Tate. But $Y$ doesn't admit a Hodge-Tate compactification as that would imply that $X$ is birational to a Tate variety and this would contradict Clemens-Griffiths' theorem saying that the intermediate jacobian of $X$ is not a direct sum of jacobians of curves as a polarized variety. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Let $E$ and $F$ be two elliptic curves and let the involution $\sigma$ act on $E\times F$ by $\sigma(e,f)=(-e,f+\alpha)$, $\alpha$ is an element of order two of $F$. Finally let $\overline{X}=(E\times F)/\sigma$ (this is a so called hyperelliptic surface). We have an inclusion $F':=0\times F/\langle\alpha\rangle\subseteq S$ and put $X:=\overline{X}\setminus F'$. Then $X$ is Hodge-Tate but all other good compactifications of $X$ are obtained by blowing ups and downs of $\overline{X}$ which means that you can never get rid of $F'$ (alternatively any good compactification $X'$ has $H^1(X')=H^1(X)$ and you need something non-Hodge-Tate at the boundary to kill that off). Addendum: This example is all wrong it took care of $H^3(X)$ but not (the more interesting) $H^1(X)$. At the moment I am less sure than I was that the answer to 1) is no. As for 2) you can just look at $\mathbb A^3\subseteq\mathbb P^3$ which is a good Hodge-Tate compactification with $\mathbb P^2$ as divisor at infinity and then blow up something non-Hodge-Tate in $\mathbb P^2$. This gives a good compactification with two components one of which (the exceptional divisor for the blowing up) is non-Hodge-Tate (as is the intersection of these two divisors). -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 43, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9532508850097656, "perplexity_flag": "head"}
http://nrich.maths.org/2665/solution	### Isosceles Triangles Draw some isosceles triangles with an area of $9$cm$^2$ and a vertex at (20,20). If all the vertices must have whole number coordinates, how many is it possible to draw? ### Route to Infinity Can you describe this route to infinity? Where will the arrows take you next? ### Eight Hidden Squares On the graph there are 28 marked points. These points all mark the vertices (corners) of eight hidden squares. Can you find the eight hidden squares? # Lost ##### Stage: 3 Challenge Level: We received a number of solutions to this problem. Stephen from Manly Selective Campus got very close to the most efficient strategy: To explain the solution, I will use an example of this problem. Say the giraffe is at (6,8). The first step would be to enter the co-ordinates (0,0). It will then say that you are 14 blocks away from the giraffe, because the two co-ordinates will add up to the number of blocks you are away from the giraffe. The next step would be to then type in the co-ordinates (7,7). It will then say that you are 2 blocks away from the giraffe. So, that leaves you with 2 possibilities: (6,8) and (8,6). It is a simple matter then to try the two and hope that the one that you say first is right. To be sure you will not need more than 3 guesses you will need to adopt the strategy worked out by a number of pupils from St Hilda's Anglican School for Girls. Gloria and Sneha suggested: 1) Start off at any corner eg. (0,0) A diagonal of possibilities will form. 2) Click on one of the two edges of the diagonal as this gives only one possible solution. If you dont click on the edge, you could have two possibilities. 3) Locate the point on the diagonal which is 'n' blocks away from the edge of the diagonal. Snap!! Emily wrote: Your first point must be on one of the 4 corners (eg. (9,9) or (0,9) etc). You should then plot out where the giraffe could possibly be using the information you have (eg. if you started from (9,9) and the giraffe is 2 blocks away, it could be at either (9,7), (8,8) or (7,9)) Choose one of the extremes or outer co-ordinates (either (9,7) or (7,9)) and, from the information you are given from that search, plot where the giraffe could be. One of these points will overlap with one of the points from the other search and this is where the giraffe is (eg. if you chose (9,7) and the giraffe was 4 blocks away, it would be at (7,9)). Ailie and Sophie summarised the strategy very clearly: Choose one of the 4 corners (9,9) (0,0) (9,0) (0,9). Then, out of the possibilities choose one that is on an edge. After you do that there will be only one possibility where the giraffe is which will be one of the possible co-ordinates from before. We also received correct solutions from Tessa, Katharine and Alarna, also pupils from St Hilda's Anglican School for Girls. Well done to you all. The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9272186756134033, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/79755-solved-vector-calculus-max-min-magnitude-acceleration.html	# Thread: 1. ## [SOLVED] [Vector Calculus]: Max, Min of magnitude of acceleration Ok so I have a vector function: r(t)= $<3cos(t) , 2sin(t)>$ Alright so I have to find the Max and Min for the magnitude of the acceleration. The steps I was undergoing were: 1. I found the velocity vector v(t)= $<-3sin(t) , 2cos(t)>$ 2. I found the acceleration vector a(t)= $<-3cos(t) , -2sin(t)>$ and this is where I'm a little confused, so I have to find the function for the magnitude of a(t) and then derive it, find the critical points and those will be my max and min. But the problem is that when I find the magnitude of acceleration it gives me a constant. (sin^2 + cos^2 = 1) But that's the equation for an ellipse so the acceleration is not constant... so what am i doing wrong? EDIT: Oh crap I just realized what a dumb mistake I was making, the magnitude is not a constant, the coefficients are different, sin squared and cos squared can't be factored.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9108413457870483, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/45487/compactness-theorem-for-first-order-logic/45559	## Compactness Theorem for First Order Logic ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Hi all, I am interested in proofs without using Goedel's completeness theorem. • Does anyone have a reference to a proof of this theorem that uses Skolem Functions? • How come Enderton's (Introduction to Logic) has a half a page proof (which looks OK to me) and Boolos (Computability And Logic) has a full chapter of it? Thanks. - See also this related MO question: mathoverflow.net/questions/9309/… – Joel David Hamkins Nov 9 2010 at 23:20 Some people prefer to present math in a concise way where you have to stare at a half-page proof for half a day to figure it out, and some prefer to present it in a hairy way which reads like novel at the expense of blowing the proof of every triviality to ten pages. Boolos is firmly in the second category. – Emil Jeřábek Mar 2 2011 at 16:42 ## 7 Answers There's a proof of compactness using Skolem functions in the book "Elements of Mathematical Logic. Model Theory" by Kreisel and Krivine. (I'm assuming here that the English version matches the French, because the latter is the one I checked.) It presupposes the compactness theorem for propositional logic. The proof runs as follows: Given a finitely satisfiable set $S$ of formulas, Skolemize them to get a set $S'$ of universal formulas that is satisfiable if and only if $S$ is. (Technical detail: If the vocabulary has no constant symbols, adjoin one, so that there are some closed terms for use in the next step.) Let $S''$ be the set of all the formulas you get from those in $S'$ by deleting the universal quantifiers and replacing the variables by closed terms in all possible ways. The formulas in $S''$ are propositional combinations of atomic sentences; by regarding the atomic sentences as propositional variables, we can view $S''$ as a set of formulas of propositional logic. Finite satisfiability of $S$ (in the sense of first-order logic) implies finite satisfiability of $S''$ (in the sense of propositional logic). By propositional compactness, there is a truth assignment satisfying $S''$. That truth assignment amounts to a description of a structure (in the first-order sense) in which every element is the value of some closed term. This structure satisfies $S'$ and therefore $S$. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. There are indeed many proofs of the Compactness theorem. As I mention in this MO answer, when I was a graduate student Leo Harrington told me that he used a different proof method for Compactness each time he taught the introductory graduate logic course in Berkeley. I am not sure for how many semesters he was able to keep this up, but when I had him, it was time for the Boolean-valued models proof. The Compactness Theorem is the assertion that if a first order theory $T$ is finitely satisfiable (all finite subtheories have a model), then $T$ itself is satisfiable. Let me describe a number of proofs. • Goedel's original proof was via the Completeness theorem, deducing it as a trivial corollary. If $T$ is inconsistent, then the proof of a contradiction is finite, so there is an inconsistent finite subtheory. This proof is deprecated by contemporary logicians, because the Compactness theorem lies completely on the semantic side of the syntax/semantic divide, and it seems beside the point to have to develop the entire syntactic theory of formal proofs and derivations in order to make a conclusion purely about the semantic notions of models and satisfiability. • The Henkin proof. The point is that the usual Henkin proof of the Completeness theorem also serves directly to prove the Compactness theorem. Suppose that every finite subset of $T$ is satisfiable. By the usual details of the Henkin argument, we may extend $T$ to a finitely-satisfiable complete Henkin theory $T^+$, in a language with new constant symbols (using the theorem on constants). That is, the new theory contains the Henkin assertions $\exists x\varphi(x)\to \varphi(c)$, where $c$ is a new constant symbol added for this purpose with $\varphi$. Now, from $T^+$ we may build a model out of the Henkin constants in the usual manner. The reduct of this model to the original language satisfies $T$, as desired. • The proof via Skolem functions (as you requested). This amounts basically just to a more complicated version of the Henkin proof. I recall Henkin giving a talk at the Berkeley Logic Colloquium in which he explained that the idea for his proof of the Completeness theorem arose to him in a dream, after considering the (at that time standard) Skolem function proof of Completeness. The point was that in that proof, one adds Skolem functions to the language to tie the formula $\varphi(x)$ to the witness $f_\varphi(x)$, so that one adds the formulas $\forall \vec y, x[\varphi(x)\to \varphi(f_\varphi(\vec y))]$, instead of the Henkin assertion (this amounts to the quantifer-reducing idea mentioned by Andreas). But otherwise, it works out similarly---one proves the analogue of the theorem on constants that allows one to add the Skolem function assertions, and then builds the model out of formal term expressions. Henkin said that he realized in his dream that there was no need to tie the witness so closely to the formula with the Skolem function, and that merely having the presence of a constant to serve as a witness sufficed. Thus was born the Henkin proof. • The ultraproduct proof. Pete has an explanation of this proof in his answer. If $T$ is finitely satisfiable, then consider the set of finite subsets $t\subset T$, each of which has a model $M_t\models t$. Let $F$ be an ultrafilter containing for each $\varphi\in T$ the set of finite $t\subset T$ with $\varphi\in t$, a collection with the finite intersection property. The ultraproduct $\Pi_t M_t/F$ satisfies every $\varphi\in T$ by \L os's theorem. • The reduced product proof. In this proof, one first develops the concept of a reduced product $\Pi_t M_t/F_0$, where $F_0$ is only a filter instead of an ultrafilter (the filter generated by the collection with FIP above). And then you can finish the job by considering a quotient of this structure, which essentially amounts to the ultraproduct. • The Boolean-valued model proof. It is similar to the ultraproduct proof and the reduced product proof (they are all essentially the same), but there is no need to quotient out by $F$ in advance. Instead, one builds a $\mathbb{B}$-valued model out of the product ${\cal M}=\Pi_t M_t$, where $\mathbb{B}$ is the Boolean algebra of all subsets of finite subsets of $T$, so that the truth-value of a statement $\varphi$ in $\cal M$ is the set of $t$ for which $M_t\models \varphi$. Then, one develops the general theory allowing one to quotient a Boolean-valued model by a filter, and the conclusion amounts to \L os in the ultraproduct proof. - Hehe. Do you remember (roughly) the year of Henkin's talk? I remember what you say there, but probably I read it somewhere; I do not think you and I ever coincided at a Logic Colloquium. – Andres Caicedo Nov 10 2010 at 0:39 Hmm... Probably read it: Leon Henkin, "The Discovery of My Completeness Proofs", The Bulletin of Symbolic Logic 2 (1996), 127-158. – Andres Caicedo Nov 10 2010 at 0:46 4 It must have been the early 90s. Henkin spoke on his dissertation results, which consisted of three chapters, one of them containing his famous proof of Completeness. I recall that he was particularly intent on communicating the other chapters---he said they were not as well remembered. (But alas, I cannot now recall exactly what they were about!) – Joel David Hamkins Nov 10 2010 at 1:07 "Ultraproduct" rather than "ultrapower" is the term I'd have used in this case, since the factors are not all equal. – Michael Hardy Nov 10 2010 at 18:03 Yes, I'll edit. – Joel David Hamkins Nov 10 2010 at 18:36 One standard modern approach is to prove the Compactness Theorem using ultraproducts. I did so at the end of a short summer course on model theory: see http://www.math.uga.edu/~pete/modeltheory2010Chapter6.pdf - 3 There is an additional advantage to this approach, from the set theoretic point of view, in that strong compactness can be characterized in terms of infinitary logic by essentially the same ultraproduct argument that gives compactness for first order. – Andres Caicedo Nov 10 2010 at 0:07 There is a nice recent book by Richard Kaye, The mathematics of logic (Cambridge, 2007). The book is centered around the completeness theorem, and treats different versions, while developing the required set theoretic and logical background along the way. It begins with a treatment of König's lemma. Then develops a "toy version" of a proof calculus, that introduces the idea of completeness of a proof system. This toy version is designed to require only König's lemma. Additional choice is added to the picture as the proof system grows (so we see Boolean algebras and then propositional calculus, Tychonov's theorem and then first-order logic). Compactness is presented topologically but not in terms of ultraproducts. Some of the exercises (from the very beginning) are challenging. I found it a very nice introduction to the topic. - For some reason I like stating the ultraproduct proof of the compactness theorem in measure-theoretic language, as follows. A filter $F$ on the index set $I$ corresponds to a "finitely additive measure", like this: $$m(A) = \begin{cases} 1 & \text{if }A \in F, \\ 0 & \text{if }I \setminus A \in F \\ \text{undefined} & \text{otherwise} \end{cases}$$ A filter is an ultrafilter if the "otherwise" case is vacuous. Then \Los's theorem says a first-order statement in the language of the model is true in the ultraproduct if and only if it's true in "almost all" of the factors. (Then as others noted above, let $I$ be the set of finite subsets of the set of statements you're trying to satisfy; let $F$ be an ultrafilter that contains every co-finite set (one exists if you believe in Zorn's lemma); then for each $i \in I$ let the $i$th factor be a model that satisfies that finite set of statements; then prove by induction on the formation of first-order formulas that the ultraproduct satisfies all of the statements. As far as I know you have to include formulas with free variables in the proof to make the induction work. I'm skipping the details since others seem to have gone into those above.) - A Category Theory Book: Locally Presentable and Accessible Categories, J. Adamek and J. Rosicky, Cambridge University Press (LMM 189) See p. 215 - See David Marker's "Model theory. An introduction." Although argument is model theoretic, you can easily transform it into syntactic one. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 52, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9370397925376892, "perplexity_flag": "head"}
http://unapologetic.wordpress.com/2009/02/16/	# The Unapologetic Mathematician ## Generalized Eigenvectors Sorry for the delay, but exam time is upon us, or at least my college algebra class. Anyhow, we’ve established that distinct eigenvalues allow us to diagonalize a matrix, but repeated eigenvalues cause us problems. We need to generalize the concept of eigenvectors somewhat. First of all, since an eigenspace generalizes a kernel, let’s consider a situation where we repeat the eigenvalue ${0}$: $\displaystyle\begin{pmatrix}{0}&1\\{0}&{0}\end{pmatrix}$ This kills off the vector $\begin{pmatrix}1\\{0}\end{pmatrix}$ right away. But the vector $\begin{pmatrix}{0}\\1\end{pmatrix}$ gets sent to $\begin{pmatrix}1\\{0}\end{pmatrix}$, where it can be killed by a second application of the matrix. So while there may not be two independent eigenvectors with eigenvalue ${0}$, there can be another vector that is eventually killed off by repeated applications of the matrix. More generally, consider a strictly upper-triangular matrix, all of whose diagonal entries are zero as well: $\displaystyle\begin{pmatrix}{0}&&*\\&\ddots&\\{0}&&{0}\end{pmatrix}$ That is, $t_i^j=0$ for all $i\geq j$. What happens as we compose this matrix with itself? I say that for $T^2$ we’ll find the $(i,k)$ entry to be zero for all $i\geq {k}+1$. Indeed, we can calculate it as a sum of terms like $t_i^jtj^k$. For each of these factors to be nonzero we need $i\leq j-1$ and $j\leq k-1$. That is, $i\leq k-2$, or else the matrix entry must be zero. Similarly, every additional factor of $T$ pushes the nonzero matrix entries one step further from the diagonal, and eventually they must fall off the upper-right corner. That is, some power of $T$ must give the zero matrix. The vectors may not have been killed by the transformation $T$, so they may not all have been in the kernel, but they will all be in the kernel of some power of $T$. Similarly, let’s take a linear transformation $T$ and a vector $v$. If $v\in\mathrm{Ker}(T-\lambda1_V)$ we said that $v$ is an eigenvector of $T$ with eigenvalue $\lambda$. Now we’ll extend this by saying that if $v\in\mathrm{Ker}(T-\lambda1_V)^n$ for some $n$, then $v$ is a generalized eigenvector of $T$ with eigenvalue $\lambda$. Posted by John Armstrong \| Algebra, Linear Algebra \| 3 Comments ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 31, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9097293615341187, "perplexity_flag": "head"}
http://psychology.wikia.com/wiki/Noisy_channel_coding_theorem?oldid=13313	# Noisy channel coding theorem Talk0 31,730pages on this wiki Revision as of 06:33, May 31, 2006 by Lifeartist (Talk \| contribs) (diff) ← Older revision \| Latest revision (diff) \| Newer revision → (diff) Assessment \| Biopsychology \| Comparative \| Cognitive \| Developmental \| Language \| Individual differences \| Personality \| Philosophy \| Social \| Methods \| Statistics \| Clinical \| Educational \| Industrial \| Professional items \| World psychology \| Cognitive Psychology: Attention · Learning · Memory · Motivation · Perception · Thinking In information theory, the noisy-channel coding theorem establishes that however contaminated with noise interference a communication channel may be, it is possible to communicate digital data (information) error-free up to a given maximum rate through the channel. This surprising result, sometimes called the fundamental theorem of information theory, or just Shannon's theorem, was first presented by Claude Shannon in 1948. The Shannon limit or Shannon capacity of a communications channel is the theoretical maximum information transfer rate of the channel, for a particular noise level. ## Overview Proved by Claude Shannon in 1948, the theorem describes the maximum possible efficiency of error-correcting methods versus levels of noise interference and data corruption. The theory doesn't describe how to construct the error-correcting method, it only tells us how good the best possible method can be. Shannon's theorem has wide-ranging applications in both communications and data storage applications. This theorem is of foundational importance to the modern field of information theory. The Shannon theorem states that given a noisy channel with information capacity C and information transmitted at a rate R, then if $R < C \,$ there exists a coding technique which allows the probability of error at the receiver to be made arbitrarily small. This means that theoretically, it is possible to transmit information without error up to a limiting rate, C. The converse is also important. If $R > C \,$ an arbitrarily small probability of error is not achievable. So, information cannot be guaranteed to be transmitted reliably across a channel at rates beyond the channel capacity. The theorem does not address the rare situation in which rate and capacity are equal. Simple schemes such as "send the message 3 times and use at best 2 out of 3 voting scheme if the copies differ" are inefficient error-correction methods, unable to asymptotically guarantee that a block of data can be communicated free of error. Advanced techniques such as Reed-Solomon codes and, more recently, Turbo codes come much closer to reaching the theoretical Shannon limit, but at a cost of high computational complexity. With Turbo codes and the computing power in today's digital signal processors, it is now possible to reach within 1/10 of one decibel of the Shannon limit. ## Mathematical statement Theorem (Shannon, 1948): 1. For every discrete memoryless channel, the channel capacity $C = \max_{P_X} \,I(X;Y)$ has the following property. For any ε > 0 and R < C, for large enough N, there exists a code of length N and rate ≥ R and a decoding algorithm, such that the maximal probability of block error is ≤ ε. 2. If a probability of bit error pb is acceptable, rates up to R(pb) are achievable, where $R(p_b) = \frac{C}{1-H_2(p_b)} .$ and $H_2(p_b)$ is the binary entropy function $H_2(p_b)=- \left[ p_b \log {p_b} + (1-p_b) \log ({1-p_b}) \right]$ 3. For any pb, rates greater than R(pb) are not achievable. (MacKay (2003), p. 162; cf Gallager (1968), ch.5; Cover and Thomas (1991), p. 198; Shannon (1948) thm. 11) ## Outline of Proof This section is a stub. You can help by adding to it</span>. As with several other major results in information theory, the proof of the noisy channel coding theorem includes an achievability result and a matching converse result. These two components serve to bound, in this case, the set of possible rates at which one can communicate over a noisy channel, and matching serves to show that these bounds are tight bounds. The following outlines are only one set of many different styles available for study in information theory texts. #### Achievability for discrete memoryless channels This particular proof of achievability follows the style of proofs that make use of the Asymptotic equipartition property(AEP). Another style can be found in information theory texts using Error Exponents. Both types of proofs make use of a random coding argument where the codebook used across a channel is randomly constructed - this serves to reduce computational complexity while still proving the existence of a code satisfying a desired low probability of error at any data rate below the Channel capacity. By an AEP-related argument, given a channel, length n strings of source symbols $X_1^{n}$, and length n strings of channel outputs $Y_1^{n}$, we can define a jointly typical set by the following: $A_\epsilon^{(n)} = \{(x^n, y^n) \in \mathcal X^n \times \mathcal Y^n$ $2^{-n(H(X)+\epsilon)} \le p(X_1^n) \le 2^{-n(H(X) - \epsilon)}$ $2^{-n(H(Y) + \epsilon)} \le p(Y_1^n) \le 2^{-n(H(Y)-\epsilon)}$ $2^{-n(H(X,Y) + \epsilon)} \le p(X_1^n, Y_1^n) \le 2^{-n(H(X,Y) -\epsilon)} \}$ We say that two sequences $X_1^n$ and $Y_1^n$ are jointly typical if they lie in the jointly typical set defined above. Steps 1. In the style of the random coding argument, we randomly generate $2^{nR}$ codewords of length n from a probability distribution Q. 2. This code is revealed to the sender and receiver. It is also assumed both know the transition matrix $p(y\|x)$ for the channel being used. 3. A message W is chosen according to the uniform distribution on the set of codewords. That is, $Pr(W = w) = 2^{-nR}, w = 1, 2, ..., 2^{nR}$. 4. The message W is sent across the channel. 5. The receiver receives a sequence according to $P(y^n\|x^n(w))= \prod_{i = 1}^np(y_i\|x_i(w))$ 6. Sending these codewords across the channel, we receive $Y_1^n$, and decode to some source sequence if there exists exactly 1 codeword that is jointly typical with Y. If there are no jointly typical codewords, or if there are more than one, an error is declared. An error also occurs if a decoded codeword doesn't match the original codeword. This is called typical set decoding. The probability of error of this scheme is divided into two parts: 1. First, error can occur if no jointly typical X sequences are found for a received Y sequence 2. Second, error can occur if an incorrect X sequence is jointly typical with a received Y sequence. • By the randomness of the code construction, we can assume that the average probability of error averaged over all codes does not depend on the index sent. Thus, without loss of generality, we can assume W = 1. • From the joint AEP, we know that the probability that no jointly typical X exists goes to 0 as n grows large. We can bound this error probability by $\epsilon$. • Also from the joint AEP, we know the probability that a particular $X_1^n(i)$ and the $Y_1^n$ resulting from W = 1 are jointly typical is $\le 2^{-n(I(X;Y) - 3\epsilon)}$. Define: $E_i = \{(X_1^n(i), Y_1^n) \in A_\epsilon^{(n)}\}, i = 1, 2, ..., 2^{nR}$ as the event that message i is jointly typical with the sequence received when message 1 is sent. $P(error) = P(error\|W=1) \le P(E_1^c) + \sum_{i=2}^{2^{nR}}P(E_i)$ $\le \epsilon + 2^{-n(I(X;Y)-R-3\epsilon)}$ We can observe that as n goes to infinity, if $R < I(X;Y)$ for the channel, the probability of error will go to 0. Finally, given that the average codebook is shown to be "good," we know that there exists a codebook whose performance is better than the average, and so satisfies our need for arbitrarily low error probability communicating across the noisy channel. #### Converse for discrete memoryless channels Suppose a code of $2^{nR}$ codewords. Let W be drawn uniformly over this set as an index. Let $X^n$ and $Y^n$ be the codewords and received codewords, respectively. 1. $nR = H(W) = H(W\|Y^n) + I(W;Y^n)\;$ using identities involving entropy and mutual information 2. $\le H(W\|Y^n) + I(X^n(W);Y^n)$ since X is a function of W 3. $\le 1 + P_e^{(n)}nR + I(X^n(W);Y^n)$ by the use of Fano's Inequality 4. $\le 1 + P_e^{(n)}nR + nC$ by the fact that capacity is maximized mutual information. The result of these steps is that $P_e^{(n)} \ge 1 - \frac{1}{nR} - \frac{C}{R}$. As the block length n goes to infinity, we obtain $P_e^{(n)}$ is bounded away from 0 if R is greater than C - we can only get arbitrarily low rates of error if R is less than C. ## Channel coding theorem for non-stationary memoryless channels We assume that the channel is memoryless, but its transition probabilities change with time, in a fashion known at the transmitter as well as the receiver. Then the channel capacity is given by $C=\lim\;\inf\;\;\max_{p^(X_1),p^(X_2),...}\frac{1}{n}\sum_{i=1}^nI(X_i;Y_i)$ The maximum is attained at the capacity achieving distributions for each respective channel. That is, $C=\lim\;\inf\;\;\frac{1}{n}\sum_{i=1}^n C_i$ where $C_i$ is the capacity of the ith channel. ### Outline of the proof The proof runs through in almost the same way as that of channel coding theorem. Achievability follows from random coding with each symbol chosen randomly from the capacity achieving distribution for that particular channel. Typicality arguments use the definition of typical sets for non-stationary sources defined in Asymptotic Equipartition Property. The technicality of lim inf comes into play when $\frac{1}{n}\sum_{i=1}^n C_i$ does not converge. ## References • C. E. Shannon, The Mathematical Theory of Information. Urbana, IL:University of Illinois Press, 1949 (reprinted 1998). • David J. C. MacKay. Information Theory, Inference, and Learning Algorithms Cambridge: Cambridge University Press, 2003. ISBN 0521642981 • Thomas Cover, Joy Thomas, Elements of Information Theory. New York, NY:John Wiley & Sons, Inc., 1991. ISBN 0471062596 # Photos Add a Photo 6,465photos on this wiki • by Dr9855 2013-05-14T02:10:22Z • by PARANOiA 12 2013-05-11T19:25:04Z Posted in more... • by Addyrocker 2013-04-04T18:59:14Z • by Psymba 2013-03-24T20:27:47Z Posted in Mike Abrams • by Omaspiter 2013-03-14T09:55:55Z • by Omaspiter 2013-03-14T09:28:22Z • by Bigkellyna 2013-03-14T04:00:48Z Posted in User talk:Bigkellyna • by Preggo 2013-02-15T05:10:37Z • by Preggo 2013-02-15T05:10:17Z • by Preggo 2013-02-15T05:09:48Z • by Preggo 2013-02-15T05:09:35Z • See all photos See all photos >	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 40, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.900934100151062, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/106382?sort=oldest	## Finding an axis-aligned ellipsoid of minimal volume which contains a given ellipsoid ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) A friend asked me to post the following question. He's not an MO user and felt it would be better received if asked by someone who was already known to the community. This is not my area, but I'll do my best to answer questions raised in the comments. Also, please feel free to retag if you want. "Let $\Sigma$ and $\Lambda$ be $n \times n$ positive semi-definite matrices. We say that $\Lambda$ is "conservative" for $\Sigma$ if, for all $n$-dimensional vectors $x$, we have $x^\top \Sigma^{-1} x \leq x^\top \Lambda^{-1} x$. Our problem is given $\Sigma$, can we find (and if so how) a conservative diagonal matrix $\Lambda$ that minimizes $\det \Lambda$? To think about it geometrically, we have an ellipsoid defined by the positive semi-definite matrix $\Sigma$, and we want to bound it with an axis-aligned ellipsoid defined by the diagonal matrix $\Lambda$ that has minimal volume. I'd also be interested in knowing, for example, whether we can bound $\frac{\det \Lambda}{\det \Sigma}$ in terms of the spectrum of $\Sigma$. One trivial $\Lambda$ that works is $\lambda_n I$, which defines a sphere with radius equal to the largest eigenvalue of $\Sigma$. It's easy to see that this can be far too large than need, however, for example if $\Sigma$ is already diagonal (say, Diag $[1, \varepsilon, \varepsilon, \dotsc \varepsilon]$ for small $\varepsilon$)." EDIT: I finally found time to chat with my friend about the answers here. It turns out that there was a typo, but not the one people (myself included) guessed. The original question should have read $x^\top \Sigma^{-1} x \geq x^\top \Lambda^{-1} x$ in the second paragraph, i.e. find the smallest axis aligned ellipsoid defined by $\Lambda$ containing the one defined by $\Sigma$. I've left the original question in place to avoid confusion. Those answering thought the typo was that the goal is to minimize det $\Lambda^{-1}$ instead of det $\Lambda$, and based on this they solved the related problem of finding the largest axis aligned ellipsoid contained inside the ellipsoid defined by $\Sigma$. Anyway, the good news is that the answers here did help my friend with his problem (which is highly related to the problem the answers solved), so I'm upvoting everyone and accepting the answer which I think helps most. Thanks to everyone for comments and answers! - 4 If your friend wants an answer, he or she should post the question. – Bill Johnson Sep 5 at 0:54 ## 5 Answers If you can generate sufficiently many points $S$ to "delimit" the ellipsoid defined by $\Sigma$, then the results of this paper yield an approximation algorithm: • Kumar & Yildirim, "Computing Minimum Volume Enclosing Axis-Aligned Ellipsoids," Journal of Optimization Theory and Applications, 136 (2), pp. 211 - 228 (2008) [journal link]: Abstract: Given a set of points $S = {x^1,\ldots,x^m} \subset \mathbb{R}^n$ and $\epsilon > 0$, we propose and analyze an algorithm for the problem of computing a $(1 + \epsilon)$-approximation to the the minimum volume axis-aligned ellipsoid enclosing $S$. We establish that our algorithm is polynomial for fixed $\epsilon$. In addition, the algorithm returns a small core set $X \subseteq S$, whose size is independent of the number of points $m$, with the property that the minimum volume axis-aligned ellipsoid enclosing $X$ is a good approximation of the minimum volume axis-aligned ellipsoid enclosing $S$. Our computational results indicate that the algorithm exhibits significantly better performance than that indicated by the theoretical worst-case complexity result. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. So you want $$\forall x,\, x^t (\Lambda^{-1} - \Sigma^{-1}) x \geq 0$$ $(\Lambda^{-1} - \Sigma^{-1})$ is a symmetric matrix, and the condition expresses that it must be a positive matrix too. Let $T = \Sigma^{-1}$ and $M = \Lambda^{-1}$ we want to minimize $\det(M)$ with the constraint that $M-T$ is semi-definite positive. This constraint can be expressed by calculating the Cholesky decomposition of $M-T$, giving $n$ inequality constraints. At this point, I would suggest resorting to quadratic programming to solve the KKT system. (See for instance this). The Cholesky algorithm can be adapted to compute derivatives in the constraints. Edit: removed a system of equation that only relied on the (necessary) constraint $\det(M-T)\geq 0$ We can deduce a couple simple bounds. For instance, the non negativity of $M-T$ implies that all its element are themselves non negative, thus $\frac{1}{\lambda_i} \geq \Sigma^{-1}_{i,i}$ and $$\prod_i^{n} \Sigma^{-1}_{i,i} \leq \det(\Lambda^{-1}) \leq \hbox{Tr}(\Sigma^{-1})^n$$ - +1. Nice catch regarding that typo. – David White Sep 5 at 16:54 Yes, I imagine the type of bound in your edited question is more the sort my friend is looking for (I feel like I want to call him the original original poster, or OOP). How does this bound of yours compare to what he suggests in the final paragraph, i.e. the eigenvalue bound? – David White Sep 5 at 18:44 It's not as good (but simpler to calculate)... I also provide a bound in the other way – Arthur B Sep 5 at 19:14 In the way it is currently written, this problem does not have a solution. Probably minimizing $\det(\Lambda^{-1})$ was meant (as mentioned by Arthur.B in his comments below). Minimizing this is equivalent to minimizing $-\log\det(\Lambda)$ subject to the prescribed constraints. This should be doable easily using standard software. Reasoning for the original problem being not solvable. The reason is as follows. Since we are talking in terms of $\Lambda^{-1}$, we need the constraint $\Lambda \succ 0$. The other given constraint is $x^T\Sigma^{-1}x \le x^T\Lambda^{-1}x$, which is equivalent to $\Sigma^{-1} \preceq \Lambda^{-1}$, which in turn is equivalent to (for invertible matrices) $\Sigma \succeq \Lambda$. Now, you wish to minimize $\det \Lambda$ for a diagonal matrix $\Lambda$. We see that setting $\Lambda = \epsilon I$ (wlog), and letting $\epsilon \to 0$ (though always strictly positive) we minimize the determinant (because for a posdef matrix the determinant is always strictly positive), while easily satisfying the constraint $\Lambda \preceq \Sigma$. In other words, the infimum is $0$, but the minimum is not attained. - 1 why the negative vote? maybe you'd like to point out the flaw in the argument?? – S. Sra Sep 5 at 10:30 Because the condition is $\Sigma-\Lambda \geq 0$ and you're satisfying $\Sigma-\Lambda \leq 0$? You're suggesting that a point at the origin will be contained in the ellipsoid. True but irrelevant. – Arthur B Sep 5 at 11:28 @Arthur: The condition is $\Sigma^{-1} \le \Lambda^{-1}$, and since the map $X \mapsto X^{-1}$ is order-reversing, this constraint translates into $\Sigma \ge \Lambda$, which is the same as $\Lambda \le \Sigma$. And for $\epsilon$ sufficiently small, for every positive definite $\Sigma$, we can find a $\Lambda$ that is smaller than $\Sigma$. So where is the catch? – S. Sra Sep 5 at 12:26 2 Oh I see, the original poster wrote minimize $\det(\Lambda)$, it's clearly a typo, one want to minimize $\det(\Lambda^{-1})$ – Arthur B Sep 5 at 14:41 1 Reading over the question, and especially the example at the end I tend to agree that it was probably a typo. He was typing into an email browser, so probably didn't see the missing $^{-1}$ – David White Sep 5 at 16:49 show 3 more comments For the reasons discussed in the comments to Suvrit's answer, I will assume your friend would like to minimize $\det(\Lambda^{-1})$. The problem in question is a convex optimization problem: \begin{align} \text{minimize } & -\log(\det(\Lambda)) \\ \text{subject to } & 0 \preceq \Lambda \preceq \Sigma, \\ & \Lambda \text{ diagonal}. \end{align} Moreover, the objective function is the natural barrier functional used for interior point methods to enforce the semi-definiteness constraint on $\Lambda$. Therefore, although strictly speaking a semidefinite program (SDP) must have a linear objective, SDP solvers typically have a mode to solve problems like this. (They may also prefer to formulate the problem equivalently as the convex program maximizing $(\det\Lambda)^{1/n}$). In MATLAB, I would suggest using an SDP parser like CVX or Yalmip to call an SDP solver like SeDuMi or SDPT3 to solve this problem numerically to your desired level of precision. To see how many variants of this problem can also be posed as SDPs, see Section 2 of Boyd and Vandenberghe's survey. - This isn't my field, so I hope this isn't a stupid comment, but knowing my friend I imagine he's interested in a theoretical solution which would hold in more generality. My guess, knowing his work, is that he needs this bound as a part of a larger paper where he's trying to bound something, perhaps asymptotically in $n$. Can SDP solvers do that sort of thing or are they just for specific problems with actual numerical inputs? – David White Sep 5 at 16:53 You are correct that then this method is only applicable to actual numerical instances. As such it wouldn't be of much help as far as symbolic or asymptotic results, except perhaps for forming and testing conjectures about what to try to prove. – Noah Stein Sep 5 at 17:31 I added a simple bound to my answer, is this the kind of result he's looking for? – Arthur B Sep 5 at 17:47 Suppose your "ellipsoid" is a line segment from $(1, 1, \dots, 1)$ to $(-1, -1, \dots, -1).$ Any "axis-aligned ellipsoid" must contain a cube of volume $2^n,$ and pretty obviously (although I haven't checked) must contain a ball or radius $1.$ while the maximal semiaxis of your ellipsoid is $\sqrt{n},$ and all the other semiaxes are of length $0.$ The volume is, of course, also $0.$ This means that your "stupid" algorithm is not so stupid -- the maximal semi axis in some sense controls the size of the smallest axis-aligned ellipsoid, up to a $\sqrt{n}$ multiplicative factor. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 86, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9517155289649963, "perplexity_flag": "head"}
http://cms.math.ca/Events/summer12/abs/pmp	2012 CMS Summer Meeting Regina Inn and Ramada Hotels (Regina~Saskatchewan), June 2 - 4, 2012 www.cms.math.ca//Events/summer12 Perspectives in Mathematical Physics Org: Yvan Saint-Aubin and Luc Vinet (Montréal) [PDF] FERENC BALOGH, Concordia University and CRM Reduction of planar orthogonality to non-hermitian orthogonality on contours [PDF] The asymptotic behaviour of orthogonal polynomials associated with measures supported in the whole complex plane is of essential importance in obtaining various limiting spectral statistics for certain non-hermitian random matrix models. The standard tools to analyse the asymptotics for orthogonal polynomials on the real line or on a complex contour, including the celebrated Riemann-Hilbert method, are not readily avaliable in the planar setting. For certain special measures, however, the two-dimensional hermitian orthogonality relations can be shown to be equivalent to a set of non-hermitian orthogonality relations with respect to an analytic weight function integrated on a combination of contours in the complex plane. This reduction amounts to solving a family of d-bar problems and, in certain cases, it simplifies the asymptotic analysis considerably since it allows the the Riemann-Hilbert approach to obtain strong asymptotics for the corresponding orthogonal polynomials. The method will be illustrated on an example based on a joint work with M. Bertola, S.Y. Lee and K. McLaughlin. VINCENT BOUCHARD, University of Alberta The (un)reasonable effectiveness of string theory in mathematics [PDF] In recent decades, an impressive number of fascinating results (many of them still conjectural) in various areas of mathematics, such as geometry, topology and number theory, have been obtained via string theory and its dualities. In this talk, I will focus on a new mysterious recursive structure that appears to unify seemingly unrelated counting problems in geometry, with far-reaching and mostly unexplored consequences. We conjectured this new structure through a careful study of topological string theory and mirror symmetry, putting to work "the (un)reasonable effectiveness of string theory in mathematics". TIAGO DINIS DA FONSECA, Centre de recherches mathématiques Higher spin 6-Vertex model and Macdonald polynomials [PDF] It is known that the $6$-Vertex model is a quantum integrable model, therefore we know, at least in theory, everything about it. For example, in the case of Domain Wall Boundary Conditions, the partition function is a relatively simple determinant (Izergin, 1987) and it is related to a Schur polynomial. In a more recent work, Caradoc, Foda and Kitanine (2006) tell us how to generalize this result for higher spins. Based in their work, one can prove that the new partition function is related to a Macdonald polynomial (dF and Balogh, to appear). In this talk, I will describe the 6-vertex model, explain how to create the higher spin model from the original model. And finally, I will sketch how one can prove that this is indeed a Macdonald polynomial. PATRICK DESROSIERS, IMAFI, Universidad de Talca Beta-ensembles of random matrices [PDF] Most models in Random Matrix Theory are solved by using techniques that depend on their symmetry properties. In the last two decades, however, general families of matrix models that provide a unifying framework for random matrices have been developed; they are called the beta-ensembles. In this talk, I will briefly introduce these ensembles and review recent advances. The presentation will focus on the interrelationships between beta-ensembles, stochastic differential equations, and remarkable special functions in many variables, such as Jack polynomials. DAVID FEDER, University of Calgary Graph Theory in Quantum Many-Body Physics [PDF] The Hamiltonian for bosonic and fermionic particles hopping on lattices can be interpreted as the adjacency matrix of an undirected and generally weighted graph. The properties of these quantum many-body systems can therefore be analyzed in terms of graph theory. For example, the simple graph for non-interacting distinguishable particles is the Cartesian product of each particle's adjacency matrix; if these particles become indistinguishable, the graph collapses via a graph equitable partition. In the presence of strong interactions between the particles, the graphs are generally decomposable as weak products (i.e. they are the Kronecker products of adjacency matrices). Under various circumstances, these techniques can allow for the efficient calculation of the eigenstates (and therefore the properties) of physically interesting quantum many-body systems. DANIEL GOTTESMAN, Perimeter Institute for Theoretical Physics Quantum Error Correction [PDF] As logic circuits become smaller and smaller, it makes sense to consider what happens once they become so small they start to behave according to quantum mechanics. What we get is a quantum computer, for which the state can be described as a vector in a finite-dimensional Hilbert space. Quantum computers offer the possibility of some dramatic computational speedups, the most famous being Shor's factoring algorithm. However, quantum states are very delicate, and we won't be able to realize the benefits of quantum computation without a way to correct errors. I will describe how to create quantum error-correcting codes which can protect even very large quantum states against noise and decoherence. The key to understanding a quantum error-correcting code lies in the stabilizer, a finite group encapsulating many of the symmetries of the code. GABOR KUNSTATTER, University of Winnipeg Quantum mechanics on the discretized half-line [PDF] We investigate nonrelativistic quantum mechanics on the discretized half-line, constructing a one-parameter family of Hamiltonians that are analogous to the Robin family of boundary conditions in continuum half-line quantum mechanics. For classically singular Hamiltonians, the construction provides a singularity avoidance mechanism that has qualitative similarities with singularity avoidance encountered in loop quantum gravity. Applications include the free particle, the attractive Coulomb potential, the scale invariant potential and a black hole described in terms of the Einstein-Rosen wormhole throat. The spectrum is analyzed by analytic and numerical techniques. In the continuum limit, the full Robin family of boundary conditions can be recovered via a suitable fine-tuning but the Dirichlet-type boundary condition emerges as generic. RAYMOND LAFLAMME, Institute for Quantum Computing, UWaterloo Quantum Error Correction: from theory to practice [PDF] The Achilles' heel of quantum information processors is the fragility of quantum states and processes. Without a method to control imperfection and imprecision of quantum devices, the probability that a quantum computation succeed will decrease exponentially in the number of gates it requires. In the last fifteen years, building on the discovery of quantum error correction, accuracy threshold theorems were proved showing that errors can be controlled using a reasonable amount of resources as long as the error rate is smaller than a certain threshold. We thus have a scalable theory describing how to control quantum systems. I will describe a variety of mathematical techniques that have been developed to turn this theorem into a useful tool in the laboratory and will sum up with a quick overview of where we are at controlling quantum systems in practice. JOSH LAPAN, McGill University Black Hole Greybody Factors and Monodromies [PDF] Black hole thermal radiation has captivated part of the theoretical physics community since its discovery by Hawking. In light of holography, deep connections have emerged between details of the black hole and a holographically dual field theory. Recent observations demonstrate that for a large class of black holes --- as well as five-dimensional black rings and black strings --- the product of inner and outer horizon areas is independent of mass; a simple connection with left- and right-moving temperatures of a dual field theory is proposed here. This directly connects to the observation that inner horizons play crucial roles in computing reflection and transmission coefficients, despite the fact that the scattering problem is set up without regard to behavior at the inner horizon. Drawing on ongoing work, we will suggest a new interpretation for how to understand the role of inner horizons in scattering problems; as a byproduct, this may lead to a different way of computing greybody factors from the classic method of matched asymptotic expansions. SHUNJI MATSUURA, McGill University Protected boundary states in gapless topological phases [PDF] I will talk about gapless topological phases of (semi-)metals and nodal superconductors. Using both K-theory and dimensional reduction procedures, a classification of topologically stable Fermi surfaces in (semi-)metals and nodal lines in superconductors is derived. We discuss a generalized bulk-boundary correspondence that relates the topological features of the Fermi surfaces and superconducting nodal lines to the presence of protected zero-energy states at the boundary of the system. STEFAN MENDEZ-DIEZ, University of Alberta Elliptic curves, KR-theory and T-duality [PDF] There are several variants of string theory. We will explore how the different string theories compactified on an elliptic curve are related via T-duality by studying the classification of elliptic curves with involution through the topological lens of KR-theory. MARCO MERKLI, Mathematics, Memorial University Repeated Interaction Quantum Systems [PDF] Consider a quantum system interacting sequentially in time, one by one, with a chain of infinitely many independent, identical other quantum systems. One may think of a scattering experiment. The entire system is an open system, due to the infinite size of the chain. We give an overview of recent results on the dynamics of such repeated interaction quantum systems. Among them are the existence and the construction of an asymptotic state and its thermodynamic properties, the influence of possible randomness, and an analysis of multi-time quantum measurements performed on chain elements after interaction. The results are based on collaborations with L. Bruneau and A. Joye, and with M. Penney. ROBERT MILSON, Dalhousie University A Conjecture on Exceptional Orthogonal Polynomials [PDF] Exceptional orthogonal polynomials (so named because they span a non-standard polynomial flag) are defined as polynomial eigenfunctions of Sturm-Liouville problems. By allowing for the possibility that the resulting sequence of polynomial degrees admits a number of gaps, we extend the classical families of Hermite, Laguerre and Jacobi. In recent years the role of the Darboux (or the factorization) transformation has been recognized as essential in the theory of orthogonal polynomials spanning a non-standard flag. In this talk we present the conjecture that ALL such polynomial systems are derived as multi-step factorizations of classical operators and offer some supporting evidence. ALEXI MORIN-DUCHESNE, Université de Montréal Periodic loop models and extended XXZ Hamiltonians [PDF] Non hermitian Hamiltonians play an important role in the description of two dimensional statistical models such as the Fortuin-Kasteleyn model and the $Q$-Potts spin model. The loop Hamiltonians, as elements of the periodic Temperley-Lieb algebra $TLP_N(\beta)$, are examples of such Hamiltonians: their eigenvalues are real and they are not diagonalizable for specific values of the parameter $\beta$. Loop Hamiltonians are known to be related to XXZ Hamiltonians and a great deal can be learned from this correspondence. In my talk, I will introduce the twist'' representations of the periodic Temperley-Lieb algebra and show how one can study the Jordan structure of the loop Hamiltonian in these representations, using tools from the XXZ models. MANU PARANJAPE, Université de Montréal Solitons and instantons in an effective model of CP violation [PDF] CP violation in the standard model is generated in the weak interactions and the CKM mass matrix. However, asymptotic states involved in such processes almost invariably involve hadronic mesonic asymptotic states, for example in recent experiments decays of B mesons to 2 K mesons and 2 $\pi$ mesons are most important. Thus it should be possible to describe CP violation entirely in terms of scalar fields. We, however, study a 1+1 dimensional analog of a 3+1 dimensional model describing CP violating decays entirely in terms of effective scalar fields. Although the equations of motion are non-linear, we find exact soliton and instanton solutions. The solitons are of the Q-ball type and represent particle states in the quantum theory while the instantons have finite action and should mediate tunnelling transitions in the theory. We speculate as to the 3+1 dimensional analogs of the exact solutions that we have found. SARAH POST, Centre de Recherches Mathematiques, Univ. de Montreal Recent advances in the theory of superintegrable systems [PDF] In this talk, I will discuss new results in the theory of superintegrable systems: Hamiltonian systems with more integrals of motion than degrees of freedom. The talk will consist of a brief survey of known results along with a focus on recent advances. In particular on the relationship between such systems and orthogonal polynomials. Topics to be discussed: new infinite families of superintegrable systems and recurrence relations for orthogonal polynomials, representations of symmetry algebras and the Askey scheme, and superintegrable spin chains. WALID ABOU SALEM, University of Saskatchewan Adiabatic evolution of coupled surface and internal waves [PDF] I discuss the dynamics of interacting surface and internal water waves over a slowly varying random bottom. Signature of such waves has been observed in pictures from the space station. The motion of the interacting waves is described by a system of coupled Schroedinger-Korteweg-de Vries equations. In the presence of a slowly varying random bottom, the coupled waves evolve adiabatically over a long time scale. The analysis covers the cases when the surface wave is a stable bound state or a long-lived metastable state. ## Event Sponsors Support from these sponsors is gratefully acknowledged.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8884389400482178, "perplexity_flag": "head"}
http://physics.aps.org/story/print/v2/st19	# Focus: Nobel Focus: Chemistry by Computer Published October 21, 1998 \| Phys. Rev. Focus 2, 19 (1998) \| DOI: 10.1103/PhysRevFocus.2.19 The 1998 Nobel Prize in chemistry recognizes two researchers whose work has allowed chemists to calculate the properties of molecules and solids on computers, without performing experiments in the lab. The basic principles of the calculation scheme were first described in Physical Review in the 1960s, and solid state physicists used them for decades before they became important in the chemistry world. The scheme drastically simplifies the solution of the quantum mechanical equations for a system of many electrons, and although approximate, the solutions are accurate enough that chemists can learn about large molecules without getting their hands wet. All of the chemical properties of molecules and electrical properties of solids are determined by electrons interacting with each other and with atomic nuclei. By 1930 physicists were fully aware of the quantum mechanical equations governing systems of many electrons, but were incapable of exactly solving them in all but the simplest cases. They developed several approximation schemes, but none was very successful. In 1964 Walter Kohn, now of the University of California at Santa Barbara, and Pierre Hohenberg, now of Yale University, proved an idea that was central to their solution scheme, which is now called density functional theory (DFT). They showed that knowing the average density of electrons at all points in space is enough to uniquely determine the total energy, and therefore all of the other properties of the system. Other available methods at the time required wave functions, which depend on the positions of every electron in the system and are far more complicated functions than the density. Kohn’s 1965 publication, with Lu Sham of the University of California at San Diego, described a procedure for deriving the electron density and energy, based on solving equations for a corresponding system of noninteracting electrons, which is much easier to manage. The trickiest part of the scheme is the starting point: One has to assume a relationship (called the “density functional”) between energy and the density function. The explicit form of the density depends on the problem, but Hohenberg and Kohn proved in the earlier paper that in principle there exists a universal density functional, good for all problems. Kohn and Sham provided a simple approximation of this relationship, and it has served solid state physics for decades, allowing for most modern calculations of the “band structure” of electrons in solids, for example. But the approximate density functional used by Kohn and Sham was not accurate enough to calculate chemical bond energies and structures of molecules, so experimental chemists had little use for it. Beginning in the late 1980s, however, physicists began modifying the Kohn-Sham density functional in ways that improved its accuracy. In 1992 John Pople, of Northwestern University in Evanston, IL, added DFT–including the latest functionals–to his widely-used chemistry computer program GAUSSIAN. The program was already recognized as the premier tool for studying structures and chemical reactions, at least for small molecules, but with the addition of DFT, chemists were immediately impressed with its accuracy and speed. Since then biochemists, drug designers, atmospheric chemists, and even astrophysicists have benefited from studying complex molecules and reactions by computer. Kohn and Pople received the Nobel Prize in chemistry for their contributions to computational chemistry, which is now used routinely as a tool by scientists in many fields. Kieron Burke, of Rutgers University in Camden, NJ, says Kohn’s DFT was a brilliant simplification of the quantum mechanical mathematics. “This is one of the greatest free lunches ever,” he says, because it saves computer time (allowing for calculations with larger molecules) and requires in exchange only that the user provide a single, approximate density functional. Announcement and background from the Nobel Foundation ### Highlighted article #### Self-Consistent Equations Including Exchange and Correlation Effects W. Kohn and L. J. Sham Published November 15, 1965 #### Inhomogeneous Electron Gas P. Hohenberg and W. Kohn Published November 9, 1964 ### Figures ISSN 1943-2879. Use of the American Physical Society websites and journals implies that the user has read and agrees to our Terms and Conditions and any applicable Subscription Agreement.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9528812766075134, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/83932/enumerating-all-non-planar-embeddings-of-a-generic-connected-graph-in-the-plane	## Enumerating all non-planar embeddings of a generic connected graph in the plane ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Hi There I've been tackling a problem in computer science, which relies quite heavily on graph theoretic concepts. I'd appreciate it if anybody could provide insight into how to enumerate all non-planar embeddings of a generic connected graph in the plane. Assume we are given an undirected connected graph $G=(V,E)$ with straight edges, $n$ nodes, and a symmetric distance matrix $D$ of size $n \times n$ that weights the edges in $E$. The weight is simply the straight line Euclidean distance between the two vertices in the corresponding edge. The 2D Graph Realization problem seeks to find a non-planar (edges may cross) embedding of $V$ in the plane, such that the measured distances are "as close to" the observed distances $D$. In general, this problem is NP-Hard. However, if we ignore degenerate cases in which there is an algebraic relation between vertices (three or more vertices are collinear) the problem becomes tractable. From here on we will assume generality. Depending on the number and configuration of edges that are known, the embedding may not be unique. The certification problem seeks to determine whether it is impossible to find a unique embedding, given the edges that are known. One is able to certify the uniqueness of an embedding in the plane using graph rigidity theory. (Jackson and Jordan,2005) proved that a graph has a unique embedding in the plane if and only if it is redundantly rigid and triconnected. The first condition relies on a concept known as Generic Rigidity, and it can be tested with the 2D Pebble Game proposed by (Hendrikson, 1992). The second condition can be tested by building a SPQR tree (Hopcraft and Tarjan, 1973). (Mutzel and Weiskircher, 1999) showed how to enumerate all combinatorial planar embeddings of a biconnected graph using an SPQR-tree. This is the closest result that I can find to my problem. However, my problem differs in the following two ways: 1. A planar graph has the distinct property that edges only intersect at the vertices. I'm looking to find a way of optimizing over combinatorial non-planar embeddings of a graph. 2. Their model assumes that the graph is biconnected. I'd prefer to drop this assumption in favour of connectedness. So far, I've thought of the following crude algorithm to achieve what I'm looking for. It begins by using the 2D pebble game to find all generically rigid clusters in a given graph. These clusters are rigid against small continuous forces, but may move relative to one another. I then parameterise the K degrees of freedom with real angles $r_1 \ldots r_K$. For clusters with greater than 4 nodes I have to also parameterise the possible reflections. In order to do this I obtain an SPQR tree for each cluster, assuming that generic rigidity implies biconnectivity (a requirement of the SPQR tree data structure). The P-nodes in the SPQR tree indicate where reflection is possible, and I encode all L possible reflections with binary variables $b_1 \ldots b_L$. The space of possible embeddings of the graph is given by $b_1 \ldots b_L$ and $r_1 \ldots r_K$. I the perform some optimization algorithm over this search space. Does this appear to be a reasonable approach, or am I making any problematic assumptions? Any comments or extra information would help me greatly! - 3 I am a little confused about what you're assuming about the input. In particular, it is generically rigid or not? If not (e.g., assuming only connectivity), then this isn't really an enumeration question. If yes, you'll still have to deal with the case of $3$-connected and minimally rigid, which can have exponentially many embeddings (see Borcea and Streinu's paper "The Number of Embeddings of Minimally Rigid Graphs"). – Louis Theran Dec 20 2011 at 11:50 Hi Louis Thank you for your comment and for correctly pointing out that this problem is not an enumeration problem. I have removed the incorrect tag accordingly. Thank you for the paper reference, which is exactly what I was looking for. I had wondered how to deal with the case of a 3-connected component that is not redundantly rigid. Thank you. Regards Andrew – Andrew Symington Dec 20 2011 at 12:34 ## 1 Answer As I mentioned in my comment, there are two kinds of problems here: • If the graph is generically rigid, this is an enumeration problem, and quite a hard one. • If the graph is generically flexible, then this is a question of the configuration space of the framework. For the rigid case, I'm not aware of a better exact answer than doing something based on Groebner basis computations, which will, in general, be very slow. I assume that some formal negative results about framework realizability (e.g., Owen and Power http://www.ams.org/journals/tran/2007-359-05/S0002-9947-06-04049-9/) translate to enumeration if you're given one realization, but I can't think of a reference. There is, however, a lot of heuristic work in this area. For example, Meera Sitharam has a method based on hierarchical decompositions that's in the spirit of what you want (http://www.cise.ufl.edu/~sitharam/skeleton.pdf). It seems like you're not interested in realizability only, but if the graph is globally rigid and you can take multiple measurements, there is a nice method due to Zhu, Gortler, and Thurston (http://dl.acm.org/citation.cfm?id=1921630). -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9286216497421265, "perplexity_flag": "head"}
http://crypto.stackexchange.com/questions/2683/is-this-a-secure-multiparty-protocol?answertab=active	# Is this a secure multiparty protocol? I am facing the following situation and i am trying to break down the problem into some specific cryptogrphic primitives. There is a function F that takes as input a bit string and produces a fingerprint.In my protocol there is a user $u$ who holds the input $D$ and $n$ other parties. I want the parties to compute $F(D)$ and all the parties to agree on that value. For example if $t$ out of $n$ where $t$ > $n/2$ agree on value $Y_1$ then the winner is $Y_1$. If there is no majority i want the protocol to run again. I want also secrecy at the results. I do not want the parties to learn the results and i want the parties to learn anything about the given input $D$. They will compute $F(D)$ without learning D. So they will receive an encrypted $E(D)$ and they will evalute $F(E(D))$ but only to evaluate the function. So my question is: Can we say that this is a secure multiparty computation ? Or it is a verifiable secret sharing? - Wait... you say you "don't care about the confidentiality of inputs", but if I'm reading your description right, anyone who knows the function $F$ (which I'd normally assume to be public knowledge unless specified otherwise) and the input $D$ can obviously compute the output $F(D)$. So I must be missing something, but what? – Ilmari Karonen May 23 '12 at 20:13 my fault. I correct it. We do care about inputs. There are secret. i update the question with sth else. I want operation on encrypted data.Sorry i am thinking while i am writing – curious May 23 '12 at 20:19 1 "I do not want the parties to learn the results and i want the parties to learn anything about the given $\hspace{.2 in}$ input $D$." $\:$ How does either part of that sentence make sense? $\;\;$ – Ricky Demer May 24 '12 at 0:11	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 21, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9244534373283386, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/173025/question-about-continuous-function-in-terms-of-limits-of-sequences	# Question about continuous function in terms of limits of sequences I am reading about continuous function, in this site http://en.wikipedia.org/wiki/Continuous_function specifically the section "Definition in terms of limits of sequences". My question is, let be $c\in \mathbb{R}$ an arbitrary element belonging to domain of $f$, where $f$ is in a closet and bounded set. Are there always a sequence $x_n$ which converges to $c$?. - 1 Your question doesn't make much sense. You were asking about continuity and then you switched to closed sets. If $f$ a function or a subset? – Asaf Karagila Jul 19 '12 at 22:30 1 What about $x_n = c$ for all $n$? – Dylan Moreland Jul 19 '12 at 22:42 2 Why is $f$ in the closet? I thought we as a society were past that... – Quinn Culver Jul 19 '12 at 23:24 ## 1 Answer There is always the constant sequence $x_n = c$ for all $n$. In the extreme case that the domain of the function is $\{c\}$, there isn't anything else. In general there is a dichotomy to be made according to whether $c$ is an isolated point of the domain $X$ of $f$. (Let's assume that $X$ is a subset of the real numbers $\mathbb{R}$, which seems to be the context of the question. In fact, with only mild notational change, this answer works in the context of arbitrary metric spaces.) We say that a point $c \in X$ is isolated if there is a $\delta > 0$ such that if $y \in X$ and $\|x-y\| < \delta$, then $y= c$. For instance, if $f(x) = \sqrt{x^2(x-1)}$, the natural domain is $\{0\} \cup [1,\infty)$ and $0$ is an isolated point. Now, for $X \subset \mathbb{R}$ and $c \in X$, the following are equivalent: (i) $c$ is an isolated point of $X$. (ii) Every sequence $\{x_n\}$ in $X$ which converges to $c$ is eventually constant: that is, $x_n = c$ for all sufficiently large $n$. (iii) Every function $f: X \rightarrow \mathbb{R}$ is continuous at $c$. In other words, an isolated point is something of a trivial case. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 36, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9516674876213074, "perplexity_flag": "head"}
http://gilkalai.wordpress.com/2009/02/24/a-little-more-on-boolean-functions-and-bounded-depth-circuits/?like=1&source=post_flair&_wpnonce=836876f408	Gil Kalai’s blog ## A Little More on Boolean Functions and Bounded Depth Circuits Posted on February 24, 2009 by Boolean circuits This post (in a few parts) contains a quick introduction to Boolean circuits. It is related to the recent news post about the solution of Braverman to the Linial-Nisan conjecture. In particular, we will describe very quickly a formulation of the $NP \ne P$ problem and discuss some issues about bounded depth circuits. I hope that this gives a nice introduction to this area for non-experts (written also by a non-expert). Any comments and corrections (especially from experts!) are welcome. I want to mention, in particular, a result by Benjamin Rossman about finding cliques in graphs for bounded depth circuits that I found exciting. ## A. Boolean functions, Boolean circuits and the NP versus P problem. ### 1. Boolean functions A Boolean function of n variables is simply a function $f(x_1,x_2,\dots,x_n)$ where the variables $x_k$ take the values +1 and -1 and the value of $f$ itself is also either +1 and -1. ### 2. Boolean circuits A Boolean circuit is a gadget that computes Boolean functions. It is built from inputs, gates and an output. We can think about these circuits as follows: On level 0 there are the variables. On level 1 there are gates acting on the variables. On level 2 there are gates acting on the outputs of the gates on level 1. And in level $r$ is a single gate leading to the output of the circuit. The depth of the circuit is this number $r$. The size of the circuit is the total number of gates. The gates perform Boolean operations: They can take an input bit and negate it. They can take several input bits and take their OR – this means that the output will be ’1′ iff one of them is ’1′. They can take several input bits and take their AND – this means that the output will be ’-1′ iff one of them is ‘-1′. ### 3. the $NP \ne P$ problem Consider a graph $G$ on $v$ vertices. The question “Does $G$ have a Hamiltonian cycle?” (a Hamiltonian cycle is a simple cycle containing all the vertices of the graph) is known to be “NP complete”. Now, suppose that $n= {{k} \choose {2}}$ and associate to every edge of the complete graph on $k$ vertices a variable $x_k$. Every graph $G$ corresponds to an assignment of Boolean values to the variables. This correspondence is crucial here and in various other places: the variables associated to the edges of the graph get value ´1´and the variables associated to edges not in the graph get value ´-1´. The property of graphs on $k$ vertices to be Hamiltonian is described by the following Boolean function $f_n$ on $n$ variables. Every assigenment of values to the variables corresponds to a graph $G$, and the value of $f_n$ is 1 if and only if the graph $G$ contains a Hamiltonian cycle. To prove that $NP \ne P$ just prove that $f_n$ cannot be described by a Boolean circuit of size $s(n)$ which is bounded above by a polynomial in $n$. Is it really the famous $P \ne NP$ problem, you may ask? Well, what we have stated here is a little bit stronger. So if you prove that the Boolean functions representing the graph property “To contain a Hamiltonian cycle” cannot be described by a polynomial size circuit you will prove that $NP \ne P$ and you will have a little change to spare! (The assertion that there is no polynomial time algorithm for deciding if a graph with $n$ vertices is Hamiltonian is equivalent to the $NP \ne P$ problem. In the setting of circuits, we talk about the size of the circuit as a substitute to the notion of running time of an algorithm.) ### 4. A little more on $NP \ne P$. What is NP? The previous paragraph tells you a formulation of the problem. So you can go ahead and solve it and not worry too much about this paragraph. But the previous paragraph does not tell a few important aspects of the problems and I will mention one aspect now. NP stands for “non-deterministically polynomial” (and not for “not-polynomial” as I thought in my early youth). What does it mean? The Wikipedea explanation is “a problem is NP as long as a given solution can be verified as correct in polynomial time”. In the world of circuits we have the following description: Suppose you want to express the property that a graph has a Hamiltonian cycle in the following way: You have a Boolean circuit with $n+m$ variables. The $n$ variables correspond to the edges of the complete graph as before. The $m$ variables are new variables $y_1,y_2,\dots, y_m$ and we assume that $m$ is bounded from above by a polynomial function of $n$. Now we want to describe the property ”The graph G is Hamiltonian” by a polynomial size circuit in these $n+m$ variables as follows: If we start with a Hamiltonian graph, we can find an assignment to the variables $y_1,y_2,\dots, y_m$ such that the output of the circuit will be 1. If we start with a non-Hamiltonian graph for every assignment of values to the variables $y_1,y_2, \dots, y_m$ the output of the circuit will be -1. Now come the crucial observation: We can do it! We can find a circuit with this property. As before, we let $x_1,x_2,\dots, x_n$ be variables representing the edges of the complete graph with the understanding that $x_i=1$ if and only if the $i$th edge is present in our graph $G$. , But now we take $n$ more variables $y_1,y_2,\dots,y_n$ again representing all possible edges of the complete graph. The $y_i$s which equal to 1 are supposed to correspond to edges of a Hamiltonian cycle in our graph $G$. So now our circuit in the $x_i$s and the $y_i$s will represent the property that the graph $H$ represented by the $y_i$s is a Hamiltonian subgraph of the graph $G$ represented by the $x_i$s. This can easily be done. This is (more or less) what is meant by saying that the property “to be Hamiltonian” is in NP. So to say that the graph property “G contains a Hamiltonian cycle” is in NP roughly means that it is possible to prove (or to verify) using a polynomial time algorithm or using a polynomial-size circuit that the graph has a Hamiltonian cycle. The proof consists of presenting a Hamiltonian cycle in the graph. (Again, the circuit notion and the polynomial-time notions are not exactly the same, but let´s ignore it.) The property “The graph does not have a Hamiltonian cycle” is not known to be in NP and indeed is believed not to be in NP. 5. But what is so special about Hamiltonian cycles. The remarkable thing about the Hamiltonian cycle property is that it is a complete problem in NP. A polynomial time algorithm for deciding if a graph is Hamiltonian will give a polynomial algorithm for every problem in NP. (In other words, any problem in NP can be reduced to the problem of deciding if a graph has a Hamiltonian cycle.) There are many, many problems with the remarkable property that they are complete in NP, so there is nothing very special about the problem we have chosen. ### 6. Reductions If somebody asks you to describe in one word what theoretical computer science is about, a good word to choose is “reductions”. The art of reductions, the science of reductions, the practice and engineering of reductions, and the philosophy of reductions. ### 7. More general circuits. We can talk about various other circuits. We can have more sophisticated gates and not just Boolean gates. We can let the variables be real numbers, or elements in some other field, and talk about algebraic gates and algebraic circuits, etc. ### 8. Asymptotically speaking The notions we have described are asymptotic. We consider a problem (like the graph property of containing a Hamiltonian cycle) and we study its computational complexity when the number of variables goes to infinity. ## B. Bounded depth circuits Bounded depth circuits are circuits where the number of levels (the depth) is bounded by a constant. We try to understand the computational power of circuits where $r$ is bounded, the number of variables $n$ goes to infinity, and the number of overall gates is polynomial in $n$. Much can be said about the computational power of bounded depth circuits. Beyond this class things are much harder. Bounded depth circuits are also fascinating mathematical objects. —to be continued— ### Like this: This entry was posted in Computer Science and Optimization. Bookmark the permalink. ### 5 Responses to A Little More on Boolean Functions and Bounded Depth Circuits 1. aravind says: Hi Gil, I look forward to your take on Rossman’s proof. (He has a shorter, informal version of the proof on his homepage which I think I understand.) Thanks. 2. Gil says: Dear Aravind, I will look at the homepage. I dont promise to explain Rossman’s proof (would you?) but I will explain that I was surprised to learn (and this may also refer to an old reasult by Paul Beam that I did not know) that the Hastad switching lemma is capable to distinguish between polynomial size bounded depth circuits and quasi polynomial size bounded (even larger) depth circuits. (This may have been clear to experts but I did not know of it and I still don’t know it.) 3. Gil Kalai says: (Two remarks after talking with Avi) I tried to give in the post a very clear and simple mathematical statement which describes the $NP \ne P$ problem. (Or in fact a “slightly” stronger statement.) If you want to think about problems which are not as strong as $NP \ne P$ but still very very very hard, you can replace “Boolean circuits” by the simpler and familiar notion of “a formula”: Proving that the property “The graph contains no Hamiltonian cycle” cannot be described by a polynomial size Boolean formula in the variables is much beyond reach. Indeed, showing that certain functions that can be described by a quasi-polynomial circuits of small depth (even depth two) cannot be described by a polynomial-size bounded-depth circuits go back even to the work of Furst, Saxe, and Sipser. I am not aware of a simple property of a Boolean function that manifests such a separation. 4. Gil Kalai says: Some further remarks about the connection between algorithms complexity and circuit complexity can be found here: http://rjlipton.wordpress.com/2009/02/12/is-np-too-big-or-p-too-small/ 5. Pingback: Noise Stability and Threshold Circuits « Combinatorics and more • ### Blogroll %d bloggers like this:	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 57, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9133583307266235, "perplexity_flag": "head"}
http://en.wikisource.org/wiki/On_the_Theory_of_Radiation_of_Moving_Bodies	# On the Theory of Radiation of Moving Bodies From Wikisource On the Theory of Radiation of Moving Bodies (1904) by Friedrich Hasenöhrl, translated by Wikisource In German: Zur Theorie der Strahlung bewegter Körper, Wiener Sitzungsberichte, 113 IIa: 1039-1055 On the Theory of Radiation of Moving Bodies Friedrich Hasenöhrl Wikisource 1904 On the Theory of Radiation of Moving Bodies. by Dr. Fritz Hasenöhrl. (With 2 text figures.) (Presented in the meeting on June 23, 1904). If a radiating surface moves with uniform velocity in the same direction as the radiation emitted by it, then work must be constantly applied to overcome the pressure exerted by the latter. Now, this work is transformed into radiation energy as well, so that more energy of that amount is emanated from a surface moving in this sense, than from a stationary one. However, if the surface is moving in the opposite direction of the emanated radiation, then the latter steadily performs work, thus less energy of that amount is emanated from the moving surface than from a stationary one. On the other hand, if an absorbing surface is moving, so that it recedes from the incident radiation, then the latter steadily performs work; the surface is therefore only capable of absorbing – i.e. transforming into inner heat – less radiation by the equivalent of the latter. However, if the absorbing surface is approaching the incident radiation, then work from outside must be steadily performed against the pressure of radiation. This work can only occur as heat in the absorbing surface. The effect of motion is thus in this case, that the surface is absorbing more heat by the amount of performed work. These modes of conception were already spoken out by different authors.[1] In the following I've tried now, to apply these theorems upon the radiation in a moving cavity. Except the radiation provided by the walls, also radiation energy must be present within it, which is gained from mechanical work and which is transformed into such work again. It is essentially determined by the motion of the cavity; its amount is, as it will be shown, proportional to the square of the system's velocity (in first approximation), thus it apparently increases the kinetic energy of the system. Circumstances are thus present, that are quite analogous to the motion of an electron. In the same way as the concept of "electromagnetic mass" is introduced there, one could also speak about an "apparent mass" here, which is caused by radiation. In the same way, as the electromagnetic mass is proportional to the static energy of the stationary electron, also the apparent mass caused by radiation is proportional to the energy content of the stationary cavity. Namely, the proportionality factor is in both cases of the same order of magnitude. Now, since the heat content of every body partly consists of radiation energy, then every body must possess such an apparent mass depending on the temperature, and which is added to the mass in the ordinary sense. 1. Now, we consider a cylindric cavity $R$, moving with constant velocity $c$ in the direction of the arrow (Fig. 1). Let $\mathfrak{B}$ be the velocity of light and $\sigma=\tfrac{c}{\mathfrak{B}}$. Furthermore, let $A$ and $B$ be two black surfaces. The shell surface of the cavity as well as the outer boundary of the black body shall be formed by surfaces which are perfectly reflecting into the interior. Let the cross-section of space $R$ be equal to 1, its height be equal to $D$. Let the outer space be completely free of radiation, thus at absolute temperature zero, while a certain temperature shall be attributed to surfaces $A$ and $B$. Now we have to distinguish between absolute and relative direction of radiation;[2] it is more convenient, to base our consideration upon the latter. Let $2\pi\ i_{0}\ \cos\phi\ \sin\phi\ d\phi\,$ be the energy quantity, emanated by $A$ in the unit of time into the relative direction between $\phi$ and $\phi+d\phi$, where $\phi$ is thus the angle enclosed by the relative beam direction with the normal (and with the direction of velocity $c$). $i_{1}$ then must be a constant (with respect to $\phi$, as I already have emphasized in an earlier work.[3]) This radiation now exerts a pressure upon $A$, whose component that coincides with the direction of the normal (in the sense of $-c$), shall have the value $2\pi\ p_{1}\ \cos\phi\ \sin\phi\ d\phi\,$ If we multiply this expression with $c$, then we obtain the work performed against this pressure in one second from the outside, which is now also transformed into radiation, so that the total radiation leaving $A$ in the given direction, has the value $2\pi\ (i_{1}+p_{1}c)\ \cos\phi\ \sin\phi\ d\phi = 2\pi\ i\ \cos\phi\ sin\phi\ d\phi$ This expression now gives us also the amount of the radiation falling upon $B$ in the unit of time. There, one fraction of it is absorbed, and one fraction is transformed into work. If $B$ would be at the absolute temperature zero, then the radiation just considered would be the only one present in $R$. Since in this case, no force of resistance against the motion of our system is to be expected, then the same pressure into opposite directions must be effective in $A$ and $B$, so that no work is performed in the whole. Thus in $B$, the energy quantity $2\pi\ i_{0}\ \cos\phi\ \sin\phi\ d\phi\,$ is absorbed, and the energy quantity $2\pi\ p_{1}\ \cos\phi\ \sin\phi\ d\phi\cdot c\,$ is transformed into mechanical work. If we now imagine that $B$ has the same temperature as $A$, then also $B$ emanates energy $2\pi\ i_{0}\ \cos\phi\ \sin\phi\ d\phi\,$ in a certain direction. If this radiation exerts the pressure $2\pi p_{2}\cos\phi\ \sin\phi\ d\phi\,$, then it performs work by which amount the energy provided by $B$ has to be diminished, so that also $B$ is left by the radiation quantity $2\pi(i_{0}-p_{2}c)\cos\phi\ \sin\phi\ d\phi=2\pi i'\cos\phi\ \sin\phi\ d\phi\,$ The same energy quantity also occurs in $A$, and a quite similar consideration as earlier teaches, that the energy quantity $2\pi i_{0}\cos\phi\ \sin\phi\ d\phi$ is absorbed there, because the work $2\pi p_{2}\cos\phi\ \sin\phi\ d\phi\,$ must also be performed against the incident radiation from outside, which is also transformed into work. Thus the same quantity is absorbed by both bodies $A$ and $B$, as it is emitted; upon both surfaces (in opposite direction) the same pressure $2\pi(p_{1}+p_{2})\cos\phi\ \sin\phi\ d\phi,\,$ is exerted, so that no work is performed altogether. Thus when no external force is acting, then the system maintains its velocity. In case one of the bodies $A$ or $B$ is replaced by a perfect mirror, the radiation condition in $R$ must remain exactly the same,[4] thus besides other things, $2\pi(p_{1}+p_{2})\cos\phi\ \sin\phi\ d\phi,\,$ is the pressure upon the mirror as well. We ask ourselves about the energy content of space $R$. The radiation traveling from $A$ to $B$, has the relative velocity:[5] $\mathfrak{B}(-\sigma\cos\phi+\sqrt{1-\sigma^{2}\sin^{2}\phi}=\mathfrak{B}_{-}.$ The radiation coming from $A$ to $B$, has the velocity $\mathfrak{B}(+\sigma\cos\phi+\sqrt{1-\sigma^{2}\sin^{2}\phi}=\mathfrak{B}_{+}.$ both have to travel the path $\tfrac{D}{\cos\phi}$, thus the total energy content of space $R$ is $\overset{\pi/2}{\int_{0}}\frac{D}{\cos\phi}\left[\frac{2\pi(i_{0}+p_{1}c)\cos\phi\sin\phi\ d\phi}{\mathfrak{B}_{-}}+\frac{2\pi(i_{0}-p_{2}c)\cos\phi\sin\phi\ d\phi}{\mathfrak{B}_{+}}\right]$ $=2\pi i_{0}D\int_{0}^{\pi/2}\sin\phi\ d\phi\left(\frac{1}{\mathfrak{B}_{-}}+\frac{1}{\mathfrak{B}_{+}}\right)+$ $+2\pi cD\int_{0}^{\pi/2}\sin\phi\ d\phi\left(\frac{p_{1}}{\mathfrak{B}_{-}}-\frac{p_{2}}{\mathfrak{B}_{+}}\right).$ (1) The first of the two summands gives the part of the energy of space $R$, which was provided by the black bodies. My paper already cited dealt with it. The second summand, however, represents the radiation energy gained from mechanical work. Since no work is performed at uniform motion as we have seen, this must be a work which must be performed when our system is accelerated. One can easily determine the amount of this energy quantity in the following way. Let us imagine a system in absolute rest, and the radiation of both black surfaces $A$ and $B$ is somehow hindered: space $R$ is thus totally free of radiation. Now, in one instant the radiation from $A$ and $B$ shall be left free, and the system shall simultaneously brought to velocity $c$. From that instant on, the work $2\pi p_{1}\cos\phi\ \sin\phi\ d\phi\cdot c$ in the unit of time must be performed against the radiation emitted from $A$. Now, time $\tfrac{D}{\mathfrak{B}_{-}\cos\phi}$ is passing until this radiation arrives in $B$, and there it provides the same quantity of work itself. During this time, the work provided from outside is thus uncompensated. Also the radiation emitted from $B$ performs right from the start the work $2\pi p_{2}\cos\phi\ \sin\phi\ d\phi\cdot c$, and now the time $\tfrac{D}{\mathfrak{B}_{+}\cos\phi}$ is passing until this radiation arrives in $A$, and there the same work from outside must be performed in the opposite sense. Thus, more work was performed from the outside by $\frac{D}{\cos\phi\cdot\mathfrak{B}_{-}}2\pi p_{1}\cos\phi\sin\phi\ d\phi\cdot c\frac{D}{\cos\phi\cdot\mathfrak{B}_{+}}2\pi p_{2}\cos\phi\sin\phi\ d\phi\cdot c$ i.e., it was gained from the radiation. If we integrate this amount over all beam directions, then we obtain the expression above. If our system is at rest in the beginning, yet the surfaces $A$ and $B$ are not hindered of emanating radiation, then radiation energy of certain amount is present at the beginning also in $A$. The previous consideration indicates, that this energy is completely absorbed and not transformed into work. Namely this can also be directly shown, when one introduces the values for $p_{1}$ and $p_{2}$, which we will give in the following section. In order of not interrupting the progression of the investigation, I transfer the proof of this theorem into § 4 of this treatise. Thus in expression (1), the first summand is the energy provided by the radiating bodies, the second one is the energy stemming from work. We denote the latter by $L$, thus we set $L=2\pi cD\int_{0}^{\pi/2}\sin\phi\ d\phi\left(\frac{p_{1}}{\mathfrak{B}_{-}}-\frac{p_{2}}{\mathfrak{B}_{+}}\right).$ (2) 2. We now want to introduce the values for magnitudes $p_{1}$ and $p_{2}$, which we preliminarily left undetermined. It was surely introduced at first by Abraham in his treatises already cited. Namely, these values have been derived from Lorentz's theory of electromagnetism. Since one can arrive at the same expressions in another way (see § 3 of the present work), their correctness seem to be even more plausible. According to Abraham, the radiation pressure upon a surface is numerically equal to the incident or emanated radiation divided by the speed of light, namely this pressure acts in the absolute direction of radiation. Let $\varphi$ be the angle, which is enclosed by the absolute direction of the radiation (emanated under the relative angle $\phi$) with the direction of $c$. Then the magnitude denoted earlier as $p_1$, is $p_{1}=\frac{1}{\mathfrak{B}}i\ \cos\varphi,$ since we understood under $p_1$ the – solely effective – normal pressure component. A relation between $\varphi$ and $\phi$ can be easily obtained by the following figure 2, which is essentially identical with Fig. 1 of my cited treatise, and which surely needs nor further explanation. It is $\mathfrak{B}_{-}=\sqrt{\mathfrak{B}^{2}+c^{2}-2\mathfrak{B}c\ \cos\ \varphi}=\mathfrak{B}\sqrt{1+\sigma^{2}-2\sigma\ \cos\ \varphi}=$ $=\mathfrak{B}\left(-\sigma\ \cos\phi+\sqrt{1-\sigma^{2}\sin^{2}\ \phi}\right),$ from which the relations $\cos\phi=\frac{1-\sigma\ \cos\varphi}{\sqrt{1+\sigma^{2}-2\sigma\ \cos\ \varphi}}$ and $\cos\ \varphi=\sigma\sin^{2}\phi+\cos\phi\sqrt{1-\sigma^{2}\sin^{2}\ \phi}$ (3) are given. Thus it is $p_{1}=\frac{i}{\mathfrak{B}}\left(\sigma\sin^{2}\phi+\cos\phi\sqrt{1-\sigma^{2}\sin^{2}\phi}\right);$ (4) if we insert this into equation $i=i_{0}+p_{1}c$, then it follows $i=i_{0}\frac{\mathfrak{B}}{\mathfrak{B}_{-}\sqrt{1-\sigma^{2}\sin^{2}\phi}}$ (5) and $p_{1}=i_{0}\frac{\sigma\sin^{2}\phi+\cos\phi\sqrt{1-\sigma^{2}\sin^{2}\phi}}{\mathfrak{B}_{-}\sqrt{1-\sigma^{2}\sin^{2}\phi}}=$ $=i_{0}\frac{\cos\phi+\sigma\sqrt{1-\sigma^{2}\sin^{2}\phi}}{\mathfrak{B}(1-\sigma^{2})\sqrt{1-\sigma^{2}\sin^{2}\phi}}.$ (6) Quite similar it is given: $i'=i_{0}\frac{\mathfrak{B}}{\mathfrak{B}_{+}\sqrt{1-\sigma^{2}\sin^{2}\phi}}$ (7) and $p_{2}=\frac{i'}{B}\left(-\sigma\sin^{2}\phi+\cos\phi\sqrt{1-\sigma^{2}\sin^{2}\phi}\right)$ (8) or $p_{2}=i_{0}\frac{-\sigma\sin^{2}\phi+\cos\phi\sqrt{1-\sigma^{2}\sin^{2}\phi}}{\mathfrak{B}_{+}\sqrt{1-\sigma^{2}\sin^{2}\phi}}=$ $=i_{0}\frac{\cos\phi-\sigma\sqrt{1-\sigma^{2}\sin^{2}\phi}}{\mathfrak{B}(1-\sigma^{2})\sqrt{1-\sigma^{2}\sin^{2}\phi}}.$ (9) If we insert this into (2), then: $L=2\pi cDi_{0}\cdot\int_{0}^{\pi/2}\frac{\sin\phi\ d\phi}{\mathfrak{B}(1-\sigma^{2})\sqrt{1-\sigma^{2}\sin^{2}\phi}}$ $\left[\cos\ \phi\left(\frac{1}{\mathfrak{B}_{-}}-\frac{1}{\mathfrak{B}_{+}}\right)+\sigma\sqrt{1-\sigma^{2}\sin^{2}\phi}\left(\frac{1}{\mathfrak{B}_{-}}+\frac{1}{\mathfrak{B}_{+}}\right)\right]=$ $=\frac{4\pi cDi_{0}\sigma}{\mathfrak{B}^{2}(1-\sigma^{2})^{2}}\int_{0}^{\pi/2}\sin\phi\ d\phi\frac{1+\cos^{2}\phi-\sigma^{2}\sin^{2}\phi}{\sqrt{1-\sigma^{2}\sin^{2}\phi}}$ If we additionally put the energy content of the resting cavity $R$: $\frac{4\pi i_{0}D}{\mathfrak{B}}=E_{0},$ then after execution of the integration $L=E_{0}\frac{c^{2}}{\mathfrak{B}^{2}}\frac{1}{(1-\sigma^{2})^{2}}\left[\frac{1}{2}+\frac{1}{2\sigma^{2}}-\frac{(1-\sigma^{2})^{2}}{4\sigma^{3}}\log\frac{1+\sigma}{1-\sigma}\right].$ If we neglect herein magnitudes of order $\sigma^3$, then it becomes $L=E_{0}\frac{c^{2}}{B^{2}}\cdot\frac{4}{3}.$ This expression has now the form and the dimension of a kinetic energy. Thus one can say, that the kinetic energy of our system was apparently increased by $L$, or that to the mechanical mass of our system, also an apparent mass $\mu=\frac{8}{3}\frac{E_{0}}{\mathfrak{B}^{2}}$ has been added. Namely, the introduction of that mass is quite similar to that of the electromagnetic mass. If we (for the moment) denote the energy of a resting electron with $\epsilon_{0}$, then the electromagnetic mass of it is equal to $\tfrac{4}{3}\tfrac{\epsilon_{0}}{\mathfrak{B}^{2}}$ or $\tfrac{8}{5}\tfrac{\epsilon_{0}}{\mathfrak{B}^{2}}$,[6] depending on whether one deals with surface or volume charge. The relation is thus the same in terms of order of magnitude. Also the exact expressions have a certain similarity, since the quantity $\log\tfrac{1+\sigma}{1-\sigma}$ plays a role in both of them. 3. In this section, I still want to make same remarks concerning the value of the radiation pressure. The total pressure upon one of the surfaces $A$ or $B$ – it is irrelevant as to whether it is imagined as black of reflecting – has the value $\begin{array}{ll} P & =2\pi(p_{1}+p_{2})\cos\phi\ \sin\phi\ d\phi\\ \\ & =4\pi i_{0}\frac{\cos^{2}\phi\ \sin\phi\ d\phi}{\mathfrak{B}(1-\sigma^{2})\sqrt{1-\sigma^{2}\sin^{2}\phi}}.\end{array}$ Herein, we want to introduce the radiation density $\rho$, which for example falls upon $B$. It is $\rho=\frac{2\pi i\ \sin\phi\ d\phi}{\mathfrak{B}_{-}}=\frac{2\pi i_{0}\ \sin\phi\ d\phi\cdot\mathfrak{B}}{\mathfrak{B}_{-}^{2}\sqrt{1-\sigma^{2}\sin^{2}\phi}}$ (this is given from (5)). Thus it is $P=2\rho\frac{\mathfrak{B}_{-}^{2}\cos^{2}\phi}{\mathfrak{B}^{2}(1-\sigma^{2})}.$ If we want to introduce herein the absolute beam direction again, then we use the relation immediately given from Fig. 2 $\mathfrak{B}_{-}cos\phi=\mathfrak{B}\ cos\varphi-c=\mathfrak{B}(\cos\ \varphi-\sigma).$ And it becomes: $P=2\rho\frac{(\cos\ \varphi-\sigma)^{2}}{1-\sigma^{2}},$ an expression already given by Abraham[7] as well. Now, by generalization of the thought that was first spoken out by Larmor[8], the same result can be derived from the standpoint of the elastic theory of light, which I would like to show soon. We consider a light source, moving under the (absolute) angle $i$ against the $X$-axis. It is given by $\zeta=A\cos\ m(x\ \cos\ i+y\ \sin\ i+\mathfrak{B}t).$ If this wave falls upon a mirror lying perpendicular to the $X$-axis, then a reflected wave is formed, which is given by $\zeta'=A'\cos\ m'(x\ \cos\ r-y\ \sin\ r-\mathfrak{B}t).$ Herein, $r$ is the angle of reflection. At the surface of the mirror it must be $\zeta+\zeta'=0$. If it is moving with velocity $c$ in the direction of the normal, i.e., in the direction of the positive $X$-axis, then it must be $x=ct,\ \zeta+\zeta'=0$ This gives $A\ cos\ m(t(\mathfrak{B}+c\ \cos\ i)+y\ \sin\ i]+$ $+A'\cos\ m'[t(c\ \cos\ r-\mathfrak{B})-y\ \sin\ r]=0.$ This equation can only then be satisfied for all values of $y$ and $t$, when $\begin{array}{rl} A= & -A',\\ m(\mathfrak{B}+c\ \cos\ i)= & m'(\mathfrak{B}-c\ \cos\ r),\\ m\ \sin i= & m'\ \sin r.),\end{array}.$ These equations give us the law of reflection $\frac{\sin\ i}{1+\sigma\ \cos\ i}=\frac{\sin\ r}{1-\sigma\ \cos\ r}$ (10) and the Doppler effect $\frac{m'}{m}=\frac{1+\sigma\ \cos\ i}{1-\sigma\ \cos\ r}$ in full agreement with Abraham. Incidentally, equation (10) is also easily given from equation (12) of my earlier treatise.[9] Now, to derive the value of the pressure from it, we have to assume that the energy density of a light wave is proportional (at equal amplitude) to the -2. power of the wave length, thus to the quantity $m^2$. Thus when we denote by $\rho$ the density of the incident wave, by $\rho'$ that of the reflecting one, then it is $\frac{\rho'}{\rho}=\left(\frac{m'}{m}\right)^{2}=\left(\frac{1+\sigma\ \cos\ i}{1-\sigma\ \cos\ r}\right)^{2}.$ Now, the energy falling upon the mirror in unit time, is contained in space $\mathfrak{B}\ cos\ i+c$; the reflected energy in space $\mathfrak{B}\ cos\ r-c$. Thus the amount the incident energy per second is: $\rho(\mathfrak{B}\ \cos\ i+c)=\rho\mathfrak{B}(\cos\ i+\sigma)$ and the amount of the energy reflected in the same time $\rho'(\mathfrak{B}\ \cos\ r-c)=\rho'\mathfrak{B}(\cos\ r-\sigma).$ The difference of these two expressions must be equal to the work of the radiation pressure per unit time. Thus it must be $Pc=\rho'\mathfrak{B}(\cos\ r-\sigma)-\rho\mathfrak{B}(\cos\ i+\sigma)$ Or $cP=\mathfrak{B}\rho(\cos\ i+\sigma)\left[\left(\frac{1+\sigma\cos\ i}{1-\sigma\cos\ r}\right)^{2}\frac{\cos\ r-\sigma}{\cos\ i+\sigma}-1\right].$ With the aid of the easily derivable relation, already given by Abraham:[10] $\frac{1+\sigma\cos\ i}{1-\sigma\cos\ r}=\frac{\sigma+\cos\ i}{\sigma-\cos\ r}=\frac{\sin\ i}{\sin\ r}$ and the equation:[11] $\sin\ r=\frac{\sin\ i(1-\sigma^{2})}{1+\sigma^{2}+2\sigma\ \cos\ i}$ it becomes $cP=\mathfrak{B}\rho(\cos\ i+\sigma)\left[\frac{1+\sigma^{2}+2\sigma\ \cos\ i}{\ i-\sigma^{2}}-1\right].$ and $P=2\rho\frac{(\cos\ i+\sigma)^{2}}{1-\sigma^{2}}.$ The agreement with the value given by Abraham is thus a complete one (here, $\sigma$ has the opposite sign as earlier.) Of course, upon this foundation one can calculate the values $p_{1}$ and $p_{2}$, when one assumes that a moving body is emanating waves, whose amplitude is the same as in the stationary state, while the density of the radiation energy is inversely proportional to the square of the wavelength. Incidentally, the values of $p_{1}$ and $p_{2}$ have been derived by Poynting[12] by a peculiar consideration. However, Poynting confines himself to the case of perpendicular emission. If one accordingly put $\cos\varphi = \cos\phi = 1$ in the expressions given here, then the agreement is complete. 4. Now, we have to provide the proof of the assertion made in § 1. Our system given by Fig. 1 shall be at rest at the beginning. Then the energy quantity $E_{0}=\frac{4\pi i_{0}}{\mathfrak{B}}\cdot D.$ is in space $R$. Now it is the question, what is happening with the energy, when the system is suddenly brought to velocity $c$. We can assume from the outset, that only a fraction of it becomes absorbed, while another fraction is transformed into mechanical work. Now we have to notice in particular, that this energy is uniformly distributed into all absolute directions. Thus when we maintain our earlier mode of expression, then the density of energy whose absolute direction of propagation encloses an angle between $\phi\,$ and $\phi+d\phi\,$ with the normal, is: $\frac{2\pi i_{0}}{\mathfrak{B}}\sin\varphi\ d\varphi.$ As soon as the system is in motion, it is about the distribution in terms of the relative propagation direction. Then the density of energy, whose relative propagation direction encloses an angle between $\phi\,$ and $\phi+d\phi\,$ with the normal, is evidently: $\frac{2\pi i_{0}}{\mathfrak{B}}\sin\varphi\ \frac{d\varphi}{d\phi}d\phi=-\frac{2\pi i_{0}}{\mathfrak{B}}\ \frac{d\ \cos\varphi}{d\phi}d\phi.$ This becomes by differentiation of equation (3): $=\frac{2\pi i_{0}}{\mathfrak{B}}\sin\phi\ d\phi\frac{\left(\sqrt{1-\sigma^{2}\sin^{2}\phi}-\sigma\ \cos\phi\right)^{2}}{\sqrt{1-\sigma^{2}\sin^{2}\phi}}=$ $=\frac{2\pi i_{0}}{\mathfrak{B}}\frac{\mathfrak{B}_{-}^{2}}{\mathfrak{B}^{2}\sqrt{1-\sigma^{2}\sin^{2}\phi}}\sin\phi\ d\phi$ Since this energy is located in a cylinder of height $D$, then the energy quantity falling upon $B$ in this direction is: $\frac{2\pi i_{0}D}{\mathfrak{B}^{3}}\cdot\frac{\mathfrak{B}_{-}^{2}}{\sqrt{1-\sigma^{2}\sin^{2}\phi}}\sin\phi\ d\phi$ Previously (§ 1) we saw, that from the incident radiation $2\pi\ i\ cos\phi\ sin\phi\ d\phi$, the fraction $2\pi\ i_{0}\ \cos\phi \sin\phi\ d\phi$ is absorbed, while the fraction $2\pi\ p_{1}\ \cos\phi \sin\phi d\phi$ is transformed into work. Thus from the totality of the incident energy quantity, the fraction $\tfrac{i_{0}}{i}$ is absorbed and the fraction $\tfrac{p_{1}c}{i}$ is transformed into work. Thus from the just incident energy quantity, the fraction $\frac{2\pi i_{0}D}{\mathfrak{B}^{3}}\cdot\frac{\mathfrak{B}_{-}^{2}}{\sqrt{1-\sigma^{2}\sin^{2}\phi}}\sin\phi\ d\phi\cdot\frac{i_{0}}{i}$ is absorbed; the fraction $\frac{2\pi i_{0}D}{\mathfrak{B}^{3}}\cdot\frac{\mathfrak{B}_{-}^{2}}{\sqrt{1-\sigma^{2}\sin^{2}\phi}}\sin\phi\ d\phi\cdot\frac{p_{1}c}{i}$ is transformed into work. In a similar way one recognizes, that the energy quantity $\frac{2\pi i_{0}D}{\mathfrak{B}^{3}}\cdot\frac{\mathfrak{B}_{+}^{2}}{\sqrt{1-\sigma^{2}\sin^{2}\phi}}\sin\phi\ d\phi$ falls upon $A$, and that we have to multiply this energy quantity with $\frac{i_{0}}{i}$ and $-\frac{p_{2}c}{i}$ respectively, to obtain the fraction of them which becomes absorbed or transformed into work. Altogether, the energy quantity $\int_{0}^{\pi/2}\frac{2\pi i_{0}D}{\mathfrak{B}^{3}}\frac{\sin\phi\ d\phi}{\sqrt{1-\sigma^{2}\sin^{2}\phi}}\left(\mathfrak{B}_{-}^{2}\frac{i_{0}}{i}+\mathfrak{B}_{+}^{2}\frac{i_{0}}{i'}\right)=M$ is absorbed, while the energy quantity $\int_{0}^{\pi/2}\frac{2\pi i_{0}D}{\mathfrak{B}^{3}}\frac{\sin\phi\ d\phi}{\sqrt{1-\sigma^{2}\sin^{2}\phi}}\left(\mathfrak{B}_{-}^{2}\frac{p_{1}c}{i}-\mathfrak{B}_{+}^{2}\frac{p_{2}c}{i'}\right)=N$ is transformed into work. One can convince oneself, that exactly the same expressions are valid, when for example the black surface $B$ is replaced by a mirror. Namely, the radiation coming from $A$ to $B$ is not absorbed in $B$, but is partly reflected, i.e., from the totality of the energy incident in $B$, the fraction $\tfrac{i'}{i}$ is reflected; it comes back to $A$, and the fraction $\frac{i_{0}}{i'}$ is absorbed there, so that eventually the fraction $\tfrac{i'}{i}\cdot\tfrac{i_{0}}{i'}=\tfrac{i_{0}}{i}$ of the radiation propagating at the beginning in the direction from $A$ to $B$, is absorbed again. Thus $M$ and therefore also $N$ remain unchanged. Now, if one uses equations (4), (5), (7) and (8), then $M=\frac{2\pi i_{0}D}{\mathfrak{B}^{4}}\int_{0}^{\pi/2}\sin\phi\ d\phi(\mathfrak{B}_{-}^{3}+\mathfrak{B}_{+}^{3})$ and $N=\frac{2\pi i_{0}D}{\mathfrak{B}^{3}}\int_{0}^{\pi/2}\frac{\sin\phi\ d\phi}{\sqrt{1-\sigma^{2}\sin^{2}\phi}}$ $\cdot\left[\sigma^{2}\sin^{2}\phi(\mathfrak{B}_{-}^{2}+\mathfrak{B}_{+}^{2})+\sigma\cos\phi\sqrt{1-\sigma^{2}\sin^{2}\phi}(\mathfrak{B}_{-}^{2}-\mathfrak{B}_{+}^{2})\right]$ By execution of this integration one can convince oneself, that it is quite exactly $M=\frac{4\pi i_{0}D}{\mathfrak{B}}=E_{0};\ N=0;$ Thus the total energy initially present in $R$, is absorbed: Thus the earlier assertions concerning the work necessary for the acceleration, are confirmed. On the other hand we also see, that from the energy content of the cavity, i.e. in expression (1), only the first part is provided by the heat content of the radiating body. This is an assumption, at which I started in my cited treatise concerning the dimension changes of matter due to the motion through the aether[13], and which was perhaps not sufficiently justified at that place. 1. See J. H. Poynting, Phil. Trans. 202, p. 525, 1904; Vl. v. Türin, Ann. d. Naturphil., III, p. 270, 1904; M. Abraham, Boltzmann-Festschrift, p. 84, and Drude, Ann. XIV, p. 236, 1904. 2. See about that, F. Hasenöhrl, these proceedings, CXIII., p. 469; — M. Abraham, l. c. 3. L. c. p. 474. 4. See J. Larmor, Boltzmann-Festschrift, p. 595, 1904. 5. See F. Hasenöhrl, l. c. 6. See M. Abraham, Drude's Ann., X, p. 151, 1903. 7. Boltzmann-Festschrift, p. 91, eq. 9. 8. Larmor. Report of British Association, 1900. 9. These proceedings., CXIII, p. 489. 10. L. c. eq. 7e. 11. See F. Hasenöhrl, l. c. p. 489, eq. 12. 12. L. c. p. 551. 13. L. c. p. 474, footnote. This is a translation and has a separate copyright status from the original text. The license for the translation applies to this edition only. Original: This work is in the public domain in the United States because it was published before January 1, 1923. The author died in 1915, so this work is also in the public domain in countries and areas where the copyright term is the author's life plus 80 years or less. This work may also be in the public domain in countries and areas with longer native copyright terms that apply the rule of the shorter term to foreign works. Translation: This work is released under the Creative Commons Attribution-ShareAlike 3.0 Unported license, which allows free use, distribution, and creation of derivatives, so long as the license is unchanged and clearly noted, and the original author is attributed.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 213, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9428515434265137, "perplexity_flag": "head"}
http://leftcensored.skepsi.net/	Where to begin… Pearson’s r: Not a good measure of electoral persistence Posted on December 30, 2012 by Pearson’s product-moment correlation, $$r$$, is an incredibly useful tool for getting some idea about how two variables are (linearly) related. But there are times when using Pearson’s $$r$$ is not appropriate and, even if linearity and all other assumptions hold, using it may lead one astray. I recently ran into such a situation when looking at the persistence of municipal-level vote shares over the last four Polish parliamentary elections. The plot below presents an example of some of the data I was looking at. It shows the share of the vote going to Samoobrona (Self-Defense of the Republic of Poland) by municipality in 2005 (horizontal axis) and 2007 (vertical axis). The linear regression line is shown in red and the 45-degree line is the dashed grey line. From the plot it is clear that support for SRP collapsed between 2005 and 2007: had SRP’s performance held steady across the two elections, we would expect the data to cluster around the 45-degree line. Instead, in all but one or two municipalities, support was lower in 2007 than 2005. Accordingly, the data fall below the 45-degree line. However, if I had ignored the plot and used Pearson’s $$r$$ as the primary measure of persistence in vote share across years — something that is not uncommon in the study party politics — I would have been lead to believe that there was, in fact, a high degree of persistence between the two elections. As reported in the plot, $$r=0.72$$, which certainly seems to be good evidence of a high degree of persistence in vote shares between 2005 and 2007. So, what’s going on? Jason Wittenberg explains in a recent paper why Pearson’s $$r$$ is not an appropriate measure in this case. The problem is the fact that Pearson’s $$r$$ is a measure of linearity, not persistence — it captures how far from a linear regression line the data fall (the red line in the above plot). But what we really want is a measure of how far the data are from the 45-degree line. For this, Wittenberg suggests using Lin’s concordance correlation coefficient, $$r_c$$. From the Wikipedia article, Lin’s $$r_c$$ is defined as follows: \( r_c = \cfrac{2s_{xy}}{s_x^2 + s_y^2 + (\bar{x} – \bar{y})^2}. \) The relationship between $$r$$ and $$r_c$$ is straightforward. From the original paper: 1. $$-1 \leq -\|r\| \leq r_c \leq \|r\| \leq 1$$; 2. $$r_c = 0$$ if and only if $$r=0$$; 3. $$r_c = r$$ if and only if $$s_x = s_y$$; and 4. $$r_c = \pm 1$$ if and only if $$r = \pm 1$$, $$s_x = s_y$$, and $$\bar{x} = \bar{y}$$. (These come from Lin’s 1989 paper, p. 258, cited below. Note I have left out a couple equivalent conditions from (iv) and have changed the notation to match the definition of $$r_c$$.) As Wittenberg notes, condition (i) is potentially quite problematic for studies of electoral persistence. It says that the magnitude of $$r_c$$ is always less than or equal to $$r$$. In other words, Pearson’s $$r$$ (almost) always inflates the amount of persistence present in the data. Evidence of this is seen the in next plot, which compares electoral performance of each major party in the 2005 and 2007 Polish parliamentary elections. Clearly, Pearson’s $$r$$ and Lin’s $$r_c$$ point to significantly different interpretations of the data when it comes to the issue of electoral persistence. As expected the magnitude of $$r_c$$ is always lower than $$r$$ — sometimes dramatically lower. In the case of SRP, for instance, $$r_c = 0.06$$ compared to $$r = 0.72$$ reported above. Also note, as expected given the conditions above, $$r_c$$ is closest to $$r$$ when regression line is near a 45-degree line. So, what’s the take-away? First, Wittenberg is right in calling into question the use of Pearson’s $$r$$ as a measure of electoral persistence (anyone know a better measure?). It simply doesn’t properly capture the concept. Second, always look at your data. In this case, I had never heard of Lin’s concordance correlation coefficient until I went looking for some way to measure the discordance I was seeing in the data. Had I just decided construct a table of correlations, I may have believed that there was a great deal more persistence in electoral votes shares than actually existed. R Code Lin’s $$r_c$$ can be calculated with the `epi.ccc` function in the `epiR` package. For those who may care, here is the code I used to make the second plot above: ```xyplot(vote_share.2007 ~ vote_share.2005 \| factor(party), data=PartyW[party %!in% c("Other", "RP")], as.table=TRUE, between=list(x=1,y=1), aspect=1, col=grey[5], cex=0.75, xlab="2005", ylab="2007", xlim=c(-0.1,1), ylim=c(-0.1,1), strip=strip.custom(bg=grey[3]), panel=function(x, y, ...) { r <- round(cor(x, y, use="pairwise.complete.obs"), 2) rc <- round(epi.ccc(x, y)[["rho.c"]][1,1], 2) panel.xyplot(x, y, ...) panel.abline(a=0, b=1, lty=2, col=grey[4], lwd=2) panel.abline(lm(y ~ x), col=red[6], lwd=2) panel.text(x=0, y=0.9, labels=bquote(r == .(r)), cex=0.75, pos=4) panel.text(x=0, y=0.8, labels=bquote(r[c] == .(rc)), cex=0.75, pos=4) })``` References • Lin, Lawrence I-Kuei. (1989). “A Concordance Correlation Coefficient to Evaluate Reproducibility.” Biometrics 45: 255–268. • Lin, Lawrence I-Kuei. (2000). “Correction: A Note on the Concordance Correlation Coefficient.” Biometrics 56: 324–325. • Wittenberg, Jason. (Oct. 2011). “How Similar Are They? Rethinking Electoral Congruence.” Quality & Quantity. Posted in Data analysis, Graphics, Poland, Political parties, R \| R Tip: Avoid using T and F as synonyms for TRUE and FALSE Posted on December 11, 2012 by By default when you start R, `T` and `F` are defined as `TRUE` and `FALSE`. When I review other people’s code, I often see functions defined with arguments set to these values by default. This is a very bad idea. `T` and `F` are symbols that can be redefined, so it shouldn’t be assumed that they will always evaluate to `TRUE` and `FALSE`. Making that assumption can introduce bugs to the code that are very hard to track down. For example, imagine you have defined a function to sample from a vector after transforming the data in some way: ```my_sample <- function(x, size, rep=F) { x <- x^2 # a simple transform sample(x, size, replace=rep) }``` When you just start R, `my_sample` will work as intended because `F` is `FALSE`: ```> my_sample(1:10, 8) [1] 4 1 9 100 49 16 36 81``` But if you call this function from inside another function, or after hours of hacking at the console, this may not be the case. For instance, what if at some point in your R session you redefined `F <- 2`: ```> my_sample(1:10, 8) [1] 9 49 100 4 64 64 81 36``` This type of bug can be very dangerous because they are very hard to replicate and, worse yet, you may never even notice you have them. Luckily, such bugs are easy to avoid: never use `T` and `F` as synonyms for `TRUE` and `FALSE`. It doesn’t take much more effort to type out `TRUE` and `FALSE`, but if you find it onerous, get an editor with tab-completion. Update Via email, Ananda Mahto suggests adding the following bit of code to the top of your scripts to provide a warning when `T` and `F` have been redefined. ```if (!identical(T, TRUE)) stop("'T' has been remapped to '", T, "'") if (!identical(F, FALSE)) stop("'F' has been remapped to '", F, "'")``` This doesn’t solve the problem, of course, but at least you get some warning. Posted in Hacking, R, Tip \| Closures in R: A useful abstraction Posted on December 2, 2012 by People who have been using R for any length of time have probably become accustomed to passing functions as arguments to other functions. From my experience, however, people are much less likely to return functions from their own custom code. This is too bad because doing so can open up a whole new world of abstraction that can greatly decrease the quantity and complexity of the code necessary to complete certain types of tasks. Here I provide some brief examples of how R programmers can utilize lexical closures to encapsulate both data and methods. To begin with a simple example, suppose you want a function that adds `2` to its argument. You would likely write something like this: `add_2 <- function(y) { 2 + y }` Which does exactly what you would expect: ```> add_2(1:10) [1] 3 4 5 6 7 8 9 10 11 12``` Now suppose you need another function that instead adds `7` to its argument. The natural thing to do would be to write another function, just like `add_2`, where the `2` is replaced with a `7`. But this would be grossly inefficient: if in the future you discover that you made a mistake and you in fact need to multiply the values instead of add them, you would be forced to change the code in two places. In this trivial example, that may not be much trouble, but for more complicated projects, duplicating code is a recipe for disaster. A better idea would be to write a function that takes one argument, `x`, that returns another function which adds its argument, `y`, to `x`. In other words, something like this: ```add_x <- function(x) { function(y) { x + y } }``` Now, when you call `add_x` with an argument, you will get back a function that does exactly what you want: ```add_2 <- add_x(2) add_7 <- add_x(7) > add_2(1:10) [1] 3 4 5 6 7 8 9 10 11 12 > add_7(1:10) [1] 8 9 10 11 12 13 14 15 16 17``` So far, this doesn’t appear too earth-shattering. But if you look closely at the definition of `add_x`, you may notice something odd: how does the return function know where to find `x` when it’s called at a later point? It turns out that R is lexically scoped, meaning that functions carry with them a reference to the environment within which they were defined. In this case, when you call `add_x`, the `x` argument you provide gets attached to the environment for the return function. In other words, in this simple example, you can think of R as just replacing all instances of the x variable in the function to be returned with the value you specify when you called add_x. Ok, so this may be a neat trick, but how this can be used more productively? For a slightly more complicated example, suppose you are performing some complex bootstrapping and, for efficiency, you pre-allocate container vectors to store the results. This is straightforward when you have just a single vector of results—all you have to do is remember to iterate an index counter each time you add a result to the vector. ```for (i in 1:nboot) { bootmeans[i] <- mean(sample(data, length(data), replace=TRUE)) } > mean(data) [1] 0.0196 > mean(bootmeans) [1] 0.0188``` But suppose you want to track several different statistics, each requiring you to keep track of a different index variable. If your bootstrapping routine is even a little bit complicated, this can be tedious and prone to error. By using closures, you can abstract away all of this bookkeeping. Here is a constructor function that wraps a pre-allocated container vector: ```make_container <- function(n) { x <- numeric(n) i <- 1 function(value=NULL) { if (is.null(value)) { return(x) } else { x[i] <<- value i <<- i + 1 } } }``` When you call `make_container` with an argument, it pre-allocates a numeric vector of the specified length, `n`, and returns a function that allows you to add data to that vector without having to worry about keeping track of an index. If you don’t the argument to that return function is `NULL`, the full vector is returned. ```bootmeans <- make_container(nboot) for (i in 1:nboot) bootmeans(mean(sample(data, length(data), replace=TRUE))) > mean(data) [1] 0.0196 > mean(bootmeans()) [1] 0.0207``` Here `make_container` is relatively simple, but it can be as complicated as you need. For instance, you may want to have the constructor function perform some expensive calculations that you would rather not do every time the function is called. In fact, this is what I have done in the boolean3 package to minimize the number of calculations done at every iteration of the optimization routine. Posted in R \| Filtering a list with the Filter higher-order function Posted on January 26, 2012 by Last week markbulling over at Drunks & Lampposts posted a method of using `sapply` to filter a list by a predicate. Today the @RLangTip tip of the day was to use `sapply` similarly. This made makes me wonder if R‘s very useful higher-order functions aren’t as well known as they should be. In this case, the `Filter` higher-order function would be the tool to use. `Filter` works more or less like the `apply` family of functions, but it performs the subsetting (the filtering) of a list based on a predicate in a single step. As an example, let’s say we have a list of 1000 vectors, each of length 2 with $$x_1,\,x_2 \in [0,\,1]$$, and we want to select only those vectors where the elements of the list sum to a value greater than 1. With `Filter`, this is all we have to do: ```mylist <- lapply(1:1000, function(i) c(runif(1), runif(1))) method.1 <- Filter(function(x) sum(x) > 1, mylist)``` Which is at least a bit more transparent than the `sapply` alternative: `method.2 <- mylist[sapply(mylist, function(x) sum(x) > 1)]` In some very quick tests, I found no performance difference between the two approaches. There are other useful higher-order functions. If you are interested, check out `?Filter`. Posted in R \| Announcing boolean3 (beta) Posted on January 24, 2012 by After entirely too long, I am happy to announce the beta release of `boolean3`, an R package for modeling causal complexity. The package can be downloaded at the following links: • Unix/Linux: boolean3_3.0.20.tar.gz • Windows: boolean3_3.0.20.zip (Please let me know if you have any trouble installing the Windows version. I didn’t have a Windows system handy when I built the package.) The theoretical foundation for the package was developed by Bear Braumoeller in a 2003 Political Analysis piece: ```Braumoeller, Bear F. (2003) 'Causal Complexity and the Study of Politics'. Political Analysis 11(3): 209-233.``` Additional theoretical content can be found at the Boolean Statistics homepage. Braumoeller and Carson have a 2011 paper titled “Political Irrelevance, Democracy, and the Limits of Militarized Conflict” in the Journal of Conflict Resolution that provides a useful example of the approach. Summarizing from the `boolean3` documentation and package description: ````boolean3` provides a means of estimating partial-observability binary response models following boolean logic. `boolean3` was developed by Jason W. Morgan under the direction of Bear Braumoeller with support from The Ohio State University's College of Social and Behavioral Sciences. The package represents a significant re-write of the original boolean implementation developed by Bear Braumoeller, Ben Goodrich, and Jacob Kline.``` As is typically the case with “significant re-writes”, `boolean3` breaks compatibility with previous versions (which can still be downloaded and installed from CRAN). Some of the many changes and enhancements include: • Removed dependence on Zelig. • Changed the method of specifying the model to me more consistent with the style of other R packages. • Improved performance. Many models can be estimated in 1/10th the time, or better, when compared to the original `boolean` package. • Added support for the optimizing methods available in the optimx package. This includes the ability to apply box constraints, which can be useful in maximizing boolean likelihoods. • Added multiprocessor/cluster support for bootstrapping using the snow package. • Genetic optimization using rgenoud also provides support for clustering with `snow`. • Plotting of predicted probabilities and likelihood profiles is now done with lattice. Some things that are still missing from the package: • Documentation is still thin. There is enough in the help pages to get people started using the package, but more is needed. • Skewed logit (scobit) is not yet implemented. Most of the code has been written, but there are a few complications yet to be worked out. If you are a current user of `boolean`, we’d love for you to try out `boolean3`. While a lot of bug-squashing has been done, there has been fairly little “real-world” testing. Bug reports and suggestions for changes and enhancements would be quite welcome, and can be emailed to Bear Braumoeller, me, or posted below in the comments. I will continue to post updates to the package as they are available. Posted in Code, R \| Welford’s method for calculating the sample variance: An implementation in Scheme Posted on December 17, 2011 by John Cook has three entries up on his blog discussing the pitfalls of calculating the sample variance using the mathematical textbook definitions. He provides a Monte Carlo comparison of methods here, and a theoretical discussion here. He also provides a C++ implementation of Welford’s method here. I recently started hacking on Scheme, writing a basic statistics library as a way to better understand the language (which is the most fun I have ever had programming). Below is a Scheme implementation of Welford’s method. For comparison, I have included a version that uses the textbook definition \( s^2 = \cfrac{1}{n-1} \sum_{i=0}^n (x_i – \bar{x})^2. \) Here is the code: ```;;; Textbook definition (define (variance x) (let ((mu (mean x))) (/ (sum (f64vector-map (lambda (x) (expt (- x mu) 2)) x)) (- (f64vector-length x) 1)))) ;;; Welford's method (define (variance-welford x) (let loop ((n (f64vector-length x)) (k 1) (m (f64vector-ref x 0)) (s 0)) (if (= k n) (/ s (fx- n 1)) (let ((xk (f64vector-ref x k)) (mk (+ m (/ (- xk m) (fx+ k 1)))) (sk (+ s (* (- xk m) (- xk mk))))) (loop n (fx+ k 1) mk sk)))))``` I implemented these in Chicken Scheme, though the code should work in any Scheme that supports SRFI-4 homogeneous numeric vectors. The code could easily be modified to support standard Scheme vectors or lists. I make no claims that this is the most efficient code, but it seems to work. Posted in Code, Data analysis, Scheme \| Three free books for better programming in R (and any other language) Posted on September 19, 2011 by Like many users and producers of R packages, I have never had any formal training in computer science. I’ve come to to the conclusion that this is a serious omission in a professional researcher’s training. Computer scientists and professional hackers have learned a lot about effective, efficient programming over the last five decades and it’s past time academic researchers begin to learn from this experience. To that end, here are three books which I think all users of R could benefit from reading: • Practical Common Lisp by Peter Seibel. Free online version. • Structure and Interpretation of Computer Programs (SICP) by Harold Abelson, Gerald Jay Sussman, and Julie Sussman. Free online version. Yes, I realize all of these are books that use Lisp/Scheme, and I realize the structure of Lisp can be a bit off-putting `((at '(first)))`. But, unless you want to learn Lisp, you shouldn’t try to read these books with the goal of understanding ever nuance of the syntax; rather, you should read them with an eye toward understand how a programming language should work for and not against you. For instance, each of these books shows, in a step-by-step approach, how seemingly complex programs can be effectively broken down into simple procedures that anyone with some programming experience can understand. A great deal of emphasis in these books is the important concepts of abstraction and non-duplication. If you’re like me and spend any significant amount of time going through various R packages, you’ll know what I mean when I say that the R community could benefit from less duplication and more abstraction in contributed packages (not to mention more comments!). Posted in Hacking, R \| The performance cost of a for-loop, and some alternatives Posted on August 21, 2011 by I’ve recently been spending a lot of time running various simulations in R. Because I often use snow to perform simulations across several computers/cores, results typically come back in the form of a list object. Summarizing the results from a list is simple enough using a `for`-loop, but it’s much “sexier” to use a functional style of programming that takes advantage of higher order functions or the `apply`-family of functions (R is, after all, a functional language at its core). But what’s the performance cost of using, for instance, `lapply` with a large list, particularly when the results from `lapply` have to be further manipulated to get them into a useful form? I recently did some performance testing on three of the `apply` functions and compared them to an equivalent `for`-loop approach. The results follow. First, I set up a large list with some fake data. Here I make a list, `Sims`, containing 30000 $$5 \times 5$$ matrices. (This is a data structure similar to one I recently found myself analyzing.) ```set.seed(12345) S <- 30000 Sims <- Map(function(x) matrix(rnorm(25), nrow=5, ncol=5), 1:S)``` Now I define four functions that extract the fifth column of each matrix in `Sims` and create a $$30000 \times 5$$ matrix of results. ```for_method <- function(S, Sims) { R <- matrix(NA, ncol=5, nrow=S) for (i in 1:S) R[i,] <- Sims[[i]][,5] R } sapply_method <- function(Sims) { t(sapply(Sims, function(x) x[,5])) } lapply_method <- function(Sims) { do.call(rbind, lapply(Sims, function(x) x[,5])) } vapply_method <- function(Sims) { t(vapply(Sims, function(x) x[,5], rep(0, 5))) }``` Notice that the `for_method` first creates a container matrix and inserts the results directly into it. Using something like `rbind` would force R to copy the matrix in each for-iteration, which is very inefficient. Also notice that in each of the apply methods, the result has to be manipulated before it is returned. It’s often possible to eliminate this step during the initial simulation process. But if you’re like me, you sometimes start your simulations before you realize what you intend on doing with the results. Now for some timing results: ```> system.time(replicate(20, R.1 <<- for_method(S, Sims))) user system elapsed 4.340 0.072 4.414 > system.time(replicate(20, R.2 <<- sapply_method(Sims))) user system elapsed 3.452 0.076 3.528 > system.time(replicate(20, R.3 <<- lapply_method(Sims))) user system elapsed 4.393 0.044 4.433 > system.time(replicate(20, R.4 <<- vapply_method(Sims))) user system elapsed 2.588 0.040 2.628 > all(identical(R.1, R.2), identical(R.1, R.3), identical(R.1, R.4)) [1] TRUE``` These results surprised me—I never expected `vapply` to be so much quicker than the other three methods. This may have to do with that fact that, since `vapply` requires you to explicitly specify the resultant data structure, R is able allocate the memory ahead of time. As a simple extension, I thought I would retest taking advantage of byte-compiled code: ```library(compiler) for_method <- cmpfun(for_method) sapply_method <- cmpfun(sapply_method) lapply_method <- cmpfun(lapply_method) vapply_method <- cmpfun(vapply_method)``` And here are the results: ```> system.time(replicate(10, R.1 <<- for_method(S, Sims))) user system elapsed 2.084 0.024 2.110 > system.time(replicate(10, R.2 <<- sapply_method(Sims))) user system elapsed 1.624 0.016 1.642 > system.time(replicate(10, R.3 <<- lapply_method(Sims))) user system elapsed 2.152 0.012 2.164 > system.time(replicate(10, R.4 <<- vapply_method(Sims))) user system elapsed 1.204 0.004 1.207 > all(identical(R.1, R.2), identical(R.1, R.3), identical(R.1, R.4)) [1] TRUE``` These results are also surprising. I really thought byte-compiling the functions would make the results much closer (doesn’t each function compile into what is effectively a `for`-loop?). However, in this simple example at least, that isn’t the case. `vapply` still has, by a significant margin, the best performance. `sapply` also performs noticeably better than `lapply` and the `for`-loop, which is an important result since `sapply` doesn’t require a pre-specified data structure. Assuming they hold in the presence of more complicated data structures (and I don’t see what they wouldn’t), these results seem to suggest that a functional approach can provide benefits in terms of coding efficiency and run-time performance. I welcome any feedback or suggestions on how these tests could be improved or extended. Update A couple things have come up in the comments. First, Henrik provided a more elegant (and efficient) version of `vapply_method`: ```vapply_method2 <- function(Sims) { t(vapply(Sims, "[", rep(0, 5), i = 1:5, j = 5)) }``` Second, several people have reported that the `for_method` on their system is just as fast, if not faster, than the `vapply_method` when compiled. I have retested my results, adding Henrik's `vapply_method2`. Here are the results for compiled code (note that I have increased the number of replications): ```> library(compiler) > set.seed(12345) > S <- 30000 > Sims <- Map(function(x) matrix(rnorm(25), nrow=5, ncol=5), 1:S) > > for_method <- function(S, Sims) { + R <- matrix(NA, ncol=5, nrow=S) + for (i in 1:S) R[i,] <- Sims[[i]][,5] + R + } > > sapply_method <- function(Sims) { + t(sapply(Sims, function(x) x[,5])) + } > > lapply_method <- function(Sims) { + do.call(rbind, lapply(Sims, function(x) x[,5])) + } > > vapply_method <- function(Sims) { + t(vapply(Sims, function(x) x[,5], rep(0, 5))) + } > > vapply_method2 <- function(Sims) { + t(vapply(Sims, "[", rep(0, 5), i = 1:5, j = 5)) + } > for_method <- cmpfun(for_method) > sapply_method <- cmpfun(sapply_method) > lapply_method <- cmpfun(lapply_method) > vapply_method <- cmpfun(vapply_method) > vapply_method2 <- cmpfun(vapply_method2) > system.time(replicate(50, R.1 <<- for_method(S, Sims))) user system elapsed 6.052 0.136 6.191 > system.time(replicate(50, R.2 <<- sapply_method(Sims))) user system elapsed 8.361 0.116 8.474 > system.time(replicate(50, R.3 <<- lapply_method(Sims))) user system elapsed 10.548 0.144 10.692 > system.time(replicate(50, R.4 <<- vapply_method(Sims))) user system elapsed 5.813 0.124 5.936 > system.time(replicate(50, R.5 <<- vapply_method2(Sims))) user system elapsed 4.268 0.084 4.352``` For whatever reason, the `vapply` methods remain the fastest on my system (particularly Henrik’s version). I also tested on another Linux machine and got the same results (I have not yet tested on Windows). Of course, this is a very simple example, so the performance differences may not hold up to the increasing complexity of the input function. I’d be interested to know if anyone has an idea as to why `vapply` would perform better in Linux than in Windows/Mac. A library thing? Posted in R \| Emacs-fu: The place to go for useful Emacs tips Posted on August 19, 2011 by After 5 years or so of using Emacs almost exclusively as my “text editor”, somehow I just ran across Emacs-fu for the first time today. It’s an incredibly useful site with numerous Emacs tips and tricks. If you are an Emacs user, you should check the site out if you don’t already know about it. And if you aren’t an Emacs user, you should still check out the site to see what your editor is missing. Posted in Emacs, Hacking \| Stronger instruments by design: Polmeth 2011 poster Posted on July 31, 2011 by I just got back from Polmeth 2011 at Princeton, where I presented a poster for a project Luke Keele and I are working on. It deals with the problem of statistical inference in the presence of a weak instrument and the non-random assignment of the instrument. We employ a technique developed by Baoicchi, Small, Lorch, & Rosenbaum in a recent JASA article to effectively strengthen the instrument, thus increasing the robustness of effect estimates to deviations from random assignment. I think the poster turned out pretty well and the feedback I received will be very useful as we move forward with project. I have included a link to the poster below the draft abstract. We hope to complete the draft paper in the next few months, but we would appreciate any comments people may have. There is growing interest in natural experiments in political science. Natural experiments are often analyzed with instrumental variable estimators reflecting a belief that combining the power of natural random assignment with an instrumental variable approach will solve many of the research design problems endemic to social science. Here, we highlight how weak instruments can interact with the assumption of random assignment of the instrument. When the instrument is not randomly assigned, weak instruments produce bias that is not alleviated by additional data. We demonstrate how matching combined with a reverse caliper can be used to strengthen an instrument within a subset of the overall study. We start by presenting an alternative non-parametric instrumental variable estimator first proposed by Rosenbaum (1996) that allows us to combine matching with an IV estimator. Unlike the standard 2SLS IV estimator, this non-parametric approach provides accurate confidence intervals and consistent causal estimates even when the instrument is weak. A further advantage of this non-parametric method is the opportunity it provides to probe the random assignment assumption with a sensitivity test. We provide substantive examples of the proposed approach with a reevaluation of a recent paper that uses rainfall as an instrument for voter turnout in US counties Hansford & Gomez (2010). The Polmeth 2011 poster: Stronger Instruments by Design. \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 36, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9327273368835449, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/76551/moving-the-origin-of-an-elliptic-curve	## Moving the origin of an elliptic curve ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Suppose one has an elliptic curve $E = (C,O)$ over a field $k$ where $C$ is a non-singular genus one curve over $k$ and $O$ is a $k$-rational point on $C$. By moving $O$ on $C$ one gets a family of elliptic curves. Is this family trivial in the sense that all the curves are mutually isomorphic? If not (as I suspect), what are the isomorphism classes of elliptic curves one gets this way and how are they related to the original curve $E$? Special cases of $k = \mathbb{C}$ and $\mathbb{Q}$ would be specially helpful to know about, for $\mathbb{C}$ in terms of lattices, and for $\mathbb{Q}$ in terms of the associated modular forms after Wiles et al. (Depending on the answer) should CM curves be distinguished? Thanks! - 5 Yes, they're isomorphic, and this is true for any algebraic group, because translation by a given ($k$-rational) group element is an isomorphism of algebraic varieties. This should be particularly easy to visualize in the case of ${\bf C}/L$ (translation by some complex number modulo the lattice $L$). – Noam D. Elkies Sep 27 2011 at 20:01 $C(k)$ is a group, and it acts on itself by multiplication. So if you have $E=(C,O)$ and $E'=(C,O')$ as you suggest on your question, wouldn't it be enough to look at the action of $C' - O$ ? – Reimundo Heluani Sep 27 2011 at 20:05 1 Is this a fishing expedition? – András Bátkai Sep 27 2011 at 22:01	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 21, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9405685663223267, "perplexity_flag": "head"}
http://mathhelpforum.com/advanced-algebra/203821-please-help-check-my-solutions-vector-space-questions.html	1Thanks • 1 Post By Deveno # Thread: 2. ## Re: Please help check my solutions for vector space questions From looking at the images you posted it seems that you are proving: $\|V\cdot U\|\le\\|V\\|\\|U\\|$ The standard proof follows from the fact $\frac{V\cdot U}{\\|V\\|\\|U\\|}=\cos(\theta)$ where $\theta$ is the angle between the two non-zero vectors $V~\&~U$. Then note that $0\le \|\cos(\theta)\|\le 1~.$ 3. ## Re: Please help check my solutions for vector space questions first of all, you have $\\|U\\|$ and $\\|V\\|$ wrong. $\\|U\\| = \sqrt{U \cdot U} = \sqrt{\int_0^{\frac{\pi}{2}} \sin^2x\ dx}$ $= \sqrt{\int_0^{\frac{\pi}{2}}(1 - \cos^2x)\ dx}$ $= \sqrt{\frac{\pi}{2} - \int_0^{\frac{\pi}{2}} \cos^2x\ dx}$ $= \sqrt{\frac{\pi}{2} - \int_0^{\frac{\pi}{2}} \frac{1 + \cos 2x}{2}\ dx}$ $= \sqrt{\frac{\pi}{2} - \frac{\pi}{4} - \frac{1}{4}\int_0^{\frac{\pi}{2}} \cos 2x\ 2dx}$ $= \sqrt{\frac{\pi}{4} - \frac{1}{4}(\sin (\pi) - \sin (0))}$ $= \frac{\sqrt{\pi}}{2}$ while: $\\|V\\| = \sqrt{V \cdot V} = \sqrt{\int_0^{\frac{\pi}{2}} \cos^2x\ dx}$ $= \sqrt{\int_0^{\frac{\pi}{2}} \frac{1+\cos 2x}{2}\ dx}$ $= \sqrt{\frac{\pi}{4} + \frac{1}{4}(\sin(\pi) - \sin(0))}$ $= \frac{\sqrt{\pi}}{2}$, as well. thus: $\\|U\\|\\|V\\| = \frac{\pi}{4}$. on the other side of the inequality we have: $\|U \cdot V\| = \left\|\int_0^{\frac{\pi}{2}} \sin x \cos x\ dx \right\|$ $= \left\| \frac{\sin^2(\frac{\pi}{2})}{2} - \frac{\sin^2(0)}{2} \right\| = \left\|\frac{1}{2} \right\| = \frac{1}{2}$ since $1 < \frac{\pi}{2}$, we indeed have for these particular U,V: $\|U \cdot V\| \leq \\|U\\|\\|V\\|$. ********* for your second problem, you are also calculating the norms of U and V incorrectly. use the definition of the inner product you are given! 4. ## Re: Please help check my solutions for vector space questions Thank you so much!! Deveno! I missed the obvious point Could you please re-check my second answer again? 5. ## Re: Please help check my solutions for vector space questions still incorrect. for starters, you have an extra 2-n term in $\\|U\\|$ and $\\|V\\|$. secondly, you don't seem to get that the inner product of two vectors should be a SCALAR, not an "expression". so for U, you should have: $\\|U\\| = \sqrt{\sum_{n=1}^{\infty} \frac{1}{2^n}} = 1$ and for V, you should have: $\\|V\\| = \sqrt{\sum_{n=1}^{\infty} \frac{1}{8^n}} = \frac{1}{\sqrt{7}}$ while for the dot product, you should have: $\|U \cdot V\| = \left\| \sum_{n=1}^{\infty} \frac{1}{4^n} \right\| = \frac{1}{3}$	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 28, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9051805734634399, "perplexity_flag": "middle"}
http://mathhelpforum.com/differential-geometry/171721-coordinate-charts-inverse-functions-print.html	# Coordinate Charts and inverse of functions. Printable View • February 18th 2011, 05:11 AM Bongo Coordinate Charts and inverse of functions. Hello, i have a question about this situation. Let denote $T^2=S^1 \times S^1$ the Torus. And let $f:\mathbb{R}^2 \rightarrow T^2$ be the projection defined by $f(x,y)=(e^{i2\pi x},e^{i2\pi y})$. The claim is now, if $\phi$ is a chart for $\mathbb{R}^2$ , s.t. $f_{\|dom\phi}$ is injective, then $\phi \circ (f_{\|dom\phi})^{-1}$ is a chart for $T^2$. i don't understand why the composition is a chart? Can you help me? Regards • February 18th 2011, 05:32 AM xxp9 As a chart means it's a 1-1 diffeomorphism in the domain that defined the map. Let V=dom(\phi), U=f(V), W=\phi^{-1}(V) Since f is injective in V, f is a 1-1 diffeomorphism between V and f(V). And \phi is defined to be a 1-1 diffeomorphism between W and \phi(W)=V. Compose two 1-1 diffeomorphism we get a 1-1 diffeomorphism. All times are GMT -8. The time now is 12:32 AM. Copyright © 2005-2013 Math Help Forum. All rights reserved. Copyright © 2005-2013 Math Help Forum. All rights reserved.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.921599805355072, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/137057/is-the-set-of-conditions-for-devaneys-definition-of-chaos-minimal	# Is the set of conditions for Devaney's definition of chaos minimal? I am reading Devaney's definition of chaos. Which says: Let $V$ be a set. $f:V \rightarrow V$ is said to be chaotic on V if 1. $f$ has sensitive dependence on initial conditions 2. $f$ is topologically transitive 3. periodic points are dense in $V$ It seems that conditions 2 and 3 implies 1. Also, condition 2 seems to imply 1. Am I missing something here? - ## 2 Answers Brooks, Cairns, Davis and Stacey proved that 2. and 3. imply 1. on metric spaces with an infinite number of points. There are no more redundancies in general metric spaces. However, if $V=[a,b]\subset\mathbb{R}$ and $f$ is continuous, then 2. implies 1. and 3. This is a result of Vellekop and Berglund, published in the American Mathematical Monthly, vol. 101, 1994. - Thanks for the reply. However, I have some more related doubts. I would post them as a separate question so that it's easier to lookup later. – Abhishek Chanda Apr 26 '12 at 18:23 over a metric space see the following http://www2.math.uu.se/~warwick/vt04/DynSyst/reading/BanksEtAl.pdf -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 7, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9223995804786682, "perplexity_flag": "head"}
http://quant.stackexchange.com/questions/tagged/risk-models+correlation	Tagged Questions 1answer 209 views What weights should be used when adjusting a correlation matrix to be positive definite? I have a correlation matrix $A$ for an equity market that is not positive definite. Higham (2002) proposes the Alternating Projections Method, minimising the weighted Frobenius norm $\|\|A-X\|\|_W$ where ... 1answer 663 views Cleansing covariance matrices via Random matrix theory I am exploring de-noising and cleansing of covariance matrices via Random Matrix Theory. RMT is a competitor to shrinkage methods of covariance estimation. There are various methods expressed usually ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8457196950912476, "perplexity_flag": "middle"}
http://mathinsight.org/function_machine_inverse	# Math Insight • Top • In threads When printing from Firefox, select the SVG renderer in the MathJax contextual menu (right click any math equation) for better printing results ### The function machine inverse #### Suggested background If $f$ is a function, we can represent it by a function machine that can take any input from its domain and transform it to an element from its codomain. In this illustration, we represent elements of the domain of $f$ by octahedra and elements of its codomain by spheres. An octahedron enters through the function machine's input funnel, and the function machine spits out a particular sphere, where the identity of the sphere depends on which octahedron went in. Now, imagine that for each unique octahedron, the function machine yielded a different sphere. Then, by examining the sphere that came out of the function machine's chute, it would be possible to determine which octahedron went into the funnel. (We'd be in trouble if multiple distinct octahedron yielded the same sphere, as there would be no way to distinguish which octahedron went into the funnel by examining the sphere.) In the case where we can examine the sphere and determine which octahedron went in, it makes sense that we could run $f$'s function machine backwards. We could put the sphere back in, and the backwards function machine could undo whatever $f$'s function machine did to produce the sphere. Then, the octahedron that started it all would pop out of the funnel. To show that we are running the machine backwards, let's flip it over so that the output chute is at the top. This backwards machine takes in spheres into the chute and spits out octahedron out of the funnel. Since this backwards machine undoes whatever processing $f$'s function machine did, we call it the inverse of $f$, or simply “$f$ inverse.” We give $f$ inverse a special symbol, too: $f^{-1}$. The superscript here does not denote exponentiation; it simply indicates we are running the function machine backwards to undo the processing $f$ did. If we denote a particular octahedron by the variable $x$, then we can denote by $f(x)$ the resulting sphere that $f$'s function machine spits out. Or, to make it shorter, we could just use the variable $y$, where $y=f(x)$. By the definition of $f^{-1}$, if we stick the sphere $y$ into the $f^{-1}$ backwards function machine, we must get the octahedron $x$ back, i.e., $f^{-1}(y)=x$. If we composed $f$ and $f^{-1}$ by connecting their function machines, then the resulting composition would not do anything to the octahedron $x$. We would put the octahedron in the combined function machine and the same octahedron would come out. In math symbols, we would find that $f^{-1}(f(x))=x$, which we could also write as $(f^{-1} \circ f)(x)=x$. You can read some examples of calculating the inverse function. #### Thread navigation ##### Math 1241 • Previous: Function machine parameters • Next: Inverse function examples #### Cite this as Nykamp DQ, “The function machine inverse.” From Math Insight. http://mathinsight.org/function_machine_inverse Keywords: function, function machine, inverse, inverse function	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 25, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9191346764564514, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/95362-taylor-maclaurin-series.html	# Thread: 1. ## Taylor and Maclaurin series Find the Taylor and Maclaurin series of the given function with the given point $z_0$ as center and determine the radius of convergence. $e^{\frac{z^2}{2}}\int e^{-t^2} dt$ By the way, integrate from 0 to z and center at $z_0 = 0$ I cannot imagine, how can I solve this. 2. Originally Posted by noppawit Find the Taylor and Maclaurin series of the given function with the given point $z_0$ as center and determine the radius of convergence. $e^{\frac{z^2}{2}}\int e^{-t^2} dt$ By the way, integrate from 0 to z and center at $z_0 = 0$ I cannot imagine, how can I solve this. centered at 0 ... $e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + ...$ $e^{-t^2} = 1 - t^2 + \frac{t^4}{2!} - \frac{t^6}{3!} + ...$ $\int_0^z e^{-t^2} \, dt = \left[t - \frac{t^3}{3} + \frac{t^5}{5 \cdot 2!} - \frac{t^7}{7 \cdot 3!} + ... \right]_0^z = z - \frac{z^3}{3} + \frac{z^5}{5 \cdot 2!} - \frac{z^7}{7 \cdot 3!} + ...$	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 9, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8938313722610474, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/60538/probability-that-two-poker-hands-have-no-aces/60541	# Probability that two poker hands have no aces The probability of no two hands from the same poker deck having no aces is $$\frac{{48 \choose 5} + {48 \choose 5} - {48 \choose 10}}{{52 \choose 5}}$$ I am not sure why this is the answer, as the second draw is conditional on the first draw so it should be ${43 \choose 5}$. - 2 I think the first "no" is in excess? – joriki Aug 29 '11 at 15:37 Are you sure this is the right formula? In its current form, the numerator is negative. – Fixee Aug 29 '11 at 15:41 Where did you get that formula? The numerator is a negative number, and the expression evaluates to around - 2515. So that absolutely cannot be a probability... – Willie Wong♦ Aug 29 '11 at 15:47 1 (BTW, what's the difference between "two poker hands" and "a draw of ten cards"?) – Willie Wong♦ Aug 29 '11 at 15:49 ## 3 Answers This is not a probability, since $\binom{48}{10}$ is a lot bigger than the other ones and would make this a large negative number. It looks like someone was trying to work out the probability that at least one of the two hands has no aces using the inclusion-exclusion principle. The probability that one hand has no aces is $$\frac{\binom{48}5}{\binom{52}5}\;,$$ and the probability that one or the other hand has no aces is twice that, except that would overcount the case where they both have no aces, which is being counted twice, so we have to subtract the probability of both hands having no aces, which is $$\frac{\binom{48}{10}}{\binom{52}{10}}\;.$$ Note the $10$ in the denominator, which makes all the difference. So the probability of at least one hand having no aces is $$\frac{{48 \choose 5} + {48 \choose 5}}{52 \choose 5} - \frac{{48 \choose 10}}{{52 \choose 10}}=\frac{576}{637}\approx0.90\;.$$ - Nice answer as usual, but it's called subtract, not substract. – Hans Lundmark Aug 29 '11 at 16:53 1 @Hans: Thanks; fixed. – joriki Aug 29 '11 at 21:02 The formula in the question cannot be right (it has a negative numerator). Here's how I would find the probability: The number of sets of 10 cards (two hands) with no aces is ${48 \choose 10}$. The number of set of 10 cards from the deck is ${52 \choose 10}$. So the overall answer is $$\frac{{48 \choose 10}}{{52 \choose 10}} = \frac{246}{595} \approx 0.41$$ - I agree that it is not clear whether this is the question being asked or the one joriki answered. In any case, this is a good answer to this one. – Ross Millikan Aug 29 '11 at 19:25 I will assume, like Fixee did, that you want to find the probability that neither hand has an ace. Then Fixee's solution is the most efficient. A mild variant of it computes probabilities directly, bypassing binomial coefficients. The probability that the first card is not an ace is $48/52$. Given that the first card is not an ace, the probability that the second card is not an ace is $47/51$. So the probability that neither of the first two cards is an ace is $(48/52)(47/51)$. Continue in this way. We find that the required probability is $$\frac{48}{52}\cdot \frac{47}{51}\cdot \frac{46}{50}\cdots\cdot\frac{40}{44}\cdot \frac{39}{43}.$$ There is a good deal of easy cancellation. After a while we arrive at $246/595$. We can also find the answer by an argument that uses your idea. Suppose that one hand is dealt, then the other. The probability that the first hand has no aces is $$\frac{\binom{48}{5}}{\binom{52}{5}}.$$ The probability that the second hand has no aces, given that the first hand has no aces, is $$\frac{\binom{43}{5}}{\binom{47}{5}}$$ (there are $47$ cards left, of which $43$ are non-aces). Thus the required probability is $$\frac{\binom{48}{5}}{\binom{52}{5}}\cdot\frac{\binom{43}{5}}{\binom{47}{5}}.$$ - So what is the point of using the inclusion/exclusion principle in this problem? Conditional probability makes much more sense in this problem. I still cannot see how you use the inclusion exclusion principle in this problem because you will have fewer possibilities to choose the second hand after you have chosen the first hand. Any ideas as to how inclusion/exclusion intuitively makes sense if this is a conditional probability problem? – lord12 Sep 2 '11 at 15:44 The people posting answers are uncertain about the exact question you are asking. Fixee and I interpret the question as asking for the probability that none of the cards (in either hand) is an Ace. The interpretation of joriki is that you are asking for the probability that at least one hand has no Ace. These are different problems, with different answers. It would be helpful if you clarified which of these two interpretations is the one you intended. For the joriki interpretation, inclusion-exclusion is natural to use. (Continued) – André Nicolas Sep 2 '11 at 16:33 Finding the probability that hand $A$ has no Ace is easy, same for hand $B$. But if we add these two probabilities, we are double-counting the situations in which neither $A$ nor $B$ has an Ace, so we have to subtract. We could also solve the problem with joriki's interpretation by conditional probabilities, by finding first the probability that both hands have at least one Ace. The problem is that this divides into cases: $A$ has exactly $1$, and $B$ has $1$. $2$, or $3$; $A$ has exactly $2$, and $B$ has $1$ or $2$; $A$ has $3$, and $B$ has $1$. Much more work! – André Nicolas Sep 2 '11 at 16:41 Even in the inclusion/exclusion case, why would the probability of no Ace in hand B be the same as no ace in A. Meaning, why isn't it {48\choose 10} as opposed to {48 \choose 3}. If A1 = {Hand 1 has no ace} and A2 = {Hand 2 has no ace}, the number of possibilities cannot be equal to the number of possibilities of hand 1. I may have a hard time understanding this. Could you give me a similar problem that is similar to this that uses the inclusion/exclusion principle? – lord12 Sep 2 '11 at 19:16 The cards don't "know" which hand is being dealt first. The probability that the second card is the Ace of Spades is the same as the probability that the ninth card is the Ace of Spades. Problem: A bin has 30 identically wrapped candies, 10 chocolate and 20 strawberry. $A$ and $B$ each take $3$ candies. What is the probability at least one of them gets no chocolate candy? – André Nicolas Sep 2 '11 at 19:50 show 2 more comments	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 33, "mathjax_display_tex": 9, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9641643762588501, "perplexity_flag": "head"}
http://en.wikibooks.org/wiki/Engineering_Acoustics/Bass_Reflex_Enclosure_Design	# Engineering Acoustics/Bass Reflex Enclosure Design Authors · Print · License Edit this template Part 1: Lumped Acoustical Systems – 1.1 – 1.2 – 1.3 – 1.4 – 1.5 – 1.6 – 1.7 – 1.8 – 1.9 – 1.10 – 1.11 Part 2: One-Dimensional Wave Motion – 2.1 – 2.2 – 2.3 Part 3: Applications – 3.1 – 3.2 – 3.3 – 3.4 – 3.5 – 3.6 – 3.7 – 3.8 – 3.9 – 3.10 – 3.11 – 3.12 – 3.13 – 3.14 – 3.15 – 3.16 – 3.17 – 3.18 – 3.19 – 3.20 – 3.21 – 3.22 – 3.23 – 3.24 ## Introduction Bass-reflex enclosures improve the low-frequency response of loudspeaker systems. Bass-reflex enclosures are also called "vented-box design" or "ported-cabinet design". A bass-reflex enclosure includes a vent or port between the cabinet and the ambient environment. This type of design, as one may observe by looking at contemporary loudspeaker products, is still widely used today. Although the construction of bass-reflex enclosures is fairly simple, their design is not simple, and requires proper tuning. This reference focuses on the technical details of bass-reflex design. General loudspeaker information can be found here. ## Effects of the Port on the Enclosure Response Before discussing the bass-reflex enclosure, it is important to be familiar with the simpler sealed enclosure system performance. As the name suggests, the sealed enclosure system attaches the loudspeaker to a sealed enclosure (except for a small air leak included to equalize the ambient pressure inside). Ideally, the enclosure would act as an acoustical compiance element, as the air inside the enclosure is compressed and rarified. Often, however, an acoustic material is added inside the box to reduce standing waves, dissipate heat, and other reasons. This adds a resistive element to the acoustical lumped-element model. A non-ideal model of the effect of the enclosure actually adds an acoustical mass element to complete a series lumped-element circuit given in Figure 1. For more on sealed enclosure design, see the Sealed Box Subwoofer Design page. Figure 1. Sealed enclosure acoustic circuit. In the case of a bass-reflex enclosure, a port is added to the construction. Typically, the port is cylindrical and is flanged on the end pointing outside the enclosure. In a bass-reflex enclosure, the amount of acoustic material used is usually much less than in the sealed enclosure case, often none at all. This allows air to flow freely through the port. Instead, the larger losses come from the air leakage in the enclosure. With this setup, a lumped-element acoustical circuit has the following form. Figure 2. Bass-reflex enclosure acoustic circuit. In this figure, $Z_{RAD}$ represents the radiation impedance of the outside environment on the loudspeaker diaphragm. The loading on the rear of the diaphragm has changed when compared to the sealed enclosure case. If one visualizes the movement of air within the enclosure, some of the air is compressed and rarified by the compliance of the enclosure, some leaks out of the enclosure, and some flows out of the port. This explains the parallel combination of $M_{AP}$, $C_{AB}$, and $R_{AL}$. A truly realistic model would incorporate a radiation impedance of the port in series with $M_{AP}$, but for now it is ignored. Finally, $M_{AB}$, the acoustical mass of the enclosure, is included as discussed in the sealed enclosure case. The formulas which calculate the enclosure parameters are listed in Appendix B. It is important to note the parallel combination of $M_{AP}$ and $C_{AB}$. This forms a Helmholtz resonator (click here for more information). Physically, the port functions as the “neck” of the resonator and the enclosure functions as the “cavity.” In this case, the resonator is driven from the piston directly on the cavity instead of the typical Helmholtz case where it is driven at the “neck.” However, the same resonant behavior still occurs at the enclosure resonance frequency, $f_{B}$. At this frequency, the impedance seen by the loudspeaker diaphragm is large (see Figure 3 below). Thus, the load on the loudspeaker reduces the velocity flowing through its mechanical parameters, causing an anti-resonance condition where the displacement of the diaphragm is a minimum. Instead, the majority of the volume velocity is actually emitted by the port itself instead of the loudspeaker. When this impedance is reflected to the electrical circuit, it is proportional to $1/Z$, thus a minimum in the impedance seen by the voice coil is small. Figure 3 shows a plot of the impedance seen at the terminals of the loudspeaker. In this example, $f_B$ was found to be about 40 Hz, which corresponds to the null in the voice-coil impedance. Figure 3. Impedances seen by the loudspeaker diaphragm and voice coil. ## Quantitative Analysis of Port on Enclosure The performance of the loudspeaker is first measured by its velocity response, which can be found directly from the equivalent circuit of the system. As the goal of most loudspeaker designs is to improve the bass response (leaving high-frequency production to a tweeter), low frequency approximations will be made as much as possible to simplify the analysis. First, the inductance of the voice coil, $\it{L_E}$, can be ignored as long as $\omega \ll R_E/L_E$. In a typical loudspeaker, $\it{L_E}$ is of the order of 1 mH, while $\it{R_E}$ is typically 8$\Omega$, thus an upper frequency limit is approximately 1 kHz for this approximation, which is certainly high enough for the frequency range of interest. Another approximation involves the radiation impedance, $\it{Z_{RAD}}$. It can be shown [1] that this value is given by the following equation (in acoustical ohms): $Z_{RAD} = \frac{\rho_0c}{\pi a^2}\left[\left(1 - \frac{J_1(2ka)}{ka}\right) + j\frac{H_1(2ka)}{ka}\right]$ Where $J_1(x)$ and $H_1(x)$ are types of Bessel functions. For small values of ka, $J_1(2ka) \approx ka$ and $H_1(2ka) \approx \frac{8(ka)^2}{3\pi}$ $\Rightarrow Z_{RAD} \approx j\frac{8\rho_0\omega}{3\pi^2a} = jM_{A1}$ Hence, the low-frequency impedance on the loudspeaker is represented with an acoustic mass $M_{A1}$ [1]. For a simple analysis, $R_E$, $M_{MD}$, $C_{MS}$, and $R_{MS}$ (the transducer parameters, or Thiele-Small parameters) are converted to their acoustical equivalents. All conversions for all parameters are given in Appendix A. Then, the series masses, $M_{AD}$, $M_{A1}$, and $M_{AB}$, are lumped together to create $M_{AC}$. This new circuit is shown below. Figure 4. Low-Frequency Equivalent Acoustic Circuit Unlike sealed enclosure analysis, there are multiple sources of volume velocity that radiate to the outside environment. Hence, the diaphragm volume velocity, $U_D$, is not analyzed but rather $U_0 = U_D + U_P + U_L$. This essentially draws a “bubble” around the enclosure and treats the system as a source with volume velocity $U_0$. This “lumped” approach will only be valid for low frequencies, but previous approximations have already limited the analysis to such frequencies anyway. It can be seen from the circuit that the volume velocity flowing into the enclosure, $U_B = -U_0$, compresses the air inside the enclosure. Thus, the circuit model of Figure 3 is valid and the relationship relating input voltage, $V_{IN}$ to $U_0$ may be computed. In order to make the equations easier to understand, several parameters are combined to form other parameter names. First, $\omega_B$ and $\omega_S$, the enclosure and loudspeaker resonance frequencies, respectively, are: $\omega_B = \frac{1}{\sqrt{M_{AP}C_{AB}}}$ $\omega_S = \frac{1}{\sqrt{M_{AC}C_{AS}}}$ Based on the nature of the derivation, it is convenient to define the parameters $\omega_0$ and h, the Helmholtz tuning ratio: $\omega_0 = \sqrt{\omega_B\omega_S}$ $h = \frac{\omega_B}{\omega_S}$ A parameter known as the compliance ratio or volume ratio, $\alpha$, is given by: $\alpha = \frac{C_{AS}}{C_{AB}} = \frac{V_{AS}}{V_{AB}}$ Other parameters are combined to form what are known as quality factors: $Q_L = R_{AL}\sqrt{\frac{C_{AB}}{M_{AP}}}$ $Q_{TS} = \frac{1}{R_{AE}+R_{AS}}\sqrt{\frac{M_{AC}}{C_{AS}}}$ This notation allows for a simpler expression for the resulting transfer function [1]: $\frac{U_0}{V_{IN}} = G(s) = \frac{(s^3/\omega_0^4)}{(s/\omega_0)^4+a_3(s/\omega_0)^3+a_2(s/\omega_0)^2+a_1(s/\omega_0)+1}$ where $a_1 = \frac{1}{Q_L\sqrt{h}}+\frac{\sqrt{h}}{Q_{TS}}$ $a_2 = \frac{\alpha+1}{h}+h+\frac{1}{Q_L Q_{TS}}$ $a_3 = \frac{1}{Q_{TS}\sqrt{h}}+\frac{\sqrt{h}}{Q_L}$ ## Development of Low-Frequency Pressure Response It can be shown [2] that for $ka < 1/2$, a loudspeaker behaves as a spherical source. Here, a represents the radius of the loudspeaker. For a 15” diameter loudspeaker in air, this low frequency limit is about 150 Hz. For smaller loudspeakers, this limit increases. This limit dominates the limit which ignores $L_E$, and is consistent with the limit that models $Z_{RAD}$ by $M_{A1}$. Within this limit, the loudspeaker emits a volume velocity $U_0$, as determined in the previous section. For a simple spherical source with volume velocity $U_0$, the far-field pressure is given by [1]: $p(r) \simeq j\omega\rho_0 U_0 \frac{e^{-jkr}}{4\pi r}$ It is possible to simply let $r = 1$ for this analysis without loss of generality because distance is only a function of the surroundings, not the loudspeaker. Also, because the transfer function magnitude is of primary interest, the exponential term, which has a unity magnitude, is omitted. Hence, the pressure response of the system is given by [1]: $\frac{p}{V_{IN}} = \frac{\rho_0s}{4\pi}\frac{U_0}{V_{IN}} = \frac{\rho_0Bl}{4\pi S_DR_EM_AS}H(s)$ Where $H(s) = sG(s)$. In the following sections, design methods will focus on $\|H(s)\|^2$ rather than $H(s)$, which is given by: $\|H(s)\|^2 = \frac{\Omega^8}{\Omega^8 + \left(a^2_3 - 2a_2\right)\Omega^6 + \left(a^2_2 + 2 - 2a_1a_3\right)\Omega^4 + \left(a^2_1 - 2a_2\right)\Omega^2 + 1}$ $\Omega = \frac{\omega}{\omega_0}$ This also implicitly ignores the constants in front of $\|H(s)\|$ since they simply scale the response and do not affect the shape of the frequency response curve. ## Alignments A popular way to determine the ideal parameters has been through the use of alignments. The concept of alignments is based upon filter theory. Filter development is a method of selecting the poles (and possibly zeros) of a transfer function to meet a particular design criterion. The criteria are the desired properties of a magnitude-squared transfer function, which in this case is $\|H(s)\|^2$. From any of the design criteria, the poles (and possibly zeros) of $\|H(s)\|^2$ are found, which can then be used to calculate the numerator and denominator. This is the “optimal” transfer function, which has coefficients that are matched to the parameters of $\|H(s)\|^2$ to compute the appropriate values that will yield a design that meets the criteria. There are many different types of filter designs, each which have trade-offs associated with them. However, this design is limited because of the structure of $\|H(s)\|^2$. In particular, it has the structure of a fourth-order high-pass filter with all zeros at s = 0. Therefore, only those filter design methods which produce a low-pass filter with only poles will be acceptable methods to use. From the traditional set of algorithms, only Butterworth and Chebyshev low-pass filters have only poles. In addition, another type of filter called a quasi-Butterworth filter can also be used, which has similar properties to a Butterworth filter. These three algorithms are fairly simple, thus they are the most popular. When these low-pass filters are converted to high-pass filters, the $s \rightarrow 1/s$ transformation produces $s^8$ in the numerator. More details regarding filter theory and these relationships can be found in numerous resources, including [5]. ## Butterworth Alignment The Butterworth algorithm is designed to have a maximally flat pass band. Since the slope of a function corresponds to its derivatives, a flat function will have derivatives equal to zero. Since as flat of a pass band as possible is optimal, the ideal function will have as many derivatives equal to zero as possible at s = 0. Of course, if all derivatives were equal to zero, then the function would be a constant, which performs no filtering. Often, it is better to examine what is called the loss function. Loss is the reciprocal of gain, thus $\|\hat{H}(s)\|^2 = \frac{1}{\|H(s)\|^2}$ The loss function can be used to achieve the desired properties, then the desired gain function is recovered from the loss function. Now, applying the desired Butterworth property of maximal pass-band flatness, the loss function is simply a polynomial with derivatives equal to zero at s = 0. At the same time, the original polynomial must be of degree eight (yielding a fourth-order function). However, derivatives one through seven can be equal to zero if [3] $\|\hat{H}(\Omega)\|^2 = 1 + \Omega^8 \Rightarrow \|H(\Omega)\|^2 = \frac{1}{1 + \Omega^8}$ With the high-pass transformation $\Omega \rightarrow 1/\Omega$, $\|H(\Omega)\|^2 = \frac{\Omega^8}{\Omega^8 + 1}$ It is convenient to define $\Omega = \omega/\omega_{3dB}$, since $\Omega = 1 \Rightarrow \|H(s)\|^2 = 0.5$ or -3 dB. This definition allows the matching of coefficients for the $\|H(s)\|^2$ describing the loudspeaker response when $\omega_{3dB} = \omega_0$. From this matching, the following design equations are obtained [1]: $a_1 = a_3 = \sqrt{4+2\sqrt{2}}$ $a_2 = 2+\sqrt{2}$ ## Quasi-Butterworth Alignment The quasi-Butterworth alignments do not have as well-defined of an algorithm when compared to the Butterworth alignment. The name “quasi-Butterworth” comes from the fact that the transfer functions for these responses appear similar to the Butterworth ones, with (in general) the addition of terms in the denominator. This will be illustrated below. While there are many types of quasi-Butterworth alignments, the simplest and most popular is the 3rd order alignment (QB3). The comparison of the QB3 magnitude-squared response against the 4th order Butterworth is shown below. $\left\|H_{QB3}(\omega)\right\|^2 = \frac{(\omega/\omega_{3dB})^8}{(\omega/\omega_{3dB})^8 + B^2(\omega/\omega_{3dB})^2 + 1}$ $\left\|H_{B4}(\omega)\right\|^2 = \frac{(\omega/\omega_{3dB})^8}{(\omega/\omega_{3dB})^8 + 1}$ Notice that the case $B = 0$ is the Butterworth alignment. The reason that this QB alignment is called 3rd order is due to the fact that as B increases, the slope approaches 3 dec/dec instead of 4 dec/dec, as in 4th order Butterworth. This phenomenon can be seen in Figure 5. Figure 5: 3rd-Order Quasi-Butterworth Response for $0.1 \leq B \leq 3$ Equating the system response $\|H(s)\|^2$ with $\|H_{QB3}(s)\|^2$, the equations guiding the design can be found [1]: $B^2 = a^2_1 - 2a_2$ $a_2^2 + 2 = 2a_1a_3$ $a_3 = \sqrt{2a_2}$ $a_2 > 2 + \sqrt{2}$ ## Chebyshev Alignment The Chebyshev algorithm is an alternative to the Butterworth algorithm. For the Chebyshev response, the maximally-flat passband restriction is abandoned. Now, a ripple, or fluctuation is allowed in the pass band. This allows a steeper transition or roll-off to occur. In this type of application, the low-frequency response of the loudspeaker can be extended beyond what can be achieved by Butterworth-type filters. An example plot of a Chebyshev high-pass response with 0.5 dB of ripple against a Butterworth high-pass response for the same $\omega_{3dB}$ is shown below. Figure 6: Chebyshev vs. Butterworth High-Pass Response. The Chebyshev response is defined by [4]: $\|\hat{H}(j\Omega)\|^2 = 1 + \epsilon^2C^2_n(\Omega)$ $C_n(\Omega)$ is called the Chebyshev polynomial and is defined by [4]: \| \| \| \| \|------------------------------------------------------------------------------\|------------------------------------------------------------------------------\|-----------------------------------------------------------------------------\| \| $C_n(\Omega) = \big\lbrace$ \| $\rm{cos}[\it{n}\rm{cos}^{-1}(\Omega)]$ \| $\|\Omega\| < 1$ \| \| $\rm{cosh}[\it{n}\rm{cosh}^{-1}(\Omega)]$ \| $\|\Omega\| > 1$ \| \| Fortunately, Chebyshev polynomials satisfy a simple recursion formula [4]: $C_0(x) = 1$ $C_1(x) = x$ $C_n(x) = 2xC_{n-1} - C_{n-2}$ For more information on Chebyshev polynomials, see the Wolfram Mathworld: Chebyshev Polynomials page. When applying the high-pass transformation to the 4th order form of $\|\hat{H}(j\Omega)\|^2$, the desired response has the form [1]: $\|H(j\Omega)\|^2 = \frac{1+\epsilon^2}{1+\epsilon^2C^2_4(1/\Omega)}$ The parameter $\epsilon$ determines the ripple. In particular, the magnitude of the ripple is $10\rm{log}[1+\epsilon^2]$ dB and can be chosen by the designer, similar to B in the quasi-Butterworth case. Using the recursion formula for $C_n(x)$, $C_4\left(\frac{1}{\Omega}\right) = 8\left(\frac{1}{\Omega}\right)^4 - 8\left(\frac{1}{\Omega}\right)^2 + 1$ Applying this equation to $\|H(j\Omega)\|^2$ [1], $\Rightarrow \|H(\Omega)\|^2 = \frac{\frac{1 + \epsilon^2}{64\epsilon^2}\Omega^8}{\frac{1 + \epsilon^2}{64\epsilon^2}\Omega^8 + \frac{1}{4}\Omega^6 + \frac{5}{4}\Omega^4 - 2\Omega^2 + 1}$ $\Omega = \frac{\omega}{\omega_n}$ $\omega_n = \frac{\omega_{3dB}}{2}\sqrt{2 + \sqrt{2 + 2\sqrt{2+\frac{1}{\epsilon^2}}}}$ Thus, the design equations become [1]: \| \| \| \| \|------------------------------------------------------------------------------\|------------------------------------------------------------------------------\|------------------------------------------------------------------------------\| \| $\omega_0 = \omega_n\sqrt[8]{\frac{64\epsilon^2}{1+\epsilon^2}}$ \| $k = \rm{tanh}\left[\frac{1}{4}\rm{sinh}^{-1}\left(\frac{1}{\epsilon}\right)\right]$ \| $D = \frac{k^4 + 6k^2 + 1}{8}$ \| \| $a_1 = \frac{k\sqrt{4 + 2\sqrt{2}}}{\sqrt[4]{D}},$ \| $a_2 = \frac{1 + k^2(1+\sqrt{2})}{\sqrt{D}}$ \| $a_3 = \frac{a_1}{\sqrt{D}}\left[1 - \frac{1 - k^2}{2\sqrt{2}}\right]$ \| ## Choosing the Correct Alignment With all the equations that have already been presented, the question naturally arises, “Which one should I choose?” Notice that the coefficients $a_1$, $a_2$, and $a_3$ are not simply related to the parameters of the system response. Certain combinations of parameters may indeed invalidate one or more of the alignments because they cannot realize the necessary coefficients. With this in mind, general guidelines have been developed to guide the selection of the appropriate alignment. This is very useful if one is designing an enclosure to suit a particular transducer that cannot be changed. The general guideline for the Butterworth alignment focuses on $Q_L$ and $Q_{TS}$. Since the three coefficients $a_1$, $a_2$, and $a_3$ are a function of $Q_L$, $Q_{TS}$, h, and $\alpha$, fixing one of these parameters yields three equations that uniquely determine the other three. In the case where a particular transducer is already given, $Q_{TS}$ is essentially fixed. If the desired parameters of the enclosure are already known, then $Q_L$ is a better starting point. In the case that the rigid requirements of the Butterworth alignment cannot be satisfied, the quasi-Butterworth alignment is often applied when $Q_{TS}$ is not large enough.. The addition of another parameter, B, allows more flexibility in the design. For $Q_{TS}$ values that are too large for the Butterworth alignment, the Chebyshev alignment is typically chosen. However, the steep transition of the Chebyshev alignment may also be utilized to attempt to extend the bass response of the loudspeaker in the case where the transducer properties can be changed. In addition to these three popular alignments, research continues in the area of developing new algorithms that can manipulate the low-frequency response of the bass-reflex enclosure. For example, a 5th order quasi-Butterworth alignment has been developed [6]. Another example [7] applies root-locus techniques to achieve results. In the modern age of high-powered computing, other researchers have focused their efforts in creating computerized optimization algorithms that can be modified to achieve a flatter response with sharp roll-off or introduce quasi-ripples which provide a boost in sub-bass frequencies [8]. Back to Engineering Acoustics ## References [1] Leach, W. Marshall, Jr. Introduction to Electroacoustics and Audio Amplifier Design. 2nd ed. Kendall/Hunt, Dubuque, IA. 2001. [2] Beranek, L. L. Acoustics. 2nd ed. Acoustical Society of America, Woodbridge, NY. 1993. [3] DeCarlo, Raymond A. “The Butterworth Approximation.” Notes from ECE 445. Purdue University. 2004. [4] DeCarlo, Raymond A. “The Chebyshev Approximation.” Notes from ECE 445. Purdue University. 2004. [5] VanValkenburg, M. E. Analog Filter Design. Holt, Rinehart and Winston, Inc. Chicago, IL. 1982. [6] Kreutz, Joseph and Panzer, Joerg. "Derivation of the Quasi-Butterworth 5 Alignments." Journal of the Audio Engineering Society. Vol. 42, No. 5, May 1994. [7] Rutt, Thomas E. "Root-Locus Technique for Vented-Box Loudspeaker Design." Journal of the Audio Engineering Society. Vol. 33, No. 9, September 1985. [8] Simeonov, Lubomir B. and Shopova-Simeonova, Elena. "Passive-Radiator Loudspeaker System Design Software Including Optimization Algorithm." Journal of the Audio Engineering Society. Vol. 47, No. 4, April 1999. ## Appendix A: Equivalent Circuit Parameters Name Electrical Equivalent Mechanical Equivalent Acoustical Equivalent Voice-Coil Resistance $R_E$ $R_{ME} = \frac{(Bl)^2}{R_E}$ $R_{AE} = \frac{(Bl)^2}{R_ES^2_D}$ Driver (Speaker) Mass See $C_{MEC}$ $M_{MD}$ $M_{AD} = \frac{M_{MD}}{S^2_D}$ Driver (Speaker) Suspension Compliance $L_{CES} = (Bl)^2C_{MS}$ $C_{MS}$ $C_{AS} = S^2_DC_{MS}$ Driver (Speaker) Suspension Resistance $R_{ES} = \frac{(Bl)^2}{R_{MS}}$ $R_{MS}$ $R_{AS} = \frac{R_{MS}}{S^2_D}$ Enclosure Compliance $L_{CEB} = \frac{(Bl)^2C_{AB}}{S^2_D}$ $C_{MB} = \frac{C_{AB}}{S^2_D}$ $C_{AB}$ Enclosure Air-Leak Losses $R_{EL} = \frac{(Bl)^2}{S^2_DR_{AL}}$ $R_{ML} = S^2_DR_{AL}$ $R_{AL}$ Acoustic Mass of Port $C_{MEP} = \frac{S^2_DM_{AP}}{(Bl)^2}$ $M_{MP} = S^2_DM_{AP}$ $M_{AP}$ Enclosure Mass Load See $C_{MEC}$ See $M_{MC}$ $M_{AB}$ Low-Frequency Radiation Mass Load See $C_{MEC}$ See $M_{MC}$ $M_{A1}$ Combination Mass Load $C_{MEC} = \frac{S^2_DM_{AC}}{(Bl)^2}$ $= \frac{S^2_D(M_{AB} + M_{A1}) + M_{MD}}{(Bl)^2}$ $M_{MC} = S^2_D(M_{AB} + M_{A1}) + M_{MD}$ $M_{AC} = M_{AD} + M_{AB} + M_{A1}$ $= \frac{M_{MD}}{S^2_D} + M_{AB} + M_{A1}$ ## Appendix B: Enclosure Parameter Formulas Figure 7: Important dimensions of bass-reflex enclosure. Based on these dimensions [1], $C_{AB} = \frac{V_{AB}}{\rho_0c^2_0}$ $M_{AB} = \frac{B\rho_{eff}}{\pi a}$ $B = \frac{d}{3}\left(\frac{S_D}{S_B}\right)^2\sqrt{\frac{\pi}{S_D}} + \frac{8}{3\pi}\left[1 - \frac{S_D}{S_B}\right]$ $\rho_0 \leq \rho_{eff} \leq \rho_0\left(1 - \frac{V_{fill}}{V_B}\right) + \rho_{fill}\frac{V_{fill}}{V_B}$ $V_{AB} = V_B\left[1-\frac{V_{fill}}{V_B}\right]\left[1 + \frac{\gamma - 1}{1 + \gamma\left(\frac{V_B}{V_{fill}} - 1\right)\frac{\rho_0c_{air}}{\rho_{fill}c_{fill}}}\right]$ $V,= hwd$ (inside enclosure volume) $S_B = wh$ (inside area of the side the speaker is mounted on) $c_{air} =$specific heat of air at constant volume $c_{fill} =$specific heat of filling at constant volume ($V_{filling}$) $\rho_0 =$mean density of air (about 1.3 kg/$\rm m^3$) $\rho_{fill} =$ density of filling $\gamma =$ ratio of specific heats for air (1.4) $c_0 =$ speed of sound in air (about 344 m/s) $\rho_{eff}$ = effective density of enclosure. If little or no filling (acceptable assumption in a bass-reflex system but not for sealed enclosures), $\rho_{eff} \approx \rho_0$	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 185, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.90074622631073, "perplexity_flag": "middle"}
http://mathhelpforum.com/discrete-math/32092-self-contradictory-recursive-problem.html	# Thread: 1. ## self-contradictory recursive problem "Let the barber of Seville shave every man of Seville who does not shave himself. Who shall shave the barber?" So my book says that this is self-contradictory. but how? From my reading of it, it's just that the barber of Seville shaves every man in Seville who doesn't shave themselves. Well the barber doesn't have to shave himself (or he could shave himself). So then someone else can shave the barber along with every man of Seville (and those who do not shave themselves can be shaved twice...) 2. Who shaves him? If he shaves himself that is impossible for he shaves everyone who does not shave himself. If he does not shave himself then he has to shave himself because he shaves everyone who does not shave himself. 3. Originally Posted by inquilinekea "Let the barber of Seville shave every man of Seville who does not shave himself. Who shall shave the barber?" So my book says that this is self-contradictory. but how? From my reading of it, it's just that the barber of Seville shaves every man in Seville who doesn't shave themselves. Well the barber doesn't have to shave himself (or he could shave himself). So then someone else can shave the barber along with every man of Seville (and those who do not shave themselves can be shaved twice...) Another possibility is that nobody shaves the barber because she doesn't grow a beard. This question ilustrates the difference between mathematics and real life. A mathematical model of a real-life problem often contains many unstated assumptions. In this case, it is assumed that the men of Seville form a precisely identified set which contains the barber and on which the relation "shaves" is a function. That is to say, each man in the set is shaved by exactly one man in the set. 4. Self-referential statements can lead to all kinds of fun paradoxes. Whether or not a statement is actually paradoxical and how one can resolve the paradox is a topic of much discussion in philosophy and mathematics. Your problem is an example of Russel's Paradox: Does the set of all those sets that do not contain themselves contain itself? One of my favorites is: "The smallest integer not definable in fewer than twelve English words." This takes a bit of thought to see why it is a paradox. 5. Originally Posted by iknowone Your problem is an example of Russel's Paradox: Does the set of all those sets that do not contain themselves contain itself? But it is not really a paradox in math. The point of this paradox is to show that the set of all sets cannot exist. Suppose that $\mathcal{S}$ is the set of all sets then by the Axiom Schema of Compresension that set $\{ x \in \mathcal{S} \| x\not \in x \}$ will exist. But this set will lead to a contradiction. Thus, the set of all sets cannot exist.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9600068926811218, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/21044/image-charges-laplace-equation-and-uniqueness-theorem/21068	# Image charges, laplace equation and uniqueness theorem Consider a well-known problem of the electric field generated by a system composed of a point charge in proximity of a large earthed conductor. It is said that the potential due to an image charge satifies all the boundary conditions - it this case, constant potential on the surface of a conductor - and therefore, by the uniqueness theorem, it is the only possible potential distribution. However, we can of course imagine other non-trivial fields satisfying the boundary condition of constant potential on the surface, for example generated by three imaginary charges such that the whole system is again symmetric. So my question is: what is the full set of boundary conditions needed to find the unique potential distribution? This specific example is just an illustration of a more general question: how to use Laplace/Poisson equations to describe the field of point charges? The divergence of electric field has a singularity at the point charge. Another incarnation of this problem occurs when we want to use the formula for the energy stored in the electric field - we find the energy of the electric field generated by a point charge to be infinite. Does this mean that the concept of a point charge is too idealised? - Where would these three imaginary charges be placed? – David Zaslavsky♦ Feb 15 '12 at 23:46 Energy of a point charge is infinite. If you had to create a pair of point charges from a charge neutral area, you would have to (classically) supply infinite energy because of the $\frac{-kq^2}{r=0}$ infinite negative potential energy of two point charges making up a neutral point. – Manishearth♦ Feb 16 '12 at 7:42 The opposite also holds:If you destroy a pair of charges by bringing them together and making an overall charge-neutral point, then the infinite energy of the field will be radiated. Note that this is all by classical mechanics. Oh, and currently, we believe that electrons(leptons), W particles, and possibly quarks are true point charges.. – Manishearth♦ Feb 16 '12 at 7:45 ## 2 Answers However, we can of course imagine other non-trivial fields satisfying the boundary condition of constant potential on the surface, for example generated by three imaginary charges such that the whole system is again symmetric. I guess you mean that putting another set of image charges on both sides of the conductor plane so that the system is symmetric again. But point charges at specific places are also part of the boundary condition! Image charges cannot be put in the space where your electric field is to be calculated. In summary, there is only one way to put image charges, and there is only one solution of the electric field. how to use Laplace/Poisson equations to describe the field of point charges? The divergence of electric field has a singularity at the point charge. Point charges can be mathematically described by Dirac Delta function. Does this mean that the concept of a point charge is too idealised? Apparently yes, by classical electromagnetism. As electrons are recognized nowadays as true point charges, and they obviously don't have infinite energy, I guess there is some quantum effect when the radius goes down (vacuum polarization, blah blah blah). - – Willie Wong Feb 16 '12 at 15:11 There is no other solution of Laplace's equation satisfying the boundary condition of zero potential on the conductor and at infinity. This can be proved as follows: supposing there are two solutions $\phi_1$ and $\phi_2$ on one side of the conducting plate, with a point charge q at a given position. Then $-\phi_2$ has a point charge of $-q$ at the same position, so that the difference $\phi_1-\phi_2$ satisfies the free Laplace equation on the domain, with no charges. But the solution of Laplace's equation is the minimum of the field energy $$\int \|\nabla \phi\|^2$$ So that the minimum of the free equation is achieved only when the gradient is everywhere zero, and this means that $\phi$ is constant. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 7, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9389435052871704, "perplexity_flag": "head"}
http://medlibrary.org/medwiki/Linear_map	Linear map Welcome to MedLibrary.org. For best results, we recommend beginning with the navigation links at the top of the page, which can guide you through our collection of over 14,000 medication labels and package inserts. For additional information on other topics which are not covered by our database of medications, just enter your topic in the search box below: In mathematics, a linear map (also called a linear mapping, linear transformation, linear operator or, in some contexts, linear function) is a function between two modules (including vector spaces) that preserves (in the sense defined below) the operations of module (or vector) addition and scalar multiplication. As a result, it always maps linear subspaces to linear subspaces, like straight lines to straight lines or to a single point. The expression "linear operator" is commonly used for a linear map from a vector space to itself (i.e., endomorphism). Sometimes the definition of a linear function coincides with that of a linear map, while in analytic geometry it does not. In the language of abstract algebra, a linear map is a homomorphism of modules. In the language of category theory it is a morphism in the category of modules over a given ring. Definition and first consequences[] Let V and W be vector spaces over the same field K. A function f: V → W is said to be a linear map if for any two vectors x and y in V and any scalar α in K, the following two conditions are satisfied: $f(\mathbf{x}+\mathbf{y}) = f(\mathbf{x})+f(\mathbf{y}) \!$ additivity $f(\alpha \mathbf{x}) = \alpha f(\mathbf{x}) \!$ homogeneity of degree 1 This is equivalent to requiring the same for any linear combination of vectors, i.e. that for any vectors x1, ..., xm ∈ V and scalars a1, ..., am ∈ K, the following equality holds: $f(a_1 \mathbf{x}_1+\cdots+a_m \mathbf{x}_m) = a_1 f(\mathbf{x}_1)+\cdots+a_m f(\mathbf{x}_m). \!$ Denoting the zero elements of the vector spaces V and W by 0V and 0W respectively, it follows that f(0V) = 0W because letting α = 0 in the equation for homogeneity of degree 1, $f(\mathbf{0}_{V}) = f(0 \cdot \mathbf{0}_{V}) = 0 \cdot f(\mathbf{0}_{V}) = \mathbf{0}_{W} .$ Occasionally, V and W can be considered to be vector spaces over different fields. It is then necessary to specify which of these ground fields is being used in the definition of "linear". If V and W are considered as spaces over the field K as above, we talk about K-linear maps. For example, the conjugation of complex numbers is an R-linear map C → C, but it is not C-linear. A linear map from V to K (with K viewed as a vector space over itself) is called a linear functional. These statements generalize to any left-module RM over a ring R without modification. Examples[] • The identity map and zero map are linear. • The map $x\mapsto cx$, where c is a constant, is linear. • For real numbers, the map $x\mapsto x^2$ is not linear. • For real numbers, the map $x\mapsto x+1$ is not linear (but is an affine transformation, and also a linear function, as defined in analytic geometry.) • If A is a real m × n matrix, then A defines a linear map from Rn to Rm by sending the column vector x ∈ Rn to the column vector Ax ∈ Rm. Conversely, any linear map between finite-dimensional vector spaces can be represented in this manner; see the following section. • The (definite) integral is a linear map from the space of all real-valued integrable functions on some interval to R • The (indefinite) integral (or antiderivative) is not considered a linear transformation, as the use of a constant of integration results in an infinite number of outputs per input. • Differentiation is a linear map from the space of all differentiable functions to the space of all functions. • If V and W are finite-dimensional vector spaces over a field F, then functions that send linear maps f : V → W to dimF(W) × dimF(V) matrices in the way described in the sequel are themselves linear maps. • The expected value of a random variable is linear, as for random variables X and Y we have E[X + Y] = E[X] + E[Y] and E[aX] = aE[X], but the variance of a random variable is not linear, as it violates the second condition, homogeneity of degree 1: V[aX] = a2V[X]. Matrices[] Main article: Transformation matrix If V and W are finite-dimensional, and one has chosen bases in those spaces, then every linear map from V to W can be represented as a matrix; this is useful because it allows concrete calculations. Conversely, matrices yield examples of linear maps: if A is a real m × n matrix, then the rule f(x) = Ax describes a linear map Rn → Rm (see Euclidean space). Let {v1, ..., vn} be a basis for V. Then every vector v in V is uniquely determined by the coefficients c1, ..., cn in $c_1 \mathbf{v}_1+\cdots+c_n \mathbf{v}_n.$ If f: V → W is a linear map, $f(c_1 \mathbf{v}_1+\cdots+c_n \mathbf{v}_n)=c_1 f(\mathbf{v}_1)+\cdots+c_n f(\mathbf{v}_n),$ which implies that the function f is entirely determined by the values of f(v1), ..., f(vn). Now let {w1, ..., wm} be a basis for W. Then we can represent the values of each f(vj) as $f(\mathbf{v}_j)=a_{1j} \mathbf{w}_1 + \cdots + a_{mj} \mathbf{w}_m.$ Thus, the function f is entirely determined by the values of aij. If we put these values into an m × n matrix M, then we can conveniently use it to compute the value of f for any vector in V. For if we place the values of c1, ..., cn in an n × 1 matrix C, we have MC = the m × 1 matrix whose ith element is the coordinate of f(v) which belongs to the base wi. A single linear map may be represented by many matrices. This is because the values of the elements of the matrix depend on the bases that are chosen. Examples of linear transformation matrices[] In two-dimensional space R2 linear maps are described by 2 × 2 real matrices. These are some examples: • rotation by 90 degrees counterclockwise: $\mathbf{A}=\begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}$ • rotation by θ degrees counterclockwise: $\mathbf{A}=\begin{pmatrix}\cos(\theta) & -\sin(\theta)\\ \sin(\theta) & \cos(\theta)\end{pmatrix}$ • reflection against the x axis: $\mathbf{A}=\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}$ • reflection against the y axis: $\mathbf{A}=\begin{pmatrix}-1 & 0\\ 0 & 1\end{pmatrix}$ • scaling by 2 in all directions: $\mathbf{A}=\begin{pmatrix}2 & 0\\ 0 & 2\end{pmatrix}$ • horizontal shear mapping: $\mathbf{A}=\begin{pmatrix}1 & m\\ 0 & 1\end{pmatrix}$ • squeeze mapping: $\mathbf{A}=\begin{pmatrix}k & 0\\ 0 & 1/k\end{pmatrix}$ • projection onto the y axis: $\mathbf{A}=\begin{pmatrix}0 & 0\\ 0 & 1\end{pmatrix}.$ Forming new linear maps from given ones[] The composition of linear maps is linear: if f: V → W and g: W → Z are linear, then so is their composition g o f: V → Z. It follows from this that the class of all vector spaces over a given field K, together with K-linear maps as morphisms, forms a category. The inverse of a linear map, when defined, is again a linear map. If f1: V → W and f2: V → W are linear, then so is their sum f1 + f2 (which is defined by (f1 + f2)(x) = f1(x) + f2(x)). If f : V → W is linear and a is an element of the ground field K, then the map af, defined by (af)(x) = a (f(x)), is also linear. Thus the set L(V, W) of linear maps from V to W itself forms a vector space over K, sometimes denoted Hom(V, W). Furthermore, in the case that V = W, this vector space (denoted End(V)) is an associative algebra under composition of maps, since the composition of two linear maps is again a linear map, and the composition of maps is always associative. This case is discussed in more detail below. Given again the finite-dimensional case, if bases have been chosen, then the composition of linear maps corresponds to the matrix multiplication, the addition of linear maps corresponds to the matrix addition, and the multiplication of linear maps with scalars corresponds to the multiplication of matrices with scalars. Endomorphisms and automorphisms[] A linear transformation f: V → V is an endomorphism of V; the set of all such endomorphisms End(V) together with addition, composition and scalar multiplication as defined above forms an associative algebra with identity element over the field K (and in particular a ring). The multiplicative identity element of this algebra is the identity map id: V → V. An endomorphism of V that is also an isomorphism is called an automorphism of V. The composition of two automorphisms is again an automorphism, and the set of all automorphisms of V forms a group, the automorphism group of V which is denoted by Aut(V) or GL(V). Since the automorphisms are precisely those endomorphisms which possess inverses under composition, Aut(V) is the group of units in the ring End(V). If V has finite dimension n, then End(V) is isomorphic to the associative algebra of all n × n matrices with entries in K. The automorphism group of V is isomorphic to the general linear group GL(n, K) of all n × n invertible matrices with entries in K. Kernel, image and the rank–nullity theorem[] If f : V → W is linear, we define the kernel and the image or range of f by $\operatorname{\ker}(f)=\{\,x\in V:f(x)=0\,\}$ $\operatorname{im}(f)=\{\,w\in W:w=f(x),x\in V\,\}$ ker(f) is a subspace of V and im(f) is a subspace of W. The following dimension formula is known as the rank–nullity theorem: $\dim(\ker( f ))+ \dim(\operatorname{im}( f ))= \dim( V ).$ The number dim(im(f)) is also called the rank of f and written as rank(f), or sometimes, ρ(f); the number dim(ker(f)) is called the nullity of f and written as null(f) or ν(f). If V and W are finite-dimensional, bases have been chosen and f is represented by the matrix A, then the rank and nullity of f are equal to the rank and nullity of the matrix A, respectively. Cokernel[] Main article: Cokernel A subtler invariant of a linear transformation is the cokernel, which is defined as $\mathrm{coker}\,f := W/f(V) = W/\mathrm{im}(f).$ This is the dual notion to the kernel: just as the kernel is a subspace of the domain, the co-kernel is a quotient space of the target. Formally, one has the exact sequence $0 \to \ker f \to V \to W \to \mathrm{coker}\,f \to 0.$ These can be interpreted thus: given a linear equation f(v) = w to solve, • the kernel is the space of solutions to the homogeneous equation f(v) = 0, and its dimension is the number of degrees of freedom in a solution, if it exists; • the co-kernel is the space of constraints that must be satisfied if the equation is to have a solution, and its dimension is the number of constraints that must be satisfied for the equation to have a solution. The dimension of the co-kernel and the dimension of the image (the rank) add up to the dimension of the target space. For finite dimensions, this means that the dimension of the quotient space W/f(V) is the dimension of the target space minus the dimension of the image. As a simple example, consider the map f: R2 → R2, given by f(x, y) = (0, y). Then for an equation f(x, y) = (a, b) to have a solution, we must have a = 0 (one constraint), and in that case the solution space is (x, b) or equivalently stated, (0, b) + (x, 0), (one degree of freedom). The kernel may be expressed as the subspace (x, 0) < V: the value of x is the freedom in a solution – while the cokernel may be expressed via the map W → R, $(a,b) \mapsto (a):$ given a vector (a, b), the value of a is the obstruction to there being a solution. An example illustrating the infinite-dimensional case is afforded by the map f: R∞ → R∞, $\{a_n\} \mapsto \{b_n\}$ with b1 = 0 and bn + 1 = an for n > 0. Its image consists of all sequences with first element 0, and thus its cokernel consists of the classes of sequences with identical first element. Thus, whereas its kernel has dimension 0 (it maps only the zero sequence to the zero sequence), its co-kernel has dimension 1. Since the domain and the target space are the same, the rank and the dimension of the kernel add up to the same sum as the rank and the dimension of the co-kernel ( $\aleph_0 + 0 = \aleph_0 + 1$ ), but in the infinite-dimensional case it cannot be inferred that the kernel and the co-kernel of an endomorphism have the same dimension (0 ≠ 1). The reverse situation obtains for the map h: R∞ → R∞, $\{a_n\} \mapsto \{c_n\}$ with cn = an + 1. Its image is the entire target space, and hence its co-kernel has dimension 0, but since it maps all sequences in which only the first element is non-zero to the zero sequence, its kernel has dimension 1. Index[] For a linear operator with finite-dimensional kernel and co-kernel, one may define index as: $\mathrm{ind}\,f := \dim \ker f - \dim \mathrm{coker}\,f,$ namely the degrees of freedom minus the number of constraints. For a transformation between finite-dimensional vector spaces, this is just the difference dim(V) − dim(W), by rank–nullity. This gives an indication of how many solutions or how many constraints one has: if mapping from a larger space to a smaller one, the map may be onto, and thus will have degrees of freedom even without constraints. Conversely, if mapping from a smaller space to a larger one, the map cannot be onto, and thus one will have constraints even without degrees of freedom. The index comes of its own in infinite dimensions: it is how homology is defined, which is a central theory in algebra and algebraic topology; the index of an operator is precisely the Euler characteristic of the 2-term complex 0 → V → W → 0. In operator theory, the index of Fredholm operators is an object of study, with a major result being the Atiyah–Singer index theorem. Algebraic classifications of linear transformations[] No classification of linear maps could hope to be exhaustive. The following incomplete list enumerates some important classifications that do not require any additional structure on the vector space. Let V and W denote vector spaces over a field, F. Let T: V → W be a linear map. • T is said to be injective or a monomorphism if any of the following equivalent conditions are true: • T is one-to-one as a map of sets. • kerT = {0V} • T is monic or left-cancellable, which is to say, for any vector space U and any pair of linear maps R: U → V and S: U → V, the equation TR = TS implies R = S. • T is left-invertible, which is to say there exists a linear map S: W → V such that ST is the identity map on V. • T is said to be surjective or an epimorphism if any of the following equivalent conditions are true: • T is onto as a map of sets. • coker T = {0W} • T is epic or right-cancellable, which is to say, for any vector space U and any pair of linear maps R: W → U and S: W → U, the equation RT = ST implies R = S. • T is right-invertible, which is to say there exists a linear map S: W → V such that TS is the identity map on W. • T is said to be an isomorphism if it is both left- and right-invertible. This is equivalent to T being both one-to-one and onto (a bijection of sets) or also to T being both epic and monic, and so being a bimorphism. • If T: V → V is an endomorphism, then: • If, for some positive integer n, the n-th iterate of T, Tn, is identically zero, then T is said to be nilpotent. • If T2 = T, then T is said to be idempotent • If T = kI, where k is some scalar, then T is said to be a scaling transformation or scalar multiplication map; see scalar matrix. Change of basis[] Given a linear map whose matrix is A, in the basis B of the space it transforms vectors coordinates [u] as [v] = A[u]. As vectors change with the inverse of B, its inverse transformation is [v] = B[v']. Substituting this in the first expression $B[v'] = AB[u']$ hence $[v'] = B^{-1}AB[u'] = A'[u'].$ Therefore the matrix in the new basis is A′ = B−1AB, being B the matrix of the given basis. Therefore linear maps are said to be 1-co 1-contra -variant objects, or type (1, 1) tensors. Continuity[] Main article: Discontinuous linear map A linear transformation between topological vector spaces, for example normed spaces, may be continuous. If its domain and codomain are the same, it will then be a continuous linear operator. A linear operator on a normed linear space is continuous if and only if it is bounded, for example, when the domain is finite-dimensional. An infinite-dimensional domain may have discontinuous linear operators. An example of an unbounded, hence discontinuous, linear transformation is differentiation on the space of smooth functions equipped with the supremum norm (a function with small values can have a derivative with large values, while the derivative of 0 is 0). For a specific example, sin(nx)/n converges to 0, but its derivative cos(nx) does not, so differentiation is not continuous at 0 (and by a variation of this argument, it is not continuous anywhere). Applications[] A specific application of linear maps is for geometric transformations, such as those performed in computer graphics, where the translation, rotation and scaling of 2D or 3D objects is performed by the use of a transformation matrix. Another application of these transformations is in compiler optimizations of nested-loop code, and in parallelizing compiler techniques. References[] • Halmos, Paul R. (1974), Finite-dimensional vector spaces, New York: Springer-Verlag, ISBN 978-0-387-90093-3 [Amazon-US \| Amazon-UK] • Lang, Serge (1987), Linear algebra, New York: Springer-Verlag, ISBN 978-0-387-96412-6 [Amazon-US \| Amazon-UK] Content in this section is authored by an open community of volunteers and is not produced by, reviewed by, or in any way affiliated with MedLibrary.org. Licensed under the Creative Commons Attribution-ShareAlike 3.0 Unported License, using material from the Wikipedia article on "Linear map", available in its original form here: http://en.wikipedia.org/w/index.php?title=Linear_map • Finding More You are currently browsing the the MedLibrary.org general encyclopedia supplement. To return to our medication library, please select from the menu above or use our search box at the top of the page. In addition to our search facility, alphabetical listings and a date list can help you find every medication in our library. • Questions or Comments? If you have a question or comment about material specifically within the site’s encyclopedia supplement, we encourage you to read and follow the original source URL given near the bottom of each article. You may also get in touch directly with the original material provider. • About This site is provided for educational and informational purposes only and is not intended as a substitute for the advice of a medical doctor, nurse, nurse practitioner or other qualified health professional.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 30, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9001452326774597, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/190656/geometric-proof-that-left-fraci-ziz-right1-for-z-in-mathbbh	Geometric proof that $\left\|\frac{i-z}{i+z}\right\|<1$ for $z\in \mathbb{H}$ I just read a proof in a complex analysis book that says there exists a biholomorphic (conformal) map between $\mathbb{H}$ and $\mathbb{D}$ and I don't understand one step in the proof, namely the norm of the map (in the title) is less than $1$, using a purely "geometric" argument (i.e. a picture was in the book). I don't understand "geometric division", so I don't immediately see the proof "geometrically", please I do NOT want an algebraic answer (this is fairly straight forward anyway). - 5 Answers Note that $\|i-z\|$ is the distance between $z$ and $i$, and that $\|i+z\|$ is the distance between $z$ and $-i$. The inequality says that $z$ lies closer to $i$ than to $-i$, which is true for points in the upper halfplane. To make this even more clear, write $\bigl\|{i-z \over i+z}\bigr\|={\|i-z\|\over \|i+z\|}<1$, and multiply through by the denominator $\|i+z\|$. - The real axis is in the middle between $i$ and $-i$, hence is the locus of all points having the same distance from $i$ and $-i$. The rest of $\mathbb C$ splits into those points closer to $i$ than to $-i$ (the upper half plane) and those closer to $-i$ (lower half plane). - Basically you want to prove that any point in $\mathbb{H}$ is closer to $i$ than to $-i$. Draw the triangle with vertices at $i, -i, z$ and observe that the angle at $-i$ is smaller than the angle at $i$. he easiest way to prove this is by extending the line through $i z$ till it intersects the real axix at $y$ and compare the angles in the triangle $i,-i,z$ with the angles in $i,-i,y$. - The norms of $i - z$ and $i+z$ are the squares of the distances between $z$ and $\pm i$. A point $z$ lies in the upper half plane if and only if it is strictly closer to $i$ than to $-i$, so that this ratio has norm strictly less than $1$ and so lies in the unit disk. - \|z-i\|=\|z+i\| is the set of points equidistant from $i$ and $-i$, the real line. \|z-i\|<\|z+i\| is the set of points closer to $i$ than $-i$, the upper half plane. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 46, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9620217084884644, "perplexity_flag": "head"}
http://mathematica.stackexchange.com/questions/5362/finding-a-fit-to-a-multi-dimensioned-function?answertab=oldest	# Finding a fit to a multi-dimensioned function I have a model function $f:\mathbb{R}^2\rightarrow\mathbb{R}^2$, and a bunch of data points for which I'd like Mathematica to fit for me. Unfortunately `FindFit` seems to only deal with functions $\mathbb{R}^n\rightarrow\mathbb{R}$. I guess I could make my own square difference from the data to the model and use `NMinimize` on that, but I wondering if there was an easier way? ### Edit - toy example with answers $g\left(x,y\right)=\left<abx+(2c+d)y,(2a+b)x+cdy\right>$ ````g[{x_, y_}, {a_, b_, c_, d_}] := {a b x + (2 c + d) y, (2 a + b) x + c d y}; points = Flatten[Table[{{x, y}, g[{x, y}, {1, 2, 3, 4}]}, {x, -1, 1, 0.5}, {y, -1, 1, 0.5}], 1]; ```` Finding a fit for each component separately doesn't work: ````data1 = points /. {{x_, y_}, {u_, v_}} -> {x, y, u}; data2 = points /. {{x_, y_}, {u_, v_}} -> {x, y, v}; FindFit[data1, g[{x, y}, {a, b, c, d}][[1]], {a, b, c, d}, {x, y}] FindFit[data2, g[{x, y}, {a, b, c, d}][[2]], {a, b, c, d}, {x, y}] ```` `{a -> 0.729723, b -> 2.74077, c -> 3.35659, d -> 3.28681}` `{a -> 1.48555, b -> 1.02891, c -> 2.81936, d -> 4.25629}` Jens suggestion works well: ````data3 = points /. {{x_, y_}, {u_, v_}} -> Sequence[{x, y, 0, u}, {x, y, 1, v}]; FindFit[data3, g[{x, y}, {a, b, c, d}].{1 - s, s}, {a, b, c, d}, {x, y, s}] ```` `{a -> 1., b -> 2., c -> 3., d -> 4.}` My original suggestion (which is similar to user840's links) ````error = Plus @@ (points /. {{x_, y_}, {u_, v_}} -> Norm[g[{x, y}, {a, b, c, d}] - {u, v}]^2); NMinimize[error, {a, b, c, d}] ```` Gives a different, but still correct, solution (or at least a close approximation thereof) `{1.4175410^-13, {a -> 1.00004, b -> 1.99993, c -> 2., d -> 6.}}` - How about using NonlinearModelFit? Take a look into the documentation and the examples. You will see that under Scope the first example shows how one can fit a model of more than one variable. If you provide example data, it would be possible to help you more detailed. – partial81 May 9 '12 at 5:53 @partial81 More than one variable, not but not a vector value. That example is for $\mathbb{R}^2 \rightarrow \mathbb{R}$ while the OP wants $\mathbb{R}^2 \rightarrow \mathbb{R}^2$. `NonlinearModelFit` doesn't seem to support this. – Szabolcs May 9 '12 at 8:11 ok, I assume you are right. I would love to try it with `NonlinearModelFit`, but without example data, I do not want to spend more time on this problem. – partial81 May 9 '12 at 10:54 Why not two fits, one for each dependent variable? – Daniel Lichtblau May 9 '12 at 18:30 @DanielLichtblau Each variable depends on all the model parameters, so I end up with different parameter values for the different fits. – wxffles May 9 '12 at 21:55 show 2 more comments ## 3 Answers Without a specific example, I can only make general suggestions here. The first thing that comes to mind is that you could fit each of the two components of your vector field independently. I.e., if $\vec{f}: \mathbb{R}^2\to\mathbb{R}^2$ is split up into $\vec{f} = \{f_x, f_y\}$ with $f_{1,2}: \mathbb{R}^2\to\mathbb{R}$, then `FindFit` would work on each of these component functions. If you don't want the fits to be determined independently, it could still be possible to use `FindFit` by introducing an auxiliary variable $s$ that labels the two component functions above, $f_s$, and then provide the fitting data with this variable `s` included. Here, `s` can only have two discrete values (I borrowed the idea from spin-half quantum mechanics). Let's choose `s = 0` for the function $f_x$ and `s = 1` for $f_y$. So you'd have data in the form ````data = {{x1, y1, 0, fx1}, {x1, y1, 1, fy1}, ...} ```` where `fx1` is the first value of $f_x$ and `fy1` the first value of $f_y$, both at the point `x1, y1`. Your model could then look something like this: ````model[x_, y_, s_] := modely[x, y]s + modelx[x, y]*(1 - s) ```` where `modely[x, y]` is the model for $f_y$ and `modelx[x, y]` is the model for $f_x$. The variables in the `FindFit` call would be `x, y, s`, and the parameters in the models `modelx`, `modely` could be the same (or different). There are probably many other reasonable ways to do such a fit, but these are some ideas. - Thanks. This is working well for my problem. `FindFit[data /. {{a_, b_}, {p_, q_}} -> Sequence[{a, b, 0, p}, {a, b, 1, q}], model.{1 - s, s}, ...]` – wxffles May 9 '12 at 21:33 Multi-layer perceptrons have this ability. The Neural network package from Wolfram offers this functionality. Neural Networks - Since that isn't part of the standard suite of packages, could you provide a link? – rcollyer May 9 '12 at 13:14 True, but now you have the additional problem of structuring the neural net appropriately (which is by no means trivial if you don't have any experience with neural nets). And after that you have to train it, which is equivalent to the original problem... – Oleksandr R. May 9 '12 at 13:46 Yes there is some effort involved, but it offers a different approach to the solution, which I thought might fill in parts of the landscape of the large selection of approximation techniques that might be deployed outside of the half a dozen or so that `FindFit` offers and MLPs have inherent support for output vectors in higher dimensions. – image_doctor May 9 '12 at 15:35 You can check Fitting Multiple-Response Data from Bruce Miller. He gives a step-by-step example of how to perform a simultaneous fit to multiple functions, and provides a fairly general function for performing such fits. Alternately, Multiple-Response Fitting notes from wolframThis package contains functions for simultaneously fitting multiple functions to data with a possibly shared set of parameters. It has flexible output options and the first part of the accompanying demonstration notebook can serve as an introduction to writing one's own fitting codes using FindMinimum. - lang-mma	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 16, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.905121386051178, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/tagged/quantum-mechanics+research-level	# Tagged Questions 1answer 32 views ### Name of a state with $d-1$ excitations, distributed uniformly among $n$ qudits Is there a particular name for a quantum state of the form (up to the normalization): $$\sum_{i_1+\ldots+i_n = d-1} \|i_1\rangle \|i_2\rangle \ldots \|i_n\rangle$$ or was it studied is some papers? ... 0answers 101 views ### Looking for modern results in semiclassical physics and relevant references What are some important approximations, especially those that are state-of-the-art, used to approximate the many-body dynamics of atoms and molecules in the semiclassical regime? To be clear, I'm not ... 2answers 100 views ### How is the energy/eigenvalue gap plot drawn for adiabatic quantum computation? I was going through arXiv:quant-ph/0001106v1, the first paper by Farhi on adiabatic quantum computation. Equation 2.24 says, $$\tilde{H}(s) = (1-s)H_B + sH_P$$ which means the adiabatic evolution ... 0answers 110 views ### Local explanation of the Aharonov-Bohm effect in terms of force fields Here is an interesting paper for the Physics SE community: On the role of potentials in the Aharonov-Bohm effect, Lev Vaidman, published in PHYSICAL REVIEW A 86, 040101(R) (2012). You should check it ... 1answer 164 views ### Are Born-Oppenheimer energies analytic functions of nuclear positions? I am looking for references to bibliography that explores the smoothness and analyticity of eigenvalues and eigenfunctions (and matrix elements in general) of a hamiltonian that depends on some ... 1answer 331 views ### Schwinger representation of operators for n-particle 2-mode symmetric states A bosonic (i.e. permutation-symmetric) state of $n$ particles in $2$ modes can be written as a homogenous polynomial in the creation operators, that is \left(c_0 \hat{a}^{\dagger n} + c_1 ... 1answer 563 views ### Is resonating valence bond (RVB) states long-range entangled? Quantum liquid is at the core of condensed matter theory study, examples include superfluid in Bose Hubbard model, quantum spin liquid around the RK point of a quantum dimer model, string-net ... 10answers 1k views ### What is spontaneous symmetry breaking in QUANTUM systems? Most descriptions of spontaneous symmetry breaking, even for spontaneous symmetry breaking in quantum systems, actually only give a classical picture. According to the classical picture, spontaneous ... 0answers 109 views ### What is a Hilbert space filter? In a recent paper, Side-Channel-Free Quantum Key Distribution, by Samuel L. Braunstein and Stefano Pirandola. Phys. Rev. Lett. 108, 130502 (2012). doi:10.1103/PhysRevLett.108.130502, ... 1answer 158 views ### partial trace with sparse matrices Let $\rho_{ABCD}$ be a sparse matrix of 4 systems each in a $d$-dimensional Hilbert space. For $d<7$ in a reasonable time (few seconds) I able to perform the partial trace $\rho_{AD}$ using the ... 3answers 99 views ### What is the physical difference between states and unital completely positive maps? Mathematically, completely positive maps on C-algebras generalize positive linear functionals in that every positive linear functional on a C-algebra $A$ is a completely positive map of $A$ into ... 2answers 56 views ### Heuristics for definitions of open and closed quantum dynamics I've been reading some of the literature on "open quantum systems" and it looks like the following physical interpretations are made: Reversible dynamics of a closed quantum system are represented ... 6answers 105 views ### Multiqubit state tomography by performing measurement in the same basis For a $n$-qubit state $\rho$ we perform all projective measurement consisting of one-particle measurements in the same basis, that is, p_{i_1i_2\ldots i_n}(\theta,\varphi) = \text{Tr}\left \{ \rho ... 1answer 109 views ### Monte Carlo integration over space of quantum states I am currently facing the problem of calculating integrals that take the general form $\int_{R} P(\sigma)d\sigma$ where $P(\sigma)$ is a probability density over the space of mixed quantum states, ... 1answer 452 views ### Entanglement in time Quantum entanglement links particles through time, according to this study that received some publicity last year: New Type Of Entanglement Allows 'Teleportation in Time,' Say Physicists at The ... 1answer 34 views ### Tip of a spreading wave-packet: asymptotics beyond all orders of a saddle point expansion This is a technical question coming from mapping of an unrelated problem onto dynamics of a non-relativistic massive particle in 1+1 dimensions. This issue is with asymptotics dominated by a term ... 1answer 46 views ### Unknown quantum state with promise of classical data I am trying to solve a problem in the measurement and identification of quantum states with a promise as to what states it could be. Here is the problem. Imagine a system that produces qubits in ... 3answers 68 views ### Constructing a Hamiltonian (as a polynomial of $q_i$ and $p_i$) from its spectrum For a countable sequence of positive numbers $S=\{\lambda_i\}_{i\in N}$ is there a construction producing a Hamiltonian with spectrum $S$ (or at least having the same eigenvalues for $i\leq s$ for ... 0answers 41 views ### Spectrum of a quantum relativistic “distance squared” operator This question disusses the same concepts as that question (this time in quantum context). Consider a relativistic system in spacetime dimension $D$. Poincare symmetry yields the conserved charges $M$ ... 1answer 86 views ### random matrix ensembles from BMN model My friends working on Thermalization of Black Holes explained solutions to their matrix-valued differential equations (from numerical implementation of the Berenstein-Maldacena-Nastase matrix model) ... 1answer 73 views ### Many body quantum states analyzed as probabilistic sequences Measurements of consecutive sites in a many body qudit system (e.q. a spin chain) can be interpreted as generating a probabilistic sequence of numbers $X_1 X_2 X_3 \ldots$, where \$X_i\in ... 3answers 130 views ### POVMs that do not require enlargement of the Hilbert space The usual justification for regarding POVMs as fundamental measurements is via Neumark's theorem, i.e., by showing that they can always be realized by a projective measurement in a larger Hilbert ... 1answer 37 views ### Optimality of the CHSH strategy The maximum achievable probability of the Clauser-Horne-Shimony-Holt game is $\cos^2(\pi/8)\approx85.355\%,$ which can be proved with Tsirelson's inequality. But I don't imagine that this remained ... 1answer 48 views ### Spekkens Toy Model, Internal Comonoids I have been thinking about Spekkens Toy model in terms of interfaces. The Spekkens paper concerns a physics based on only being able to receive answers to half the number of questions necessary to ... 1answer 35 views ### Convexity — reference request I've been reading a few papers on generalized probabilistic theories, and have been struggling through proofs of some results that involve use of convexity and group theory, e.g. this paper on bit ... 3answers 134 views ### Hilbert-Schmidt basis for many qubits - reference Every density matrix of $n$ qubits can be written in the following way \hat{\rho}=\frac{1}{2^n}\sum_{i_1,i_2,\ldots,i_n=0}^3 t_{i_1i_2\ldots i_n} ... 1answer 65 views ### How does one geometrically quantize the Bloch equations? I've just now rated David Bar Moshe's post (below) as an "answer", for which appreciation and thanks are given. Nonetheless there's more to be said, and in hopes of stimulating further posts, I've ... 3answers 364 views ### Rigorous proof of Bohr-Sommerfeld quantization Bohr-Sommerfeld quantization provides an approximate recipe for recovering the spectrum of a quantum integrable system. Is there a mathematically rigorous explanation why this recipe works? In ... 0answers 424 views ### Experimental test of the non-statisticality theorem? Context: The recent paper The quantum state cannot be interpreted statistically by Pusey, Barrett and Rudolph shows under suitable assumptions that the quantum state cannot be interpreted as a ... 1answer 45 views ### Accurate quantum state estimation via “Keeping the experimentalist honest” Bob has a black-box, with the label "V-Wade", which he has been promised prepares a qubit which he would like to know the state of. He asks Alice, who happens also to be an experimental physicist, to ... 1answer 81 views ### Quantum mechanical gravitational bound states The quantum mechanics of Coloumb-force bound states of atomic nuclei and electrons lead to the extremely rich theory of molecules. In particular, I think the richness of the theory is related to the ... 1answer 69 views ### Quantum mechanics as a Markov process I am currently involved in some understanding on this matter with a colleague of mine. I know all the literature about but I do not know the state of art. Please, could you provide some relevant ... 1answer 42 views ### Allowed states vis-a-vis allowed dynamics in generalized probabilistic theories (GPTs) In his work on information processing in GPTs http://arxiv.org/abs/quant-ph/0508211 Barrett speculates that the trade-off between allowed states and the allowed dynamics in a GPT is optimal in quantum ... 2answers 123 views ### Geometric quantization of identical particles Background: It is well known that the quantum mechanics of $n$ identical particles living on $\mathbb{R}^3$ can be obtained from the geometric quantization of the cotangent bundle of the manifold ... 3answers 91 views ### Which are the simplest known contextual inequalities? It is well-known that quantum mechanics does not admit a noncontextual ontological model, and there are countless different proofs of it. I'm interested in the simplest proofs that can be cast as an ... 1answer 111 views ### Majorana-like representation for mixed symmetric states? Is there a generalization of the Majorana representation of pure symmetric $n$-qubit states to mixed states (made of pure symmetric $n$-qubit)? By Majorana representation I mean the decomposition of ... 1answer 129 views ### What proof techniques have failed for solving the SIC-POVM problem and what new insights have been gleaned from them? The SIC-POVM problem is remarkably easy to state given that it has not yet been solved. It goes like this. With dim($\mathcal H$) $=d$, find states $\|\psi_k\rangle\in\mathcal H$, $k=1,\ldots,d^2$ ... 1answer 55 views ### Explicit construction for unitary extensions of CPTP maps? Given a completely positive and trace preserving map $\Phi : \textrm{L}(\mathcal{H})\to\textrm{L}(\mathcal{G})$, it is clear by the Kraus representation theorem that there exist \$A_k \in ... 1answer 81 views ### What is the Holevo-Schumacher-Westmoreland capacity of a Pauli channel? Suppose you are given an $n$-qubit quantum channel defined as $\mathcal{E}(\rho) = \sum_{i} p_i X_i \rho X_i^\dagger$, where $X_i$ denotes an $n$-fold tensor product of Pauli matrices and $\{p_i\}$ is ... 2answers 117 views ### Relevance of SIC-POVMs to quantum information What is the real relevance of SIC-POVMs (symmetric informationally complete POVMs) to concrete tasks in quantum information theory? A lot of work has been put into giving explicit constructions, and ... 1answer 31 views ### Connections of iterative solvers for large systems of equation in Physics? I am trying to find the domains in physics where solving large systems of equations is computationally expensive. The sparse systems are of my particular interest, where the input matrix A is in GBs ... 4answers 208 views ### direct sum of anyons? In the topological phase of a fractional quantum Hall fluid, the excitations of the ground state (quasiparticles) are anyons, at least conjecturally. There is then supposed to be a braided fusion ... 11answers 827 views ### Negative probabilities in quantum physics Negative probabilities are naturally found in the Wigner function (both the original one and its discrete variants), the Klein paradox (where it is an artifact of using a one-particle theory) and the ... 6answers 252 views ### Is there a theorem that says that QFT reduces to QM in a suitable limit? A theorem similar to Ehrenfest's theorem? Is there a theorem that says that QFT reduces to QM in a suitable limit? Of course, it should be, as QFT is relativisitc quantum mechanics. But, is there a more manifest one? such as Ehrenfest's ... 0answers 174 views ### Orbits of maximally entangled mixed states It is well known (Please, see for example Geometry of quantum states by Bengtsson and Życzkowski ) that the set of $N-$dimensional density matrices is stratified by the adjoint action of $U(N)$, where ... 1answer 49 views ### Principle behind fidelity balance in quantum cloning If we do optimal state estimation on an unknown qubit, we can recreate a state with fidelity $F_c=2/3$ with respect to the original. Let us define the "quantum information content" $I_q=1-2/3=1/3$ as ... 1answer 61 views ### Quasiparticles in Bohmian mechanics My questions are about de Broglie-Bohm "pilot wave" interpretation of quantum mechanics (a.k.a. Bohmian mechanics). Do quasiparticles have any meaning in Bohmian mechanics, or not? Specifically, is ... 2answers 50 views ### Counting complete sets of mutually unbiased bases composed of stabilizer states Consider $N$ qubits. There are many complete sets of $2^N+1$ mutually unbiased bases formed exclusively of stabilizer states. How many? Each complete set can be constructed as follows: partition the ... 1answer 23 views ### Sub and super multiplicativity of norms for understanding non-locality In relation to various problems in understanding entanglement and non-locality, I have come across the following mathematical problem. It is most concise by far to state in its most mathematical form ... 1answer 51 views ### Stabilizer formalism for symmetric spin-states? This question developed out of conversation between myself and Joe Fitzsimons. Is there a succinct stabilizer representation for symmetric states, on systems of n spin-1/2 or (more generally) n higher ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 47, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9068542718887329, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/31637/list	## Return to Answer 3 pedantic TeX correction Since I could use the reputation... letting $\psi: G \to Diff(M)$ \operatorname{Diff}(M)$be the action then$X' = \psi_* X$, where we identify$Lie(Diff(M))$\operatorname{Lie}(\operatorname{Diff}(M))$ with vector fields on $M$. This is because they both correspond to the one-parameter subgroup $t\mapsto \psi(\exp tX)$ of $Diff(M)$. \operatorname{Diff}(M)$. Then$[X', Y'] = [X,Y]'\$ is immediate. 2 added 107 characters in body Since I could use the reputation... letting $\psi: G \to Diff(M)$ be the action then $X' = \psi_* X$, where we identify $Lie(Diff(M))$ with vector fields on $M$. This is because they both correspond to the one-parameter subgroup $t\mapsto \psi(\exp tX)$ of $Diff(M)$. Then $[X', Y'] = [X,Y]'$ is immediate. 1 Since I could use the reputation... letting $\psi: G \to Diff(M)$ be the action then $X' = \psi_* X$, where we identify $Lie(Diff(M))$ with vector fields on $M$. Then $[X', Y'] = [X,Y]'$ is immediate.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 20, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8569443821907043, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/107938/list	## Return to Question 2 added 8 characters in body A number of sources concerning Speiser's 1934 result state that the Riemann Hypothesis (RH) implies $\zeta'(s)\neq 0$ for all $0<\text{Re}(s)<1/2$. But I have seen some (possibly less reliable) sources without proof suggesting this is an if and only if relationship, i.e. RH$\Leftrightarrow\zeta'(s)\neq RH$\Longleftrightarrow\zeta'(s)\neq 0$. However, those (perhaps more reliable) sources state only forward implication, i.e. RH$\Rightarrow\zeta'(s)\neq RH$\Longrightarrow\zeta'(s)\neq 0$. My question is this: is Speiser's result an if and only if relationship or not? 1 # A question about Speiser's 1934 result on the Riemann hypothesis A number of sources concerning Speiser's 1934 result state that the Riemann Hypothesis (RH) implies $\zeta'(s)\neq 0$ for all $0<\text{Re}(s)<1/2$. But I have seen some (possibly less reliable) sources without proof suggesting this is an if and only if relationship, i.e. RH$\Leftrightarrow\zeta'(s)\neq 0$. However, those (perhaps more reliable) sources state only forward implication, i.e. RH$\Rightarrow\zeta'(s)\neq 0$. My question is this: is Speiser's result an if and only if relationship or not?	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 9, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9003283977508545, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/tagged/convergence+calculus	# Tagged Questions 2answers 33 views ### Checking $\displaystyle \int_{0}^{\infty} {\frac{\sin^2x}{\sqrt[3]{x^7 + 1}} dx}$ for convergence Given $\displaystyle\int_{0}^{\infty} {\frac{\sin^2x}{\sqrt[3]{x^7 + 1}} dx}$, prove that it converges. So of course, I said: We have to calculate \$\displaystyle \lim_{b \to \infty} ... 2answers 18 views ### Develop the next function:$f(x)=\frac{4x+53}{x^2-x-30}$ into power series, Find the radius on convergence and find $f^{(20)}(0)$ Develop the next function:$\displaystyle f(x)=\frac{4x+53}{x^2-x-30}$ into power series, Find the radius on convergence and find $f^{(20)}(0).$ For the first part: \$\displaystyle\frac ... 7answers 124 views ### How to prove that $1/n!$ is less than $1/n^2$? I want to prove $$\sum_{n=0}^{\infty} \frac{1}{n!}$$ is a converging series. So I want to compare it with $\sum_{n=0}^{\infty} \frac{1}{n^2}$. I want to do direct comparison test. How to prove ... 1answer 16 views ### Absolute convergence. Determine if absolutely convergent or not; Justify. $$\sum_{n=1}^\infty (-1)^n n^2 3^{1-n} x^n \text{ s.t }\|x\|<3$$ if we take the abs value of $(-1)^n$ we are left with $n^{2} 3^{1-n} x^{n}$ now ... 3answers 28 views ### behavior of a sequence Imaging a sequence $a_{k} \in \Omega$ with $\Omega \subset \Bbb{R}$ closed, $\lim\limits_{k \to \infty} \\| a_{k+1} - a_{k} \\| = 0$. My Professor said that because of this the sequence would ... 2answers 32 views ### What's the radius of convergence of the next sum: $\sum_{n=0}^\infty (\int_o^n\frac{\sin^2t}{\sqrt[3]{t^7+1}}dt)x^n$ What's the radius of convergence of the next sum: $$\sum_{n=0}^\infty \left(\int_0^n\frac{\sin^2t}{\sqrt[3]{t^7+1}}dt\right)x^n$$ I know that $$\int_0^\infty\frac{\sin^2t}{\sqrt[3]{t^7+1}}dt$$ does ... 1answer 49 views ### Conditions for taking a limit into an infinite sum Suppose $f\left(x\right)={\displaystyle \sum_{n=0}^{\infty}g_{n}\left(x\right)}$ under what conditions is it true that: \lim_{x\to c}f\left(x\right)={\displaystyle \sum_{n=0}^{\infty}\lim_{x\to ... 1answer 36 views ### Radius of convergence of a power series with Bernoulli numbers Say, we use the definition: Bernoulli numbers arise in Taylor series in the expansion $$\frac{x}{e^x-1}=\sum_{k=0}^\infty B_k \frac{x^k}{k!}$$ and then derive power series representations of the ... 1answer 30 views ### Calculate (& determine if converges): $\sum_{n=1}^\infty (-1)^n\frac 1n(0.5)^n$ Calculate (& determine if converges): $\sum_{n=1}^\infty (-1)^n\frac 1n(0.5)^n$ The above is a specific equation for $x_0=0.5$, but from here i'm pretty much stuck. 1answer 67 views ### Radius of convergence of: $\sum_{n=1}^\infty\frac {x^{n!}}{n!}$? What the Radius of convergence of: $\sum\limits_{n=1}^\infty\frac {x^{n!}}{n!}$? So far I tried finding the ... 2answers 63 views ### A question on the convergence of a Taylor series of some prominent function The function $f:\mathbb{R}\to\mathbb{R}$ defined by $$f(x)=\begin{cases}e^{-\frac{1}{x^2}} &if &x\neq 0\\ 0 & else \end{cases}$$ is a prominent example of a function whose Taylor series ... 3answers 91 views ### How can I prove that $\int_1^\infty \left\lvert\frac{\sin x}{x}\right\rvert dx$ diverges? I know a start could be to try and prove that $\int_1^\infty \frac{\sin^2x}{x} dx$ diverges since $\frac{\sin^2x}{x} \le \left\lvert\frac{\sin x}{x}\right\rvert$ in this interval, but I wouldn't know ... 1answer 45 views ### Radius of convergence of $\sum_{n=1}^{\infty} { (n \sin{\frac{1}{n}})^{n} x^n }$ We need to calculate the radius of convergence $R$ of: $$\sum_{n=1}^{\infty} {\left(n \sin{\frac{1}{n}}\right)^{n} x^n }.$$ Here's what I did: \lim_{n\to\infty} { \left\| ... 3answers 78 views ### Radius of convergence for $\sum_{n=1}^{\infty} { \frac{1}{n!} \cdot x^{n!}}$ How do we find the radius of convergence $R$ of this power sum? $$\sum_{n=1}^{\infty} { \frac{1}{n!} \cdot x^{n!}}$$ How do we handle the $n!$ as the power of $x$? 2answers 43 views ### Finding the radius of convergence and what does it mean We have just started learning this and I do not really understand it fully. Given: $$\sum_{n=1}^{\infty} {(\pi^n + n + 1) \cdot x^n}$$ We should check what is the radius of convergence $R$ and what ... 3answers 68 views ### Show that $(x_n-y_n)$ converges to $x-y$. Given $(x_n)$ and $(y_n)$ are sequences of real number which converge to $x$ and $y$ respectively. Show that $(x_n-y_n)$ converges to $x-y$. If it's asking about $(x_n+y_n)$. I know that I can ... 1answer 29 views ### Study the convergence of this sequence of functions I have the following sequence of function: $$f_n(\lambda)=\bigg[\alpha-i\bigg(\lambda+\frac{1}{n}\bigg)\bigg]^{-1}-\bigg[\alpha-i\lambda\bigg]^{-1},\,\,\,\alpha\neq 0$$ and I have to study its ... 1answer 57 views ### Series of Functions - Pointwise and Uniform Convergence. I was hoping for some help for the following questions. Prove that the series $\sum_{n=1}^\infty x^n(1-x)$ converges pointwise but not uniformly on $[0,1]$. Prove that the series \$\sum_{n=1}^\infty ... 3answers 66 views ### Show that the sequence $(x_n)=\left( \sum_{i=1}^n\frac 1 i\right)$ diverge by epsilon delta definition. Show that the sequence $\displaystyle (x_n)=\left( \sum_{i=1}^n\frac 1 i\right)$ diverge by epsilon delta definition. I'm not familiar with proving divergent sequence. Do anyone have any des? ... 1answer 49 views ### Prove convergence of improper integral using change of variable. This may be trivial, but I could use some help... Consider a real function $f: (0,1) \rightarrow \mathbb{R}$, continuous, positive, but not necessarily bounded. Let $g: [0,1] \rightarrow [0,1]$ be a ... 2answers 48 views ### Prove or disprove a result for a double sequence. Suppose that a double sequence $\{a_{n,k}\}=\left\{\frac{1}{n^{\frac{k-1}{k}}}\right\}$. Prove or disprove $\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{k}a_{n,k}=0$. 2answers 88 views ### What is $\lim_{n\to\infty}\frac{1}{e^n}\Bigl(1+\frac1n\Bigr)^{n^2}$? How to solve the following limit question? $$\lim_{n\to\infty}\frac{1}{e^n}\Bigl(1+\frac1n\Bigr)^{n^2}$$ Thanks a lot. 1answer 62 views ### When does $\lim\limits_{n\to\infty}\int_{b}^{a_n}f_n(x)dx=\lim\limits_{n\to\infty}\int_b^\infty f_n(x)dx$ hold? Let $\{a_n\}\subset \mathbb{R}$ be sequence and $$f_n:[b,\infty)\longrightarrow \mathbb{R}, \qquad n=1,2,\dots .$$ Assume that $$\lim_{n\longrightarrow\infty}a_n=+\infty.$$ Obviously, from the ... 3answers 125 views ### Evaluating $\lim \limits_{n\to \infty} \left( n \int_{0}^{\frac \pi 2} 1-\sqrt [n]{\sin x} \,\mathrm dx \right)$ Evaluate the following limit: $$\lim \limits_{n\to \infty} \;\; n \int_{0}^{\frac \pi 2} \left(1-\sqrt [n]{\sin x} \right)\,\mathrm dx$$ I have done the problem . How I solved is First I ... 2answers 85 views ### Interval of convergence of $\sum_{n=4}^\infty x^n/n^5$ Find the interval of convergence $$\sum_{n=4}^\infty x^n/n^5$$ I'm lost here. My intuition was to use the ratio test. $$\lim_{n \to \infty} \frac{x^{n+1}}{(n+1)^5} \times \frac{n^5}{x^n}$$ ... 3answers 97 views ### How to solve $\sum_{k=2}^\infty {\frac{1}{k^2-1}}$ I'm using the integral test to determine if this series converges. From what I have so far it seems that it diverges, but according to wolfram alpha it converges. Where is my mistake? ... 5answers 135 views ### Convergence of a sequence $c_n$ Suppose that $(a_n)$ and $(b_n)$ be sequences such that $\lim (a_n)=0$ and $\displaystyle \lim \left( \sum_{i=1}^n b_i \right)$ exists. Define $c_n = a_1 b_n + a_2 b_{n-1} + \dots + a_n b_1$. Prove ... 1answer 189 views ### binomial expansion formula proof, bases on Lagrange form of Taylor series remainder Another exercise from Bartle/Sherbert Introduction to Real Analysis book (this one is exercise 9.4.14): Use the Lagrange form of the remainder to justify the general Binomial Expansion ... 0answers 36 views ### Is this pointwise convergence sequence also uniform convergence? $f_{n}$ and $f$ are continuous functions and $f_{n}\rightarrow f$ pointwise. Which of the following are correct? $\int _{0}^{x}F_{n}\left( t\right) dt\rightarrow\int _{0}^{x}F\left( t\right) dt$ ... 1answer 47 views ### convergence of complex series. For what values of $(a,b\in \mathbb{C})$ does this series converge or diverge? $\sum\frac{(k-a)^2}{(k-b)^3}$ if $a,b\in\mathbb{R}$ by the Limit comparison test (with $\sum\frac{1}{k}$ ) I know ... 3answers 67 views ### convergence of $\sum\frac{a_{n}}{n}$ if $\sum_{k=1}^{n}a_{k}\le Mn^{r}$ where $r<1$ Show that if the partial sums $s_{n}$ of the series $\sum_{k=1}^{\infty}a_{k}$ satisfy $\|s_{n}\|\le Mn^{r}$ for some $r<1$, then the series $\sum_{n=1}^{\infty}\frac{a_{n}}{n}$ converges. 2answers 30 views ### Radius of convergence in a series. Ratio test. I am having a hard time with this question. $$\sum_{k=0}^{\infty} \frac{-(1)^k (4^k -3)x^{2k}}{k^4+3}$$ I used the ratio test and got stuck here: x^2 \lim_{k\to\infty} \frac ... 1answer 100 views ### Proof that $\sum\limits_{k=1}^n\frac{\sin(kx)}{k^2}$ convergences uniformly using the Cauchy criterion I'd like to use the Cauchy criterion to show that $$f_n(x)=\sum\limits_{k=1}^n\frac{\sin(kx)}{k^2} \mbox{convergences uniformly}$$ Here is what I did: We want to show that \$\forall \, ... 3answers 89 views ### Does this series converge or diverge? I have a series here, and I'm supposed to determine whether it converges or diverges. I've tried the different tests, but I can't quite get the answer. ... 1answer 49 views ### Composition Taylor Series Is there any theorem that specifies when we are allowed to compose the taylor series of two functions? Does it have a name? Thanks. 1answer 31 views ### Two quick questions about convergence (in the context of pointwise vs. uniform convergence) I found this example online: Let $\{f_{n}\}$ be the sequence of functions on $(0,\infty)$ defined by $f_{n}(x)=\frac{nx}{1+n^{2}x^{2}}$ .This function converges pointwise to zero. ... 3answers 64 views ### Representing Functions as Power Series Rewrite $$f(x)=(1+x)/(1-x)^2$$ as a power series. Work thus far: I separated it into two parts: $$1/(1-x)^2 + x/(1-x)^2$$ I realize that the first expression is the derivative of $1/(1-x)$ and ... 1answer 29 views ### Specify conditions for $\alpha$ so that the iteration $x_{n+1} = x_n - \alpha f(x_n)$ converges to root of f. Specify conditions on $\alpha$ so that the iterative process $x_{n+1} = x_n - \alpha f(x_n)$ converges to root of f if started with $x_0$ close to the root. It is suggested that the proof should ... 1answer 56 views ### Value of limsup i? This is a part of my question. $\lim \sup \cos(n\pi/12)$ as n goes to infinity What is the value of this limit? 1answer 81 views ### Cauchy but not fast cauchy As the title indicates, I am trying to find a Cauchy sequence that is not fast (or rapidly) Cauchy. Could anyone suggest something? A sequence $\{a_n\}_{n \in \Bbb N}$is termed fast (or rapidly) ... 3answers 55 views ### Convergence of definite integral I have to find out the convergence of the next integral: $$\int^{\pi/2}_0{\frac{\ln(\sin(x))}{\sqrt{x}}}dx$$ Any help? Thanks 2answers 29 views ### What does $\sum_{n=0}^{\infty} (-1)^n .\frac{x^{2n+1}}{2n+1}$ converge to at x= 1 and x = -1 This is what I did. $\sum_{n=0}^{\infty} (-1)^n .\frac{1^{2n+1}}{2n+1}=\sum_{n=0}^{\infty}(-1)^n .\frac{1}{2n+1}$ Now I broke it up to positive and negative. \$\sum_{n=0}^{\infty}(-1)^n ... 1answer 89 views ### A question on uniform convergence Let $f,g:[0,1]\rightarrow \mathbb{R}$ be continuous functions. Define $$\displaystyle x_{n}(t)=f(t)+ \int_{0}^{t}x_{n-1}(s)ds\quad 0\leq t\leq 1,\;n=1,2,\cdots ,$$ where $x_{0}(t)=g(t),0\leq t\leq 1$ ... 2answers 68 views ### Checking convergence of an improper integral I did a quick search here but couldn't find a similar problem (it's probably out there somewhere...) I'm stuck with this rather simple improper integral: $\int_{1}^{\infty} \frac{1}{x^{\alpha}-1}dx$ ... 2answers 85 views ### Question about convergence of series if $\{n a_n\} \to 0$ This is part of Rudin's PMA Exercise 3.14 (d). If I understand correctly, it would be helpful to prove the following: Let $a_n$ be some sequence. Assume that $\lim_{n\to\infty} na_n = 0$. Prove ... 3answers 87 views ### Uniform convergence of a sequence of functions Let $(u_n(x))$ be a sequence of functions in $(0,\infty)$ such that: $u_1(x)=x$, $u_{n+1}=\frac12\left(u_n(x)+\frac1{u_n(x)}\right)$ for $n\in\mathbb{N}$. Check if $u_n(x)$ converge unfiormly in ... 0answers 42 views ### Convergence of integrals over divergent parts I'm wondering if it is possible for an integral which diverges in the limits $1$ to $\infty$ to converge in the limits from $0$ to $\infty$. And if so: how could I find this out? For example ... 2answers 121 views ### $\sum\limits_{n=1}^\infty \|a_n\|$ converges implies $\sum\limits_{n=1}^\infty \|a_n\|^2$ converges? [duplicate] Possible Duplicate: Prove that $\sum_{n=1}^{\infty}\ a_n^2$ is convergent if $\sum_{n=1}^{\infty}\ a_n$ is absolutely convergent If $\sum\limits_{n=1}^\infty \|a_n\|$ converges, the ... 1answer 38 views ### Convergence integral causal function I have an exercise where there is the following given: $f$ is a causal function. $f$ is Laplace transformable:$\int_{0}^{\infty} f(t)e^{-zt} \, dt = L(z)$ with $Real(z)> -1$ I have to ... 1answer 59 views ### Why one comparison test and not the other I'm studying integral comparison tests and I come to this one. $$f(x)=\int_{1}^{\infty}\frac{\sqrt{x}}{x^5+\sqrt[3]{x}}dx$$ The solution provided is to do \$\displaystyle ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 120, "mathjax_display_tex": 23, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9022445678710938, "perplexity_flag": "head"}
http://cs.stackexchange.com/questions/tagged/linear-algebra	# Tagged Questions The linear-algebra tag has no wiki summary. 0answers 38 views ### how to represent Sparse Matrices I have been using Harwell Boeing format, also known as Compressed Column Strorage (CCS) in order to store Sparse Matrices. Could you please suggest me some other way to store/represent sparse ... 2answers 62 views ### Finding the required value of an algebric expression I have an expression $$Ax+By+Cz.$$ where $A$, $B$ and $C$ are positive constants $\ge1$. The variables $x$, $y$ and $z$ are non-negative integers. I am also given a number $T$. I want to find the ... 3answers 30 views ### Solving system of linear inequalities I am trying to solve a system of inequalities in the following form: $\ x_i - x_j \leq w$ I know these inequalities can be solved using Bellman-Ford algorithm. ... 2answers 78 views ### Can you complete a basis in polynomial time? Here is the problem: we are given vectors $v_1, \ldots, v_k$ lying in $\mathbb{R}^n$ which are orthogonal. We assume that the entries of $v_i$ are rational, with numerator and denominator taking $K$ ... 1answer 41 views ### Counting solutions to system of linear equations modulo prime I have implemented Gaussian elimination for solving system of linear equations in the field of modulo prime remainders. If there is a pivot equal to zero I assume the system has no solution but how to ... 2answers 95 views ### What is the complexity of this matrix transposition? I'm working on some exercises regarding graph theory and complexity. Now I'm asked to give an algorithm that computes a transposed graph of $G$, $G^T$ given the adjacency matrix of $G$. So basically ... 0answers 28 views ### multigrid method to solve PDE [closed] I need explanation of the Multigrid Method or some literature. I am familiar with iterational methods including BiCGStab,CG,GS,Jacobi and preconditioning, but I am a beginner with multigrid method. ... 1answer 46 views ### LU decomposition with pivoting I have to solve system of linear algebraic equations $AX=B$, where $A$ is a two-dimensional matrix with all elements of main diagonal equal to zero. How to solve this problem? Iterational methods are ... 1answer 121 views ### Machine Learning: how to correctly calculate gradient descent for simple linear problem So, I was trying to learn machine learning, and, after watching a couple of Andrew Ng's lectures decided to try and write a simple piece of code to determine what someone's salary would be based on ... 0answers 22 views ### Maximum feasible subsystem problem (MaxFS) in 2 variables Topic: The maximum feasible subsystem problem, which is generally NP-hard [1]. Question: Are there special algorithms in case of only 2 variables (2D linear constraints)? The problem seems to be a ... 0answers 103 views ### Alternatives to SVD for rank factorization I have rank-deficient matrix $M \in \mathbb{R}^{n\times m}$ with $\text{rank}(M) = k$ and I want to find a rank factorization $M = PQ$ with $P \in \mathbb{R}^{n \times k}$ and \$Q \in \mathbb{R}^{k ... 0answers 40 views ### Time - Complexity Convex Optimization and Eigen Decomposition Say I had the choice of choosing one out of the following two optimization problems which I could use to solve my problem. Which choice is the fastest? How much of a trade-off would it be? Is the ... 1answer 117 views ### Probabilistic test of matrix multiplication with one-sided error Given three matrices $A, B,C \in \mathbb{Z}^{n \times n}$ we want to test whether $AB \neq C$. Assume that the arithmetic operations $+$ and $-$ take constant time when applied to numbers from ... 1answer 352 views ### Complexity of checking whether linear equations have a positive solution Consider a system of linear equations $Ax=0$, where $A$ is a $n\times n$ matrix with rational entries. Assume that the rank of $A$ is $<n$. What is the complexiy to check whether it has a solution ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 31, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.920265793800354, "perplexity_flag": "head"}
http://en.wikisource.org/wiki/On_General_Relativity	# On General Relativity From Wikisource On General Relativity (1918) by Harry Bateman Philosophical Magazine 6, 37 (218), 219-229, Online On General Relativity Harry Bateman 1918 On General Relativity. To the Editors of the Philosophical Magazine. Gentlemen,— THE appearance of Dr. Silberstein's recent article[1] on "General Relativity without the Equivalence Hypothesis" encourages me to restate my own views on the subject. I am perhaps entitled to do this as my work on the subject of General Relativity was published before that of Einstein and Kottler, and appears to have been overlooked by recent writers. In 1909 I proposed a scheme of electromagnetic equations[2] which are covariant for all transformations of co-ordinates which are biuniform in the domain we are interested in. These equations were similar to Maxwell's equations, except that the familiar relations B = μH, D = kE of Maxwell's theory were replaced by more general equations, which implied that two fundamental integral forms were reciprocals with regard to a quadratic differential form $\sum\sum g_{m,n}dx_{m}dx_{n}$ which was assumed to be invariant for all transformations of co-ordinates. The coefficients of the quadratic form were regarded as characteristics of the medium supporting the electromagnetic field and of the motion of the medium and its parts. The vanishing of the quadratic form was regarded as the condition that two neighbouring particles should be in positions such that a disturbance starting from one at the associated time should arrive at the other at its associated time[3]. The idea that the coefficients of the quadratic form might be considered as characteristics of the mind interpreting the phenomena was also entertained[4], and it was suggested that a correspondence or transformation of co-ordinates might be considered as a crude mathematical symbol for a mind. The phenomena here considered were those occurring in the brain and body; and although the correspondence by which the universe is reconstructed, so to speak, may be totally different[5] from the type contemplated here, yet it was thought that some of the general conclusions might still be valid if a transformation of co-ordinates was adopted as a working model of the correspondence. It was thought, for instance, that there was an analogy between the relativity principle that the earth's motion in space cannot be detected from experiments with terrestrial objects: and the interesting fact that we are unaware of the flow of blood and other processes taking place in our own bodies so long as they take place in the normal way. It was thought that the correspondence by which the elementary processes in the brain are interpreted may be adjusted in such a way that some of the changes are obscured. Again, if we assume that the nature of an electromagnetic field depends on the type of fundamental quadratic form which determines the constitutive relations, and thus depends indirectly on a transformation which filters the coefficients of this quadratic form, this dependence may be a symbol for the relation between physical and mental phenomena instead of giving the influence of gravitation on light as in Einstein's theory. Einstein and others have attempted to formulate a set of equations of motion which will cover all physical phenomena; but the present writer does not feel inclined to accept them as final, because in his opinion the true equations of motion should be capable of accounting for the phenomena of life, which after all are the most important physical phenomena. To make my position more definite, let us consider one of the methods by which the equations of motion of an electron are obtained in the usual electromagnetic theory. The principle is adopted that at each instant the integral over the electron of the total force on each element must be zero. Now before this principle can be used to write down the equations of motion we must know the design of the electron, and we must know the way in which the motions of the different elements are co-ordinated. This co-ordination or organization of the motions of the elements may be represented mathematically by a sequence of infinitesimal transformations, by which some of the features of the design are preserved. The design of the electron and the co-ordinated motion of its parts may, perhaps, be specified by a quadratic differential form in four variables, which determines a mapping of the interior of the electron on the interior of a stationary sphere; but I doubt if this is sufficiently general. A knowledge of this quadratic differential form is necessary then before we can write down the equations of motion of the electron as a whole. What we usually regard as the equations of motion of matter need then to be supplemented by geometrical conditions which specify the design and organization of each elementary portion of matter. Furthermore, when this design and organization is assumed to be known, the ordinary equations of motion may be regarded as a consequence of the electromagnetic laws and the above-mentioned principle. It must be confessed, however, that this principle does not seem satisfactory for a fundamental principle, and is probably a consequence of some deep underlying principles which are the true equations of motion. These new principles should indicate the reason for a similarity of design of the different electrons. One of the fundamental facts of life is that a good design is copied, and that there is a certain characteristic of the design of an object and its surrounding medium, depending perhaps on the closeness of fit of an imperfect correspondence, which determines the extent to which the design of the object is copied and preserved in the surrounding medium. This may be called the value of the design in relation to the medium, and it is a quantity which I feel must be taken into account in the true equations of motion, and a number assigned to it at each instant. As an example of standardization, the Ford motor-car is not in it with the electron; and, according to the above view, we must regard the design of the electron as one of very great value in relation to the surrounding medium. Returning to our generalized scheme of electromagnetic equations, and looking at matters from the point of view of physical optics, it may be remarked that the scheme of constitutive relations mentioned above is not sufficiently general to cover the case of a doubly-refracting crystalline medium[6]. To remedy this defect we may use a biquadratic integral form instead of a quadratic differential form to specify the constitutive relations. The vanishing of the biquadratic integral form may perhaps be regarded as the condition for action of a moving curve on a particle, a type of condition that seems natural if we regard moving Faraday tubes as fundamental. With this generalized theory it is possible for the elementary wave surface in a medium to be a general Kummer surface, a surface of which Fresnel's wave surface is a particular case. It is doubtful whether this generalized theory is sufficiently general for all purposes, and the above example is given just to emphasize that the absolute calculus of Ricci and Levi Civita can be used to develop a theory of generalized relativity on many lines in addition to that adopted by Einstein. Going back to the case in which a quadratic form is sufficient to determine the optical properties of a medium, we may remark that if Einstein's idea of the gravitational equations is accepted, it is still by no means certain that his quadratic form from which the gravitational equations are derived is the same as the quadratic form which determines the optical properties of the medium. Indeed, the example which I considered on p. 262 of my first paper would seem to indicate that this was not the case. It should be mentioned that in the first seven equations in this example there is a misprint, $\epsilon\mu$ should be replaced by $(\epsilon\mu)^{-1}$. On the above view Einstein's idea of an influence of gravitation on light is simply an hypothesis, but a very interesting and reasonable one. It may be remarked, however, that in the theory of surfaces there are two fundamental quadratic forms, and we may perhaps expect something similar in general relativity. With regard to possible extensions of the idea of relativity it may be worth while to consider transformations analogous to the contact transformations of dynamics in which the co-ordinates x, y, z, t and the component velocities u, v, w correspond to a new set $\left(x_{1}y_{1}z_{1}t_{1}u_{1}v_{1}w_{1}\right)$ in such a way that the differential equations $\frac{dx_{1}}{u_{1}}=\frac{dy_{1}}{v_{1}}=\frac{dz_{1}}{w_{1}}=dt_{1}$ are a consequence of the equations $\frac{dx}{u}=\frac{dy}{v}=\frac{dz}{w}=dt$ This may be secured by making a single quadratic form, such as $\begin{array}{cc} \left(dx^{2}+dy^{2}+dz^{2}-c^{2}dt^{2}\right)\left(c^{2}-u^{2}-v^{2}-w^{2}\right)\\ +\left(c^{2}dt-udx-vdy-wdz\right)^{2}, & \left(c^{2}>u^{2}+v^{2}+w^{2}\right)\end{array}$ an invariant[7]. Various other quadratic forms consisting of sums of squares may, of course, be adopted instead. H. Bateman. Throop College. Pasandena, Cal. Aug. 10th, 1918. 1. Phil. Mag. July 1918 2. Loc. cit. p. 225. See also Amer. Journ. of Math. vol. xxxiv. p. 340 (1912). 3. The term correspondence is used here in a very general sense, and is by no means restricted to the familiar one to one correspondence of entities of the same type, such as points. We should say, for instance, that there is a correspondence between the disturbance running along a telephone-wire and the sound-waves which produce it, because we can pass from one to the other by mathematical equations of a definite type, or rather by solving the equations and the boundary conditions. A correspondence is, moreover, regarded as an entity which may have real existence and be capable of growth and variation. 4. Proc. London Math. Soc. ser. 2, vol. viii. p. 375. See also p. 261 of my first paper. 5. This is a positive definite quadratic form in the variables dx - udt, dy - vdt, dz - wdt, and so can only vanish when all these quantities are zero. This work is in the public domain in the United States because it was published before January 1, 1923. The author died in 1946, so this work is also in the public domain in countries and areas where the copyright term is the author's life plus 60 years or less. This work may also be in the public domain in countries and areas with longer native copyright terms that apply the rule of the shorter term to foreign works.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 7, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9534021019935608, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/177125/holomorphic-function-on-the-complex-plane?answertab=votes	# holomorphic function on the complex plane let $f(z) \in Hol(\mathbb{C})$ holomorphic function such that for each $z_0 \in \mathbb{C}$ there exists $N(z_0)$ such that $f^{(N(z_0))}(z_0) = 0$ Prove: that f is a polynom - ## 2 Answers Modifying Davide's argument but avoiding Baire's category theorem: Let $\overline{\mathbb E}$ be the closed unit ball and $$F_n := \{ z \in \overline{\mathbb E} : f^{(n)}(z) = 0 \}.$$ Then, by assumption, $\overline{\mathbb E} = \bigcup_n F_n$. Since $\overline{\mathbb E}$ is uncountable, not all $F_n$ can be finite. Let $n \in \mathbb N$ s.t. $F_n$ is infinite. Since $\overline{\mathbb E}$ is compact, $F_n$ has a limit point in $\overline{\mathbb E}$, i.e. there is a $z_0 \in \overline{\mathbb E}$ and a sequence $(z_k)$ in $F_n \setminus \{z_0\}$ that converges to $z_0$. We have $f^{(n)}(z_k) = 0$ for all $k$, hence $f^{(n)} = 0$ by the identity theorem, so $f$ is a polynomial. - +1 Damn, that was beautiful! – DonAntonio Aug 2 '12 at 21:06 $f^n=0$ means $f$ is a polynomial? how? – Taxi Driver Aug 3 '12 at 4:15 $f^{(n)}$ means $f$ power n or what? – Taxi Driver Aug 3 '12 at 4:27 $f^{(n)}$ denotes the $n$-th derivative of $f$. I assumed that you know that since you used that notation in your question. To see that $f^{(n)} = 0$ implies that $f$ is a polynomial just look at the Taylor expansion of $f$. – marlu Aug 3 '12 at 15:19 Let $F_n:=\{z\in\Bbb C, f^{(n)}(z)=0\}$. Since $f^{(n)}$ is continuous, $F_n$ is closed and by hypothesis, $\Bbb C=\bigcup_{n\in\Bbb N}F_n$. By Baire's theorem, there exists a $n_0$ such that $F_{n_0}$ has a non-empty interior. Hence $f^{(n_0)}$ vanishes on a ball, and it's necessarily the null function. This proves that $f$ is a polynomial. Note that there is a similar result for $C^{\infty}$ functions from $\Bbb R$ to $\Bbb R$, but it's harder to show. It has been discussed here. - is there a proof without baire theorem? i didnt learn that in the complex analysis course – bar Jul 31 '12 at 11:33 And what did you learn? In which chapter such an exercise appear? Actually, Baire is more a theorem from topology than complex analysis. – Davide Giraudo Jul 31 '12 at 11:38 i think the relevant theorems for this exercise are: uniqeness theorem: if $f\in Hol(D)$ D is a domain and there is U a subset of D such that U has accumulation point and $\forall z \in U f(z)=0$ then $\forall z\in D f(z) = 0$ – bar Jul 31 '12 at 11:43	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 45, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9520666003227234, "perplexity_flag": "head"}
http://quant.stackexchange.com/questions/2148/how-to-detect-regime-change-when-estimating-asset-correlation-from-historical-ti?answertab=votes	# How to detect regime change when estimating asset correlation from historical time series? Suppose I have two asset time series, $X_t$ and $Y_t$, and I'm estimating their correlation from historical data. I'd like to apply some systematic criterion to estimate what time window I should use to estimate the correlation reliably, and also to spot "regime changes" (when the correlation jumps suddenly) after which I should discard old data completely (as opposed to rolling the window in a continuous manner). Can you recommend some approaches which have a decent theoretical background? - ## 3 Answers You can use changepoint analysis to identify regime change. You can also look at large angle differences in the eigenvectors between your most up-to-date/recent covariance matrix and the covariance matrix from the prior window. Another way to identify regime change is using a factor model. If the returns on a particular set of factors is X standard deviations from its usual terrain for a sustained period then you can call this regime change. I do not believe you will find a single time window that is best. Regime duration is variable. Key here is identifying an estimation procedure for a covariance matrix that produces reasonable out-of-time forecasts. You will need to do some empirical testing, or develop a rule to re-estimate your model based on the how you identify regimes, or use Garch (or other dynamic model) as Patrick suggests. Technical side note: You probably don't want to discard old data completely but instead weight more recent data with exponential weighting, or re-scale the covariance matrix to reflect current volatility. The eigenvectors of the correlation matrix (after the 1st eigenvector which is the market factor) will correspond to sector and industry groups. These correlations will persist. When market flip from bull to bear market (let's call this a 1st order approximation of regime - as opposed to style and industry changes) what is happening is that the variance explained by the largest eigenvector has increased substantially. - Do you have any references about changepoint analysis? – quant_dev Oct 14 '11 at 6:50 1 – strimp099 Oct 14 '11 at 15:32 Yep - those links look great – Quant Guy Oct 14 '11 at 17:47 @strimp099 Are there any resources in these search results you find particularly instructive and interesting? Introductions, surveys, papers, books? – vanguard2k Dec 27 '12 at 15:38 @vanguard2k I found the first two links interesting but in fair disclosure have done very little in the field area. I have not seen any papers either... – strimp099 Jan 7 at 18:23 I would suggest a multivariate garch model as a possibility. We aren't exactly overrun with wonderful software for that, but with just bivariate data I would think that the in-sample correlation estimates would be reasonably robust over models and estimation. It would be good to try two or three ways of doing it to make sure I'm right about that. You may find that the garch route is good enough to use as your solution if you aren't forecasting very far ahead. - Could I use R for that? – quant_dev Oct 12 '11 at 15:14 1 of course, garch is implemented in several R packages – RockScience Oct 13 '11 at 7:05 One approach would be Engle (2002) dynamic conditional correlations. Taking your $Y_t$ and $X_t$, I will make the simplifying assumption that the mean equation of these is: $$\boxed{Y_t = \mu_y + \varepsilon_{y,t}}$$ $$\boxed{X_t = \mu_x + \varepsilon_{x,t}}$$ with $\varepsilon_{y,t} = z_{y,t} \sigma_{y,t} \sim N(0,\sigma_{y,t})$, $\varepsilon_{x,t} = z_{x,t} \sigma_{x,t} \sim N(0,\sigma_{x,t})$. In practice you might want to specify a GARCH-M, or a GARCH with exogenous variables inside the mean equation. For example if $Y_t$ and $X_t$ are individual stocks in the same market, you might want to include $R_{M,t}$ so that you don't detect any correlation due to this shared factor. If you're looking across borders GARCH-M terms may become important if you want to control for portfolio rebalancing due to changing relative risks or something. The mean equations can be written in vector form as: $$\boxed{ Z_t = \mathbf{\mu} + \mathbf{\varepsilon}_t}$$ with $Z_t := [Y_t,X_t]'$, $\mathbf{\varepsilon} \sim N(0,H_t)$, $H_t := D_t R_t D_t$. Here, $R_t$ is the (possibly) time-varying correlation matrix and $D_t$ is simply $\text{diag}(\sigma_{y,t},\sigma_{x,t})$ when we assume no news or variance spillovers in the variance equation. The DCC estimator in the `rmgarch` package will provide you with the dynamics of $R_t$ with no effort on your part. You can then visually inspect for a break in the correlation. However, for an objective approach withing the DCC framework, take a look at this paper for the ability to hypothesis test structural breaks in the correlation. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 16, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9419596791267395, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?t=514850	Physics Forums ## Diagonalization 1. The problem statement, all variables and given/known data I think my teacher made a mistake in his homework answer. I need to verify this for practice. The answer I got is below. The answer the teacher has is in the pdf. 2. Relevant equations 3. The attempt at a solution So there is two eigenvalues= 4 and 2 but the eigenvalue 2 has 2 eigenvectors [-1 1 0]T and [0 0 1]T but my teacher has only one [-1 1 0]T. That's why he says A is not diagonalizable. Do you think it's correct? Attached Thumbnails PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug I get the same result as your teacher for $$\lambda = 2$$ $$A= \begin{bmatrix} 3 & 1 & 0\\ 0 & 2 & 1\\ 1 & 1 & 3 \end{bmatrix}$$ so for lamda = 2, $$(A-2I)\vec{v}=\vec{0}$$ $$\begin{bmatrix} 1 & 1 & 0\\ 0 & 0 & 1\\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} v_1\\ v_2\\ v_3 \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\\ \end{bmatrix}$$ I've personally always found it easier to not do row operations here and just jump straight in. From that you get $$v_3 = 0$$ and $$v_1 = -v_2$$ so therefore $$\vec{v} = \begin{bmatrix} 1\\ -1\\ 0 \end{bmatrix}$$ Since it is a 3x3 matrix, it needs 3 eigenvectors to be diagonalizable. Hope this helps. Hi, thanks for replying. Attached is how I got my vectors, do you think my steps are correct. Attached Thumbnails Mentor ## Diagonalization No. Why do you think (0, 0, 1) is an eigenvector? You seem to just pull that out of thin air. You seem to have made a mistake in the step $$(A-2I)\vec{v}=\vec{0}$$ Why is your second row 1 0 1 instead of 0 0 1 ? Have you said $$v_2= \delta$$ ? I'm not sure if I have read that correctly. If it is a delta, you can't say that unless the whole row equals zero. i.e. $$\begin{bmatrix} 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} v_1 \end{bmatrix} = \begin{bmatrix} 0 \end{bmatrix}$$ Tags algebra, diagonalization, eigenvalue, linear Thread Tools \| \| \| \| \|--------------------------------------\|----------------------------\|---------\| \| Similar Threads for: Diagonalization \| \| \| \| Thread \| Forum \| Replies \| \| \| Calculus & Beyond Homework \| 0 \| \| \| Calculus & Beyond Homework \| 3 \| \| \| Calculus & Beyond Homework \| 5 \| \| \| Quantum Physics \| 1 \| \| \| Quantum Physics \| 2 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 10, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9221944808959961, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/292533/convergent-or-divergent	# Convergent or Divergent Determine if the series converges or diverges: $$\sum_{n=1}^\infty \frac{\ln(n)}{n \left\|\sin(n)\right\|} =$$ We know that $$\sum_{n=1}^\infty \frac{1}i = \text{Divergent}$$ $$\sum_{n=1}^1 \ln(i )= 0$$ How to solve this? - 1 you need to remove 1 from right hand side of first eq. The other two can be reworded too. – Maesumi Feb 2 at 3:26 ## 2 Answers Another way: also note $\frac{\log n }{n\|\sin n\|} \geq \frac{\log n}{n}$. Now look at $\sum_{n=1}^{\infty}\frac{\log n }{n}$. By intergral test if the integral diverges, then the sum diverges too. $\int_{1}^{\infty} \frac{\log x dx}{x} = \left. \frac{ \log^{2} x}{2} \|^{\infty}_{1}\right.=\infty$. The integral diverges, hence the sum diverges too. Since the original sum is lower-bounded by this one, by comparison test it diverges too. - Hint: Since $\|\sin n\|\le 1$, and $\ln n \gt 1$ for $n\ge 3$, the $n$-th term of our sequence is greater than $\frac{1}{n}$ if $n\ge 3$. Now use the Comparison Test. - Got it thanks! That seems easier now – user1730308 Feb 2 at 3:18	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 9, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8597381114959717, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/139549/how-to-show-path-connectedness	# How to show path-connectedness Well, I am not getting any hint how to show $GL_n(\mathbb{C})$ is path connected. So far I have thought that let $A$ be any invertible complex matrix and $I$ be the idenity matrix, I was trying to show a path from $A$ to $I$ then define $f(t)=At+(1-t)I$ for $t\in[0,1]$ which is possible continous except where the $\operatorname{det}{f(t)}=0$ i.e. which has $n$ roots and I can choose a path in $\mathbb{C}\setminus\{t_1,\dots,t_n\}$ where $t_1,\dots,t_n$ are roots of $\operatorname{det}{f(t)}=0$, is my thinking was correct? Could anyone tell me the solution? - This path can fail in liying on $GL_n(\mathbb{C})$. For example, if you take $A=-I$, at the middle point $t=1/2$ you will get $f(t)=0$ – matgaio May 1 '12 at 19:45 +1 Looks good to me. @matgaio, the idea seems to be that you go around that troublesome $t=1/2$ - there is room for that, when you let $t$ be a complex number! It might have been clearer to define $f(t)$ for all complex $t$, exclude the finite set of points, and then select a path from $0$ to $1$. – Jyrki Lahtonen May 1 '12 at 19:47 Uhm, ok. I was understanding he was trying to go straight from $A$ to $I$. – matgaio May 1 '12 at 19:50 2 – Antonio Vargas May 1 '12 at 20:02 ## 3 Answers • If $P$ is a polynomial of degree $n$, the set $\{\lambda,P(\lambda)\neq 0\}$ is path connected (because its complement is finite, so you can pick a polygonal path). • Let $P(t):=\det(A+t(I-A))$. We have that $P(0)=\det A\neq 0$, and $P(1)=\det I=1\neq 0$, so we can find a path $\gamma\colon[0,1]\to\mathbb C$ such that $\gamma(0)=0$, $\gamma(1)=1$, and $P(\gamma(t))\neq 0$ for all $t$. Finally, put $\Gamma(t):=A+\gamma(t)(I-A)$. • If $B_1$ and $B_2$ are two invertible matrices, consider $\gamma(t):=B_2\cdot\gamma(t)$, where we chose $\gamma$ for $A:=B_2^{-1}B_1$. - what is $f(\gamma(t))$, will it be $P(\gamma(t))$? – Taxi Driver Feb 28 at 11:19 Since any matrix $A\in GL_n(\mathbb C)$ has only finitely many eigenvalues, and 0 isn't one of them, there is a point $z\in S^1$ such that the line through the origin containing $z$ doesn't intersect any of the eigenvalues of $A$. Now, consider the path $f(t)=At+z(1-t)I$. This has determinant 0 iff $z(t-1)$ is an eigenvalue of $At$, which happens iff $z(1-1/t)$ is an eigenvalue of $A$ (this doesn't work when $t=0$, but then it is clear that the determinant is non-zero). By construction, it isn't for any $t\in[0,1]$ so this defines a path form $A$ to $zI$. now there is a path not passing through 0 from $z$ to 1, and this gives rise to a path from $zI$ to $I$, and so concatenating the two paths, we get a path from $A$ to $I$, showing that $GL_n(\mathbb C)$ is path connected. - Use $\Gamma(t) = e^{t \log A + (1-t) \log B}$. This is well defined since $A,B$ are invertible. $\Gamma(t)$ is clearly invertible for all $t$, $\Gamma(0) = B$, $\Gamma(1) = A$. - This idea is not mine, I just can't remember where I saw it, it was in a control theory context. – copper.hat May 1 '12 at 20:20	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 68, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9570044279098511, "perplexity_flag": "head"}

http://mathhelpforum.com/statistics/167501-probability-check.html

# Thread: 1. ## Probability check I came across this task, and I could not solve it furthermore. Can anyone check my equation and help me solve it? -The probability to succeed in a random test is 0.8. It's known that if N people attending to the test the probability that all of them will succeed is 0.4096. A- Find N. I was only able to do this equation (Bernouli's table) $(N/K)*(0.8)^k*(0.2)$ $^n^-K=0.4096$ Can anyone help me solving furthermore with this equation please. that's ^n-k 2. Originally Posted by Mathematicsfan I came across this task, and I could not solve it furthermore. Can anyone check my equation and help me solve it? -The probability to succeed in a random test is 0.8. It's known that if N people attending to the test the probability that all of them will succeed is 0.4096. A- Find N. I was only able to do this equation (Bernouli's table) $(N/K)*(0.8)^k*(0.2)$ $^n^-K=0.4096$ Can anyone help me solving furthermore with this equation please. that's ^n-k $(0.8)^N=0.4096$ Can you solve for N? 3. No, I don't know I tried desperately. can you guide me ? 4. Originally Posted by mathematicsfan no, i don't know i tried desperately. Can you guide me ? $(0.8)^N=0.4096$ $(0.8)^N=(0.8)^4$ Find N. 5. How did you get 4? cause that's the answer . and where did the 0.2 ^n-k go ? and N\K ? 6. Originally Posted by Mathematicsfan How did you get 4? cause that's the answer . and where did the 0.2 ^n-k go ? and N\K ? This is a very simple question which does not require the use of Bernoulli's tables. You're probably overthinking it. 7. then i don't know how to solve it. can you help me with what to do next ? 8. Originally Posted by Mathematicsfan then i don't know how to solve it. can you help me with what to do next ? I solved it for you in post #4. N=4. 9. lolllllllllllll I got it lollll this is so easy wow.w I just had to play with the Powers / lol thanks a lot #### Search Tags View Tag Cloud Copyright © 2005-2013 Math Help Forum. All rights reserved.

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9689220190048218, "perplexity_flag": "middle"}

http://nrich.maths.org/public/leg.php?code=-38&cl=2&cldcmpid=6745

# Search by Topic #### Resources tagged with Permutations similar to Weekly Problem 45 - 2009: Filter by: Content type: Stage: Challenge level: ##### Other tags that relate to Weekly Problem 45 - 2009 Place value. Mathematical reasoning & proof. Working systematically. Addition & subtraction. Algebra - generally. Number - generally. Factors and multiples. Combinations. Properties of numbers. Permutations. ### There are 13 results Broad Topics > Decision Mathematics and Combinatorics > Permutations ### Six Times Five ##### Stage: 3 Challenge Level: How many six digit numbers are there which DO NOT contain a 5? ### Even Up ##### Stage: 3 Challenge Level: Consider all of the five digit numbers which we can form using only the digits 2, 4, 6 and 8. If these numbers are arranged in ascending order, what is the 512th number? ### And So on and So On ##### Stage: 3 Challenge Level: If you wrote all the possible four digit numbers made by using each of the digits 2, 4, 5, 7 once, what would they add up to? ### Flagging ##### Stage: 3 Challenge Level: How many tricolour flags are possible with 5 available colours such that two adjacent stripes must NOT be the same colour. What about 256 colours? ### Permute It ##### Stage: 3 Challenge Level: Take the numbers 1, 2, 3, 4 and 5 and imagine them written down in every possible order to give 5 digit numbers. Find the sum of the resulting numbers. ### Power Crazy ##### Stage: 3 Challenge Level: What can you say about the values of n that make $7^n + 3^n$ a multiple of 10? Are there other pairs of integers between 1 and 10 which have similar properties? ### Painting Cubes ##### Stage: 3 Challenge Level: Imagine you have six different colours of paint. You paint a cube using a different colour for each of the six faces. How many different cubes can be painted using the same set of six colours? ### Bell Ringing ##### Stage: 3 Challenge Level: Suppose you are a bellringer. Can you find the changes so that, starting and ending with a round, all the 24 possible permutations are rung once each and only once? ### Ding Dong Bell ##### Stage: 3, 4 and 5 The reader is invited to investigate changes (or permutations) in the ringing of church bells, illustrated by braid diagrams showing the order in which the bells are rung. ### Card Shuffle ##### Stage: 3 and 4 This article for students and teachers tries to think about how long would it take someone to create every possible shuffle of a pack of cards, with surprising results. ### Shuffle Shriek ##### Stage: 3 Challenge Level: Can you find all the 4-ball shuffles? ### Master Minding ##### Stage: 3 Challenge Level: Your partner chooses two beads and places them side by side behind a screen. What is the minimum number of guesses you would need to be sure of guessing the two beads and their positions? ### Euromaths ##### Stage: 3 Challenge Level: How many ways can you write the word EUROMATHS by starting at the top left hand corner and taking the next letter by stepping one step down or one step to the right in a 5x5 array? The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8663181066513062, "perplexity_flag": "middle"}

http://physics.stackexchange.com/questions/52524/is-it-part-of-special-relativity-that-mass-possessing-energy-is-more-dense/52526

# Is it part of special relativity that mass possessing energy is more dense? I was reading http://www.edge.org/3rd_culture/hillis/hillis_p2.html and it says that a charged battery weighs more than a dead one or a rotating object weighs more than a stationary one (i.e. mass containing energy weighs more than mass that doesn't). Is this true in special relativity? Because the energy would be confined to the area of the original mass, wouldn't it become more dense if it is true? Also, if the mass we are speaking of does become more gravity-sensitive on energy gain, does it actually gain physical mass (protons, electrons and neutrons)?? - ## 2 Answers Is this true in special relativity? Well, it is a consequence of special relativity. So yes, I guess. Because the energy would be confined to the area of the original mass, wouldn't it become more dense if it is true? It is not true in the general that the volume of the mass remains the same. There are shortening effects as well. For example in the linear case: if you just consider a linear mass with rest mass $m_0$ and rest length $l_0$, then its linear mass density is $\mu_0=m_0/l_0$. And if the mass is now moving along its length at velocity $v$, then its new mass is, $m = \gamma m_0$ and new length is $l = l_0/\gamma$, where $\gamma=\frac{1}{\sqrt{1-v^2/c^2}}$ is the Lorentz factor. Which makes its new mass density: $\mu = \gamma^2 m_0/l_0$. So the mass density has increased by a quadratic factor. Since energy is simply $mc^2$, the energy density has also increased quadratically. But for a linear mass but with some other shape (say in the shape of a U), it gets more complicated. Say its moving along the $x$-directions, then its length along $x$ will decrease but its length along $y$ and $z$ will stay the same. So the total length might change in some complicated way. Other issues in higher dimensions is that the Lorentz factor for the mass will use the total velocity while the Lorentz factor for length along each direction will use the velocity along that direction. I have no idea how to do the calculation for a rotating body. A rotating sphere would be a really interesting calculation to do. Also, if the mass we are speaking of does become more gravity-sensitive on energy gain, does it actually gain physical mass (protons, electrons and neutrons)? It doesn't actually gain more protons, electrons, neutrons. What happens is that the fundamental particles themselves get 'heavier'. Are you wondering whether energy gains might contribute to gravitational mass but not to inertial mass (I guess that's what you are referring to as physical mass)? As far as we know, those two things are the same, but we don't know why yet. - The reason a rotating object weighs more than a non-rotating one is that weight is a quantity that indicates how strongly gravity is coupling to (interacting with) the object in question. When the object rotates, its energy increases, and since what determines the strength of the gravitational interaction is energy and momentum of the object, not just rest energy (mass), the weight of the object increases. This might be confusing at first because Newton's law of gravitation indicates that the strength of the gravitational interaction between two objects goes like the product of their masses and says nothing about energy, but this law is only approximately true, and general relativity tells us otherwise. The "mass we are speaking of" as you say, does not change. The term "mass" in modern physics is used to describe what people often call "rest mass." The number of atoms in a spinning object does not change. This is true both in special, and in general relativity. -

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9540835022926331, "perplexity_flag": "head"}

http://www.haskell.org/haskellwiki/index.php?title=User:Michiexile/MATH198/Lecture_1&diff=46142&oldid=30053

# User:Michiexile/MATH198/Lecture 1 ### From HaskellWiki (Difference between revisions) | | | | | |----------------------------------------|----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|-----------------------------------------------------------------|----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------| | | | Current revision (02:50, 22 June 2012) (edit) (undo)m ( Typo.) | | | (10 intermediate revisions not shown.) | | | | | Line 1: | | Line 1: | | | - | IMPORTANT NOTE: THESE NOTES ARE STILL UNDER DEVELOPMENT. PLEASE WAIT UNTIL AFTER THE FIRST LECTURE WITH HANDING ANYTHING IN, OR TAKING THE NOTES AS READY TO READ. | + | This page now includes additional information based on the notes taken in class. Hopefully this will make the notes reasonably complete for everybody. | | | | | | | - | ==Welcome, administrativia== | + | | | | | + | ==Welcome, administratrivia== | | | | | | | | I'm Mikael Vejdemo-Johansson. I can be reached in my office 383-BB, especially during my office hours; or by email to mik@math.stanford.edu. | | I'm Mikael Vejdemo-Johansson. I can be reached in my office 383-BB, especially during my office hours; or by email to mik@math.stanford.edu. | | Line 32: | | Line 33: | | | | * Matrices | | * Matrices | | | * Homomorphisms | | * Homomorphisms | | | | + | | | | | + | ===Good references=== | | | | + | | | | | + | On reserve in the mathematics/CS library are: | | | | + | * ''Mac Lane'': '''Categories for the working mathematician''' | | | | + | :This book is technical, written for a mathematical audience, and puts in more work than is strictly necessary in many of the definitions. When Awodey and Mac Lane deviate, we will give Awodey priority. | | | | + | * ''Awodey'': '''Category Theory''' | | | | + | :This book is also available as an ebook, accessible from the Stanford campus network. The coursework webpage has links to the ebook under Materials. | | | | + | | | | | + | ===Monoids=== | | | | + | | | | | + | In order to settle notation and ensure everybody's seen a definition before: | | | | + | | | | | + | | | | | + | A ''semigroup'' is a monoid without the requirement for an identity element. | | | | + | | | | | + | A function <math>f:M\to N</math> is a monoid homomorphism if the following conditions hold: | | | | + | * <math>f(\emptyset) = \emptyset</math> | | | | + | * <math>f(m*m') = f(m)*f(m')</math> | | | | + | | | | | + | '''Examples''' | | | | + | * Any group is a monoid. Thus, specifically, the integers with addition is a monoid. | | | | + | * The natural numbers, with addition. | | | | + | * Strings <math>L^*</math> in an alphabet <math>L</math> is a monoid with string concatenation forming the operation and the empty string the identity. | | | | + | * Non-empty strings form a semigroup. | | | | + | | | | | + | | | | | + | Awodey: p. 10. | | | | + | | | | | + | ===Partially ordered set=== | | | | + | | | | | + | '''Definition''' A ''partially ordered set'', or a ''partial order'', or a ''poset'' is a set <math>P</math> equipped with a binary relation <math>\leq</math> which is: | | | | + | * Reflexive: <math>x\leq x</math> for all <math>x\in P</math> | | | | + | * Anti-symmetric: <math>x\leq y</math> and <math>y\leq x</math> implies <math>x=y</math> for all <math>x,y\in P</math>. | | | | + | * Transitive: <math>x\leq y</math> and <math>y\leq z</math> implies <math>x\leq z</math> for all <math>x,y,z\in P</math>. | | | | + | | | | | + | If <math>x\leq y</math> or <math>y\leq x</math>, we call <math>x,y</math> ''comparable''. Otherwise we call them ''incomparable''. A poset where all elements are mutually comparable is called a ''totally ordered set'' or a ''total order''. | | | | + | | | | | + | If we drop the requirement for anti-symmetry, we get a ''pre-ordered set'' or a ''pre-order''. | | | | + | | | | | + | If we have several posets, we may indicate which poset we're comparing ''in'' byindicating the poset as a subscript to the relation symbol. | | | | + | | | | | + | A ''monotonic'' map of posets is a function <math>f:P\to Q</math> such that <math>x\leq_P y</math> implies <math>f(x)\leq_Q f(y)</math>. | | | | + | | | | | + | '''Examples''' | | | | + | * The reals, natural numbers, integers all are posets with the usual comparison relation. A poset in which all elements are comparable. | | | | + | * The natural numbers, excluding 0, form a poset with <math>a\leq b</math> if <math>a|b</math>. | | | | + | * Any family of subsets of a given set form a poset with the order given by inclusion. | | | | + | | | | | + | | | | | + | Awodey: p. 6. Preorders are defined on page 8-9. | | | | | | | | ==Category== | | ==Category== | | | | + | | | | | + | Awodey has a slightly different exposition. Relevant pages in Awodey for this lecture are: sections 1.1-1.4 (except Def. 1.2), 1.6-1.8. | | | | | | | | ===Graphs=== | | ===Graphs=== | | Line 44: | | Line 101: | | | | | | | | | We shall not, in general, require either of the collections to be a set, but will happily accept larger collections; dealing with set-theoretical paradoxes as and when we have to. A graph where both nodes and arrows are sets shall be called ''small''. A graph where either is a class shall be called ''large''. | | We shall not, in general, require either of the collections to be a set, but will happily accept larger collections; dealing with set-theoretical paradoxes as and when we have to. A graph where both nodes and arrows are sets shall be called ''small''. A graph where either is a class shall be called ''large''. | | - | | + | 0 | | | If both <math>G_0</math> and <math>G_1</math> are finite, the graph is called ''finite'' too. | | If both <math>G_0</math> and <math>G_1</math> are finite, the graph is called ''finite'' too. | | | | | | | Line 74: | | Line 131: | | | | ===Categories=== | | ===Categories=== | | | | | | | - | We now are ready to define a category. A ''category'' is a graph <math>C</math> equipped with an associative composition operation <math>\circ:G_2\to G_1</math>, and an identity element for composition <math>1_x</math> for each node <math>x</math> of the graph. | + | We now are ready to define a category. A ''category'' is a graph <math>G</math> equipped with an associative composition operation <math>\circ:G_2\to G_1</math>, and an identity element for composition <math>1_x</math> for each node <math>x</math> of the graph. | | | | | | | | Note that <math>G_2</math> can be viewed as a subset of <math>G_1\times G_1</math>, the set of all pairs of arrows. It is intentional that we define the composition operator on only a subset of the set of all pairs of arrows - the composable pairs. Whenever you'd want to compose two arrows that don't line up to a path, you'll get nonsense, and so any statement about the composition operator has an implicit "whenever defined" attached to it. | | Note that <math>G_2</math> can be viewed as a subset of <math>G_1\times G_1</math>, the set of all pairs of arrows. It is intentional that we define the composition operator on only a subset of the set of all pairs of arrows - the composable pairs. Whenever you'd want to compose two arrows that don't line up to a path, you'll get nonsense, and so any statement about the composition operator has an implicit "whenever defined" attached to it. | | Line 109: | | Line 166: | | | | | | | | | A category with objects a collection of sets and morphisms a selection from all possible set-valued functions such that the identity morphism for each object is a morphism, and composition in the category is just composition of functions is called ''concrete''. Concrete categories form a very rich source of examples, though far from all categories are concrete. | | A category with objects a collection of sets and morphisms a selection from all possible set-valued functions such that the identity morphism for each object is a morphism, and composition in the category is just composition of functions is called ''concrete''. Concrete categories form a very rich source of examples, though far from all categories are concrete. | | | | + | | | | | + | Again, the Wikipedia page on Category (mathematics) [[http://en.wikipedia.org/wiki/Category_%28mathematics%29]] is a good starting point for many things we will be looking at throughout this course. | | | | | | | | ===New Categories from old=== | | ===New Categories from old=== | | Line 137: | | Line 196: | | | | # We can also form the category with objects <math>\langle A,B\rangle</math> for every pair of objects <math>A\in C, B\in D</math>. A morphism in <math>Hom(\langle A,B\rangle,\langle A',B'\rangle)</math> is simply a pair <math>\langle f:A\to A',g:B\to B'\rangle</math>. Composition is defined componentwise. This category is the categorical correspondent to the cartesian ''product'', and we denot it by <math>C\times D</math>. | | # We can also form the category with objects <math>\langle A,B\rangle</math> for every pair of objects <math>A\in C, B\in D</math>. A morphism in <math>Hom(\langle A,B\rangle,\langle A',B'\rangle)</math> is simply a pair <math>\langle f:A\to A',g:B\to B'\rangle</math>. Composition is defined componentwise. This category is the categorical correspondent to the cartesian ''product'', and we denot it by <math>C\times D</math>. | | | | | | | - | In these three constructions - the dual, the product and the coproduct - he arrows in the categories are formal constructions, not functions; even if the original category was given by functions, the result is no longer given by a function. | + | In these three constructions - the dual, the product and the coproduct - the arrows in the categories are formal constructions, not functions; even if the original category was given by functions, the result is no longer given by a function. | | | | | | | | Given a category <math>C</math> and an object <math>A</math> of that category, we can form the ''slice category'' <math>C/A</math>. Objects in the slice category are arrows <math>B\to A</math> for some object <math>B</math> in <math>C</math>, and an arrow <math>\phi:f\to g</math> is an arrow <math>s(f)\to s(g)</math> such that <math>f=g\circ\phi</math>. Composites of arrows are just the composites in the base category. | | Given a category <math>C</math> and an object <math>A</math> of that category, we can form the ''slice category'' <math>C/A</math>. Objects in the slice category are arrows <math>B\to A</math> for some object <math>B</math> in <math>C</math>, and an arrow <math>\phi:f\to g</math> is an arrow <math>s(f)\to s(g)</math> such that <math>f=g\circ\phi</math>. Composites of arrows are just the composites in the base category. | | Line 143: | | Line 202: | | | | Notice that the same arrow <math>\phi</math> in the base category <math>C</math> represents potentially many different arrows in <math>C/A</math>: it represents one arrow for each choice of source and target compatible with it. | | Notice that the same arrow <math>\phi</math> in the base category <math>C</math> represents potentially many different arrows in <math>C/A</math>: it represents one arrow for each choice of source and target compatible with it. | | | | | | | - | There is a dual notion: the ''coslice category'', where the objects are paired with maps <math>A\to B</math>. | + | There is a dual notion: the ''coslice category'' <math>A\backslash C</math>, where the objects are paired with maps <math>A\to B</math>. | | | | | | | - | Slice categories can be used, among other things, to specify the idea of parametrization. The slice category <math>(C\downarrow A)</math> gives a sense to the idea of ''objects from <math>C</math> labeled by elements of <math>A</math>''. | + | Slice categories can be used, among other things, to specify the idea of parametrization. The slice category <math>C/A</math> gives a sense to the idea of ''objects from <math>C</math> labeled by elements of <math>A</math>''. | | | | | | | | We get this characterization by interpreting the arrow representing an object as representing its source and a ''type function''. Hence, in a way, the <hask>Typeable</hask> type class in Haskell builds a slice category on an appropriate subcategory of the category of datatypes. | | We get this characterization by interpreting the arrow representing an object as representing its source and a ''type function''. Hence, in a way, the <hask>Typeable</hask> type class in Haskell builds a slice category on an appropriate subcategory of the category of datatypes. | | Line 151: | | Line 210: | | | | Alternatively, we can phrase the importance of the arrow in a slice categories of, say, Set, by looking at preimages of the slice functions. That way, an object <math>f:B\to A</math> gives us a family of (disjoint) subsets of <math>B</math> ''indexed'' by the elements of <math>A</math>. | | Alternatively, we can phrase the importance of the arrow in a slice categories of, say, Set, by looking at preimages of the slice functions. That way, an object <math>f:B\to A</math> gives us a family of (disjoint) subsets of <math>B</math> ''indexed'' by the elements of <math>A</math>. | | | | | | | - | Finally, any graph yields a category by just filling in the arrows that are missing. The result is called the ''free category generated by the graph'', and is a concept we will return to in some depth. Free objects have a strict categorical definition, and they serve to give a model of thought for the things they are free objects for. Thus, categories are essentially graphs, possibly with restrictions or relations imposed; and monoids are essentially strings in some alphabet, with restrictions or relatinos. | + | Finally, any graph yields a category by just filling in the arrows that are missing. The result is called the ''free category generated by the graph'', and is a concept we will return to in some depth. Free objects have a strict categorical definition, and they serve to give a model of thought for the things they are free objects for. Thus, categories are essentially graphs, possibly with restrictions or relations imposed; and monoids are essentially strings in some alphabet, with restrictions or relations. | | | | | | | | ===Examples=== | | ===Examples=== | | Line 168: | | Line 227: | | | | ** Sets for objects. Arrows are pairs <math>(S'\subseteq S,f:S'\to T)\in PFn(S,T)</math>. | | ** Sets for objects. Arrows are pairs <math>(S'\subseteq S,f:S'\to T)\in PFn(S,T)</math>. | | | ** <math>PFn(A,B)</math> is a partially ordered set. <math>(S_f,f)\leq(S_g,g)</math> precisely if <math>S_f\subseteq S_g</math> and <math>f=g|_{S_f}</math>. | | ** <math>PFn(A,B)</math> is a partially ordered set. <math>(S_f,f)\leq(S_g,g)</math> precisely if <math>S_f\subseteq S_g</math> and <math>f=g|_{S_f}</math>. | | | | + | ** The exposition at Wikipedia uses the construction here: [[http://en.wikipedia.org/wiki/Partial_function]]. | | | * There is an alternative way to define a category of partial functions: For objects, we take sets, and for morphisms <math>S\to T</math>, we take subsets <math>F\subseteq S\times T</math> such that each element in <math>S</math> occurs in at most one pair in the subset. Composition is by an interpretation of these subsets corresponding to the previous description. We'll call this category <math>PFn'</math>. | | * There is an alternative way to define a category of partial functions: For objects, we take sets, and for morphisms <math>S\to T</math>, we take subsets <math>F\subseteq S\times T</math> such that each element in <math>S</math> occurs in at most one pair in the subset. Composition is by an interpretation of these subsets corresponding to the previous description. We'll call this category <math>PFn'</math>. | | | * Every partial order is a category. Each hom-set has at most one element. | | * Every partial order is a category. Each hom-set has at most one element. | | | ** Objects are the elements of the poset. Arrows are unique, with <math>A\to B</math> precisely if <math>A\leq B</math>. | | ** Objects are the elements of the poset. Arrows are unique, with <math>A\to B</math> precisely if <math>A\leq B</math>. | | - | * Every monoid is a category. Only one object. | + | * Every monoid is a category. Only one object. The elements of the monoid correspond to the endo-arrows of the one object. | | - | *** Kleene closure. Free monoids. | + | | | | * The category of Sets and injective functions. | | * The category of Sets and injective functions. | | | | + | ** Objects are sets. Morphisms are injective functions between the sets. | | | * The category of Sets and surjective functions. | | * The category of Sets and surjective functions. | | | | + | ** Objects are sets. Morphisms are surjective functions between the sets. | | | * The category of <math>k</math>-vector spaces and linear maps. | | * The category of <math>k</math>-vector spaces and linear maps. | | | * The category with objects the natural numbers and <math>Hom(m,n)</math> the set of <math>m\times n</math>-matrices. | | * The category with objects the natural numbers and <math>Hom(m,n)</math> the set of <math>m\times n</math>-matrices. | | | | + | ** Composition is given by matrix multiplication. | | | * The category of Data Types with Computable Functions. | | * The category of Data Types with Computable Functions. | | | ** Our ideal programming language has: | | ** Our ideal programming language has: | | Line 189: | | Line 251: | | | | ** This allows us to model a functional programming language with a category. | | ** This allows us to model a functional programming language with a category. | | | * The category with objects logical propositions and arrows proofs. | | * The category with objects logical propositions and arrows proofs. | | - | | + | * The category Rel has objects finite sets and morphisms <math>A\to B</math> being subsets of <math>A\times B</math>. Composition is by <math>(a,c)\in g\circ f</math> if there is some <math>b\in B</math> such that <math>(a,b)\in f, (b,c)\in g</math>. Identity morphism is the diagonal <math>(a,a): a\in A</math>. | | | | | | | | | | | | Line 197: | | Line 259: | | | | ===Homework=== | | ===Homework=== | | | | | | | - | For a passing mark, a written, acceptable solution to at least 2 of the 5 questions should be given no later than midnight before the next lecture. | + | For a passing mark, a written, acceptable solution to at least 3 of the 6 questions should be given no later than midnight before the next lecture. | | | | | | | | For each lecture, there will be a few exercises marked with the symbol *. These will be more difficult than the other exercises given, will require significant time and independent study, and will aim to complement the course with material not covered in lectures, but nevertheless interesting for the general philosophy of the lecture course. | | For each lecture, there will be a few exercises marked with the symbol *. These will be more difficult than the other exercises given, will require significant time and independent study, and will aim to complement the course with material not covered in lectures, but nevertheless interesting for the general philosophy of the lecture course. | | | | | | | - | # Prove the general associative law: that for any path, and any bracketing of that path, the same composition may be found. | + | # Prove the general associative law: that for any path, and any bracketing of that path, the same composition results. | | | | + | # Which of the following form categories? Proof and disproof for each: | | | | + | #* Objects are finite sets, morphisms are functions such that <math>|f^{-1}(b)|\leq 2</math> for all morphisms f, objects B and elements b. | | | | + | #* Objects are finite sets, morphisms are functions such that <math>|f^{-1}(b)|\geq 2</math> for all morphisms f, objects B and elements b. | | | | + | #* Objects are finite sets, morphisms are functions such that <math>|f^{-1}(b)|<\infty</math> for all morphisms f, objects B and elements b. | | | | + | :Recall that <math>f^{-1}(b)=\{a\in A: f(a)=b\}</math>. | | | # Suppose <math>u:A\to A</math> in some category <math>C</math>. | | # Suppose <math>u:A\to A</math> in some category <math>C</math>. | | | ## If <math>g\circ u=g</math> for all <math>g:A\to B</math> in the category, then <math>u=1_A</math>. | | ## If <math>g\circ u=g</math> for all <math>g:A\to B</math> in the category, then <math>u=1_A</math>. | | | ## If <math>u\circ h=h</math> for all <math>h:B\to A</math> in the category, then <math>u=1_A</math>. | | ## If <math>u\circ h=h</math> for all <math>h:B\to A</math> in the category, then <math>u=1_A</math>. | | - | ## These two results characterize the objects in a category by the properties of their corresponding identity arrows completely. | + | ## These two results characterize the objects in a category by the properties of their corresponding identity arrows completely. Specifically, there is a way to rephrase the definition of a category such that everything is stated in terms of arrows. | | | # For as many of the examples given as you can, prove that they really do form a category. Passing mark is at least 60% of the given examples. | | # For as many of the examples given as you can, prove that they really do form a category. Passing mark is at least 60% of the given examples. | | | #* Which of the categories are subcategories of which other categories? Which of these are wide? Which are full? | | #* Which of the categories are subcategories of which other categories? Which of these are wide? Which are full? | | Line 211: | | Line 278: | | | | ## For which sets is the free monoid on that set commutative. | | ## For which sets is the free monoid on that set commutative. | | | ## Prove that for any category <math>C</math>, the set <math>Hom(A,A)</math> is a monoid under composition for every object <math>A</math>. | | ## Prove that for any category <math>C</math>, the set <math>Hom(A,A)</math> is a monoid under composition for every object <math>A</math>. | | | | + | :For details on the construction of a free monoid, see the Wikipedia pages on the Free Monoid [[http://en.wikipedia.org/wiki/Free_monoid]] and on the Kleene star [[http://en.wikipedia.org/wiki/Kleene_star]]. | | | # * Read up on <math>\omega</math>-complete partial orders. Suppose <math>S</math> is some set and <math>\mathfrak P</math> is the set of partial functions <math>S\to S</math> - in other words, an element of <math>\mathfrak P</math> is some pair <math>(S_0,f:S_0\to S)</math> with <math>S_0\subseteq S</math>. We give this set a poset structure by <math>(S_0,f)\leq(S_1,g)</math> precisely if <math>S_0\subseteq S_1</math> and <math>f(s)=g(s)\forall s\in S_0</math>. | | # * Read up on <math>\omega</math>-complete partial orders. Suppose <math>S</math> is some set and <math>\mathfrak P</math> is the set of partial functions <math>S\to S</math> - in other words, an element of <math>\mathfrak P</math> is some pair <math>(S_0,f:S_0\to S)</math> with <math>S_0\subseteq S</math>. We give this set a poset structure by <math>(S_0,f)\leq(S_1,g)</math> precisely if <math>S_0\subseteq S_1</math> and <math>f(s)=g(s)\forall s\in S_0</math>. | | | #* Show that <math>\mathfrak P</math> is a strict <math>\omega</math>-CPO. | | #* Show that <math>\mathfrak P</math> is a strict <math>\omega</math>-CPO. | | Line 216: | | Line 284: | | | | #*# <math>k(0) = 1</math> | | #*# <math>k(0) = 1</math> | | | #*# <math>k(n)</math> is defined only if <math>h(n-1)</math> is defined, and then by <math>k(n)=n*h(n-1)</math>. | | #*# <math>k(n)</math> is defined only if <math>h(n-1)</math> is defined, and then by <math>k(n)=n*h(n-1)</math>. | | - | :Describe <math>\phi(n\mapsto n^2)</math> and <math>\phi(n\mapsto n^3)</math>. Show that <math>\phi</math> is ''continuous''. Find a fixpoint <math>(S_0,f)</math> of <math>\phi</math> such that any other fixpoint of the same function is less than this one. | + | :Describe <math>\phi(n\mapsto n^2)</math> and <math>\phi(n\mapsto n^3)</math>. Show that <math>\phi</math> is ''continuous''. Find a fixpoint <math>(S_0,f)</math> of <math>\phi</math> such that any other fixpoint of the same function is larger than this one. | | | :Find a continuous endofunction on some <math>\omega</math>-CPO that has the fibonacci function <math>F(0)=0, F(1)=1, F(n)=F(n-1)+F(n-2)</math> as the least fixed point. | | :Find a continuous endofunction on some <math>\omega</math>-CPO that has the fibonacci function <math>F(0)=0, F(1)=1, F(n)=F(n-1)+F(n-2)</math> as the least fixed point. | ## Current revision This page now includes additional information based on the notes taken in class. Hopefully this will make the notes reasonably complete for everybody. ## 1 Welcome, administratrivia I'm Mikael Vejdemo-Johansson. I can be reached in my office 383-BB, especially during my office hours; or by email to mik@math.stanford.edu. I encourage, strongly, student interactions. I will be out of town September 24 - 29. I will monitor forum and email closely, and recommend electronic ways of getting in touch with me during this week. I will be back again in time for the office hours on the 30th. ## 2 Introduction ### 2.1 Why this course? An introduction to Haskell will usually come with pointers toward Category Theory as a useful tool, though not with much more than the mention of the subject. This course is intended to fill that gap, and provide an introduction to Category Theory that ties into Haskell and functional programming as a source of examples and applications. ### 2.2 What will we cover? The definition of categories, special objects and morphisms, functors, natural transformation, (co-)limits and special cases of these, adjunctions, freeness and presentations as categorical constructs, monads and Kleisli arrows, recursion with categorical constructs. Maybe, just maybe, if we have enough time, we'll finish with looking at the definition of a topos, and how this encodes logic internal to a category. Applications to fuzzy sets. ### 2.3 What do we require? Our examples will be drawn from discrete mathematics, logic, Haskell programming and linear algebra. I expect the following concepts to be at least vaguely familiar to anyone taking this course: • Sets • Functions • Permutations • Groups • Partially ordered sets • Vector spaces • Linear maps • Matrices • Homomorphisms ### 2.4 Good references On reserve in the mathematics/CS library are: • Mac Lane: Categories for the working mathematician This book is technical, written for a mathematical audience, and puts in more work than is strictly necessary in many of the definitions. When Awodey and Mac Lane deviate, we will give Awodey priority. • Awodey: Category Theory This book is also available as an ebook, accessible from the Stanford campus network. The coursework webpage has links to the ebook under Materials. ### 2.5 Monoids In order to settle notation and ensure everybody's seen a definition before: Definition A monoid is a set M equipped with a binary associative operation * (in Haskell: mappend ) and an identity element $\emptyset$ (in Haskell: mempty ). A semigroup is a monoid without the requirement for an identity element. A function $f:M\to N$ is a monoid homomorphism if the following conditions hold: • $f(\emptyset) = \emptyset$ • f(m * m') = f(m) * f(m') Examples • Any group is a monoid. Thus, specifically, the integers with addition is a monoid. • The natural numbers, with addition. • Strings L * in an alphabet L is a monoid with string concatenation forming the operation and the empty string the identity. • Non-empty strings form a semigroup. Awodey: p. 10. ### 2.6 Partially ordered set Definition A partially ordered set, or a partial order, or a poset is a set P equipped with a binary relation $\leq$ which is: • Reflexive: $x\leq x$ for all $x\in P$ • Anti-symmetric: $x\leq y$ and $y\leq x$ implies x = y for all $x,y\in P$. • Transitive: $x\leq y$ and $y\leq z$ implies $x\leq z$ for all $x,y,z\in P$. If $x\leq y$ or $y\leq x$, we call x,y comparable. Otherwise we call them incomparable. A poset where all elements are mutually comparable is called a totally ordered set or a total order. If we drop the requirement for anti-symmetry, we get a pre-ordered set or a pre-order. If we have several posets, we may indicate which poset we're comparing in byindicating the poset as a subscript to the relation symbol. A monotonic map of posets is a function $f:P\to Q$ such that $x\leq_P y$ implies $f(x)\leq_Q f(y)$. Examples • The reals, natural numbers, integers all are posets with the usual comparison relation. A poset in which all elements are comparable. • The natural numbers, excluding 0, form a poset with $a\leq b$ if a | b. • Any family of subsets of a given set form a poset with the order given by inclusion. Awodey: p. 6. Preorders are defined on page 8-9. ## 3 Category Awodey has a slightly different exposition. Relevant pages in Awodey for this lecture are: sections 1.1-1.4 (except Def. 1.2), 1.6-1.8. ### 3.1 Graphs We recall the definition of a (directed) graph. A graph G is a collection of edges (arrows) and vertices (nodes). Each edge is assigned a source node and a target node. $source \to target$ Given a graph G, we denote the collection of nodes by G0 and the collection of arrows by G1. These two collections are connected, and the graph given its structure, by two functions: the source function $s:G_1\to G_0$ and the target function $t:G_1\to G_0$. We shall not, in general, require either of the collections to be a set, but will happily accept larger collections; dealing with set-theoretical paradoxes as and when we have to. A graph where both nodes and arrows are sets shall be called small. A graph where either is a class shall be called large. 0 If both G0 and G1 are finite, the graph is called finite too. The empty graph has $G_0 = G_1 = \emptyset$. A discrete graph has $G_1=\emptyset$. A complete graph has $G_1 = \{ (v,w) | v,w\in G_0\}$. A simple graph has at most one arrow between each pair of nodes. Any relation on a set can be interpreted as a simple graph. • Show some examples. A homomorphism $f:G\to H$ of graphs is a pair of functions $f_0:G_0\to H_0$ and $f_1:G_1\to H_1$ such that sources map to sources and targets map to targets, or in other words: • s(f1(e)) = f0(s(e)) • t(f1(e)) = f0(t(e)) By a path in a graph G from the node x to the node y of length k, we mean a sequence of edges $(f_1,f_2,\dots,f_k)$ such that: • s(f1) = x • t(fk) = y • s(fi) = t(fi − 1) for all other i. Paths with start and end point identical are called closed. For any node x, there is a unique closed path () starting and ending in x of length 0. For any edge f, there is a unique path from s(f) to t(f) of length 1: (f). We denote by Gk the set of paths in G of length k. ### 3.2 Categories We now are ready to define a category. A category is a graph G equipped with an associative composition operation $\circ:G_2\to G_1$, and an identity element for composition 1x for each node x of the graph. Note that G2 can be viewed as a subset of $G_1\times G_1$, the set of all pairs of arrows. It is intentional that we define the composition operator on only a subset of the set of all pairs of arrows - the composable pairs. Whenever you'd want to compose two arrows that don't line up to a path, you'll get nonsense, and so any statement about the composition operator has an implicit "whenever defined" attached to it. The definition is not quite done yet - this composition operator, and the identity arrows both have a few rules to fulfill, and before I state these rules, there are some notation we need to cover. #### 3.2.1 Backwards! If we have a path given by the arrows (f,g) in G2, we expect $f:A\to B$ and $g:B\to C$ to compose to something that goes $A\to C$. The origin of all these ideas lies in geometry and algebra, and so the abstract arrows in a category are supposed to behave like functions under function composition, even though we don't say it explicitly. Now, we are used to writing function application as f(x) - and possibly, from Haskell, as f x . This way, the composition of two functions would read g(f(x)). On the other hand, the way we write our paths, we'd read f then g. This juxtaposition makes one of the two ways we write things seem backwards. We can resolve it either by making our paths in the category go backwards, or by reversing how we write function application. In the latter case, we'd write x.f, say, for the application of f to x, and then write x.f.g for the composition. It all ends up looking a lot like Reverse Polish Notation, and has its strengths, but feels unnatural to most. It does, however, have the benefit that we can write out function composition as $(f,g) \mapsto f.g$ and have everything still make sense in all notations. In the former case, which is the most common in the field, we accept that paths as we read along the arrows and compositions look backwards, and so, if $f:A\to B$ and $g:B\to C$, we write $g\circ f:A\to C$, remembering that elements are introduced from the right, and the functions have to consume the elements in the right order. The existence of the identity map can be captured in a function language as well: it is the existence of a function $u:G_0\to G_1$. Now for the remaining rules for composition. Whenever defined, we expect associativity - so that $h\circ(g\circ f)=(h\circ g)\circ f$. Furthermore, we expect: 1. Composition respects sources and targets, so that: • $s(g\circ f) = s(f)$ • $t(g\circ f) = t(g)$ 2. s(u(x)) = t(u(x)) = x In a category, arrows are also called morphisms, and nodes are also called objects. This ties in with the algebraic roots of the field. We denote by HomC(A,B), or if C is obvious from context, just Hom(A,B), the set of all arrows from A to B. This is the hom-set or set of morphisms, and may also be denoted C(A,B). If a category is large or small or finite as a graph, it is called a large/small/finite category. A category with objects a collection of sets and morphisms a selection from all possible set-valued functions such that the identity morphism for each object is a morphism, and composition in the category is just composition of functions is called concrete. Concrete categories form a very rich source of examples, though far from all categories are concrete. Again, the Wikipedia page on Category (mathematics) [[3]] is a good starting point for many things we will be looking at throughout this course. ### 3.3 New Categories from old As with most other algebraic objects, one essential part of our tool box is to take known objects and form new examples from them. This allows us generate a wealth of examples from the ones that shape our intuition. Typical things to do here would be to talk about subobjects, products and coproducts, sometimes obvious variations on the structure, and what a typical object looks like. Remember from linear algebra how subspaces, cartesian products (which for finite-dimensional vectorspaces covers both products and coproducts) and dual spaces show up early, as well as the theorems giving dimension as a complete descriptor of a vectorspace. We'll go through the same sequence here; with some significant small variations. A category D is a subcategory of the category C if: • $D_0\subseteq C_0$ • $D_1\subseteq C_1$ • D1 contains 1X for all $X\in D_0$ • sources and targets of all the arrows in D1 are all in D0 • the composition in D is the restriction of the composition in C. Written this way, it does look somewhat obnoxious. It does become easier though, with the realization - studied closer in homework exercise 2 - that the really important part of a category is the collection of arrows. Thus, a subcategory is a subcollection of the collection of arrows - with identities for all objects present, and with at least all objects that the existing arrows imply. A subcategory $D\subseteq C$ is full if D(A,B) = C(A,B) for all objects A,B of D. In other words, a full subcategory is completely determined by the selection of objects in the subcategory. A subcategory $D\subseteq C$ is wide if the collection of objects is the same in both categories. Hence, a wide subcategory picks out a subcollection of the morphisms. The dual of a category is to a large extent inspired by vector space duals. In the dual C * of a category C, we have the same objects, and the morphisms are given by the equality C * (A,B) = C(B,A) - every morphism from C is present, but it goes in the wrong direction. Dualizing has a tendency to add the prefix co- when it happens, so for instance coproducts are the dual notion to products. We'll return to this construction many times in the course. Given two categories C,D, we can combine them in several ways: 1. We can form the category that has as objects the disjoint union of all the objects of C and D, and that sets $Hom(A,B)=\emptyset$ whenever A,B come from different original categories. If A,B come from the same original category, we simply take over the homset from that category. This yields a categorical coproduct, and we denote the result by C + D. Composition is inherited from the original categories. 2. We can also form the category with objects $\langle A,B\rangle$ for every pair of objects $A\in C, B\in D$. A morphism in $Hom(\langle A,B\rangle,\langle A',B'\rangle)$ is simply a pair $\langle f:A\to A',g:B\to B'\rangle$. Composition is defined componentwise. This category is the categorical correspondent to the cartesian product, and we denot it by $C\times D$. In these three constructions - the dual, the product and the coproduct - the arrows in the categories are formal constructions, not functions; even if the original category was given by functions, the result is no longer given by a function. Given a category C and an object A of that category, we can form the slice category C / A. Objects in the slice category are arrows $B\to A$ for some object B in C, and an arrow $\phi:f\to g$ is an arrow $s(f)\to s(g)$ such that $f=g\circ\phi$. Composites of arrows are just the composites in the base category. Notice that the same arrow φ in the base category C represents potentially many different arrows in C / A: it represents one arrow for each choice of source and target compatible with it. There is a dual notion: the coslice category $A\backslash C$, where the objects are paired with maps $A\to B$. Slice categories can be used, among other things, to specify the idea of parametrization. The slice category C / A gives a sense to the idea of objects from C labeled by elements of A. We get this characterization by interpreting the arrow representing an object as representing its source and a type function. Hence, in a way, the Typeable type class in Haskell builds a slice category on an appropriate subcategory of the category of datatypes. Alternatively, we can phrase the importance of the arrow in a slice categories of, say, Set, by looking at preimages of the slice functions. That way, an object $f:B\to A$ gives us a family of (disjoint) subsets of B indexed by the elements of A. Finally, any graph yields a category by just filling in the arrows that are missing. The result is called the free category generated by the graph, and is a concept we will return to in some depth. Free objects have a strict categorical definition, and they serve to give a model of thought for the things they are free objects for. Thus, categories are essentially graphs, possibly with restrictions or relations imposed; and monoids are essentially strings in some alphabet, with restrictions or relations. ### 3.4 Examples • The empty category. • No objects, no morphisms. • The one object/one arrow category 1. • A single object and its identity arrow. • The categories 2 and 1 + 1. • Two objects, A,B with identity arrows and a unique arrow $A\to B$. • The category Set of sets. • Sets for objects, functions for arrows. • The catgeory FSet of finite sets. • Finite sets for objects, functions for arrows. • The category PFn of sets and partial functions. • Sets for objects. Arrows are pairs $(S'\subseteq S,f:S'\to T)\in PFn(S,T)$. • PFn(A,B) is a partially ordered set. $(S_f,f)\leq(S_g,g)$ precisely if $S_f\subseteq S_g$ and $f=g|_{S_f}$. • The exposition at Wikipedia uses the construction here: [[4]]. • There is an alternative way to define a category of partial functions: For objects, we take sets, and for morphisms $S\to T$, we take subsets $F\subseteq S\times T$ such that each element in S occurs in at most one pair in the subset. Composition is by an interpretation of these subsets corresponding to the previous description. We'll call this category PFn'. • Every partial order is a category. Each hom-set has at most one element. • Objects are the elements of the poset. Arrows are unique, with $A\to B$ precisely if $A\leq B$. • Every monoid is a category. Only one object. The elements of the monoid correspond to the endo-arrows of the one object. • The category of Sets and injective functions. • Objects are sets. Morphisms are injective functions between the sets. • The category of Sets and surjective functions. • Objects are sets. Morphisms are surjective functions between the sets. • The category of k-vector spaces and linear maps. • The category with objects the natural numbers and Hom(m,n) the set of $m\times n$-matrices. • Composition is given by matrix multiplication. • The category of Data Types with Computable Functions. • Our ideal programming language has: • Primitive data types. • Constants of each primitive type. • Operations, given as functions between types. • Constructors, producing elements from data types, and producing derived data types and operations. • We will assume that the language is equipped with • A do-nothing operation for each data type. Haskell has id . • An empty type 1, with the property that each type has exactly one function to this type. Haskell has () . We will use this to define the constants of type t as functions $1\to t$. Thus, constants end up being 0-ary functions. • A composition constructor, taking an operator $f:A\to B$ and another operator $g:B\to C$ and producing an operator $g\circ f:A\to C$. Haskell has (.) . • This allows us to model a functional programming language with a category. • The category with objects logical propositions and arrows proofs. • The category Rel has objects finite sets and morphisms $A\to B$ being subsets of $A\times B$. Composition is by $(a,c)\in g\circ f$ if there is some $b\in B$ such that $(a,b)\in f, (b,c)\in g$. Identity morphism is the diagonal $(a,a): a\in A$. ### 3.5 Homework For a passing mark, a written, acceptable solution to at least 3 of the 6 questions should be given no later than midnight before the next lecture. For each lecture, there will be a few exercises marked with the symbol *. These will be more difficult than the other exercises given, will require significant time and independent study, and will aim to complement the course with material not covered in lectures, but nevertheless interesting for the general philosophy of the lecture course. 1. Prove the general associative law: that for any path, and any bracketing of that path, the same composition results. 2. Which of the following form categories? Proof and disproof for each: • Objects are finite sets, morphisms are functions such that $|f^{-1}(b)|\leq 2$ for all morphisms f, objects B and elements b. • Objects are finite sets, morphisms are functions such that $|f^{-1}(b)|\geq 2$ for all morphisms f, objects B and elements b. • Objects are finite sets, morphisms are functions such that $|f^{-1}(b)|<\infty$ for all morphisms f, objects B and elements b. Recall that $f^{-1}(b)=\{a\in A: f(a)=b\}$. 1. Suppose $u:A\to A$ in some category C. 1. If $g\circ u=g$ for all $g:A\to B$ in the category, then u = 1A. 2. If $u\circ h=h$ for all $h:B\to A$ in the category, then u = 1A. 3. These two results characterize the objects in a category by the properties of their corresponding identity arrows completely. Specifically, there is a way to rephrase the definition of a category such that everything is stated in terms of arrows. 2. For as many of the examples given as you can, prove that they really do form a category. Passing mark is at least 60% of the given examples. • Which of the categories are subcategories of which other categories? Which of these are wide? Which are full? 3. For this question, all parts are required: 1. For which sets is the free monoid on that set commutative. 2. Prove that for any category C, the set Hom(A,A) is a monoid under composition for every object A. For details on the construction of a free monoid, see the Wikipedia pages on the Free Monoid [[5]] and on the Kleene star [[6]]. 1. * Read up on ω-complete partial orders. Suppose S is some set and $\mathfrak P$ is the set of partial functions $S\to S$ - in other words, an element of $\mathfrak P$ is some pair $(S_0,f:S_0\to S)$ with $S_0\subseteq S$. We give this set a poset structure by $(S_0,f)\leq(S_1,g)$ precisely if $S_0\subseteq S_1$ and $f(s)=g(s)\forall s\in S_0$. • Show that $\mathfrak P$ is a strict ω-CPO. • An element x of S is a fixpoint of $f:S\to S$ if f(x) = x. Let $\mathfrak N$ be the ω-CPO of partially defined functions on the natural numbers. We define a function $\phi:\mathfrak N\to\mathfrak N$ by sending some $h:\mathbb N\to\mathbb N$ to a function k defined by 1. k(0) = 1 2. k(n) is defined only if h(n − 1) is defined, and then by k(n) = n * h(n − 1). Describe $\phi(n\mapsto n^2)$ and $\phi(n\mapsto n^3)$. Show that φ is continuous. Find a fixpoint (S0,f) of φ such that any other fixpoint of the same function is larger than this one. Find a continuous endofunction on some ω-CPO that has the fibonacci function F(0) = 0,F(1) = 1,F(n) = F(n − 1) + F(n − 2) as the least fixed point. Implement a Haskell function that finds fixed points in an ω-CPO. Implement the two fixed points above as Haskell functions - using the ω-CPO fixed point approach in the implementation. It may well be worth looking at Data.Map to provide a Haskell context for a partial function for this part of the task.

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 104, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.869941234588623, "perplexity_flag": "middle"}

http://math.stackexchange.com/questions/131921/very-interesting-sequence-problem

# Very Interesting Sequence Problem I need help primarily with the finding a solution, I already came up with an answer. We define an Eden sequence to be a subset of the set $\{1,2,3,4,\ldots ,N\}$. The Eden sequence has three conditions. 1. each of its terms is an element of the set of consecutive integers $\{1,2,3,4,\ldots ,N\}$, 2. the sequence is increasing, and 3. the terms in odd numbered positions are odd and the terms in even numbered positions are even. We then define a function $e(N)$ such that $e(N)$ denotes the number of Eden sequences of the set $\{1,2,3,4,\ldots ,N\}$. If we are given that $e(17)=4180$ and $q(20)=17710$, how would we find $e(18)$ and $e(19)$ using a mathematical approach? I am pretty sure the answers are that $e(18) = 6764$ and $e(19) = 10945$. Thanks for your help in advance! - ## 3 Answers Ok, either you or I are off by one, but here goes. The fact that the sequence is increasing but the parity stays the same means that the missing numbers go in consecutive blocks of two. So the number of Eden sequences is the same as the number of ways of tiling the numbers from 1-n with dominoes of one size one or two. These are well known to be the Fibonacci numbers. It appears that the sequence itself is not counted as the numbers you quote are one greater than the corresponding Fibonacci number. As will has observed, this does not count the sequences where an odd number of numbers are left out of the tail of the Eden sequence. These are eqivalent to the tilings that have a 2-domino extending one unit past the end. There are $F(n-1)$ of these (the sequences of length $n-1$ with a 2 domino slapped on the end). So there are $F(n)+F(n-1)=F(n+1)$ possible sequences, and it appears that the empty sequence is the missing one. - Hi thanks for the response! my calculation was based on trial and error it has no legit proof behind it. I tested it for small numbers and say this: q(1)=1, q(2)=q(1)+1=2,q(3)=q(2)+2=4,q(4)=q(3)+3=7,q(5)=q(4)+5=12,q(6)=q(5)+8=20. So obviously there is some relation to fibonacci numbers, but if you go along in this pattern my answer comes up and so does the conditions given in the problem. How can we show that q(n)=q(n-1)+F_n . Where F_n is the nth fibonacci number. Once again thanks for the help! – ahuang Apr 15 '12 at 3:53 Also can you please explain why the missing numbers go in consecutive blocks of two, and the domino idea. Thanks! – ahuang Apr 15 '12 at 3:59 @alan If you have an odd number followed by a larger even number there are an even number of integers skipped, Likewise for an even number followed by an odd number. – deinst Apr 15 '12 at 4:21 I believe the quoted numbers are one less than the corresponding Fibonacci numbers, most likely because the empty sequence is not counted. (More about this in my answer.) Nice observation about the dominoes, by the way. – Will Orrick Apr 15 '12 at 15:51 An observation related to domino tilings: use "." and "=" to represent dominoes of lengths 1 and 2. The tilings we get for the first few $N$ are {.} ($N=1$), {.., =} ($N=2$), {...,=.,.=} ($N=3$), {....,=..,.=.,..=,==} ($N=4$). These correspond to the Eden sequences {{1}}, ($N=1$), {{1,2},{}} ($N=2$), {{1,2,3},{3},{1}} ($N=3$), {{1,2,3,4},{3,4},{1,4},{1,2},{}} ($N=4$). Observe that (1) for $N$ even, only even-length Eden sequences correspond to domino tilings, while for $N$ odd, only odd-length Eden sequences correspond to domino tilings; – Will Orrick Apr 15 '12 at 16:33 show 2 more comments The answers for q(18) and q(19) are correct. The formula that can be established is that q(N)=q(n-1) + q(N-2) + 1, N>2. I found that examining q(N) separately for N odd and for N even helped. Also, for each N up to 6 I looked separately at the sequences that ended in an even number and those that ended in an odd number; this led me to the above equation. Using the formula along with the given values for q(17) and q(20) it was relatively easy to calculate q(18) and then q(19). I could go into more detail but if I can establish the above formula using this approach I'm sure you'll have no problem. - I tried to do just as you said but I still am not able to come up with the formula I dont know what it is that im doing wrong. Can you please go in a little bit more detail. Once again thank you very much! – ahuang Apr 15 '12 at 4:22 I was thinking similarly to the above answer using o(N) and e(N) but I did get q(18) = 6764 so I'll quickly outline my approach. Starting with N odd, we have q(N) = o(N) + e(N). Now for q(N+1), we can add (N+1) to the end of each odd sequence to form o(N) more sequences giving us q(N+1) = 2*o(N) + e(N). For q(N+2), with (N+2) being odd we can add it to the end of the e(N) sequences and the o(N) sequences we just made even, plus have (N+2) as a single digit sequence, giving us q(N+2) = 3*o(N) + 2 *e(N) + 1 = q(N+1) + q(N) + 1, which can be rewritten as q(N) = q(N-1) + q(N-2) + 1. – Brian Apr 15 '12 at 5:51 The same thing can be done starting with N even, yielding the same formula. I hope my approach isn't too muddled for you. – Brian Apr 15 '12 at 5:55 I think the discrepancy between our answers comes down to whether the empty sequence is allowed or not. I did not check, although I should have, whether the original poster's numbers for $q(17)$ and $q(20)$ included the empty sequence. Since in fact they do not, your recurrences are the correct ones. I will add a note about this to my answer. – Will Orrick Apr 15 '12 at 15:38 Denote $\{1,2,\ldots,N\}$ by $[N]$. Let $e(N)$ and $o(N)$ be the numbers of Eden sequences from $[N]$ of even length and of odd length. So $q(N)=e(N)+o(N)$. If $N$ is even, then $e(N)=e(N-1)+o(N-1)$ since an even-length sequence from $[N-1]$ is also an even-length sequence from $[N]$, while one may add $N$ to an odd-length sequence from $[N-1]$ to get an even-length sequence from $[N]$. Furthermore, $o(N)=o(N-1)$ since an odd-length sequence from $[N]$ cannot have $N$ as its last element. The $N$ odd case can be handled similarly. Using these recurrences, one may express $e(N)$ and $o(N)$, and hence $q(N)$, for $N>17$, in terms of $e(17)$ and $o(17)$. Since $q(17)=e(17)+o(17)$, one may use the known values of $q(17)$ and $q(20)$ to solve for $e(17)$ and $o(17)$. One may then compute $q(18)$ and $q(19)$. I find the same value of $q(19)$ that you found, but a different value of $q(18)$. Edit: The analysis above assumes that the empty sequence counts as one of the even sequences. This is not the assumption that was made in the original poster's question since $q(17)=4180$ and $q(20)=17710$ are the correct numbers when the empty sequence is omitted. We can modify the analysis above to account for this by leaving the definitions of $e(N)$ and $o(N)$ as is (that is, $e(N)$ includes the empty sequence), but redefining $q(N)$ as $q(N)=e(N)-1+o(N)$ (so that $q(N)$ does not include the empty sequence). With this modification our numbers are in perfect agreement. -

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 67, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9429876804351807, "perplexity_flag": "head"}

http://mathhelpforum.com/number-theory/212816-rearrangements-infinite-series.html

1Thanks • 1 Post By johng # Thread: 1. ## Rearrangements of an infinite Series If ak>=0 for all k in N, and Σak= prove that Σa'k= for any rearrangement Σa'k of Σak I think I understand this proof, I'm just having trouble making it rigorous. I was going to go the route that since ak is positive, then its sum is monotone increasing. In order to be convergent it would need to be bounded, so I can assume it's unbounded since it diverges. Then any rearrangement of the series is also going to be unbounded and then it will diverge. Is that correct thinking? 2. ## Re: Rearrangements of an infinite Series For a formal proof, I think you have to explicitly use the definition of series rearrangement. Here's my version: 3. ## Re: Rearrangements of an infinite Series to help decipher what johng has done: we know we can find a partial sum for the original series that is greater than any given M. this is a finite sum of terms of the original series. now in the "re-arranged" series, these terms may be scattered apart, but we can surely find a partial sum of the re-arranged series that contains ALL of them (even if we have to go out much further than we did in the original series), first we find "how far out (the largest index in the re-arragement)" we went, and include all those terms in our partial sum for the re-arranged series. this includes every term in our original partial sum, plus perhaps a lot more, but certainly no less. so this partial sum is at least the same as the partial sum of the original series (but probably a lot bigger). in any case, its bigger than M, and M can be as large as we like. EDIT: a note about the formalism involved. with infinite things, we can't really explicitly write them out. so, for example, to say a quantity (our series, in this instance) is infinite, we have to say something like: for every (finite) real number M, there exists a partial sum Sn, with Sn > M. this lets us say informally: $\sum_{k = 0}^{\infty} a_k = \infty$ but it must be kept in mind that the "=" sign here is just a SHORTHAND for the formal definition above (which doesn't even mention infinity, because infinity is NOT a number). re-arranging "infinite" things can have some counter-intuitive properties (it is possible with alternating series to get finite AND infinite "re-arrangements", which justifies the cautious approach taken here), so we want to take "behavior at infinity" back to the realm of the finite whenever possible, or else the reasoning can get a little "soft".

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9545174241065979, "perplexity_flag": "head"}

http://math.stackexchange.com/questions/tagged/q-analogs

# Tagged Questions Use this tag for questions pretaining to q-analogs of functions, for example q-Binomials, $q$-derivatives, the q-theta function, the q-Pochammer symbol, etc. 1answer 61 views ### What does the $q$-Catalan Numbers count? I had completed a paper describing the $q$-Catalan numbers, which is the $q$-analog of the Catalan numbers. The $n$-th Catalan numbers can be represented by: $$C_n=\frac{1}{n+1}{2n \choose n}$$ and ... 1answer 54 views ### Combinatorial identity. Using echelon matrices. Determine the exponents $e_i$ s.t. the following identity is correct. $$\sum\limits_{i=0}^k q^{e_i} {\binom mi}_q {\binom{n}{k-i}}_q = {\binom{n+m}{k}}_q$$ Note: When $q=1$ the equation reduces to a ... 3answers 101 views ### How to prove it? (one of the Rogers-Ramanujan identities) Prove the following identity (one of the Rogers-Ramanujan identities) on formal power series by interpreting each side as a generating function for partitions: ... 1answer 182 views ### Which families of groups have interesting formulas for the number of elements of given order? Suppose that $G$ is a group and that $n$ is a positive integer diving the order of $G$. Let $f_n(G)$ be the number of elements satisfying $x^n = 1$ in $G$. According to a theorem of Frobenius, then we ... 0answers 60 views ### A general Combinatorics problem (Coefficients of the q factorial) I was solving a combinatorics problem when I encountered difficulties. The problem was: $x_1 \in \{0,1\}$ $x_2 \in \{0,1,2\}$ . . $x_{n-1}\in\{0,1,2..,n-1\}$ We have to find the number of ways ... 4answers 337 views ### Proving q-binomial identities I was wondering if anyone could show me how to prove q-binomial identities? I do not have a single example in my notes, and I can't seem to find any online. For example, consider: \${a + 1 + b \brack ... 0answers 133 views ### Different notions of q-numbers It seems that most of the literature dealing with q-analogs defines q-numbers according to $$[n]_q\equiv \frac{q^n-1}{q-1}.$$ Even Mathematica uses this definition: with the built-in function QGamma ... 1answer 128 views ### Reciprocity Law of the Gaussian (or $q$-Binomial) Coefficient It is a standard exercise in combinatorics to show that the binomial coefficient satisfies the reciprocity law $\binom{-n}{k} = (-1)^k \binom{n+k-1}{k}$ for $n, k \geqslant 0$, which is the multiset ... 0answers 53 views ### Necessary and sufficient condition for $f(q^n)$ to be in $\mathbb{Z}[q,q^{-1}]$ when $f\in\mathbb{Q}(q)[x]$? In this question, user bgins shows that for each $k$ there is a unique polynomial $P_k(x)$ of degree $k$ whose coefficients are in $\mathbb{Q}(q)$, the field of rational functions, such that ... 2answers 217 views ### q-Analogue of the formula $x^n=\sum_k\left\{n\atop k\right\}(x)_k$. The stirling numbers of the second kind satisfy the formula $x^n=\sum_k\left\{n\atop k\right\}(x)_k$, where $(x)_k$ is the falling factorial. Consider the $q$-analog recursive definition of the ... 2answers 188 views ### Recurrence for $q$-analog for the Stirling numbers? I read in some papers that the Stirling numbers (of the second kind) have a natural $q$-analog $S_q(n,k)$, which satisfy the recurrence $$S_q(n,k)=(k)_qS_q(n-1,k)+q^{k-1}S_q(n-1,k-1)$$ with the ... 1answer 150 views ### Inferring Jacobi Triple product from $q$-binomial theorem? Quite a while ago I asked a question about deducing the Jacobi triple product from the $q$-binomial theorem. In fact, from the $q$-binomial theorem ... 1answer 148 views ### Is the following product of $q$-binomial coefficients a polynomial in $q$? $$\frac{\binom{n}{j}_q\binom{n+1}{j}_q \cdots\binom{n+k-1}{j}_q}{\binom{j}{j}_q\binom{j+1}{j}_q\cdots\binom{j+k-1}{j}_q}$$ where $n,j,k$ are non-negative integers. 1answer 219 views ### Intermediate step in deducing Jacobi's triple product identity. An intermediate step deduces Jacobi's triple product identity by taking the $q$-binomial theorem $$\prod_{i=1}^{m-1}(1+xq^i)=\sum_{j=0}^m\binom{m}{j}_q q^{\binom{j}{2}}x^j$$ and deducing ... 2answers 243 views ### Representing the $q$-binomial coefficient as a polynomial with coefficients in $\mathbb{Q}(q)$? Trying a bit of combinatorics this winter break, and I don't understand a certain claim. The claim is that for each $k$ there is a unique polynomial $P_k(x)$ of degree $k$ whose coefficients are ... 1answer 119 views ### Primitive roots as roots of a $q$-multinomial. If $n$ is divisible by $m$, why is it the case that the $m$th primitive roots of unity are also roots of $\binom{n}{k}_q$ if and only if $m$ does not divide $k$? I'm viewing $\binom{n}{k}_q$ as a ... 1answer 152 views ### Deriving Cauchy's identity from the $q$-binomial theorem? Cauchy's identity states that $$\prod_{i\geq 0}\frac{1-axq^i}{1-xq^i}=\sum_{n\geq 0}\frac{(1-a)(1-aq)\cdots(1-aq^{n-1})}{(1-q)(1-q^2)\cdots(1-q^n)}x^n.$$ Is it possible to somehow derive this ... 1answer 130 views ### What's the reasoning for this recurrence on $q$-multinomial coefficients? I'm familiar with the recurrence for binomial coefficients based on Pascal's triangle. However, in general, there is the recurrence for $q$-multinomial coefficients given by ... 1answer 114 views ### Closed form for $\sum_{m \geq 1} (-1)^m q^{m(m+1)/2 + m \Delta}$? Is there a useful closed form for the following series ($|\Delta|$ is a small integer)? $$f(q,\Delta) =\sum_{m=1}^{\infty} (-1)^m q^{m(m+1)/2 + m \Delta}$$ It is a large-$n$ approximation of the ... 4answers 447 views ### Intriguing polynomials coming from a combinatorial physics problem For real $0<q<1$, integer $n >0$ and integer $k\ge 0$, define $$[k, n]_q \equiv -\sum_{m=1}^{n} q^{m(k+1)} (q^{-n}; q)_m = -\sum_{m=1}^{n} q^{m(k+1)} \prod_{l=0}^{m-1} (1-q^{l-n})$$ ... 1answer 105 views ### Interpolation of a sequence of polynomials (viewed in terms of q-analogue powers) I'm trying to find closed-form expressions for a sequence of coefficients, such that the index of the coefficient occurs as number such that I can later interpolate to fractional indexes as well. ... 1answer 245 views ### Is there a gamma-like function for the q-factorial? I'm looking at quantum calculus and just trying to understand what is going with this subject. Looking at the q-factorial made me wonder if this function could take all real or even complex numbers in ... 2answers 239 views ### What is Eulerian? I encountered an interesting function which is called "Eulerian" by the Wolfram's MathWorld: $$\phi(q)=\prod_{k=1}^{\infty} (1-q^{k})$$ It is interesting because it seems that roots of any ...

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 65, "mathjax_display_tex": 10, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9183622002601624, "perplexity_flag": "head"}

http://www.physicsforums.com/showthread.php?t=611510

Physics Forums Why electric field is zero inside a conductor? 1.Why electric field is zero inside a conductor? Can anyone explain it with Gauss's law and without Gauss's law? 2.Is it possible to create electric field inside a conductor? If so,how and in which condition it is created? PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus 1. In electrostatics the electric field is 0 inside a perfect conductor because otherwise there would be moving charges. The charges must redistribute themselves to make a net electric field inside the conductor 0. 2. It is possible to have an electric field inside a conductor. You can put one there by e.g. using a battery. The trick is that this prompts charges to move inside the conductor and therefore a current to develop. This is no longer electrostatics. Well,I got this.But how you will explain this with Gauss's law?Explain why electric field is zero inside a conductor by the help of Gauss's law. Now if i ask,why charges are always distributed on the surface?Why there is no charge inside the conductor?I prepared like this... Consider a gaussian surface inside the conductor A (see the figure attached).As electric field is zero inside the conductor,Gauss's law requires that there could be no net charge inside the surface.Now imagine shrinking the surface like a collapsing balloon until it encloses a region so small that we may consider it as a point.Then the charge at that point must be zero.We can do this anywhere inside the conductor,so there could be no excess charge an any point within a solid conductor.Any excess charge must reside on the conductor surface. Attached Thumbnails Why electric field is zero inside a conductor? Gauss' Law can not be used since as stated that is for an electrostatic condition. In other words it is not used when you have a moving charge. If you have a charge within an electric field that puts a force on that charge. If it is free to move it will move due to that force. In your illustration each charge would create its own electric field that other charges will react to. All of the charges push against all of the other charges and they end up on the outer boundary of the conductor where they cannot move away from each other any further. Ohm's law (differential form) $\textbf{j}=\sigma\textbf{E}$ If E is not zero, then a current will very quickly form in a conductor. If the conductor is surrounded by insulator, then charge can't escape from the conductor, so you can't have net current escaping from the conductor. Therefore, the flux of E over the surface of the conductor is 0. There's no law that says E is zero inside the conductor, but it goes to 0 very quickly in steady state or slowly varying conditions. Internal E field will generate internal currents, which will quickly dissipate electrical energy into heat via internal resistance. Tags electric field, electricity, gauss's law Thread Tools | | | | |---------------------------------------------------------------------|-------------------------------|---------| | Similar Threads for: Why electric field is zero inside a conductor? | | | | Thread | Forum | Replies | | | Classical Physics | 12 | | | Introductory Physics Homework | 3 | | | Classical Physics | 37 | | | Introductory Physics Homework | 1 | | | Introductory Physics Homework | 1 |

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9228706955909729, "perplexity_flag": "middle"}

http://mathoverflow.net/revisions/75096/list

## Return to Answer 2 reversed inclusion for $H^1$ and $L^4$ I am not going to try to find the most general conditions under which lower semi-continuity holds but for that I suggest the standard reference for all of this is "Direct methods in the Calculus of Variations" by Bernard Dacorogna which covers all of this in full detail. I will give a brief outline of the answer however. In order for $\int f(x,u,du)dx$ to be lower semi-continuous with respect to weak convergence one does not in general need any sort of convexity in the $u$ variable and what is more important generally is 1) Coercivity (see below) of the functional $f$ in the $du$ variable. 2) The right embedding along with continuity of $f$ in the $u$ variable with respect to the topology of strong convergence in this space. Example: A good example is $E(u) = \int_0^1 |\nabla u |^2 + (1-u^2)^2$ with $u = 0$ on $\partial \Omega$ in $\mathbb{R}^d$. Observe that when $d \leq 4$ one has $L^4 H^1(\Omega) \subset \subset H^1(\Omega)$ L^4(\Omega)$and consequently the non-convex term depending on$u$is lower order. Therefore one can pass to the limit in any minimizing sequence even though$(1-u^2)^2$is very non convex (but it is continuous with respect to strong convergence). Notice however that for the energy$\bar E(u) = \int_0^1 |\nabla u|^2 + u^4$in$\mathbb{R}^5$with$u$prescribed on$\partial \Omega\$, the second term is not lower order but here we may use convexity to conclude lower semi-continuity of the second term. The function $p \mapsto f(x,z,p)$ is coercive if there exists some constant $C > 0$ so that $f(x,z,p) \geq C|p|^q$ for some range of $q$. It then depends on what spaces one is working in but the goal is to use an embedding theorem such as $L^q \subset\subset W^{1,p}$ for $1 \leq q < p^*$ and to conclude that for a minimizing sequence $u_n$, there is in fact a strongly convergent subsequence in $L^q$ for some $q$. Then one will generally expect some sort of continuity of $f$ in the $u$ variable. There are many more interesting examples but in almost all cases in practice the goal is to show that one has strong convergence of the $u_n$s in your minimizing sequence in some $L^p$ space. The book by Braides focuses mostly on asymptotics of functionals which depend on some large (or small) parameter and I'm not sure how much he talks about the assumptions needed in the direct method. 1 I am not going to try to find the most general conditions under which lower semi-continuity holds but for that I suggest the standard reference for all of this is "Direct methods in the Calculus of Variations" by Bernard Dacorogna which covers all of this in full detail. I will give a brief outline of the answer however. In order for $\int f(x,u,du)dx$ to be lower semi-continuous with respect to weak convergence one does not in general need any sort of convexity in the $u$ variable and what is more important generally is 1) Coercivity (see below) of the functional $f$ in the $du$ variable. 2) The right embedding along with continuity of $f$ in the $u$ variable with respect to the topology of strong convergence in this space. Example: A good example is $E(u) = \int_0^1 |\nabla u |^2 + (1-u^2)^2$ with $u = 0$ on $\partial \Omega$ in $\mathbb{R}^d$. Observe that when $d \leq 4$ one has $L^4 \subset \subset H^1(\Omega)$ and consequently the non-convex term depending on $u$ is lower order. Therefore one can pass to the limit in any minimizing sequence even though $(1-u^2)^2$ is very non convex (but it is continuous with respect to strong convergence). Notice however that for the energy $\bar E(u) = \int_0^1 |\nabla u|^2 + u^4$ in $\mathbb{R}^5$ with $u$ prescribed on $\partial \Omega$, the second term is not lower order but here we may use convexity to conclude lower semi-continuity of the second term. The function $p \mapsto f(x,z,p)$ is coercive if there exists some constant $C > 0$ so that $f(x,z,p) \geq C|p|^q$ for some range of $q$. It then depends on what spaces one is working in but the goal is to use an embedding theorem such as $L^q \subset\subset W^{1,p}$ for $1 \leq q < p^*$ and to conclude that for a minimizing sequence $u_n$, there is in fact a strongly convergent subsequence in $L^q$ for some $q$. Then one will generally expect some sort of continuity of $f$ in the $u$ variable. There are many more interesting examples but in almost all cases in practice the goal is to show that one has strong convergence of the $u_n$s in your minimizing sequence in some $L^p$ space. The book by Braides focuses mostly on asymptotics of functionals which depend on some large (or small) parameter and I'm not sure how much he talks about the assumptions needed in the direct method.

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 64, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9544402956962585, "perplexity_flag": "head"}

http://physics.stackexchange.com/questions/tagged/work?sort=faq&pagesize=30

# Tagged Questions The work tag has no wiki summary. 7answers 3k views ### Why does holding something up cost energy while no work is being done? I read the definition of work as $$W ~=~ \vec{F} \cdot \vec{d}$$ $$\text{ Work = (Force) $\cdot$ (Distance)}.$$ If a book is there on the table, no work is done as no distance is covered. If I ... 4answers 337 views ### Violation of Newton's Second Law (?) Here the big circle denotes the circular path of a stone(small circle on path) tied to a string from the centre of the circular path . This is COMPLETELY HORIZONTAL At an instant the velocity in ... 2answers 555 views ### Why is there a $\frac 1 2$ in $\frac 1 2 mv^2$? For elastic collisions of n particles, we know that momentum in the three orthogonal directions are independently conserved:$$\frac{d}{dt}\sum\limits_i^n m_iv_{ij} =0,\quad j=1,2,3$$ From this, it ... 1answer 304 views ### A Question about Virtual Work related to Newton's Third Law In describing D'Alembert's principle, the lecture note I was provided with states that the total force $\mathbb F_l$ acting on a particle can be taken as, $$\mathbb F_l=F_l+\sum_mf_{ml}+C_l,$$ ... 3answers 534 views ### Is the normal force a conservative force? Most of the time the normal force doesn't do any work because it's perpendicular to the direction of motion but if it does do work, would it be conservative or non-conservative? For example, consider ... 2answers 241 views ### Intuitively Understanding Work and Energy It is easy to understand the concepts of momentum and impulse. The formula $mv$ is simple, and easy to reason about. It has an obvious symmetry to it. The same cannot be said for kinetic energy, ... 6answers 324 views ### Electrostatic Potential Energy Derivation How is the boxed step , physically as well as mathematically justified and correct ? Source:Wiki http://en.wikipedia.org/wiki/Electric_potential_energy As work done = $- \Delta U$. for Conservative ... 3answers 170 views ### Why does it require such little energy to create the fastest thing in the universe? I have noticed when I turn on the light switch in my house light comes from the bulb. How is this light created?(process occurring in the bulb) and why is this small amount of electricity enough to ... 4answers 601 views ### Work done by the Magnetic Force The magnetic part of the Lorentz force acts perpendicular to the charge's velocity, and consequently does zero work on it. Can we extrapolate this statement to say that such a nature of the force ... 1answer 997 views ### How much work is needed to compress a certain volume of gas? I want to know the formula (and what does the symbols stand for) for how much work is needed to compress a certain volume of gas?

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9234683513641357, "perplexity_flag": "middle"}

http://mathoverflow.net/questions/97100/jordan-algebra-and-quaternionic-projective-space/97101

## Jordan algebra and quaternionic projective space ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) How we can define quaternionic projective space and metric on it using Jordan algebra? Thank you in advance! - ## 2 Answers I am not much educated in this subject so please take my answer with a grain of salt. The appropriate Jordan algebra is $J_n(\mathbb{H}) := \{ A \in M(n,\mathbb{H})\,|\, \overline{A}^T=A \}$ with multiplication defined as $A\circ B = (AB+BA)/{2}$. The metric is defined by $\mathrm{Tr}(A^2)$. The quaternionic projective space can be then defined as the space of elements of $J_n(\mathbb{H})$ of rank one and trace one (think of these matrices as projectors to one-dimensional subspaces) or alternatively as the (real) projectivization of rank one matrices. I am little bit more familiar with the case of octonionic projective plane, so let me explain that case. There the relevant algebra is $J_3(\mathbb{O})$ and the plane $\mathbb{OP}^2$ and its metric is defined in the same way[1] as in the quaternionic case. Now there is a well defined cubic form on $J_3(\mathbb{O})$ which is basically the determinant. The group that preserves this cubic form is $E_6$. In fact one defines some sort of cross product (I think it is called Jordan cross product.) out of the cubic form which then gives the incidence relation of the projective geometry of $\mathbb{OP}^2$. If I remember correctly, the product $A\times B$ is defined by the relation $(A\times B,C) = (A,B,C)$, where on the left hand side the brackets denote the polarization of the quadratic form while on the right hand side the brackets denote the polarization of the cubic form. The group $E_6$ is then the group of collineations of $\mathbb{OP}^2$ The group that preserves the determinant and the quadratic form (the norm) is the group $F_4$. This sheds some light on the fact that $F_4$ is in fact the isometry group of $\mathbb{OP}^2$. It can be proven that $F_4$ is in fact the automorphism group of the Jordan algebra $J_3(\mathbb{O})$! In the quaternionic case one finds that $E_6$ is replaced by $Sl(n,\mathbb{H})$ and $F_4$ by $Sp(n)$. The projective plane $\mathbb{OP}^2$ was much studied by Freudenthal and others but I do not know of any reference where the quaternionic case is treated via Jordan algebras. [1] Of course one needs to be a little careful with the definition of rank because of the nonassociativity of $\mathbb{O}$. Either one uses the fact that any element of $J_3(\mathbb{O})$ can be mapped by the action of $F_4$ to a diagonal matrix and then one defines the rank by the number of nonzero elements in this diagonalization. Or one can experiment a little bit and discover that it is possible to define determinants of 2x2 minors in such a way that their vanishing is equivalent to matrix of being of rank one according to first definition. - 1 How does the structure of the Jordan algebra comes into play ? Is the space of rank 1 closed with respect to Jordan's product ? I would be very thankful for further comments... For the long time I keep some papers on Jordan algs nearby me but always does not see key how to open them :) – Alexander Chervov May 16 2012 at 10:28 @robot Still I am not clear about the role of Jordan algebra structure. If you will omit this word and just write matrices A^t=A , norm Tr(A^2)... nothing changes.. Any way thank you for yours answer, it useful (at least for me). – Alexander Chervov May 16 2012 at 12:00 I do not know of a nice direct way to connect Jordan algebra structure and the geometry of the projective space. In the octonionic case, the incidence relation is given by the cross product which can be expressed in terms of Jordan product and traces. (See arxiv.org/abs/0902.0431) – robot May 16 2012 at 12:05 You are right, the conditions $\overline{A}^t=A$ and being of rank one and trace one are sufficient to define the projective space and its metric. The Jordan multiplication lurks in the background, because - at least in the octonionic case - the group of isometries is the group of automorphisms of the Jordan algebra. – robot May 16 2012 at 12:08 Is the condition $\overline{A}^{T}=A$ here because these matrices preserve $H$-inner product? Is there any connection between $Sp(n)$ and and $J_{n}(H)$? – Mirjana May 17 2012 at 9:33 show 3 more comments ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. (Perhaps my answer will be too general for your purposes.) The description of the projective space that you mention, is only one special case of a very general construction (due to O. Loos) of a projective space from a Jordan pair, which is a pair of vector spaces $(V^+, V^-)$ equipped with certain quadratic operators. The notion of a Jordan pair generalizes the notion of a Jordan algebra, in the sense that any Jordan algebra $J$ can be made into a Jordan pair (but the converse is not true). If $V = (V^+, V^-)$ is such a Jordan pair, then the projective space $X(V)$ of $V$ is the quotient of $V^+ \times V^-$ by projective equivalence, defined as $$(x, y) \sim (x', y') \iff (x, y - y') \text{ is quasi-invertible, and } x' = x^{y-y'},$$ where $x^{y-y'}$ denotes the quasi-inverse of the pair $(x, y-y')$. In the case where $V$ is the Jordan pair arising from the Jordan algebra of hermitian matrices over any skew field $D$ (of which $D = \mathbb{H}$ is just a particular case), the projective space $X(V)$ is the classical projective space coordinatized by the skew field $D$. - what is "projective space" ? Some nice manifold ? Symmetric space ? homogeneous space ? – Alexander Chervov May 16 2012 at 12:29 For general fields, it's just an algebraic object, but for instance if $k = \mathbb{C}$, then one obtains the compact hermitian symmetric spaces. – Tom De Medts May 16 2012 at 12:33 @Tom thank you ! Do all compact hermitian spaces can be obtained in this way ? – Alexander Chervov May 16 2012 at 13:00 I guess so (but I'm not very familiar with symmetric spaces though). You might be interested in the lecture notes "Jordan pairs and bounded symmetric domains" by O. Loos, available on the website homepage.uibk.ac.at/~c70202/jordan/archive/irvine/…. – Tom De Medts May 16 2012 at 13:10

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 45, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9127271175384521, "perplexity_flag": "head"}

http://mathoverflow.net/questions/91939?sort=votes

## Higher computability : Constructive ordinal and $\Delta^1_1$ predicates ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Everything I know on this subject comes from Sacks book : "Higher recursion theory" Let $\mathcal{O^Y}$ be the set of codes for ordinals constructive in $Y$. We should have the result that $A \subseteq \omega \times 2^\omega$ is $\Delta^1_1$ iff $\exists a \in \mathcal{O}\ \ \exists e \in \omega$ such that $A(n, Y) \leftrightarrow \varphi_e^{H_a^Y}(n) \downarrow$ where $H_a^Y$ is the $|a|$-th iteration of the turing jump of $Y$. Two things are now in contradiction in my mind : The first one : $X$ is $\Delta^1_1(Y)$ iff $\exists a \in \mathcal{O^Y}\ \ \exists e \in \omega$ such that $n \in X \leftrightarrow \varphi_e^{H_a^Y}(n) \downarrow$. Potentially we can have $|a| \geq \omega_1^{ck}$ if $\omega_1^{Y} \geq \omega_1^{ck}$ The second one : $X$ is $\Delta^1_1(Y)$ iff then there exists a $\Delta^1_1$ predicate $A \subseteq \omega \times 2^\omega$ and $\exists a \in \mathcal{O}\ \ \exists e \in \omega$ such that $n \in X \leftrightarrow A(n, Y) \leftrightarrow \varphi_e^{H_a^Y}(n) \downarrow$ This time, the code $a$ for the constructive ordinal is always smaller than $\omega_1^{ck}$. Can anyone see where I made a mistake ? Thanks in advance - ## 1 Answer I finaly got an answer from another forum. The answer is simple, I was assuming that if $A \subseteq \omega$ is $\Delta^1_1(Y)$ it means that there is a $\Pi^1_1$ predicate $F \subseteq \omega \times 2^\omega$ and a $\Sigma^1_1$ predicate $E \subseteq \omega \times 2^\omega$ such that $\forall n\ \ A(n) \leftrightarrow F(n, Y) \leftrightarrow E(n, Y)$ and $\forall X\ \ \forall n\ \ F(n, X) \leftrightarrow E(n, X)$. But it does just mean that $\forall n\ \ F(n, Y) \leftrightarrow E(n, Y)$ which is indeed very different. -

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 33, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9488143920898438, "perplexity_flag": "head"}

http://climateaudit.org/2007/05/04/more-on-hegerl-et-al-2006-non-confidence-intervals/?like=1&source=post_flair&_wpnonce=134bd2d71b

by Steve McIntyre ## More on Hegerl et al 2006 Non-Confidence Intervals There are a few other blogs that from time to time do detailed analyses of what people are doing, not dissimilar in format to what I do. Last year, in Apr 2006 shortly after publication, I observed here that the upper and lower confidence intervals of Hegerl et al crossed over. In Feb 2007, Tapio Schneider published a Comment in Nature observing that the confidence intervals in Hegerl et al were wrong. Hegerl published a Reply and replaced the Supplementary Information with new data (I kept the old version in case anyone wants a comparison.) James Annan recently discussed the matter , linking to my graph, acknowledging it in a business-like way. About the new Supplementary Information, he said: There is now a file giving the reconstruction back to 1500 with new confidence intervals, which no longer vanish or swap over. This new data doesn’t match the description of their method, or the results they plotted in their Fig 1 (which is almost surely a smoothed version of the original supplementary data). He went on to say: Hegerl et al used a regression to estimate past temperatures anomalies as a function of proxy data, and estimated the uncertainty in reconstructed temperature as being entirely due to the uncertainty in the regression coefficient. The problem with this manifests itself most clearly when the tree ring anomaly is zero, as in this event the uncertainty in the reconstructed temperature is also zero! Maybe UC (or Jean S who we haven’t heard from for a while) can comment on this. This comment still doesn’t seem right to me as I can’t think of why the uncertainty would be zero merely because all of the uncertainty was allocated to the regression coefficient. I still can’t get a foothold on what they’re doing here; Annan said that Tapio Schneider had been unsuccessful in getting Hegerl to document what they did in any of the calculations. I’ll write but I’m not optimistic about my chances. I’m up to about 20 emails with Crowley trying to find out how they got their Mongolia and Urals series, without any success. Eli Rabett observed that Huang et al appeared to have done the same thing in a borehole study. In the caption, Huang refer to Bayesian methods being used, so maybe there’s a clue for someone. Whatever these folks are doing, it’s not a totally isolated incident. Who knows – one day,we might even find out how the MBH99 confidence intervals were calculated – presently one of the 21st Century Hilbert Problems in climate science. References: Hegerl JClim 2006 here Hegerl Nature 2006 here Hegerl SI here ### Like this: Like Loading... This entry was written by , posted on May 4, 2007 at 11:06 AM, filed under Hegerl 2006. Bookmark the permalink. Follow any comments here with the RSS feed for this post. Post a comment or leave a trackback: Trackback URL. ### 25 Comments 1. John Hekman Posted May 4, 2007 at 11:32 AM | Permalink | Reply if y = a + bx + e, and you want to know the variance of the estimated value, y^, that variance is not zero when you are comparing the actual time series with the fitted values and one of the fitted values is equal to the actual value. If that is what he is saying here, he’s nuts. You just have to look at the ANOVA. http://en.wikipedia.org/wiki/ANOVA 2. Posted May 4, 2007 at 4:18 PM | Permalink | Reply Confidence intervals, regardless of method, cannot vanish or cross over. If one does end up with such confidence intervals, it is a sure sign that something is seriously wrong in the way calculations have been done. If it happened to me, I would withdraw to an isolated room, not get out until I figured out my error and I would be ashamed to tell anyone about whatever stupid mistake I made. However, this paper got published. Sad. Here is what I mean: The width of the confidence interval for mean response in a simple OLS regression with one independent variable depends on the standard error of the mean response (which, in turn, depends on the x-value at which mean response is being evaluated). We have, from the Intro Stats textbook I use in the undergrad stats class I teach, $\mathrm{SE}_{\hat{\mu}} = s\sqrt{\frac{1}{n} + \frac{(x^*-\bar{x})^2}{\Sum (x_i-\bar{x})^2}}$ If $x^* = \bar{x}$, this reduces to $s\sqrt{\frac{1}{n}}$ which cannot be zero (unless the regression line is a perfect fit) because s is the square root of MSE. Now, confidence intervals for mean response will be narrower than prediction confidence intervals (for the same confidence) but they cannot vanish and the lower and upper bounds cannot cross. I am baffled. Sinan 3. 2dogs Posted May 4, 2007 at 4:21 PM | Permalink | Reply It looks like the uncertainty is all assigned to the regression co-effecient of the anomoly (i.e. to b in y = bx + e); consequently, the upper and lower bounds flip over when the anomoly passes zero. 4. Posted May 4, 2007 at 4:22 PM | Permalink | Reply Correction to #2: In the formula for the standard error of mean response, there is a sum $\Sigma$ sign missing from the denominator of the second term in the square root. Also, I mention the undergrad intro stats textbook to underline the fact that this is basic, elementary stuff. Vanishing confidence intervals or crossing upper and lower bounds is an indication that the author of the article ought not be trusted to balance his own checkbook. Sinan 5. Steve McIntyre Posted May 4, 2007 at 10:07 PM | Permalink | Reply I sent the following inquiry to Gabi Hegerl: I have not received the requested information on the Mongolian and Urals series from Tom. I realize that you have other concerns but this has been going on far too long. Also can you provide a replicable description of how you calculated your confidence intervals. I realize that the original SI required correction, but in the wake of this there is no formal description of your procedures or statistical references. If it would save you time to merely provide source code for the results, that would be fine with me. 6. Steve McIntyre Posted May 5, 2007 at 7:46 AM | Permalink | Reply #5. Gabi Hegerl replied as follows: Tom, can you send Steve some more information, please? Steve, Tom produced the individual timeseries, but not with fortran code but with other software, so I am not sure there is a source code. The correction in the SI was minimal and it would have been totally sufficient to change order 3 words in the caption, but I thought it was more helpful to instead link the reconstructions with full uncertainty] range. The description in the SI is very complete on that, particularly together with my J Climate paper. It should make it completely reproducable. Gabi 7. Nicholas Posted May 5, 2007 at 8:42 AM | Permalink | Reply Did you plot that diagram yourself? If so, is the R code easily available somewhere? I’m asking because they look an awful lot like they are merely scaled versions of each other. It would be interesting to check if that’s true. 8. Steve McIntyre Posted May 5, 2007 at 9:12 AM | Permalink | Reply I’ll try to post up the code today. It might be an interesting project for statistically-interested people to decode their uncertainty methodology. I’ll post up as much relevant data as I have. 9. Henry Posted May 5, 2007 at 5:45 PM | Permalink | Reply It looks to me as if they originally used a statistical recipe in the wrong context. Trying to imagine a proper context: suppose you tried to regress national GDP against national population using data for 100 countries. $y_i=\alpha +\beta x_i +\varepsilon_i$ would be a bad model since it is obvious that a country with a tiny population would have a tiny GDP. $y_i=\beta x_i +\varepsilon_i$ would be almost as bad as the absolute values of errors would clearly be related to population. $y_i=\beta \varepsilon_i x_i$ starts to make more sense. But that model only makes sense if all the values are positive and if errors tend to increase with the independent variable. That doesn’t apply here, so it was the wrong method to apply. It probably stemmed from a desire to have everything at zero in the base period for calculating anomalies. And the graph should have been spotted by the peer reviewers. 10. Vince Causey Posted May 6, 2007 at 12:06 PM | Permalink | Reply I have heard it said before, that a lot of these results that involve statistical techniques lack the input of suitably qualified statisticians. Am I right in thinking that elementary statistical errors are being made due to lack of competence? If this is the case, then the quality of a lot of this kind of research must be suspect. 11. Posted May 6, 2007 at 1:04 PM | Permalink | Reply Very briefly: something odd in the manuscript figure 2 as well, cross overs.. BTW, ‘negative bias of the variance of the reconstruction’ in supplementary refers to vS04. I believe that this negative bias is a result of ‘natural calibration’ assumption, i.e. optimal solution in the case where P and T are obtained from joint normal distribution. I’ll return to this topic later ( and if I’m wrong I just disappear ) Recommended reading: Confidence and Conflict in Multivariate Calibration (PJ Brown, R Sundberg, Journal of the Royal Statistical Society, Ser B, Vol 49, No 1, 1987) Multivariate Calibration – Direct and Indirect Regression Methodology (R Sundberg) Those papers contain some ideas how to deal with uncertainties in this context (filtering theory is another way, but it doesn’t seem to interest climate people ). 12. Tim Posted May 8, 2007 at 7:22 PM | Permalink | Reply Here is how one can get “confidence” intervals of zero width in temperature reconstructions: 1) Estimate a regression model T – [T] = b(p – [p] + eta) + eps (*) where T-[T] are temperature anomalies (mean temperature [T]), p-[p] are proxy anomalies, and eta and eps are error terms. 2) “Reconstruct” temperature anomalies by taking expected values in the regression model (*) and plugging in estimated regression coefficients b’ (and estimated mean values [T] and [p]). -> The “reconstructed” temperature anomalies T-[T] are zero for any value b’ of the estimated regression coefficient if the proxy anomaly p-[p] is zero. 3) Vary the estimated regression coefficient b’ within estimated confidence limits and “reconstruct” temperature anomalies as in (2) for each value of b’. -> The “reconstructed” temperature anomalies will still be zero for all values of b’ whenever p-[p] is zero. 4) If one infers confidence intervals from the reconstructed temperature anomalies for the different b’ for each year, they will have zero width whenever p-[p] is zero. Of course, it makes no sense to estimate confidence intervals in this way. This seems to be what Hegerl et al. were doing and what led to their manifestly wrong confidence intervals for the reconstructed temperatures. Schneider’s point in the Nature comment, however, seems to be more general. Hegerl et al. base their inference about climate sensitivity on the residual difference between the “reconstructed” temperature anomalies and temperature anomalies simulated with an EBM. Schneider points out that What should enter the calculation of the likelihoods of the temperature-anomaly time series T-[T] is the estimated variance of the residuals r, not just the sample variance proportional to their sum of squares, sum(r^2). While the error terms eta and eps do not enter the expected value of the difference between EBM temperature anomalies and reconstructed temperature anomalies, they do enter the variance of the residual difference. The variance of the estimated regression coefficients contributes to the residual variance, but so do the variances of eta and eps. This is elementary regression analysis: The variances of eta and eps are generally greater (by a factor of order sample size) than the variance of b’. 13. Posted May 9, 2007 at 6:59 AM | Permalink | Reply #12 Clarifying posts are appreciated, thks (and posters should not be worried about making mistakes, if you make an explicit mistake, gavin drops by and corrects it, so no worries ). It is 1959 in climate science in statistics sense, Williams just published Regression Analysis. Next year will be interesting, as Kalman will publish his A New Approach to Linear Filtering and Prediction Problems. 14. Steve McIntyre Posted May 9, 2007 at 7:06 AM | Permalink | Reply #12. That makes total sense as an explanation. That’s unbelievable. What a joke. I’ll make a separate post on it in a few days. 15. bender Posted May 15, 2007 at 10:56 PM | Permalink | Reply Any updates on this? It’s hard to imagine an error of this magnitude being made by an author and then not being caught by co-authors, internal reviewers, peer reviewers, co-editors, editor. The crossing over of the curves is an obvious sign something’s gone wrong. An undergraduate could tell you that. 16. bender Posted May 15, 2007 at 11:11 PM | Permalink | Reply Steve M, look at the original confidence interval in the Nature 2006 paper, the region in gray. Why is the confidence interval so thick in those problem areas where the upper and lower bounds cross over? The answer, which you get in the form of a delay in graphics redrawing when you zoom in close, is that the authors have heavily thickened the lines of the gray graphic object. (If they were thin, the pinch would be obvious.) Now why would they do that? To hide the fact that the confidence region pinches as the bounds cross over? Anyways, this is probably why reviewers never caught the error. But it would be nice to know to what degree this deception was intentional. • bender Posted Jul 9, 2010 at 9:03 AM | Permalink | Reply “Hide the crossover”. 17. bender Posted May 15, 2007 at 11:39 PM | Permalink | Reply The new SI at least answers john lichtenstein’s #5 from the previous thread: Am I the only one perplexed with the confidence interval around 1400, 1650, and 1750? The new SI (which starts from 1505) shows the CI’s around 1650 and 1750 to be very wide. Still, as Annan argues, the CI’s are wrong. In order to get the uncertainty on the inferred value, the uncertainty in the observed value (the proxy value on which the temp reconstruction is based) has to propagate through the uncertainty in the regression coefficients. It makes no sense to assume all the uncertainty comes from the standard errors in the estimated regression coefficients. Some of it must come from sampling error from the estimation of the means. It is very important to know exactly what these folks are doing. 18. Posted May 16, 2007 at 1:46 AM | Permalink | Reply Some notes / questions: TLS uses first principal component, minimizes perpendicular distances from data points to the line ( a new line to my calibration-line-plot ) ? In Hegerl J. Climate paper Note that if the uncertainties in the paleo reconstruction are much larger than in instrumental data, an alternative is the use of inverse regression, neglecting error in instrumental data (Coehlo et al., 2004). I think this is an misunderstanding. IMO Coehlo explains CCE and ICE quite well, and Hegerl et al didn’t read carefully. ICE requires prior distribution for temperatures, and another viewpoint of ICE was just developed in http://www.climateaudit.org/?p=1515#comment-108922 and following comments. If I were a reviewer, I’d stop reading that manuscript right there, page 7. ( I don’t have the final version, was anything corrected? ). I’m not surprised that their CIs went wrong.. 19. bender Posted May 16, 2007 at 7:41 AM | Permalink | Reply Suppose: inferred Temperature = A * Proxy measure + B where A, B are regression parameters, each with standard errors, and Proxy P is subject to sampling error. Then the error in the inference T is the quadrature sum of the errors in A*P and B, and the error in A*P here is the straight sum of the relative errors in A and P. Is that not the correct way to compute the confidence level for a quantity inferred from an error-prone calibration? That’s what I learned in high school physics, anyways. 20. Steve McIntyre Posted May 16, 2007 at 7:48 AM | Permalink | Reply Gabi Hegerl has sent me the following email last week in response to my request for further particulars on her method: if you tell me which part you find hard to understand, I can send you an algorithm. The records are processed by a number of programs. Tom;s teaching is over for the semester so I think you’ll get more detail out of him soon. Just so that I meet the requirements of other readers, UC and bender, can you summarize any questions? BTW I got a copy of the Hegerl J CLim article as published posted up here Ther references ot an SI have been removed and there is no SI at the J Climate website. 21. Earle Williams Posted May 16, 2007 at 1:43 PM | Permalink | Reply Re #20 Steve M, The URL for the Hegerl paper is a bit off. It looks to be http://data.climateaudit.org/pdf/Hegerll07.jclim.pdf 22. Steve McIntyre Posted May 21, 2007 at 10:17 PM | Permalink | Reply James Annan writes: Zero-width confidence intervals are not necessarily wrong. They are rather unconventional, perhaps, but that doesn’t make them incorrect. 23. Willis Eschenbach Posted May 22, 2007 at 2:55 PM | Permalink | Reply Zero-width confidence intervals are not necessarily wrong. They are rather unconventional, perhaps, but that doesn’t make them incorrect. Man, that’s just plain and fancy footwork. I can’t offhand think of any physical measurement that has a confidence interval of zero. Examples, anyone? w. 24. bender Posted May 22, 2007 at 3:10 PM | Permalink | Reply Re #23 That remark baffled me, too. (I didn’t say anything because I’ve been commenting too much lately.) • ### Tip Jar (The Tip Jar is working again, via a temporary location) • ### NOTICE Click on the "Reply" link to respond to a comment. Frequent visitors will want the CA Assistant. Sort/hide comments; improved reply box, etc. • ### Blog Stats • 10,052,573 hits since 2010-Sep-12 • ### Recent Comments %d bloggers like this:

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 7, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9246657490730286, "perplexity_flag": "middle"}

http://mathhelpforum.com/calculus/75425-reduction-formula.html

# Thread: 1. ## reduction formula hey there, first time poster, long time struggler...so i need some help with a problem. it says to find the reduction formula for the integral (e^x)(sin x)^n in terms of the integral (e^x)(sin x)^n-2. i know its integration by parts but i have no idea what to do for my substitutions because (sin x)^n on its own is difficult enough so how do I incorporate the e^x? cheers! 2. Originally Posted by jackm7 hey there, first time poster, long time struggler...so i need some help with a problem. it says to find the reduction formula for the integral (e^x)(sin x)^n in terms of the integral (e^x)(sin x)^n-2. i know its integration by parts but i have no idea what to do for my substitutions because (sin x)^n on its own is difficult enough so how do I incorporate the e^x? cheers! Integration by parts twice. For the first use $u = \sin^n x,\; dv = e^x\,dx$ so $du = n \sin ^{n-1}x \cos x\, dx \;\;v = e^x$ and $e^x \sin^n x - n \int e^x \sin ^{n-1}x \cos x\,dx$ For the second by parts $u = \sin ^{n-1}x \cos x,\; dv = e^x\,dx$ and use some identites - see how that goes.

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9163363575935364, "perplexity_flag": "middle"}

http://math.stackexchange.com/questions/1706/does-set-mathbbr-include-zero/1740

# Does set $\mathbb{R}^+$ include zero? I've been trying to find answer to this question for some time but in every document I've found so far it's taken for granted that reader know what $\mathbf ℝ^+$ is. - 19 It depends on the choice of the person using the notation: sometimes it does, sometimes it doesn't. It is just a variant of the situation with $\mathbb N$, which half the world (the mistaken half!) considers to include zero. – Mariano Suárez-Alvarez♦ Aug 6 '10 at 9:05 8 It is just as Mariano says. In this case, the "correct" convention is that it should not include zero (after all, zero is not positive), but you certainly can't count on this: about half the time, the author means to include $0$. – Pete L. Clark Aug 6 '10 at 9:17 7 You will often find $\mathbf R^+$ for the positive reals, and $\mathbf R^+_0$ for the positive reals and the zero. – zar Aug 6 '10 at 10:20 2 Let @zar post it as an answer and we are done. :) – Pratik Deoghare Aug 6 '10 at 13:38 5 I tend to use $\mathbb{R}_{> 0}$ or $\mathbb{R}_{\geq 0}$ and avoid that notation altogether. – Andrea Ferretti Aug 6 '10 at 14:22 show 8 more comments ## 5 Answers It depends on the choice of the person using the notation: sometimes it does, sometimes it doesn't. It is just a variant of the situation with $\mathbb N$, which half the world (the mistaken half!) considers to include zero. - You will often find $\mathbf R^+$ for the positive reals, and $\mathbf R^+_0$ for the positive reals and the zero. - As a rule of thumb most mathematicians of the anglo saxon school consider that positive numbers (be it $\mathbb{N}$ or $\mathbb{R}^{+}$) do not include while the latin (French, Italian) and russian schools make a difference between positive and strictly positive and between negative and strictly negative. This means by the way that $0$ is the intersection of positive and negative numbers. One needs to know upfront the convention. - Mh, here in Italy when we say "positive" we mean "$>0$", not "$\ge 0$". – zar Aug 6 '10 at 17:10 3 Hmm. As above, for me $\mathbb{R}^+ = (0,\infty)$, but $\mathbb{N}$ includes zero. Hence I wouldn't call the latter the set of positive integers: for that I use $\mathbb{Z}^+$. Seems logical... – Pete L. Clark Aug 6 '10 at 17:35 As well in France the anglo saxon school is taking over, but if you read Italian mathematicians of the beginning of the 20 th century you will realise that like Bourbaki slightly later in France positive included zero – marwalix Aug 6 '10 at 19:14 2 Early editions of Bourbaki indeed defined zero to be both positive and negative, but by the 1930s even Bourbaki changed their mind. As a general rule, $\mathbb{N}$ excludes zero if and only if you are a number theorist. – JeffE Aug 23 '10 at 19:54 1 In my first semester of graduate school (in the U.S.), my algebra professor told us on the first day that $\mathbb{N}$ includes zero and $\mathbb{Z}_+$ does not. That afternoon, my analysis professor told us the exact opposite. How tedious! I prefer to say "positive" or "nonnegative" rather than use these symbols, but have also come to appreciate $\mathbb{Z}_{\geq 0}$ and $\mathbb{Z}_{>0}$ where brevity is desired. – Jonas Meyer Aug 27 '10 at 23:06 show 1 more comment I write, e.g., $\mathbb R_{>0}$, $\mathbb R_{\geq0}$, $\mathbb N_{>0}$. - I met (in IBDP programme, UK and Poland) the followin notation: $$\mathbb{R}^{+} = \{ x | x \in \mathbb{R} \land x > 0 \}$$ $$\mathbb{R}^{+} \cup {0} = \{ x | x \in \mathbb{R} \land x \geq 0 \}$$ With the explanation that $\mathbb{R}^{+}$ denotes set of positive reals and $0$ is neither positive nor negative. $\mathbb{N}$ is possibly a sligthly different case and it usually differ from branch of mathematics to branch of mathematics. I belive that is usually includes $0$ but I belive theory of numbers is easier without it. It can be easilly extended in such was to have $\mathbb{N}^+ = \mathbb{Z}^+$ denoting positive integers/naturals. Of course as noted before it is mainly a question of notation. -

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 32, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9524615406990051, "perplexity_flag": "middle"}

http://mathhelpforum.com/algebra/45224-how-factorise.html

# Thread: 1. ## how to factorise this Trying to jog my memory here, how would I go about factorising this, as I'm trying to find its limit as x approaches 3 from the left. \[ \frac{{x^2 - x + 12}}{{x + 3}} \] 2. fyi, the numerator will not factor, but why factor at all? there is no discontinuity at x = 3 ... just plug in 3 for x and calculate the limit straight up. 3. sorry should be as x approaches -3 (not 3). How would I go about finding the limit then in this case? 4. Originally Posted by Craka sorry should be as x approaches -3 (not 3). How would I go about finding the limit then in this case? It's already been pointed out that the numerator does not factorise. So you can't cancel a common factor of x + 3. The limit approaches -oo as x --> -3 from the left. If you don't like this answer, I suggest you go back to the original question and check whether you made a mistake in copying it. If you haven't made a mistake in the equation posted, then I'm afraid you're stuck with the answer I've given. 5. Sorry to clarify, the question ask to find the value for the for the following expression. $ \mathop {\lim }\limits_{x \to ( - 3)} \frac{{x^2 - x + 12}}{{x + 3}} $ The answer given is -7. 6. Originally Posted by Craka Sorry to clarify, the question ask to find the value for the for the following expression. $ \mathop {\lim }\limits_{x \to ( - 3)} \frac{{x^2 - x + 12}}{{x + 3}} $ The answer given is -7. Did you check the question like I asked? To get that answer the numerator must be $x^2 - x {\color{red}-} 12$ NOT $x^2 - x {\color{red}+} 12$ as you've posted several times now. $x^2 - x - 12 = (x - 4)(x + 3)$. 7. Yes I have checked the question several times. That is the question being asked. From what you are saying it appears there is a mistake in the question being asked on the paper. 8. Originally Posted by Craka Yes I have checked the question several times. That is the question being asked. From what you are saying it appears there is a mistake in the question being asked on the paper. Looks like it. 9. Apply the L'HOSPITAL RULE differentiate with respect rule u will get =lim x-->-3 (2x-1) =2(-3)-1 =-6-1 =-7 10. Originally Posted by Craka Trying to jog my memory here, how would I go about factorising this, as I'm trying to find its limit as x approaches 3 from the left. \[ \frac{{x^2 - x + 12}}{{x + 3}} \] Originally Posted by Craka sorry should be as x approaches -3 (not 3). How would I go about finding the limit then in this case? Originally Posted by tutor Apply the L'HOSPITAL RULE differentiate with respect rule u will get =lim x-->-3 (2x-1) =2(-3)-1 =-6-1 =-7 NO!! The limit does NOT have an indeterminant form 0/0. Using l'Hopital's Rule here is totally wrong. This is the problem with l'Hopital's Rule - it's a lazy approach that is easily abused. (And in a pre-algebra forum I doubt that l'Hopital's Rule has ever even been heard of).

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.967134416103363, "perplexity_flag": "middle"}

http://www.physicsforums.com/showthread.php?p=4065657

Physics Forums Page 2 of 4 < 1 2 3 4 > ## The problem with infinity. Quote by bahamagreen Thanks for mentioning those, I had never run across the Ant On A Rubber Rope before. Might Hilbert's Hotel have a flaw in the premise that relates to problems with infinity? If an infinite number of rooms are each occupied by a guest, where does the new guest come from? Some think that an infinite collection must necessarily contain all instances... It was Georg Cantor that demonstrated that not only was actual infinities perfectly logical but that it necessarily entailed orders of infinity, called aleph numbers $\aleph$ or cardinality (a powerset). Cardinality is basically the size of an infinite set. This was controversial in Cantor's day since the only infinity acceptable prior to that was unbounded sets, or potential infinities. You can look up his work, including his diagonal argument, for a more quantitative treatment. I'll just give a more intuitive description. If you have a finite interval, i.e., distance between two points, then logically you can divide it into an infinite set of infinitesimal points. To answer the above question one way, it cannot be said that the infinite set of points between 0 and 1 contain all possible numbers. There is not only more than one infinite set, there are an infinite set of infinite sets. There is no flaw in Hilbert's Hotel. There is also countably infinite sets and uncountable sets. There is also open sets and closed sets, which differ only in whether they contain the infinitesimal boundary points or not. There are also dense sets, which play a role in many quantum foundation arguments, including EPR and the scalability of quantum computing. The one thing you cannot do is make a priori generalized statements, like infinity must contain the entirety of the whole Universe to be infinite. Any finite subset of the Universe also contains an infinite set of infinitesimal points. Neither can you, for the same reason, make the claim that two infinite sets must be the same size. You must restrict your statements about infinity to those statements that can be mathematically demonstrated to be consistent, and avoid the intuitively implied and inconsistent properties Zeno's arguments depended on. Quote by my_wan To answer the above question one way, it cannot be said that the infinite set of points between 0 and 1 contain all possible numbers. There is not only more than one infinite set, there are an infinite set of infinite sets. There is no flaw in Hilbert's Hotel. I'm not thinking that the infinite set of points between 0 and 1 contains all possible numbers, only that it contains all possible numbers between 0 and 1. It seems to me by definition, the set of points between 0 and 1 must include every point between 0 and 1. Are you suggesting otherwise? I'm thinking that any arbitrary number I specify between 0 and 1 must already be included in the set of points between 0 and 1; so I don't see how any possible number between 0 and 1 is not already a member of the set of points between 0 and 1. The "new guest" coming to Hilbert's Hotel's is like a point between 0 and 1 that is not a member of the set of points between 0 and 1... I see this as a flaw in the premise. If the 0 to 1 range is problematic, we can do the same with the set of natural numbers... I'm thinking that the set of natural numbers must include any and all arbitrary natural numbers that I may specify... this seems clear by definition. If each occupied room is mapped to a natural number, an infinite number of rooms means all the natural numbers are mapped, as are their corresponding guests... the "new guest" would need to represent an unmapped natural number, but there are none, by definition. Maybe I'm missing something...? Quote by bahamagreen The "new guest" coming to Hilbert's Hotel's is like a point between 0 and 1 that is not a member of the set of points between 0 and 1... I see this as a flaw in the premise. How does it remotely resemble this? EDIT: Upon closer inspection, the paradox is simply mapping natural numbers to new natural numbers. The set ##\mathbb{Z}\cup\left[1,\infty\right)## has the same cardinality as the set ##\mathbb{Z}\cup\left[2,\infty\right)##. This should solve the paradox quite easily. Quote by Whovian Actually, in most cases, infinite means unbounded, so there would be no such edge. or finite without boundaries, like a torus. Quote by bahamagreen I'm not thinking that the infinite set of points between 0 and 1 contains all possible numbers, only that it contains all possible numbers between 0 and 1. It seems to me by definition, the set of points between 0 and 1 must include every point between 0 and 1. Are you suggesting otherwise? In the original post I addressed it was implied that maybe if there was an infinite number of hotel rooms, then these rooms being occupied implied infinite guest such that there could be no new guest. If there are an infinite set of point between the two points, [0,1], and each of these correspond to a hotel room occupied by a point, then the original suggestion to work around the hotel hotel paradox implies that this infinity of points, [0,1], contains all points that might occupy the infinity of hotel rooms. Hence I made the suggestion in order to provide proof by contradiction that the hotel paradox was not flawed. I'm thinking that any arbitrary number I specify between 0 and 1 must already be included in the set of points between 0 and 1; so I don't see how any possible number between 0 and 1 is not already a member of the set of points between 0 and 1. The "new guest" coming to Hilbert's Hotel's is like a point between 0 and 1 that is not a member of the set of points between 0 and 1... I see this as a flaw in the premise. If the points between 0 and 1 are an infinite set of occupied hotel rooms, and yet "new guest" are still available from members that are not member of the set of points between 0 and 1, why is this a special case? The original suggestion was that an infinite number of guest implied no more guest exist, but here you add a special case to say there are more guest available from sets other that [0,1]. If the 0 to 1 range is problematic, we can do the same with the set of natural numbers... Precisely. The infinity problem is just as big in the interval [0,1] as it is in the interval [0,∞]. I'm thinking that the set of natural numbers must include any and all arbitrary natural numbers that I may specify... this seems clear by definition. If each occupied room is mapped to a natural number, an infinite number of rooms means all the natural numbers are mapped, as are their corresponding guests... the "new guest" would need to represent an unmapped natural number, but there are none, by definition. Maybe I'm missing something...? Only problem is that I can pull new guest from the infinite set of real number which you didn't included here. Note that the numbers are merely name tags on the guest, and it make no difference which ones you label with which numbers. I can relabel an infinite number of guest labeled with even numbers with odd numbers, and visa versa, and the count remains the same. I can also relabel all natural numbers as real numbers simply by multiplying their name tags with an irrational number and assigning them that number. It changes nothing about the total number of guest. Cantor showed that all these infinities existed, but we should not lose sight of the fact that they are mathematical infinities. Mathematically, what does it mean to say that something exists? If a mathematician can write down a set of non-contradictory axioms, and set down rules for deducing mathematically true statements from them, then those statements can be said to ‘exist’. This existence requires only logical self-consistency. Physical existence is completely unnecessary. If there can be a profound difference between physical and mathematical “existence” then it seems reasonable to identify a similar difference between physical and mathematical “truth”. Cantor’s infinities were all mathematical infinities, as are the rooms and guests in Hilbert’s Hotel. They may bear no relation to any possible physical infinity, which would include an infinite universe. The actual, ancient, fear of infinity was not removed; it was just that Cantor provided the world with a “label” that could be attached to infinity, which reads: “this is a mathematical infinity – it doesn’t bite”. my_wan, Your responses support my thinking that the Hilbert Hotel premise is flawed. A "complete" mapping to the natural numbers would be to include all of them by taking the natural numbers in order... 1,2,3... any other mapping scheme like 2,4,6... is an obvious mechanism to skip some numbers, yet there would be objection to a scheme that skipped points between 0 and 1. Claiming that the new guest could be from the set of real numbers when the set of guests is represented by the natural numbers misses the whole point. If the points between 0 and 1 represent the rooms, it is the guests that are mapped to the natural numbers. The paradox is based on the assumption that all guests, including any possible new guests, are all of the same class of thing - natural numbers, and violating that by suggesting a new guest could be a real number is like solving the question of the origin of the new guest by finding a chair and checking that chair into the hotel as a new guest... no, the new guest has to be a person like all the other guests. Quote by bahamagreen Claiming that the new guest could be from the set of real numbers when the set of guests is represented by the natural numbers misses the whole point. It doesn't miss the point any more than saying that the set of all hotel customers must consist of all possible hotel customers, and that is the only way you can claim there is nobody remaining to request a room in the hotel. If the real numbers represent the set of all present hotel customers, and the natural numbers represent the set of all people that might request a room, then there are an infinity of people that may request a room even after an infinite number of people have already filled the hotel. To assume otherwise is effectively an attempt to impose a boundary condition on an unbounded variable. Quote by Endervhar Cantor showed that all these infinities existed, but we should not lose sight of the fact that they are mathematical infinities. Very true, and so far the justification for "actual infinities" is fairly slim. However, attempting to avoid them has a number of problems. Avoiding actual infinities is just as problematic and paradoxical as accepting them. Modern mathematics didn't select the axioms simply to avoid contradictions with infinities, they where selected to avoid mathematical contradiction as a result of attempting to avoid them. In the physical sciences the scalability of quantum computers actually depends on these mathematical properties associated with mathematical infinities. This comes from the fact that a Hilbert space must be "complete", i.e., is a complete metric space. The very thing that allows the calculus of limits, and from which we derive our mathematical justifications for infinities. Quote by StJohnRiver There are physicists who insist that the universe is finite and has a distinct geometry. So what'd be the problem if the universe were infinite? Quote by my_wan It doesn't miss the point any more than saying that the set of all hotel customers must consist of all possible hotel customers, and that is the only way you can claim there is nobody remaining to request a room in the hotel. If the real numbers represent the set of all present hotel customers, and the natural numbers represent the set of all people that might request a room, then there are an infinity of people that may request a room even after an infinite number of people have already filled the hotel. To assume otherwise is effectively an attempt to impose a boundary condition on an unbounded variable. The set of natural numbers must include all possible natural numbers. The set of points between 0 and 1 must include all possible points between o and 1. The set of all hotel customers must comprise all possible hotel customers. You are defining: The reals is the set of present hotel customers. The naturals is the set of people that might request a room (not presently hotel customers). I'm suggesting that is a flaw because you are defining some persons as both a non-customer and a customer - because the naturals and reals share some members in common (all naturals are members of the reals, some reals are members of the naturals). But maybe I'm still missing something? Quote by bahamagreen The set of all hotel customers must comprise all possible hotel customers. This can't be justified. The set of all actual hotel customers anywhere in the world does not comprise all possible hotel customers. If that was necessarily true then in order for hotels to have any customers they must have every person on earth as a customer. Conversely, by this logic, since I am not a customer, either hotels have no customers or I am not a potential customer. Neither of which is true. Quote by MY Wan This can't be justified. The set of all actual hotel customers anywhere in the world does not comprise all possible hotel customers. As ObsessiveMathsFreak pointed out, this "depends on what you mean by infinite." In terms of mathematical infinities your assertion is undoubtedly true, but it is a mathematical "truth" and has no significance in reality. You cannot have an infinite number of rooms, an infinite number of people, or an infinite number of anything. Earlier, someone suggested that infinite and boundless might be synonymous; this is not so. Anything that is infinite is boundless, but not everything that is boundless is infinite. Quote by nikkoo Actually, infinite means unbounded Sorry, Nikkoo, I missed your quote when I was looking for it. You will probably want to take issue with my previous post. :) Quote by Endervhar As ObsessiveMathsFreak pointed out, this "depends on what you mean by infinite." In terms of mathematical infinities your assertion is undoubtedly true, but it is a mathematical "truth" and has no significance in reality. I have already gone over how it is relevant to the physical sciences, by way of Hilbert space. You can call Hilbert space a mathematical fiction, or slide rule of sorts, used to calculate. But a rejection of mathematically defined infinities has very real empirical consequences. The inability to scale quantum computers being a major one. Others involves issues surrounding Bell's theorem, and other no-go theorems. You cannot escape the issues mathematicians have worked around with a "mathematical fiction" clause. Finite mathematics is inconsistent without mathematical infinities. How do you propose the reinstate consistency if you reimpose a finiteness condition on the physical world? You can't simply close your eyes and pretend it has no physical consequences, whether those consequences ultimately justify actual infinities or not. Quote by Pianoasis Infinity is a number that cannot be divided, cannot be measured, and cannot be contained. This infinite universe obviously does not exist due to the fact that all pieces of space are made of this ultimately small unit. Every quantity can be described by this unit, thus making the concept of infinity null. That's an absurd assertion. The number line is infinite, but we can still use integers to measure and divide segments of the line. Quote by my_wan This can't be justified. The set of all actual hotel customers anywhere in the world does not comprise all possible hotel customers. If that was necessarily true then in order for hotels to have any customers they must have every person on earth as a customer. Conversely, by this logic, since I am not a customer, either hotels have no customers or I am not a potential customer. Neither of which is true. How about addressing my previous post #28 first? Page 2 of 4 < 1 2 3 4 > Thread Tools | | | | |-------------------------------------------------|----------------------------|---------| | Similar Threads for: The problem with infinity. | | | | Thread | Forum | Replies | | | Calculus & Beyond Homework | 6 | | | Calculus & Beyond Homework | 8 | | | Cosmology | 28 | | | General Math | 13 |

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9594243168830872, "perplexity_flag": "head"}

http://math.stackexchange.com/questions/84231/how-do-i-set-up-the-following-problem-to-arrive-at-the-answer?answertab=votes

# How do I set up the following problem to arrive at the answer? A warehouse has 10 unlabelled rows of pallets. Each row of pallets contains thousands of cell phones destined for different countries. Each 100 gram cell phone is exactly the same except for those in the row destined for Japan, which have a “special” 2 gram chip encased within each phone to make sure it works on the Japanese networks. How can the warehouse manager make sure the right phones go to Japan in the quickest time possible? All the warehouse manager has at his disposal is a digital balance. - If this is homework, please tag it as 'homework'. Regardless, please share with us what you have attempted so far. What is the maximum capacity of the digital scale? We don't want to break the manager's only scale, do we? – jthetzel Nov 21 '11 at 1:29 1 How does statistics enter into this problem? It seems to be similar to one of several math puzzles involving minimum number of steps to identify the odd piece. If so, this question may be off-topic here and perhaps on topic on math.SE. – varty Nov 21 '11 at 4:57 1 Do the phones intended for Japan weigh $102$ grams instead of $100$ grams? I don't get this from the problem statement, e.g. the $2$ gram chip in phones intended for the Japanese market might be replacing a $1.5$ gram chip, which would make @PeterFlom's solution not work.... Please edit your problem statement to state the weight of the special phones explicitly. – Dilip Sarwate Nov 21 '11 at 15:55 ## 1 Answer This is an old old puzzle that can be found in many books. Solution: Weigh 1 cell phone from row 1, 2 from row 2 and so on, all at the same time. Divide the weight by 100. Divide the remainder by 2. That's the row number with the Japanese phones. - You probably want to subtract $5500$, not divide by $100$. A total of $55$ phones are being weighed, and so the total weight is $5500 + 2n$ where the $n$ phones from pallet $n$ are contributing an extra $2n$ grams of weight. – Dilip Sarwate Nov 21 '11 at 19:53

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 11, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9435410499572754, "perplexity_flag": "middle"}

http://stats.stackexchange.com/questions/22940/bayesian-prior-corresponding-to-penalized-regression-coefficients

# Bayesian prior corresponding to penalized regression coefficients I'm working on a Bayesian Regression problem where I would like to estimate the beta coefficients subject to this constraint (penalty): $\sum|\beta_i|<C$ or similarly $\sum \beta_i^2<C$ Which is basically a Lasso or L2 Penalty. Now, if I understand correctly, we constrain the coefficients through the prior in Bayesian analysis. Therefore my question is what would an appropriate prior be for the Betas? I should note, that for my case, the betas are restricted to be positive, they cannot be negative. - ## 4 Answers L2 penalty penalizes the sum of squared betas but not via a constraint such as $< C$. The L1 penalty is the lasso. For the Bayesian lasso see the 2008 JASA paper by Trevor Park and George Cassella. - Thanks, I wasn't sure what modifications I need however if I want to restrict the prior so only positive values are possible. – Glen Feb 16 '12 at 4:46 @Frank: Do you mind explaining your first comment? For each value of the regularization parameter $\lambda$, there is, indeed a constant $C$ corresponding to the constrained optimization problem of the OP. And vice versa. (Just think about the Lagrangian formulation.) – cardinal Feb 16 '12 at 15:13 My understanding of $L_2$ penalty is that large values of sum of squared $\beta$s are penalized against but there is no cutoff on how large this sum can be. – Frank Harrell Feb 17 '12 at 3:22 @FrankHarrell - this is true, but characterising the maximum $\beta$ is 1-to-1 between $C$ and $\lambda$. For any $C$ which leads to a maximum $\hat{\beta}_C$ there is a corresponding $\lambda(C)$ which leads to the same maximum $\hat{\beta}_{\lambda(C)}=\hat{\beta}_{C}$ – probabilityislogic May 6 '12 at 0:12 @prob: The correspondence is not quite bijective, unless restricted to subintervals of the positive half-line. See my previous comment to this effect above. – cardinal May 6 '12 at 0:16 For the lasso penalty this corresponds to a double exponential prior - so long as you are taking the posterior mode as your estimate. If you constraint the betas to be positive then you have an exponential prior. The parameter of the exponential distribution $\lambda$ has a correspondence with your $C$ in that you can choose a value of each such that the same naximum is achieved. The bayesian prior is the langrangian form (on log scale) of the constraint. For the ridge penalty constraning the parameter to be positive is just a truncated normal distribution. - The $L_2$ constraint on the coefficients is Tikhonov regularization. It turns out that for the case where the prior is multivariate normal and the model is linear, the posterior is also multivariate normal. It turns out that the mean of the posterior distribution occurs at a point in parameter space that can also be obtained by Tikhonov regularization and the relationship between the Tikhonov regularization parameter and the equivalent multivariate normal prior is pretty simple. This is textbook material that can be found in Tarantola's textbook among other places. When the model being fit is nonlinear, or there are additional constraints on the parameters (such as your nonnegativity constraint), or the prior isn't MVN, it all becomes much more complicated and this simple equivalence breaks down. - If you want your $\beta$s to be non-negative and sum to a given value then it seems a scaled Dirichlet prior would make sense. -

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 18, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9094456434249878, "perplexity_flag": "head"}

http://mathoverflow.net/questions/61829?sort=newest

## Best introduction to probability spaces, convergence, spectral analysis ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I'm not sure if this stuff all falls under what most would just term "probability", but I'm researching applied macroeconomics and need to get a handle on the following concepts ASAP: • probability spaces and sigma algebras • Borel sets • convergence • stationarity/ergodicity • martingales • laws of large numbers • spectral analysis The text I'm using right now, Fabio Canova's Methods for Applied Macroeconomic Research, touches on all these things briefly in the space of about twenty-five pages, but it's pretty impenetrable. Does anyone know anything off-hand that presents these topics in a more practical, easier-to-digest way? This is a little bit better, but does anyone have any other suggestions? - As a mathematician, I really like the book by Durrett. But if you are not familiar with sigma algebras... maybe you should begin with a book on integration (for example the book by Rudin "Real and complex analysis"). Note that "ASAP" will probably require some time (a couple of month ?). Note also that you may get more answers on a forum devoted to teaching. – camomille Apr 15 2011 at 15:38 Jacod and Protter's book is a reasonably friendly introduction, and doesn't require any background beyond multivariable calculus. As camomille said, you're looking at a couple of months of work to take all this stuff on board. – Simon Lyons Apr 15 2011 at 16:00 Try "Measure, integral and probability", too. It doesn't cover all the topics you mentioned, but you should be able to work through it quite quickly, and it will establish a good base to build on. – Simon Lyons Apr 16 2011 at 14:51 ## 11 Answers There is no royal road to probability. The closest is W. Feller's book, which has many (but not all) of the topics you mention, but I strongly advise reading (at least parts of) it first. Otherwise, you will go through life hopelessly confused. - After looking at most of the other books listed it seems like Feller's book is the easiest for the non-mathematician to pick up. Thanks for the recommendation. – jefflovejapan Apr 16 2011 at 14:09 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. I just started up a blog and have put a link to my own alternative introduction to advanced probability theory. It could be useful as an introduction to Probability With Martingales, and might help you see the route ahead, as it were. http://fermatslastspreadsheet.wordpress.com/2011/11/30/introduction-to-probability-theory/ - Any thoughts on how Cinlar compares to Klenke, "Probability Theory A Comprehensive Course"? They both appear to be excellent recent alternatives to the likes of Billingsley and Chung. Also what would be good or required prep for Cinlar? Capinski and Kopp (seems to have a lot of overlap with Cinlar and Klenke)? Royden and Fitzpatrick (certain chapters)? - The book "Probability and Stochastics" by Professor Erhan Cinlar is among the best introductory (actually it touches deeper things too) book on probability theory, assuming a little background on integration in general measure spaces. It starts with basic things (probability spaces, convergence, conditioning), and ends with some deeper waters like Levy-Ito for Levy processes, Hunt processes. Professor Cinlar is an expert in the theory of Markov processes (e.g. he co-authored papers that characterize Markov semimartingales), among many other things. The book uses very precise language to describe things, in a very streamlined way. The author guides the reader through more difficult materials very easily, without making the reader feel the difficulty. In other words, pedagogically (in addition to the broad spectrum of things and the depth of the treatment) this book is among the best IMHO. I would recommend this book to anyone with an interest in probability theory and stochastic processes. In terms of coverage of the OP's list, the book covers at least the following: probability spaces and sigma algebras, Borel sets, convergence, martingales, laws of large numbers - Many of the really good introduction-type books have already been mentioned. As a current grad student I encounter many of them on an almost daily basis and would suggest the following: • Billingsley - "Probability and measure", although I would skip the first part about the dyadic intervals. • Durrett - "Probability: Theory and examples". I used the 3rd version when I was taught from this book and then it did not have that much measure theory in the, sense that it was confined to the appendix. As I understand it this is not the case for the 4th edition and I really love the way Durrett presents the material so this is a good starting point. • Shiryayev - "Probability". Great book from one of the current masters. It starts with an intuitive discussion about probability theory and then moves on to develop the mathematical theory needed (sample spaces and so on) in order to "do real probability". I haven't read the entire thing so that's why I can't put this as my #1 choice, I seems as if it has the potential though. • Williams - "Probability with martingales". Very nicely written account of measure theoretic probability. It is quite concise though and depending on your mathematical maturity it could perhaps be a bit difficult to follow completely. One possibility is to keep the author's "Weighing the odds" at your side to get the elementary theory as a complement to the more advanced book. • Feller - There is a reason why it's considered a classic. Another, perhaps somewhat odd and unconventional, choice might also be Tomas Björk - "Arbitrage theory in continuous time". Now it's not an introduction to probability but depending on what type of economics you are interested this could be a nice reference book. It has an appendix devoted to measure theory and probability and Prof. Björk being one of the best teachers I've had in terms of providing you with an intuitive feeling for the subject at hand, I'm sure that such sections would work very well for someone who is just trying to get a basic understanding of the subject. - OK ... "probability spaces" or "probability theory"?? Most of the other responses tell you texts for probability theory, where probability spaces are mentioned only in passing if at all. This would include the last 4 of your topics. On the other hand, "probability spaces" would be a branch of "measure and integration" theory. - That's an important distinction, but it certainly looks from the list of topics like the OP is interested in learning about probability theory, and most books on probability theory do devote a chapter to probability spaces before those recede to the background. – Mark Meckes Apr 17 2011 at 23:54 Given your starting point, specific interests, and desire to get to the bottom line fast, I recommend starting with Probability with Martingales by David Williams, which is yet another first-year graduate text, but more concise than the others. - I agree with tipanverella that you're probably looking for textbooks designed for first-year graduate students in this field. I think you'd be hard-pressed to find one book that covers all the topics listed, but there's nothing wrong with reading parts of several books to get you up to speed. For everything except ergodic theory and spectral analysis, I highly recommend Probability Theory and Elements of Measure Theory by Heinz Bauer. This book has a fantastic first chapter and really develops $\sigma$-algebras from scratch (most books I've seen just throw the definitions at you). Throughout the development of measure theory, connections are made to probability and part II of the book is all probability theory. There's a whole chapter on the Law of Large Numbers and another on Martingales. By the way, I would NOT read Rudin for this. Rudin's great if you already know the material and want a reference with the shortest, most elegant proofs possible (with many details skipped), but for actually learning the material and for connecting it to probability theory his book is not the greatest. Williams' Probability with Martingales really emphasizes exercises, so if you want examples with all the details worked out they will not be as prevalent. But the examples and exercises he gives are really great if you work through them yourself. I don't know much about the other books mentioned. As for Ergodic Theory, I really like An Outline to Ergodic Theory by Steven Kalikow and Randall McCutcheon. If you click that link you can check out the table of contents and the introduction, which includes great examples to get you started as well as statements of all the big theorems you will need. I know this wasn't part of your question, but you may also want to add entropy to your list of things to learn. This book would help with that as well. Another thing you may find yourself needing is the Borel-Cantelli Lemma (found in Bauer's book). Good Luck! - Those are probability topics, so I would go with what graduate students in probability are trained with. These past few years it seems that the standard is Rick Durrett's "Probability: Theory and Examples" (4th edition, Cambridge U. Press, 2010). MOst of what you want are in the first 5 chapters. Also a version of the book is provided in PDF format by the author, http://www.math.duke.edu/~rtd/PTE/PTE4_Jan2010.pdf. You could also get Shiryaev's "Probability" from Springer-Verlag or Stroock's "Probability Theory, an analytic view. Hope that helps. Cheers, Tipan - These don't have anything on spectral analysis, but cover the rest pretty well: Infinite Dimensional Analysis: A Hitchhiker's Guide by Aliprantis and Border, and Stochastic Limit Theory by Davidson. The first book was written for economic theorists, the second one for econometricians. So they specialize quite nicely to what you seem to be working on. However, they are both rather advanced. A great introduction to measure theoretic probability is Probability with Martingales by Williams. It's quite chatty and fun, but does still require some mathematical sophistication. A book that I think is a bit dry but that proceeds in small and easy steps with all the details included is A Probability Path by Resnick. - 1 +1 for mentioning Williams' Probability with martingales, aka the blue book. – Didier Piau Apr 16 2011 at 5:56 I looked at Probability with Martingales, and I really enjoy the style, but it's a little out of my reach. I'm going to check out Hitchhiker's Guide on Monday. Davidson and Resnick aren't available at my library, but I put requests in for them. Thanks for the advice. – jefflovejapan Apr 16 2011 at 14:11 Hitchhiker's guide is considerably more technical than Williams. It's very complete, but there's a danger of information overload there. – Simon Lyons Apr 16 2011 at 14:49 The first two books are both great pure math texts, and its odd that they both come out of economics. – arsmath Jun 30 2011 at 12:39 I think Probability and Measure by Patrick Billingsley is what you're looking for. It contains all of the topics you're asking about (except maybe spectral theory), and I feel it does a good job introducing the world of probability to the reader without assuming you already understand it, which Durrett's book tends to do. - It looks good, but the starting point's too advanced for me. Thanks for the advice. – jefflovejapan Apr 16 2011 at 14:19

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9563212990760803, "perplexity_flag": "middle"}

http://mathhelpforum.com/geometry/1637-geometry-help-needed-please.html

# Thread: 1. ## geometry help needed, please Why is line segment ab = line segment ab? what is the name of the reason/postulate/whatever it is called? Also, if you know that you have 2 triangles and you are trying to prove they are congruent and you know that 2 sides are equal and 1 angle is equal, which reason/law/postulate do you cite? Thank you! 2. Originally Posted by e-mom Why is line segment ab = line segment ab? what is the name of the reason/postulate/whatever it is called? Also, if you know that you have 2 triangles and you are trying to prove they are congruent and you know that 2 sides are equal and 1 angle is equal, which reason/law/postulate do you cite? Thank you! $AB=AB$ Called the Reflexsive Postulate. When two triangles have the same sides and and included angle they are congruent. This is called S.A.S. which means SIDE ANGLE SIDE. With ANGLE in between SIDE and SIDE, which was required. Thank you! 4. ## geometry help needed, again If you have 2 line segments, which form the 2 sides of a traingle, and have been given the fact that corresponding segments of the sides are equal, what rule could you cite to prove that the entire length of the 2 sides are the same. I am eventually trying to prove that the triangle is isosceles. A diagram is attached. Thanks! Attached Files • math help document.doc (19.0 KB, 59 views) 5. The way how I understand the problem there is no way to prove what you are saying because it is wrong. I understand it as that you know the small one is isoseles and and to show the big one is also isoseles but there is not enough infromation. Perhaps you wanted to say that line down the middle is a bisector? 6. ## thanks Thank you for the info that my reasoning wasn't going to prove it that way. I'll look for another way. Thank you! #### Search Tags View Tag Cloud Copyright © 2005-2013 Math Help Forum. All rights reserved.

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9681377410888672, "perplexity_flag": "middle"}

http://mathoverflow.net/questions/52075/solution-of-plateau-problem-for-a-simple-smooth-closed-curve-on-a-riemannian-man/52168

## Solution of Plateau Problem for a simple, smooth closed curve on a Riemannian Manifold (Kahler) gives a surface that can be parametrized by a closed disk? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Hi, Perhaps it's a stupid question, in that case i'll delete it. Let M be a compact orientable smooth (Kahler if changes things) manifold of dimension $dim_{\mathbb{R}}(M)=2n$ with $n\geq1$, let $\gamma$ be a simple smooth closed curve that lies in a (holomorphic) coordinate chart and that can be taken as small as necessary (one can choose $\gamma$ such $diam(\gamma)<\epsilon$ with $\epsilon>0$). If i want to solve Plateau problem for a $\gamma$ so small such that the solution is contained in the coordinate chart, do i get something that can be parametrized by closed disc? Thank you in advance. Edit: I'll try to clarify my question, this is what i wanted to know. Let $U$ be a sufficiently small geodesically convex set of a manifold $M$ and $\gamma$ a smooth simple closed curve lying in $U$ (no other assumptions on $\gamma$). 1) Can $\gamma$ be the boundary of an embedded closed disk? 2) If $\gamma$ can be the boundary of a closed disk, then can it be the boundary of a minimal (as a surface, not only among the disks that it bounds) embedded closed disk? I anticipate that i couldn't see works of Douglas so i don't know if the answer to my question is there. My suspect was that the answer could be yes for dimension 2 (i think about jordan curve theorem), Professor Thusrston example of the knotted curve suggests me that in dimension 3 i need additional assumptions on the curve not only on the linking number. But what happens for dimension $n\geq4$? - I think this is a good question. I haven't worked out the details myself, but I am pretty sure that you can prove this on any Riemannian manifold using standard tools in differential geometry. I'm hoping that a real expert will provide a reference, more explicit details, or a counterexample. The idea is to use the inverse function theorem applied to the appropriate functional to demonstrate the existence of a minimal surface parameterized by the closed disk. Then use a comparison argument to show existence and minimality. – Deane Yang Jan 14 2011 at 16:53 Just to echo what has been said, you should be more precise in your question. Namely, given your $\gamma$ there is a solution that is an (immersed) minimal disk spanning $\gamma$ namely a Douglas-Rado solution (see Bill's great answer for some details). This is because the $\gamma$ you describe is spanned by some disk and hence by direct methods in the calculus of variations is spanned by a minimal disk (of course easier said then proved). However, there may be many other minimal surfaces (or weaker concepts) spanning the curve so it is not correct to speak of THE solution. – Rbega Jan 15 2011 at 18:01 I edited the question, i hope it is more clear now – Italo Jan 16 2011 at 0:39 Bill Thurston's answer (which is quite good) shows that my comment is not right. At best the approach outlined by me might work if the initial curve is "sufficiently flat", whatever that means. – Deane Yang Jan 16 2011 at 3:24 ## 1 Answer I've been waiting for someone with more expertise than me to answer, but since they haven't (so far) I'll say something. There are different versions of Plateau-like problems; I'm not sure if there's a specific single one that's generally accepted as "the Plataeu problem". One can ask for a mapped-in disk with minimal area having the given boundary, a mapped-in surface of minimal area, a current of minimal area, an integral current of minimal area, an embedded "minimal surface" meaning that it's just a critical point for area among embedded (or mapped in if you prefer) surfaces, a minimal disk ... A lot is known about these different questions, and the answers aren't the same. First: even for a curve in Euclidean space, there might not be an embedded minimal disk. The easiest examples are for a knotted curve in R^3. However, there are also unknotted curves in R^3 that do not bound a disk in their convex hull, so they do not bound any embedded minimal disk. The minimum area of a disk bounding an unknotted curve grows exponentially in an appropriate measure of the complexity of the curve; the genus of a minimum area surface also grows exponentially. (Fred Almgren and I once wrote a paper about this). The same examples can be transported to any higher dimension, e.g. taking the product with another manifold. Second: Jesse Douglas showed how to find mapped-in minimal disks in great generality. This will work locally within any manifold of dimension. The basic technique is that for any parametrization of the curve by the boundary of a disk, first find an energy-minimizing map of the disk that extends this parametrization --- a harmonic map. Now consider the energy as a function of the parametrization (which is a kind of Teichmuller space). The critical points of the harmonic energy with respect to the Teichmuller space are minimal surfaces: the basic insight is that critical points of the harmonic maps within the Teichmuller space are when the harmonic map is conformal. When it is not conformal, you can change the conformal structure on the disk (which is equivalent to giving a reparametrization, by the uniformization theorem) and reduce energy. Third: minimal surfaces of any type (inclding currents) whose boundary is inside a convex set are always contained in the convex set. Here convex means that geodesic arcs between nearby points on the boundary are contained within the set. Even weaker conditions are sufficient, but this is good enough for your purposes --- a metric ball of small radius in any Riemannian manifold is convex in this sense. Fourth: there always exist minimizing objects of some sort, but they might not be things that you are happy with: one thing that happens is that $k$ times a curve might bound a surface of area less than $k$ times that for the original curve (in dimensions > 3). A limit of $1/k$ times the minimum area for $k$ times the curve might be a diffuse current, spread out in an entire region. I think this can happen even locally in a Kahler manifold, but I'm not sure. Even without using fractional weights, the minimizing object would in general be an integral current that is like a surface but with singularities. I don't know the classification for the 2-d case, but i'm sure the experts do. On the other hand, if you take a nearly round circle in a small coordinate chart, there should be an embedded minimal disk --- basically because nearly flat minimal surfaces are stable, so changing the metric a little bit gives you a new minimal surface by implicit function theorem type arguments (in the space of surfaces, which in this case can be described as graphs of functions). In R^3, there's a theorem that any curve whose total curvature is less than 4$\pi$ bounds an embedded minimal disk; I suspect there are known estimates likke this in Riemannian manifolds of higher dimension, using curvature assumptions about some convex set contining the curve, but I don't actually know. Is there an expert who can correct or extend what I've said? - Thank you very much for the detailed answer! I'll wait a little more before accepting it because i want to see if you or someone has anything else to say. – Italo Jan 16 2011 at 0:45 Not an expert, but I'll add one bit to (part 4 of) this excellent answer: Given a boundary, you can always find an area-minimizing rectifiable current with the given boundary. Almgren's big regularity paper says that the interior of this minimizer is an embedded smooth manifold away from a Hausdorff codimension two singular set. In the case where the boundary is 1-dimensional, Sheldon Chang further showed that the singular points are isolated classical branch points. – Dan Lee Jan 24 2011 at 22:08

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 23, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9383137226104736, "perplexity_flag": "head"}

http://mathhelpforum.com/calculus/122296-help-particle-motion-derivatives-etc-print.html

# Help with particle motion, derivatives, etc. Printable View • January 3rd 2010, 12:04 PM macroecon Help with particle motion, derivatives, etc. I have never seen any examples of these problems, and they showed up on a review sheet for the final exam in May. If you can help me through these or explain to me what I should do for each step it would help me a lot. Thanks. 1. A particle moves along a line so that at any time (t) its position is given by x(t)=πt+cos(πt). btw, π is pi. a) Find the velocity at time t. b) Find the acceleration at time t. c) What are all the values of t, 0≤t≤3, for which the particle is at rest? d) What is the maximum velocity over the interval 0≤t≤3? 2. Let f be the function defined by f(x)=(x^2 +1)e^-x for all x such that -4≤x≤4. a) For what value of x does f reach its absolute maximum? Justify. b) Find the x-coordinates of all points of inflection of f. Justify. 3. Let p and q be real numbers and let f be the function defined by: f(x)={1+3p(x-2)+(x-2)^2 for x≤2 f(x)={qx+p for x>2 a) Find the value of q, in terms of p, for which f is continuous at x=2. b) Find the values of p and q for which f is differentiable at x=2. c) If p and q have the values determined in part b, is f" a continuous function? Justify. • January 3rd 2010, 01:09 PM ANDS! The position, velocity and acceleration functions are all related in that: Velocity is the derivative of the position function, and acceleration is the derivative of the velocity function, and the second derivative of the position function. Therefore if the position function is defined as $x(t)=\pi t+cos(\pi t)$ (taking $\pi$ as a constant): A. $x'(t)=v(t)=\pi-\pi sin(\pi t)$ B. $x''(t)=v'(t)=a(t)=-\pi^2 cos(\pi t)$ C. For this problem, understand that for an object to be at rest, its velocity must be equal to 0. Therefore solve the equation given in A for 0: $v(t)=\pi-\pi sin(\pi t) \Longrightarrow 0=\pi-\pi sin(\pi t) \Longrightarrow \pi sin(\pi t)=\pi \Longrightarrow sin (\pi t)=1$ For what value(s) of t, between 0 and 3 will make the above equation 1? D. You have your velocity function - now you are being asked for what values of t will you have maximum velocity. In others words, optimize the velocity function. • January 3rd 2010, 01:21 PM ANDS! For your second problem, you are doing much of the same, so I will not go into detail in that, simply note that to find the maximum (or minimum of a fucntion) you need to find the derivatve and set it to zero. If you function has multiply zeros, then test (in the original function) for the critical point that has the largest value. For the inflection point, the method is the same, except you are take the second derivative, and finding the critical points. Once your critical points have been established, test points around them to see if your values of f(x) are changing sign. If so, then the critical point is an inflection point. Review your notes on this if you are having trouble but it is a simple exercise. For your third problem, remember that a function is continuous at a point if and only if the limit as x approaches a from either direction (the left or the right) is the same, and f(a) actually exists. Assuming it does exist: First equation: $f(2)=1+3p(2-)+(2-2)^2 \Longrightarrow f(2)=1$ Second equation: $f(2)=2q+p$ Equating the two equations: $1=2q+p \Longrightarrow \frac{1-p}{2}=q$ I will leave parts B and C to you. If you want a check on your answer post work here. • January 3rd 2010, 07:00 PM macroecon We aren't supposed to use a calculator on any of these. So for par c of number 1, what work am I supposed to show to justify my answers of t=1/2 and 2 1/2? Thanks. • January 3rd 2010, 07:10 PM pickslides $v(t) = 0$ $0= \pi-\pi\sin(\pi t)$ $\pi\sin(\pi t)= \pi$ $\sin(\pi t)= 1$ Form the grpah of the sine function $\pi t= \frac{\pi}{2}$ $t= \frac{1}{2}$ As the period of the fucntion is 2 the next solution will be at $t= \frac{5}{2} = 2.5$ • January 3rd 2010, 08:08 PM macroecon And on the second one, I get stuck at f'(x)=(-e^-x)((x-1)^2) and I'm not sure exactly how to progress from there. Any help? • January 3rd 2010, 08:46 PM ANDS! The great thing about optimization problems, is that when it comes time to set our differentiated function equal to zero, ANYTHING in the denominator dies a swift death at the hands of Mr. Multiplication-By-Zero: $f'(x)=\frac{(x-1)^2}{-e^x} \Longrightarrow 0=\frac{(x-1)^2}{-e^x} \Longrightarrow 0=(x-1)^2$ From here it should be simple to see what your solution is. Perhaps you weren't seeing the next step because you were writing $-e^-x$ instead of $\frac{1}{-e^x}$ All times are GMT -8. The time now is 01:20 PM.

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 18, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9258739352226257, "perplexity_flag": "middle"}

http://stats.stackexchange.com/questions/1430/example-of-a-2nd-order-stationary-but-not-strictly-stationary-process

# Example of a 2nd order stationary, but not strictly stationary process Does anybody have a nice example of a stochastic process that is 2nd-order stationary, but is not strictly stationary? - I would like to add a stochastic processes tag, but don't have the reputation, perhaps somebody could edit this for me? – Robby McKilliam Aug 9 '10 at 6:51 retag done ! – robin girard Aug 9 '10 at 6:59 ## 1 Answer Take any process $(X_t)_t$ with independent components that has a constant first and second moment and put a varying third moment. It is second order stationnary because $E[ X_t X_{t+h} ]=0$ and it is not strictly stationnary because $P( X_t \geq x_t, X_{t+1} \geq x_{t+1})$ depends upon $t$ - Perhaps I am a little confused, or we are using different definitions? Am I correct in thinking that a process is 2nd order stationary if the joint marginal cdfs $F_{X(t),X(t+τ)}$ are equal for all $\tau$? Similarly for a process to be 1st-order stationary the marginal cdfs $F_{X(t)}$ need to be the same for every $t$. So all the moments of the $X(t)$ must be equal. 2nd-order stationary implies 1st-order stationary, correct? – Robby McKilliam Aug 9 '10 at 7:44 Extending this, a process is $N$th order stationary if for every $t_1, t_2, \dots, t_N$ the marginal cdfs $F_{X(t_1 + \tau), X(t_2 + \tau)\dots,X(t_N+\tau)}$ are the same for all $\tau$. Strictly stationary is Nth order stationary for all N. – Robby McKilliam Aug 9 '10 at 7:45 – robin girard Aug 9 '10 at 8:05 – Robby McKilliam Aug 9 '10 at 8:24 @robby OK... I didn't know this "second order stationnarity" I think you should not say second order stationnarity in the question and give the definintion instead. For more clarity you should ask another question. do you have a paper that refers to this second order stationnarity ? – robin girard Aug 9 '10 at 8:32 show 1 more comment

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9149402379989624, "perplexity_flag": "middle"}

http://mathoverflow.net/questions/121260/isometric-embeddings-of-cayley-graphs-in-nice-spaces/121286

isometric embeddings of Cayley graphs in “nice” spaces Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) This is from a physicist I know and as may be expected, I am threading my way between poorly defined and poorly translated. What groups have Cayley graphs (w.r.t. a fixed finite generating set, and metrized so that each edge has length 1) that can be isometrically embedded in some reasonably nice space? Let us start with either f.d. Euclidean space, or f.d. hyperbolic space as the list of reasonably nice spaces. This fails to mention whether the embedding should be a quasi-isometry, but let us bring in that question only if it seems to be looming in importance. Yes, Cartesian products of finitely many copies of $\mathbf Z$ did occur to me, as well as (in dimension 2) triangle groups, but the length restrictions will severely restrict which ones work. Helpful contributions will be either a few examples, or a few families, or pointers to what references might be relevant and anything else that might be relevant. - 1 If you are talking about the geodesic metric on the Cayley graph, there will never be even a locally isometric embedding in a Riemannian manifold. – Lee Mosher Feb 9 at 1:34 The question is too vague, so I do not think you will get any answer. One only can say something related. First of all isometric should mean quasiisometric (otherwise the question has no sense). Say if you are interested in Euclidean space then the group has polynomial growth --- it is sufficient, but that is for an other question. Asymptotic dimension is yet an other thing which should forbid embeddings in Hyperbolic spaces as well as Euclidean spaces. I am sure it can be continued forever... – Anton Petrunin Feb 9 at 2:31 For embedding groups into Euclidean spaces, try googling "hilbert compression exponents". But then again, this is very far from isometry, since in this theory you can get, at best, only Lipschitz embeddings. The requirement of embeddings being isometries, rather than some weaker classes of maps, seems very stringent. Is that really what you want? – Marcin Kotowski Feb 9 at 2:38 Okay, so the word metric does not survive so well in the ambient target space. My apologies. Now that I am brought to a bit more sense in this, I would have thought the taxi metric more useful than the max metric suggested below, but I don't trust myself at the moment. I will discuss all this with my friend. – Matt Brin Feb 9 at 3:56 1 What Lee said is 99.9% correct, since cycles embed isometrically into $S^1$ and the doubly infinite path embeds isometrically into $\mathbb{R}$. But these are the only cases. Indeed, a 3-pointed star in a graph (with 3 vertices at distance 1 of a central vertex) cannot be realized in a Riemannian manifold, using standard properties of geodesics. So a regular graph isometrically embeddable in a Riemannian manifold, necessarily has degree 2. – Alain Valette Feb 9 at 9:42 4 Answers You can get a general answer avoiding the discussion on the choice of the generating set or of the metric on the larger space: which finitely generated (f.g.) groups have a bilipschitz embedding into a Euclidean space? The answer is: virtually abelian f.g. groups (i.e. having a f.g. abelian group of finite index). Indeed such groups admit proper cocompact actions on $\mathbf{R}^n$ for some $n$, giving rise to a quasi-isometric embedding which can easily be deformed to a bilipschitz embedding (alternatively, if you like to stay with an action, you can add some extra dimensions to make the action [whose kernel is finite] faithful but you lose cocompactness). Conversely, if a group $\Gamma$ admits a bilipschitz embedding (coarse embedding --aka uniform embedding-- would be enough) into a Euclidean space, by a growth argument, the group has polynomial growth. So it is virtually nilpotent, by Gromov's polynomial growth Theorem. Then a result of general result of Pauls (2001) shows that a virtually nilpotent f.g. group admits a bilipschitz embedding into a CAT(0) space (or a uniformly convex Banach space). However, I guess that this follows, in this special case (Euclidean target) from Pansu's 1989 Annals paper about metric differentiability. The argument consists of observing that a bilipschitz map $\Gamma\to\mathbf{R}^n$ induces a bilipschitz embedding from the asymptotic cone (which is a simply connected nilpotent group $G$ with a Carnot-Caratheodory metric, by an older result of Pansu) into $\mathbf{R}^n$, to show that generically it is metrically differentiable and the differential is generically an injective homomorphism; in particular $G$ is abelian, and this holds iff the original discrete group is virtually abelian. To go back to the original question: if you stick to isometric embeddings, what you get as a corollary is that if $\Gamma$ is a group with a finite generating set $S$, and $E$ is a finite-dimensional Banach space, then if the Cayley graph of $\Gamma$ w.r.t. $S$ embeds isometrically into $E$ then $\Gamma$ is virtually abelian (which is an extremely restrictive condition). The converse, however, depends on the specific choice of $S$ and the norm on $E$, as pointed out in other answers. - Even more, Scott Pauls proved in his thesis that a nilpotent group QI embeds in a CAT(0) space if and only if the group is virtually abelian. – Misha Feb 9 at 14:44 @Misha: it's the same: you can trivially change a QI (from a f.g. group) into a bilipschitz map (into a geodesic space not reduced to a point) by a little perturbation. – Yves Cornulier Feb 9 at 19:27 You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. I'd like to suggest to study the embeddings of Cayley graphs into $\mathbf Z^n$ with the max metrics $d$: $$d(x \ y) := \max_{k=1...n}\ |x_k-y_k|$$ My short note, on (the degree of) the universality of the $\mathbf R^n$ space with the max distance function (metrics), published in AMS Notices (in the late 1970s I'd think), may motivate the above approach. Let me stress that a finite metric space with integer distances is isometrically embeddable in $\mathbf R^n\quad\Leftrightarrow\quad$ it is embeddable into $\mathbf Z^n$, both spaces considered with the max metrics. The max metrics makes embeddings easy. - Could you please give a more precise reference to your short note? It is impossible for me to guess either title of the paper or its author from the information on this page. – Martin Feb 9 at 3:41 I am sorry. My name is Włodzimierz Holsztyński. The title was something like "$\mathbf R^n$ as a universal metric space". (It's not easy for me to access a mathematical library, and I don't have my reprints at this time, and some of them I don't have at all). I'll post my name in my profile (thought that it was easily available to the participants of OM). – Wlodzimierz Holsztynski Feb 9 at 5:44 The taxi metrics is inferior to the max metrics in the context of your question. Indeed, you need ease of embeddings. The max product preserves the injectivity. – Wlodzimierz Holsztynski Feb 9 at 7:30 Hm, your question admits more than one interpretation. I was concerned with the finite metric space which has the group as its set of points (there would be no other points), and the distance function induced by the Cayley graph. I was not concerned with the edges as parts of the space to be embedded. – Wlodzimierz Holsztynski Feb 9 at 7:38 Thank you. I found the article: W. Holsztyński, $\mathbf R^n$ as a universal metric space, Notices AMS 25 (3) (1978) A- 367. – Martin Feb 9 at 10:27 In my first answer above I considered the embeddings of the group only (according to the Cayley graph). The requirement to embed isometrically the whole graph (an infinite metric space) in a sense does not add any complication at all when you embed the whole graph into $\mathbf R^n$ with the max metrics, while you may insist on embedding the vertices into $\mathbf Z^n\subseteq\mathbf R^n$ (if you want to). Indeed, once vertices are embedded, the embedding can be extended to an isometric embedding of the whole graph (injectivity); furthermore, there is exactly one canonical (the simplest) isometric extension. While considering other distance functions in the Euclidean spaces can be of some interest, it's clear to me that max is special, is "the right one"--"the best". This is also what the theory of categories strongly suggests. - The Cayley graph of arbitrary cyclic group, for any $1$-element generating set, can be isometrically embedded in $R^{\lceil\frac n2\rceil}$ with the distance function given by maximum (while the group itself gets embedded in $Z^3$). On the other hand it is not difficult to show that the cyclic group $Z/5$, with a 1-generator presentation, cannot be isometrically embedded in $R^3$ with the max distance. - How do you that, say, for the cyclic group of order 3 ? – Alain Valette Feb 17 at 9:01 $Z/3$ is embeddable in $Z^^2$ with the max metric: $$\\{(0\ 0)\ \ (0\ 1)\ \ (1\ \ 0)\\}$$ while mapping onto the unit vectors of $R^3$ is nicer. – Wlodzimierz Holsztynski Feb 17 at 9:34 A typo correction: that "Z^^2" above is really $Z^2$. – Wlodzimierz Holsztynski Feb 17 at 9:36

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 37, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9221734404563904, "perplexity_flag": "head"}

http://math.stackexchange.com/questions/279003/full-hessian-spherical-coordinates

# full hessian, spherical coordinates The question itself is pretty simple. I am running into confusion. Seems like there is a typo in the book. I wanna check myself. Maybe I am doing something wrong. Suppose we have the function (which I term full hessian): $$w=\sum_{i,j}f_if_{ij}f_j$$ with $$i,j=x,y,z$$ where $f_{ij}$ and $f_i$ means partial derivatives with respect to $i$ and $j$, and $i$, respectively. How do I write $w$ for some function $\rho(r)$ expressed in spherical coordinates? Note that $\rho$ is independent of $\phi$ and $\theta$. Let's take $\rho(r)=e^{-ar}$ as an example. Thanks in advance:) - ## 1 Answer You are free to call it full Hessian, but the established name for $w$ is $\infty$-Laplacian. Here is a nice expository article, though being 10 years old, it's a little out of date in the fast-developing subject. Recall that for any vector $v$ the expression $\sum_{i,j} v_i f_{ij} v_j$ is the directional second derivative of $f$, that is, $\frac{d^2}{dt^2}f(x+tv)$ evaluated at $t=0$. When we plug in $v=\nabla f(x)$, the result is the second derivative along the gradient. For your function $f(x)=\rho(|x|)$, the gradient is $\rho\,'(|x|)\frac{x}{|x|}$; that is, a radial vector with magnitude $|\rho\,'|$. The second derivative along this vector is $\rho\,''(\rho\,')^2$. This is the $\infty$-Laplacian of a spherically symmetric function. In particular, we observe that spherically symmetric $\infty$-harmonic functions are precisely the cones $f(x)=a|x|+b$. Comparison with cones is an effective method of gaining control over $\infty$-harmonic functions, as seen in the aforementioned article by Aronsson, Crandall and Juutinen. - 1 thnx for your nice answer!:) It was really helpful for me. – molkee Jan 26 at 22:47

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 27, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9418821930885315, "perplexity_flag": "head"}

http://mathoverflow.net/questions/38825/what-kind-of-lagrangians-can-we-have/39185

## What kind of Lagrangians can we have? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) In any physics book I've read the Lagrangian is introuced as as a functional whose critical points govern the dynamics of the system. It is then usually shown that a finite collection of non-interacting particles has a Lagrangian $\frac{1}{2}(m_1\dot{x}_1^2 + \cdots + m_n \dot{x}_n^2)$. It is then generally argued that $L=T-U$. I feel like something is missing here. What exactly are the physical hypotheses that go into this? Can we have other forms of the Lagrangian? How do we know those are "right"? Do we always have to compare them to the form of the equations we derived previously? For example, the Lagrangian formalism seems to be justified usually in so far as it 'works' for a finite collection of particles. Then you can solve any dynamics problem involving a collection of particles. I have been vague so let me try to be more precise in my question. Is the principle of least action an experimental hypothesis? Is it always true that $L=T-U$? When we don't know what the Lagrangian is, do we have to just guess and hope it is compatible with the dynamical equations we had already? Or can we perhaps start with the ansatz of a Lagrangian in some cases? I hope this is sufficiently precise. - 1 It seems to me you have to guess which forms of energies come into play in your physical system, i.e. guess the Lagrangian/Hamiltonian, from which you can derive equations of motion (through the principle of least action), which allows you to test whether particular guesses are suitable or not. – Sam Derbyshire Sep 15 2010 at 14:54 3 @Dorian This is a question that I have often thought of or asked but with no good reply. I haven't even seen a book which explain this point clearly. One can frame the question in various parts. Q1. Given a function on the phase space when is the function a valid Lagrangian? Q2: Given a "physical system" when does it have a Lagrangian? Q3: When is a Lagrangian system also Hamiltonian? (issues about dissipative forces). – Anirbit Sep 15 2010 at 14:56 1 Also I would like to know how does the theory of "Lagrangian submanifolds" (as understood in symplectic geometry) connect to the Physicist's intuitions. Thanks for motivating this topic. I really hope enlightening answers come up. – Anirbit Sep 15 2010 at 14:57 @Anirbit, you may find an answer to your final question in Arnol'ds book on mechanics. I think Q2 is a bit too vague to be answered, as it's basically asking "when does a physical model admit a variational formulation at one level or another". – jc Nov 5 2010 at 17:11 ## 12 Answers This question has a very wide scope, judging by the comments which have appeared already. I will try to narrow the scope a little in my answer. Classical mechanics is all about how a system evolves in time. The system itself is defined by specifying a configuration space, which we take to be a smooth manifold $M$. (The coordinates in the configuration space are usually known as "generalized coordinates" in the physics literature, to distinguish them from the standard coordinates in $\mathbb{R}^n$.) A lagrangian is then simply a differentiable function $L: TM \to \mathbb{R}$ on the tangent bundle. Relative to local coordinates $(x,v)$ for $TM$ (where $x$ --- the "positions" --- are local coordinates on $M$ and $v$ --- the "velocities" --- are the fibre coordinates), the lagrangian $L$ determines the equations of motion via the principle of least action, which results in the Euler-Lagrange equations: $$\frac{d}{dt} \frac{\partial L}{\partial v} = \frac{\partial L}{\partial x}$$ evaluated on curves $(x,\dot x)$. In addition one would like the fibre derivative $\frac{\partial L}{\partial v}$ to be nondegenerate, so that we get a fibrewise isomorphism $TM \to T^*M$. Then $p = \frac{\partial L}{\partial v}$ are the canonical momenta and nondegeneracy means that we can invert this relation and write $v$ as a function of $p$. The hamiltonian $H : T^*M \to \mathbb{R}$ is defined as the Legendre transform of the lagrangian $$H(x,p) = p v - L$$ where $v$ is given as a function of $p$. The cotangent bundle $T^*M$ has a natural symplectic form $\omega$ and the hamiltonian vector field associated to $H$ defines a flow which evolves the state $(x,p)$ in the same way as do the Euler-Lagrange equations. In general there is no canonical way to choose $L$ (or $H$): it ultimately depends on the physical system one is trying to model. Of course, we have had so much experience, that there are many heuristics one can use to guess the appropriate lagrangian. In doing so, symmetries play an important rôle and nowhere is this more evident than in the business of "model building" in particle physics. Here we are dealing with field theories, whence the configuration space consists of sections of certain homogeneous vector (or more general fibre) bundles on the spacetime and there are a number of symmetries (both Lie and discrete) that one might want our lagrangian to have. More interesting perhaps are the theories for which no lagrangians are known, or at least no lagrangians which are manifestly invariant under the desired symmetry. There are many examples of such theories, some very interesting ones in six dimensions which are intimately linked to the geometric Langlands program as described in this paper of Witten's, for istance. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Let me comment on your second question: When we don't know what the Lagrangian is, do we have to just guess and hope it is compatible with the dynamical equations we had already? If you want the variational derivative of your Lagrangian to yield your equations of motion (or a system which is equivalent to your equations of motion) you should solve the inverse problem of calculus of variations (googling will land you plenty of references; for the "mechanical" case which is apparently your primary interest, you can start here; see also this book by Ian Anderson and Gerard Thompson). In general, there are certain necessary conditions for your equations of motion to satisfy in order that the Lagrangian exists at all. If these are satisfied, finding the Lagrangian $L$ essentially boils down to using an appropriate homotopy operator to reconstruct $L$ from its variational derivative, see e.g. Ch. 5 of the book Applications of Lie groups to Differential Equations by Peter Olver. As a historical remark, it is interesting to note that one of the first Fileds medalists, Jesse Douglas, has made significant contributions to this field of research. Sorry for being rather sketchy, maybe I'll expand this answer a bit later. - As a theoretical physicist who shifted to pure mathematics, I think to answer this question and as a clarification to previous posts, we should not forget the historical side of the evolution and origins of the terms involved in physics theory modeling. The origin of the principle of stationary ('least' most of the time) action comes as a variational formulation using a functional $S[q(t)]:=\int_{t_1}^{t_2}L(q^i (t),\dot{q}^i (t);t)dt$ for the Lagrange equations of the real motion of a system with generalized coordinates $q^i (t)$ (obtained this way they are called Euler-Lagrange equations): $$\delta S[q_{real}(t)]=0\Rightarrow \frac{d}{dt}\frac{\partial L}{\partial \dot{q}^i}-\frac{\partial L}{\partial q^i}=0$$ The original Lagrange equations of motion were obtained as a reformulation of Newton's second law of mechanics for the case of generalized coordinates (the remaining degrees of freedom after introducing constraints) as an easier to handle version of D'Alembert's principle. D'Alembert wrote the 'virtal work' principle as a way to state a general equation for statics, which in a way amounts to postulate that constraint forces do not exert work (as they should be internal and the weak 3rd law applies). Now given Newton's second law for a system of particles, D'Alembert reformulates constrained dynamics as statics introducing the inertial force (you can see details at wikipedia) $$\sum_i (\vec{F}_{i}^{ext}-\frac{d\vec{p}_i}{dt})\cdot\delta \vec{r_i}=0$$ Lagrange wanted to work only with generalized coordinates which changing variables in D'Alembert's principle led him to the original Lagrange's equations of motion in terms of a kinetic generalised energy $T$ and generalized forces $Q_i$; furthermore he absorbed those conservative forces which derived from a potential $Q_i^c=-\partial U/\partial q_i$ (or more generally those which even depended on speed or time but had the required form of the left-hand side) along with the kinetic energy, into a function $L=T-U$ now called the Lagrangian $$\frac{d}{dt}\frac{\partial T}{\partial \dot{q}^i}-\frac{\partial T}{\partial q^i}=Q_i\Rightarrow \frac{d}{dt}\frac{\partial L}{\partial \dot{q}^i}-\frac{\partial L}{\partial q^i}=Q_i^{no-c.}$$ That is the origin of the form of the lagrangian as $L=T-U$. Now, the non-conservative forces $Q_i^{no-c.}$ appear because we are considering general 'open systems' which interchange energy with their enviroment, other systems as sources of energy, etc. To check this form in detail, first consdier that a closed system is always of the form $L=T-U$ because of a constructive reasoning as follows. For a free particle, at least locally, because of the homogeneity of space and time, and isotropy of space, we must have $L(q_i,\dot{q}_i;t)=T(v^2)$, that is a scalar which only depends on the particle and its speed (but no direction); since a free particle must still be free in another inertial system by def. the relation must be linear $L=a\cdot v^2$ (this is because any other Lagrangian giving the same equations of motion has to be of the form $L'=A\cdot L+\frac{d}{dt}\Omega(q;t)$ for which an inertial transformation $\vec{v}'=\vec{v}-\vec{\epsilon}$ must imply at first order $-2\vec{v}\cdot\vec{\epsilon}\partial L(v^2)/\partial (v^2)\propto d\Omega(q;t)/dt\Leftrightarrow \partial L/\partial (v^2)=0$); now the constant $a$ characterizes the particle, and must reduce to the free Newtonian second law ($p_i=m\cdot v_i =0$) therefore $a=m/2$. For a system of free particles without interactions the same applies summing over the individual kinetic energies $L_{free}=T=\frac{1}{2}\sum_i m_i v_i^2=\frac{1}{2}\sum_{i,j} A_{ij}\dot{q}^i\dot{q}^j$ (when $T$ is a cuadratic form the system is called 'natural' which happens when $\partial \vec{r}_k(q^i(t),\dot{q}^i(t);t)/\partial t = 0$ for all particles $k$, common for the natural case of constraints not dependent on time). To add interactions between particles in a closed system, their degrees of freedom must be coupled, which just means that the relative energies are not lost to the exterior, and this is satisfactorily implemented by a potential function $U=U(q_i,\dot{q}_i)$. By this method we obtain a general framework for mechanics using Lagrange's equations: postulating different potentials we obtain different models of interactions which shall be contrasted with experiment to constrain the form of $U$ (this may be done to obtain just effective phenomenological models, nevertheless all fundamental interactions of physics seem to fit this framework very well). Finally, when part B of the system A+B is fixed (like considered external) its motion can be treated like already 'solved', which opens the previous closed system, separating some of the degrees of freedom $q_B^i(t)$; therefore the Lagrangian decouples $L_{open}=T_A+T_B(\dot{q}^i_B(t))-U(q_A,q_B(t),\dot{q}_A,\dot{q}_B(t))$, in this case $T_B=d\Omega(q_B,t)/dt$ for some $\Omega$, so we can neglect that part since does not contribute to the equations of motion and therefore $L_{open}=T_A-U_{open}(q_A,\dot{q}_A,t)$. This way $U$ creates the effect of non-conservative forces which are just those that add or subtract energy from the system without taking into account where this energy goes. Therefore we can consider the Lagrange equation (without non-conservative forces) as the general more fundamental equation given individual interactions between the particles of the Universe, at least in principle. This is why the form $L=T-U$ is the usual approach in theoretical physics to deal with dynamics, since the very concept of instantaneous force at a distance à-la Newton is anti-natural (above all after special-general relativity) and the energetic local approach is not only mathematically better but philosophically more satisfactory. The Hamiltonian approach can be taken also as a starting point, or as a Legendre transformation of Lagrange dynamics, but that way one makes a non-relativistic break down of the coordinates $x^i, t$ which is useful for non-relativistic quantum mechanics. Now in general, physics is a theory of fields so we want to add these to the framework. This is easily done by considering them as continuous systems of degrees of freedom, providing generalizations of kinetic and potential energies of those. Furthermore one wishes for a (special or general) relativistic invariant theory which forces the $Ldt$ to be a Lorentz scalar in order to get Lorentz-covariant ($SO(3,1)$) equations of motion (i.e. covariant as tensor equations in a pseudo-Riemannian manifold). This way the free case reasoned above leads us to $ds=Ldt$ (where $ds$ is the relativistic space-time interval or 'proper time' measured by the particle) which indeed reduces to $ds\approx \frac{1}{2}mv^2$ for speeds $v\ll c$, and therefore can be taken as a relativistic generalization of the framework. Nevertheless point-particles are idealizations and both the classical and quantum theories require a field-theoretic treatment of matter (in the classical case, particles appear as small density lumps, and in the quantum case particle behaviour arises for discrete-like energy levels of the quantum field states). In the wikipedia article about Lagrangians different examples can be seen from point particles to fields, their kinetic terms and so on. In any case, the traditional form $L=T-U$ has its roots in the very nature of mechanics and any modern field theory is created by building up possible $T-U$ functionals from the field's degrees of freedom, i.e. the fields $\phi_a(x^\mu)$ and their "velocities" $\partial \phi_a/\partial x^\nu$ (for relativistic invariance reasons time-like speed $\frac{\partial}{\partial t}$ is not enough and the whole $\partial_\nu$ must be used). Hence one constructs kinetic terms like (Einstein's summation convenction) $T_\phi=\frac{1}{2}\partial_\mu\phi_a\partial^\mu\phi_a$ for scalar fields or $T_A=F_{\mu\nu}F^{\mu\nu}$ for vector fields ($F_{\mu\nu}:=\partial_\mu A_\nu-\partial_\nu A_\mu$ is in this case the tensor constructed from the field used to define its kinetic energy because we want to build invariant 'speed-like' $\partial_\mu$ scalars that are as well Gauge invariant which is another physical symmetry typically required besides Lorentz; furthermore, requiring gauge invariance for different Lie groups, tyipcally $SU(N)$, one forces the automatic appearance of fields and couplings between the matter fields responsible for the interactions, which is just working with connections and curvatures of fiber bundles). Since any term which makes the field equation of motion non-homogeneous is considered a source of perturbation, or force, one sees the corresponding couplings in the Lagrangian like a potential energy, giving as always $\mathcal{L}=T_\phi+T_A-U(A_\mu,\phi)$. With this, the coupled physical equations of motion are deduced for the principle of stationary action, where now the integration must be over the whole space and an interval of time (i.e. $\mathcal{L}$ is a density), the field Euler-Lagrange equations of motion: $$\partial_\mu\frac{\partial\mathcal{L}}{\partial (\partial_\mu\phi)}-\frac{\partial\mathcal{L}}{\partial\phi}=0$$ Finally these equations give the solutions of motion, that is the real field configurations at any point in space-time, which contribute the most in a quantum-theoretic framework, where one weighs complex transition amplitudes by Feynman operational methods using the action for each possible field-configuration ('motion'): $e^{\frac{i}{\hbar}S[\phi,A_\mu]}$. As remarked in other comment, the semi-classical approximation $\hbar\rightarrow 0$ makes the classical solution of Euler-Lagrange the predominant one, hence deducing the principle of stationary action. (Also, in the non-relativistic hamiltonian quantum-mechanics Ehrenfest's theorem provides a classical Newton-like equation of motion for the average degrees of freedom) Indeed as science makes progress we must see previous theories and models as limiting cases of new more precise theories, but in the beginning any physical theory must be constructed by physical insight, intuition and permanent comparison with experiment. Even though Feynman titled his thesis that way, he does not develop his approach to quantum theory starting from the action-principle, on the contrary he works with standard (hamiltonian) quantum mechanics and obtains a novel method for computing quantum probability amplitudes in which the classical action appears. You could start with quantum mechanics as an axiomatic system, develop Feynman's approach to quantum propagators of motion and deduce that classical approximate solutions of motion must obey, at first order of quantum corrections, the classical Euler-Lagrange equations of motion. At the same time, you need the input from the experimentally succsessful classical mechanics to get to quantum mechanics... and so forth. In the end at any moment of time in the history of science we are getting better and better mathematical structures that model reality to different degrees of precision; the important point is that a new one must contain an old one as an approximation in some regime, and explain even more. This way, theoretical physics tends to get better structures to encompass more phenomena from Nature in a simpler and simpler manner as it explains more effective laws with less theories (currently almost all the phenomena observed is explained by general relativity and quantum field theory, and unifying both is the neverending quest for the holy grail of physics). - 1 This is for sure the most comprehensive and enlightening answer! I really enjoyed reading this response and it should have more up-votes. – Samuel Reid Mar 23 2012 at 6:37 As far as the priniciples are concerned that enter the choice of a Lagrangian it is important to distinguish between classical systems and quantum systems. In the case of classical particle systems, or classical field theories, the primary input has historically been the dynamics of the system, as encoded in the equations of motion. The choice of Lagrangian, or rather the class of equivalent Lagrangians, then is determined by the constraint that the variational principle needs to reproduce this known dynamics. In the case of quantum field theories the strategy is somewhat different, and Feynman's approach to the computation of physical observables via the path integral puts more emphasis on the Lagrangian as the fundamental quantity. Here a powerful guide are symmetries that constrain the possible forms of the Lagrangian. Generally, symmetries are not enough to determine its form, and more experimental constraints are necessary to specify the theory. The final justification of any given Lagrangian is that its physical consequences are consistent with the observed phenomena. - I believe the standard answer to your question is that if one starts from Newton's Laws of Motion for a system of particles, then the second law says essentially $F = Ma$, giving the equations of motion as a second order ODE that is completely determined if we know the forces between particles. Then, and this is the important point that is often un-appreciated, Newton's Third Law (action and reaction are equal and opposite) IMPLIES that the forces are derivable from a potential function $U$. [The latter statement is WRONG ! See the comments and correction below.] It then follows by easy formal manipulation that $F=Ma$ is equivalent to the Euler Lagrange Equations for the Lagrangian $L= K - U$ where the kinetic energy $K$ is given by the formula you gave above for the Lagrangian without interactions. If you are interested in further details, I have a recent book (published by AMS and co-authored with my son Robert) called "Differential Equations, Mechanics, and Computation" in which I go to great lengths to give a conceptual approach to exactly the sort of question you are asking. There is a companion website at http://ode-math.com/ at which you can download more than half the material in the book as pdf files. - 1 I don't understand how Newton's third law can imply that forces are conservative. It seems to me that the consideration of a two-particle system provides a counterexample: Let the force on the first particle be an arbitrary function of the two positions and just let the force on the second be the precise opposite. Did you have further hypotheses in mind for this claim? – Harald Hanche-Olsen Sep 15 2010 at 16:19 4 Thank you Harald. Yes, you are correct (and your example is excellent); Newton's Third Law does NOT imply conservation of energy but rather conservation of linear momentum. To get conservation of energy, or that the force is a gradient of a potential, you must add to Newton's Laws a "no free lunch" axiom: The work done (line integral of the force) around a closed loop is zero. Sorry---another example of "If you say something complicated too early, you may have too eat your words for breakfast". :-) – Dick Palais Sep 15 2010 at 17:14 "Is the principle of least action an experimental hypothesis?" A physicist views classical mechanics as a semiclassical limit of quantum mechanics, valid in the limit that $\hbar \rightarrow 0$. In the path integral approach to quantum mechanics one studies integrals over an infinite dimensional space of paths (yes,yes I know these are not completely rigorous) of the form $\int {\cal D}x e^{i S[x]/\hbar}$ with $S= \int {\cal L} dt$ where ${\cal L}$ is the Lagrangian of the system. The principle of least action can then be viewed as a stationary phase approximation to the path integral, valid in the semiclassical limit. Thus rather than a hypothesis I would say the principle of least action is derived as part of the process of taking the classical limit of quantum mechanics. - 1 But is this not actually circular? I thought the path integral was arrived at departing from the principle of least action. In fact, Feynman's PhD thesis was titled "The principle of least action in quantum mechanics." (I confess it's been a while since I looked at it, though.) – José Figueroa-O'Farrill Nov 13 2010 at 14:37 I don't think it is circular, except perhaps in some historical sense. You can't derive QM from classical mechanics, but you can definitely take a classical limit of QM. – Jeff Harvey Dec 9 2010 at 16:27 Here are two nice, natural, examples of Lagrangians not of the form $T-U$ which occur naturally in physics. For a relativistic particle of charge e, mass m, travelling under the influence of an electromagnetic field F, with one-form potential A (so $F = dA$), the Lagrangian is $L = mds - eA$ where $ds$ is the proper time, or element of arc-length in Minkowski space (or more generally, the space-time within which said particle is travelling). [References: Feynman, vol. II, p. 19-7; Folland `Quantum Field Theory: A tourist's guide, ch. 2, p. 29. Most intermediate level physics texts] This can be parsed so as to have the form $L(q, \dot q)$ in two different ways. Way 1: choose an arbitrary parameter $u$ for the time and parameterize the space-time path with $u$. Then the $\dot q$ is $dq/du$. Way 2: choose to parameterize with time, which in one of the space-time coordinates. The non-relativistic limit of this $L$ is the $L$ of Alex R.s answer. (This variational principle is particularly interesting for photons, where $ds = 0$.) The oldest variational principle in mechanics (excluding Fermat's for optics) is that attributed to Maupertuis. He says that solutions to Newton's equations at fixed energy E, extremize $L = \sqrt{(E-U)T}$, using your terminology above. This Lagrangian represents a metric conformal to the Euclidean metric, as represented by $T$. (The geodesic time of these extremals is not Newtonian time.) I agree with Moduli that symmetry is one of the key ingredients that goes into the choice of a Lagrangian. In the symmetry light, it is worth noting that the two above Lagrangians are homogeneous of degree 1 in `velocities', so invariant under reparameterization of paths. - The most valuable discovery of the 20th century physics is symmetry. When you try to write down a Lagrangian for a system, usually there are two "principles": (1) the symmetries this system has and (2) experience. So no, L is not always T-U. For example, in General Relativity the Lagrangian (or rather Lagrangian density) is R, the scalar curvature of the spacetime manifold. As another example, in WZW models, the Lagrangian has two parts: a "kinetic" term plus a Wess–Zumino term. It is in quantum field theory that all kinds of Lagrangians pop up. What is more interesting is that the best you can do is to write down an “effective Lagrangian” at a certain energy scale, which may (and usually does) get quantum corrections at higher energy scales. - 2 "For example, in General Relativity the Lagrangian (or rather Lagrangian density) is R, the scalar curvature of the spacetime manifold." My knowledge of general relativity is admittedly limited but I don't see the contradiction. General relativity is a theory of statics (so there is no kinetic term) and the scalar curvature is a potential energy of intrinsic bending. It's analogous to how the energy of a static membrane is the integral of $|\nabla u|^2$ where u is the scalar displacement. – Per Vognsen Sep 18 2010 at 5:44 Good point. But energy in GR has never been well defined, so it is tricky to talk about "energy" in GR. You can say R is like a potential energy by some sort of analogies, but personally I do not think this is a good argument. – Moduli Sep 18 2010 at 15:33 1 The statement above that GR is a theory of statics is wrong. Static spacetimes are very special solutions to the Einstein equations (see Wald 6.1). Generic solutions are not static. If you expand the metric about flat spacetime you will find that the perturbations, which describe gravitational waves, have kinetic terms. etc. etc. – Jeff Harvey Nov 5 2010 at 16:49 It seems wildly arrogant to say `symmetry' is a 20th century discovery. I seem to recall a certain Galileo. Symmetry is central to Newton's formulation of the N-body equations and all the great mechanicians -Euler, Lagrange, Hamilton - used symmetry all the time. – Richard Montgomery Mar 22 2012 at 23:13 Is $L=T-V$ ? Depends what you mean by "$V$". Try finding $V$ for a system with nonconservative forces. There's a sleight of hand here of course, since when we talk about $V$ we really mean $V=V(q)$, a function of the coordinates $q$. This doesn't mean we can't find a Lagrangian. Take electromagnetism for example, say a charged particle moving through an electromagnetic field. It's lagrangian can we written as: $L=\frac{1}{2}m\dot{\mathbf{q}}^2-e\phi(\mathbf{q},t)+e\dot{\mathbf{q}}\cdot \mathbf{A}(\mathbf{q},t)$ where $\phi$ and $\mathbf{A}$ are the scalar and vector potentials respectively. How did we get this Lagrangian? One way is working backwards from the Lorenz force $\mathbf{F}=e\left(E+\frac{1}{c}\mathbf{v}\times\mathbf{B}\right)$, which gives us the equations of motion, and then we go back to our course in electromagnetism to remember that electromagnetism comes from scalar and vector potentials. If you are handed the equations of motion of a system (EOM), then there can be multiple Lagrangians that give rise to these EOM. In fact, at least locally, when two Lagrangians $L_1,L_2$ give rise to the same EOM, then $L_1-L_2=d\Phi/dt$, where $\Phi$ is a function on the position coordinates $\mathbb{Q}$. The point is, if you already have equations of motion, then finding a Lagrangian becomes a derivative matching game. - The Lagrangians in physics usually take the form $L=m\dot{x}^2/2+...$ (or $\mathcal{L}=(\partial \phi)^2/2+...$ in field theory), i.e. the velocity dependence is quadratic. An important consequence of this form can be seen in the path integral formulation of quantum mechanics. The amplitude as a path integral in the phase space is $\int \mathcal{D}x \ \mathcal{D}p \ e^{i\int(p \ dx - H \ dt)}$ where the Hamlitonian $H$ is a function of $p, x$ and $t$. The thing is, this amplitude is in general NOT equal to the form we usually use $\int \mathcal{D}x \ e^{i\int L \ dt}$. The two forms are equal when $H$ depends on $p$ as $p^2$, i.e. when $L$ depends on $\dot{x}$ as $\dot{x}^2$. (the derivation can be found in Polchinski, Peskin etc., basically $e^{ip^2}$ will give a constant by Gaussian integration) I am not sure if this is the reason that physical Lagrangian must take the form $L=m\dot{x}^2/2+...$, but at least I feel this shows some deep connection between the Lagrangian formulation and Hamiltonian formulation in quantum theory. - Here are two nice, natural, examples of Lagrangians not of the form $T-U$ which occur naturally in physics. For a relativistic particle of charge e, mass m, travelling under the influence of an electromagnetic field F, with one-form potential A (so $F = dA$), the Lagrangian is $L = mds - eA$ where $ds$ is the proper time, or element of arc-length in Minkowski space (or more generally, the space-time within which said particle is travelling). [References: Feynman, vol. II, p. 19-7; Folland `Quantum Field Theory: A tourist's guide, ch. 2, p. 29. Most intermediate level physics texts] This can be parsed so as to have the form $L(q, \dot q)$ in two different ways. Way 1: choose an arbitrary parameter $u$ for the time and parameterize the space-time path with $u$. Then the $\dot q$ is $dq/du$. Way 2: choose to parameterize with time, which in one of the space-time coordinates. The non-relativistic limit of this $L$ is the $L$ of Alex R.s answer. (This variational principle is particularly interesting for photons, where $ds = 0$.) The oldest variational principle in mechanics (excluding Fermat's for optics) is that attributed to Maupertuis. He says that solutions to Newton's equations at fixed energy E, extremize $L = \sqrt{(E-U)T}$, using your terminology above. This Lagrangian represents a metric conformal to the Euclidean metric, as represented by $T$. (The geodesic time of these extremals is not Newtonian time.) I agree with Moduli that symmetry is one of the key ingredients that goes into the choice of a Lagrangian. In the symmetry light, it is worth noting that the two above Lagrangians are homogeneous of degree 1 in `velocities', so invariant under reparameterization of paths. - 1 Somehow you posted this answer twice. – arsmath Dec 9 2010 at 12:26 In general, I would say that you have to check that a variational principle implies some physical law (usually in the form of a differential equation). In classical mechanics, in particular, Hamilton's principle of least (or stationary) action implies Newton's second law of motion (in Cartesian coordinates). But then you can use Hamilton's principle in any system of coordinates because the Euler-Lagrange equation is valid in any system of coordinates. -

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 148, "mathjax_display_tex": 6, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9398801326751709, "perplexity_flag": "head"}

http://physics.stackexchange.com/questions/52527/can-low-gravity-planets-sustain-a-breathable-atmosphere

# Can low-gravity planets sustain a breathable atmosphere? If astronauts could deliver a large quantity of breathable air to somewhere with lower gravity, such as Earth's moon, would the air form an atmosphere, or would it float away and disappear? Is there a minimum amount of gravity necessary to trap a breathable atmosphere on a planet? - ## 3 Answers Gravity is a major factor in planets retaining atmospheres over the eons. But there are other factors that must be taken into consideration to consider the volatility of an atmosphere. Solar wind is the main factor of erosion on any atmosphere. But a healthy magnetic field can deflect most of the solar radiation and decrease the erosion. It has been a matter of debate recently if exo-moons of jovian planets in habitable zones of their host stars would be able to sustain atmospheres: such moons are most likely tidally-locked, so their magnetic fields are not expected to be high, but their host planets will likely have strong radiation belts. But is not clear at the moment if the radiation belts will protect or erode the atmosphere. Saturn has a benign level of radiation, so we have Titan, which has an atmosphere that is thicker than earth's - 2 I'm not so sure that Solar wind is always the dominant factor. There is also the possibility that thermal fluctuations in the molecules' speeds can take them above the escape velocity. This would be the case on Earth if our atmosphere was made of $\text{H}_2$, IIRC. – Nathaniel Jan 30 at 14:05 good point, @Nathaniel – lurscher Jan 30 at 19:22 I guess the devil is in the details. For example, if the celestial body in question is far from its star, so its temperature is very low, it is easier to retain low-temperature air around the body. On the other hand, very cold air is not breathable anyway. There is another way though. If the astronauts can bring so much air to the body, why don't they arrange a membrane around the body to keep the air? Furthermore, they don't need the membrane around the entire body, they can arrange it just over some limited area where they want to live. On the other hand, they would need to protect such a membrane from meteorites... So I guess there is a lot they can do and a lot of factors that could make their life miserable:-) - The speed of oxygen at room temperature (293k) is 1720km per hour so if the escape velocity of the moon or planet is greater than that then at least you will have oxygen. If you want some nitrogen in the mix then you will have to google it's speed like I did for oxygen;-) - 2 The speed you give is (I think) the RMS speed, but there's a distribution of speeds around it. Some molecules will have higher speeds, and may escape, and if the temperature is maintained some other molecules will gain energy to fill those high velocity states in the distribution... leading to a gradual evaporation of the atmosphere. – Kyle Jan 30 at 14:30

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9504743218421936, "perplexity_flag": "middle"}

http://mathhelpforum.com/pre-calculus/96095-how-find-maximum-minimum-points.html

# Thread: 1. ## How to find Maximum and Minimum Points? 1. $y=(1-x^2)(4-x^2)$ 2. $y=3x^4-16x^3+18x^2$ Answer: 1. $(-\sqrt{\frac{5}{2}}, -2) (\sqrt{\frac{5}{2}}, -2)$ 2. $(1, 5) (3, -27)$ 2. The condition is : dy/dx = 0 but i get three values for each question, not two 3. ## Differentiation The only way I know to solve this question is to employ some calculus and differentiate. Differentiate to find $\frac{dy}{dx}$, then set $\frac{dy}{dx}=0$, and solve the resulting equation. Eg. No.2 $\frac{dy}{dx}=12x^3-48x^2+36x$ $=12x(x^2-4x+3)$ $\frac{dy}{dx}=12x(x^2-4x+3)=0$ $=(12x)(x-1)(x-3)=0$ $x=0,1,3$ Place these results into the original equation When $x=1$ $y=3(1)^4-16(1)^3+18(1)^2=5$ When $x=3$ $y=3(3)^4-16(3)^3+18(3)^2=-27$ When $x=0,y=0$ $(0,0)$

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 17, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.848802924156189, "perplexity_flag": "middle"}

http://mathhelpforum.com/advanced-algebra/174018-orthogonal-matrices-help.html

# Thread: 1. ## orthogonal matrices help using the definition of an orthogonal matrix prove : P is an orthogonal matrix if and only if P^t is an orthogonal matrix. I know P is orthogonal if we can multiply P by P^t and get the identity matrix but dont know where to go from there!! any help appreciated!! 2. Originally Posted by eleahy using the definition of an orthogonal matrix prove : P is an orthogonal matrix if and only if P^t is an orthogonal matrix. I know P is orthogonal if we can multiply P by P^t and get the identity matrix but dont know where to go from there!! any help appreciated!! Just use the fact that $\left(P^{\top}\right)^{-1}=\left(P^{-1}\right)^{-1}=P$ and so $\left(P^{\top}\right)^{-1}=P=\left(P^{\top}\right)^{\top}$. 3. still not exactly sure but have stuff to work with ill figure it thanks so much

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9092798829078674, "perplexity_flag": "middle"}

http://math.stackexchange.com/questions/23729/what-is-the-condition-for-a-field-to-make-the-degree-of-its-algebraic-closure-ov

# What is the condition for a field to make the degree of its algebraic closure over it infinite? As we all know, the algebraic closure often has an infinite degree. Also, this shows the necessary and sufficient condition for a Galois extension to be a finite extension of fields. However, we may want to characterize the cases of the case where the extension is not Galois, which is actually my question. And this question is related to this question. - 1 Geez. The second link is to Pete Clark's notes which are 73 pages long; I'm supposed to go hunting through it to see what specifically you are refering to? Perhaps you might consider quoting it instead of sending the reader on a scavenger hunt through a 73 page long document... – Arturo Magidin Feb 25 '11 at 16:22 @Arturo Magidin: Sorry, according to the last link, it is in section 12.5, thanks in any case. – awllower Feb 25 '11 at 16:25 5 As far as I can see, all the questions asked here are already answered in $\S 12.4$ of my notes. – Pete L. Clark Feb 25 '11 at 19:31 I am very sorry. – awllower Feb 26 '11 at 2:55 ## 2 Answers Let me summarize the results of $\S 12.4$ of my notes. (This involves making explicit some things which were left as "exercises", but I'm okay with that.) Let $K$ be any field, let $K^{\operatorname{sep}}$ be any separable closure and let $\overline{K}$ be any algebraic closure containing $K^{\operatorname{sep}}$. Then: 1) Suppose first that $K = K^{\operatorname{sep}}$, i.e., $K$ is separably closed. Then either: 1a) $K$ is algebraically closed, or 1b) It isn't, i.e., $K$ has positive characteristic $p$ and there exists $a \in K$ such that the polynomial $t^p - a$ is irreducible. Then $t^{p^n}-a$ is irreducible for all $n$, so $[\overline{K}:K]$ is infinite. (For this, see Lemma 32 in $\S 6.1$.) 2) Suppose that $K$ is not separably closed, so $G = \operatorname{Aut}(K^{\operatorname{sep}}/K)$ is nontrivial. Then by the Artin-Schreier theorem, either 2a) $\# G = 2$, in which case $K$ is formally real and $K(\sqrt{-1})$ is algebraically closed, or 2b) $\# G > 2$, in which case $G$ is infinite, which implies $[\overline{K}:K]$ is infinite. Note that some of the exercises outline further extensions of the Artin-Schreier Theorem, i.e., if $\overline{K}/K$ is "small" in other ways then $K$ is either real closed or algebraically closed. - Thanks very much, @Pete L. Clark. – awllower Feb 26 '11 at 3:25 I suddenly found that I forgot to accept it as an answer, hope you won't mind, @Pete L. Clark... – awllower Mar 12 '11 at 3:20 The Artin-Schreier theorem asserts that, if the field $k$ is not algebraically closed and $[\bar{k} : k]$ is finite, then in fact it equals $2$ and $\bar{k} = k[i]$ and $k$ is real closed. So I guess the condition you're looking for is: if and only if $k$ is not algebraically closed and also not real closed. I am answering the title question, not the body question. The body question is not equivalent to the title question; the answer to it is that the algebraic closure is Galois if and only if it's separable, hence if and only if $k$ is perfect. Perhaps you have only seen the definition of Galois extension for finite extensions. This is not the definition which works in maximal generality. The definition which does work in maximal generality is the following: Definition: An algebraic extension $k \to L$ is Galois if the fixed field of $\text{Aut}_k(L)$ is $k$. The separable closure $k_s$ of $k$ is always Galois (in fact it is the maximal Galois extension of $k$), and agrees with the algebraic closure if and only if $k$ is perfect. - Thanks very much. – awllower Feb 25 '11 at 16:33

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 41, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9470527768135071, "perplexity_flag": "head"}

http://mathoverflow.net/questions/20381/crystalline-cohomology-of-abelian-varieties/20385

Crystalline cohomology of abelian varieties Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I am trying to learn a little bit about crystalline cohomology (I am interested in applications to ordinariness). Whenever I try to read anything about it, I quickly encounter divided power structures, period rings and the de Rham-Witt complex. Before looking into these things, it would be nice to have an idea of what the cohomology that you construct at the end looks like. The l-adic cohomology of abelian varieties has a simple description in terms of the Tate module. My question is: is there something similar for crystalline cohomology of abelian varieties? More precisely, let $X$ be an abelian scheme over $\mathbb{Z}_p$. Is there a concrete description of $H^1(X_0/\mathbb{Z}_p)$? (or just `$H^1(X_0/\mathbb{Z}_p) \otimes_{\mathbb{Z}_p} \mathbb{Q}_p$`?) I think that this should consist of three things: a $\mathbb{Z}_p$-module $M$, a filtration on `$M \otimes_{\mathbb{Z}_p} \mathbb{Q}_p$` (which in the case of an abelian variety has only one term which is neither 0 nor everything) and a Frobenius-linear morphism $M \to M$. I believe that the answer has something to do with Dieudonné modules, but I don't know what they are either. - 3 The $\ell$-adic Tate mod. of an ab. var. $A$ over field $k$ of char. distinct from is not the etale cohom. (with $\mathbf{Z}_ {\ell}$ coefficients) of the $A$ in general (unless $k = k_s$), but rather of $A_{k_s}$ (with such coefficients). The etale coh. of $A$ itself is rather more complicated! Just as you probably learned about Tate modules and $\ell$-adic Galois representations before etale cohomology, you should learn about Dieudonne modules and their relation with $p$-divisible groups before learning crystalline cohomology. – BCnrd Apr 5 2010 at 12:25 Indeed, I meant the l-adic cohomology of the abelian variety over $k_s$. – Martin Orr Apr 6 2010 at 21:40 1 Answer To add a bit more to Brian's comment: the crystalline cohomology of an abelian variety (over a finite field of characteristic p, say) is canonically isomorphic to the Dieudonné module of the p-divisible group of the abelian variety (which is a finite free module over the Witt vectors of the field with a semi-linear Frobenius). If you start with an abelian scheme over the Witt vectors of this field then the crystalline cohomology of the special fibre is canonically isomorphic to the algebraic de Rham cohomology of the thing upstairs, hence receives a Hodge filtration also. A good starting place to understand Dieudonné modules is Demazure's 'Lectures on p-divisible groups', which appears in Springer LNM. In particular, he gives a nice description of the analogy with Tate modules (and the relation between the various Frobenii that appear). For a general picture of crystalline cohomology, and the various structures that can be placed on it, I would look at Illusie's survey in the Motives volumes (this is a little out of date now, but gives a good description of the basic theory). -

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9087642431259155, "perplexity_flag": "head"}

http://math.stackexchange.com/questions/112217/velocity-radius-and-angular-speed-finding-units

Velocity, Radius and Angular Speed - Finding Units I have been trying to figure this one out for two days, and I really have no idea what to do. To begin with, I need to know what units will be produced in the following situations: • Velocity's units when angular speed and radius are rpm and feet, respectively. (I think it is feet/minute) • Anuglar speed's units when radius and velocity are foot and feet per minute, respectively. (It stays as feet/minute?) • Radius' units when velocity and angular speed are mph and rpm, respectively. (I am guessing it will come out as miles, according to what my teacher tried to communicate) Those are just a few issues, but there are more. I believe that, if I am correct/find why I am incorrect, I will be able to apply the concept to the other problems. The second problem I am having is that I have a problem where: Velocity is 50 feet/second, Angular Speed is 100 revolutions/second and I need to find the radius in miles. According to what I know, velocity is angular speed times theta, so I just have to divide by angular speed. However, I do not know where to go when there are different units. I know it is a lot of text, but I appreciate anyone who can help me; my teacher said she is preoccupied and cannot answer any more of my questions today, so I'm stuck. - 2 Answers I always think of this sort of stuff in terms of forcing the the result to have the units you want and/or the only units that can possibly make sense. In the first case, you are given $\frac{rotations}{minutes}$ and $feet=\frac{feet}{1}$ and you want to end up with velocity, which has the general form $\frac{distance}{time}$. In this case, the pertinent units for distance and time are feet and minutes, respectively. The key, though, is the following: one rotation equals $2\pi$ times the radius (in feet). That is, $$\frac{rotations}{minutes}=2\pi\cdot radius\cdot\frac{feet}{minutes}.$$ But remember that in the end we just want an expression for velocity; that is, $\frac{feet}{minutes}=$ something. That's simple, though. Just divide both sides of the above equation by $2\pi\cdot radius$. The second and third ones are very similar, but angular velocity has units $\frac{rotations}{minutes}$. The last thing you asked is a specific instance of this sort of dimensional analysis, except that you have to convert everything to miles in the end. Alternatively, you can think of it as follows: if the velocity is 50 feet/second and the angular speed is 100 revolutions/second, that means in one second, the apparatus is making 100 revolutions - and furthermore, that these 100 revolutions = 50 feet. From this you get that one revolution is 1/2 a foot. That means that the circumference of the circle drawn out by the motion of the apparatus is 1/2 a foot, meaning that the radius of the circle is $\frac{1}{2}\cdot\frac{1}{2\pi}$ feet. To turn this into miles, just divide by 5280. Hope that helps. - Yay! Much obliged. – nmagerko Feb 22 '12 at 23:52 Angular speed has to be some measure of angle divided by some measure of time. In your second bullet point it will probably be revolutions per minute (rpm). You presumably have some formula saying velocity is related the product of radius and angular speed. Here you probably have $$50 \text{ feet/ second} = r \times 2\pi \text{ radians/revolution }\times 100 \text{ revolutions/ second}$$ which gives a radius of $r = 0.25 / \pi \text{ feet}$, not very much. If you must convert it into miles, then use $1 \text{ mile} = 5280 \text{ feet}$, though that will give you an even smaller number. - That is why I don't understand why we do problems like this! – nmagerko Feb 23 '12 at 0:42 1 It does get slightly easier using SI units. – Henry Feb 23 '12 at 0:54

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 10, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.962917685508728, "perplexity_flag": "head"}

http://mathoverflow.net/questions/75964?sort=votes

## Extending a vector bundle to a torsion free sheaf ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let's say that $X$ is an integral scheme of finite type over a field and $Y \subset X$ is a closed subscheme. Given a vector bundle $E$ on $Y$, is $E$ the restriction to $Y$ of a vector bundle on a neighborhood $U$ of $Y$ in $X$? - Isn't this exactly the Leff(X,Y) condition? This is true if $Y$ is a complete intersection of dimension at least 2 in a smooth projective variety $X$. Given $E$ on $Y$, you take $\hat{E}$ on the completion $\hat{X}$ of $X$ along $Y$, and Leff(X,Y) allows you to lift it to a bundle $E'$ in a neighborhood of $Y$ in $X$. – Parsa Apr 8 2012 at 17:14 What is Leff(X,Y)? – Ian Shipman Apr 8 2012 at 17:24 Ok, I found it. Even if the pair (X,Y) does not satisfy Leff, is there a way to understand when E on Y extends to a bundle on the formal completion? Some kind of obstruction perhaps to extending to various infinitesimal thickenings of Y... – Ian Shipman Apr 8 2012 at 17:29 ## 1 Answer No, this is not even true for line bundles. For an example, let $X = \mathbb{P}^2$ and $Y$ a smooth curve in $X$ of genus $>0$ (over an algebraically closed field). Since $X$ is smooth, any line bundle on an open set $U$ extends to a line bundle on $X$ so the map $\operatorname{Pic}(X) \to \operatorname{Pic}(U)$ is surjective. Since $\operatorname{Pic}(X) \cong \mathbb{Z}$, it follows that the image of $\operatorname{Pic}(U)$ in $\operatorname{Pic}(Y)$ is of rank $1$ and is independent of $U \supset Y$. Since $\operatorname{Pic}(Y)$ is not even finitely generated we see that there exist (many) line bundles on $Y$ which do not extend to any open $U \supset Y$. -

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 36, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9326929450035095, "perplexity_flag": "head"}

http://unapologetic.wordpress.com/2009/09/01/row-echelon-form/?like=1&source=post_flair&_wpnonce=d91e0e6d9a

# The Unapologetic Mathematician ## Row Echelon Form For now, I only want to focus on elementary row operations. That is, transformations of matrices that can be effected by multiplying by elementary matrices on the left, not on the right. Row echelon form is a (non-unique) simplification of the form of a matrix that can be reached by elementary row operations. A matrix is in row echelon form if all the empty (all-zero) rows are below all the nonzero rows, and the leftmost nonzero entry in a given row is to the right of the leftmost nonzero entry in the row above it. For example, an upper-triangular matrix is in row echelon form. We put a matrix into row echelon form by a method called “Gaussian elimination”, which always moves a matrix closer and closer to row echelon form. First, we look at the first column. If it’s empty we leave it alone and move on the next column. If there are any nonzero entries, there must be only one of them at the top of the column, or the matrix can’t be in row echelon form. So, if this isn’t the case then we pick one of the nonzero entries and call it our “pivot”. Swap its whole row up to the top row, and then use shears to eliminate all the other nonzero entries in the column before moving on to the second column. Before we continue, I’ll point out that the pivot is now the leading entry on a nonzero row at the top of the matrix. We’ll regard it as fixed, so whenever I say “all rows” I mean all the rows that we haven’t locked down yet. Now that we’ve moved on to the next column, we again look for any nonzero entries. If there are none, we can move on. But if there are, we choose one to be a pivot, swap it to the top (below the previously locked rows) and use shears to remove all the other nonzero entries in the column (other than the locked rows). We continue like this, choosing pivots where there are nonzero entries to deal with, moving them to the top, and eliminating all the other nonzero entries below them, until we’ve locked down all the rows or run out of columns. Let’s look at this method applied to the matrix $\displaystyle\begin{pmatrix}2&1&-1&8\\-3&-1&2&-11\\-2&1&2&-3\end{pmatrix}$ in the first column we have the nonzero value $2$ in the first row, so we may as well leave it where it is. We add $\frac{3}{2}$ times the first row to the second to eliminate the $-3$ there, and add the first row to the third to eliminate the $-2$ there. At this point our matrix looks like $\displaystyle\begin{pmatrix}2&1&-1&8\\{0}&\frac{1}{2}&\frac{1}{2}&1\\{0}&2&1&5\end{pmatrix}$ Now in the second row we have $\frac{1}{2}$ at the top (since the very top row is finished). We leave it where it is and add $-4$ times the second row to the third to eliminate the $2$ there. Now the matrix looks like $\displaystyle\begin{pmatrix}2&1&-1&8\\{0}&\frac{1}{2}&\frac{1}{2}&1\\{0}&0&-1&1\end{pmatrix}$ now there are no empty rows, so “all” of them are below all the nonzero rows. Further, within the nonzero rows the leading entries move from left to right as we move down the rows. The matrix is now in row echelon form. Now let’s try this one: $\displaystyle\begin{pmatrix}2&1&-1&8\\-4&-2&2&-11\\4&4&-3&6\\-2&-3&2&-3\end{pmatrix}$ Again, we’ll start with using the upper-left $2$ as our first pivot. We use shears to eliminate the rest of the entries in the first column and get $\displaystyle\begin{pmatrix}2&1&-1&8\\{0}&0&0&5\\{0}&2&-1&-10\\{0}&-2&1&5\end{pmatrix}$ Now there are nonzero entries in the second column, but none of them are at the top. So we swap the second and third rows to bring a pivot to the top $\displaystyle\begin{pmatrix}2&1&-1&8\\{0}&2&-1&-10\\{0}&0&0&5\\{0}&-2&1&5\end{pmatrix}$ and eliminate using a shear $\displaystyle\begin{pmatrix}2&1&-1&8\\{0}&2&-1&-10\\{0}&0&0&5\\{0}&0&0&-5\end{pmatrix}$ the third column has no nonzero entries (other than the two locked-in entries at the top), so we skip it. In the fourth column we use $5$ as a pivot and eliminate, getting $\displaystyle\begin{pmatrix}2&1&-1&8\\{0}&2&-1&-10\\{0}&0&0&5\\{0}&0&0&0\end{pmatrix}$ Now the empty row is on the bottom, and the leading entries of the other rows move from left to right as we move down the rows. Thus, the matrix is now in row echelon form. ### Like this: Posted by John Armstrong | Algebra, Linear Algebra ## 2 Comments » 1. [...] Equations with Gaussian Elimination The row echelon form of a matrix isn’t unique, but it’s still useful for solving systems of linear [...] Pingback by | September 2, 2009 | Reply 2. [...] Row Echelon Form Let’s take row echelon form and push it a little further into a form that is unique: reduced row echelon [...] Pingback by | September 3, 2009 | Reply « Previous | Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 17, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8943586349487305, "perplexity_flag": "head"}

http://math.stackexchange.com/questions/132062/asymptotic-notation

# Asymptotic notation [THIS IS HOMEWORK so please, don't just provide solutions, help me reasoning] I'm dealing with the following problem: Let $f(n)$ be the worst case time complexity of an algorithm A. Assume that there exist $c > 0$ and $n_{0}$, such that for each $n > n_{0}$ there exists an input of size $n$ for which A runs in exactly $c\cdot{g(n)}$ steps. a. Is it necessarily true that $f(n) = \Omega(g(n))$? b. Assume $f(n) = O(g(n))$. Is it necessarily true that $f(n) = \Theta(g(n))$? For a. I think it must be true since the $\Omega$ notation has on it all the elements that are either $f(n) = c\cdot g(n)$ or $f(n) > c\cdot g(n)$ so following the assumption leads to $f(n) = c\cdot g(n)$ being always true, therefore making the whole sentence true. (hope I'm right). For b. I assume $f(n) = O(g(n))$ and $f(n)=c\cdot{g(n)}$ ($\forall{n>n_{0}}$... of course) then I concluded that since (a) is true then $f(n) = \Theta(g(n))$. any help will be appreciated, thanks. - Part b. seems a little strange. If you're already given that $f(n) = c\,g(n)$ when $n$ is large enough why do you need to assume that $f(n) = O(g(n))$? Is that exactly how the question was written? – Antonio Vargas Apr 15 '12 at 13:15 @Antonio yes by 100% – JanosAudron Apr 15 '12 at 13:27 1 In that case then I think your reasoning is correct. Everything is easy if you know that $f(n) = c \,g(n)$ eventually. – Antonio Vargas Apr 15 '12 at 14:07

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 21, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9411389231681824, "perplexity_flag": "head"}

http://math.stackexchange.com/questions/160577/population-change-differential-equations?answertab=active

# Population change differential equations Lake has 400 fish, estimated carrying capacity is 10,000 for that year. Fish tripled in first year. a) Assuming size of the fish population satisfies the logisitic equation find an expression for the size of population after t years. b) How long will it take for population for increase to 5000? I do not know which logisitical equation to use or how to put the numbers in it, this is not like the practice problems. - 5 Am away from your book. The DE will have shape $\frac{dy}{dt}=ky(10000-y)$. – André Nicolas Jun 20 '12 at 0:18 ## 1 Answer As Andre commented above, the logistic equation has the form $$\frac{dy}{dt}=ry(10000-y),$$ and your initial condition $y(0)=400$. This is a separable equation, whose solution is given by $$y(t)=\frac{1000 e^{10000rt}}{24+e^{10000 rt}}.$$ You also have the condition that $y(1)=3\cdot y(0)=1200$, from which it is possible to estimate $r$: $$r\approx 0.0001185.$$ Now you can find the moment $t$ when $y(t)=5000$ (it is $2.6805$). -

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 7, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9436778426170349, "perplexity_flag": "middle"}

http://math.stackexchange.com/questions/102107/combinatorial-proof-to-the-identity-sum-k-0n-binomxkk-binomxn1n

Combinatorial proof to the identity $\sum_{k=0}^n\binom{x+k}{k}=\binom{x+n+1}{n}$? [duplicate] Possible Duplicate: Combinatorial proof for two identities The identity $$\sum_{k=0}^n\binom{x+k}{k}=\binom{x+n+1}{n}$$ can be verified by a straightforward induction, but is there some nice combinatorial argument which proves it? Here $x$ is a nonnegative integer, by the way. I've always found combinatorial arguments make it easier to remember such identities and intuitively grasp them so I would appreciate seeing one. Thank you. - Sorry, it was hard to tell out of the questions with similar titles if one was a duplicate, since the identities weren't in the title. I tried to include it in my title. – Threepio Jan 24 '12 at 21:35 marked as duplicate by Arturo Magidin, Byron Schmuland, Aryabhata, Fabian, Grumpy ParsnipJan 24 '12 at 22:02 This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question. 1 Answer Rewrite it as $$\sum_{k=0}^n\binom{x+k}x=\binom{x+n+1}{x+1}\;.$$ The righthand side clearly counts the subsets of $[x+n+1]$ of size $x+1$. The lefthand side breaks that count down according to the largest element of the subset. That is, if the largest element is $x+k+1$, the other $x$ elements must be chosen from the set $[x+k]$, and this can be done in $\binom{x+k}k$ ways. (Here I use $[n]$ for $\{1,\dots,n\}$.) -

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 10, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9427751302719116, "perplexity_flag": "head"}

http://cs.stackexchange.com/questions/9831/context-free-grammar-for-language-l

# Context Free Grammar for language L Can someone help with this: $L=\{a^ib^j \mid i,j \ge 1 \text{ and } i \ne j \text{ and } i<2j\}$ I'm trying to write a grammar for this language? I tried this: $S \to S_1 \mid S_2 \\ S_1 \to aXb \\ X \to aXb \mid aaXb \mid aab \\ S_2 \to aYb \\ Y \to aYb \mid Yb \mid b \\$ What do you think? - 1 Seems correct. Is there a problem? – Karolis Juodelė Feb 16 at 11:46 – Paresh Feb 16 at 13:37 ## 1 Answer The solution in the question is correct. The constraint $i\ne j$ is the one that gives us trouble. To get around it we have to split into two cases: (i) $i<j$, and (ii) $i>j$ (but still $i<2j$) thus $S\to S_{(i)} \mid S_{(ii)}$ as the first production splits into this two cases. Now, divide and conquer: case (i) is very simple, since we only need $i<j$, $S_{(i)} \to aS_{(i)}b \mid B$ $B \to Bb \mid b$ This is your $S_2$. For case (ii), we need to $j< i < 2j$, so for every single $a$, the variable $C$ will generate exactly one $b$, and at a certain point we switch to the variable $D$ that will generate two $a$'s for each $b$. $S_{(ii)} \to aCb$ $C \to aCb \mid D$ $D \to aaDb \mid aab$ This is your $S_1$. - To get $i < j$ you need $S_{(i)} \rightarrow a S_{(i)} b \mid B b$, $B \rightarrow B b \mid \epsilon$. The $B$ alternative adds at least one $b$ here. – vonbrand Mar 20 at 13:29 yes, you're right. – Ran G. Mar 20 at 16:19

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 27, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8735233545303345, "perplexity_flag": "middle"}

http://mathoverflow.net/questions/55383?sort=oldest

## Impact of discrepancy between Kunen’s and Jech’s definition of iterated forcing on full support iterations ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) One of the first things we usually learn when we study iterated forcing is that we can force over a model of ZFC + GCH to make the continuum function ($\lambda \mapsto 2^{\lambda}$) restricted to some set $S$ of regular cardinals follow any nondecreasing pattern satisfying König's Theorem ($\text{cof}(2^{\lambda}) > \lambda$). This is accomplished by an Easton product of adding the desired number of Cohen subsets to each cardinal in $S$. For example, assuming that the GCH holds in $V$, we can force $2^{\aleph_n} = \aleph_{n+2}$ for all Natural numbers $n$ with the full support $\omega$-iteration where we force with the ground model poset adding $\aleph_{n+2}$ many Cohen subsets of $\aleph_n$ at every stage $n$. However, if we instead force with the posets adding $\aleph_{n+2}$ Cohen subsets of $\aleph_n$ from the successive extensions, then the iteration will preserve the GCH. Such a phenomenon is easily explained away by the fact that after forcing to add Cohen subsets of $\aleph_n$, we have necessarily changed the definition of the forcing adding Cohen subsets of $\aleph_{n+1}$ in the extension. Now consider a full support $\omega$-iteration $\mathbb{P}$ where at every stage $n$, we force to add a Cohen real. Since the poset adding a Cohen real is the set of finite partial binary sequences with domain $\omega$, all transitive models of ZFC agree on its definition. Therefore, the aforementioned concern disappears in this context. Because the forcing to add a Cohen real satisfies the countable chain condition (c.c.c.), it is proper. Then since proper forcing is closed under countable support iterations using Jech's definition of a forcing iteration, $\mathbb{P}$ will be proper and hence $\omega_1$-preserving. However, using Kunen's definition, the iteration can collapse $\omega_1$. For example, from Chapter VIII, Exercise (E4): Let $\mathbb{P}_{\omega}$ be defined by countable supports, and let each $\pi_n$ be $(\text{Fn}(\omega, 2)){}^{\check{}}$). Show that $\mathbb{P}_{\omega}$ collapses $\omega_1$. Kunen then notes that this problem goes away when full names are used. On the other hand, Jech seems to circumvent the problem altogether by defining $\mathbb{P}_n * \dot{\mathbb{Q}}_n$ to be the set of pairs $\langle p, \dot{q}\rangle \in \mathbb{P} \times \text{dom}(\dot{\mathbb{Q}}_n)$ such that all conditions from $\mathbb{P}_n$ force $\dot{q}$ to be in $\dot{\mathbb{Q}}_n$ rather than just requiring that the condition $p$ forces this as Kunen does. What I would like to see then is an elaboration of the explanation of why these forcing iterations are different including the intuition behind these differences. Specifically, my question is as follows: Why does the full support $\omega$-iteration forcing with $(\text{Fn}(\omega, 2)){}^{\check{}}$ (forcing to add a Cohen real) at every stage collapse $\omega_1$ under Kunen's definition but not under Jech's? For example, how is it that every binary $\omega$-sequence from the ground model is coded in the generic filter using Kunen's definition but not Jech's? As a side request related to how this question came up, can you provide an example of a full support $\omega$-iteration that must have size at least $\aleph_2$, but each individual stage of forcing is necessarily proper and of size at most $\aleph_1$? - For the side request, please also assume CH. – Jason Feb 14 2011 at 6:11 1 Jech is hiding the use of full names in his construction. The assertion that every element of a PO forces a statement is exactly the statement that 1 forces that statement, which when viewed from the confines of Kunens construction is exactly his definition of full-name. There is a nice intuition and kinda picture that I imagine when thinking about both not all that sure I can describe it adequately though. – Michael Blackmon Feb 14 2011 at 11:21 ## 1 Answer I hope I can answer this question in a reasonable way. The natural way to define the iteration $\mathbb P*\dot{\mathbb Q}$ is to consider all pairs $(p,\dot q)$ with $p\in\mathbb P$ and $p\Vdash\dot q\in\dot{\mathbb Q}$. Unfortunately this is a proper class. (The definition in Jech's book also gives a proper class.) The problem in Kunen's definition of the iteration $\mathbb P*\dot{\mathbb Q}$ is not so much that he defines $(p,\dot q)\in\mathbb P*\dot{\mathbb Q}$ if $p$ forces $\dot q$ to be in $\dot{\mathbb Q}$, but that he only considers names $\dot q$ in the domain of $\dot{\mathbb Q}$. He does this to make sure that the iteration $\mathbb P*\dot{\mathbb Q}$ is a set rather than a proper class. But there are other ways to make sure that the iteration is a set, for example by allowing arbitrary names for $\dot q$ that are in some sufficiently large $H_\chi$ (sets with transitive closure of size $<\chi$). "Sufficiently large" depends on $\mathbb P$ and $\dot{\mathbb Q}$. Now Jech only considers pairs $(p,\dot q)$ such that $\dot q$ is forced by $1_{\mathbb P}$ to be in $\dot{\mathbb Q}$. But this doesn't really make a difference since by the existential completeness lemma, if $p\Vdash\dot q\in\dot{\mathbb Q}$, then there is a name $\dot r$ which is forced by $1_{\mathbb P}$ to be a condition in $\dot{\mathbb Q}$ and such that $p\Vdash\dot q=\dot r$. Now in the case of a two step iteration, the definitions by Jech and Kunen are equivalent, and both are equivalent to the one that I suggested (arbitrary names in the second coordinate, but cutting off somewhere). The right way to see this is to first show that if you cut off Jech's iteration at a sufficiently large $H_\chi$, then you get a dense subset of Jech's iteration. This cut off version of Jech's iteration is dense in my version (with the same sufficiently large $H_\chi$). Now Kunen's iteration is equivalent to mine since given $(p,\dot q)$ in my iteration, there are $s\leq p$ and $\dot t$ in the domain of $\dot Q$ such that $s\Vdash\dot q=\dot t$. We have $(s,\dot t)\leq(p,\dot q)$, showing that Kunen's iteration is dense in mine. Things start to fall apart if we go to longer iterations, and iterations with infinite support. My argument above still shows that Kunen's iteration of arbitrary length, with finite supports, is equivalent to my version for long iterations. This is because given a finitely supported condition in the long iteration in my sense, say with support of length $n$, you can decrease the first $n-1$ coordinates to force the last coordinate to be something given by a name in the domain of the respective name of a partial order. Then you decrease the first $n-2$ coordinates to take care of the $n-1$-th coordinate and so on. This argument clearly does not go through with countable supports. Hence the countable support iteration in Kunen's sense is not necessarily equivalent to my version or Jech's version. It is equivalent, however, if all the iterands are given by full names. Now, what is the problem in terms of properness? The proof of properness of countable support iterations of proper forcing (and I am referring to the proof in Goldstern's "Tools for your forcing construction") makes frequent use of the existential completeness lemma to cook up names for conditions. Now in the Kunen version of the iteration it may happen that these names cannot be used since they may not be in the domain of the respective name for a partial order. And in fact, as the exercise shows, even the countable support iteration of Cohen forcing of length $\omega$ in Kunen's sense depends on which names you choose for the iterands: Full names yield a proper forcing (which would be equivalent to the iteration in my sense), but taking canonical names yields a forcing notion that collapses $\omega_1$. I hope this helps. - I still want to think through some stuff, but this is very helpful. Thank you very much. – Jason Mar 11 2011 at 20:06

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 80, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9434811472892761, "perplexity_flag": "head"}

http://cms.math.ca/10.4153/CJM-2012-042-9

Canadian Mathematical Society www.cms.math.ca | | | | | |----------|----|-----------|----| | | | | | | | | Site map | | | CMS store | | location: Publications → journals → CJM Abstract view # Cumulants of the $q$-semicircular Law, Tutte Polynomials, and Heaps Read article [PDF: 242KB] Published:2012-11-13 • Matthieu Josuat-Vergès, CNRS and Institut Gaspard Monge, Université Paris-Est Marne-la-Vallée, 5 Boulevard Descartes, Champs-sur-Marne, 77454 Marne-la-Vallée cedex 2, France Features coming soon: Citations (via CrossRef) Tools: Search Google Scholar: Format: LaTeX MathJax PDF ## Abstract The $q$-semicircular distribution is a probability law that interpolates between the Gaussian law and the semicircular law. There is a combinatorial interpretation of its moments in terms of matchings where $q$ follows the number of crossings, whereas for the free cumulants one has to restrict the enumeration to connected matchings. The purpose of this article is to describe combinatorial properties of the classical cumulants. We show that like the free cumulants, they are obtained by an enumeration of connected matchings, the weight being now an evaluation of the Tutte polynomial of a so-called crossing graph. The case $q=0$ of these cumulants was studied by Lassalle using symmetric functions and hypergeometric series. We show that the underlying combinatorics is explained through the theory of heaps, which is Viennot's geometric interpretation of the Cartier-Foata monoid. This method also gives a general formula for the cumulants in terms of free cumulants. Keywords: moments, cumulants, matchings, Tutte polynomials, heaps MSC Classifications: 05A18 - Partitions of sets 05C31 - Graph polynomials 46L54 - Free probability and free operator algebras

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.7995302081108093, "perplexity_flag": "middle"}

http://www.all-science-fair-projects.com/science_fair_projects_encyclopedia/Moment_(physics)

# All Science Fair Projects ## Science Fair Project Encyclopedia for Schools! Search Browse Forum Coach Links Editor Help Tell-a-Friend Encyclopedia Dictionary # Science Fair Project Encyclopedia For information on any area of science that interests you, enter a keyword (eg. scientific method, molecule, cloud, carbohydrate etc.). Or else, you can start by choosing any of the categories below. # Moment (physics) See also moment (mathematics) for a more abstract concept of moments that evolved from this concept of physics. In physics, the moment is a quantity that represents the magnitude of force applied to a rotational system at a distance from the axis of rotation. The concept of the moment arm, this characteristic distance, is key to the operation of the lever, pulley, gear, and most other simple machines capable of generating mechanical advantage. Contents ## Overview In general, the moment M of a vector B is $\mathbf{M_A} = \mathbf{r} \times \mathbf{B} \,$ where r is the position where quantity B is applied. If r is a vector relative to point A, then the moment is the "moment M with respect to the axis that goes through the point A", or simply "moment M around A". If A is the origin, one often omits A and says simply moment. ## Parallel axis theorem Since the moment is dependent on the given axis, the moment expression possess a common property when the observation axis is changed. If MA is the moment around A, then the moment around the axis that goes through a point B is $\mathbf{M_B} = \mathbf{M_A} + \mathbf{R} \times \mathbf{B} \,$ where R is the vector from point B to point A. This expression is usually referred to as the parallel axis theorem. For cases when the moment is the sum of individual "submoments", such as in rigid body dynamics where each particle of the body contribute to a moment, the axis change is the sum of a macroscopic and microscopic quantity, $\mathbf{M_B} = \mathbf{R} \times \mathbf{B} + \sum_{i=0}{\mathbf{r_i} \times \mathbf{b_i}} \,$ where $\mathbf{B} = \sum_{i=0}{\mathbf{b_i}} \,$ or alternatively, $\mathbf{M_B} = \mathbf{R} \times \mathbf{B} + \mathbf{M_A} \,$ ## Related quantities Some notable physical quantities arise from the application of moments: • Angular momentum (L = Iω), which is typically the cause of rotational motion of a body. • Moment of inertia (I = mω×r), which is analogous to mass in discussions of rotational motion. • Torque (τ = rF), which is a force applied on a position of a body. When no net torque is applied, angular momentum is conserved. ## History The principle of moments is derived from Archimedes' discovery of the operating principle of the lever. In the lever one applies a force, in his day most often human muscle, to an arm, a beam of some sort. Archimedes noted that the amount of force applied to the object, the moment of force, is defined as M = rF, where F is the applied force, and r is the distance from the applied force to object. 03-10-2013 05:06:04

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 5, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8755354881286621, "perplexity_flag": "middle"}

http://math.stackexchange.com/questions/294443/would-it-also-be-useful-to-include-an-ordered-pair-function-in-first-order-logic

# Would it also be useful to include an ordered pair function in first order logic? Typically, first-order logic is assumed to include an equality relation $=$, even though this is "non-logical," together with some postulates about equality. Would it also be useful to include an ordered pair function $(*,*)?$ One could assume that $(x,y)=(x',y')$ precisely when $x=x$ and $y=y'$. Or perhaps it would be best to add denumerably many such functions, $(*)$, $(*,*)$, $(*,*,*)$, etc. The upshot of doing (either) of these is that relations can now all be unary. This would prettify a lot of notation. For instance, the axiom schema of replacement would look much neater. Thoughts, anyone? EDIT: Replacement would read as follows. $$\forall w \forall A[(\forall x \in A \exists ! y \phi(x,y,w,A)) \Rightarrow \exists B\forall y(y \in B \Leftrightarrow \exists x \in A \phi(x,y,w,A))]$$ - 1 I am very confused about how the Axiom Schema of Replacement would "look much neater" with the introduction of pairing/tripling/$n$-tupling operations in the syntax itself. Please express this Schema in your new syntax so I/we can judge. – Arthur Fischer Feb 4 at 12:14 So now every function has to be defined on tuples of every arity? Not to mention on tuples whose elements are themselves tuples? This doesn't sound very neat to me. – Chris Eagle Feb 4 at 12:31 @ArthurFischer Done. – user18921 Feb 4 at 20:48 ## 4 Answers The main issue with this that I can see would be that it wouldn't allow for any finite models, because if a model has $n$ distinct elements $a_i$, then it also has to have $n^2$ distinct elements of the form $(a_i, a_j)$. I think this on its own is reason enough to avoid it for first order logic in general. Having said that, there are several individual theories where it is convenient to have a pairing operator. Set theory and number theory both have conservative extensions with pairing operators (that is, there are already "implicit" pairing operators in the theory). Because of this, in these theories someone will often use brackets $(,)$ in their formulas as if there is a pairing operator. Like you said, this does make the notation much nicer. - The fact that you mention Replacement leads me to believe that you're especially interested in first-order theories which are meant to capture some sort of set theory. In a set theory, we have a way of identifying ordered pairs with elements of the universe - the standard choice in ZFC is to code $(a,b)$ by the element $\{\{a\},\{a,b\}\}$. Now we have a definable predicate Pair($x$) which determines whether some set $x$ codes a pair (it says "This set has two elements. One of these has one element, $a$. The other has at most two elements, one of which is $a$.") and definable functions First($x$) and Second($x$) which return $a$ and $b$. The theory then proves $\forall a \forall b \exists x\, \text{Pair}(x) \land (\text{First}(x) = a) \land (\text{Second}(x) = b)$. By adding the ordered pair function, you're adding a Skolem function for the sentence above. This eliminates the existential quantifier, so you now have $\forall a \forall b \text{Pair}((a,b)) \land (\text{First}((a,b)) = a) \land (\text{Second}((a,b)) = b)$. This is a totally standard thing to do in first-order logic - it simplifies syntax without changing the strength of the theory or its class of models (this is what the user aws means by saying that set theory has a "conservative" extension with a pairing operator). It's analogous to adding the inverse symbol ($^{-1}$) to the theory of groups. You can read more details on the wikipedia link. But set theory is somehow special in that there is a canonical way to code a pair of elements as a single element. That is, there is a definable injection $M^2 \rightarrow M$ for any model $M$. In general first-order logic, you don't have this kind of structure, and adding a pairing function to a general theory could really change the theory. However, there is a "safe" way to add pairing (and n-tuple) functions to an arbitrary first-order theory, which I'll describe now. We move from single-sorted logic to many-sorted logic, with one sort for each natural number. The $n^{th}$ sort represents the cartesian power $M^n$. Now for each $n$, we can also add an $n$-ary function $f_n$ which takes the $n$ elements $x_1,\dots,x_n$ to the element in the $n^{th}$ sort representing the $n$-tuple ($x_1,\dots,x_n$). We also add axioms stating that $f_n$ is a bijection for each $n$. One can check that this is construction gives a "conservative extension" in some sense. The advantage of this situation is that we can now view all definable functions and relations as unary, as you wanted. The expense is that we're in a many-sorted context, which you have to get used to. Model theorists often go further, adding new sorts not just for all $M^n$, but for all quotients of $M^n$ by definable equivalence relations - the resulting theory is called $T^{eq}$, and it is a nice place to work for a variety of reasons. - In fact, there are lots of pairing functions in set theory – one benefit of using an explicit pairing function is to get rid of the "theorems" that depend on the precise implementation of pairing. – Zhen Lin Feb 4 at 19:14 2 @ZhenLin Yes, that's a good point - there's another approach, which I didn't talk about, where you add a pairing function to the language and an axiom saying that it's injective, but you don't specify the form of output. This is attractive in that you don't get "meaningless theorems" like $\{a,b\} \in (a,b)$. On the other hand, it might be unattractive to people who want set theory to be "as close to complete as possible" - it makes lots of very simple sentences independent! Of course, this might support the point of view that this desire for completeness is misguided... – Alex Kruckman Feb 4 at 19:38 A comment on an incidental (but interesting) feature of the question, rather than a direct answer. According to the OP: Typically, first-order logic is assumed to include an equality relation =, even though this is "non-logical", ... But why suppose identity is non-logical? We need a demarcation criterion for distinguishing logical from non-logical operators. Let's consider two: A. There's the idea going back to Gentzen that a logical operator is one that is fully defined by introduction rules and (harmonious) elimination rules. This idea is explored in e.g. a famous paper by Ian Hacking 'What is Logic?' (1979) reprinted in e.g. Dov Gabbay (ed) What is a Logical System?. B. There's the idea going back to Tarski that a logical operator is one that exhibits a certain invariance under arbitrary permutations of the domain of objects. This idea is explored in e.g. Tarski's 'What are Logical Notions?' History and Philosophy of Logic (1986). (Transcript of a 1966 talk.) Both ideas A and B, for a start, make identity a logical operator. - I was implicitly defining non-logical as follows: A logical symbol can be viewed as acting on formulae and returning a formula, whereas a non-logical symbol can be viewed as acting on acting on expressions and returning a formula. However, the possible definitions of "logical operator" you mention are probably better definitions than what I had in mind. – user18921 Feb 4 at 20:58 Pairing does exist in first-order logic. It is implicit in untyped logic, as pairing is managed through the way we organize formulae. e.g. in the theory of real closed fields (arithmetic of real numbers), we can't have a variable $P$ that represents a point of the Euclidean plane $\mathbb{R}^2$, but we can systematically use two variables $x$ and $y$ always quantified in pairs and used together in relations and functions. In untyped logic, we can introduce pairing at a purely syntactic level as well. We can define a new language that allows variables that range over tuples of objects, and provide a purely syntactic translation from this new language into the old language where all variables are merely objects. There is also typed logic, which generally have explicit product types. e.g. taking again the theory of real closed fields, we can let $R$ denote the type comprising numbers, and there are product types such as $R \times R$ comprising pairs of numbers. -

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 51, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9543578028678894, "perplexity_flag": "head"}

http://math.stackexchange.com/questions/237246/find-orthonormal-basis-for-image-plane-given-camera-matrix

# Find Orthonormal Basis for Image Plane Given Camera Matrix Given a projection matrix $P = [M | p_4]$, ($M \in 3 \times 3$, $p_4 \in 3 \times 1$), the principal axis (the vector that passes through the center of projection and is perpendicular to the image plane) $v$ is $$v = det(M)m_3$$ where $m_3$ is the third row of $M$. Is this correct? Now I want to find two more vectors, say $v_1$ and $v_2$, that form an orthonormal basis with $v$ and span the image plane. I know that $v_1$ and $v_2$ are not unique so I want them to be equal to the $x$- and $y$-axes once the principal axis coincides with the $z$-axis (of some world coordinate system). If the principal axis differs from the $z$-axis due to some transformation I want that same transformation to give me $v_1$ and $v_2$ if applied to the $x$- and $y$-axes. Hope you get what I mean by that! Thanks in advance! - Your requirement still does not force a unique representation: you can rotate about the $z$ axis while maintaining alignment of that axis with your principal axis. As a consequence, “that same transformation” isn't well defined. Basically you want two more vectors orthogonal to $m_3$, right? Simply compute the cross product between that axis and any vector, and the result will be orthogonal. If may be zero, so take the cross product with the three unit vectors and choose the result with maximal length as $v_1$. You can then compute $v_2=m_3\times v_1$. – MvG Nov 14 '12 at 17:54 Hmmmmm i see :( ... what if I know the extrinsic parameters of my camera? Doesn't that give me that transformation? – Yasmin Nov 15 '12 at 9:29

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 23, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9151535630226135, "perplexity_flag": "head"}

http://mathoverflow.net/questions/42189/number-of-edges-in-low-complexity-graphs

## Number of edges in low complexity graphs ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Is the number of edges in the 3-connected graph with the minimum number of spanning trees always $\lceil {\frac{3}{2}}n\rceil$? - 1 What are some examples of such 3c-min-span graphs? Presumably you know some classes that achieve the min... – Joseph O'Rourke Oct 14 2010 at 19:40 It seems there are no classes which achieve the minimum always. An example of a class which achieve a low number of spanning trees are Prism Graphs, but they do not always have the minimum number of spanning trees. – utdiscant Oct 14 2010 at 20:50 For what values of n do you know this to be true? – jc Oct 15 2010 at 0:18 ## 1 Answer Are the minimally 3-connected graphs (edge removal pushes you into a 2-connected graph) the same as the class of 3-connected graphs with the minimal number of spanning trees? There is a nice classification of the minimal 3-connected graphs (or maximal not-3-connected graphs, I forget which direction he does it in) in Jonsson's book "Simplicial Complexes of Graphs". I think there's also a classification of minimal 3-connected graphs inside a paper by David Fisher, Kathryn Fraughnaugh, and Larry Langley which could help ["3-connected graphs of minimal size".] At the very least, they note that all 3-connected graphs on n vertices have at least 3/2(n-2) edges, and produce graphs achieving this bound. All of your graphs with the minimal number of spanning trees must be one of these minimal 3-connected graphs (as if your hypothetical "minimal spanning tree" graph wasn't also a minimal 3-connected graph, removing edges to a minimal 3-connected graph would reduce the number of spanning trees.) - It is true that the 3-connected graph(s) with the minimum number of spanning trees must be found among the 3-connected graphs with the minimal number of edges, but this is not enough to draw a conclusion, since it is easy to find examples showing that minimal 3-connected graphs have more than ceil(3n/2) edges. For instance the Wheel Graph on 6 vertices is a minimal 3-connected graph, but has 10 edges. – utdiscant Oct 15 2010 at 1:57 Also, I can't seem to find the paper ["3-connected graphs of minimal size".] which you are referring too, do you have a link? – utdiscant Oct 15 2010 at 2:01 The actual title appears to be "P_3 - connected graphs of minimal size" in Ars Combinatorica. I found this by putting one of the author's names into my favorite search engine. This led me to the personal website of the author, which included a CV, which included a list of papers. (MathSciNet would have been even quicker.) – JBL Oct 15 2010 at 2:31 Yeah, that's a good point. Since the complex of not 3-connected graphs is homotopic to a wedge of spheres of the same dimension, I was under the impression that all of the minimal 3-connected graphs were of the same size - not the case. The critical graphs left over after an appropriately chosen discrete Morse matching are though... Do you have these graphs classified for small n? – Gwyn Whieldon Oct 15 2010 at 5:00

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9455713033676147, "perplexity_flag": "middle"}

http://physics.stackexchange.com/questions/15098/linear-acceleration-vs-angular-acceleration-equation?answertab=oldest

# Linear acceleration vs angular acceleration equation I'm learning about angular velocity, momentum, etc. and how all the equations are parallel to linear equations such as velocity or momentum. However, I'm having trouble comparing angular acceleration to linear acceleration. Looking at each equation, they are not as similar as some of the other equations are: • Anglular acceleration = velocity squared / radius • Linear acceleration = force/ mass I would think angular acceleration would take torque into consideration. How is Vsquared similar in relation to force, and how is radius's relation to Vsquared match the relationship between mass and force? I suppose the root of this misunderstanding is how I'm thinking of angular acceleration, which is only an vector representing an axis's direction, and having a magnitude equal to the number of radians rotated per second. I also am confused on what exactly 'V' (tangential velocity) represents and how it's used. Is it a vector whose magnitude is equal to the number of radians any point on a polygon should rotate? What is the explanation? - ## 3 Answers You made a mistake in assuming that the angular acceleration ($\alpha$) is equal to $v^2/r$ which actually is the centripetal acceleration. In simple words, angular acceleration is the rate of change of angular velocity, which further is the rate of change of the angle $\theta$. This is very similar to how the linear acceleration is defined. $$a=\frac{d^2x}{dt^2} \rightarrow \alpha=\frac{d^2\theta}{dt^2}$$ Like the linear acceleration is $F/m$, the angular acceleration is indeed $\tau/I$, $\tau$ being the torque and I being moment of inertia (equivalent to mass). I also am confused on what exactly 'V' (tangential velocity) represents and how it's used. Is it a vector who's magnitude is equal to the number of radians any point on a polygon should rotate? The tangential velocity in case of a body moving with constant speed in a circle is same as its ordinary speed. The name comes from the fact that this speed is along the tangent to the circle (the path of motion for the body). Its magnitude is equal to the rate at which it moves along the circle. Geometrically you can show that $v = r\omega$. - $a_c = \frac{v^2}{r}$ isn't angular acceleration. It's the magnitude of the linear acceleration towards the centre of an object following a circular path at constant angular velocity. Angular acceleration is the derivative of angular velocity, and the analogue of Newton's second law is that angular acceleration equals torque divided by moment of inertia. - Always start with the units. They'll tell you a lot about the equations, and allow you to fix consistency errors. Incidentally, this is why I prefer Leibniz's notation over Newton's for derivatives, the units are immediately determined by examining the derivative, e.g. $dx/dt$ has units of distance over time assuming the usual definition of $x$ and $t$. In this case, the angle, $\theta$, is the equivalent of distance traveled in linear kinematics, and it has units of radians (${rad}$). (Radians, being unit less, are to some extent a place holder, but place holders can be very useful, so keep them in mind.) So, then the rate of change of angle with respect to time, $\omega$, has the units of ${rad}/s$. Angular acceleration, $\alpha$, will then have units of ${rad}/s^2$. With those in mind, you can immediately tell that $a_c = \frac{v^2}{r}$ is not an angular acceleration, but a linear acceleration, as described by Peter. Similarly, angular acceleration is not directly related to force, but to torque, $\tau = I \alpha$, where $I$ is the moment of inertia. (From a mathematical perspective, the moment of inertia is the second moment of the mass distribution where the center of mass is the first moment.) Torque has the units ${kg}\ m^2/s^2$, where the radians were dropped. Note, it has units of energy, or $(Force)(distance)$, and $\tau = r \times F$. On any single parameter curve in $\mathbb{R}^n$, $n\geq2$, the derivative with respect to that parameter always lies tangent to the curve. The derivative is literally showing us how the position is going to change. From a physics perspective, you can think of this as attaching the velocity and acceleration vectors to the moving object, itself, as in drawing a free body diagram. To be concrete, for uniform circular motion, the position is $$r(t) = R( \cos(t) \hat{i} + \sin(t) \hat{j} )$$ where $R$ is the radius of the circle, $\hat{i}$ and $\hat{j}$ are the unit vectors in the $x$ and $y$ directions, respectively, and the velocity is $$v(t) = R( -\sin(t) \hat{i} + \cos(t) \hat{j} ).$$ Note, that the velocity is perpendicular to the position which is a property of circular motion. From this, you should be able to mathematically demonstrate that the acceleration is perpendicular to the velocity and anti-parallel to the position. I'll leave you with the problem of understanding why this makes sense physically, also. - +1 Nice point on examining the units – Griffin Dec 13 '11 at 6:38

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 30, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9479528069496155, "perplexity_flag": "head"}

http://mathoverflow.net/revisions/59846/list

## Return to Answer 1 [made Community Wiki] Well, there are of course many examples but perhaps the Dirac $\delta$-function is the most striking one. With quite some violence to the math of usual calculus, Dirac's idea turned out to be exremely useful: the "function" everywhere zero and very high at $0$ so that the integral is $1$. Applications are ranging from Fourier analysis, (linear) PDE, quantum mechanics, and many more. Now of course we know how to make things rigorous in the framework of Schwartz's test function and distribution theory. But the intuition is clearly from physics. Similar and also in this framework is the notion of oscillatory integrals...

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9420386552810669, "perplexity_flag": "middle"}

http://bmet.wikia.com/wiki/Optics

# Optics Talk0 1,606pages on this wiki ## About Optics is the study of the behavior and properties of light, including its interactions with matter and the construction of optical instruments that use or Photodetector it. Optics usually describes the behavior of visible, ultraviolet, and infrared light. Because light is an electromagnetic wave, other forms of electromagnetic radiation such as X-rays, microwaves, and radio waves exhibit similar properties.[1] Optics is regarded within theoretical physics as a subfield of electromagnetism, and the vast majority of optical phenomena can be accounted for using the classical electromagnetic description of light. Complete electromagnetic descriptions of light are often difficult to apply in practice, however, so practical optics is usually done using simplified models. The most common of these, geometric optics, treats light as a collection of Ray (optics)s that travel in straight lines and bend when they pass through or reflect from surfaces. Physical optics is a more comprehensive model of light, which includes wave effects such as diffraction and interference that cannot be accounted for in geometric optics. Not all optical phenomena can be treated using classical electromagnetism. Some phenomena depend on the fact that light has both wave-particle duality. Explanation of these effects requires quantum mechanics. In quantum mechanics, light is treated as a collection of particles called photons. Quantum optics deals with the application of quantum mechanics to optical systems. Optical science is relevant to and studied in many related disciplines including astronomy, electrical engineering, photography, and medicine (particularly ophthalmology and optometry). Practical applications of optics are found in a variety of technologies and everyday objects, including mirrors, lenses, optical telescope, microscopes, lasers, and fiber optics. ## Early history of optics The earliest known lenses were made from polished crystal, often quartz, and have been dated as early as 700 BC for Assyrian lenses such as the Layard / Nimrud lens. There are many similar lenses from ancient Egypt, Greece and Babylon. The ancient Romans and Greeks filled glass spheres with water to make lenses. However, glass lenses were not thought of until the Middle Ages. Some lenses fixed in ancient Egyptian statues are much older than those mentioned above. There is some doubt as to whether or not they qualify as lenses, but they are undoubtedly glass and served at least ornamental purposes. The statues appear to be anatomically correct schematic eyes. In ancient India, the philosophical schools of Samkhya and Vaisheshika, from around the 6th–5th century BC, developed theories on light. According to the Samkhya school, light is one of the five fundamental "subtle" elements (tanmatra) out of which emerge the gross elements. In contrast, the Vaisheshika school gives an atomic theory of the physical world on the non-atomic ground of ether, space and time. (See Indian atomism.) The basic atoms are those of earth (prthivı), water (apas), fire (tejas), and air (vayu), that should not be confused with the ordinary meaning of these terms. These atoms are taken to form binary molecules that combine further to form larger molecules. Motion is defined in terms of the movement of the physical atoms. Light rays are taken to be a stream of high velocity of tejas (fire) atoms. The particles of light can exhibit different characteristics depending on the speed and the arrangements of the tejas atoms. Around the first century BC, the Vishnu Purana refers to sunlight as the "the seven rays of the sun". In the fifth century BC, Empedocles postulated that everything was composed of four elements; fire, air, earth and water. He believed that Aphrodite made the human eye out of the four elements and that she lit the fire in the eye which shone out from the eye making sight possible. If this were true, then one could see during the night just as well as during the day, so Empedocles postulated an interaction between rays from the eyes and rays from a source such as the sun. In his Optics Greek mathematician Euclid observed that "things seen under a greater angle appear greater, and those under a lesser angle less, while those under equal angles appear equal". In the 36 propositions that follow, Euclid relates the apparent size of an object to its distance from the eye and investigates the apparent shapes of cylinders and cones when viewed from different angles. Pappus believed these results to be important in astronomy and included Euclid's Optics, along with his Phaenomena, in the Little Astronomy, a compendium of smaller works to be studied before the Syntaxis (Almagest) of Ptolemy. In 55 BC, Lucretius, a Roman who carried on the ideas of earlier Greek atomists, wrote: "The light and heat of the sun; these are composed of minute atoms which, when they are shoved off, lose no time in shooting right across the interspace of air in the direction imparted by the shove." —Lucretius , On the nature of the Universe Despite being similar to later particle theories of light, Lucretius's views were not generally accepted and light was still theorized as emanating from the eye. In his Catoptrica, Hero of Alexandria showed by a geometrical method that the actual path taken by a ray of light reflected from a plane mirror is shorter than any other reflected path that might be drawn between the source and point of observation. In a twelfth-century translation assigned to Roman mathematician Claudius Ptolemy, a study of refraction, including atmospheric refraction, was described. It was suggested that the angle of refraction is proportional to the angle of incidence. Later in 499, Aryabhata, who proposed a heliocentric solar system of gravitation in his Aryabhatiya, wrote that the planets and the Moon do not have their own light but reflect the light of the Sun. The Indian Buddhists, such as Dignāga in the 5th century and Dharmakirti in the 7th century, developed a type of atomism that is a philosophy about reality being composed of atomic entities that are momentary flashes of light or energy. They viewed light as being an atomic entity equivalent to energy, similar to the modern concept of photons, though they also viewed all matter as being composed of these light/energy particles. ## Classical optics In pre-quantum-mechanical optics, light is an electromagnetic wave composed of oscillating electric and magnetic fields. These fields continually generate each other, as the wave propagates through space and oscillates in time. The frequency of a light wave is determined by the period of the oscillations. The frequency does not normally change as the wave travels through different materials ("media"), but the speed of the wave depends on the medium. The speed, frequency, and wavelength of a wave are related by the formula $v=\lambda\,f ,$ where v is the speed, λ is the wavelength and f is the frequency. Because the frequency is fixed, a change in the wave's speed produces a change in its wavelength. The speed of light in a medium is typically characterized by the index of refraction, n, which is the ratio of the speed of light in vacuum, c, to the speed in the medium: $n=c/v.$ The speed of light in vacuum is a constant, which is exactly 299,792,458 meters per second. Thus, a light ray with a wavelength of λ in a vacuum will have a wavelength of λ / n in a material with index of refraction n. The amplitude of the light wave is related to the intensity of the light, which is related to the energy stored in the wave's electric and magnetic fields. Traditional optics is divided into two main branches: geometrical optics and physical optics. ### Geometrical optics Main article: Geometrical optics As light wave travels through space, it oscillates in amplitude. In this image, each maximum amplitude crest is marked with a plane to illustrate the wavefront. The ray is the arrow perpendicular to these parallel surfaces. Geometrical optics, or ray optics, describes light propagation in terms of "rays". The "ray" in geometric optics is an abstraction, or "instrument", that can be used to predict the path of light. A light ray is a ray that is perpendicular to the light's wavefronts (and therefore collinear with the wave vector). Light rays bend at the interface between two dissimilar media and may be curved in a medium in which the refractive index changes. Geometrical optics provides rules for propagating these rays through an optical system, which indicates how the actual wavefront will propagate. This is a significant simplification of optics that fails to account for optical effects such as diffraction and polarization. It is a good approximation, however, when the wavelength is very small compared with the size of structures with which the light interacts. Geometric optics can be used to describe the geometrical aspects of imaging, including optical aberrations. A slightly more rigorous definition of a light ray follows from Fermat's principle which states that the path taken between two points by a ray of light is the path that can be traversed in the least time. #### Approximations Geometrical optics is often simplified by making the paraxial approximation, or "small angle approximation." The mathematical behavior then becomes linear, allowing optical components and systems to be described by simple matrices. This leads to the techniques of Gaussian optics and paraxial ray tracing, which are used to find basic properties of optical systems, such as approximate image and object positions and magnifications. #### Reflections Main article: Reflection (physics) Reflections can be divided into two types: specular reflection and diffuse reflection. Specular reflection describes glossy surfaces such as mirrors, which reflect light in a simple, predictable way. This allows for production of reflected images that can be associated with an actual (real) or extrapolated (virtual) location in space. Diffuse reflection describes matte surfaces, such as paper or rock. The reflections from these surfaces can only be described statistically, with the exact distribution of the reflected light depending on the microscopic structure of the surface. Many diffuse reflectors are described or can be approximated by Lambert's cosine law, which describes surfaces that have equal luminance when viewed from any angle. In specular reflection, the direction of the reflected ray is determined by the angle the incident ray makes with the surface normal, a line perpendicular to the surface at the point where the ray hits. The incident and reflected rays lie in a single plane, and the angle between the reflected ray and the surface normal is the same as that between the incident ray and the normal. This is known as the Law of Reflection. For flat mirrors, the law of reflection implies that images of objects are upright and the same distance behind the mirror as the objects are in front of the mirror. The image size is the same as the object size. (The magnification of a flat mirror is unity.) The law also implies that mirror images are parity inverted, which we perceive as a left-right inversion. Images formed from reflection in two (or any even number of) mirrors are not parity inverted. Corner reflectors retro-reflect light, producing reflected rays that travel back in the direction from which the incident rays came. Mirrors with curved surfaces can be modeled by ray-tracing and using the law of reflection at each point on the surface. For mirrors with parabolic surfaces, parallel rays incident on the mirror produce reflected rays that converge at a common focus. Other curved surfaces may also focus light, but with aberrations due to the diverging shape causing the focus to be smeared out in space. In particular, spherical mirrors exhibit spherical aberration. Curved mirrors can form images with magnification greater than or less than one, and the magnification can be negative, indicating that the image is inverted. An upright image formed by reflection in a mirror is always virtual, while an inverted image is real and can be projected onto a screen. #### Refractions Main article: Refraction Refraction occurs when light travels through an area of space that has a changing index of refraction; this principle allows for lenses and the focusing of light. The simplest case of refraction occurs when there is an interface between a uniform medium with index of refraction n1 and another medium with index of refraction n2. In such situations, Snell's Law describes the resulting deflection of the light ray: $n_1\sin\theta_1 = n_2\sin\theta_2\$ where θ1 and θ2 are the angles between the normal (to the interface) and the incident and refracted waves, respectively. This phenomenon is also associated with a changing speed of light as seen from the definition of index of refraction provided above which implies: $v_1\sin\theta_2\ = v_2\sin\theta_1$ where v1 and v2 are the wave velocities through the respective media. Various consequences of Snell's Law include the fact that for light rays traveling from a material with a high index of refraction to a material with a low index of refraction, it is possible for the interaction with the interface to result in zero transmission. This phenomenon is called total internal reflection and allows for fiber optics technology. As light signals travel down a fiber optic cable, it undergoes total internal reflection allowing for essentially no light lost over the length of the cable. It is also possible to produce polarized light rays using a combination of reflection and refraction: When a refracted ray and the reflected ray form a right angle, the reflected ray has the property of "plane polarization". The angle of incidence required for such a scenario is known as Brewster's angle. Snell's Law can be used to predict the deflection of light rays as they pass through "linear media" as long as the indexes of refraction and the geometry of the media are known. For example, the propagation of light through a prism results in the light ray being deflected depending on the shape and orientation of the prism. Additionally, since different frequencies of light have slightly different indexes of refraction in most materials, refraction can be used to produce dispersion spectra that appear as rainbows. The discovery of this phenomenon when passing light through a prism is famously attributed to Isaac Newton. Some media have an index of refraction which varies gradually with position and, thus, light rays curve through the medium rather than travel in straight lines. This effect is what is responsible for mirages seen on hot days where the changing index of refraction of the air causes the light rays to bend creating the appearance of specular reflections in the distance (as if on the surface of a pool of water). Material that has a varying index of refraction is called a gradient-index (GRIN) material and has many useful properties used in modern optical scanning technologies including photocopiers and scanners. The phenomenon is studied in the field of gradient-index optics. A ray tracing diagram for a converging lens. A device which produces converging or diverging light rays due to refraction is known as a lens. Thin lenses produce focal points on either side that can be modeled using the lensmaker's equation. In general, two types of lenses exist: convex lenses, which cause parallel light rays to converge, and concave lenses, which cause parallel light rays to diverge. The detailed prediction of how images are produced by these lenses can be made using ray-tracing similar to curved mirrors. Similarly to curved mirrors, thin lenses follow a simple equation that determines the location of the images given a particular focal length (f) and object distance (S1): $\frac{1}{S_1} + \frac{1}{S_2} = \frac{1}{f}$ where $S_2$ is the distance associated with the image and is considered by convention to be negative if on the same side of the lens as the object and positive if on the opposite side of the lens. The focal length f is considered negative for concave lenses. Incoming parallel rays are focused by a convex lens into an inverted real image one focal length from the lens, on the far side of the lens. Rays from an object at finite distance are focused further from the lens than the focal distance; the closer the object is to the lens, the further the image is from the lens. With convex lenses, incoming parallel rays diverge after going through the lens, in such a way that they seem to have originated at an upright virtual image one focal length from the lens, on the same side of the lens that the parallel rays are approaching on. Rays from an object at finite distance are associated with a virtual image that is closer to the lens than the focal length, and on the same side of the lens as the object. The closer the object is to the lens, the closer the virtual image is to the lens. Likewise, the magnification of a lens is given by $M = - \frac{S_2}{S_1} = \frac{f}{f - S_1}$ where S2 is the distance associated with the image and is considered by convention to be negative if on the same side of the lens as the object and positive if on the opposite side of the lens.[39] The focal length f is considered negative for concave lenses. Incoming parallel rays are focused by a convex lens into an inverted real image one focal length from the lens, on the far side of the lens. Rays from an object at finite distance are focused further from the lens than the focal distance; the closer the object is to the lens, the further the image is from the lens. With convex lenses, incoming parallel rays diverge after going through the lens, in such a way that they seem to have originated at an upright virtual image one focal length from the lens, on the same side of the lens that the parallel rays are approaching on. Rays from an object at finite distance are associated with a virtual image that is closer to the lens than the focal length, and on the same side of the lens as the object. The closer the object is to the lens, the closer the virtual image is to the lens. Lenses suffer from aberrations that distort images and focal points. These are due to both to geometrical imperfections and due to the changing index of refraction for different wavelengths of light (chromatic aberration). ### Physical optics Main article: Physical optics hysical optics or wave optics builds on Huygens's principle, which states that every point on an advancing wavefront is the center of a new disturbance. When combined with the superposition principle, this explains how optical phenomena are manifested when there are multiple sources or obstructions that are spaced at distances similar to the wavelength of the light.[40] Complex models based on physical optics can account for the propagation of any wavefront through an optical system, including predicting the wavelength, amplitude, and phase of the wave. Additionally, all of the results from geometrical optics can be recovered using the techniques of Fourier optics which apply many of the same mathematical and analytical techniques used in acoustic engineering and signal processing. Using numerical modeling on a computer, optical scientists can simulate the propagation of light and account for most diffraction, interference, and polarization effects. Such simulations typically still rely on approximations, however, so this is not a full electromagnetic wave theory model of the propagation of light. Such a full model is computationally demanding and is normally only used to solve small-scale problems that require extraordinary accuracy. Gaussian beam propagation is a simple paraxial physical optics model for the propagation of coherent radiation such as laser beams. This technique partially accounts for diffraction, allowing accurate calculations of the rate at which a laser beam expands with distance, and the minimum size to which the beam can be focused. Gaussian beam propagation thus bridges the gap between geometric and physical optics. #### Superposition and interference Main article: Superposition principle In the absence of nonlinear effects, the superposition principle can be used to predict the shape In the absence of nonlinear effects, the superposition principle can be used to predict the shape of interacting waveforms through the simple addition of the disturbances. This interaction of waves to produce a resulting pattern is generally termed "interference" and can result in a variety of outcomes. If two waves of the same wavelength and frequency are in phase, both the wave crests and wave troughs align. This results in constructive interference and an increase in the amplitude of the wave, which for light is associated with a brightening of the waveform in that location. Alternatively, if the two waves of the same wavelength and frequency are out of phase, then the wave crests will align with wave troughs and vice-versa. This results in destructive interference and a decrease in the amplitude of the wave, which for light is associated with a dimming of the waveform at that location. See below for an illustration of this effect Since Huygens's principle states that every point of a wavefront is associated with the production of a new disturbance, it is possible for a wavefront to interfere with itself constructively or destructively at different locations producing bright and dark fringes in regular and predictable patterns. Interferometry is the science of measuring these patterns, usually as a means of making precise determinations of distances or angular resolutions. The Michelson interferometer was a famous instrument which used interference effects to accurately measure the speed of light. The appearance of thin films and coatings is directly affected by interference effects. Antireflective coatings use destructive interference to reduce the reflectivity of the surfaces they coat, and can be used to minimize glare and unwanted reflections. The simplest case is a single layer with thickness one-fourth the wavelength of incident light. The reflected wave from the top of the film and the reflected wave from the film/material interface are then exactly 180° out of phase, causing destructive interference. The waves are only exactly out of phase for one wavelength, which would typically be chosen to be near the center of the visible spectrum, around 550 nm. More complex designs using multiple layers can achieve low reflectivity over a broad band, or extremely low reflectivity at a single wavelength. Constructive interference in thin films can create strong reflection of light in a range of wavelengths, which can be narrow or broad depending on the design of the coating. These films are used to make dielectric mirrors, interference filters, heat reflectors, and filters for color separation in color television cameras. This interference effect is also what causes the colorful rainbow patterns seen in oil slicks. #### Diffraction and optical resolution Diffraction is the process by which light interference is most commonly observed. The effect was first described in 1665 by Francesco Maria Grimaldi, who also coined the term from the Latin diffringere, 'to break into pieces'. Later that century, Robert Hooke and Isaac Newton also described phenomena now known to be diffraction in Newton's rings while James Gregory recorded his observations of diffraction patterns from bird feathers. The first physical optics model of diffraction that relied on Huygens' Principle was developed in 1803 by Thomas Young in his accounts of the interference patterns of two closely spaced slits. Young showed that his results could only be explained if the two slits acted as two unique sources of waves rather than corpuscles. In 1815 and 1818, Augustin-Jean Fresnel firmly established the mathematics of how wave interference can account for diffraction. The simplest physical models of diffraction use equations that describe the angular separation of light and dark fringes due to light of a particular wavelength (λ). In general, the equation takes the form $m \lambda = d \sin \theta$ where d is the separation between two wavefront sources (in the case of Young's experiments, it was two slits), θ is the angular separation between the central fringe and the mth order fringe, where the central maximum is m = 0. This equation is modified slightly to take into account a variety of situations such as diffraction through a single gap, diffraction through multiple slits, or diffraction through a diffraction grating that contains a large number of slits at equal spacing. More complicated models of diffraction require working with the mathematics of Fresnel or Fraunhofer diffraction. X-ray diffraction makes use of the fact that atoms in a crystal have regular spacing at distances that are on the order of one angstrom. To see diffraction patterns, x-rays with similar wavelengths to that spacing are passed through the crystal. Since crystals are three-dimensional objects rather than two-dimensional gratings, the associated diffraction pattern varies in two directions according to Bragg reflection, with the associated bright spots occurring in unique patterns and d being twice the spacing between atoms. Diffraction effects limit the ability for an optical detector to optically resolve separate light sources. In general, light that is passing through an aperture will experience diffraction and the best images that can be created (as described in diffraction-limited optics) appear as a central spot with surrounding bright rings, separated by dark nulls; this pattern is known as an Airy pattern, and the central bright lobe as an Airy disk. The size of such a disk is given by $\sin \theta = 1.22 \frac{\lambda}{D}$ where θ is the angular resolution, λ is the wavelength of the light, and D is the diameter of the lens aperture. If the angular separation of the two points is significantly less than the Airy disk angular radius, then the two points cannot be resolved in the image, but if their angular separation is much greater than this, distinct images of the two points are formed and they can therefore be resolved. Rayleigh defined the somewhat arbitrary "Rayleigh criterion" that two points whose angular separation is equal to the Airy disk radius (measured to first null, that is, to the first place where no light is seen) can be considered to be resolved. It can be seen that the greater the diameter of the lens or its aperture, the finer the resolution. Interferometry, with its ability to mimic extremely large baseline apertures, allows for the greatest angular resolution possible. For astronomical imaging, the atmosphere prevents optimal resolution from being achieved in the visible spectrum due to the atmospheric scattering and dispersion which cause stars to twinkle. Astronomers refer to this effect as the quality of astronomical seeing. Techniques known as adaptive optics have been utilized to eliminate the atmospheric disruption of images and achieve results that approach the diffraction limit. #### Dispersion and scattering Main article: Dispersion (optics) Refractive processes take place in the physical optics limit, where the wavelength of light is similar to other distances, as a kind of scattering. The simplest type of scattering is Thomson scattering which occurs when electromagnetic waves are deflected by single particles. In the limit of Thompson scattering, in which the wavelike nature of light is evident, light is dispersed independent of the frequency, in contrast to Compton scattering which is frequency-dependent and strictly a quantum mechanical process, involving the nature of light as particles. In a statistical sense, elastic scattering of light by numerous particles much smaller than the wavelength of the light is a process known as Rayleigh scattering while the similar process for scattering by particles that are similar or larger in wavelength is known as Mie scattering with the Tyndall effect being a commonly observed result. A small proportion of light scattering from atoms or molecules may undergo Raman scattering, wherein the frequency changes due to excitation of the atoms and molecules. Brillouin scattering occurs when the frequency of light changes due to local changes with time and movements of a dense material Dispersion occurs when different frequencies of light have different phase velocities, due either to material properties (material dispersion) or to the geometry of an optical waveguide (waveguide dispersion). The most familiar form of dispersion is a decrease in index of refraction with increasing wavelength, which is seen in most transparent materials. This is called "normal dispersion". It occurs in all dielectric materials, in wavelength ranges where the material does not absorb light. In wavelength ranges where a medium has significant absorption, the index of refraction can increase with wavelength. This is called "anomalous dispersion". Dispersion occurs when different frequencies of light have different phase velocities, due either to material properties (material dispersion) or to the geometry of an optical waveguide (waveguide dispersion). The most familiar form of dispersion is a decrease in index of refraction with increasing wavelength, which is seen in most transparent materials. This is called "normal dispersion". It occurs in all dielectric materials, in wavelength ranges where the material does not absorb light. In wavelength ranges where a medium has significant absorption, the index of refraction can increase with wavelength. This is called "anomalous dispersion". Material dispersion is often characterized by the Abbe number, which gives a simple measure of dispersion based on the index of refraction at three specific wavelengths. Waveguide dispersion is dependent on the propagation constant. Both kinds of dispersion cause changes in the group characteristics of the wave, the features of the wave packet that change with the same frequency as the amplitude of the electromagnetic wave. "Group velocity dispersion" manifests as a spreading-out of the signal "envelope" of the radiation and can be quantified with a group dispersion delay parameter: $D = \frac{1}{v_g^2} \frac{dv_g}{d\lambda}$ where $v_g$ is the group velocity.[2] For a uniform medium, the group velocity is $v_g = c \left( n - \lambda \frac{dn}{d\lambda} \right)^{-1}$ where n is the index of refraction and c is the speed of light in a vacuum.[3] This gives a simpler form for the dispersion delay parameter: $D = - \frac{\lambda}{c} \, \frac{d^2 n}{d \lambda^2}.$ If D is less than zero, the medium is said to have positive dispersion or normal dispersion. If D is greater than zero, the medium has negative dispersion. If a light pulse is propagated through a normally dispersive medium, the result is the higher frequency components slow down more than the lower frequency components. The pulse therefore becomes positively chirped, or up-chirped, increasing in frequency with time. This causes the spectrum coming out of a prism to appear with red light the least refracted and blue/violet light the most refracted. Conversely, if a pulse travels through an anomalously (negatively) dispersive medium, high frequency components travel faster than the lower ones, and the pulse becomes negatively chirped, or down-chirped, decreasing in frequency with time. The result of group velocity dispersion, whether negative or positive, is ultimately temporal spreading of the pulse. This makes dispersion management extremely important in optical communications systems based on optical fibers, since if dispersion is too high, a group of pulses representing information will each spread in time and merge together, making it impossible to extract the signal. #### Polarization Main article: Polarization Polarization is a general property of waves that describes the orientation of their oscillations. For transverse waves such as many electromagnetic waves, it describes the orientation of the oscillations in the plane perpendicular to the wave's direction of travel. The oscillations may be oriented in a single direction (linear polarization), or the oscillation direction may rotate as the wave travels (circular or elliptical polarization). Circularly polarized waves can rotate rightward or leftward in the direction of travel, and which of those two rotations is present in a wave is called the wave's chirality. The typical way to consider polarization is to keep track of the orientation of the electric field vector as the electromagnetic wave propagates. The electric field vector of a plane wave may be arbitrarily divided into two perpendicular components labeled x and y (with z indicating the direction of travel). The shape traced out in the x-y plane by the electric field vector is a Lissajous figure that describes the polarization state. The following figures show some examples of the evolution of the electric field vector (blue), with time (the vertical axes), at a particular point in space, along with its x and y components (red/left and green/right), and the path traced by the vector in the plane (purple): The same evolution would occur when looking at the electric field at a particular time while evolving the point in space, along the direction opposite to propagation. In all other cases, where the two components either do not have the same amplitudes and/or their phase difference is neither zero nor a multiple of 90°, the polarization is called elliptical polarization because the electric vector traces out an ellipse in the plane (the polarization ellipse). This is shown in the above figure on the right. Detailed mathematics of polarization is done using Jones calculus and is characterized by the Stokes parameters. Media that have different indexes of refraction for different polarization modes are called birefringent. Well known manifestations of this effect appear in optical wave plates/retarders (linear modes) and in Faraday rotation/optical rotation (circular modes). If the path length in the birefringent medium is sufficient, plane waves will exit the material with a significantly different propagation direction, due to refraction. For example, this is the case with macroscopic crystals of calcite, which present the viewer with two offset, orthogonally polarized images of whatever is viewed through them. It was this effect that provided the first discovery of polarization, by Erasmus Bartholinus in 1669. In addition, the phase shift, and thus the change in polarization state, is usually frequency dependent, which, in combination with dichroism, often gives rise to bright colors and rainbow-like effects. In mineralogy, such properties, known as pleochroism, are frequently exploited for the purpose of identifying minerals using polarization microscopes. Additionally, many plastics that are not normally birefringent will become so when subject to mechanical stress, a phenomenon which is the basis of photoelasticity. ## Modern optics Main article: Optical physics Modern optics encompasses the areas of optical science and engineering that became popular in the 20th century. These areas of optical science typically relate to the electromagnetic or quantum properties of light but do include other topics. A major subfield of modern optics, quantum optics, deals with specifically quantum mechanical properties of light. Quantum optics is not just theoretical; some modern devices, such as lasers, have principles of operation that depend on quantum mechanics. Light detectors, such as photomultipliers and channeltrons, respond to individual photons. Electronic image sensors, such as CCDs, exhibit shot noise corresponding to the statistics of individual photon events. Light-emitting diodes and photovoltaic cells, too, cannot be understood without quantum mechanics. In the study of these devices, quantum optics often overlaps with quantum electronics. Specialty areas of optics research include the study of how light interacts with specific materials as in crystal optics and metamaterials. Other research focuses on the phenomenology of electromagnetic waves as in singular optics, non-imaging optics, non-linear optics, statistical optics, and radiometry. Additionally, computer engineers have taken an interest in integrated optics, machine vision, and photonic computing as possible components of the "next generation" of computers. Today, the pure science of optics is called optical science or optical physics to distinguish it from applied optical sciences, which are referred to as optical engineering. Prominent subfields of optical engineering include illumination engineering, photonics, and optoelectronics with practical applications like lens design, fabrication and testing of optical components, and image processing. Some of these fields overlap, with nebulous boundaries between the subjects terms that mean slightly different things in different parts of the world and in different areas of industry. A professional community of researchers in nonlinear optics has developed in the last several decades due to advances in laser technology. ## Applications Optics is part of everyday life. The ubiquity of visual systems in biology indicate the central role optics plays as the science of one of the five senses. Many people benefit from eyeglasses or contact lenses, and optics are integral to the functioning of many consumer goods including cameras. Rainbows and mirages are examples of optical phenomena. Optical communication provides the backbone for both the Internet and modern telephony. ### Human eye Main article: Human eye The human eye functions by focusing light onto an array of photoreceptor cells called the retina, which covers the back of the eye. The focusing is accomplished by a series of transparent media. Light entering the eye passes first through the cornea, which provides much of the eye's optical power. The light then continues through the fluid just behind the cornea—the anterior chamber, then passes through the pupil. The light then passes through the lens, which focuses the light further and allows adjustment of focus. The light then passes through the main body of fluid in the eye—the vitreous humor, and reaches the retina. The cells in the retina cover the back of the eye, except for where the optic nerve exits; this results in a blind spot. There are two types of photoreceptor cells, rods and cones, which are sensitive to different aspects of light. Rod cells are sensitive to the intensity of light over a wide frequency range, thus are responsible for black-and-white vision. Rod cells are not present on the fovea, the area of the retina responsible for central vision, and are not as responsive as cone cells to spatial and temporal changes in light. There are, however, twenty times more rod cells than cone cells in the retina because the rod cells are present across a wider area. Because of their wider distribution, rods are responsible for peripheral vision. In contrast, cone cells are less sensitive to the overall intensity of light, but come in three varieties that are sensitive to different frequency-ranges and thus are used in the perception of color and photopic vision. Cone cells are highly concentrated in the fovea and have a high visual acuity meaning that they are better at spatial resolution than rod cells. Since cone cells are not as sensitive to dim light as rod cells, most night vision is limited to rod cells. Likewise, since cone cells are in the fovea, central vision (including the vision needed to do most reading, fine detail work such as sewing, or careful examination of objects) is done by cone cells. Ciliary muscles around the lens allow the eye's focus to be adjusted. This process is known as accommodation. The near point and far point define the nearest and farthest distances from the eye at which an object can be brought into sharp focus. For a person with normal vision, the far point is located at infinity. The near point's location depends on how much the muscles can increase the curvature of the lens, and how inflexible the lens has become with age. Optometrists, ophthalmologists, and opticians usually consider an appropriate near point to be closer than normal reading distance—approximately 25 cm. Defects in vision can be explained using optical principles. As people age, the lens becomes less flexible and the near point recedes from the eye, a condition known as presbyopia. Similarly, people suffering from hyperopia cannot decrease the focal length of their lens enough to allow for nearby objects to be imaged on their retina. Conversely, people who cannot increase the focal length of their lens enough to allow for distant objects to be imaged on the retina suffer from myopia and have a far point that is considerably closer than infinity. A condition known as astigmatism results when the cornea is not spherical but instead is more curved in one direction. This causes horizontally extended objects to be focused on different parts of the retina than vertically extended objects, and results in distorted images. All of these conditions can be corrected using corrective lenses. For presbyopia and hyperopia, a converging lens provides the extra curvature necessary to bring the near point closer to the eye while for myopia a diverging lens provides the curvature necessary to send the far point to infinity. Astigmatism is corrected with a cylindrical surface lens that curves more strongly in one direction than in another, compensating for the non-uniformity of the cornea. The optical power of corrective lenses is measured in diopters, a value equal to the reciprocal of the focal length measured in meters; with a positive focal length corresponding to a converging lens and a negative focal length corresponding to a diverging lens. For lenses that correct for astigmatism as well, three numbers are given: one for the spherical power, one for the cylindrical power, and one for the angle of orientation of the astigmatism #### Visual effects Main article: Optical illusions Optical illusions (also called visual illusions) are characterized by visually perceived images that differ from objective reality. The information gathered by the eye is processed in the brain to give a percept that differs from the object being imaged. Optical illusions can be the result of a variety of phenomena including physical effects that create images that are different from the objects that make them, the physiological effects on the eyes and brain of excessive stimulation (e.g. brightness, tilt, color, movement), and cognitive illusions where the eye and brain make unconscious inferences. Cognitive illusions include some which result from the unconscious misapplication of certain optical principles. For example, the Ames room, Hering, Müller-Lyer, Orbison, Ponzo, Sander, and Wundt illusions all rely on the suggestion of the appearance of distance by using converging and diverging lines, in the same way that parallel light rays (or indeed any set of parallel lines) appear to converge at a vanishing point at infinity in two-dimensionally rendered images with artistic perspective. This suggestion is also responsible for the famous moon illusion where the moon, despite having essentially the same angular size, appears much larger near the horizon than it does at zenith. This illusion so confounded Ptolemy that he incorrectly attributed it to atmospheric refraction when he described it in his treatise, Optics Another type of optical illusion exploits broken patterns to trick the mind into perceiving symmetries or asymmetries that are not present. Examples include the café wall, Ehrenstein, Fraser spiral, Poggendorff, and Zöllner illusions. Related, but not strictly illusions, are patterns that occur due to the superimposition of periodic structures. For example transparent tissues with a grid structure produce shapes known as moiré patterns, while the superimposition of periodic transparent patterns comprising parallel opaque lines or curves produces line moiré patterns #### Optical instruments Main article: Optical instruments Single lenses have a variety of applications including photographic lenses, corrective lenses, and magnifying glasses while single mirrors are used in parabolic reflectors and rear-view mirrors. Combining a number of mirrors, prisms, and lenses produces compound optical instruments which have practical uses. For example, a periscope is simply two plane mirrors aligned to allow for viewing around obstructions. The most famous compound optical instruments in science are the microscope and the telescope which were both invented by the Dutch in the late 16th century. Microscopes were first developed with just two lenses: an objective lens and an eyepiece. The objective lens is essentially a magnifying glass and was designed with a very small focal length while the eyepiece generally has a longer focal length. This has the effect of producing magnified images of close objects. Generally, an additional source of illumination is used since magnified images are dimmer due to the conservation of energy and the spreading of light rays over a larger surface area. Modern microscopes, known as compound microscopes have many lenses in them (typically four) to optimize the functionality and enhance image stability. A slightly different variety of microscope, the comparison microscope, looks at side-by-side images to produce a stereoscopic binocular view that appears three dimensional when used by humans. The first telescopes, called refracting telescopes were also developed with a single objective and eyepiece lens. In contrast to the microscope, the objective lens of the telescope was designed with a large focal length to avoid optical aberrations. The objective focuses an image of a distant object at its focal point which is adjusted to be at the focal point of an eyepiece of a much smaller focal length. The main goal of a telescope is not necessarily magnification, but rather collection of light which is determined by the physical size of the objective lens. Thus, telescopes are normally indicated by the diameters of their objectives rather than by the magnification which can be changed by switching eyepieces. Because the magnification of a telescope is equal to the focal length of the objective divided by the focal length of the eyepiece, smaller focal-length eyepieces cause greater magnification. Since crafting large lenses is much more difficult than crafting large mirrors, most modern telescopes are reflecting telescopes, that is, telescopes that use a primary mirror rather than an objective lens. The same general optical considerations apply to reflecting telescopes that applied to refracting telescopes, namely, the larger the primary mirror, the more light collected, and the magnification is still equal to the focal length of the primary mirror divided by the focal length of the eyepiece. Professional telescopes generally do not have eyepieces and instead place an instrument (often a charge-coupled device) at the focal point instead. ### Atmospheric optics The unique optical properties of the atmosphere cause a wide range of spectacular optical phenomena. The blue color of the sky is a direct result of Rayleigh scattering which redirects higher frequency (blue) sunlight back into the field of view of the observer. Because blue light is scattered more easily than red light, the sun takes on a reddish hue when it is observed through a thick atmosphere, as during a sunrise or sunset. Additional particulate matter in the sky can scatter different colors at different angles creating colorful glowing skies at dusk and dawn. Scattering off of ice crystals and other particles in the atmosphere are responsible for halo (optical phenomenon), afterglows, Corona (meteorology), rays of sunlight, and sun dogs. The variation in these kinds of phenomena is due to different particle sizes and geometries.[4] Mirages are another sort of optical phenomena due to variations in the refraction of light through the atmosphere. Other dramatic optical phenomena associated with this include the Novaya Zemlya effect where the sun appears to rise earlier than predicted with a distorted shape. A spectacular form of refraction occurs with a temperature inversion called the Fata Morgana (mirage) where objects on the horizon or even beyond the horizon, such as islands, cliffs, ships or icebergs, appear elongated and elevated, like "fairy tale castles".[5] Rainbows are the result of a combination of optical effects: total internal reflection and dispersion of light in raindrops. A single reflection off the backs of an array of raindrops produces a coherent rainbow with an angular size on the sky that ranges from 40 to 42 degrees with red on the outside. Double rainbows are produced by two internal reflections with angular size of 50.5 to 54 degrees with violet on the outside. Because rainbows must be seen with the sun 180 degrees away from the center of the rainbow, rainbows are more prominent the closer the sun is to the horizon. ## References 1. ↑ 2. ↑ Rajiv Ramaswami and Kumar N. Sivarajan, Optical Networks: A Practical Perspective (Academic Press: London 1998). 3. ↑ Brillouin, Léon. Wave Propagation and Group Velocity. Academic Press Inc., New York (1960) 4. ↑ Ahrens,C. Donald, Meteorology Today: an introduction to weather, climate, and the environment, West Publishing Company,1994 5. ↑ • M. Born and E.Wolf. Principles of Optics (7th ed.). Pergamon Press., 1999 • E. Hecht. Optics (4th ed.). Pearson Education. 2001. isbn=0805385665 • R. A. Serway and J. W. Jewett. Physics for Scientists and Engineers (6th ed.). Brooks/Cole. 2004. isbn=0534408427 • P. Tipler. Physics for Scientists and Engineers: Electricity, Magnetism, Light, and Elementary Modern Physics (5th ed.). W. H. Freeman. 2004. isbn=0716708108 • S. G. Lipson. Optical Physics (3rd ed.). Cambridge University Press. 1995. isbn=0521436311 ## Read more • Formulas • The Tyndall effect, also known as Tyndall scattering, is the scattering of light by colloidal... Tyndall effect • Rayleigh scattering (named after the English physicist Lord Rayleigh) is the elastic scattering... Rayleigh scattering # Photos Add a Photo 1,165photos on this wiki • by BiomedGuy 2013-02-26T00:32:14Z Posted in Radiation Safety • by BiomedGuy 2013-02-10T00:57:16Z • by BiomedGuy 2013-02-10T00:55:59Z • by BiomedGuy 2013-02-10T00:54:57Z • by BiomedGuy 2013-02-10T00:53:55Z • by BiomedGuy 2013-02-10T00:53:14Z • by BiomedGuy 2013-02-10T00:51:45Z • by BiomedGuy 2013-02-02T00:52:21Z • by BiomedGuy 2013-01-27T19:09:40Z Posted in Main Page • by BiomedGuy 2013-01-27T19:08:35Z Posted in Main Page • by BiomedGuy 2013-01-27T19:07:48Z Posted in Main Page • See all photos See all photos >

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 13, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.933448076248169, "perplexity_flag": "middle"}

http://mathhelpforum.com/discrete-math/120872-not-understanding.html

Thread: 1. not understanding have to do a couple problems for homework and im not understanding exactly what is going on and why its the answer 27. Let m, n, and d be integers. Show that if d \ m and d \ n, then d\ (m - n). my professor gave us the answer to this but i do not understand why it works answer: m = dq1 n = dq2 m - n = dq1 - dq2 = d(q1 - q2) therefore d \ (m - n) these are the other problems i have to do 28. Let m, n, and d be integers. Show that if d \ m, then d \ mn. 31. Let a, b, and c be integers. Show that if a \ b and b \ c, then a \ c. 33. Give an example of consecutive primes p1 = 2, p2 ..., pn where p1p2... pn + 1 is not prime. any help would be much appreciated 2. Originally Posted by pooshipple have to do a couple problems for homework and im not understanding exactly what is going on and why its the answer 27. Let m, n, and d be integers. Show that if d \ m and d \ n, then d\ (m - n). If $d\mid m,d\mid n$ then $m=dz,n=dz'$ for some $z,z'\in\mathbb{Z}$. Clearly then $m-n=dz-dz'=d\left(z-z'\right)$ and since $z,z'\in\mathbb{Z}\implies z-z'\in\mathbb{Z}$ the conclusion follows. my professor gave us the answer to this but i do not understand why it works answer: m = dq1 n = dq2 m - n = dq1 - dq2 = d(q1 - q2) therefore d \ (m - n) Sorry, I just noticed this. What it means to say that $d\mid n$ is that $d$ divides $n$ or that $\frac{n}{d}\in\mathbb{Z}$. Does that make a little more sense? So if $m-n=d\left(z-z'\right)$ then $\frac{m-n}{d}=\frac{d\left(z-z'\right)}{d}=z-z'\in\mathbb{Z}$ these are the other problems i have to do 28. Let m, n, and d be integers. Show that if d \ m, then d \ mn. $d\mid m\implies m=dz$ so then $mn=dzn$ or $\frac{mn}{d}=zn$ and since the integers are closed under multiplication the conclusion follows. 31. Let a, b, and c be integers. Show that if a \ b and b \ c, then a \ c. These all can be done the same. $a\mid b\implies b=za$ and $b\mid c\implies c=bz'$. So then $c=bz'=\left(az\right)z'$ and the conclusion follows. 33. Give an example of consecutive primes p1 = 2, p2 ..., pn where p1p2... pn + 1 is not prime. What does "consecutive" mean? That $p_2=p_1+1$ or that is merely the next prime in the list of primes? I'm assuming the latter. What do you think?

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 18, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9095651507377625, "perplexity_flag": "middle"}

http://mathhelpforum.com/differential-equations/173762-tricky-equation-first-order-print.html

# A tricky equation of the first order Printable View • March 7th 2011, 10:29 AM getsallad A tricky equation of the first order My teacher gave me a "bonus" assignment the other day, and I've been racking my brain trying to solve it. Here it is: $xy'(ln(\frac{x}{y}) +1) = y(ln(\frac{x}{y}) - 1)$ So far I've figured that I should substitute $ln(\frac{x}{y})$ for $u(x)$, thus giving me $y = \frac{x}{e^{u(x)}}$, but from there I'm kind of lost. Any ideas? • March 7th 2011, 10:35 AM topsquark Quote: Originally Posted by getsallad My teacher gave me a "bonus" assignment the other day, and I've been racking my brain trying to solve it. Here it is: $xy'(ln(\frac{x}{y}) +1) = y(ln(\frac{x}{y}) - 1)$ So far I've figured that I should substitute $ln(\frac{x}{y})$ for $u(x)$, thus giving me $y = \frac{x}{e^{u(x)}}$, but from there I'm kind of lost. Any ideas? I haven't done anything with it, but note that if you divide both sides by x you have an equation in y/x... -Dan • March 7th 2011, 10:41 AM getsallad I'm sorry, my knowledge of differential equations in itself is quite shaky, but how would that help me? If I get what you're saying, is it that I get $y'(ln(\frac{x}{y}) + 1) = \frac{x}{y}(ln(\frac{x}{y}) - 1)$? If that's the case, are you suggesting I could substitute u(x) for $\frac{x}{y}$ instead? • March 7th 2011, 10:53 AM Ackbeet The substitution $u = x/y$ does render the equation separable (the original DE is homogeneous). The problem is, the resulting integral in $u$ is horrendous. I suppose you could say, at that point, that you've "reduced the DE to quadratures", but if your professor is looking for a closed-form solution, that won't do. The substitution $u=\ln(x/y)$ is much better. You have to translate the DE over to the $u$ domain: $u'=\dfrac{y}{x}\cdot\dfrac{y-xy'}{y^{2}}=\dfrac{y-xy'}{xy}.$ Solve this for $y',$ and plug everything into the DE. A few things should cancel, leaving you with a much nicer separable equation. • March 7th 2011, 11:00 AM Ackbeet Very cute problem, by the way! Thanks for posting! • March 7th 2011, 11:05 AM getsallad Thanks for the help. I'm trying it out right now, doing all the steps myself to see if I get it. • March 7th 2011, 11:53 AM topsquark I just did the integral using Akbeet's method as well as using the homogeneous method and I'd rate them to be about the same level of difficulty. For the "connoisseurs" out there, the homogeneous solution includes a nice little integral: $\displaystyle \int \frac{ln(u)~du}{u}$ which I haven't seen done in a long time. (it's not particularly hard to do, just a nice little piece of work.) -Dan • March 7th 2011, 12:04 PM getsallad OK, so I've gotten somewhere. Now I'm stuck on the actual separable differential equation. I have $\frac{2}{u + 1} = xu'$ which leads me to $\frac{2}{u + 1} du = x dx$ but that approach seems to give me a completely different solution from what Wolfram Alpha gives me when I feed it the equation. So I figure I must be doing something wrong, since Wolfram Alpha's solution also seems more in tune with the final solution of the problem. • March 7th 2011, 12:06 PM Ackbeet • March 7th 2011, 01:23 PM getsallad OK, so I've hit one last roadblock and it's mighty frustrating. In the end I get $\frac{u^2}{2} + u = 2ln(x) + c$ and Wolfram Alpha gives the solution $\frac{1}{2}*ln^2(\frac{x}{y}) - ln(\frac{x}{y}) = 2ln(x) + c$ Which is basically the same thing with - instead of +. Should I show you step by step what I am doing? This is driving me nuts. • March 7th 2011, 01:30 PM Ackbeet No, WolframAlpha is not giving you that solution, it's giving you this solution. Notice that the x and y in the WolframAlpha solution are flipped from what you have. Because the left-most logarithm is squared, the change in sign goes unnoticed. That, of course, doesn't happen with the second term. I'm saying your solution is correct. • March 7th 2011, 01:32 PM Ackbeet Incidentally, you could, if you wanted to, use the quadratic formula on log(x/y) and then solve for y. You get a multi-valued "function" then, but it is an explicit formula for y, which is nice. • March 7th 2011, 01:39 PM getsallad Ah, can't believe I missed that. Thanks a million for the help on this one! I'll be sure to include the explicit formula for y just to be on the safe side. You've been most helpful! =) • March 7th 2011, 03:43 PM Ackbeet You're very welcome! Have a good one! All times are GMT -8. The time now is 03:54 PM.

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 21, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9576013684272766, "perplexity_flag": "middle"}

http://psychology.wikia.com/wiki/Least_square

Least squares Talk0 31,725pages on this wiki Redirected from Least square Assessment | Biopsychology | Comparative | Cognitive | Developmental | Language | Individual differences | Personality | Philosophy | Social | Methods | Statistics | Clinical | Educational | Industrial | Professional items | World psychology | Statistics: Scientific method · Research methods · Experimental design · Undergraduate statistics courses · Statistical tests · Game theory · Decision theory In regression analysis, least squares, also known as ordinary least squares analysis, is a method for linear regression that determines the values of unknown quantities in a statistical model by minimizing the sum of the squared residuals (the difference between the predicted and observed values). This method was first described by Carl Friedrich Gauss around 1794.[1] Today, this method is available in most statistical software packages. The least-squares approach to regression analysis has been shown to be optimal in the sense that it satisfies the Gauss-Markov theorem. A related method is the least mean squares (LMS) method. It occurs when the number of measured data is 1 and the gradient descent method is used to minimize the squared residual. LMS is known to minimize the expectation of the squared residual, with the smallest number of operations per iteration). However, it requires a large number of iterations to converge. Many other types of optimization problems can be expressed in a least squares form, by either minimizing energy or maximizing entropy. History Context The method of least squares grew out of the fields of astronomy and geodesy as scientists and mathematicians sought to provide solutions to the challenges of navigating the Earth's oceans during the Age of Exploration. The accurate description of the behavior of celestial bodies was key to enabling ships to sail in open seas where before sailors had to rely on land sightings to determine the positions of their ships. The method was the culmination of several advances that took place during the course of the eighteenth century[2]: • The combination of different observations taken under the same conditions as opposed to simply trying one's best to observe and record a single observation accurately. This approach was notably used by Tobias Mayer while studying the librations of the moon. • The combination of different observations as being the best estimate of the true value; errors decrease with aggregation rather than increase, perhaps first expressed by Roger Cotes. • The combination of different observations taken under different conditions as notably performed by Roger Joseph Boscovich in his work on the shape of the earth and Pierre-Simon Laplace in his work in explaining the differences in motion of Jupiter and Saturn. • The development of a criterion that can be evaluated to determine when the solution with the minimum error has been achieved, developed by Laplace in his Method of Situation. The method itself Carl Friedrich Gauss is credited with developing the fundamentals of the basis for least-squares analysis in 1795 at the age of eighteen. An early demonstration of the strength of Gauss's method came when it was used to predict the future location of the newly discovered asteroid Ceres. On January 1st, 1801, the Italian astronomer Giuseppe Piazzi discovered Ceres and was able to track its path for 40 days before it was lost in the glare of the sun. Based on this data, it was desired to determine the location of Ceres after it emerged from behind the sun without solving the complicated Kepler's nonlinear equations of planetary motion. The only predictions that successfully allowed Hungarian astronomer Franz Xaver von Zach to relocate Ceres were those performed by the 24-year-old Gauss using least-squares analysis. Gauss did not publish the method until 1809, when it appeared in volume two of his work on celestial mechanics, Theoria Motus Corporum Coelestium in sectionibus conicis solem ambientium. In 1829, Gauss was able to state that the least-squares approach to regression analysis is optimal in the sense that in a linear model where the errors have a mean of zero, are uncorrelated, and have equal variances, the best linear unbiased estimators of the coefficients is the least-squares estimators. This result is known as the Gauss-Markov theorem. The idea of least-squares analysis was also independently formulated by the Frenchman Adrien-Marie Legendre in 1805 and the American Robert Adrain in 1808. Problem statement The objective consists of adjusting a model function to best fit a data set. The chosen model function has adjustable parameters. The data set consist of n points $(y_i,\vec{x}_i)$ with $i = 1, 2,\dots, n$. The model function has the form $y=f(\vec{x},\vec{a})$, where $y$ is the dependent variable, $\vec{x}$ are the independent variables, and $\vec{a}$ are the model adjustable parameters. We wish to find the parameter values such that the model best fits the data according to a defined error criterion. The least sum square method minimizes the sum square error equation $S = \sum_{i=1}^n (y_i - f(\vec{x}_i,\vec{a}))^2$ with respect to the adjustable parameters $\vec{a}$. For an example, the data is height measurements over a surface. We choose to model the data by a plane with parameters for plane mean height, plane tip angle, and plane tilt angle. The model equation is then $y = f ( x_1, x_2 ) = a_1 + a_2 x_1 + a_3 x_2$, the independent variables are $\vec{x}=(x_1,x_2)$, and the adjustable parameters are $\vec{a}=(a_1,a_2,a_3)$. Solving the least squares problem Least square optimization problems can be divided into linear and non-linear problems. The linear problem has a closed form solution. The optimization problem is said to be a linear optimization problem if the first order partial derivatives of S with respect to the parameters $\vec{a}$ results in a set of equations that is linear in the parameter variables. The general, non-linear, unconstrained optimization problem has no closed form solution. In this case recursive methods, such as Newton's method, combined with the gradient descent method, or specialized methods for least squares analysis, such as the Gauss-Newton algorithm or the Levenberg-Marquardt algorithm can be used. Least squares and regression analysis In regression analysis, one replaces the relation $f(x_i)\approx y_i$ by $f(x_i) = y_i + \varepsilon_i,$ where the noise term ε is a random variable with mean zero. Note that we are assuming that the $x$ values are exact, and all the errors are in the $y$ values. Again, we distinguish between linear regression, in which case the function f is linear in the parameters to be determined (e.g., f(x) = ax2 + bx + c), and nonlinear regression. As before, linear regression is much simpler than nonlinear regression. (It is tempting to think that the reason for the name linear regression is that the graph of the function f(x) = ax + b is a line. But fitting a curve like f(x) = ax2 + bx + c when estimating a, b, and c by least squares, is an instance of linear regression because the vector of least-square estimates of a, b, and c is a linear transformation of the vector whose components are f(xi) + εi. Parameter estimates By recognizing that the $y_i = \alpha + \beta x_i + \varepsilon_i$ regression model is a system of linear equations we can express the model using data matrix X, target vector Y and parameter vector $\delta$. The ith row of X and Y will contain the x and y value for the ith data sample. Then the model can be written as $\begin{bmatrix} y_1\\ y_2\\ \vdots\\ y_n \end{bmatrix}= \begin{bmatrix} 1 & x_1\\ 1 & x_2\\ \vdots & \vdots\\ 1 & x_n \end{bmatrix} \begin{bmatrix} \alpha \\ \beta \end{bmatrix} + \begin{bmatrix} \varepsilon_1\\ \varepsilon_2\\ \vdots\\ \varepsilon_n \end{bmatrix}$ which when using pure matrix notation becomes $Y = X \delta + \varepsilon \,$ where ε is normally distributed with expected value 0 (i.e., a column vector of 0s) and variance σ2 In, where In is the n×n identity matrix. The least-squares estimator for $\delta$ is $\widehat{\delta} = (X^T X)^{-1}\; X^T Y \,$ (where XT is the transpose of X) and the sum of squares of residuals is $Y^T (I_n - X (X^T X)^{-1} X^T)\, Y.$ One of the properties of least-squares is that the matrix $X\widehat{\delta}$ is the orthogonal projection of Y onto the column space of X. The fact that the matrix X(XTX)−1XT is a symmetric idempotent matrix is incessantly relied on in proofs of theorems. The linearity of $\widehat{\delta}$ as a function of the vector Y, expressed above by saying $\widehat{\delta} = (X^TX)^{-1}X^TY,\,$ is the reason why this is called "linear" regression. Nonlinear regression uses nonlinear methods of estimation. The matrix In − X (XT X)−1 XT that appears above is a symmetric idempotent matrix of rank n − 2. Here is an example of the use of that fact in the theory of linear regression. The finite-dimensional spectral theorem of linear algebra says that any real symmetric matrix M can be diagonalized by an orthogonal matrix G, i.e., the matrix G′MG is a diagonal matrix. If the matrix M is also idempotent, then the diagonal entries in G′MG must be idempotent numbers. Only two real numbers are idempotent: 0 and 1. So In − X(XTX) -1XT, after diagonalization, has n − 2 1s and two 0s on the diagonal. That is most of the work in showing that the sum of squares of residuals has a chi-square distribution with n−2 degrees of freedom. Regression parameters can also be estimated by Bayesian methods. This has the advantages that • confidence intervals can be produced for parameter estimates without the use of asymptotic approximations, • prior information can be incorporated into the analysis. Suppose that in the linear regression $y = \alpha + \beta x + \varepsilon \,$ we know from domain knowledge that alpha can only take one of the values {−1, +1} but we do not know which. We can build this information into the analysis by choosing a prior for alpha which is a discrete distribution with a probability of 0.5 on −1 and 0.5 on +1. The posterior for alpha will also be a discrete distribution on {−1, +1}, but the probability weights will change to reflect the evidence from the data. In modern computer applications, the actual value of $\beta$ is calculated using the QR decomposition or slightly more robust methods when $X^TX$ is near singular. The code for the MATLAB backslash function, "`\`", is an excellent example of a robust method. Summarizing the data We sum the observations, the squares of the Xs and the products XY to obtain the following quantities. $S_X = x_1 + x_2 + \cdots + x_n \,$ $S_Y = y_1 + y_2 + \cdots + y_n \,$ $S_{XX} = x_1^2 + x_2^2 + \cdots + x_n^2 \,$ $S_{XY} = x_1 y_1 + x_2 y_2 + \cdots + x_n y_n. \,$ Estimating beta (the slope) We use the summary statistics above to calculate $\widehat\beta$, the estimate of β. $\widehat\beta = {n S_{XY} - S_X S_Y \over n S_{XX} - S_X S_X}. \,$ Estimating alpha (the intercept) We use the estimate of β and the other statistics to estimate α by: $\widehat\alpha = {S_Y - \widehat\beta S_X \over n}. \,$ A consequence of this estimate is that the regression line will always pass through the "center" $(\bar{x},\bar{y}) = (S_X/n, S_Y/n)$. Limitations Least squares estimation for linear models is notoriously non-robust to outliers. If the distribution of the outliers is skewed, the estimates can be biased. In the presence of any outliers, the least squares estimates are inefficient and can be extremely slow. When outliers occur in the data, methods of robust regression are more appropriate. References 1. ↑ Bretscher, Otto (1995). Linear Algebra With Applications, 3rd ed., Upper Saddle River NJ: Prentice Hall. 2. ↑ Stigler, Stephen M. (1986). The History of Statistics: The Measurement of Uncertainty Before 1900, Cambridge, MA: Belknap Press of Harvard University Press. Photos Add a Photo 6,465photos on this wiki • by Dr9855 2013-05-14T02:10:22Z • by PARANOiA 12 2013-05-11T19:25:04Z Posted in more... • by Addyrocker 2013-04-04T18:59:14Z • by Psymba 2013-03-24T20:27:47Z Posted in Mike Abrams • by Omaspiter 2013-03-14T09:55:55Z • by Omaspiter 2013-03-14T09:28:22Z • by Bigkellyna 2013-03-14T04:00:48Z Posted in User talk:Bigkellyna • by Preggo 2013-02-15T05:10:37Z • by Preggo 2013-02-15T05:10:17Z • by Preggo 2013-02-15T05:09:48Z • by Preggo 2013-02-15T05:09:35Z • See all photos See all photos >

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 37, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9054163694381714, "perplexity_flag": "head"}

http://mathoverflow.net/questions/43319/do-convex-and-decreasing-functions-preserve-the-semimartingale-property/43389

## Do convex and decreasing functions preserve the semimartingale property? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Some time ago I spent a lot of effort trying to show that the semimartingale property is preserved by certain functions. Specifically, that a convex function of a semimartingale and decreasing function of time is itself a semimartingale. This was needed for a result which I was trying to prove (more details below) and eventually managed to work around this issue, but it was not easy. For twice continuously differentiable functions this is an immediate consequence of Ito's lemma, but this cannot be applied in the general case. After failing at this task, I also spent a considerable amount of time trying to construct a counterexample, also with no success. So, my question is as follows. 1) Let ƒ: ℝ+×ℝ → ℝ be such that ƒ(t,x) is convex in x and continuous and decreasing in t. Then, for any semimartingale X, is ƒ(t,Xt) necessarily a semimartingale? Actually, it can be assumed here that X is both continuous and a martingale which, with some work, would imply the general case. As it turns out, this can be phrased purely as a real-analysis question. 2) Let ƒ: ℝ+×ℝ → ℝ be such that ƒ(t,x) is convex in x and decreasing in t. Can we write ƒ = g − h where g(t,x) and h(t,x) are both convex in x and increasing in t ? Stated like this, maybe someone with a good knowledge of convex functions would be able to answer the question one way or the other. For ƒ(t,x) convex in x and increasing in t then approximation by smooth functions and applying Ito's lemma allows us to express ƒ(t,Xt) as the sum of a stochastic integral and an increasing process $$f(t,X_t)=\int_0^t\frac{\partial f}{\partial x}(s,X_s)\,dX_s+V_t,\qquad(*)$$ so it is a semimartingale. If, instead, ƒ is decreasing in t, then an affirmative answer to question 2 will reduce it to the easier case where it is increasing in t, also giving a positive answer to the first question. Explaining why question 1 implies 2 is a bit trickier. If 2 was false, then it would be possible to construct martingales X such that the decomposition (*) holds where the variation of V explodes at some positive time. This problem arose while I was trying to prove the following: is a continuous martingale uniquely determined by its one dimensional marginals? For arbitrary continuous martingales this is false, but is known to be true for diffusions dXt = σ(t,Xt) dW for Brownian motion W and smooth parameter σ. The idea is to back out σ from the Kolmogorov forward equation. This is well-known in finance as the local volatility model. However, I was trying to show rather more than this. All continuous and strong Markov martingales are uniquely determined by the one dimensional marginals. I was able to prove this, and the relation between the marginals and joint distributions of the martingale has many nice properties (I wrote a paper on this, submitted to the arXiv, but not published as I am still working on changes asked for by the referees). The method was to reformulate the Kolmogorov backward equation in terms of the marginals. This does use Ito's lemma, requiring twice differentiability, but can be circumvented with a bit of integration by parts as long as ƒ(t,Xt) is a semimartingale for the kinds of functions mentioned above. The question above arose from trying, and failing, to prove this. Without an answer to this question, the problem becomes much much harder, as many of the techniques from stochastic calculus can no longer be applied (and, approximating by semimartingales didn't seem to help either). The work around was very involved, part of which I published as a standalone paper here and the rest forms most of a paper I submitted to the arXiv here. Adding together those papers, it comes to maybe 50 pages of maths and a lot of effort to work around the question above. - ## 3 Answers I still have no idea what the answer to this question is. However, it is possible to attack the problem in several different ways, and there are various different (but logically equivalent) ways of stating it. I'll post some of these as an answer now, as it seems rather long to fit in the original statement of the question. Maybe one of the reformulations below will help lead to a resolution of the problem. This answer is already very long, and I've been trying to shorten it as much as I can. I can't see any way of giving proper proofs of all the statements below without making it a lot longer. So, I'll only give very few details of he proofs here. I will, however, list each of the equivalent formulations H1-H6 below in the logical order, so that each statement can be proven to be equivalent to the preceding one without too much work. Let's start from statement 2 of the original question, which I will refer to as Hypothesis (H1) for the purposes of this answer. Hypothesis (H1): Let ƒ: ℝ+×ℝ → ℝ be such that ƒ(t,x) is convex in x and decreasing in t. Then, ƒ = g − h where g(t,x) and h(t,x) are both convex in x and increasing in t . The decomposition in (H1) exists if and only if it exists locally. Letting I = [0,1] denote the unit interval, we get the following equivalent statement. Hypothesis (H2): Let ƒ: I2 → ℝ be such that ƒ(t,x) is convex and Lipschitz continuous in x and decreasing in t. Then, ƒ = g − h where g(t,x) and h(t,x) are convex in x and increasing in t. The truth of this statement remains unchanged if it is restricted to functions ƒ which are zero on the three edges I × {0}, I × {1}, {0} × I of the unit square. I'll use D to denote the set of such functions satisfying the conditions of (H2). Then, whenever the decomposition in (H2) exists, it is always possible to choose g, h to be zero on I × {0}, I × {1} and {1} × I. From now on, whenever the decomposition ƒ = g − h in (H2) is referred to, it will be assumed that g, h are chosen to satisfy these conditions. We can strengthen (H2) by also placing a uniform bound on the terms g, h in the decomposition. Here, ‖g‖ denotes the supremum norm and ƒx denotes the partial derivative. Hypothesis (H3): There is a constant K > 0 such that, for all ƒ ∈ D, the decomposition ƒ = g − h as in H2 exists and can be chosen such that ‖g‖,‖h‖ ≤ K‖ƒx‖. Statement (H3) is particularly convenient because of the following: to prove that (H3) holds, it is enough to look at a dense subset of functions in D. Taking limits of the decompositions would then extend to the result to all of D. So, it is enough to concentrate on, say, smooth functions or piecewise linear functions. Next, it is useful to choose the decomposition in (H2) to minimize ‖g‖ and ‖h‖. Lemma 1: Suppose that ƒ ∈ D and that the decomposition ƒ = g − h as in (H2) exists. Then, there is a unique maximal choice for g, h. That is, if ƒ = g1 − h1 is any other such decomposition then g ≥ g1 and h ≥ h1. I'll refer to the decomposition in Lemma 1 as the optimal decomposition. As it not clear that any such decomposition should exist, I'll briefly sketch the argument now. The idea is to discretize time, using a partition of the unit interval 0 = t0 < t1 < … < tr = 1. Denote the convex hull of a function u: I → ℝ by v = H(u), which is the maximum convex function v: I → ℝ bounding u from below, $$\begin{align} v(x) &= \sup\{w(x)\colon w\le u{\rm\ is\ convex}\}\\ &=\inf\{\left((b-x)u(a)+(x-a)u(b)\right)/(b-a)\colon a\le x\le b\}. \end{align}$$ The optimal decomposition in discrete time can be constructed as functions hk: I → ℝ, starting at the final time k = r and working backwards to k = 0, $$h_r(x) = 0,\ h_{k-1}={\rm H}\left(h_k+f(t_k,\cdot)-f(t_{k-1},\cdot)\right).$$ Interpolate this to be piecewise constant in time, defining h(t,x) = hk(x) for tk−1 < t ≤ tk. Then, h(t,x) and g ≡ ƒ + h are convex in x and increasing in time t, restricting to times in the partition. Lemma 2: Suppose that ƒ ∈ D and let 0 = tn,0 < tn,1 <…< tn,rn = 1 be a sequence of partitions of the unit interval. For each n let hn(t,x) be the function corresponding to the partition and piecewise constant in t, as constructed above. We also suppose that the partitions have mesh going to zero and eventually include all times at which ƒ is discontinuous. Then one of the following holds. • ƒ decomposes as in (H2) and hn(t,x) → h(t,x) pointwise on I2, where ƒ = g − h is the optimal decomposition. • ƒ does not have a decomposition as in (H2) and hn(0,x) → -∞ for all 0 < x < 1. The idea is that, if hn has any limit point h, then h(t,x) and g = ƒ + h will be convex in x and increasing in t, giving the decomposition required by (H2). Furthermore, by construction, if ƒ = g' − h' is any other such decomposition, then hn ≥ h' at times in the partition, so h ≥ h'. This shows that the decomposition is optimal and, as the optimal decomposition is unique, all limit limit points of hn are the same, so hn → h. The only alternative is that hn has no limit points, in which case the second statement of the Lemma holds. Using this construction of the optimal decomposition, (H3) can be shown to be equivalent to the following. Hypothesis (H4): There is a constant K > 0 such that, for all smooth functions ƒ,g: I2 → ℝ with ‖ƒ‖, ‖g‖, ‖ƒx‖, ‖gx‖ bounded by 1 and ƒ(t,x), g(t,x) convex in x and respectively decreasing and increasing in t, then, $$\int_0^1\int_0^1 f_{xx}g_t\,dxdt \le K.$$ As this statement is quite different from the preceding ones, I'd better give some explanation now. The idea is to use integration by parts, $$\begin{align} \int_0^1\int_0^1(f_{xx}g_t+g_{xx}f_t)\,dxdt &= \left[\int_0^1(f_xg_t+g_xf_t)\,dt\right]_{x=0}^1-\left[\int_0^1f_xg_x\,dx\right]_{t=0}^1\\ &\le 6(\Vert f_x\Vert \Vert g\Vert + \Vert g_x\Vert \Vert f\Vert). \end{align}$$ If ƒ and g are increasing in time then the terms on the left hand side are both positive, so we get bounds for the integrals of ƒxxgt and gxxƒt individually. Hypothesis (H3) extends this to the case where ƒ is decreasing in time, implying (H4). Conversely, suppose that (H4) holds. Letting f = g − h be the decomposition computed along a partition as described above, we can use the fact that the convex hull v = H(u) of a function u satisfies vxx(u −v) = 0 to get the equality (hk−1)xx(hk − hk−1 + ƒ(tk,.) − ƒ(tk−1,.)) = 0. This leads to the following inequalities, $$\begin{align} \frac12\Vert h\Vert^2&\le\frac12\int_0^1 h_x(0,x)^2\,dx\le\sum_{k=1}^r \int_0^1(h_{k-1})_{xx}(h_{k-1}-h_k)\,dx\\ &=-\sum_{k=1}^r\int_0^1 (h_{k-1})_{xx}(f_k-f_{k-1})\,dx. \end{align}$$ Hypothesis (H4) can be used to bound the final term, showing that h cannot diverge as the mesh of the partition goes to zero, so we get convergence to an optimal decomposition satisfying a bound as in (H3). The hypothesis can now be formulated as a statement about martingales. Hypothesis (H5): There is a constant K > 0 such that, for all ƒ ∈ D and martingales 0 ≤ Xt ≤ 1, ƒ(t,Xt) decomposes as M + V where M is a martingale and V has variation $$\mathbb{E}\left[\int_0^1\,\vert dV\vert\right]\le K\Vert f_x\Vert.$$ The idea is that g(t,x) ≡ E[(Xt − x)+] is convex in x and increasing in t. In the case where ƒ, g are smooth and X is a continuous martingale, Ito's formula can be used to split ƒ(t,Xt) into a martingale term plus the sum of the increasing process ½∫ƒxx(t,X)d[X] and the decreasing process ∫ƒt(t,X)dt, which have expectations ∫∫ƒxxgtdxdt and ∫∫ƒtgxxdxdt respectively. Finally, by adding a randomly occurring jump term to any semimartingale, and with a bit more work, it is possible to reduce this to the martingale case. This gives the statement asked in the original question. Hypothesis (H6): Let ƒ:ℝ+×ℝ → ℝ be such that ƒ(t,x) is convex in x and continuous and decreasing in t. Then, for any semimartingale X, ƒ(t,Xt) is a semimartingale. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. What follows is too long to fit in a comment. Some thoughts on the second problem. Let us consider the problem for $t$ taking values in the compact interval $[0,T]$. The general case perhaps can be approximated by this case. If the $t$ variable were discrete and finite the problem would be: we have a sequence of functions $f_1 \ge f_2 \ge f_3 \cdots \ge f_n$ (let us make the simplifying assumption that $f_i$ are all positive; otherwise, replace $f_i$ with $f_i +C$ where $C$ is $C = -\min_{i,x} f_i(x)$), can one find convex and increasing ${h_i}$ and ${g_i}$ such that $f_i = h_i - g_i$? The answer to this question is obviously yes. For example, $h_i = \sum_{j\le i} f_i$ and $g_i = \sum_{j < i} f_i$ is one possible solution. The problem with this solution when $t$ is continuous is that the $h_i$ and $g_i$ would explode as the discretization of $t$ is refined. Thus, one needs to choose $h$ and $g$ in a way that they increase slowly. How slowly can this increase be? We can start with $h_1 = f_1$ and $g_1 = 0$. $h_2$ and $g_2$ will be of the following form: $$h_2 = h_1 + S_2 = f_1 + S_2,$$ and $$g_2 = g_1 + R_2 = R_2$$ such that: 1) $S_2, R_2 \ge 0$, 2) $h_2 = f_1 + S_2$, $g_2 = R_2$ are convex and 3) $h_2 - g_2 = f_1 + S_2 - R_2 = f_2$. The last of these is equivalent to $R_2 = f_1 - f_2 + S_2$. Thus, what we are looking for is a function $S_2$ satisfying the above conditions. There will be many such $S_2$, the goal is to choose $S_2$ in a minimal way so that $h_i$ and $g_i$ grow slowly.The best would be: $S_2 = 0$, which is indeed a solution when $f_1 - f_i$ is convex. If $f_0 - f_t$ is convex for all $t \in [0,T]$ then this solution directly generalizes to the original continuous time problem (i.e., if $f_0-f_t$ is convex then $h(t,x) = f(0,x)$ and $g(t,x) = f(0,x) - f(t,x)$ is a solution). Now, let us consider the case when $f(t,x)$ is such that $\frac{\partial^2}{\partial x^2}f(t,x)$ is continuous in $(t,x)$ and $x$ also takes values in a compact set $K$. The following type of argument quickly comes to mind. Define $$\tau \doteq \{t: \exists g, h:[0,t]\times K\rightarrow {\mathbb R}, f = h-g, h,g \text{ convex in } x \text{ and increasing in } t \}.$$ $0$ is clearly a member of this set. One can perhaps argue that $\sup \tau$ must be $T$ as follows. If $t_0 = \sup \tau < T$ then one can slightly modify $h(t_0,x)$ and $g(t_0,x)$ to obtain functions $h(t_0+\delta,x)$ and $g(t_0+\delta,x)$ whose difference will be $f(t_0+\delta,x)$ [the slight modification is made possible by the fact that the second derivative of $f$ with respect to $x$ barely changes between $t_0$ and $t_0 + \delta$]. - Thanks for looking at this. I agree that discretizing it is a good first step (it also helps to restrict to a compact interval for x). However, it works out a bit better if you solve for g,h by starting at the last time and working backwards. Doing this, you can show that it converges to the continuous time decomposition if and only if there is a continuous time decomposition. I'm going to post an answer myself, when I have time, describing a few different ways to reformulate this problem. – George Lowther Oct 30 2010 at 21:38 Also, if f is either twice differentiable in x or once in t, then it will have the required decomposition. The problem appears with nondifferentiable functions. You can always approximate by smooth functions but, then, it is possible that the decompositions of these smooth approximations diverge when you take the limit. – George Lowther Oct 30 2010 at 21:42 You are welcome; your questions above and your related results are very interesting, thanks for sharing. What makes the problem difficult seems to be that there are many things that influence how $g$ and $h$ are to evolve in time and there seems to be many choices. The optimization problem in the discrete formulation (the choice of $S_2$) referred to in my answer seems to be nontrivial (I think it can be formulated as a control problem). – has2 Oct 31 2010 at 7:45 Another thing that seem relevant is the following: the region where $x \rightarrow f(x,0) - f(x,t)$ is not convex must be small, because otherwise $f(x,0)$ could not dominate $f(x,t)$. – has2 Oct 31 2010 at 7:50 Hi, Maybe I haven't fully understood your question but I think there is a counter-example (in a simpler context) given in Protter's book on Stochastic Integration in the form of a theorem. It's theorem 71, chapter IV, section 7 (Local Times). It's claimed there that for any $\alpha \in(0,1/2)$ and any continuous local martingale $X_t$, then $Y_t=|X_t|^\alpha$ is not a semi-martingale unless it is a null process. The proof is based on local time and Tanaka-Meyer's theorem. Hope it helps, (but if not I would appreciate that you explain where I went wrong). Best regards. - The function $f(x)=\vert x\vert^\alpha$ is not convex for $\alpha < 1$. Convex functions do preserve the semimartingale property (see the Ito-Tanaka formula for example). It is only when you have convex in space and decreasing in time that it gets difficult. – George Lowther Oct 24 2010 at 15:56 Actually, I think what I just called the Ito-Tanaka formula is the same thing as Tanaka-Meyer's theorem that you just mentioned. It is also called the Ito-Tanaka-Meyer formula. – George Lowther Oct 24 2010 at 15:59 Sorry to insit but $f$ is concave though (right ?),so taking $-f$ should do the trick. I must miss something about the convex function definition here. Regards – The Bridge Oct 24 2010 at 16:19 No. f is concave individually on the intervals $[0,\infty)$ and $(-\infty,0]$. It is neither convex nor concave on any open interval about zero. And that is the point of the result you refer to. In any case, a local martingale restricted to being either nonnegative everywhere or nonpositive everywhere must remain at 0 once it hits zero. It can only escape 0 if it can only take both positive and negative values, so the behaviour of f in a neighbourhood of 0 is what is important here. – George Lowther Oct 24 2010 at 16:29 One further point: I don't have Protter's book to hand right now but, if you look at the proof of the result you just quoted, I expect that it is writing the drift term of f(X) as $\int L^x_t f^{\prime\prime}(x)\,dx$ where $L^x_t$ is the local time at x. This explodes if $L^0>0$ in your example but, if f was either convex or concave, $f^{\prime\prime}$ would be locally integrable and you get that f(X) is indeed a semimartingale. – George Lowther Oct 24 2010 at 16:39 show 2 more comments

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 78, "mathjax_display_tex": 10, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9474689364433289, "perplexity_flag": "head"}

http://math.stackexchange.com/questions/189399/expected-number-of-balls-withdrawn-to-get-equal-numbers-of-black-and-white-balls/189413

# expected number of balls withdrawn to get equal numbers of black and white balls There are $n$ black balls and $n$ white balls in a bin. I withdraw the balls one at a time without replacement until I have an equal number of white and black balls. What is the expected number of balls that I have to withdraw? It appears that the answer should be $4^n\left/{2n \choose n}\right.$. So for $n = 3$ it would be: $$4^3\left/{6 \choose 3}\right. = \frac{64}{20} = \frac{16}{5}$$ I have verified the answers for $n = 2$ and $n = 3$, but I am not able to prove the general result. - Of course it is possible that it will never occur. There is a positive probability that you will exhaust say the white balls without ever getting the number of black equal to the number of white. So the expectation only makes sense conditional on being successful. – Michael Chernick Aug 31 '12 at 20:14 4 But after $2n$ drawing, the number balls of each color is bound to be equal. – Sasha Aug 31 '12 at 20:15 Since we have equal numbers of black and white balls, it will always occur. In the worst case, after I have drawn 2n balls, I will have equal numbers of white and black balls. If I have exhausted all the white balls, I would continue by drawing the remaining n black balls so that when I have drawn all 2n balls, I will have equality. In general, I will achieve equality after drawing 2k balls with 1 <= k <= n – keng5 Aug 31 '12 at 20:20 At the start, you have zero balls of each color. The expected number of draws to have equal numbers is zero. – Ross Millikan Aug 31 '12 at 20:33 1 Nice problem, by the way! – Byron Schmuland Aug 31 '12 at 21:02 show 1 more comment ## 3 Answers Your conjecture is true. The following argument is not elegant, but it works! The number of arrangements of $k$ white and $k$ black balls where the first equalization occurs at $2k$ is $2 C_{k-1}$, where $C_{k-1}={1\over k}{2(k-1)\choose k-1}$ is the $k-1$th Catalan number. The number of arrangements of $n$ white and $n$ black balls where the first equalization occurs at $2k$ is therefore $${2\over k}{2(k-1)\choose k-1} {2(n-k)\choose n-k}.\tag 1$$ All ${2n\choose n}$ arrangements are equally likely, so the chance that equalization first occurs at time $2k$ is $$\mathbb{P}(T=2k)= {1\over{2n\choose n}} {2\over k}{2(k-1)\choose k-1} {2(n-k)\choose n-k}.\tag 2$$ The expected time to equalization is therefore $$\mathbb{E}(T)={1\over{2n\choose n}} \sum_{k=1}^n 2k\, {2\over k}\,{2(k-1)\choose k-1} {2(n-k)\choose n-k}.\tag 3$$ Cancelling the $k$'s in (3) and using the known identity $$\sum_{k=1}^n {2(k-1)\choose k-1} {2(n-k)\choose n-k}=4^{n-1}\tag4$$ gives the result. - Sorry: you must have posted while I was writing. – Brian M. Scott Aug 31 '12 at 21:22 No problem. Additional explanations are always welcome. – Byron Schmuland Aug 31 '12 at 21:22 Suppose that the first ball drawn is white and that you stop on draw number $2k$. Then the $2k$-th ball drawn was black, and the $2k-2$ balls drawn in positions $2$ through $2k-1$ form a Dyck word of length $2k-2$. Conversely, any sequence of $2k-2$ white and black balls in which the number of black balls never exceeds the number of white balls can occupy those $2k-2$ positions. Thus, there are $C_{k-1}$ sequences of draws beginning with a white ball that terminate with draw $2k$. If we imagine continuing until the bin is empty, there are $\binom{2n-2k}{n-k}$ ways to complete the draw, for a total of $C_{k-1}\binom{2n-2k}{n-k}$ full draws that start with a white ball and first balance (at equal numbers of white and black balls) on draw $2k$. There is an equal number starting with a black ball, so $$2C_{k-1}\binom{2n-2k}{n-k}=\frac2k\binom{2k-2}{k-1}\binom{2n-2k}{n-k}$$ of the $\binom{2n}n$ possible full draws first balance on draw $2k$. The expected number of draws to the first balanced sample is therefore $$\binom{2n}n^{-1}\sum_{k=1}^n\frac2k\binom{2k-2}{k-1}\binom{2n-2k}{n-k}(2k)=4\binom{2n}n^{-1}\sum_{k=1}^n\binom{2k-2}{k-1}\binom{2n-2k}{n-k}\;,$$ and your conjecture is equivalent to $$4^{n-1}=\sum_{k=1}^n\binom{2k-2}{k-1}\binom{2n-2k}{n-k}=\sum_{k=0}^{n-1}\binom{2k}k\binom{2n-2k-2}{n-k-1}$$ or, after replacing $n-1$ by $n$, to $$4^n=\sum_{k=0}^n\binom{2k}k\binom{2n-2k}{n-k}\;.$$ You can find a proof of this identity in this question together with an outline of a combinatorial proof; this answer gives a full combinatorial proof. - Byron and Brian, Thanks for the proofs. – keng5 Sep 1 '12 at 14:28 I would start by defining $f(n,m), n \ge m$ as the expected number of draws starting from $n$ white balls and $m$ black balls with $n-m$ black balls in hand. Then we have $f(n,n)=1+f(n,n-1)$ $f(n,0)=n$ $f(n,n-1)=\frac n{2n-1}+\frac{n-1}{2n-1}(1+f(n,n-2))$ $f(n,m)=1+\frac m{n+m}f(n,m-1)+\frac n{n+m}f(n-1,m)$ where the cases are checked in this order and seek a solution. No guarantees that this will work. - – leonbloy Sep 1 '12 at 0:06

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 44, "mathjax_display_tex": 9, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9402660727500916, "perplexity_flag": "head"}

http://mathoverflow.net/questions/5761/change-rates-of-the-2nd-chebychevs-function/6064

## change rates of the 2nd Chebychev´s function ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $\psi(x):=\sum_{n\leq x}\Lambda(n)$ denotes the 2nd Chebyshev function, where $\Lambda$ stands for the von-Mangoldt function. Are there any known (and 'nice') estimates for the change rates $\psi(x+h)-\psi(x)$ for general or special $x$ and $h$? Thanks in advance, efq - ## 2 Answers There is the asymptotic estimate $\psi(x+h) - \psi(x) \sim h$ for $x^{7/12 + \epsilon} \leq h \leq x$, valid for any $\epsilon > 0$. This is due to M. N. Huxley, and dates to 1972. I am not aware of any better range for $h$ if you want asymptotic equality. But if you are satisfied with an order of magnitude result, you can have $c_1h \leq \psi(x+h) - \psi(x) \leq c_2h$ with $c_1$ and $c_2$ positive constants and $x^{\theta} \leq h \leq x$ for some $\theta$ slightly larger than $0.5$. I can't give references offhand, but you should be able to find such papers by searching on R. C. Baker, G. Harman, J. Pintz in Mathscinet. - Thanks a lot for you answer! I believe this order-of-magnitude-result may in fact work out for me. I´ll definitely have a look into the reference you mention. – ex falso quodlibet Nov 20 2009 at 19:09 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Also, I think Selberg's result assert that for "almost all x" we have $\psi(x + h) - \psi(x) \sim h$ for $h \asymp (\log x)^2$ this was shown to not be true "pointwise" by Mayer. - That is interesting. Is it known whether this null set is finite? – ex falso quodlibet Nov 20 2009 at 20:49 there is a sequence of x_k -> oo such that psi(x_k + (log x_k)^2) - psi(x_k) > (1+epsilon) (log x_k)^2 ... (I hope this answers your question). – maki Nov 20 2009 at 23:16 yes, of course! ignore my last comment. – ex falso quodlibet Nov 21 2009 at 23:19

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9134442210197449, "perplexity_flag": "middle"}

http://mathhelpforum.com/differential-geometry/122635-fractional-part-integration-print.html

# fractional part and integration Printable View • January 6th 2010, 08:27 AM Shanks fractional part and integration Let $\{x\}$ be the positive fractional part of real number $x$, for example, $\{\pi\}=0.1415926...$ . Prove that the set $S=\{\{nv\}:n\in Z^+\}$ is dense in [0,1] if $v$ is positive irrational number. And for any $f(x)\in C[0,1]$(the class of all real-valued continious function on [0,1]), and any positive irrational number $v$, we have $\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^{N}f(\{nv\} )=\int_0^1f(x)dx$. I know how to do this problem, but i can't write down the formal solution and some details completely. I need a complete and elaborate solution, thanks in advance! • January 6th 2010, 10:47 AM Laurent Quote: Originally Posted by Shanks I know how to do this problem, but i can't write down the formal solution and some details completely. I need a complete and elaborate solution, thanks in advance! You should tell us more precisely which details are lacking in your solution, it would spare some time. • January 7th 2010, 01:38 AM Shanks notice the fact that if $t\in S$, then the fractional part of the multiply of t is also in S. To prove S is dense in [0,1], it is suffice to prove that there is a decreasing subsequence in S such that it converges to 0. Or in another word, 0 is the inf limit of S. This can be easily proved by contradiction! To prove the limit is equal to the integration, we need the definition of Riemann Integral and the uniform continuity of f(x). that is, it is suffice to prove that for any $\delta >0$, there exist N, such that if k > N, then the longest distance of two consecutive( the oder of number, not the index n) among the previous k points is less than $\delta$, Or in another word, the distance of any two consecutive points of the previous k points is less than $\delta$. Although I understand what is actually going on, I can't express by word, can't write down the details here!(Doh) • January 8th 2010, 12:33 AM Laurent Quote: Originally Posted by Shanks To prove the limit is equal to the integration, we need the definition of Riemann Integral and the uniform continuity of f(x). that is, it is suffice to prove that for any $\delta >0$, there exist N, such that if k > N, then the longest distance of two consecutive( the oder of number, not the index n) among the previous k points is less than $\delta$, Or in another word, the distance of any two consecutive points of the previous k points is less than $\delta$. Although I understand what is actually going on, I can't express by word, can't write down the details here!(Doh) You need more than just that: not only need the points of the sequence to be close enough, but they need to be "uniformly distributed" (asymptotically) for the sum to converge to the integral. I give you a magic trick: first prove the limit when $f(x)=e^{2i\pi nx}$ for some non-zero integer $n$ (direct computation), then for trigonometric polynomials, i.e. linear combinations of the previous functions (direct consequence), and finally for general $f$ (using approximation by a trigonometric polynomial). This is the usual way to procede (not very intuitive, I admit), but there may be others. • January 8th 2010, 02:52 AM Shanks Quote: Originally Posted by Laurent You need more than just that: not only need the points of the sequence to be close enough, but they need to be "uniformly distributed" (asymptotically) for the sum to converge to the integral. I give you a magic trick: first prove the limit when $f(x)=e^{2i\pi nx}$ for some non-zero integer $n$ (direct computation), then for trigonometric polynomials, i.e. linear combinations of the previous functions (direct consequence), and finally for general $f$ (using approximation by a trigonometric polynomial). This is the usual way to procede (not very intuitive, I admit), but there may be others. I know that this is the way written in G.polya book. But Here I don't want to using approximation of continious function by trigonometric polynomial. I want to prove it by using the definition of Riemann integral. • January 16th 2010, 03:03 AM Shanks The Problem occurs again in Rudin's analysis book, therefore it is a impressive problem. All times are GMT -8. The time now is 12:49 AM.

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 21, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9343944787979126, "perplexity_flag": "head"}

http://math.stackexchange.com/questions/202879/prove-the-existence-of-a-c-infty-bounded-function-with-arbitrary-derivatives

# Prove the existence of a $C^\infty$ bounded function with arbitrary derivatives at a fixed point. (Borel?) I found this problem on an Italian forum and since then I struggled to solve it. The autor there claims it was proposed by Borel. At any rate the problem is as follows (Borel?) For any $\{a_n\}_{n=0}^{\infty}\subseteq \mathbb R,\: x_0\in\mathbb R\,$ and $\,\varepsilon>0$ does there exist a $C^\infty$ function $f\colon\mathbb R\to\mathbb R$ such that $$f^{(n)}(x_0)=a_n,\; \forall n\in\mathbb N\cup\{0\}\tag{1}$$ and moreover $$\left|f(x)-a_0\right|<\varepsilon,\;\forall x\in\mathbb R?$$ Clearly even reference about problem are welcomed, but I strongly encourage anybody to think about it because really, when I met it for the first time, I thought: "this is a wonderful problem". I do not have any clue towards the solution. I hope you will have fun with this. Cheers. - – user42754 Sep 26 '12 at 15:56 – user42754 Sep 26 '12 at 15:58 Thanks for the references – uforoboa Sep 27 '12 at 9:53 ## 1 Answer Here is a counterexample. For arbitrary interval $I\subset\mathbb{R}$ denote $M_k(f,I)=\max_{t\in I}|f^{(k)}(t)|$, then from exercise 14 Chapter 5 in Rudin's Principles of Mathematical analysis we know that $$M_1^2(f,I)\leq 4M_0(f,I)M_2(f,I)$$ for any interval $I$ and $f\in C^3(I)$ Let's take $a_1>2\sqrt{(|a_0|+\varepsilon)(|a_2|+\varepsilon)}$ and the rest whatever you want. Assume that there exist a smooth function $f$ with predescribed properties. Since $f\in C^\infty(\mathbb{R})$ then we can find neighbourhood $U$ of $x_0$ where $$M_2(f,U)<|f^{(2)}(x_0)|+\varepsilon=|a_2|+\varepsilon$$ By construction $|f(x)-a_0|<\varepsilon$ for all $x\in\mathbb{R}$, so $$M_0(f,U)<|a_0|+\varepsilon$$ Finally $$M_1^2(f,U)>|f'(x_0)|^2=a_1^2>4(|a_0|+\varepsilon)(|a_2|+\varepsilon)>4M_0(f,U)M_2(f,U)$$ Contradiction. - Good! This show that boundedness is restrictive in this case, as we should have $|a_n|\leq 4(|a_n|+\varepsilon)(|a_{n+1}|+\varepsilon)$ for all $n$. – Davide Giraudo Sep 26 '12 at 17:10 @DavideGiraudo, thanks! – Norbert Sep 26 '12 at 17:37 @Norbert I see your point. I was aware of that result you quoted however I never linked it to the solution of this problem. Thank you. – uforoboa Sep 27 '12 at 9:52

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 18, "mathjax_display_tex": 6, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9293622374534607, "perplexity_flag": "head"}

http://stats.stackexchange.com/questions/3438/calculating-percentile-of-normal-distribution

# Calculating percentile of normal distribution http://en.wikipedia.org/wiki/Binomial_proportion_confidence_interval#Agresti-Coull_Interval To get the Agresti-Coull Interval, one needs to calculate a percentile of the normal distribution, called $z$. How do I calculate tha percentile? Is there a ready-made function that does this in Wolfram Mathematica and/or Python/NumPy/SciPy? - 1 The integral expression in the "normal cdf I got exactly from Wiki" is unfortunately off by a factor of $1/\sqrt{\pi}$. There is no known exact formula for the normal cdf or its inverse using a finite number of terms involving standard functions ($\exp, \log, \sin \cos$ etc) but both the normal cdf and its inverse have been studied a lot and approximate formulas for both are programmed into many calculator, spreadsheets, not to mention statistical packages. I am not familiar with R but I would be astounded if it did not have what you are looking for built in already. – Dilip Sarwate Feb 14 '12 at 20:52 @DilipSarwate, it's fixed! I am doing this using inverse tranformation, also "not allowed" to use too much built in. It's for the sake of learning I suppose. – user1061210 Feb 14 '12 at 21:26 1 @Dilip: Not only is there no known exact formula, better yet, it is known that no such formula can exist! – cardinal Feb 14 '12 at 21:30 1 The Box-Muller method generates samples from a joint distribution of independent standard normal random variables. So histograms of the values generated will resemble standard normal distributions. But the Box-Muller method is not a method for computing values of $\Phi(x)$ except incidentally as in "I generated $10^4$ standard normal samples of which $8401$ has value $1$ or less, and so $\Phi(1) \approx 0.8401$, and $\Phi^{-1}(0.8401) \approx 1$. – Dilip Sarwate Feb 14 '12 at 22:29 1 I just chose $8401$ as an example of the kinds of numbers you might expect. $\Phi(1) = 0.8413\ldots$ and so if you generate $10^4$ samples of a standard normal distribution, you should expect close to $8413$ of the $10000$ samples to have value $\leq 1$. You are implementing the Box-Muller method correctly, but are not understanding the results that you are getting and are not relating them to the cdf etc. – Dilip Sarwate Feb 15 '12 at 0:27 show 12 more comments ## 5 Answers For Mathematica, `$VersionNumber > 5` you can use ````Quantile[NormalDistribution[mu,sigma], 100 q] ```` for the q-th percentile. Otherwise, you have to load the appropriate Statistics package first. - (I have version 7.) I have no problem loading the Statistics package. But what's the function in there called? Because I get the impression that this `Quantile` line will do the calculation manually instead of using a formula. – Ram Rachum Oct 9 '10 at 14:13 Evaluate it with symbolic parameters (i.e. don't assign values to `mu`, `sigma`, and `q`); you should get an expression involving the inverse error function. – J. M. Oct 9 '10 at 14:24 John Cook's page, Distributions in Scipy, is a good reference for this type of stuff: ````In [15]: import scipy.stats In [16]: scipy.stats.norm.ppf(0.975) Out[16]: 1.959963984540054 ```` - Thanks for adding a real illustration with ppf. – chl♦ Oct 9 '10 at 16:22 Well, you didn't ask about R, but in R you do it using ?qnorm (It's actually the quantile, not the percentile, or so I believe) ````> qnorm(.5) [1] 0 > qnorm(.95) [1] 1.644854 ```` - 1 Quantile vs. percentile (it's merely a matter of terminology), j.mp/dsYz9z. – chl♦ Oct 9 '10 at 14:22 1 While we are in, in R Wald-adjusted CIs (e.g. Agresti-Coull) are available in the `PropCIs` package. Wilson's method is the default in `Hmisc::binconf` (as suggested by Agresti and Coull). – chl♦ Oct 9 '10 at 14:36 Thanks for the comments chl – Tal Galili Oct 10 '10 at 4:11 In Python, you can use the stats module from the scipy package (look for `cdf()`, as in the following example). (It seems the transcendantal package also includes usual cumulative distributions). - You can use the inverse erf function, which is available in MatLab and Mathematica, for instance. For the normal CDF, starting from $$y=\Phi\left(x\right)=\frac{1}{2}\left[1+\text{erf}\left(\frac{x}{\sqrt{2}}\right)\right]$$ We get $$x=\sqrt{2}\ \text{erf}^{-1}\left(2y-1\right)$$ For the log-normal CDF, starting from $$y=F_{x}(x;\mu,\sigma)=\frac{1}{2}\text{erfc}\left(\frac{-\log x-\mu}{\sigma\sqrt{2}}\right)$$ We get $$-\log \left(x\right)=\mu+\sigma\sqrt{2}\ \text{erfc}^{-1}\left(2y\right)$$ - 1 isn't this more of a comment than an answer? – Macro Feb 15 '12 at 22:18 My idea was that if you have inverses for the erf and erfc functions, then the problem is solved. MatLab, for instance, has such preprogrammed functions. – Jean-Victor Côté Mar 5 '12 at 19:31 @Jean-VictorCôté Please, develop your ideas in your reply. Otherwise, it merely looks like a comment as suggested above. – chl♦ Mar 5 '12 at 22:31 The lognormal calculation doesn't look right. After all, its inverse CDF should be identical to the inverse CDF for the normal apart for the use of $\log(x)$ instead of $x$. – whuber♦ Mar 6 '12 at 19:15 default

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8805760741233826, "perplexity_flag": "middle"}

http://physics.stackexchange.com/questions/52469/differentiation-and-delta-function

# Differentiation and delta function Need help doing this simple differentiation. Consider 4 d Euclidean(or Minkowskian) spacetime. \begin{equation} \partial_{\mu}\frac{(a-x)_\mu}{(a-x)^4}= ? \end{equation} where $a_\mu$ is a constant vector and the indices are summed over since one really doesn't need to bother about upper and lower indices in flat space. Also $(a-x)^2=(a-x)_\mu(a-x)_\mu$. A simple minded calculation gives me the result $0$. But I think the answer may contain Dirac Delta function like the relation below in 3 dimension \begin{equation} {\bf{\nabla}} . \frac{\hat{r}}{r^2}=4\pi \delta ^3(r) \end{equation} A corollary question. What will be $\partial_{\mu}\frac{1}{(a-x)^2}$ ? Is it $\frac{2(a-x)_\mu}{(a-x)^4}$ or does it also involve the Delta function? - – Qmechanic♦ Jan 29 at 15:31 ## 2 Answers 1. In a $d$-dimensional Euclidean space (with positive definite norm), one has $$\vec{\nabla} \cdot \frac{\vec{r}}{r^d} ~=~{\rm Vol}(S^{d-1})~\delta^d(\vec{r}),$$ cf. the divergence theorem and arguments involving either test functions and integration by part, or $\epsilon$-regularization, similar to methods applied in this Phys.SE answer. Here ${\rm Vol}(S^{d-1})$ is the surface area of the $(d-1)$-dimensional units sphere $S^{d-1}$. 2. By similar arguments one may show that the identity $$\vec{\nabla}(r^{2-d}) ~=~(2-d)\frac{\vec{r}}{r^d} , \qquad d\neq 2,$$ contains no distributional contributions in $d$-dimensional Euclidean space. 3. For the related questions in Minkowski space, one suggestion is to introduce an $\epsilon$-regularization in the Euclidean formulation, and then perform a Wick rotation, and at the end of the calculation, let $\epsilon\to 0^+$. - Thanks Qmechanic. Writing down your equation in the index notation gives \begin{equation} \partial_\mu \frac{x_\mu}{x^2}=2\pi^2\delta^4(x) \end{equation} in 4d which is the result I wanted. Now it looks a bit silly to have asked the question in the first place. -

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8474989533424377, "perplexity_flag": "head"}

http://mathoverflow.net/questions/tagged/harmonic-analysis

## Tagged Questions 1answer 58 views ### Plancherel formula for non-second-countable (non-unimodular) groups The Plancherel formula for unimodular, second-countable, type 1 groups can be found in A Course in Abstrict Harmonic Analysis by Gerald Folland (theorem 7.44) or here. It states th … 1answer 150 views ### Littlewood-Paley theory and norm estimation In the paper "A Convolution Inequality Concerning Cantor-Lebesgue Measures", the Littlewood-Paley theory is used to estimate the norm of multiplier operator in Lemma 1. It is claim … 0answers 75 views ### Laplacian type operator on compact Lie group Consider the operator $S = \sum_{i,j} X_{ij}^2$ on $L^2(SO(n+1))$, where $X_{ij}$ generates the rotation of the sphere $S^n$ in the $ij$-plane keeping the $(n - 2)$ complementary d … 0answers 187 views ### Fourier coefficients as spectrum Let $\mathbb{T}=[0,1]$ be identified with the circle $\{ e^{2 \pi it} : t \in [0,1] \}$, $\delta_0 \in M(\mathbb{T})$ be the Dirac measure at $0 \in \mathbb{T}$. Suppose \$f \in L^1 … 0answers 74 views ### Character amenability Hello 1)IS ANY relationship between character amenability and weak amenability of Banach algebras?.... If a Banach $A$ is amenable then $A$ is $\phi$-amenable for every \$\phi\i … 1answer 202 views ### For what spaces is the Hardy-Littlewood maximal operator of strong type $(p,p)$ if and only if $p > p_0 > 1$? (This is essentially a continuation of my previous question, here.) Let $(X,d,\mu)$ be a metric measure space, i.e. $\mu$ is a Borel measure on the metric space $(X,d)$. Further a … 1answer 205 views ### Could we interpolate the compactness of compact operators? Classical theorems of Marcinkiewicz and Riesz and their extensions to general Banach spaces by Calderón, Lions, Peetre, et al. allow us to interpolate the continuity of two operato … 0answers 82 views ### Question about a oscillatory integrals on manifold Let $M$ be a compact oriented Riemannian manifold without boundary. Set $f(x)=a(x)+\sqrt{-1}b(x)$ be a complex-valued function on $M$, where $a(x),b(x)$ are real-valued function o … 2answers 544 views ### Is there a “right” proof of Riemann’s Theta Relation? Let $\theta$ denote the usual Jacobi Theta function (with auxiliary parameter $\tau = i$, for simplicity), i.e. \theta(z) = \sum_{n \in \mathbb{Z}} \exp(-\pi (a + n)^2 + 2 \pi … 2answers 408 views ### Image of L^1 under the Fourier Transform The Fourier Transform $\mathcal{F}:L^1(\mathbb{R})\to C_0(\mathbb{R})$ is an injective, bounded linear map that isn't onto. It is known (if I remember correctly) that the range is … 1answer 55 views ### Description of Bessel potential spaces Hi, let $1 < p <\infty$, $0 < \alpha < 1$, and $\mathscr{L}^p_\alpha(R^n)$ be the usual Bessel potential space defined by \mathscr{L}^p_\alpha = (1-\triangle)^{- … 0answers 48 views ### Notation for a functional L2 matrix norm Hi, Let $v(z)$ be a 2x2 matrix depending on a complex variable z, defined on an oriented contour $\Sigma$ in the complex plane. Can anyone tell me the meaning of the notation: … 0answers 137 views ### variant of Haar measure I found a certain trig identity in a discussion on Lie groups \[ \frac{\prod_{i < j} 2 \sin (\mu_i - \mu_j) \prod_{i< j} 2 \sin (\nu_i - \nu_j) }{\prod_{i, j} 2 \cos (\mu_ … 0answers 193 views ### What information about a locally compact group $G$ is encoded in $C_r^\ast(G)$ which is not in $L^1(G)$? Let $G$ be a locally compact group and let $C_r^\ast(G)$ denote its reduced group $C^\ast$-algebra. Many features of a $G$ can be realized from $L^1(G)$ or $C_r^\ast(G)$. For exa … 0answers 452 views ### The Fourier Transform of taking Eigenvalues The purpose of this question is to ask about the Fourier transform of the map which associate to an $n$ by $n$ matrix its $n$ eigenvalues, or some function of the $n$ eigenvalues. … 15 30 50 per page

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 46, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8081178665161133, "perplexity_flag": "middle"}

http://math.stackexchange.com/questions/118231/proving-that-11-cannot-be-written-in-the-form-p24pqq2

# Proving that 11 cannot be written in the form $p^2+4pq+q^2$ I have to prove that it is imossible to write 11 in the form $p^2+4pq+q^2$ where $p$ and $q$ are integers. I tried to start a proof by contradiction like this: $$p^2+4pq+q^2 = 11\\ 2(p+q)^2 - (p^2 + q^2) = 11\\ 2(p+q)^2 = 11 + (p^2 + q^2)$$ I'm really not sure where to go from there, though. A pointer in the right direction would be much appreciated! - 12 Try considering the problem mod 4 – Tom Cooney Mar 9 '12 at 14:34 You could simply go through all posibilities: Without loss of generality you can suppose that p >= q. If q = 0 only p is important. 1^2 = 1, 2^2 = 4, 3^2 = 9, 4^2 = 16. If q = 1. 1^2 + 4*1*1 + 1^2 = 6, 2^2 + 4*2*1 ... if q = 2. 2^2 + 4*2*2 + 2^2 = 4 + 16 + 2 != 11. – moose Mar 9 '12 at 18:50 ## 4 Answers In an introductory course in Number Theory, one of the early things one learns is that the sum of two squares cannot be congruent to $-1$ modulo $4$. This can be shown by a short enumeration of cases. Modulo $4$, the middle term $4pq$ is irrelevant, and we are finished. We can also work modulo $3$, directly by a short consideration of cases, made even shorter by the symmetry. Or else we can reluctantly break symmetry and rewrite our expression as $(p+2q)^2-3q^2$, and note that modulo $3$ we need pay no attention to $3q^2$. But the perfect square $(p+2q)^2$ is congruent to $0$ or $1$ modulo $3$, while $11$ is congruent to $-1$. Or else one can note that modulo $3$ we are looking at $p^2-2pq+q^2$, that is, $(p-q)^2$. Remark: The strategy we used is a common one. When we are given a Diophantine equation, and suspect that it has no solutions, it is worthwhile to look at the equation modulo some small numbers. If for some $m$ the equation cannot hold modulo $m$, then it cannot hold in the integers. The simplest such arguments are parity arguments (working modulo $2$). We can think of working modulo numbers greater than $2$ as generalized parity arguments. - For variety, here's a method that doesn't require guessing at which small modulus to try. (however, it does use some more sophisticated mathematics) First, we simplify our quadratic by completing the square $$(p+2q)^2 - 3q^2 = 11$$ Next, we homogenize: $$(pr+2qr)^2 = 11 r^2 + 3 (qr)^2$$ and for clarity, change variable $$z^2 = 11 x^2 + 3 y^2$$ (rational solutions to the original equation would be $q = y/x$ and $p = z/x - 2q$) Now, we invoke a two relevant sledgehammers. This equation obviously a real solution, and the only relevant primes to check are $3$ and $11$. $$(3, 11)_{3} = \left(\frac{11}{3}\right) = -1$$ $$(3, 11)_{11} = \left(\frac{3}{11}\right) = 1$$ (and from this, we can also conclude $(3,11)_2 = -1$) where the left hand side is the Hilbert symbol and the right hand side is the Legendre symbol. Since we got a $-1$, the equation has no rational solutions, and thus it has no integer solutions. Incidentally, this result explains why the other posts were successful when checking modulo $3$ and checking modulo $8$. A lowbrow summary of this method would be ````Complete the square, find a real solution, then try working modulo 8, then try working modulo every prime dividing one of the coefficients. If you found solutions in all cases, then your quadratic has rational solutions. Otherwise, it doesn't. ```` The main thing the sophisticated math tells you is that those are the only primes useful to check, and that if the check passes every test, then there truly are rational solutions. (actually proving none of the rational solutions are integer solutions is generally much trickier) - I'm a number theory noob, but taking all mod 4 keeping in mind what odd or even integers squares can yield mod 4, should solve it?! :) - Looking at $$2(p+q)^2 - (p^2 + q^2),$$ this will be even when $p$ and $q$ have the same parity. But if $p$ and $q$ have different parity then modulo $4$ it is equivalent to $2\times 1 - 1 =1$, while $11 \equiv 3 \mod 4$. So $11$ cannot be expressed in that form. -

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 41, "mathjax_display_tex": 7, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9226929545402527, "perplexity_flag": "head"}

http://mathoverflow.net/questions/28520/hodge-numbers-of-compactifications/28597

## Hodge numbers of compactifications ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $X$ be a smooth complex quasi-projective variety. We can find good compactification: a smooth proper variety $\bar{X}$ such that ${\bar X} \setminus X$ is a divisor with normal crossing. The variety $\bar{X}$ is then stratified by the singulartities of the divisor. And one can compute the mixed Hodge structure on $H^{\bullet}(X)$ in terms of the pure Hodge structures $H^{\bullet}(S_\alpha)$ of the smooth closed strata using a spectral sequence. Let's say a variety $Y$ is Hodge-Tate if $h^{p,q}(Y) = 0$ for $p\neq q$. If all the closed strata of $\bar{X}$ are Hodge-Tate then $X$ is Hodge-Tate. Question: Let $X$ be a smooth complex quasi-projective variety. Assume $X$ is Hodge-Tate. 1. Can one find a good compactification $\bar{X}$ with Hodge-Tate strata? 2. Are all good compactifications of $X$ of this type? (Edit: Answer is no, see Torsten's elementary example). - ## 2 Answers I asked the question to Claire Voisin and she immediatly gave me the following counter example: Consider Fermat's cubic of dimension 3 $X$. There are 5 cones on elliptic curves $E_i$ inside $X$ and one can verify that $Y=X\setminus U_iE_i$ is Hodge-Tate. But $Y$ doesn't admit a Hodge-Tate compactification as that would imply that $X$ is birational to a Tate variety and this would contradict Clemens-Griffiths' theorem saying that the intermediate jacobian of $X$ is not a direct sum of jacobians of curves as a polarized variety. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Let $E$ and $F$ be two elliptic curves and let the involution $\sigma$ act on $E\times F$ by $\sigma(e,f)=(-e,f+\alpha)$, $\alpha$ is an element of order two of $F$. Finally let $\overline{X}=(E\times F)/\sigma$ (this is a so called hyperelliptic surface). We have an inclusion $F':=0\times F/\langle\alpha\rangle\subseteq S$ and put $X:=\overline{X}\setminus F'$. Then $X$ is Hodge-Tate but all other good compactifications of $X$ are obtained by blowing ups and downs of $\overline{X}$ which means that you can never get rid of $F'$ (alternatively any good compactification $X'$ has $H^1(X')=H^1(X)$ and you need something non-Hodge-Tate at the boundary to kill that off). Addendum: This example is all wrong it took care of $H^3(X)$ but not (the more interesting) $H^1(X)$. At the moment I am less sure than I was that the answer to 1) is no. As for 2) you can just look at $\mathbb A^3\subseteq\mathbb P^3$ which is a good Hodge-Tate compactification with $\mathbb P^2$ as divisor at infinity and then blow up something non-Hodge-Tate in $\mathbb P^2$. This gives a good compactification with two components one of which (the exceptional divisor for the blowing up) is non-Hodge-Tate (as is the intersection of these two divisors). -

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 43, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9532508850097656, "perplexity_flag": "head"}

http://nrich.maths.org/2665/solution

### Isosceles Triangles Draw some isosceles triangles with an area of $9$cm$^2$ and a vertex at (20,20). If all the vertices must have whole number coordinates, how many is it possible to draw? ### Route to Infinity Can you describe this route to infinity? Where will the arrows take you next? ### Eight Hidden Squares On the graph there are 28 marked points. These points all mark the vertices (corners) of eight hidden squares. Can you find the eight hidden squares? # Lost ##### Stage: 3 Challenge Level: We received a number of solutions to this problem. Stephen from Manly Selective Campus got very close to the most efficient strategy: To explain the solution, I will use an example of this problem. Say the giraffe is at (6,8). The first step would be to enter the co-ordinates (0,0). It will then say that you are 14 blocks away from the giraffe, because the two co-ordinates will add up to the number of blocks you are away from the giraffe. The next step would be to then type in the co-ordinates (7,7). It will then say that you are 2 blocks away from the giraffe. So, that leaves you with 2 possibilities: (6,8) and (8,6). It is a simple matter then to try the two and hope that the one that you say first is right. To be sure you will not need more than 3 guesses you will need to adopt the strategy worked out by a number of pupils from St Hilda's Anglican School for Girls. Gloria and Sneha suggested: 1) Start off at any corner eg. (0,0) A diagonal of possibilities will form. 2) Click on one of the two edges of the diagonal as this gives only one possible solution. If you dont click on the edge, you could have two possibilities. 3) Locate the point on the diagonal which is 'n' blocks away from the edge of the diagonal. Snap!! Emily wrote: Your first point must be on one of the 4 corners (eg. (9,9) or (0,9) etc). You should then plot out where the giraffe could possibly be using the information you have (eg. if you started from (9,9) and the giraffe is 2 blocks away, it could be at either (9,7), (8,8) or (7,9)) Choose one of the extremes or outer co-ordinates (either (9,7) or (7,9)) and, from the information you are given from that search, plot where the giraffe could be. One of these points will overlap with one of the points from the other search and this is where the giraffe is (eg. if you chose (9,7) and the giraffe was 4 blocks away, it would be at (7,9)). Ailie and Sophie summarised the strategy very clearly: Choose one of the 4 corners (9,9) (0,0) (9,0) (0,9). Then, out of the possibilities choose one that is on an edge. After you do that there will be only one possibility where the giraffe is which will be one of the possible co-ordinates from before. We also received correct solutions from Tessa, Katharine and Alarna, also pupils from St Hilda's Anglican School for Girls. Well done to you all. The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9272186756134033, "perplexity_flag": "middle"}

http://mathhelpforum.com/calculus/79755-solved-vector-calculus-max-min-magnitude-acceleration.html

# Thread: 1. ## [SOLVED] [Vector Calculus]: Max, Min of magnitude of acceleration Ok so I have a vector function: r(t)= $<3cos(t) , 2sin(t)>$ Alright so I have to find the Max and Min for the magnitude of the acceleration. The steps I was undergoing were: 1. I found the velocity vector v(t)= $<-3sin(t) , 2cos(t)>$ 2. I found the acceleration vector a(t)= $<-3cos(t) , -2sin(t)>$ and this is where I'm a little confused, so I have to find the function for the magnitude of a(t) and then derive it, find the critical points and those will be my max and min. But the problem is that when I find the magnitude of acceleration it gives me a constant. (sin^2 + cos^2 = 1) But that's the equation for an ellipse so the acceleration is not constant... so what am i doing wrong? EDIT: Oh crap I just realized what a dumb mistake I was making, the magnitude is not a constant, the coefficients are different, sin squared and cos squared can't be factored.

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9108413457870483, "perplexity_flag": "middle"}

http://mathoverflow.net/questions/45487/compactness-theorem-for-first-order-logic/45559

## Compactness Theorem for First Order Logic ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Hi all, I am interested in proofs without using Goedel's completeness theorem. • Does anyone have a reference to a proof of this theorem that uses Skolem Functions? • How come Enderton's (Introduction to Logic) has a half a page proof (which looks OK to me) and Boolos (Computability And Logic) has a full chapter of it? Thanks. - See also this related MO question: mathoverflow.net/questions/9309/… – Joel David Hamkins Nov 9 2010 at 23:20 Some people prefer to present math in a concise way where you have to stare at a half-page proof for half a day to figure it out, and some prefer to present it in a hairy way which reads like novel at the expense of blowing the proof of every triviality to ten pages. Boolos is firmly in the second category. – Emil Jeřábek Mar 2 2011 at 16:42 ## 7 Answers There's a proof of compactness using Skolem functions in the book "Elements of Mathematical Logic. Model Theory" by Kreisel and Krivine. (I'm assuming here that the English version matches the French, because the latter is the one I checked.) It presupposes the compactness theorem for propositional logic. The proof runs as follows: Given a finitely satisfiable set $S$ of formulas, Skolemize them to get a set $S'$ of universal formulas that is satisfiable if and only if $S$ is. (Technical detail: If the vocabulary has no constant symbols, adjoin one, so that there are some closed terms for use in the next step.) Let $S''$ be the set of all the formulas you get from those in $S'$ by deleting the universal quantifiers and replacing the variables by closed terms in all possible ways. The formulas in $S''$ are propositional combinations of atomic sentences; by regarding the atomic sentences as propositional variables, we can view $S''$ as a set of formulas of propositional logic. Finite satisfiability of $S$ (in the sense of first-order logic) implies finite satisfiability of $S''$ (in the sense of propositional logic). By propositional compactness, there is a truth assignment satisfying $S''$. That truth assignment amounts to a description of a structure (in the first-order sense) in which every element is the value of some closed term. This structure satisfies $S'$ and therefore $S$. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. There are indeed many proofs of the Compactness theorem. As I mention in this MO answer, when I was a graduate student Leo Harrington told me that he used a different proof method for Compactness each time he taught the introductory graduate logic course in Berkeley. I am not sure for how many semesters he was able to keep this up, but when I had him, it was time for the Boolean-valued models proof. The Compactness Theorem is the assertion that if a first order theory $T$ is finitely satisfiable (all finite subtheories have a model), then $T$ itself is satisfiable. Let me describe a number of proofs. • Goedel's original proof was via the Completeness theorem, deducing it as a trivial corollary. If $T$ is inconsistent, then the proof of a contradiction is finite, so there is an inconsistent finite subtheory. This proof is deprecated by contemporary logicians, because the Compactness theorem lies completely on the semantic side of the syntax/semantic divide, and it seems beside the point to have to develop the entire syntactic theory of formal proofs and derivations in order to make a conclusion purely about the semantic notions of models and satisfiability. • The Henkin proof. The point is that the usual Henkin proof of the Completeness theorem also serves directly to prove the Compactness theorem. Suppose that every finite subset of $T$ is satisfiable. By the usual details of the Henkin argument, we may extend $T$ to a finitely-satisfiable complete Henkin theory $T^+$, in a language with new constant symbols (using the theorem on constants). That is, the new theory contains the Henkin assertions $\exists x\varphi(x)\to \varphi(c)$, where $c$ is a new constant symbol added for this purpose with $\varphi$. Now, from $T^+$ we may build a model out of the Henkin constants in the usual manner. The reduct of this model to the original language satisfies $T$, as desired. • The proof via Skolem functions (as you requested). This amounts basically just to a more complicated version of the Henkin proof. I recall Henkin giving a talk at the Berkeley Logic Colloquium in which he explained that the idea for his proof of the Completeness theorem arose to him in a dream, after considering the (at that time standard) Skolem function proof of Completeness. The point was that in that proof, one adds Skolem functions to the language to tie the formula $\varphi(x)$ to the witness $f_\varphi(x)$, so that one adds the formulas $\forall \vec y, x[\varphi(x)\to \varphi(f_\varphi(\vec y))]$, instead of the Henkin assertion (this amounts to the quantifer-reducing idea mentioned by Andreas). But otherwise, it works out similarly---one proves the analogue of the theorem on constants that allows one to add the Skolem function assertions, and then builds the model out of formal term expressions. Henkin said that he realized in his dream that there was no need to tie the witness so closely to the formula with the Skolem function, and that merely having the presence of a constant to serve as a witness sufficed. Thus was born the Henkin proof. • The ultraproduct proof. Pete has an explanation of this proof in his answer. If $T$ is finitely satisfiable, then consider the set of finite subsets $t\subset T$, each of which has a model $M_t\models t$. Let $F$ be an ultrafilter containing for each $\varphi\in T$ the set of finite $t\subset T$ with $\varphi\in t$, a collection with the finite intersection property. The ultraproduct $\Pi_t M_t/F$ satisfies every $\varphi\in T$ by \L os's theorem. • The reduced product proof. In this proof, one first develops the concept of a reduced product $\Pi_t M_t/F_0$, where $F_0$ is only a filter instead of an ultrafilter (the filter generated by the collection with FIP above). And then you can finish the job by considering a quotient of this structure, which essentially amounts to the ultraproduct. • The Boolean-valued model proof. It is similar to the ultraproduct proof and the reduced product proof (they are all essentially the same), but there is no need to quotient out by $F$ in advance. Instead, one builds a $\mathbb{B}$-valued model out of the product ${\cal M}=\Pi_t M_t$, where $\mathbb{B}$ is the Boolean algebra of all subsets of finite subsets of $T$, so that the truth-value of a statement $\varphi$ in $\cal M$ is the set of $t$ for which $M_t\models \varphi$. Then, one develops the general theory allowing one to quotient a Boolean-valued model by a filter, and the conclusion amounts to \L os in the ultraproduct proof. - Hehe. Do you remember (roughly) the year of Henkin's talk? I remember what you say there, but probably I read it somewhere; I do not think you and I ever coincided at a Logic Colloquium. – Andres Caicedo Nov 10 2010 at 0:39 Hmm... Probably read it: Leon Henkin, "The Discovery of My Completeness Proofs", The Bulletin of Symbolic Logic 2 (1996), 127-158. – Andres Caicedo Nov 10 2010 at 0:46 4 It must have been the early 90s. Henkin spoke on his dissertation results, which consisted of three chapters, one of them containing his famous proof of Completeness. I recall that he was particularly intent on communicating the other chapters---he said they were not as well remembered. (But alas, I cannot now recall exactly what they were about!) – Joel David Hamkins Nov 10 2010 at 1:07 "Ultraproduct" rather than "ultrapower" is the term I'd have used in this case, since the factors are not all equal. – Michael Hardy Nov 10 2010 at 18:03 Yes, I'll edit. – Joel David Hamkins Nov 10 2010 at 18:36 One standard modern approach is to prove the Compactness Theorem using ultraproducts. I did so at the end of a short summer course on model theory: see http://www.math.uga.edu/~pete/modeltheory2010Chapter6.pdf - 3 There is an additional advantage to this approach, from the set theoretic point of view, in that strong compactness can be characterized in terms of infinitary logic by essentially the same ultraproduct argument that gives compactness for first order. – Andres Caicedo Nov 10 2010 at 0:07 There is a nice recent book by Richard Kaye, The mathematics of logic (Cambridge, 2007). The book is centered around the completeness theorem, and treats different versions, while developing the required set theoretic and logical background along the way. It begins with a treatment of König's lemma. Then develops a "toy version" of a proof calculus, that introduces the idea of completeness of a proof system. This toy version is designed to require only König's lemma. Additional choice is added to the picture as the proof system grows (so we see Boolean algebras and then propositional calculus, Tychonov's theorem and then first-order logic). Compactness is presented topologically but not in terms of ultraproducts. Some of the exercises (from the very beginning) are challenging. I found it a very nice introduction to the topic. - For some reason I like stating the ultraproduct proof of the compactness theorem in measure-theoretic language, as follows. A filter $F$ on the index set $I$ corresponds to a "finitely additive measure", like this: $$m(A) = \begin{cases} 1 & \text{if }A \in F, \\ 0 & \text{if }I \setminus A \in F \\ \text{undefined} & \text{otherwise} \end{cases}$$ A filter is an ultrafilter if the "otherwise" case is vacuous. Then \Los's theorem says a first-order statement in the language of the model is true in the ultraproduct if and only if it's true in "almost all" of the factors. (Then as others noted above, let $I$ be the set of finite subsets of the set of statements you're trying to satisfy; let $F$ be an ultrafilter that contains every co-finite set (one exists if you believe in Zorn's lemma); then for each $i \in I$ let the $i$th factor be a model that satisfies that finite set of statements; then prove by induction on the formation of first-order formulas that the ultraproduct satisfies all of the statements. As far as I know you have to include formulas with free variables in the proof to make the induction work. I'm skipping the details since others seem to have gone into those above.) - A Category Theory Book: Locally Presentable and Accessible Categories, J. Adamek and J. Rosicky, Cambridge University Press (LMM 189) See p. 215 - See David Marker's "Model theory. An introduction." Although argument is model theoretic, you can easily transform it into syntactic one. -

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 52, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9370397925376892, "perplexity_flag": "head"}

http://unapologetic.wordpress.com/2009/02/16/

# The Unapologetic Mathematician ## Generalized Eigenvectors Sorry for the delay, but exam time is upon us, or at least my college algebra class. Anyhow, we’ve established that distinct eigenvalues allow us to diagonalize a matrix, but repeated eigenvalues cause us problems. We need to generalize the concept of eigenvectors somewhat. First of all, since an eigenspace generalizes a kernel, let’s consider a situation where we repeat the eigenvalue ${0}$: $\displaystyle\begin{pmatrix}{0}&1\\{0}&{0}\end{pmatrix}$ This kills off the vector $\begin{pmatrix}1\\{0}\end{pmatrix}$ right away. But the vector $\begin{pmatrix}{0}\\1\end{pmatrix}$ gets sent to $\begin{pmatrix}1\\{0}\end{pmatrix}$, where it can be killed by a second application of the matrix. So while there may not be two independent eigenvectors with eigenvalue ${0}$, there can be another vector that is eventually killed off by repeated applications of the matrix. More generally, consider a strictly upper-triangular matrix, all of whose diagonal entries are zero as well: $\displaystyle\begin{pmatrix}{0}&&*\\&\ddots&\\{0}&&{0}\end{pmatrix}$ That is, $t_i^j=0$ for all $i\geq j$. What happens as we compose this matrix with itself? I say that for $T^2$ we’ll find the $(i,k)$ entry to be zero for all $i\geq {k}+1$. Indeed, we can calculate it as a sum of terms like $t_i^jtj^k$. For each of these factors to be nonzero we need $i\leq j-1$ and $j\leq k-1$. That is, $i\leq k-2$, or else the matrix entry must be zero. Similarly, every additional factor of $T$ pushes the nonzero matrix entries one step further from the diagonal, and eventually they must fall off the upper-right corner. That is, some power of $T$ must give the zero matrix. The vectors may not have been killed by the transformation $T$, so they may not all have been in the kernel, but they will all be in the kernel of some power of $T$. Similarly, let’s take a linear transformation $T$ and a vector $v$. If $v\in\mathrm{Ker}(T-\lambda1_V)$ we said that $v$ is an eigenvector of $T$ with eigenvalue $\lambda$. Now we’ll extend this by saying that if $v\in\mathrm{Ker}(T-\lambda1_V)^n$ for some $n$, then $v$ is a generalized eigenvector of $T$ with eigenvalue $\lambda$. Posted by John Armstrong | Algebra, Linear Algebra | 3 Comments ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 31, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9097293615341187, "perplexity_flag": "head"}

http://psychology.wikia.com/wiki/Noisy_channel_coding_theorem?oldid=13313

# Noisy channel coding theorem Talk0 31,730pages on this wiki Revision as of 06:33, May 31, 2006 by Lifeartist (Talk | contribs) (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) Assessment | Biopsychology | Comparative | Cognitive | Developmental | Language | Individual differences | Personality | Philosophy | Social | Methods | Statistics | Clinical | Educational | Industrial | Professional items | World psychology | Cognitive Psychology: Attention · Learning · Memory · Motivation · Perception · Thinking In information theory, the noisy-channel coding theorem establishes that however contaminated with noise interference a communication channel may be, it is possible to communicate digital data (information) error-free up to a given maximum rate through the channel. This surprising result, sometimes called the fundamental theorem of information theory, or just Shannon's theorem, was first presented by Claude Shannon in 1948. The Shannon limit or Shannon capacity of a communications channel is the theoretical maximum information transfer rate of the channel, for a particular noise level. ## Overview Proved by Claude Shannon in 1948, the theorem describes the maximum possible efficiency of error-correcting methods versus levels of noise interference and data corruption. The theory doesn't describe how to construct the error-correcting method, it only tells us how good the best possible method can be. Shannon's theorem has wide-ranging applications in both communications and data storage applications. This theorem is of foundational importance to the modern field of information theory. The Shannon theorem states that given a noisy channel with information capacity C and information transmitted at a rate R, then if $R < C \,$ there exists a coding technique which allows the probability of error at the receiver to be made arbitrarily small. This means that theoretically, it is possible to transmit information without error up to a limiting rate, C. The converse is also important. If $R > C \,$ an arbitrarily small probability of error is not achievable. So, information cannot be guaranteed to be transmitted reliably across a channel at rates beyond the channel capacity. The theorem does not address the rare situation in which rate and capacity are equal. Simple schemes such as "send the message 3 times and use at best 2 out of 3 voting scheme if the copies differ" are inefficient error-correction methods, unable to asymptotically guarantee that a block of data can be communicated free of error. Advanced techniques such as Reed-Solomon codes and, more recently, Turbo codes come much closer to reaching the theoretical Shannon limit, but at a cost of high computational complexity. With Turbo codes and the computing power in today's digital signal processors, it is now possible to reach within 1/10 of one decibel of the Shannon limit. ## Mathematical statement Theorem (Shannon, 1948): 1. For every discrete memoryless channel, the channel capacity $C = \max_{P_X} \,I(X;Y)$ has the following property. For any ε > 0 and R < C, for large enough N, there exists a code of length N and rate ≥ R and a decoding algorithm, such that the maximal probability of block error is ≤ ε. 2. If a probability of bit error pb is acceptable, rates up to R(pb) are achievable, where $R(p_b) = \frac{C}{1-H_2(p_b)} .$ and $H_2(p_b)$ is the binary entropy function $H_2(p_b)=- \left[ p_b \log {p_b} + (1-p_b) \log ({1-p_b}) \right]$ 3. For any pb, rates greater than R(pb) are not achievable. (MacKay (2003), p. 162; cf Gallager (1968), ch.5; Cover and Thomas (1991), p. 198; Shannon (1948) thm. 11) ## Outline of Proof This section is a stub. You can help by adding to it. As with several other major results in information theory, the proof of the noisy channel coding theorem includes an achievability result and a matching converse result. These two components serve to bound, in this case, the set of possible rates at which one can communicate over a noisy channel, and matching serves to show that these bounds are tight bounds. The following outlines are only one set of many different styles available for study in information theory texts. #### Achievability for discrete memoryless channels This particular proof of achievability follows the style of proofs that make use of the Asymptotic equipartition property(AEP). Another style can be found in information theory texts using Error Exponents. Both types of proofs make use of a random coding argument where the codebook used across a channel is randomly constructed - this serves to reduce computational complexity while still proving the existence of a code satisfying a desired low probability of error at any data rate below the Channel capacity. By an AEP-related argument, given a channel, length n strings of source symbols $X_1^{n}$, and length n strings of channel outputs $Y_1^{n}$, we can define a jointly typical set by the following: $A_\epsilon^{(n)} = \{(x^n, y^n) \in \mathcal X^n \times \mathcal Y^n$ $2^{-n(H(X)+\epsilon)} \le p(X_1^n) \le 2^{-n(H(X) - \epsilon)}$ $2^{-n(H(Y) + \epsilon)} \le p(Y_1^n) \le 2^{-n(H(Y)-\epsilon)}$ $2^{-n(H(X,Y) + \epsilon)} \le p(X_1^n, Y_1^n) \le 2^{-n(H(X,Y) -\epsilon)} \}$ We say that two sequences $X_1^n$ and $Y_1^n$ are jointly typical if they lie in the jointly typical set defined above. Steps 1. In the style of the random coding argument, we randomly generate $2^{nR}$ codewords of length n from a probability distribution Q. 2. This code is revealed to the sender and receiver. It is also assumed both know the transition matrix $p(y|x)$ for the channel being used. 3. A message W is chosen according to the uniform distribution on the set of codewords. That is, $Pr(W = w) = 2^{-nR}, w = 1, 2, ..., 2^{nR}$. 4. The message W is sent across the channel. 5. The receiver receives a sequence according to $P(y^n|x^n(w))= \prod_{i = 1}^np(y_i|x_i(w))$ 6. Sending these codewords across the channel, we receive $Y_1^n$, and decode to some source sequence if there exists exactly 1 codeword that is jointly typical with Y. If there are no jointly typical codewords, or if there are more than one, an error is declared. An error also occurs if a decoded codeword doesn't match the original codeword. This is called typical set decoding. The probability of error of this scheme is divided into two parts: 1. First, error can occur if no jointly typical X sequences are found for a received Y sequence 2. Second, error can occur if an incorrect X sequence is jointly typical with a received Y sequence. • By the randomness of the code construction, we can assume that the average probability of error averaged over all codes does not depend on the index sent. Thus, without loss of generality, we can assume W = 1. • From the joint AEP, we know that the probability that no jointly typical X exists goes to 0 as n grows large. We can bound this error probability by $\epsilon$. • Also from the joint AEP, we know the probability that a particular $X_1^n(i)$ and the $Y_1^n$ resulting from W = 1 are jointly typical is $\le 2^{-n(I(X;Y) - 3\epsilon)}$. Define: $E_i = \{(X_1^n(i), Y_1^n) \in A_\epsilon^{(n)}\}, i = 1, 2, ..., 2^{nR}$ as the event that message i is jointly typical with the sequence received when message 1 is sent. $P(error) = P(error|W=1) \le P(E_1^c) + \sum_{i=2}^{2^{nR}}P(E_i)$ $\le \epsilon + 2^{-n(I(X;Y)-R-3\epsilon)}$ We can observe that as n goes to infinity, if $R < I(X;Y)$ for the channel, the probability of error will go to 0. Finally, given that the average codebook is shown to be "good," we know that there exists a codebook whose performance is better than the average, and so satisfies our need for arbitrarily low error probability communicating across the noisy channel. #### Converse for discrete memoryless channels Suppose a code of $2^{nR}$ codewords. Let W be drawn uniformly over this set as an index. Let $X^n$ and $Y^n$ be the codewords and received codewords, respectively. 1. $nR = H(W) = H(W|Y^n) + I(W;Y^n)\;$ using identities involving entropy and mutual information 2. $\le H(W|Y^n) + I(X^n(W);Y^n)$ since X is a function of W 3. $\le 1 + P_e^{(n)}nR + I(X^n(W);Y^n)$ by the use of Fano's Inequality 4. $\le 1 + P_e^{(n)}nR + nC$ by the fact that capacity is maximized mutual information. The result of these steps is that $P_e^{(n)} \ge 1 - \frac{1}{nR} - \frac{C}{R}$. As the block length n goes to infinity, we obtain $P_e^{(n)}$ is bounded away from 0 if R is greater than C - we can only get arbitrarily low rates of error if R is less than C. ## Channel coding theorem for non-stationary memoryless channels We assume that the channel is memoryless, but its transition probabilities change with time, in a fashion known at the transmitter as well as the receiver. Then the channel capacity is given by $C=\lim\;\inf\;\;\max_{p^(X_1),p^(X_2),...}\frac{1}{n}\sum_{i=1}^nI(X_i;Y_i)$ The maximum is attained at the capacity achieving distributions for each respective channel. That is, $C=\lim\;\inf\;\;\frac{1}{n}\sum_{i=1}^n C_i$ where $C_i$ is the capacity of the ith channel. ### Outline of the proof The proof runs through in almost the same way as that of channel coding theorem. Achievability follows from random coding with each symbol chosen randomly from the capacity achieving distribution for that particular channel. Typicality arguments use the definition of typical sets for non-stationary sources defined in Asymptotic Equipartition Property. The technicality of lim inf comes into play when $\frac{1}{n}\sum_{i=1}^n C_i$ does not converge. ## References • C. E. Shannon, The Mathematical Theory of Information. Urbana, IL:University of Illinois Press, 1949 (reprinted 1998). • David J. C. MacKay. Information Theory, Inference, and Learning Algorithms Cambridge: Cambridge University Press, 2003. ISBN 0521642981 • Thomas Cover, Joy Thomas, Elements of Information Theory. New York, NY:John Wiley & Sons, Inc., 1991. ISBN 0471062596 # Photos Add a Photo 6,465photos on this wiki • by Dr9855 2013-05-14T02:10:22Z • by PARANOiA 12 2013-05-11T19:25:04Z Posted in more... • by Addyrocker 2013-04-04T18:59:14Z • by Psymba 2013-03-24T20:27:47Z Posted in Mike Abrams • by Omaspiter 2013-03-14T09:55:55Z • by Omaspiter 2013-03-14T09:28:22Z • by Bigkellyna 2013-03-14T04:00:48Z Posted in User talk:Bigkellyna • by Preggo 2013-02-15T05:10:37Z • by Preggo 2013-02-15T05:10:17Z • by Preggo 2013-02-15T05:09:48Z • by Preggo 2013-02-15T05:09:35Z • See all photos See all photos >

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 40, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.900934100151062, "perplexity_flag": "middle"}

http://mathoverflow.net/questions/106382?sort=oldest

## Finding an axis-aligned ellipsoid of minimal volume which contains a given ellipsoid ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) A friend asked me to post the following question. He's not an MO user and felt it would be better received if asked by someone who was already known to the community. This is not my area, but I'll do my best to answer questions raised in the comments. Also, please feel free to retag if you want. "Let $\Sigma$ and $\Lambda$ be $n \times n$ positive semi-definite matrices. We say that $\Lambda$ is "conservative" for $\Sigma$ if, for all $n$-dimensional vectors $x$, we have $x^\top \Sigma^{-1} x \leq x^\top \Lambda^{-1} x$. Our problem is given $\Sigma$, can we find (and if so how) a conservative diagonal matrix $\Lambda$ that minimizes $\det \Lambda$? To think about it geometrically, we have an ellipsoid defined by the positive semi-definite matrix $\Sigma$, and we want to bound it with an axis-aligned ellipsoid defined by the diagonal matrix $\Lambda$ that has minimal volume. I'd also be interested in knowing, for example, whether we can bound $\frac{\det \Lambda}{\det \Sigma}$ in terms of the spectrum of $\Sigma$. One trivial $\Lambda$ that works is $\lambda_n I$, which defines a sphere with radius equal to the largest eigenvalue of $\Sigma$. It's easy to see that this can be far too large than need, however, for example if $\Sigma$ is already diagonal (say, Diag $[1, \varepsilon, \varepsilon, \dotsc \varepsilon]$ for small $\varepsilon$)." EDIT: I finally found time to chat with my friend about the answers here. It turns out that there was a typo, but not the one people (myself included) guessed. The original question should have read $x^\top \Sigma^{-1} x \geq x^\top \Lambda^{-1} x$ in the second paragraph, i.e. find the smallest axis aligned ellipsoid defined by $\Lambda$ containing the one defined by $\Sigma$. I've left the original question in place to avoid confusion. Those answering thought the typo was that the goal is to minimize det $\Lambda^{-1}$ instead of det $\Lambda$, and based on this they solved the related problem of finding the largest axis aligned ellipsoid contained inside the ellipsoid defined by $\Sigma$. Anyway, the good news is that the answers here did help my friend with his problem (which is highly related to the problem the answers solved), so I'm upvoting everyone and accepting the answer which I think helps most. Thanks to everyone for comments and answers! - 4 If your friend wants an answer, he or she should post the question. – Bill Johnson Sep 5 at 0:54 ## 5 Answers If you can generate sufficiently many points $S$ to "delimit" the ellipsoid defined by $\Sigma$, then the results of this paper yield an approximation algorithm: • Kumar & Yildirim, "Computing Minimum Volume Enclosing Axis-Aligned Ellipsoids," Journal of Optimization Theory and Applications, 136 (2), pp. 211 - 228 (2008) [journal link]: Abstract: Given a set of points $S = {x^1,\ldots,x^m} \subset \mathbb{R}^n$ and $\epsilon > 0$, we propose and analyze an algorithm for the problem of computing a $(1 + \epsilon)$-approximation to the the minimum volume axis-aligned ellipsoid enclosing $S$. We establish that our algorithm is polynomial for fixed $\epsilon$. In addition, the algorithm returns a small core set $X \subseteq S$, whose size is independent of the number of points $m$, with the property that the minimum volume axis-aligned ellipsoid enclosing $X$ is a good approximation of the minimum volume axis-aligned ellipsoid enclosing $S$. Our computational results indicate that the algorithm exhibits significantly better performance than that indicated by the theoretical worst-case complexity result. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. So you want $$\forall x,\, x^t (\Lambda^{-1} - \Sigma^{-1}) x \geq 0$$ $(\Lambda^{-1} - \Sigma^{-1})$ is a symmetric matrix, and the condition expresses that it must be a positive matrix too. Let $T = \Sigma^{-1}$ and $M = \Lambda^{-1}$ we want to minimize $\det(M)$ with the constraint that $M-T$ is semi-definite positive. This constraint can be expressed by calculating the Cholesky decomposition of $M-T$, giving $n$ inequality constraints. At this point, I would suggest resorting to quadratic programming to solve the KKT system. (See for instance this). The Cholesky algorithm can be adapted to compute derivatives in the constraints. Edit: removed a system of equation that only relied on the (necessary) constraint $\det(M-T)\geq 0$ We can deduce a couple simple bounds. For instance, the non negativity of $M-T$ implies that all its element are themselves non negative, thus $\frac{1}{\lambda_i} \geq \Sigma^{-1}_{i,i}$ and $$\prod_i^{n} \Sigma^{-1}_{i,i} \leq \det(\Lambda^{-1}) \leq \hbox{Tr}(\Sigma^{-1})^n$$ - +1. Nice catch regarding that typo. – David White Sep 5 at 16:54 Yes, I imagine the type of bound in your edited question is more the sort my friend is looking for (I feel like I want to call him the original original poster, or OOP). How does this bound of yours compare to what he suggests in the final paragraph, i.e. the eigenvalue bound? – David White Sep 5 at 18:44 It's not as good (but simpler to calculate)... I also provide a bound in the other way – Arthur B Sep 5 at 19:14 In the way it is currently written, this problem does not have a solution. Probably minimizing $\det(\Lambda^{-1})$ was meant (as mentioned by Arthur.B in his comments below). Minimizing this is equivalent to minimizing $-\log\det(\Lambda)$ subject to the prescribed constraints. This should be doable easily using standard software. Reasoning for the original problem being not solvable. The reason is as follows. Since we are talking in terms of $\Lambda^{-1}$, we need the constraint $\Lambda \succ 0$. The other given constraint is $x^T\Sigma^{-1}x \le x^T\Lambda^{-1}x$, which is equivalent to $\Sigma^{-1} \preceq \Lambda^{-1}$, which in turn is equivalent to (for invertible matrices) $\Sigma \succeq \Lambda$. Now, you wish to minimize $\det \Lambda$ for a diagonal matrix $\Lambda$. We see that setting $\Lambda = \epsilon I$ (wlog), and letting $\epsilon \to 0$ (though always strictly positive) we minimize the determinant (because for a posdef matrix the determinant is always strictly positive), while easily satisfying the constraint $\Lambda \preceq \Sigma$. In other words, the infimum is $0$, but the minimum is not attained. - 1 why the negative vote? maybe you'd like to point out the flaw in the argument?? – S. Sra Sep 5 at 10:30 Because the condition is $\Sigma-\Lambda \geq 0$ and you're satisfying $\Sigma-\Lambda \leq 0$? You're suggesting that a point at the origin will be contained in the ellipsoid. True but irrelevant. – Arthur B Sep 5 at 11:28 @Arthur: The condition is $\Sigma^{-1} \le \Lambda^{-1}$, and since the map $X \mapsto X^{-1}$ is order-reversing, this constraint translates into $\Sigma \ge \Lambda$, which is the same as $\Lambda \le \Sigma$. And for $\epsilon$ sufficiently small, for every positive definite $\Sigma$, we can find a $\Lambda$ that is smaller than $\Sigma$. So where is the catch? – S. Sra Sep 5 at 12:26 2 Oh I see, the original poster wrote minimize $\det(\Lambda)$, it's clearly a typo, one want to minimize $\det(\Lambda^{-1})$ – Arthur B Sep 5 at 14:41 1 Reading over the question, and especially the example at the end I tend to agree that it was probably a typo. He was typing into an email browser, so probably didn't see the missing $^{-1}$ – David White Sep 5 at 16:49 show 3 more comments For the reasons discussed in the comments to Suvrit's answer, I will assume your friend would like to minimize $\det(\Lambda^{-1})$. The problem in question is a convex optimization problem: \begin{align*} \text{minimize } & -\log(\det(\Lambda)) \\ \text{subject to } & 0 \preceq \Lambda \preceq \Sigma, \\ & \Lambda \text{ diagonal}. \end{align*} Moreover, the objective function is the natural barrier functional used for interior point methods to enforce the semi-definiteness constraint on $\Lambda$. Therefore, although strictly speaking a semidefinite program (SDP) must have a linear objective, SDP solvers typically have a mode to solve problems like this. (They may also prefer to formulate the problem equivalently as the convex program maximizing $(\det\Lambda)^{1/n}$). In MATLAB, I would suggest using an SDP parser like CVX or Yalmip to call an SDP solver like SeDuMi or SDPT3 to solve this problem numerically to your desired level of precision. To see how many variants of this problem can also be posed as SDPs, see Section 2 of Boyd and Vandenberghe's survey. - This isn't my field, so I hope this isn't a stupid comment, but knowing my friend I imagine he's interested in a theoretical solution which would hold in more generality. My guess, knowing his work, is that he needs this bound as a part of a larger paper where he's trying to bound something, perhaps asymptotically in $n$. Can SDP solvers do that sort of thing or are they just for specific problems with actual numerical inputs? – David White Sep 5 at 16:53 You are correct that then this method is only applicable to actual numerical instances. As such it wouldn't be of much help as far as symbolic or asymptotic results, except perhaps for forming and testing conjectures about what to try to prove. – Noah Stein Sep 5 at 17:31 I added a simple bound to my answer, is this the kind of result he's looking for? – Arthur B Sep 5 at 17:47 Suppose your "ellipsoid" is a line segment from $(1, 1, \dots, 1)$ to $(-1, -1, \dots, -1).$ Any "axis-aligned ellipsoid" must contain a cube of volume $2^n,$ and pretty obviously (although I haven't checked) must contain a ball or radius $1.$ while the maximal semiaxis of your ellipsoid is $\sqrt{n},$ and all the other semiaxes are of length $0.$ The volume is, of course, also $0.$ This means that your "stupid" algorithm is not so stupid -- the maximal semi axis in some sense controls the size of the smallest axis-aligned ellipsoid, up to a $\sqrt{n}$ multiplicative factor. -

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 86, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9517155289649963, "perplexity_flag": "head"}

http://cms.math.ca/Events/summer12/abs/pmp

2012 CMS Summer Meeting Regina Inn and Ramada Hotels (Regina~Saskatchewan), June 2 - 4, 2012 www.cms.math.ca//Events/summer12 Perspectives in Mathematical Physics Org: Yvan Saint-Aubin and Luc Vinet (Montréal) [PDF] FERENC BALOGH, Concordia University and CRM Reduction of planar orthogonality to non-hermitian orthogonality on contours [PDF] The asymptotic behaviour of orthogonal polynomials associated with measures supported in the whole complex plane is of essential importance in obtaining various limiting spectral statistics for certain non-hermitian random matrix models. The standard tools to analyse the asymptotics for orthogonal polynomials on the real line or on a complex contour, including the celebrated Riemann-Hilbert method, are not readily avaliable in the planar setting. For certain special measures, however, the two-dimensional hermitian orthogonality relations can be shown to be equivalent to a set of non-hermitian orthogonality relations with respect to an analytic weight function integrated on a combination of contours in the complex plane. This reduction amounts to solving a family of d-bar problems and, in certain cases, it simplifies the asymptotic analysis considerably since it allows the the Riemann-Hilbert approach to obtain strong asymptotics for the corresponding orthogonal polynomials. The method will be illustrated on an example based on a joint work with M. Bertola, S.Y. Lee and K. McLaughlin. VINCENT BOUCHARD, University of Alberta The (un)reasonable effectiveness of string theory in mathematics [PDF] In recent decades, an impressive number of fascinating results (many of them still conjectural) in various areas of mathematics, such as geometry, topology and number theory, have been obtained via string theory and its dualities. In this talk, I will focus on a new mysterious recursive structure that appears to unify seemingly unrelated counting problems in geometry, with far-reaching and mostly unexplored consequences. We conjectured this new structure through a careful study of topological string theory and mirror symmetry, putting to work "the (un)reasonable effectiveness of string theory in mathematics". TIAGO DINIS DA FONSECA, Centre de recherches mathématiques Higher spin 6-Vertex model and Macdonald polynomials [PDF] It is known that the $6$-Vertex model is a quantum integrable model, therefore we know, at least in theory, everything about it. For example, in the case of Domain Wall Boundary Conditions, the partition function is a relatively simple determinant (Izergin, 1987) and it is related to a Schur polynomial. In a more recent work, Caradoc, Foda and Kitanine (2006) tell us how to generalize this result for higher spins. Based in their work, one can prove that the new partition function is related to a Macdonald polynomial (dF and Balogh, to appear). In this talk, I will describe the 6-vertex model, explain how to create the higher spin model from the original model. And finally, I will sketch how one can prove that this is indeed a Macdonald polynomial. PATRICK DESROSIERS, IMAFI, Universidad de Talca Beta-ensembles of random matrices [PDF] Most models in Random Matrix Theory are solved by using techniques that depend on their symmetry properties. In the last two decades, however, general families of matrix models that provide a unifying framework for random matrices have been developed; they are called the beta-ensembles. In this talk, I will briefly introduce these ensembles and review recent advances. The presentation will focus on the interrelationships between beta-ensembles, stochastic differential equations, and remarkable special functions in many variables, such as Jack polynomials. DAVID FEDER, University of Calgary Graph Theory in Quantum Many-Body Physics [PDF] The Hamiltonian for bosonic and fermionic particles hopping on lattices can be interpreted as the adjacency matrix of an undirected and generally weighted graph. The properties of these quantum many-body systems can therefore be analyzed in terms of graph theory. For example, the simple graph for non-interacting distinguishable particles is the Cartesian product of each particle's adjacency matrix; if these particles become indistinguishable, the graph collapses via a graph equitable partition. In the presence of strong interactions between the particles, the graphs are generally decomposable as weak products (i.e. they are the Kronecker products of adjacency matrices). Under various circumstances, these techniques can allow for the efficient calculation of the eigenstates (and therefore the properties) of physically interesting quantum many-body systems. DANIEL GOTTESMAN, Perimeter Institute for Theoretical Physics Quantum Error Correction [PDF] As logic circuits become smaller and smaller, it makes sense to consider what happens once they become so small they start to behave according to quantum mechanics. What we get is a quantum computer, for which the state can be described as a vector in a finite-dimensional Hilbert space. Quantum computers offer the possibility of some dramatic computational speedups, the most famous being Shor's factoring algorithm. However, quantum states are very delicate, and we won't be able to realize the benefits of quantum computation without a way to correct errors. I will describe how to create quantum error-correcting codes which can protect even very large quantum states against noise and decoherence. The key to understanding a quantum error-correcting code lies in the stabilizer, a finite group encapsulating many of the symmetries of the code. GABOR KUNSTATTER, University of Winnipeg Quantum mechanics on the discretized half-line [PDF] We investigate nonrelativistic quantum mechanics on the discretized half-line, constructing a one-parameter family of Hamiltonians that are analogous to the Robin family of boundary conditions in continuum half-line quantum mechanics. For classically singular Hamiltonians, the construction provides a singularity avoidance mechanism that has qualitative similarities with singularity avoidance encountered in loop quantum gravity. Applications include the free particle, the attractive Coulomb potential, the scale invariant potential and a black hole described in terms of the Einstein-Rosen wormhole throat. The spectrum is analyzed by analytic and numerical techniques. In the continuum limit, the full Robin family of boundary conditions can be recovered via a suitable fine-tuning but the Dirichlet-type boundary condition emerges as generic. RAYMOND LAFLAMME, Institute for Quantum Computing, UWaterloo Quantum Error Correction: from theory to practice [PDF] The Achilles' heel of quantum information processors is the fragility of quantum states and processes. Without a method to control imperfection and imprecision of quantum devices, the probability that a quantum computation succeed will decrease exponentially in the number of gates it requires. In the last fifteen years, building on the discovery of quantum error correction, accuracy threshold theorems were proved showing that errors can be controlled using a reasonable amount of resources as long as the error rate is smaller than a certain threshold. We thus have a scalable theory describing how to control quantum systems. I will describe a variety of mathematical techniques that have been developed to turn this theorem into a useful tool in the laboratory and will sum up with a quick overview of where we are at controlling quantum systems in practice. JOSH LAPAN, McGill University Black Hole Greybody Factors and Monodromies [PDF] Black hole thermal radiation has captivated part of the theoretical physics community since its discovery by Hawking. In light of holography, deep connections have emerged between details of the black hole and a holographically dual field theory. Recent observations demonstrate that for a large class of black holes --- as well as five-dimensional black rings and black strings --- the product of inner and outer horizon areas is independent of mass; a simple connection with left- and right-moving temperatures of a dual field theory is proposed here. This directly connects to the observation that inner horizons play crucial roles in computing reflection and transmission coefficients, despite the fact that the scattering problem is set up without regard to behavior at the inner horizon. Drawing on ongoing work, we will suggest a new interpretation for how to understand the role of inner horizons in scattering problems; as a byproduct, this may lead to a different way of computing greybody factors from the classic method of matched asymptotic expansions. SHUNJI MATSUURA, McGill University Protected boundary states in gapless topological phases [PDF] I will talk about gapless topological phases of (semi-)metals and nodal superconductors. Using both K-theory and dimensional reduction procedures, a classification of topologically stable Fermi surfaces in (semi-)metals and nodal lines in superconductors is derived. We discuss a generalized bulk-boundary correspondence that relates the topological features of the Fermi surfaces and superconducting nodal lines to the presence of protected zero-energy states at the boundary of the system. STEFAN MENDEZ-DIEZ, University of Alberta Elliptic curves, KR-theory and T-duality [PDF] There are several variants of string theory. We will explore how the different string theories compactified on an elliptic curve are related via T-duality by studying the classification of elliptic curves with involution through the topological lens of KR-theory. MARCO MERKLI, Mathematics, Memorial University Repeated Interaction Quantum Systems [PDF] Consider a quantum system interacting sequentially in time, one by one, with a chain of infinitely many independent, identical other quantum systems. One may think of a scattering experiment. The entire system is an open system, due to the infinite size of the chain. We give an overview of recent results on the dynamics of such repeated interaction quantum systems. Among them are the existence and the construction of an asymptotic state and its thermodynamic properties, the influence of possible randomness, and an analysis of multi-time quantum measurements performed on chain elements after interaction. The results are based on collaborations with L. Bruneau and A. Joye, and with M. Penney. ROBERT MILSON, Dalhousie University A Conjecture on Exceptional Orthogonal Polynomials [PDF] Exceptional orthogonal polynomials (so named because they span a non-standard polynomial flag) are defined as polynomial eigenfunctions of Sturm-Liouville problems. By allowing for the possibility that the resulting sequence of polynomial degrees admits a number of gaps, we extend the classical families of Hermite, Laguerre and Jacobi. In recent years the role of the Darboux (or the factorization) transformation has been recognized as essential in the theory of orthogonal polynomials spanning a non-standard flag. In this talk we present the conjecture that ALL such polynomial systems are derived as multi-step factorizations of classical operators and offer some supporting evidence. ALEXI MORIN-DUCHESNE, Université de Montréal Periodic loop models and extended XXZ Hamiltonians [PDF] Non hermitian Hamiltonians play an important role in the description of two dimensional statistical models such as the Fortuin-Kasteleyn model and the $Q$-Potts spin model. The loop Hamiltonians, as elements of the periodic Temperley-Lieb algebra $TLP_N(\beta)$, are examples of such Hamiltonians: their eigenvalues are real and they are not diagonalizable for specific values of the parameter $\beta$. Loop Hamiltonians are known to be related to XXZ Hamiltonians and a great deal can be learned from this correspondence. In my talk, I will introduce the twist'' representations of the periodic Temperley-Lieb algebra and show how one can study the Jordan structure of the loop Hamiltonian in these representations, using tools from the XXZ models. MANU PARANJAPE, Université de Montréal Solitons and instantons in an effective model of CP violation [PDF] CP violation in the standard model is generated in the weak interactions and the CKM mass matrix. However, asymptotic states involved in such processes almost invariably involve hadronic mesonic asymptotic states, for example in recent experiments decays of B mesons to 2 K mesons and 2 $\pi$ mesons are most important. Thus it should be possible to describe CP violation entirely in terms of scalar fields. We, however, study a 1+1 dimensional analog of a 3+1 dimensional model describing CP violating decays entirely in terms of effective scalar fields. Although the equations of motion are non-linear, we find exact soliton and instanton solutions. The solitons are of the Q-ball type and represent particle states in the quantum theory while the instantons have finite action and should mediate tunnelling transitions in the theory. We speculate as to the 3+1 dimensional analogs of the exact solutions that we have found. SARAH POST, Centre de Recherches Mathematiques, Univ. de Montreal Recent advances in the theory of superintegrable systems [PDF] In this talk, I will discuss new results in the theory of superintegrable systems: Hamiltonian systems with more integrals of motion than degrees of freedom. The talk will consist of a brief survey of known results along with a focus on recent advances. In particular on the relationship between such systems and orthogonal polynomials. Topics to be discussed: new infinite families of superintegrable systems and recurrence relations for orthogonal polynomials, representations of symmetry algebras and the Askey scheme, and superintegrable spin chains. WALID ABOU SALEM, University of Saskatchewan Adiabatic evolution of coupled surface and internal waves [PDF] I discuss the dynamics of interacting surface and internal water waves over a slowly varying random bottom. Signature of such waves has been observed in pictures from the space station. The motion of the interacting waves is described by a system of coupled Schroedinger-Korteweg-de Vries equations. In the presence of a slowly varying random bottom, the coupled waves evolve adiabatically over a long time scale. The analysis covers the cases when the surface wave is a stable bound state or a long-lived metastable state. ## Event Sponsors Support from these sponsors is gratefully acknowledged.

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8884389400482178, "perplexity_flag": "head"}

http://physics.aps.org/story/print/v2/st19

# Focus: Nobel Focus: Chemistry by Computer Published October 21, 1998 | Phys. Rev. Focus 2, 19 (1998) | DOI: 10.1103/PhysRevFocus.2.19 The 1998 Nobel Prize in chemistry recognizes two researchers whose work has allowed chemists to calculate the properties of molecules and solids on computers, without performing experiments in the lab. The basic principles of the calculation scheme were first described in Physical Review in the 1960s, and solid state physicists used them for decades before they became important in the chemistry world. The scheme drastically simplifies the solution of the quantum mechanical equations for a system of many electrons, and although approximate, the solutions are accurate enough that chemists can learn about large molecules without getting their hands wet. All of the chemical properties of molecules and electrical properties of solids are determined by electrons interacting with each other and with atomic nuclei. By 1930 physicists were fully aware of the quantum mechanical equations governing systems of many electrons, but were incapable of exactly solving them in all but the simplest cases. They developed several approximation schemes, but none was very successful. In 1964 Walter Kohn, now of the University of California at Santa Barbara, and Pierre Hohenberg, now of Yale University, proved an idea that was central to their solution scheme, which is now called density functional theory (DFT). They showed that knowing the average density of electrons at all points in space is enough to uniquely determine the total energy, and therefore all of the other properties of the system. Other available methods at the time required wave functions, which depend on the positions of every electron in the system and are far more complicated functions than the density. Kohn’s 1965 publication, with Lu Sham of the University of California at San Diego, described a procedure for deriving the electron density and energy, based on solving equations for a corresponding system of noninteracting electrons, which is much easier to manage. The trickiest part of the scheme is the starting point: One has to assume a relationship (called the “density functional”) between energy and the density function. The explicit form of the density depends on the problem, but Hohenberg and Kohn proved in the earlier paper that in principle there exists a universal density functional, good for all problems. Kohn and Sham provided a simple approximation of this relationship, and it has served solid state physics for decades, allowing for most modern calculations of the “band structure” of electrons in solids, for example. But the approximate density functional used by Kohn and Sham was not accurate enough to calculate chemical bond energies and structures of molecules, so experimental chemists had little use for it. Beginning in the late 1980s, however, physicists began modifying the Kohn-Sham density functional in ways that improved its accuracy. In 1992 John Pople, of Northwestern University in Evanston, IL, added DFT–including the latest functionals–to his widely-used chemistry computer program GAUSSIAN. The program was already recognized as the premier tool for studying structures and chemical reactions, at least for small molecules, but with the addition of DFT, chemists were immediately impressed with its accuracy and speed. Since then biochemists, drug designers, atmospheric chemists, and even astrophysicists have benefited from studying complex molecules and reactions by computer. Kohn and Pople received the Nobel Prize in chemistry for their contributions to computational chemistry, which is now used routinely as a tool by scientists in many fields. Kieron Burke, of Rutgers University in Camden, NJ, says Kohn’s DFT was a brilliant simplification of the quantum mechanical mathematics. “This is one of the greatest free lunches ever,” he says, because it saves computer time (allowing for calculations with larger molecules) and requires in exchange only that the user provide a single, approximate density functional. Announcement and background from the Nobel Foundation ### Highlighted article #### Self-Consistent Equations Including Exchange and Correlation Effects W. Kohn and L. J. Sham Published November 15, 1965 #### Inhomogeneous Electron Gas P. Hohenberg and W. Kohn Published November 9, 1964 ### Figures ISSN 1943-2879. Use of the American Physical Society websites and journals implies that the user has read and agrees to our Terms and Conditions and any applicable Subscription Agreement.

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9528812766075134, "perplexity_flag": "middle"}

http://mathoverflow.net/questions/83932/enumerating-all-non-planar-embeddings-of-a-generic-connected-graph-in-the-plane

## Enumerating all non-planar embeddings of a generic connected graph in the plane ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Hi There I've been tackling a problem in computer science, which relies quite heavily on graph theoretic concepts. I'd appreciate it if anybody could provide insight into how to enumerate all non-planar embeddings of a generic connected graph in the plane. Assume we are given an undirected connected graph $G=(V,E)$ with straight edges, $n$ nodes, and a symmetric distance matrix $D$ of size $n \times n$ that weights the edges in $E$. The weight is simply the straight line Euclidean distance between the two vertices in the corresponding edge. The 2D Graph Realization problem seeks to find a non-planar (edges may cross) embedding of $V$ in the plane, such that the measured distances are "as close to" the observed distances $D$. In general, this problem is NP-Hard. However, if we ignore degenerate cases in which there is an algebraic relation between vertices (three or more vertices are collinear) the problem becomes tractable. From here on we will assume generality. Depending on the number and configuration of edges that are known, the embedding may not be unique. The certification problem seeks to determine whether it is impossible to find a unique embedding, given the edges that are known. One is able to certify the uniqueness of an embedding in the plane using graph rigidity theory. (Jackson and Jordan,2005) proved that a graph has a unique embedding in the plane if and only if it is redundantly rigid and triconnected. The first condition relies on a concept known as Generic Rigidity, and it can be tested with the 2D Pebble Game proposed by (Hendrikson, 1992). The second condition can be tested by building a SPQR tree (Hopcraft and Tarjan, 1973). (Mutzel and Weiskircher, 1999) showed how to enumerate all combinatorial planar embeddings of a biconnected graph using an SPQR-tree. This is the closest result that I can find to my problem. However, my problem differs in the following two ways: 1. A planar graph has the distinct property that edges only intersect at the vertices. I'm looking to find a way of optimizing over combinatorial non-planar embeddings of a graph. 2. Their model assumes that the graph is biconnected. I'd prefer to drop this assumption in favour of connectedness. So far, I've thought of the following crude algorithm to achieve what I'm looking for. It begins by using the 2D pebble game to find all generically rigid clusters in a given graph. These clusters are rigid against small continuous forces, but may move relative to one another. I then parameterise the K degrees of freedom with real angles $r_1 \ldots r_K$. For clusters with greater than 4 nodes I have to also parameterise the possible reflections. In order to do this I obtain an SPQR tree for each cluster, assuming that generic rigidity implies biconnectivity (a requirement of the SPQR tree data structure). The P-nodes in the SPQR tree indicate where reflection is possible, and I encode all L possible reflections with binary variables $b_1 \ldots b_L$. The space of possible embeddings of the graph is given by $b_1 \ldots b_L$ and $r_1 \ldots r_K$. I the perform some optimization algorithm over this search space. Does this appear to be a reasonable approach, or am I making any problematic assumptions? Any comments or extra information would help me greatly! - 3 I am a little confused about what you're assuming about the input. In particular, it is generically rigid or not? If not (e.g., assuming only connectivity), then this isn't really an enumeration question. If yes, you'll still have to deal with the case of $3$-connected and minimally rigid, which can have exponentially many embeddings (see Borcea and Streinu's paper "The Number of Embeddings of Minimally Rigid Graphs"). – Louis Theran Dec 20 2011 at 11:50 Hi Louis Thank you for your comment and for correctly pointing out that this problem is not an enumeration problem. I have removed the incorrect tag accordingly. Thank you for the paper reference, which is exactly what I was looking for. I had wondered how to deal with the case of a 3-connected component that is not redundantly rigid. Thank you. Regards Andrew – Andrew Symington Dec 20 2011 at 12:34 ## 1 Answer As I mentioned in my comment, there are two kinds of problems here: • If the graph is generically rigid, this is an enumeration problem, and quite a hard one. • If the graph is generically flexible, then this is a question of the configuration space of the framework. For the rigid case, I'm not aware of a better exact answer than doing something based on Groebner basis computations, which will, in general, be very slow. I assume that some formal negative results about framework realizability (e.g., Owen and Power http://www.ams.org/journals/tran/2007-359-05/S0002-9947-06-04049-9/) translate to enumeration if you're given one realization, but I can't think of a reference. There is, however, a lot of heuristic work in this area. For example, Meera Sitharam has a method based on hierarchical decompositions that's in the spirit of what you want (http://www.cise.ufl.edu/~sitharam/skeleton.pdf). It seems like you're not interested in realizability only, but if the graph is globally rigid and you can take multiple measurements, there is a nice method due to Zhu, Gortler, and Thurston (http://dl.acm.org/citation.cfm?id=1921630). -

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9286216497421265, "perplexity_flag": "head"}

http://crypto.stackexchange.com/questions/2683/is-this-a-secure-multiparty-protocol?answertab=active

# Is this a secure multiparty protocol? I am facing the following situation and i am trying to break down the problem into some specific cryptogrphic primitives. There is a function F that takes as input a bit string and produces a fingerprint.In my protocol there is a user $u$ who holds the input $D$ and $n$ other parties. I want the parties to compute $F(D)$ and all the parties to agree on that value. For example if $t$ out of $n$ where $t$ > $n/2$ agree on value $Y_1$ then the winner is $Y_1$. If there is no majority i want the protocol to run again. I want also secrecy at the results. I do not want the parties to learn the results and i want the parties to learn anything about the given input $D$. They will compute $F(D)$ without learning D. So they will receive an encrypted $E(D)$ and they will evalute $F(E(D))$ but only to evaluate the function. So my question is: Can we say that this is a secure multiparty computation ? Or it is a verifiable secret sharing? - Wait... you say you "don't care about the confidentiality of inputs", but if I'm reading your description right, anyone who knows the function $F$ (which I'd normally assume to be public knowledge unless specified otherwise) and the input $D$ can obviously compute the output $F(D)$. So I must be missing something, but what? – Ilmari Karonen May 23 '12 at 20:13 my fault. I correct it. We do care about inputs. There are secret. i update the question with sth else. I want operation on encrypted data.Sorry i am thinking while i am writing – curious May 23 '12 at 20:19 1 "I do not want the parties to learn the results and i want the parties to learn anything about the given $\hspace{.2 in}$ input $D$." $\:$ How does either part of that sentence make sense? $\;\;$ – Ricky Demer May 24 '12 at 0:11

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 21, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9244534373283386, "perplexity_flag": "head"}

http://math.stackexchange.com/questions/173025/question-about-continuous-function-in-terms-of-limits-of-sequences

# Question about continuous function in terms of limits of sequences I am reading about continuous function, in this site http://en.wikipedia.org/wiki/Continuous_function specifically the section "Definition in terms of limits of sequences". My question is, let be $c\in \mathbb{R}$ an arbitrary element belonging to domain of $f$, where $f$ is in a closet and bounded set. Are there always a sequence $x_n$ which converges to $c$?. - 1 Your question doesn't make much sense. You were asking about continuity and then you switched to closed sets. If $f$ a function or a subset? – Asaf Karagila Jul 19 '12 at 22:30 1 What about $x_n = c$ for all $n$? – Dylan Moreland Jul 19 '12 at 22:42 2 Why is $f$ in the closet? I thought we as a society were past that... – Quinn Culver Jul 19 '12 at 23:24 ## 1 Answer There is always the constant sequence $x_n = c$ for all $n$. In the extreme case that the domain of the function is $\{c\}$, there isn't anything else. In general there is a dichotomy to be made according to whether $c$ is an isolated point of the domain $X$ of $f$. (Let's assume that $X$ is a subset of the real numbers $\mathbb{R}$, which seems to be the context of the question. In fact, with only mild notational change, this answer works in the context of arbitrary metric spaces.) We say that a point $c \in X$ is isolated if there is a $\delta > 0$ such that if $y \in X$ and $|x-y| < \delta$, then $y= c$. For instance, if $f(x) = \sqrt{x^2(x-1)}$, the natural domain is $\{0\} \cup [1,\infty)$ and $0$ is an isolated point. Now, for $X \subset \mathbb{R}$ and $c \in X$, the following are equivalent: (i) $c$ is an isolated point of $X$. (ii) Every sequence $\{x_n\}$ in $X$ which converges to $c$ is eventually constant: that is, $x_n = c$ for all sufficiently large $n$. (iii) Every function $f: X \rightarrow \mathbb{R}$ is continuous at $c$. In other words, an isolated point is something of a trivial case. -

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 36, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9516674876213074, "perplexity_flag": "head"}

http://gilkalai.wordpress.com/2009/02/24/a-little-more-on-boolean-functions-and-bounded-depth-circuits/?like=1&source=post_flair&_wpnonce=836876f408

Gil Kalai’s blog ## A Little More on Boolean Functions and Bounded Depth Circuits Posted on February 24, 2009 by Boolean circuits This post (in a few parts) contains a quick introduction to Boolean circuits. It is related to the recent news post about the solution of Braverman to the Linial-Nisan conjecture. In particular, we will describe very quickly a formulation of the $NP \ne P$ problem and discuss some issues about bounded depth circuits. I hope that this gives a nice introduction to this area for non-experts (written also by a non-expert). Any comments and corrections (especially from experts!) are welcome. I want to mention, in particular, a result by Benjamin Rossman about finding cliques in graphs for bounded depth circuits that I found exciting. ## A. Boolean functions, Boolean circuits and the NP versus P problem. ### 1. Boolean functions A Boolean function of n variables is simply a function $f(x_1,x_2,\dots,x_n)$ where the variables $x_k$ take the values +1 and -1 and the value of $f$ itself is also either +1 and -1. ### 2. Boolean circuits A Boolean circuit is a gadget that computes Boolean functions. It is built from inputs, gates and an output. We can think about these circuits as follows: On level 0 there are the variables. On level 1 there are gates acting on the variables. On level 2 there are gates acting on the outputs of the gates on level 1. And in level $r$ is a single gate leading to the output of the circuit. The depth of the circuit is this number $r$. The size of the circuit is the total number of gates. The gates perform Boolean operations: They can take an input bit and negate it. They can take several input bits and take their OR – this means that the output will be ’1′ iff one of them is ’1′. They can take several input bits and take their AND – this means that the output will be ’-1′ iff one of them is ‘-1′. ### 3. the $NP \ne P$ problem Consider a graph $G$ on $v$ vertices. The question “Does $G$ have a Hamiltonian cycle?” (a Hamiltonian cycle is a simple cycle containing all the vertices of the graph) is known to be “NP complete”. Now, suppose that $n= {{k} \choose {2}}$ and associate to every edge of the complete graph on $k$ vertices a variable $x_k$. Every graph $G$ corresponds to an assignment of Boolean values to the variables. This correspondence is crucial here and in various other places: the variables associated to the edges of the graph get value ´1´and the variables associated to edges not in the graph get value ´-1´. The property of graphs on $k$ vertices to be Hamiltonian is described by the following Boolean function $f_n$ on $n$ variables. Every assigenment of values to the variables corresponds to a graph $G$, and the value of $f_n$ is 1 if and only if the graph $G$ contains a Hamiltonian cycle. To prove that $NP \ne P$ just prove that $f_n$ cannot be described by a Boolean circuit of size $s(n)$ which is bounded above by a polynomial in $n$. Is it really the famous $P \ne NP$ problem, you may ask? Well, what we have stated here is a little bit stronger. So if you prove that the Boolean functions representing the graph property “To contain a Hamiltonian cycle” cannot be described by a polynomial size circuit you will prove that $NP \ne P$ and you will have a little change to spare! (The assertion that there is no polynomial time algorithm for deciding if a graph with $n$ vertices is Hamiltonian is equivalent to the $NP \ne P$ problem. In the setting of circuits, we talk about the size of the circuit as a substitute to the notion of running time of an algorithm.) ### 4. A little more on $NP \ne P$. What is NP? The previous paragraph tells you a formulation of the problem. So you can go ahead and solve it and not worry too much about this paragraph. But the previous paragraph does not tell a few important aspects of the problems and I will mention one aspect now. NP stands for “non-deterministically polynomial” (and not for “not-polynomial” as I thought in my early youth). What does it mean? The Wikipedea explanation is “a problem is NP as long as a given solution can be verified as correct in polynomial time”. In the world of circuits we have the following description: Suppose you want to express the property that a graph has a Hamiltonian cycle in the following way: You have a Boolean circuit with $n+m$ variables. The $n$ variables correspond to the edges of the complete graph as before. The $m$ variables are new variables $y_1,y_2,\dots, y_m$ and we assume that $m$ is bounded from above by a polynomial function of $n$. Now we want to describe the property ”The graph G is Hamiltonian” by a polynomial size circuit in these $n+m$ variables as follows: If we start with a Hamiltonian graph, we can find an assignment to the variables $y_1,y_2,\dots, y_m$ such that the output of the circuit will be 1. If we start with a non-Hamiltonian graph for every assignment of values to the variables $y_1,y_2, \dots, y_m$ the output of the circuit will be -1. Now come the crucial observation: We can do it! We can find a circuit with this property. As before, we let $x_1,x_2,\dots, x_n$ be variables representing the edges of the complete graph with the understanding that $x_i=1$ if and only if the $i$th edge is present in our graph $G$. , But now we take $n$ more variables $y_1,y_2,\dots,y_n$ again representing all possible edges of the complete graph. The $y_i$s which equal to 1 are supposed to correspond to edges of a Hamiltonian cycle in our graph $G$. So now our circuit in the $x_i$s and the $y_i$s will represent the property that the graph $H$ represented by the $y_i$s is a Hamiltonian subgraph of the graph $G$ represented by the $x_i$s. This can easily be done. This is (more or less) what is meant by saying that the property “to be Hamiltonian” is in NP. So to say that the graph property “G contains a Hamiltonian cycle” is in NP roughly means that it is possible to prove (or to verify) using a polynomial time algorithm or using a polynomial-size circuit that the graph has a Hamiltonian cycle. The proof consists of presenting a Hamiltonian cycle in the graph. (Again, the circuit notion and the polynomial-time notions are not exactly the same, but let´s ignore it.) The property “The graph does not have a Hamiltonian cycle” is not known to be in NP and indeed is believed not to be in NP. 5. But what is so special about Hamiltonian cycles. The remarkable thing about the Hamiltonian cycle property is that it is a complete problem in NP. A polynomial time algorithm for deciding if a graph is Hamiltonian will give a polynomial algorithm for every problem in NP. (In other words, any problem in NP can be reduced to the problem of deciding if a graph has a Hamiltonian cycle.) There are many, many problems with the remarkable property that they are complete in NP, so there is nothing very special about the problem we have chosen. ### 6. Reductions If somebody asks you to describe in one word what theoretical computer science is about, a good word to choose is “reductions”. The art of reductions, the science of reductions, the practice and engineering of reductions, and the philosophy of reductions. ### 7. More general circuits. We can talk about various other circuits. We can have more sophisticated gates and not just Boolean gates. We can let the variables be real numbers, or elements in some other field, and talk about algebraic gates and algebraic circuits, etc. ### 8. Asymptotically speaking The notions we have described are asymptotic. We consider a problem (like the graph property of containing a Hamiltonian cycle) and we study its computational complexity when the number of variables goes to infinity. ## B. Bounded depth circuits Bounded depth circuits are circuits where the number of levels (the depth) is bounded by a constant. We try to understand the computational power of circuits where $r$ is bounded, the number of variables $n$ goes to infinity, and the number of overall gates is polynomial in $n$. Much can be said about the computational power of bounded depth circuits. Beyond this class things are much harder. Bounded depth circuits are also fascinating mathematical objects. —to be continued— ### Like this: This entry was posted in Computer Science and Optimization. Bookmark the permalink. ### 5 Responses to A Little More on Boolean Functions and Bounded Depth Circuits 1. aravind says: Hi Gil, I look forward to your take on Rossman’s proof. (He has a shorter, informal version of the proof on his homepage which I think I understand.) Thanks. 2. Gil says: Dear Aravind, I will look at the homepage. I dont promise to explain Rossman’s proof (would you?) but I will explain that I was surprised to learn (and this may also refer to an old reasult by Paul Beam that I did not know) that the Hastad switching lemma is capable to distinguish between polynomial size bounded depth circuits and quasi polynomial size bounded (even larger) depth circuits. (This may have been clear to experts but I did not know of it and I still don’t know it.) 3. Gil Kalai says: (Two remarks after talking with Avi) I tried to give in the post a very clear and simple mathematical statement which describes the $NP \ne P$ problem. (Or in fact a “slightly” stronger statement.) If you want to think about problems which are not as strong as $NP \ne P$ but still very very very hard, you can replace “Boolean circuits” by the simpler and familiar notion of “a formula”: Proving that the property “The graph contains no Hamiltonian cycle” cannot be described by a polynomial size Boolean formula in the variables is much beyond reach. Indeed, showing that certain functions that can be described by a quasi-polynomial circuits of small depth (even depth two) cannot be described by a polynomial-size bounded-depth circuits go back even to the work of Furst, Saxe, and Sipser. I am not aware of a simple property of a Boolean function that manifests such a separation. 4. Gil Kalai says: Some further remarks about the connection between algorithms complexity and circuit complexity can be found here: http://rjlipton.wordpress.com/2009/02/12/is-np-too-big-or-p-too-small/ 5. Pingback: Noise Stability and Threshold Circuits « Combinatorics and more • ### Blogroll %d bloggers like this:

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 57, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9133583307266235, "perplexity_flag": "head"}

http://en.wikisource.org/wiki/On_the_Theory_of_Radiation_of_Moving_Bodies

# On the Theory of Radiation of Moving Bodies From Wikisource On the Theory of Radiation of Moving Bodies (1904) by Friedrich Hasenöhrl, translated by Wikisource In German: Zur Theorie der Strahlung bewegter Körper, Wiener Sitzungsberichte, 113 IIa: 1039-1055 On the Theory of Radiation of Moving Bodies Friedrich Hasenöhrl Wikisource 1904 On the Theory of Radiation of Moving Bodies. by Dr. Fritz Hasenöhrl. (With 2 text figures.) (Presented in the meeting on June 23, 1904). If a radiating surface moves with uniform velocity in the same direction as the radiation emitted by it, then work must be constantly applied to overcome the pressure exerted by the latter. Now, this work is transformed into radiation energy as well, so that more energy of that amount is emanated from a surface moving in this sense, than from a stationary one. However, if the surface is moving in the opposite direction of the emanated radiation, then the latter steadily performs work, thus less energy of that amount is emanated from the moving surface than from a stationary one. On the other hand, if an absorbing surface is moving, so that it recedes from the incident radiation, then the latter steadily performs work; the surface is therefore only capable of absorbing – i.e. transforming into inner heat – less radiation by the equivalent of the latter. However, if the absorbing surface is approaching the incident radiation, then work from outside must be steadily performed against the pressure of radiation. This work can only occur as heat in the absorbing surface. The effect of motion is thus in this case, that the surface is absorbing more heat by the amount of performed work. These modes of conception were already spoken out by different authors.[1] In the following I've tried now, to apply these theorems upon the radiation in a moving cavity. Except the radiation provided by the walls, also radiation energy must be present within it, which is gained from mechanical work and which is transformed into such work again. It is essentially determined by the motion of the cavity; its amount is, as it will be shown, proportional to the square of the system's velocity (in first approximation), thus it apparently increases the kinetic energy of the system. Circumstances are thus present, that are quite analogous to the motion of an electron. In the same way as the concept of "electromagnetic mass" is introduced there, one could also speak about an "apparent mass" here, which is caused by radiation. In the same way, as the electromagnetic mass is proportional to the static energy of the stationary electron, also the apparent mass caused by radiation is proportional to the energy content of the stationary cavity. Namely, the proportionality factor is in both cases of the same order of magnitude. Now, since the heat content of every body partly consists of radiation energy, then every body must possess such an apparent mass depending on the temperature, and which is added to the mass in the ordinary sense. 1. Now, we consider a cylindric cavity $R$, moving with constant velocity $c$ in the direction of the arrow (Fig. 1). Let $\mathfrak{B}$ be the velocity of light and $\sigma=\tfrac{c}{\mathfrak{B}}$. Furthermore, let $A$ and $B$ be two black surfaces. The shell surface of the cavity as well as the outer boundary of the black body shall be formed by surfaces which are perfectly reflecting into the interior. Let the cross-section of space $R$ be equal to 1, its height be equal to $D$. Let the outer space be completely free of radiation, thus at absolute temperature zero, while a certain temperature shall be attributed to surfaces $A$ and $B$. Now we have to distinguish between absolute and relative direction of radiation;[2] it is more convenient, to base our consideration upon the latter. Let $2\pi\ i_{0}\ \cos\phi\ \sin\phi\ d\phi\,$ be the energy quantity, emanated by $A$ in the unit of time into the relative direction between $\phi$ and $\phi+d\phi$, where $\phi$ is thus the angle enclosed by the relative beam direction with the normal (and with the direction of velocity $c$). $i_{1}$ then must be a constant (with respect to $\phi$, as I already have emphasized in an earlier work.[3]) This radiation now exerts a pressure upon $A$, whose component that coincides with the direction of the normal (in the sense of $-c$), shall have the value $2\pi\ p_{1}\ \cos\phi\ \sin\phi\ d\phi\,$ If we multiply this expression with $c$, then we obtain the work performed against this pressure in one second from the outside, which is now also transformed into radiation, so that the total radiation leaving $A$ in the given direction, has the value $2\pi\ (i_{1}+p_{1}c)\ \cos\phi\ \sin\phi\ d\phi = 2\pi\ i\ \cos\phi\ sin\phi\ d\phi$ This expression now gives us also the amount of the radiation falling upon $B$ in the unit of time. There, one fraction of it is absorbed, and one fraction is transformed into work. If $B$ would be at the absolute temperature zero, then the radiation just considered would be the only one present in $R$. Since in this case, no force of resistance against the motion of our system is to be expected, then the same pressure into opposite directions must be effective in $A$ and $B$, so that no work is performed in the whole. Thus in $B$, the energy quantity $2\pi\ i_{0}\ \cos\phi\ \sin\phi\ d\phi\,$ is absorbed, and the energy quantity $2\pi\ p_{1}\ \cos\phi\ \sin\phi\ d\phi\cdot c\,$ is transformed into mechanical work. If we now imagine that $B$ has the same temperature as $A$, then also $B$ emanates energy $2\pi\ i_{0}\ \cos\phi\ \sin\phi\ d\phi\,$ in a certain direction. If this radiation exerts the pressure $2\pi p_{2}\cos\phi\ \sin\phi\ d\phi\,$, then it performs work by which amount the energy provided by $B$ has to be diminished, so that also $B$ is left by the radiation quantity $2\pi(i_{0}-p_{2}c)\cos\phi\ \sin\phi\ d\phi=2\pi i'\cos\phi\ \sin\phi\ d\phi\,$ The same energy quantity also occurs in $A$, and a quite similar consideration as earlier teaches, that the energy quantity $2\pi i_{0}\cos\phi\ \sin\phi\ d\phi$ is absorbed there, because the work $2\pi p_{2}\cos\phi\ \sin\phi\ d\phi\,$ must also be performed against the incident radiation from outside, which is also transformed into work. Thus the same quantity is absorbed by both bodies $A$ and $B$, as it is emitted; upon both surfaces (in opposite direction) the same pressure $2\pi(p_{1}+p_{2})\cos\phi\ \sin\phi\ d\phi,\,$ is exerted, so that no work is performed altogether. Thus when no external force is acting, then the system maintains its velocity. In case one of the bodies $A$ or $B$ is replaced by a perfect mirror, the radiation condition in $R$ must remain exactly the same,[4] thus besides other things, $2\pi(p_{1}+p_{2})\cos\phi\ \sin\phi\ d\phi,\,$ is the pressure upon the mirror as well. We ask ourselves about the energy content of space $R$. The radiation traveling from $A$ to $B$, has the relative velocity:[5] $\mathfrak{B}(-\sigma\cos\phi+\sqrt{1-\sigma^{2}\sin^{2}\phi}=\mathfrak{B}_{-}.$ The radiation coming from $A$ to $B$, has the velocity $\mathfrak{B}(+\sigma\cos\phi+\sqrt{1-\sigma^{2}\sin^{2}\phi}=\mathfrak{B}_{+}.$ both have to travel the path $\tfrac{D}{\cos\phi}$, thus the total energy content of space $R$ is $\overset{\pi/2}{\int_{0}}\frac{D}{\cos\phi}\left[\frac{2\pi(i_{0}+p_{1}c)\cos\phi\sin\phi\ d\phi}{\mathfrak{B}_{-}}+\frac{2\pi(i_{0}-p_{2}c)\cos\phi\sin\phi\ d\phi}{\mathfrak{B}_{+}}\right]$ $=2\pi i_{0}D\int_{0}^{\pi/2}\sin\phi\ d\phi\left(\frac{1}{\mathfrak{B}_{-}}+\frac{1}{\mathfrak{B}_{+}}\right)+$ $+2\pi cD\int_{0}^{\pi/2}\sin\phi\ d\phi\left(\frac{p_{1}}{\mathfrak{B}_{-}}-\frac{p_{2}}{\mathfrak{B}_{+}}\right).$ (1) The first of the two summands gives the part of the energy of space $R$, which was provided by the black bodies. My paper already cited dealt with it. The second summand, however, represents the radiation energy gained from mechanical work. Since no work is performed at uniform motion as we have seen, this must be a work which must be performed when our system is accelerated. One can easily determine the amount of this energy quantity in the following way. Let us imagine a system in absolute rest, and the radiation of both black surfaces $A$ and $B$ is somehow hindered: space $R$ is thus totally free of radiation. Now, in one instant the radiation from $A$ and $B$ shall be left free, and the system shall simultaneously brought to velocity $c$. From that instant on, the work $2\pi p_{1}\cos\phi\ \sin\phi\ d\phi\cdot c$ in the unit of time must be performed against the radiation emitted from $A$. Now, time $\tfrac{D}{\mathfrak{B}_{-}\cos\phi}$ is passing until this radiation arrives in $B$, and there it provides the same quantity of work itself. During this time, the work provided from outside is thus uncompensated. Also the radiation emitted from $B$ performs right from the start the work $2\pi p_{2}\cos\phi\ \sin\phi\ d\phi\cdot c$, and now the time $\tfrac{D}{\mathfrak{B}_{+}\cos\phi}$ is passing until this radiation arrives in $A$, and there the same work from outside must be performed in the opposite sense. Thus, more work was performed from the outside by $\frac{D}{\cos\phi\cdot\mathfrak{B}_{-}}2\pi p_{1}\cos\phi\sin\phi\ d\phi\cdot c\frac{D}{\cos\phi\cdot\mathfrak{B}_{+}}2\pi p_{2}\cos\phi\sin\phi\ d\phi\cdot c$ i.e., it was gained from the radiation. If we integrate this amount over all beam directions, then we obtain the expression above. If our system is at rest in the beginning, yet the surfaces $A$ and $B$ are not hindered of emanating radiation, then radiation energy of certain amount is present at the beginning also in $A$. The previous consideration indicates, that this energy is completely absorbed and not transformed into work. Namely this can also be directly shown, when one introduces the values for $p_{1}$ and $p_{2}$, which we will give in the following section. In order of not interrupting the progression of the investigation, I transfer the proof of this theorem into § 4 of this treatise. Thus in expression (1), the first summand is the energy provided by the radiating bodies, the second one is the energy stemming from work. We denote the latter by $L$, thus we set $L=2\pi cD\int_{0}^{\pi/2}\sin\phi\ d\phi\left(\frac{p_{1}}{\mathfrak{B}_{-}}-\frac{p_{2}}{\mathfrak{B}_{+}}\right).$ (2) 2. We now want to introduce the values for magnitudes $p_{1}$ and $p_{2}$, which we preliminarily left undetermined. It was surely introduced at first by Abraham in his treatises already cited. Namely, these values have been derived from Lorentz's theory of electromagnetism. Since one can arrive at the same expressions in another way (see § 3 of the present work), their correctness seem to be even more plausible. According to Abraham, the radiation pressure upon a surface is numerically equal to the incident or emanated radiation divided by the speed of light, namely this pressure acts in the absolute direction of radiation. Let $\varphi$ be the angle, which is enclosed by the absolute direction of the radiation (emanated under the relative angle $\phi$) with the direction of $c$. Then the magnitude denoted earlier as $p_1$, is $p_{1}=\frac{1}{\mathfrak{B}}i\ \cos\varphi,$ since we understood under $p_1$ the – solely effective – normal pressure component. A relation between $\varphi$ and $\phi$ can be easily obtained by the following figure 2, which is essentially identical with Fig. 1 of my cited treatise, and which surely needs nor further explanation. It is $\mathfrak{B}_{-}=\sqrt{\mathfrak{B}^{2}+c^{2}-2\mathfrak{B}c\ \cos\ \varphi}=\mathfrak{B}\sqrt{1+\sigma^{2}-2\sigma\ \cos\ \varphi}=$ $=\mathfrak{B}\left(-\sigma\ \cos\phi+\sqrt{1-\sigma^{2}\sin^{2}\ \phi}\right),$ from which the relations $\cos\phi=\frac{1-\sigma\ \cos\varphi}{\sqrt{1+\sigma^{2}-2\sigma\ \cos\ \varphi}}$ and $\cos\ \varphi=\sigma\sin^{2}\phi+\cos\phi\sqrt{1-\sigma^{2}\sin^{2}\ \phi}$ (3) are given. Thus it is $p_{1}=\frac{i}{\mathfrak{B}}\left(\sigma\sin^{2}\phi+\cos\phi\sqrt{1-\sigma^{2}\sin^{2}\phi}\right);$ (4) if we insert this into equation $i=i_{0}+p_{1}c$, then it follows $i=i_{0}\frac{\mathfrak{B}}{\mathfrak{B}_{-}\sqrt{1-\sigma^{2}\sin^{2}\phi}}$ (5) and $p_{1}=i_{0}\frac{\sigma\sin^{2}\phi+\cos\phi\sqrt{1-\sigma^{2}\sin^{2}\phi}}{\mathfrak{B}_{-}\sqrt{1-\sigma^{2}\sin^{2}\phi}}=$ $=i_{0}\frac{\cos\phi+\sigma\sqrt{1-\sigma^{2}\sin^{2}\phi}}{\mathfrak{B}(1-\sigma^{2})\sqrt{1-\sigma^{2}\sin^{2}\phi}}.$ (6) Quite similar it is given: $i'=i_{0}\frac{\mathfrak{B}}{\mathfrak{B}_{+}\sqrt{1-\sigma^{2}\sin^{2}\phi}}$ (7) and $p_{2}=\frac{i'}{B}\left(-\sigma\sin^{2}\phi+\cos\phi\sqrt{1-\sigma^{2}\sin^{2}\phi}\right)$ (8) or $p_{2}=i_{0}\frac{-\sigma\sin^{2}\phi+\cos\phi\sqrt{1-\sigma^{2}\sin^{2}\phi}}{\mathfrak{B}_{+}\sqrt{1-\sigma^{2}\sin^{2}\phi}}=$ $=i_{0}\frac{\cos\phi-\sigma\sqrt{1-\sigma^{2}\sin^{2}\phi}}{\mathfrak{B}(1-\sigma^{2})\sqrt{1-\sigma^{2}\sin^{2}\phi}}.$ (9) If we insert this into (2), then: $L=2\pi cDi_{0}\cdot\int_{0}^{\pi/2}\frac{\sin\phi\ d\phi}{\mathfrak{B}(1-\sigma^{2})\sqrt{1-\sigma^{2}\sin^{2}\phi}}$ $\left[\cos\ \phi\left(\frac{1}{\mathfrak{B}_{-}}-\frac{1}{\mathfrak{B}_{+}}\right)+\sigma\sqrt{1-\sigma^{2}\sin^{2}\phi}\left(\frac{1}{\mathfrak{B}_{-}}+\frac{1}{\mathfrak{B}_{+}}\right)\right]=$ $=\frac{4\pi cDi_{0}\sigma}{\mathfrak{B}^{2}(1-\sigma^{2})^{2}}\int_{0}^{\pi/2}\sin\phi\ d\phi\frac{1+\cos^{2}\phi-\sigma^{2}\sin^{2}\phi}{\sqrt{1-\sigma^{2}\sin^{2}\phi}}$ If we additionally put the energy content of the resting cavity $R$: $\frac{4\pi i_{0}D}{\mathfrak{B}}=E_{0},$ then after execution of the integration $L=E_{0}\frac{c^{2}}{\mathfrak{B}^{2}}\frac{1}{(1-\sigma^{2})^{2}}\left[\frac{1}{2}+\frac{1}{2\sigma^{2}}-\frac{(1-\sigma^{2})^{2}}{4\sigma^{3}}\log\frac{1+\sigma}{1-\sigma}\right].$ If we neglect herein magnitudes of order $\sigma^3$, then it becomes $L=E_{0}\frac{c^{2}}{B^{2}}\cdot\frac{4}{3}.$ This expression has now the form and the dimension of a kinetic energy. Thus one can say, that the kinetic energy of our system was apparently increased by $L$, or that to the mechanical mass of our system, also an apparent mass $\mu=\frac{8}{3}\frac{E_{0}}{\mathfrak{B}^{2}}$ has been added. Namely, the introduction of that mass is quite similar to that of the electromagnetic mass. If we (for the moment) denote the energy of a resting electron with $\epsilon_{0}$, then the electromagnetic mass of it is equal to $\tfrac{4}{3}\tfrac{\epsilon_{0}}{\mathfrak{B}^{2}}$ or $\tfrac{8}{5}\tfrac{\epsilon_{0}}{\mathfrak{B}^{2}}$,[6] depending on whether one deals with surface or volume charge. The relation is thus the same in terms of order of magnitude. Also the exact expressions have a certain similarity, since the quantity $\log\tfrac{1+\sigma}{1-\sigma}$ plays a role in both of them. 3. In this section, I still want to make same remarks concerning the value of the radiation pressure. The total pressure upon one of the surfaces $A$ or $B$ – it is irrelevant as to whether it is imagined as black of reflecting – has the value $\begin{array}{ll} P & =2\pi(p_{1}+p_{2})\cos\phi\ \sin\phi\ d\phi\\ \\ & =4\pi i_{0}\frac{\cos^{2}\phi\ \sin\phi\ d\phi}{\mathfrak{B}(1-\sigma^{2})\sqrt{1-\sigma^{2}\sin^{2}\phi}}.\end{array}$ Herein, we want to introduce the radiation density $\rho$, which for example falls upon $B$. It is $\rho=\frac{2\pi i\ \sin\phi\ d\phi}{\mathfrak{B}_{-}}=\frac{2\pi i_{0}\ \sin\phi\ d\phi\cdot\mathfrak{B}}{\mathfrak{B}_{-}^{2}\sqrt{1-\sigma^{2}\sin^{2}\phi}}$ (this is given from (5)). Thus it is $P=2\rho\frac{\mathfrak{B}_{-}^{2}\cos^{2}\phi}{\mathfrak{B}^{2}(1-\sigma^{2})}.$ If we want to introduce herein the absolute beam direction again, then we use the relation immediately given from Fig. 2 $\mathfrak{B}_{-}cos\phi=\mathfrak{B}\ cos\varphi-c=\mathfrak{B}(\cos\ \varphi-\sigma).$ And it becomes: $P=2\rho\frac{(\cos\ \varphi-\sigma)^{2}}{1-\sigma^{2}},$ an expression already given by Abraham[7] as well. Now, by generalization of the thought that was first spoken out by Larmor[8], the same result can be derived from the standpoint of the elastic theory of light, which I would like to show soon. We consider a light source, moving under the (absolute) angle $i$ against the $X$-axis. It is given by $\zeta=A\cos\ m(x\ \cos\ i+y\ \sin\ i+\mathfrak{B}t).$ If this wave falls upon a mirror lying perpendicular to the $X$-axis, then a reflected wave is formed, which is given by $\zeta'=A'\cos\ m'(x\ \cos\ r-y\ \sin\ r-\mathfrak{B}t).$ Herein, $r$ is the angle of reflection. At the surface of the mirror it must be $\zeta+\zeta'=0$. If it is moving with velocity $c$ in the direction of the normal, i.e., in the direction of the positive $X$-axis, then it must be $x=ct,\ \zeta+\zeta'=0$ This gives $A\ cos\ m(t(\mathfrak{B}+c\ \cos\ i)+y\ \sin\ i]+$ $+A'\cos\ m'[t(c\ \cos\ r-\mathfrak{B})-y\ \sin\ r]=0.$ This equation can only then be satisfied for all values of $y$ and $t$, when $\begin{array}{rl} A= & -A',\\ m(\mathfrak{B}+c\ \cos\ i)= & m'(\mathfrak{B}-c\ \cos\ r),\\ m\ \sin i= & m'\ \sin r.),\end{array}.$ These equations give us the law of reflection $\frac{\sin\ i}{1+\sigma\ \cos\ i}=\frac{\sin\ r}{1-\sigma\ \cos\ r}$ (10) and the Doppler effect $\frac{m'}{m}=\frac{1+\sigma\ \cos\ i}{1-\sigma\ \cos\ r}$ in full agreement with Abraham. Incidentally, equation (10) is also easily given from equation (12) of my earlier treatise.[9] Now, to derive the value of the pressure from it, we have to assume that the energy density of a light wave is proportional (at equal amplitude) to the -2. power of the wave length, thus to the quantity $m^2$. Thus when we denote by $\rho$ the density of the incident wave, by $\rho'$ that of the reflecting one, then it is $\frac{\rho'}{\rho}=\left(\frac{m'}{m}\right)^{2}=\left(\frac{1+\sigma\ \cos\ i}{1-\sigma\ \cos\ r}\right)^{2}.$ Now, the energy falling upon the mirror in unit time, is contained in space $\mathfrak{B}\ cos\ i+c$; the reflected energy in space $\mathfrak{B}\ cos\ r-c$. Thus the amount the incident energy per second is: $\rho(\mathfrak{B}\ \cos\ i+c)=\rho\mathfrak{B}(\cos\ i+\sigma)$ and the amount of the energy reflected in the same time $\rho'(\mathfrak{B}\ \cos\ r-c)=\rho'\mathfrak{B}(\cos\ r-\sigma).$ The difference of these two expressions must be equal to the work of the radiation pressure per unit time. Thus it must be $Pc=\rho'\mathfrak{B}(\cos\ r-\sigma)-\rho\mathfrak{B}(\cos\ i+\sigma)$ Or $cP=\mathfrak{B}\rho(\cos\ i+\sigma)\left[\left(\frac{1+\sigma\cos\ i}{1-\sigma\cos\ r}\right)^{2}\frac{\cos\ r-\sigma}{\cos\ i+\sigma}-1\right].$ With the aid of the easily derivable relation, already given by Abraham:[10] $\frac{1+\sigma\cos\ i}{1-\sigma\cos\ r}=\frac{\sigma+\cos\ i}{\sigma-\cos\ r}=\frac{\sin\ i}{\sin\ r}$ and the equation:[11] $\sin\ r=\frac{\sin\ i(1-\sigma^{2})}{1+\sigma^{2}+2\sigma\ \cos\ i}$ it becomes $cP=\mathfrak{B}\rho(\cos\ i+\sigma)\left[\frac{1+\sigma^{2}+2\sigma\ \cos\ i}{\ i-\sigma^{2}}-1\right].$ and $P=2\rho\frac{(\cos\ i+\sigma)^{2}}{1-\sigma^{2}}.$ The agreement with the value given by Abraham is thus a complete one (here, $\sigma$ has the opposite sign as earlier.) Of course, upon this foundation one can calculate the values $p_{1}$ and $p_{2}$, when one assumes that a moving body is emanating waves, whose amplitude is the same as in the stationary state, while the density of the radiation energy is inversely proportional to the square of the wavelength. Incidentally, the values of $p_{1}$ and $p_{2}$ have been derived by Poynting[12] by a peculiar consideration. However, Poynting confines himself to the case of perpendicular emission. If one accordingly put $\cos\varphi = \cos\phi = 1$ in the expressions given here, then the agreement is complete. 4. Now, we have to provide the proof of the assertion made in § 1. Our system given by Fig. 1 shall be at rest at the beginning. Then the energy quantity $E_{0}=\frac{4\pi i_{0}}{\mathfrak{B}}\cdot D.$ is in space $R$. Now it is the question, what is happening with the energy, when the system is suddenly brought to velocity $c$. We can assume from the outset, that only a fraction of it becomes absorbed, while another fraction is transformed into mechanical work. Now we have to notice in particular, that this energy is uniformly distributed into all absolute directions. Thus when we maintain our earlier mode of expression, then the density of energy whose absolute direction of propagation encloses an angle between $\phi\,$ and $\phi+d\phi\,$ with the normal, is: $\frac{2\pi i_{0}}{\mathfrak{B}}\sin\varphi\ d\varphi.$ As soon as the system is in motion, it is about the distribution in terms of the relative propagation direction. Then the density of energy, whose relative propagation direction encloses an angle between $\phi\,$ and $\phi+d\phi\,$ with the normal, is evidently: $\frac{2\pi i_{0}}{\mathfrak{B}}\sin\varphi\ \frac{d\varphi}{d\phi}d\phi=-\frac{2\pi i_{0}}{\mathfrak{B}}\ \frac{d\ \cos\varphi}{d\phi}d\phi.$ This becomes by differentiation of equation (3): $=\frac{2\pi i_{0}}{\mathfrak{B}}\sin\phi\ d\phi\frac{\left(\sqrt{1-\sigma^{2}\sin^{2}\phi}-\sigma\ \cos\phi\right)^{2}}{\sqrt{1-\sigma^{2}\sin^{2}\phi}}=$ $=\frac{2\pi i_{0}}{\mathfrak{B}}\frac{\mathfrak{B}_{-}^{2}}{\mathfrak{B}^{2}\sqrt{1-\sigma^{2}\sin^{2}\phi}}\sin\phi\ d\phi$ Since this energy is located in a cylinder of height $D$, then the energy quantity falling upon $B$ in this direction is: $\frac{2\pi i_{0}D}{\mathfrak{B}^{3}}\cdot\frac{\mathfrak{B}_{-}^{2}}{\sqrt{1-\sigma^{2}\sin^{2}\phi}}\sin\phi\ d\phi$ Previously (§ 1) we saw, that from the incident radiation $2\pi\ i\ cos\phi\ sin\phi\ d\phi$, the fraction $2\pi\ i_{0}\ \cos\phi \sin\phi\ d\phi$ is absorbed, while the fraction $2\pi\ p_{1}\ \cos\phi \sin\phi d\phi$ is transformed into work. Thus from the totality of the incident energy quantity, the fraction $\tfrac{i_{0}}{i}$ is absorbed and the fraction $\tfrac{p_{1}c}{i}$ is transformed into work. Thus from the just incident energy quantity, the fraction $\frac{2\pi i_{0}D}{\mathfrak{B}^{3}}\cdot\frac{\mathfrak{B}_{-}^{2}}{\sqrt{1-\sigma^{2}\sin^{2}\phi}}\sin\phi\ d\phi\cdot\frac{i_{0}}{i}$ is absorbed; the fraction $\frac{2\pi i_{0}D}{\mathfrak{B}^{3}}\cdot\frac{\mathfrak{B}_{-}^{2}}{\sqrt{1-\sigma^{2}\sin^{2}\phi}}\sin\phi\ d\phi\cdot\frac{p_{1}c}{i}$ is transformed into work. In a similar way one recognizes, that the energy quantity $\frac{2\pi i_{0}D}{\mathfrak{B}^{3}}\cdot\frac{\mathfrak{B}_{+}^{2}}{\sqrt{1-\sigma^{2}\sin^{2}\phi}}\sin\phi\ d\phi$ falls upon $A$, and that we have to multiply this energy quantity with $\frac{i_{0}}{i}$ and $-\frac{p_{2}c}{i}$ respectively, to obtain the fraction of them which becomes absorbed or transformed into work. Altogether, the energy quantity $\int_{0}^{\pi/2}\frac{2\pi i_{0}D}{\mathfrak{B}^{3}}\frac{\sin\phi\ d\phi}{\sqrt{1-\sigma^{2}\sin^{2}\phi}}\left(\mathfrak{B}_{-}^{2}\frac{i_{0}}{i}+\mathfrak{B}_{+}^{2}\frac{i_{0}}{i'}\right)=M$ is absorbed, while the energy quantity $\int_{0}^{\pi/2}\frac{2\pi i_{0}D}{\mathfrak{B}^{3}}\frac{\sin\phi\ d\phi}{\sqrt{1-\sigma^{2}\sin^{2}\phi}}\left(\mathfrak{B}_{-}^{2}\frac{p_{1}c}{i}-\mathfrak{B}_{+}^{2}\frac{p_{2}c}{i'}\right)=N$ is transformed into work. One can convince oneself, that exactly the same expressions are valid, when for example the black surface $B$ is replaced by a mirror. Namely, the radiation coming from $A$ to $B$ is not absorbed in $B$, but is partly reflected, i.e., from the totality of the energy incident in $B$, the fraction $\tfrac{i'}{i}$ is reflected; it comes back to $A$, and the fraction $\frac{i_{0}}{i'}$ is absorbed there, so that eventually the fraction $\tfrac{i'}{i}\cdot\tfrac{i_{0}}{i'}=\tfrac{i_{0}}{i}$ of the radiation propagating at the beginning in the direction from $A$ to $B$, is absorbed again. Thus $M$ and therefore also $N$ remain unchanged. Now, if one uses equations (4), (5), (7) and (8), then $M=\frac{2\pi i_{0}D}{\mathfrak{B}^{4}}\int_{0}^{\pi/2}\sin\phi\ d\phi(\mathfrak{B}_{-}^{3}+\mathfrak{B}_{+}^{3})$ and $N=\frac{2\pi i_{0}D}{\mathfrak{B}^{3}}\int_{0}^{\pi/2}\frac{\sin\phi\ d\phi}{\sqrt{1-\sigma^{2}\sin^{2}\phi}}$ $\cdot\left[\sigma^{2}\sin^{2}\phi(\mathfrak{B}_{-}^{2}+\mathfrak{B}_{+}^{2})+\sigma\cos\phi\sqrt{1-\sigma^{2}\sin^{2}\phi}(\mathfrak{B}_{-}^{2}-\mathfrak{B}_{+}^{2})\right]$ By execution of this integration one can convince oneself, that it is quite exactly $M=\frac{4\pi i_{0}D}{\mathfrak{B}}=E_{0};\ N=0;$ Thus the total energy initially present in $R$, is absorbed: Thus the earlier assertions concerning the work necessary for the acceleration, are confirmed. On the other hand we also see, that from the energy content of the cavity, i.e. in expression (1), only the first part is provided by the heat content of the radiating body. This is an assumption, at which I started in my cited treatise concerning the dimension changes of matter due to the motion through the aether[13], and which was perhaps not sufficiently justified at that place. 1. See J. H. Poynting, Phil. Trans. 202, p. 525, 1904; Vl. v. Türin, Ann. d. Naturphil., III, p. 270, 1904; M. Abraham, Boltzmann-Festschrift, p. 84, and Drude, Ann. XIV, p. 236, 1904. 2. See about that, F. Hasenöhrl, these proceedings, CXIII., p. 469; — M. Abraham, l. c. 3. L. c. p. 474. 4. See J. Larmor, Boltzmann-Festschrift, p. 595, 1904. 5. See F. Hasenöhrl, l. c. 6. See M. Abraham, Drude's Ann., X, p. 151, 1903. 7. Boltzmann-Festschrift, p. 91, eq. 9. 8. Larmor. Report of British Association, 1900. 9. These proceedings., CXIII, p. 489. 10. L. c. eq. 7e. 11. See F. Hasenöhrl, l. c. p. 489, eq. 12. 12. L. c. p. 551. 13. L. c. p. 474, footnote. This is a translation and has a separate copyright status from the original text. The license for the translation applies to this edition only. Original: This work is in the public domain in the United States because it was published before January 1, 1923. The author died in 1915, so this work is also in the public domain in countries and areas where the copyright term is the author's life plus 80 years or less. This work may also be in the public domain in countries and areas with longer native copyright terms that apply the rule of the shorter term to foreign works. Translation: This work is released under the Creative Commons Attribution-ShareAlike 3.0 Unported license, which allows free use, distribution, and creation of derivatives, so long as the license is unchanged and clearly noted, and the original author is attributed.

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 213, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9428515434265137, "perplexity_flag": "head"}

http://leftcensored.skepsi.net/

Where to begin… Pearson’s r: Not a good measure of electoral persistence Posted on December 30, 2012 by Pearson’s product-moment correlation, $$r$$, is an incredibly useful tool for getting some idea about how two variables are (linearly) related. But there are times when using Pearson’s $$r$$ is not appropriate and, even if linearity and all other assumptions hold, using it may lead one astray. I recently ran into such a situation when looking at the persistence of municipal-level vote shares over the last four Polish parliamentary elections. The plot below presents an example of some of the data I was looking at. It shows the share of the vote going to Samoobrona (Self-Defense of the Republic of Poland) by municipality in 2005 (horizontal axis) and 2007 (vertical axis). The linear regression line is shown in red and the 45-degree line is the dashed grey line. From the plot it is clear that support for SRP collapsed between 2005 and 2007: had SRP’s performance held steady across the two elections, we would expect the data to cluster around the 45-degree line. Instead, in all but one or two municipalities, support was lower in 2007 than 2005. Accordingly, the data fall below the 45-degree line. However, if I had ignored the plot and used Pearson’s $$r$$ as the primary measure of persistence in vote share across years — something that is not uncommon in the study party politics — I would have been lead to believe that there was, in fact, a high degree of persistence between the two elections. As reported in the plot, $$r=0.72$$, which certainly seems to be good evidence of a high degree of persistence in vote shares between 2005 and 2007. So, what’s going on? Jason Wittenberg explains in a recent paper why Pearson’s $$r$$ is not an appropriate measure in this case. The problem is the fact that Pearson’s $$r$$ is a measure of linearity, not persistence — it captures how far from a linear regression line the data fall (the red line in the above plot). But what we really want is a measure of how far the data are from the 45-degree line. For this, Wittenberg suggests using Lin’s concordance correlation coefficient, $$r_c$$. From the Wikipedia article, Lin’s $$r_c$$ is defined as follows: $ r_c = \cfrac{2s_{xy}}{s_x^2 + s_y^2 + (\bar{x} – \bar{y})^2}. $ The relationship between $$r$$ and $$r_c$$ is straightforward. From the original paper: 1. $$-1 \leq -|r| \leq r_c \leq |r| \leq 1$$; 2. $$r_c = 0$$ if and only if $$r=0$$; 3. $$r_c = r$$ if and only if $$s_x = s_y$$; and 4. $$r_c = \pm 1$$ if and only if $$r = \pm 1$$, $$s_x = s_y$$, and $$\bar{x} = \bar{y}$$. (These come from Lin’s 1989 paper, p. 258, cited below. Note I have left out a couple equivalent conditions from (iv) and have changed the notation to match the definition of $$r_c$$.) As Wittenberg notes, condition (i) is potentially quite problematic for studies of electoral persistence. It says that the magnitude of $$r_c$$ is always less than or equal to $$r$$. In other words, Pearson’s $$r$$ (almost) always inflates the amount of persistence present in the data. Evidence of this is seen the in next plot, which compares electoral performance of each major party in the 2005 and 2007 Polish parliamentary elections. Clearly, Pearson’s $$r$$ and Lin’s $$r_c$$ point to significantly different interpretations of the data when it comes to the issue of electoral persistence. As expected the magnitude of $$r_c$$ is always lower than $$r$$ — sometimes dramatically lower. In the case of SRP, for instance, $$r_c = 0.06$$ compared to $$r = 0.72$$ reported above. Also note, as expected given the conditions above, $$r_c$$ is closest to $$r$$ when regression line is near a 45-degree line. So, what’s the take-away? First, Wittenberg is right in calling into question the use of Pearson’s $$r$$ as a measure of electoral persistence (anyone know a better measure?). It simply doesn’t properly capture the concept. Second, always look at your data. In this case, I had never heard of Lin’s concordance correlation coefficient until I went looking for some way to measure the discordance I was seeing in the data. Had I just decided construct a table of correlations, I may have believed that there was a great deal more persistence in electoral votes shares than actually existed. R Code Lin’s $$r_c$$ can be calculated with the `epi.ccc` function in the `epiR` package. For those who may care, here is the code I used to make the second plot above: ```xyplot(vote_share.2007 ~ vote_share.2005 | factor(party), data=PartyW[party %!in% c("Other", "RP")], as.table=TRUE, between=list(x=1,y=1), aspect=1, col=grey[5], cex=0.75, xlab="2005", ylab="2007", xlim=c(-0.1,1), ylim=c(-0.1,1), strip=strip.custom(bg=grey[3]), panel=function(x, y, ...) { r <- round(cor(x, y, use="pairwise.complete.obs"), 2) rc <- round(epi.ccc(x, y)[["rho.c"]][1,1], 2) panel.xyplot(x, y, ...) panel.abline(a=0, b=1, lty=2, col=grey[4], lwd=2) panel.abline(lm(y ~ x), col=red[6], lwd=2) panel.text(x=0, y=0.9, labels=bquote(r == .(r)), cex=0.75, pos=4) panel.text(x=0, y=0.8, labels=bquote(r[c] == .(rc)), cex=0.75, pos=4) })``` References • Lin, Lawrence I-Kuei. (1989). “A Concordance Correlation Coefficient to Evaluate Reproducibility.” Biometrics 45: 255–268. • Lin, Lawrence I-Kuei. (2000). “Correction: A Note on the Concordance Correlation Coefficient.” Biometrics 56: 324–325. • Wittenberg, Jason. (Oct. 2011). “How Similar Are They? Rethinking Electoral Congruence.” Quality & Quantity. Posted in Data analysis, Graphics, Poland, Political parties, R | R Tip: Avoid using T and F as synonyms for TRUE and FALSE Posted on December 11, 2012 by By default when you start R, `T` and `F` are defined as `TRUE` and `FALSE`. When I review other people’s code, I often see functions defined with arguments set to these values by default. This is a very bad idea. `T` and `F` are symbols that can be redefined, so it shouldn’t be assumed that they will always evaluate to `TRUE` and `FALSE`. Making that assumption can introduce bugs to the code that are very hard to track down. For example, imagine you have defined a function to sample from a vector after transforming the data in some way: ```my_sample <- function(x, size, rep=F) { x <- x^2 # a simple transform sample(x, size, replace=rep) }``` When you just start R, `my_sample` will work as intended because `F` is `FALSE`: ```> my_sample(1:10, 8) [1] 4 1 9 100 49 16 36 81``` But if you call this function from inside another function, or after hours of hacking at the console, this may not be the case. For instance, what if at some point in your R session you redefined `F <- 2`: ```> my_sample(1:10, 8) [1] 9 49 100 4 64 64 81 36``` This type of bug can be very dangerous because they are very hard to replicate and, worse yet, you may never even notice you have them. Luckily, such bugs are easy to avoid: never use `T` and `F` as synonyms for `TRUE` and `FALSE`. It doesn’t take much more effort to type out `TRUE` and `FALSE`, but if you find it onerous, get an editor with tab-completion. Update Via email, Ananda Mahto suggests adding the following bit of code to the top of your scripts to provide a warning when `T` and `F` have been redefined. ```if (!identical(T, TRUE)) stop("'T' has been remapped to '", T, "'") if (!identical(F, FALSE)) stop("'F' has been remapped to '", F, "'")``` This doesn’t solve the problem, of course, but at least you get some warning. Posted in Hacking, R, Tip | Closures in R: A useful abstraction Posted on December 2, 2012 by People who have been using R for any length of time have probably become accustomed to passing functions as arguments to other functions. From my experience, however, people are much less likely to return functions from their own custom code. This is too bad because doing so can open up a whole new world of abstraction that can greatly decrease the quantity and complexity of the code necessary to complete certain types of tasks. Here I provide some brief examples of how R programmers can utilize lexical closures to encapsulate both data and methods. To begin with a simple example, suppose you want a function that adds `2` to its argument. You would likely write something like this: `add_2 <- function(y) { 2 + y }` Which does exactly what you would expect: ```> add_2(1:10) [1] 3 4 5 6 7 8 9 10 11 12``` Now suppose you need another function that instead adds `7` to its argument. The natural thing to do would be to write another function, just like `add_2`, where the `2` is replaced with a `7`. But this would be grossly inefficient: if in the future you discover that you made a mistake and you in fact need to multiply the values instead of add them, you would be forced to change the code in two places. In this trivial example, that may not be much trouble, but for more complicated projects, duplicating code is a recipe for disaster. A better idea would be to write a function that takes one argument, `x`, that returns another function which adds its argument, `y`, to `x`. In other words, something like this: ```add_x <- function(x) { function(y) { x + y } }``` Now, when you call `add_x` with an argument, you will get back a function that does exactly what you want: ```add_2 <- add_x(2) add_7 <- add_x(7) > add_2(1:10) [1] 3 4 5 6 7 8 9 10 11 12 > add_7(1:10) [1] 8 9 10 11 12 13 14 15 16 17``` So far, this doesn’t appear too earth-shattering. But if you look closely at the definition of `add_x`, you may notice something odd: how does the return function know where to find `x` when it’s called at a later point? It turns out that R is lexically scoped, meaning that functions carry with them a reference to the environment within which they were defined. In this case, when you call `add_x`, the `x` argument you provide gets attached to the environment for the return function. In other words, in this simple example, you can think of R as just replacing all instances of the x variable in the function to be returned with the value you specify when you called add_x. Ok, so this may be a neat trick, but how this can be used more productively? For a slightly more complicated example, suppose you are performing some complex bootstrapping and, for efficiency, you pre-allocate container vectors to store the results. This is straightforward when you have just a single vector of results—all you have to do is remember to iterate an index counter each time you add a result to the vector. ```for (i in 1:nboot) { bootmeans[i] <- mean(sample(data, length(data), replace=TRUE)) } > mean(data) [1] 0.0196 > mean(bootmeans) [1] 0.0188``` But suppose you want to track several different statistics, each requiring you to keep track of a different index variable. If your bootstrapping routine is even a little bit complicated, this can be tedious and prone to error. By using closures, you can abstract away all of this bookkeeping. Here is a constructor function that wraps a pre-allocated container vector: ```make_container <- function(n) { x <- numeric(n) i <- 1 function(value=NULL) { if (is.null(value)) { return(x) } else { x[i] <<- value i <<- i + 1 } } }``` When you call `make_container` with an argument, it pre-allocates a numeric vector of the specified length, `n`, and returns a function that allows you to add data to that vector without having to worry about keeping track of an index. If you don’t the argument to that return function is `NULL`, the full vector is returned. ```bootmeans <- make_container(nboot) for (i in 1:nboot) bootmeans(mean(sample(data, length(data), replace=TRUE))) > mean(data) [1] 0.0196 > mean(bootmeans()) [1] 0.0207``` Here `make_container` is relatively simple, but it can be as complicated as you need. For instance, you may want to have the constructor function perform some expensive calculations that you would rather not do every time the function is called. In fact, this is what I have done in the boolean3 package to minimize the number of calculations done at every iteration of the optimization routine. Posted in R | Filtering a list with the Filter higher-order function Posted on January 26, 2012 by Last week markbulling over at Drunks & Lampposts posted a method of using `sapply` to filter a list by a predicate. Today the @RLangTip tip of the day was to use `sapply` similarly. This made makes me wonder if R‘s very useful higher-order functions aren’t as well known as they should be. In this case, the `Filter` higher-order function would be the tool to use. `Filter` works more or less like the `*apply` family of functions, but it performs the subsetting (the filtering) of a list based on a predicate in a single step. As an example, let’s say we have a list of 1000 vectors, each of length 2 with $$x_1,\,x_2 \in [0,\,1]$$, and we want to select only those vectors where the elements of the list sum to a value greater than 1. With `Filter`, this is all we have to do: ```mylist <- lapply(1:1000, function(i) c(runif(1), runif(1))) method.1 <- Filter(function(x) sum(x) > 1, mylist)``` Which is at least a bit more transparent than the `sapply` alternative: `method.2 <- mylist[sapply(mylist, function(x) sum(x) > 1)]` In some very quick tests, I found no performance difference between the two approaches. There are other useful higher-order functions. If you are interested, check out `?Filter`. Posted in R | Announcing boolean3 (beta) Posted on January 24, 2012 by After entirely too long, I am happy to announce the beta release of `boolean3`, an R package for modeling causal complexity. The package can be downloaded at the following links: • Unix/Linux: boolean3_3.0.20.tar.gz • Windows: boolean3_3.0.20.zip (Please let me know if you have any trouble installing the Windows version. I didn’t have a Windows system handy when I built the package.) The theoretical foundation for the package was developed by Bear Braumoeller in a 2003 Political Analysis piece: ```Braumoeller, Bear F. (2003) 'Causal Complexity and the Study of Politics'. Political Analysis 11(3): 209-233.``` Additional theoretical content can be found at the Boolean Statistics homepage. Braumoeller and Carson have a 2011 paper titled “Political Irrelevance, Democracy, and the Limits of Militarized Conflict” in the Journal of Conflict Resolution that provides a useful example of the approach. Summarizing from the `boolean3` documentation and package description: ````boolean3` provides a means of estimating partial-observability binary response models following boolean logic. `boolean3` was developed by Jason W. Morgan under the direction of Bear Braumoeller with support from The Ohio State University's College of Social and Behavioral Sciences. The package represents a significant re-write of the original boolean implementation developed by Bear Braumoeller, Ben Goodrich, and Jacob Kline.``` As is typically the case with “significant re-writes”, `boolean3` breaks compatibility with previous versions (which can still be downloaded and installed from CRAN). Some of the many changes and enhancements include: • Removed dependence on Zelig. • Changed the method of specifying the model to me more consistent with the style of other R packages. • Improved performance. Many models can be estimated in 1/10th the time, or better, when compared to the original `boolean` package. • Added support for the optimizing methods available in the optimx package. This includes the ability to apply box constraints, which can be useful in maximizing boolean likelihoods. • Added multiprocessor/cluster support for bootstrapping using the snow package. • Genetic optimization using rgenoud also provides support for clustering with `snow`. • Plotting of predicted probabilities and likelihood profiles is now done with lattice. Some things that are still missing from the package: • Documentation is still thin. There is enough in the help pages to get people started using the package, but more is needed. • Skewed logit (scobit) is not yet implemented. Most of the code has been written, but there are a few complications yet to be worked out. If you are a current user of `boolean`, we’d love for you to try out `boolean3`. While a lot of bug-squashing has been done, there has been fairly little “real-world” testing. Bug reports and suggestions for changes and enhancements would be quite welcome, and can be emailed to Bear Braumoeller, me, or posted below in the comments. I will continue to post updates to the package as they are available. Posted in Code, R | Welford’s method for calculating the sample variance: An implementation in Scheme Posted on December 17, 2011 by John Cook has three entries up on his blog discussing the pitfalls of calculating the sample variance using the mathematical textbook definitions. He provides a Monte Carlo comparison of methods here, and a theoretical discussion here. He also provides a C++ implementation of Welford’s method here. I recently started hacking on Scheme, writing a basic statistics library as a way to better understand the language (which is the most fun I have ever had programming). Below is a Scheme implementation of Welford’s method. For comparison, I have included a version that uses the textbook definition $ s^2 = \cfrac{1}{n-1} \sum_{i=0}^n (x_i – \bar{x})^2. $ Here is the code: ```;;; Textbook definition (define (variance x) (let ((mu (mean x))) (/ (sum (f64vector-map (lambda (x) (expt (- x mu) 2)) x)) (- (f64vector-length x) 1)))) ;;; Welford's method (define (variance-welford x) (let loop ((n (f64vector-length x)) (k 1) (m (f64vector-ref x 0)) (s 0)) (if (= k n) (/ s (fx- n 1)) (let* ((xk (f64vector-ref x k)) (mk (+ m (/ (- xk m) (fx+ k 1)))) (sk (+ s (* (- xk m) (- xk mk))))) (loop n (fx+ k 1) mk sk)))))``` I implemented these in Chicken Scheme, though the code should work in any Scheme that supports SRFI-4 homogeneous numeric vectors. The code could easily be modified to support standard Scheme vectors or lists. I make no claims that this is the most efficient code, but it seems to work. Posted in Code, Data analysis, Scheme | Three free books for better programming in R (and any other language) Posted on September 19, 2011 by Like many users and producers of R packages, I have never had any formal training in computer science. I’ve come to to the conclusion that this is a serious omission in a professional researcher’s training. Computer scientists and professional hackers have learned a lot about effective, efficient programming over the last five decades and it’s past time academic researchers begin to learn from this experience. To that end, here are three books which I think all users of R could benefit from reading: • Practical Common Lisp by Peter Seibel. Free online version. • Structure and Interpretation of Computer Programs (SICP) by Harold Abelson, Gerald Jay Sussman, and Julie Sussman. Free online version. Yes, I realize all of these are books that use Lisp/Scheme, and I realize the structure of Lisp can be a bit off-putting `((at '(first)))`. But, unless you want to learn Lisp, you shouldn’t try to read these books with the goal of understanding ever nuance of the syntax; rather, you should read them with an eye toward understand how a programming language should work for and not against you. For instance, each of these books shows, in a step-by-step approach, how seemingly complex programs can be effectively broken down into simple procedures that anyone with some programming experience can understand. A great deal of emphasis in these books is the important concepts of abstraction and non-duplication. If you’re like me and spend any significant amount of time going through various R packages, you’ll know what I mean when I say that the R community could benefit from less duplication and more abstraction in contributed packages (not to mention more comments!). Posted in Hacking, R | The performance cost of a for-loop, and some alternatives Posted on August 21, 2011 by I’ve recently been spending a lot of time running various simulations in R. Because I often use snow to perform simulations across several computers/cores, results typically come back in the form of a list object. Summarizing the results from a list is simple enough using a `for`-loop, but it’s much “sexier” to use a functional style of programming that takes advantage of higher order functions or the `*apply`-family of functions (R is, after all, a functional language at its core). But what’s the performance cost of using, for instance, `lapply` with a large list, particularly when the results from `lapply` have to be further manipulated to get them into a useful form? I recently did some performance testing on three of the `*apply` functions and compared them to an equivalent `for`-loop approach. The results follow. First, I set up a large list with some fake data. Here I make a list, `Sims`, containing 30000 $$5 \times 5$$ matrices. (This is a data structure similar to one I recently found myself analyzing.) ```set.seed(12345) S <- 30000 Sims <- Map(function(x) matrix(rnorm(25), nrow=5, ncol=5), 1:S)``` Now I define four functions that extract the fifth column of each matrix in `Sims` and create a $$30000 \times 5$$ matrix of results. ```for_method <- function(S, Sims) { R <- matrix(NA, ncol=5, nrow=S) for (i in 1:S) R[i,] <- Sims[[i]][,5] R } sapply_method <- function(Sims) { t(sapply(Sims, function(x) x[,5])) } lapply_method <- function(Sims) { do.call(rbind, lapply(Sims, function(x) x[,5])) } vapply_method <- function(Sims) { t(vapply(Sims, function(x) x[,5], rep(0, 5))) }``` Notice that the `for_method` first creates a container matrix and inserts the results directly into it. Using something like `rbind` would force R to copy the matrix in each for-iteration, which is very inefficient. Also notice that in each of the apply methods, the result has to be manipulated before it is returned. It’s often possible to eliminate this step during the initial simulation process. But if you’re like me, you sometimes start your simulations before you realize what you intend on doing with the results. Now for some timing results: ```> system.time(replicate(20, R.1 <<- for_method(S, Sims))) user system elapsed 4.340 0.072 4.414 > system.time(replicate(20, R.2 <<- sapply_method(Sims))) user system elapsed 3.452 0.076 3.528 > system.time(replicate(20, R.3 <<- lapply_method(Sims))) user system elapsed 4.393 0.044 4.433 > system.time(replicate(20, R.4 <<- vapply_method(Sims))) user system elapsed 2.588 0.040 2.628 > all(identical(R.1, R.2), identical(R.1, R.3), identical(R.1, R.4)) [1] TRUE``` These results surprised me—I never expected `vapply` to be so much quicker than the other three methods. This may have to do with that fact that, since `vapply` requires you to explicitly specify the resultant data structure, R is able allocate the memory ahead of time. As a simple extension, I thought I would retest taking advantage of byte-compiled code: ```library(compiler) for_method <- cmpfun(for_method) sapply_method <- cmpfun(sapply_method) lapply_method <- cmpfun(lapply_method) vapply_method <- cmpfun(vapply_method)``` And here are the results: ```> system.time(replicate(10, R.1 <<- for_method(S, Sims))) user system elapsed 2.084 0.024 2.110 > system.time(replicate(10, R.2 <<- sapply_method(Sims))) user system elapsed 1.624 0.016 1.642 > system.time(replicate(10, R.3 <<- lapply_method(Sims))) user system elapsed 2.152 0.012 2.164 > system.time(replicate(10, R.4 <<- vapply_method(Sims))) user system elapsed 1.204 0.004 1.207 > all(identical(R.1, R.2), identical(R.1, R.3), identical(R.1, R.4)) [1] TRUE``` These results are also surprising. I really thought byte-compiling the functions would make the results much closer (doesn’t each function compile into what is effectively a `for`-loop?). However, in this simple example at least, that isn’t the case. `vapply` still has, by a significant margin, the best performance. `sapply` also performs noticeably better than `lapply` and the `for`-loop, which is an important result since `sapply` doesn’t require a pre-specified data structure. Assuming they hold in the presence of more complicated data structures (and I don’t see what they wouldn’t), these results seem to suggest that a functional approach can provide benefits in terms of coding efficiency and run-time performance. I welcome any feedback or suggestions on how these tests could be improved or extended. Update A couple things have come up in the comments. First, Henrik provided a more elegant (and efficient) version of `vapply_method`: ```vapply_method2 <- function(Sims) { t(vapply(Sims, "[", rep(0, 5), i = 1:5, j = 5)) }``` Second, several people have reported that the `for_method` on their system is just as fast, if not faster, than the `vapply_method` when compiled. I have retested my results, adding Henrik's `vapply_method2`. Here are the results for compiled code (note that I have increased the number of replications): ```> library(compiler) > set.seed(12345) > S <- 30000 > Sims <- Map(function(x) matrix(rnorm(25), nrow=5, ncol=5), 1:S) > > for_method <- function(S, Sims) { + R <- matrix(NA, ncol=5, nrow=S) + for (i in 1:S) R[i,] <- Sims[[i]][,5] + R + } > > sapply_method <- function(Sims) { + t(sapply(Sims, function(x) x[,5])) + } > > lapply_method <- function(Sims) { + do.call(rbind, lapply(Sims, function(x) x[,5])) + } > > vapply_method <- function(Sims) { + t(vapply(Sims, function(x) x[,5], rep(0, 5))) + } > > vapply_method2 <- function(Sims) { + t(vapply(Sims, "[", rep(0, 5), i = 1:5, j = 5)) + } > for_method <- cmpfun(for_method) > sapply_method <- cmpfun(sapply_method) > lapply_method <- cmpfun(lapply_method) > vapply_method <- cmpfun(vapply_method) > vapply_method2 <- cmpfun(vapply_method2) > system.time(replicate(50, R.1 <<- for_method(S, Sims))) user system elapsed 6.052 0.136 6.191 > system.time(replicate(50, R.2 <<- sapply_method(Sims))) user system elapsed 8.361 0.116 8.474 > system.time(replicate(50, R.3 <<- lapply_method(Sims))) user system elapsed 10.548 0.144 10.692 > system.time(replicate(50, R.4 <<- vapply_method(Sims))) user system elapsed 5.813 0.124 5.936 > system.time(replicate(50, R.5 <<- vapply_method2(Sims))) user system elapsed 4.268 0.084 4.352``` For whatever reason, the `vapply` methods remain the fastest on my system (particularly Henrik’s version). I also tested on another Linux machine and got the same results (I have not yet tested on Windows). Of course, this is a very simple example, so the performance differences may not hold up to the increasing complexity of the input function. I’d be interested to know if anyone has an idea as to why `vapply` would perform better in Linux than in Windows/Mac. A library thing? Posted in R | Emacs-fu: The place to go for useful Emacs tips Posted on August 19, 2011 by After 5 years or so of using Emacs almost exclusively as my “text editor”, somehow I just ran across Emacs-fu for the first time today. It’s an incredibly useful site with numerous Emacs tips and tricks. If you are an Emacs user, you should check the site out if you don’t already know about it. And if you aren’t an Emacs user, you should still check out the site to see what your editor is missing. Posted in Emacs, Hacking | Stronger instruments by design: Polmeth 2011 poster Posted on July 31, 2011 by I just got back from Polmeth 2011 at Princeton, where I presented a poster for a project Luke Keele and I are working on. It deals with the problem of statistical inference in the presence of a weak instrument and the non-random assignment of the instrument. We employ a technique developed by Baoicchi, Small, Lorch, & Rosenbaum in a recent JASA article to effectively strengthen the instrument, thus increasing the robustness of effect estimates to deviations from random assignment. I think the poster turned out pretty well and the feedback I received will be very useful as we move forward with project. I have included a link to the poster below the draft abstract. We hope to complete the draft paper in the next few months, but we would appreciate any comments people may have. There is growing interest in natural experiments in political science. Natural experiments are often analyzed with instrumental variable estimators reflecting a belief that combining the power of natural random assignment with an instrumental variable approach will solve many of the research design problems endemic to social science. Here, we highlight how weak instruments can interact with the assumption of random assignment of the instrument. When the instrument is not randomly assigned, weak instruments produce bias that is not alleviated by additional data. We demonstrate how matching combined with a reverse caliper can be used to strengthen an instrument within a subset of the overall study. We start by presenting an alternative non-parametric instrumental variable estimator first proposed by Rosenbaum (1996) that allows us to combine matching with an IV estimator. Unlike the standard 2SLS IV estimator, this non-parametric approach provides accurate confidence intervals and consistent causal estimates even when the instrument is weak. A further advantage of this non-parametric method is the opportunity it provides to probe the random assignment assumption with a sensitivity test. We provide substantive examples of the proposed approach with a reevaluation of a recent paper that uses rainfall as an instrument for voter turnout in US counties Hansford & Gomez (2010). The Polmeth 2011 poster: Stronger Instruments by Design. |

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 36, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9327273368835449, "perplexity_flag": "middle"}

http://mathoverflow.net/questions/76551/moving-the-origin-of-an-elliptic-curve

## Moving the origin of an elliptic curve ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Suppose one has an elliptic curve $E = (C,O)$ over a field $k$ where $C$ is a non-singular genus one curve over $k$ and $O$ is a $k$-rational point on $C$. By moving $O$ on $C$ one gets a family of elliptic curves. Is this family trivial in the sense that all the curves are mutually isomorphic? If not (as I suspect), what are the isomorphism classes of elliptic curves one gets this way and how are they related to the original curve $E$? Special cases of $k = \mathbb{C}$ and $\mathbb{Q}$ would be specially helpful to know about, for $\mathbb{C}$ in terms of lattices, and for $\mathbb{Q}$ in terms of the associated modular forms after Wiles et al. (Depending on the answer) should CM curves be distinguished? Thanks! - 5 Yes, they're isomorphic, and this is true for any algebraic group, because translation by a given ($k$-rational) group element is an isomorphism of algebraic varieties. This should be particularly easy to visualize in the case of ${\bf C}/L$ (translation by some complex number modulo the lattice $L$). – Noam D. Elkies Sep 27 2011 at 20:01 $C(k)$ is a group, and it acts on itself by multiplication. So if you have $E=(C,O)$ and $E'=(C,O')$ as you suggest on your question, wouldn't it be enough to look at the action of $C' - O$ ? – Reimundo Heluani Sep 27 2011 at 20:05 1 Is this a fishing expedition? – András Bátkai Sep 27 2011 at 22:01

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 21, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9405685663223267, "perplexity_flag": "head"}

http://mathhelpforum.com/advanced-algebra/203821-please-help-check-my-solutions-vector-space-questions.html

1Thanks • 1 Post By Deveno # Thread: 2. ## Re: Please help check my solutions for vector space questions From looking at the images you posted it seems that you are proving: $|V\cdot U|\le\|V\|\|U\|$ The standard proof follows from the fact $\frac{V\cdot U}{\|V\|\|U\|}=\cos(\theta)$ where $\theta$ is the angle between the two non-zero vectors $V~\&~U$. Then note that $0\le |\cos(\theta)|\le 1~.$ 3. ## Re: Please help check my solutions for vector space questions first of all, you have $\|U\|$ and $\|V\|$ wrong. $\|U\| = \sqrt{U \cdot U} = \sqrt{\int_0^{\frac{\pi}{2}} \sin^2x\ dx}$ $= \sqrt{\int_0^{\frac{\pi}{2}}(1 - \cos^2x)\ dx}$ $= \sqrt{\frac{\pi}{2} - \int_0^{\frac{\pi}{2}} \cos^2x\ dx}$ $= \sqrt{\frac{\pi}{2} - \int_0^{\frac{\pi}{2}} \frac{1 + \cos 2x}{2}\ dx}$ $= \sqrt{\frac{\pi}{2} - \frac{\pi}{4} - \frac{1}{4}\int_0^{\frac{\pi}{2}} \cos 2x\ 2dx}$ $= \sqrt{\frac{\pi}{4} - \frac{1}{4}(\sin (\pi) - \sin (0))}$ $= \frac{\sqrt{\pi}}{2}$ while: $\|V\| = \sqrt{V \cdot V} = \sqrt{\int_0^{\frac{\pi}{2}} \cos^2x\ dx}$ $= \sqrt{\int_0^{\frac{\pi}{2}} \frac{1+\cos 2x}{2}\ dx}$ $= \sqrt{\frac{\pi}{4} + \frac{1}{4}(\sin(\pi) - \sin(0))}$ $= \frac{\sqrt{\pi}}{2}$, as well. thus: $\|U\|\|V\| = \frac{\pi}{4}$. on the other side of the inequality we have: $|U \cdot V| = \left|\int_0^{\frac{\pi}{2}} \sin x \cos x\ dx \right|$ $= \left| \frac{\sin^2(\frac{\pi}{2})}{2} - \frac{\sin^2(0)}{2} \right| = \left|\frac{1}{2} \right| = \frac{1}{2}$ since $1 < \frac{\pi}{2}$, we indeed have for these particular U,V: $|U \cdot V| \leq \|U\|\|V\|$. ********* for your second problem, you are also calculating the norms of U and V incorrectly. use the definition of the inner product you are given! 4. ## Re: Please help check my solutions for vector space questions Thank you so much!! Deveno! I missed the obvious point Could you please re-check my second answer again? 5. ## Re: Please help check my solutions for vector space questions still incorrect. for starters, you have an extra 2-n term in $\|U\|$ and $\|V\|$. secondly, you don't seem to get that the inner product of two vectors should be a SCALAR, not an "expression". so for U, you should have: $\|U\| = \sqrt{\sum_{n=1}^{\infty} \frac{1}{2^n}} = 1$ and for V, you should have: $\|V\| = \sqrt{\sum_{n=1}^{\infty} \frac{1}{8^n}} = \frac{1}{\sqrt{7}}$ while for the dot product, you should have: $|U \cdot V| = \left| \sum_{n=1}^{\infty} \frac{1}{4^n} \right| = \frac{1}{3}$

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 28, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9051805734634399, "perplexity_flag": "middle"}

http://mathhelpforum.com/differential-geometry/171721-coordinate-charts-inverse-functions-print.html

# Coordinate Charts and inverse of functions. Printable View • February 18th 2011, 05:11 AM Bongo Coordinate Charts and inverse of functions. Hello, i have a question about this situation. Let denote $T^2=S^1 \times S^1$ the Torus. And let $f:\mathbb{R}^2 \rightarrow T^2$ be the projection defined by $f(x,y)=(e^{i2\pi x},e^{i2\pi y})$. The claim is now, if $\phi$ is a chart for $\mathbb{R}^2$ , s.t. $f_{|dom\phi}$ is injective, then $\phi \circ (f_{|dom\phi})^{-1}$ is a chart for $T^2$. i don't understand why the composition is a chart? Can you help me? Regards • February 18th 2011, 05:32 AM xxp9 As a chart means it's a 1-1 diffeomorphism in the domain that defined the map. Let V=dom(\phi), U=f(V), W=\phi^{-1}(V) Since f is injective in V, f is a 1-1 diffeomorphism between V and f(V). And \phi is defined to be a 1-1 diffeomorphism between W and \phi(W)=V. Compose two 1-1 diffeomorphism we get a 1-1 diffeomorphism. All times are GMT -8. The time now is 12:32 AM. Copyright © 2005-2013 Math Help Forum. All rights reserved. Copyright © 2005-2013 Math Help Forum. All rights reserved.

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.921599805355072, "perplexity_flag": "middle"}

http://math.stackexchange.com/questions/137057/is-the-set-of-conditions-for-devaneys-definition-of-chaos-minimal

# Is the set of conditions for Devaney's definition of chaos minimal? I am reading Devaney's definition of chaos. Which says: Let $V$ be a set. $f:V \rightarrow V$ is said to be chaotic on V if 1. $f$ has sensitive dependence on initial conditions 2. $f$ is topologically transitive 3. periodic points are dense in $V$ It seems that conditions 2 and 3 implies 1. Also, condition 2 seems to imply 1. Am I missing something here? - ## 2 Answers Brooks, Cairns, Davis and Stacey proved that 2. and 3. imply 1. on metric spaces with an infinite number of points. There are no more redundancies in general metric spaces. However, if $V=[a,b]\subset\mathbb{R}$ and $f$ is continuous, then 2. implies 1. and 3. This is a result of Vellekop and Berglund, published in the American Mathematical Monthly, vol. 101, 1994. - Thanks for the reply. However, I have some more related doubts. I would post them as a separate question so that it's easier to lookup later. – Abhishek Chanda Apr 26 '12 at 18:23 over a metric space see the following http://www2.math.uu.se/~warwick/vt04/DynSyst/reading/BanksEtAl.pdf -

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 7, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9223995804786682, "perplexity_flag": "head"}

http://quant.stackexchange.com/questions/tagged/risk-models+correlation

Tagged Questions 1answer 209 views What weights should be used when adjusting a correlation matrix to be positive definite? I have a correlation matrix $A$ for an equity market that is not positive definite. Higham (2002) proposes the Alternating Projections Method, minimising the weighted Frobenius norm $||A-X||_W$ where ... 1answer 663 views Cleansing covariance matrices via Random matrix theory I am exploring de-noising and cleansing of covariance matrices via Random Matrix Theory. RMT is a competitor to shrinkage methods of covariance estimation. There are various methods expressed usually ...

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8457196950912476, "perplexity_flag": "middle"}

http://mathinsight.org/function_machine_inverse

# Math Insight • Top • In threads When printing from Firefox, select the SVG renderer in the MathJax contextual menu (right click any math equation) for better printing results ### The function machine inverse #### Suggested background If $f$ is a function, we can represent it by a function machine that can take any input from its domain and transform it to an element from its codomain. In this illustration, we represent elements of the domain of $f$ by octahedra and elements of its codomain by spheres. An octahedron enters through the function machine's input funnel, and the function machine spits out a particular sphere, where the identity of the sphere depends on which octahedron went in. Now, imagine that for each unique octahedron, the function machine yielded a different sphere. Then, by examining the sphere that came out of the function machine's chute, it would be possible to determine which octahedron went into the funnel. (We'd be in trouble if multiple distinct octahedron yielded the same sphere, as there would be no way to distinguish which octahedron went into the funnel by examining the sphere.) In the case where we can examine the sphere and determine which octahedron went in, it makes sense that we could run $f$'s function machine backwards. We could put the sphere back in, and the backwards function machine could undo whatever $f$'s function machine did to produce the sphere. Then, the octahedron that started it all would pop out of the funnel. To show that we are running the machine backwards, let's flip it over so that the output chute is at the top. This backwards machine takes in spheres into the chute and spits out octahedron out of the funnel. Since this backwards machine undoes whatever processing $f$'s function machine did, we call it the inverse of $f$, or simply “$f$ inverse.” We give $f$ inverse a special symbol, too: $f^{-1}$. The superscript here does not denote exponentiation; it simply indicates we are running the function machine backwards to undo the processing $f$ did. If we denote a particular octahedron by the variable $x$, then we can denote by $f(x)$ the resulting sphere that $f$'s function machine spits out. Or, to make it shorter, we could just use the variable $y$, where $y=f(x)$. By the definition of $f^{-1}$, if we stick the sphere $y$ into the $f^{-1}$ backwards function machine, we must get the octahedron $x$ back, i.e., $f^{-1}(y)=x$. If we composed $f$ and $f^{-1}$ by connecting their function machines, then the resulting composition would not do anything to the octahedron $x$. We would put the octahedron in the combined function machine and the same octahedron would come out. In math symbols, we would find that $f^{-1}(f(x))=x$, which we could also write as $(f^{-1} \circ f)(x)=x$. You can read some examples of calculating the inverse function. #### Thread navigation ##### Math 1241 • Previous: Function machine parameters • Next: Inverse function examples #### Cite this as Nykamp DQ, “The function machine inverse.” From Math Insight. http://mathinsight.org/function_machine_inverse Keywords: function, function machine, inverse, inverse function

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 25, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9191346764564514, "perplexity_flag": "middle"}

http://mathhelpforum.com/calculus/95362-taylor-maclaurin-series.html

# Thread: 1. ## Taylor and Maclaurin series Find the Taylor and Maclaurin series of the given function with the given point $z_0$ as center and determine the radius of convergence. $e^{\frac{z^2}{2}}\int e^{-t^2} dt$ By the way, integrate from 0 to z and center at $z_0 = 0$ I cannot imagine, how can I solve this. 2. Originally Posted by noppawit Find the Taylor and Maclaurin series of the given function with the given point $z_0$ as center and determine the radius of convergence. $e^{\frac{z^2}{2}}\int e^{-t^2} dt$ By the way, integrate from 0 to z and center at $z_0 = 0$ I cannot imagine, how can I solve this. centered at 0 ... $e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + ...$ $e^{-t^2} = 1 - t^2 + \frac{t^4}{2!} - \frac{t^6}{3!} + ...$ $\int_0^z e^{-t^2} \, dt = \left[t - \frac{t^3}{3} + \frac{t^5}{5 \cdot 2!} - \frac{t^7}{7 \cdot 3!} + ... \right]_0^z = z - \frac{z^3}{3} + \frac{z^5}{5 \cdot 2!} - \frac{z^7}{7 \cdot 3!} + ...$

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 9, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8938313722610474, "perplexity_flag": "middle"}

http://math.stackexchange.com/questions/60538/probability-that-two-poker-hands-have-no-aces/60541

# Probability that two poker hands have no aces The probability of no two hands from the same poker deck having no aces is $$\frac{{48 \choose 5} + {48 \choose 5} - {48 \choose 10}}{{52 \choose 5}}$$ I am not sure why this is the answer, as the second draw is conditional on the first draw so it should be ${43 \choose 5}$. - 2 I think the first "no" is in excess? – joriki Aug 29 '11 at 15:37 Are you sure this is the right formula? In its current form, the numerator is negative. – Fixee Aug 29 '11 at 15:41 Where did you get that formula? The numerator is a negative number, and the expression evaluates to around - 2515. So that absolutely cannot be a probability... – Willie Wong♦ Aug 29 '11 at 15:47 1 (BTW, what's the difference between "two poker hands" and "a draw of ten cards"?) – Willie Wong♦ Aug 29 '11 at 15:49 ## 3 Answers This is not a probability, since $\binom{48}{10}$ is a lot bigger than the other ones and would make this a large negative number. It looks like someone was trying to work out the probability that at least one of the two hands has no aces using the inclusion-exclusion principle. The probability that one hand has no aces is $$\frac{\binom{48}5}{\binom{52}5}\;,$$ and the probability that one or the other hand has no aces is twice that, except that would overcount the case where they both have no aces, which is being counted twice, so we have to subtract the probability of both hands having no aces, which is $$\frac{\binom{48}{10}}{\binom{52}{10}}\;.$$ Note the $10$ in the denominator, which makes all the difference. So the probability of at least one hand having no aces is $$\frac{{48 \choose 5} + {48 \choose 5}}{52 \choose 5} - \frac{{48 \choose 10}}{{52 \choose 10}}=\frac{576}{637}\approx0.90\;.$$ - Nice answer as usual, but it's called subtract, not substract. – Hans Lundmark Aug 29 '11 at 16:53 1 @Hans: Thanks; fixed. – joriki Aug 29 '11 at 21:02 The formula in the question cannot be right (it has a negative numerator). Here's how I would find the probability: The number of sets of 10 cards (two hands) with no aces is ${48 \choose 10}$. The number of set of 10 cards from the deck is ${52 \choose 10}$. So the overall answer is $$\frac{{48 \choose 10}}{{52 \choose 10}} = \frac{246}{595} \approx 0.41$$ - I agree that it is not clear whether this is the question being asked or the one joriki answered. In any case, this is a good answer to this one. – Ross Millikan Aug 29 '11 at 19:25 I will assume, like Fixee did, that you want to find the probability that neither hand has an ace. Then Fixee's solution is the most efficient. A mild variant of it computes probabilities directly, bypassing binomial coefficients. The probability that the first card is not an ace is $48/52$. Given that the first card is not an ace, the probability that the second card is not an ace is $47/51$. So the probability that neither of the first two cards is an ace is $(48/52)(47/51)$. Continue in this way. We find that the required probability is $$\frac{48}{52}\cdot \frac{47}{51}\cdot \frac{46}{50}\cdots\cdot\frac{40}{44}\cdot \frac{39}{43}.$$ There is a good deal of easy cancellation. After a while we arrive at $246/595$. We can also find the answer by an argument that uses your idea. Suppose that one hand is dealt, then the other. The probability that the first hand has no aces is $$\frac{\binom{48}{5}}{\binom{52}{5}}.$$ The probability that the second hand has no aces, given that the first hand has no aces, is $$\frac{\binom{43}{5}}{\binom{47}{5}}$$ (there are $47$ cards left, of which $43$ are non-aces). Thus the required probability is $$\frac{\binom{48}{5}}{\binom{52}{5}}\cdot\frac{\binom{43}{5}}{\binom{47}{5}}.$$ - So what is the point of using the inclusion/exclusion principle in this problem? Conditional probability makes much more sense in this problem. I still cannot see how you use the inclusion exclusion principle in this problem because you will have fewer possibilities to choose the second hand after you have chosen the first hand. Any ideas as to how inclusion/exclusion intuitively makes sense if this is a conditional probability problem? – lord12 Sep 2 '11 at 15:44 The people posting answers are uncertain about the exact question you are asking. Fixee and I interpret the question as asking for the probability that none of the cards (in either hand) is an Ace. The interpretation of joriki is that you are asking for the probability that at least one hand has no Ace. These are different problems, with different answers. It would be helpful if you clarified which of these two interpretations is the one you intended. For the joriki interpretation, inclusion-exclusion is natural to use. (Continued) – André Nicolas Sep 2 '11 at 16:33 Finding the probability that hand $A$ has no Ace is easy, same for hand $B$. But if we add these two probabilities, we are double-counting the situations in which neither $A$ nor $B$ has an Ace, so we have to subtract. We could also solve the problem with joriki's interpretation by conditional probabilities, by finding first the probability that both hands have at least one Ace. The problem is that this divides into cases: $A$ has exactly $1$, and $B$ has $1$. $2$, or $3$; $A$ has exactly $2$, and $B$ has $1$ or $2$; $A$ has $3$, and $B$ has $1$. Much more work! – André Nicolas Sep 2 '11 at 16:41 Even in the inclusion/exclusion case, why would the probability of no Ace in hand B be the same as no ace in A. Meaning, why isn't it {48\choose 10} as opposed to {48 \choose 3}. If A1 = {Hand 1 has no ace} and A2 = {Hand 2 has no ace}, the number of possibilities cannot be equal to the number of possibilities of hand 1. I may have a hard time understanding this. Could you give me a similar problem that is similar to this that uses the inclusion/exclusion principle? – lord12 Sep 2 '11 at 19:16 The cards don't "know" which hand is being dealt first. The probability that the second card is the Ace of Spades is the same as the probability that the ninth card is the Ace of Spades. Problem: A bin has 30 identically wrapped candies, 10 chocolate and 20 strawberry. $A$ and $B$ each take $3$ candies. What is the probability at least one of them gets no chocolate candy? – André Nicolas Sep 2 '11 at 19:50 show 2 more comments

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 33, "mathjax_display_tex": 9, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9641643762588501, "perplexity_flag": "head"}

http://en.wikibooks.org/wiki/Engineering_Acoustics/Bass_Reflex_Enclosure_Design

# Engineering Acoustics/Bass Reflex Enclosure Design Authors · Print · License Edit this template Part 1: Lumped Acoustical Systems – 1.1 – 1.2 – 1.3 – 1.4 – 1.5 – 1.6 – 1.7 – 1.8 – 1.9 – 1.10 – 1.11 Part 2: One-Dimensional Wave Motion – 2.1 – 2.2 – 2.3 Part 3: Applications – 3.1 – 3.2 – 3.3 – 3.4 – 3.5 – 3.6 – 3.7 – 3.8 – 3.9 – 3.10 – 3.11 – 3.12 – 3.13 – 3.14 – 3.15 – 3.16 – 3.17 – 3.18 – 3.19 – 3.20 – 3.21 – 3.22 – 3.23 – 3.24 ## Introduction Bass-reflex enclosures improve the low-frequency response of loudspeaker systems. Bass-reflex enclosures are also called "vented-box design" or "ported-cabinet design". A bass-reflex enclosure includes a vent or port between the cabinet and the ambient environment. This type of design, as one may observe by looking at contemporary loudspeaker products, is still widely used today. Although the construction of bass-reflex enclosures is fairly simple, their design is not simple, and requires proper tuning. This reference focuses on the technical details of bass-reflex design. General loudspeaker information can be found here. ## Effects of the Port on the Enclosure Response Before discussing the bass-reflex enclosure, it is important to be familiar with the simpler sealed enclosure system performance. As the name suggests, the sealed enclosure system attaches the loudspeaker to a sealed enclosure (except for a small air leak included to equalize the ambient pressure inside). Ideally, the enclosure would act as an acoustical compiance element, as the air inside the enclosure is compressed and rarified. Often, however, an acoustic material is added inside the box to reduce standing waves, dissipate heat, and other reasons. This adds a resistive element to the acoustical lumped-element model. A non-ideal model of the effect of the enclosure actually adds an acoustical mass element to complete a series lumped-element circuit given in Figure 1. For more on sealed enclosure design, see the Sealed Box Subwoofer Design page. Figure 1. Sealed enclosure acoustic circuit. In the case of a bass-reflex enclosure, a port is added to the construction. Typically, the port is cylindrical and is flanged on the end pointing outside the enclosure. In a bass-reflex enclosure, the amount of acoustic material used is usually much less than in the sealed enclosure case, often none at all. This allows air to flow freely through the port. Instead, the larger losses come from the air leakage in the enclosure. With this setup, a lumped-element acoustical circuit has the following form. Figure 2. Bass-reflex enclosure acoustic circuit. In this figure, $Z_{RAD}$ represents the radiation impedance of the outside environment on the loudspeaker diaphragm. The loading on the rear of the diaphragm has changed when compared to the sealed enclosure case. If one visualizes the movement of air within the enclosure, some of the air is compressed and rarified by the compliance of the enclosure, some leaks out of the enclosure, and some flows out of the port. This explains the parallel combination of $M_{AP}$, $C_{AB}$, and $R_{AL}$. A truly realistic model would incorporate a radiation impedance of the port in series with $M_{AP}$, but for now it is ignored. Finally, $M_{AB}$, the acoustical mass of the enclosure, is included as discussed in the sealed enclosure case. The formulas which calculate the enclosure parameters are listed in Appendix B. It is important to note the parallel combination of $M_{AP}$ and $C_{AB}$. This forms a Helmholtz resonator (click here for more information). Physically, the port functions as the “neck” of the resonator and the enclosure functions as the “cavity.” In this case, the resonator is driven from the piston directly on the cavity instead of the typical Helmholtz case where it is driven at the “neck.” However, the same resonant behavior still occurs at the enclosure resonance frequency, $f_{B}$. At this frequency, the impedance seen by the loudspeaker diaphragm is large (see Figure 3 below). Thus, the load on the loudspeaker reduces the velocity flowing through its mechanical parameters, causing an anti-resonance condition where the displacement of the diaphragm is a minimum. Instead, the majority of the volume velocity is actually emitted by the port itself instead of the loudspeaker. When this impedance is reflected to the electrical circuit, it is proportional to $1/Z$, thus a minimum in the impedance seen by the voice coil is small. Figure 3 shows a plot of the impedance seen at the terminals of the loudspeaker. In this example, $f_B$ was found to be about 40 Hz, which corresponds to the null in the voice-coil impedance. Figure 3. Impedances seen by the loudspeaker diaphragm and voice coil. ## Quantitative Analysis of Port on Enclosure The performance of the loudspeaker is first measured by its velocity response, which can be found directly from the equivalent circuit of the system. As the goal of most loudspeaker designs is to improve the bass response (leaving high-frequency production to a tweeter), low frequency approximations will be made as much as possible to simplify the analysis. First, the inductance of the voice coil, $\it{L_E}$, can be ignored as long as $\omega \ll R_E/L_E$. In a typical loudspeaker, $\it{L_E}$ is of the order of 1 mH, while $\it{R_E}$ is typically 8$\Omega$, thus an upper frequency limit is approximately 1 kHz for this approximation, which is certainly high enough for the frequency range of interest. Another approximation involves the radiation impedance, $\it{Z_{RAD}}$. It can be shown [1] that this value is given by the following equation (in acoustical ohms): $Z_{RAD} = \frac{\rho_0c}{\pi a^2}\left[\left(1 - \frac{J_1(2ka)}{ka}\right) + j\frac{H_1(2ka)}{ka}\right]$ Where $J_1(x)$ and $H_1(x)$ are types of Bessel functions. For small values of ka, $J_1(2ka) \approx ka$ and $H_1(2ka) \approx \frac{8(ka)^2}{3\pi}$ $\Rightarrow Z_{RAD} \approx j\frac{8\rho_0\omega}{3\pi^2a} = jM_{A1}$ Hence, the low-frequency impedance on the loudspeaker is represented with an acoustic mass $M_{A1}$ [1]. For a simple analysis, $R_E$, $M_{MD}$, $C_{MS}$, and $R_{MS}$ (the transducer parameters, or Thiele-Small parameters) are converted to their acoustical equivalents. All conversions for all parameters are given in Appendix A. Then, the series masses, $M_{AD}$, $M_{A1}$, and $M_{AB}$, are lumped together to create $M_{AC}$. This new circuit is shown below. Figure 4. Low-Frequency Equivalent Acoustic Circuit Unlike sealed enclosure analysis, there are multiple sources of volume velocity that radiate to the outside environment. Hence, the diaphragm volume velocity, $U_D$, is not analyzed but rather $U_0 = U_D + U_P + U_L$. This essentially draws a “bubble” around the enclosure and treats the system as a source with volume velocity $U_0$. This “lumped” approach will only be valid for low frequencies, but previous approximations have already limited the analysis to such frequencies anyway. It can be seen from the circuit that the volume velocity flowing into the enclosure, $U_B = -U_0$, compresses the air inside the enclosure. Thus, the circuit model of Figure 3 is valid and the relationship relating input voltage, $V_{IN}$ to $U_0$ may be computed. In order to make the equations easier to understand, several parameters are combined to form other parameter names. First, $\omega_B$ and $\omega_S$, the enclosure and loudspeaker resonance frequencies, respectively, are: $\omega_B = \frac{1}{\sqrt{M_{AP}C_{AB}}}$ $\omega_S = \frac{1}{\sqrt{M_{AC}C_{AS}}}$ Based on the nature of the derivation, it is convenient to define the parameters $\omega_0$ and h, the Helmholtz tuning ratio: $\omega_0 = \sqrt{\omega_B\omega_S}$ $h = \frac{\omega_B}{\omega_S}$ A parameter known as the compliance ratio or volume ratio, $\alpha$, is given by: $\alpha = \frac{C_{AS}}{C_{AB}} = \frac{V_{AS}}{V_{AB}}$ Other parameters are combined to form what are known as quality factors: $Q_L = R_{AL}\sqrt{\frac{C_{AB}}{M_{AP}}}$ $Q_{TS} = \frac{1}{R_{AE}+R_{AS}}\sqrt{\frac{M_{AC}}{C_{AS}}}$ This notation allows for a simpler expression for the resulting transfer function [1]: $\frac{U_0}{V_{IN}} = G(s) = \frac{(s^3/\omega_0^4)}{(s/\omega_0)^4+a_3(s/\omega_0)^3+a_2(s/\omega_0)^2+a_1(s/\omega_0)+1}$ where $a_1 = \frac{1}{Q_L\sqrt{h}}+\frac{\sqrt{h}}{Q_{TS}}$ $a_2 = \frac{\alpha+1}{h}+h+\frac{1}{Q_L Q_{TS}}$ $a_3 = \frac{1}{Q_{TS}\sqrt{h}}+\frac{\sqrt{h}}{Q_L}$ ## Development of Low-Frequency Pressure Response It can be shown [2] that for $ka < 1/2$, a loudspeaker behaves as a spherical source. Here, a represents the radius of the loudspeaker. For a 15” diameter loudspeaker in air, this low frequency limit is about 150 Hz. For smaller loudspeakers, this limit increases. This limit dominates the limit which ignores $L_E$, and is consistent with the limit that models $Z_{RAD}$ by $M_{A1}$. Within this limit, the loudspeaker emits a volume velocity $U_0$, as determined in the previous section. For a simple spherical source with volume velocity $U_0$, the far-field pressure is given by [1]: $p(r) \simeq j\omega\rho_0 U_0 \frac{e^{-jkr}}{4\pi r}$ It is possible to simply let $r = 1$ for this analysis without loss of generality because distance is only a function of the surroundings, not the loudspeaker. Also, because the transfer function magnitude is of primary interest, the exponential term, which has a unity magnitude, is omitted. Hence, the pressure response of the system is given by [1]: $\frac{p}{V_{IN}} = \frac{\rho_0s}{4\pi}\frac{U_0}{V_{IN}} = \frac{\rho_0Bl}{4\pi S_DR_EM_AS}H(s)$ Where $H(s) = sG(s)$. In the following sections, design methods will focus on $|H(s)|^2$ rather than $H(s)$, which is given by: $|H(s)|^2 = \frac{\Omega^8}{\Omega^8 + \left(a^2_3 - 2a_2\right)\Omega^6 + \left(a^2_2 + 2 - 2a_1a_3\right)\Omega^4 + \left(a^2_1 - 2a_2\right)\Omega^2 + 1}$ $\Omega = \frac{\omega}{\omega_0}$ This also implicitly ignores the constants in front of $|H(s)|$ since they simply scale the response and do not affect the shape of the frequency response curve. ## Alignments A popular way to determine the ideal parameters has been through the use of alignments. The concept of alignments is based upon filter theory. Filter development is a method of selecting the poles (and possibly zeros) of a transfer function to meet a particular design criterion. The criteria are the desired properties of a magnitude-squared transfer function, which in this case is $|H(s)|^2$. From any of the design criteria, the poles (and possibly zeros) of $|H(s)|^2$ are found, which can then be used to calculate the numerator and denominator. This is the “optimal” transfer function, which has coefficients that are matched to the parameters of $|H(s)|^2$ to compute the appropriate values that will yield a design that meets the criteria. There are many different types of filter designs, each which have trade-offs associated with them. However, this design is limited because of the structure of $|H(s)|^2$. In particular, it has the structure of a fourth-order high-pass filter with all zeros at s = 0. Therefore, only those filter design methods which produce a low-pass filter with only poles will be acceptable methods to use. From the traditional set of algorithms, only Butterworth and Chebyshev low-pass filters have only poles. In addition, another type of filter called a quasi-Butterworth filter can also be used, which has similar properties to a Butterworth filter. These three algorithms are fairly simple, thus they are the most popular. When these low-pass filters are converted to high-pass filters, the $s \rightarrow 1/s$ transformation produces $s^8$ in the numerator. More details regarding filter theory and these relationships can be found in numerous resources, including [5]. ## Butterworth Alignment The Butterworth algorithm is designed to have a maximally flat pass band. Since the slope of a function corresponds to its derivatives, a flat function will have derivatives equal to zero. Since as flat of a pass band as possible is optimal, the ideal function will have as many derivatives equal to zero as possible at s = 0. Of course, if all derivatives were equal to zero, then the function would be a constant, which performs no filtering. Often, it is better to examine what is called the loss function. Loss is the reciprocal of gain, thus $|\hat{H}(s)|^2 = \frac{1}{|H(s)|^2}$ The loss function can be used to achieve the desired properties, then the desired gain function is recovered from the loss function. Now, applying the desired Butterworth property of maximal pass-band flatness, the loss function is simply a polynomial with derivatives equal to zero at s = 0. At the same time, the original polynomial must be of degree eight (yielding a fourth-order function). However, derivatives one through seven can be equal to zero if [3] $|\hat{H}(\Omega)|^2 = 1 + \Omega^8 \Rightarrow |H(\Omega)|^2 = \frac{1}{1 + \Omega^8}$ With the high-pass transformation $\Omega \rightarrow 1/\Omega$, $|H(\Omega)|^2 = \frac{\Omega^8}{\Omega^8 + 1}$ It is convenient to define $\Omega = \omega/\omega_{3dB}$, since $\Omega = 1 \Rightarrow |H(s)|^2 = 0.5$ or -3 dB. This definition allows the matching of coefficients for the $|H(s)|^2$ describing the loudspeaker response when $\omega_{3dB} = \omega_0$. From this matching, the following design equations are obtained [1]: $a_1 = a_3 = \sqrt{4+2\sqrt{2}}$ $a_2 = 2+\sqrt{2}$ ## Quasi-Butterworth Alignment The quasi-Butterworth alignments do not have as well-defined of an algorithm when compared to the Butterworth alignment. The name “quasi-Butterworth” comes from the fact that the transfer functions for these responses appear similar to the Butterworth ones, with (in general) the addition of terms in the denominator. This will be illustrated below. While there are many types of quasi-Butterworth alignments, the simplest and most popular is the 3rd order alignment (QB3). The comparison of the QB3 magnitude-squared response against the 4th order Butterworth is shown below. $\left|H_{QB3}(\omega)\right|^2 = \frac{(\omega/\omega_{3dB})^8}{(\omega/\omega_{3dB})^8 + B^2(\omega/\omega_{3dB})^2 + 1}$ $\left|H_{B4}(\omega)\right|^2 = \frac{(\omega/\omega_{3dB})^8}{(\omega/\omega_{3dB})^8 + 1}$ Notice that the case $B = 0$ is the Butterworth alignment. The reason that this QB alignment is called 3rd order is due to the fact that as B increases, the slope approaches 3 dec/dec instead of 4 dec/dec, as in 4th order Butterworth. This phenomenon can be seen in Figure 5. Figure 5: 3rd-Order Quasi-Butterworth Response for $0.1 \leq B \leq 3$ Equating the system response $|H(s)|^2$ with $|H_{QB3}(s)|^2$, the equations guiding the design can be found [1]: $B^2 = a^2_1 - 2a_2$ $a_2^2 + 2 = 2a_1a_3$ $a_3 = \sqrt{2a_2}$ $a_2 > 2 + \sqrt{2}$ ## Chebyshev Alignment The Chebyshev algorithm is an alternative to the Butterworth algorithm. For the Chebyshev response, the maximally-flat passband restriction is abandoned. Now, a ripple, or fluctuation is allowed in the pass band. This allows a steeper transition or roll-off to occur. In this type of application, the low-frequency response of the loudspeaker can be extended beyond what can be achieved by Butterworth-type filters. An example plot of a Chebyshev high-pass response with 0.5 dB of ripple against a Butterworth high-pass response for the same $\omega_{3dB}$ is shown below. Figure 6: Chebyshev vs. Butterworth High-Pass Response. The Chebyshev response is defined by [4]: $|\hat{H}(j\Omega)|^2 = 1 + \epsilon^2C^2_n(\Omega)$ $C_n(\Omega)$ is called the Chebyshev polynomial and is defined by [4]: | | | | |------------------------------------------------------------------------------|------------------------------------------------------------------------------|-----------------------------------------------------------------------------| | $C_n(\Omega) = \big\lbrace$ | $\rm{cos}[\it{n}\rm{cos}^{-1}(\Omega)]$ | $|\Omega| < 1$ | | $\rm{cosh}[\it{n}\rm{cosh}^{-1}(\Omega)]$ | $|\Omega| > 1$ | | Fortunately, Chebyshev polynomials satisfy a simple recursion formula [4]: $C_0(x) = 1$ $C_1(x) = x$ $C_n(x) = 2xC_{n-1} - C_{n-2}$ For more information on Chebyshev polynomials, see the Wolfram Mathworld: Chebyshev Polynomials page. When applying the high-pass transformation to the 4th order form of $|\hat{H}(j\Omega)|^2$, the desired response has the form [1]: $|H(j\Omega)|^2 = \frac{1+\epsilon^2}{1+\epsilon^2C^2_4(1/\Omega)}$ The parameter $\epsilon$ determines the ripple. In particular, the magnitude of the ripple is $10\rm{log}[1+\epsilon^2]$ dB and can be chosen by the designer, similar to B in the quasi-Butterworth case. Using the recursion formula for $C_n(x)$, $C_4\left(\frac{1}{\Omega}\right) = 8\left(\frac{1}{\Omega}\right)^4 - 8\left(\frac{1}{\Omega}\right)^2 + 1$ Applying this equation to $|H(j\Omega)|^2$ [1], $\Rightarrow |H(\Omega)|^2 = \frac{\frac{1 + \epsilon^2}{64\epsilon^2}\Omega^8}{\frac{1 + \epsilon^2}{64\epsilon^2}\Omega^8 + \frac{1}{4}\Omega^6 + \frac{5}{4}\Omega^4 - 2\Omega^2 + 1}$ $\Omega = \frac{\omega}{\omega_n}$ $\omega_n = \frac{\omega_{3dB}}{2}\sqrt{2 + \sqrt{2 + 2\sqrt{2+\frac{1}{\epsilon^2}}}}$ Thus, the design equations become [1]: | | | | |------------------------------------------------------------------------------|------------------------------------------------------------------------------|------------------------------------------------------------------------------| | $\omega_0 = \omega_n\sqrt[8]{\frac{64\epsilon^2}{1+\epsilon^2}}$ | $k = \rm{tanh}\left[\frac{1}{4}\rm{sinh}^{-1}\left(\frac{1}{\epsilon}\right)\right]$ | $D = \frac{k^4 + 6k^2 + 1}{8}$ | | $a_1 = \frac{k\sqrt{4 + 2\sqrt{2}}}{\sqrt[4]{D}},$ | $a_2 = \frac{1 + k^2(1+\sqrt{2})}{\sqrt{D}}$ | $a_3 = \frac{a_1}{\sqrt{D}}\left[1 - \frac{1 - k^2}{2\sqrt{2}}\right]$ | ## Choosing the Correct Alignment With all the equations that have already been presented, the question naturally arises, “Which one should I choose?” Notice that the coefficients $a_1$, $a_2$, and $a_3$ are not simply related to the parameters of the system response. Certain combinations of parameters may indeed invalidate one or more of the alignments because they cannot realize the necessary coefficients. With this in mind, general guidelines have been developed to guide the selection of the appropriate alignment. This is very useful if one is designing an enclosure to suit a particular transducer that cannot be changed. The general guideline for the Butterworth alignment focuses on $Q_L$ and $Q_{TS}$. Since the three coefficients $a_1$, $a_2$, and $a_3$ are a function of $Q_L$, $Q_{TS}$, h, and $\alpha$, fixing one of these parameters yields three equations that uniquely determine the other three. In the case where a particular transducer is already given, $Q_{TS}$ is essentially fixed. If the desired parameters of the enclosure are already known, then $Q_L$ is a better starting point. In the case that the rigid requirements of the Butterworth alignment cannot be satisfied, the quasi-Butterworth alignment is often applied when $Q_{TS}$ is not large enough.. The addition of another parameter, B, allows more flexibility in the design. For $Q_{TS}$ values that are too large for the Butterworth alignment, the Chebyshev alignment is typically chosen. However, the steep transition of the Chebyshev alignment may also be utilized to attempt to extend the bass response of the loudspeaker in the case where the transducer properties can be changed. In addition to these three popular alignments, research continues in the area of developing new algorithms that can manipulate the low-frequency response of the bass-reflex enclosure. For example, a 5th order quasi-Butterworth alignment has been developed [6]. Another example [7] applies root-locus techniques to achieve results. In the modern age of high-powered computing, other researchers have focused their efforts in creating computerized optimization algorithms that can be modified to achieve a flatter response with sharp roll-off or introduce quasi-ripples which provide a boost in sub-bass frequencies [8]. Back to Engineering Acoustics ## References [1] Leach, W. Marshall, Jr. Introduction to Electroacoustics and Audio Amplifier Design. 2nd ed. Kendall/Hunt, Dubuque, IA. 2001. [2] Beranek, L. L. Acoustics. 2nd ed. Acoustical Society of America, Woodbridge, NY. 1993. [3] DeCarlo, Raymond A. “The Butterworth Approximation.” Notes from ECE 445. Purdue University. 2004. [4] DeCarlo, Raymond A. “The Chebyshev Approximation.” Notes from ECE 445. Purdue University. 2004. [5] VanValkenburg, M. E. Analog Filter Design. Holt, Rinehart and Winston, Inc. Chicago, IL. 1982. [6] Kreutz, Joseph and Panzer, Joerg. "Derivation of the Quasi-Butterworth 5 Alignments." Journal of the Audio Engineering Society. Vol. 42, No. 5, May 1994. [7] Rutt, Thomas E. "Root-Locus Technique for Vented-Box Loudspeaker Design." Journal of the Audio Engineering Society. Vol. 33, No. 9, September 1985. [8] Simeonov, Lubomir B. and Shopova-Simeonova, Elena. "Passive-Radiator Loudspeaker System Design Software Including Optimization Algorithm." Journal of the Audio Engineering Society. Vol. 47, No. 4, April 1999. ## Appendix A: Equivalent Circuit Parameters Name Electrical Equivalent Mechanical Equivalent Acoustical Equivalent Voice-Coil Resistance $R_E$ $R_{ME} = \frac{(Bl)^2}{R_E}$ $R_{AE} = \frac{(Bl)^2}{R_ES^2_D}$ Driver (Speaker) Mass See $C_{MEC}$ $M_{MD}$ $M_{AD} = \frac{M_{MD}}{S^2_D}$ Driver (Speaker) Suspension Compliance $L_{CES} = (Bl)^2C_{MS}$ $C_{MS}$ $C_{AS} = S^2_DC_{MS}$ Driver (Speaker) Suspension Resistance $R_{ES} = \frac{(Bl)^2}{R_{MS}}$ $R_{MS}$ $R_{AS} = \frac{R_{MS}}{S^2_D}$ Enclosure Compliance $L_{CEB} = \frac{(Bl)^2C_{AB}}{S^2_D}$ $C_{MB} = \frac{C_{AB}}{S^2_D}$ $C_{AB}$ Enclosure Air-Leak Losses $R_{EL} = \frac{(Bl)^2}{S^2_DR_{AL}}$ $R_{ML} = S^2_DR_{AL}$ $R_{AL}$ Acoustic Mass of Port $C_{MEP} = \frac{S^2_DM_{AP}}{(Bl)^2}$ $M_{MP} = S^2_DM_{AP}$ $M_{AP}$ Enclosure Mass Load See $C_{MEC}$ See $M_{MC}$ $M_{AB}$ Low-Frequency Radiation Mass Load See $C_{MEC}$ See $M_{MC}$ $M_{A1}$ Combination Mass Load $C_{MEC} = \frac{S^2_DM_{AC}}{(Bl)^2}$ $= \frac{S^2_D(M_{AB} + M_{A1}) + M_{MD}}{(Bl)^2}$ $M_{MC} = S^2_D(M_{AB} + M_{A1}) + M_{MD}$ $M_{AC} = M_{AD} + M_{AB} + M_{A1}$ $= \frac{M_{MD}}{S^2_D} + M_{AB} + M_{A1}$ ## Appendix B: Enclosure Parameter Formulas Figure 7: Important dimensions of bass-reflex enclosure. Based on these dimensions [1], $C_{AB} = \frac{V_{AB}}{\rho_0c^2_0}$ $M_{AB} = \frac{B\rho_{eff}}{\pi a}$ $B = \frac{d}{3}\left(\frac{S_D}{S_B}\right)^2\sqrt{\frac{\pi}{S_D}} + \frac{8}{3\pi}\left[1 - \frac{S_D}{S_B}\right]$ $\rho_0 \leq \rho_{eff} \leq \rho_0\left(1 - \frac{V_{fill}}{V_B}\right) + \rho_{fill}\frac{V_{fill}}{V_B}$ $V_{AB} = V_B\left[1-\frac{V_{fill}}{V_B}\right]\left[1 + \frac{\gamma - 1}{1 + \gamma\left(\frac{V_B}{V_{fill}} - 1\right)\frac{\rho_0c_{air}}{\rho_{fill}c_{fill}}}\right]$ $V,= hwd$ (inside enclosure volume) $S_B = wh$ (inside area of the side the speaker is mounted on) $c_{air} =$specific heat of air at constant volume $c_{fill} =$specific heat of filling at constant volume ($V_{filling}$) $\rho_0 =$mean density of air (about 1.3 kg/$\rm m^3$) $\rho_{fill} =$ density of filling $\gamma =$ ratio of specific heats for air (1.4) $c_0 =$ speed of sound in air (about 344 m/s) $\rho_{eff}$ = effective density of enclosure. If little or no filling (acceptable assumption in a bass-reflex system but not for sealed enclosures), $\rho_{eff} \approx \rho_0$

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 185, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.90074622631073, "perplexity_flag": "middle"}

http://mathhelpforum.com/discrete-math/32092-self-contradictory-recursive-problem.html

# Thread: 1. ## self-contradictory recursive problem "Let the barber of Seville shave every man of Seville who does not shave himself. Who shall shave the barber?" So my book says that this is self-contradictory. but how? From my reading of it, it's just that the barber of Seville shaves every man in Seville who doesn't shave themselves. Well the barber doesn't have to shave himself (or he could shave himself). So then someone else can shave the barber along with every man of Seville (and those who do not shave themselves can be shaved twice...) 2. Who shaves him? If he shaves himself that is impossible for he shaves everyone who does not shave himself. If he does not shave himself then he has to shave himself because he shaves everyone who does not shave himself. 3. Originally Posted by inquilinekea "Let the barber of Seville shave every man of Seville who does not shave himself. Who shall shave the barber?" So my book says that this is self-contradictory. but how? From my reading of it, it's just that the barber of Seville shaves every man in Seville who doesn't shave themselves. Well the barber doesn't have to shave himself (or he could shave himself). So then someone else can shave the barber along with every man of Seville (and those who do not shave themselves can be shaved twice...) Another possibility is that nobody shaves the barber because she doesn't grow a beard. This question ilustrates the difference between mathematics and real life. A mathematical model of a real-life problem often contains many unstated assumptions. In this case, it is assumed that the men of Seville form a precisely identified set which contains the barber and on which the relation "shaves" is a function. That is to say, each man in the set is shaved by exactly one man in the set. 4. Self-referential statements can lead to all kinds of fun paradoxes. Whether or not a statement is actually paradoxical and how one can resolve the paradox is a topic of much discussion in philosophy and mathematics. Your problem is an example of Russel's Paradox: Does the set of all those sets that do not contain themselves contain itself? One of my favorites is: "The smallest integer not definable in fewer than twelve English words." This takes a bit of thought to see why it is a paradox. 5. Originally Posted by iknowone Your problem is an example of Russel's Paradox: Does the set of all those sets that do not contain themselves contain itself? But it is not really a paradox in math. The point of this paradox is to show that the set of all sets cannot exist. Suppose that $\mathcal{S}$ is the set of all sets then by the Axiom Schema of Compresension that set $\{ x \in \mathcal{S} | x\not \in x \}$ will exist. But this set will lead to a contradiction. Thus, the set of all sets cannot exist.

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9600068926811218, "perplexity_flag": "middle"}

http://physics.stackexchange.com/questions/21044/image-charges-laplace-equation-and-uniqueness-theorem/21068

# Image charges, laplace equation and uniqueness theorem Consider a well-known problem of the electric field generated by a system composed of a point charge in proximity of a large earthed conductor. It is said that the potential due to an image charge satifies all the boundary conditions - it this case, constant potential on the surface of a conductor - and therefore, by the uniqueness theorem, it is the only possible potential distribution. However, we can of course imagine other non-trivial fields satisfying the boundary condition of constant potential on the surface, for example generated by three imaginary charges such that the whole system is again symmetric. So my question is: what is the full set of boundary conditions needed to find the unique potential distribution? This specific example is just an illustration of a more general question: how to use Laplace/Poisson equations to describe the field of point charges? The divergence of electric field has a singularity at the point charge. Another incarnation of this problem occurs when we want to use the formula for the energy stored in the electric field - we find the energy of the electric field generated by a point charge to be infinite. Does this mean that the concept of a point charge is too idealised? - Where would these three imaginary charges be placed? – David Zaslavsky♦ Feb 15 '12 at 23:46 Energy of a point charge is infinite. If you had to create a pair of point charges from a charge neutral area, you would have to (classically) supply infinite energy because of the $\frac{-kq^2}{r=0}$ infinite negative potential energy of two point charges making up a neutral point. – Manishearth♦ Feb 16 '12 at 7:42 The opposite also holds:If you destroy a pair of charges by bringing them together and making an overall charge-neutral point, then the infinite energy of the field will be radiated. Note that this is all by classical mechanics. Oh, and currently, we believe that electrons(leptons), W particles, and possibly quarks are true point charges.. – Manishearth♦ Feb 16 '12 at 7:45 ## 2 Answers However, we can of course imagine other non-trivial fields satisfying the boundary condition of constant potential on the surface, for example generated by three imaginary charges such that the whole system is again symmetric. I guess you mean that putting another set of image charges on both sides of the conductor plane so that the system is symmetric again. But point charges at specific places are also part of the boundary condition! Image charges cannot be put in the space where your electric field is to be calculated. In summary, there is only one way to put image charges, and there is only one solution of the electric field. how to use Laplace/Poisson equations to describe the field of point charges? The divergence of electric field has a singularity at the point charge. Point charges can be mathematically described by Dirac Delta function. Does this mean that the concept of a point charge is too idealised? Apparently yes, by classical electromagnetism. As electrons are recognized nowadays as true point charges, and they obviously don't have infinite energy, I guess there is some quantum effect when the radius goes down (vacuum polarization, blah blah blah). - – Willie Wong Feb 16 '12 at 15:11 There is no other solution of Laplace's equation satisfying the boundary condition of zero potential on the conductor and at infinity. This can be proved as follows: supposing there are two solutions $\phi_1$ and $\phi_2$ on one side of the conducting plate, with a point charge q at a given position. Then $-\phi_2$ has a point charge of $-q$ at the same position, so that the difference $\phi_1-\phi_2$ satisfies the free Laplace equation on the domain, with no charges. But the solution of Laplace's equation is the minimum of the field energy $$\int |\nabla \phi|^2$$ So that the minimum of the free equation is achieved only when the gradient is everywhere zero, and this means that $\phi$ is constant. -

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 7, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9389435052871704, "perplexity_flag": "head"}

http://medlibrary.org/medwiki/Linear_map

Linear map Welcome to MedLibrary.org. For best results, we recommend beginning with the navigation links at the top of the page, which can guide you through our collection of over 14,000 medication labels and package inserts. For additional information on other topics which are not covered by our database of medications, just enter your topic in the search box below: In mathematics, a linear map (also called a linear mapping, linear transformation, linear operator or, in some contexts, linear function) is a function between two modules (including vector spaces) that preserves (in the sense defined below) the operations of module (or vector) addition and scalar multiplication. As a result, it always maps linear subspaces to linear subspaces, like straight lines to straight lines or to a single point. The expression "linear operator" is commonly used for a linear map from a vector space to itself (i.e., endomorphism). Sometimes the definition of a linear function coincides with that of a linear map, while in analytic geometry it does not. In the language of abstract algebra, a linear map is a homomorphism of modules. In the language of category theory it is a morphism in the category of modules over a given ring. Definition and first consequences[] Let V and W be vector spaces over the same field K. A function f: V → W is said to be a linear map if for any two vectors x and y in V and any scalar α in K, the following two conditions are satisfied: $f(\mathbf{x}+\mathbf{y}) = f(\mathbf{x})+f(\mathbf{y}) \!$ additivity $f(\alpha \mathbf{x}) = \alpha f(\mathbf{x}) \!$ homogeneity of degree 1 This is equivalent to requiring the same for any linear combination of vectors, i.e. that for any vectors x1, ..., xm ∈ V and scalars a1, ..., am ∈ K, the following equality holds: $f(a_1 \mathbf{x}_1+\cdots+a_m \mathbf{x}_m) = a_1 f(\mathbf{x}_1)+\cdots+a_m f(\mathbf{x}_m). \!$ Denoting the zero elements of the vector spaces V and W by 0V and 0W respectively, it follows that f(0V) = 0W because letting α = 0 in the equation for homogeneity of degree 1, $f(\mathbf{0}_{V}) = f(0 \cdot \mathbf{0}_{V}) = 0 \cdot f(\mathbf{0}_{V}) = \mathbf{0}_{W} .$ Occasionally, V and W can be considered to be vector spaces over different fields. It is then necessary to specify which of these ground fields is being used in the definition of "linear". If V and W are considered as spaces over the field K as above, we talk about K-linear maps. For example, the conjugation of complex numbers is an R-linear map C → C, but it is not C-linear. A linear map from V to K (with K viewed as a vector space over itself) is called a linear functional. These statements generalize to any left-module RM over a ring R without modification. Examples[] • The identity map and zero map are linear. • The map $x\mapsto cx$, where c is a constant, is linear. • For real numbers, the map $x\mapsto x^2$ is not linear. • For real numbers, the map $x\mapsto x+1$ is not linear (but is an affine transformation, and also a linear function, as defined in analytic geometry.) • If A is a real m × n matrix, then A defines a linear map from Rn to Rm by sending the column vector x ∈ Rn to the column vector Ax ∈ Rm. Conversely, any linear map between finite-dimensional vector spaces can be represented in this manner; see the following section. • The (definite) integral is a linear map from the space of all real-valued integrable functions on some interval to R • The (indefinite) integral (or antiderivative) is not considered a linear transformation, as the use of a constant of integration results in an infinite number of outputs per input. • Differentiation is a linear map from the space of all differentiable functions to the space of all functions. • If V and W are finite-dimensional vector spaces over a field F, then functions that send linear maps f : V → W to dimF(W) × dimF(V) matrices in the way described in the sequel are themselves linear maps. • The expected value of a random variable is linear, as for random variables X and Y we have E[X + Y] = E[X] + E[Y] and E[aX] = aE[X], but the variance of a random variable is not linear, as it violates the second condition, homogeneity of degree 1: V[aX] = a2V[X]. Matrices[] Main article: Transformation matrix If V and W are finite-dimensional, and one has chosen bases in those spaces, then every linear map from V to W can be represented as a matrix; this is useful because it allows concrete calculations. Conversely, matrices yield examples of linear maps: if A is a real m × n matrix, then the rule f(x) = Ax describes a linear map Rn → Rm (see Euclidean space). Let {v1, ..., vn} be a basis for V. Then every vector v in V is uniquely determined by the coefficients c1, ..., cn in $c_1 \mathbf{v}_1+\cdots+c_n \mathbf{v}_n.$ If f: V → W is a linear map, $f(c_1 \mathbf{v}_1+\cdots+c_n \mathbf{v}_n)=c_1 f(\mathbf{v}_1)+\cdots+c_n f(\mathbf{v}_n),$ which implies that the function f is entirely determined by the values of f(v1), ..., f(vn). Now let {w1, ..., wm} be a basis for W. Then we can represent the values of each f(vj) as $f(\mathbf{v}_j)=a_{1j} \mathbf{w}_1 + \cdots + a_{mj} \mathbf{w}_m.$ Thus, the function f is entirely determined by the values of aij. If we put these values into an m × n matrix M, then we can conveniently use it to compute the value of f for any vector in V. For if we place the values of c1, ..., cn in an n × 1 matrix C, we have MC = the m × 1 matrix whose ith element is the coordinate of f(v) which belongs to the base wi. A single linear map may be represented by many matrices. This is because the values of the elements of the matrix depend on the bases that are chosen. Examples of linear transformation matrices[] In two-dimensional space R2 linear maps are described by 2 × 2 real matrices. These are some examples: • rotation by 90 degrees counterclockwise: $\mathbf{A}=\begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}$ • rotation by θ degrees counterclockwise: $\mathbf{A}=\begin{pmatrix}\cos(\theta) & -\sin(\theta)\\ \sin(\theta) & \cos(\theta)\end{pmatrix}$ • reflection against the x axis: $\mathbf{A}=\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}$ • reflection against the y axis: $\mathbf{A}=\begin{pmatrix}-1 & 0\\ 0 & 1\end{pmatrix}$ • scaling by 2 in all directions: $\mathbf{A}=\begin{pmatrix}2 & 0\\ 0 & 2\end{pmatrix}$ • horizontal shear mapping: $\mathbf{A}=\begin{pmatrix}1 & m\\ 0 & 1\end{pmatrix}$ • squeeze mapping: $\mathbf{A}=\begin{pmatrix}k & 0\\ 0 & 1/k\end{pmatrix}$ • projection onto the y axis: $\mathbf{A}=\begin{pmatrix}0 & 0\\ 0 & 1\end{pmatrix}.$ Forming new linear maps from given ones[] The composition of linear maps is linear: if f: V → W and g: W → Z are linear, then so is their composition g o f: V → Z. It follows from this that the class of all vector spaces over a given field K, together with K-linear maps as morphisms, forms a category. The inverse of a linear map, when defined, is again a linear map. If f1: V → W and f2: V → W are linear, then so is their sum f1 + f2 (which is defined by (f1 + f2)(x) = f1(x) + f2(x)). If f : V → W is linear and a is an element of the ground field K, then the map af, defined by (af)(x) = a (f(x)), is also linear. Thus the set L(V, W) of linear maps from V to W itself forms a vector space over K, sometimes denoted Hom(V, W). Furthermore, in the case that V = W, this vector space (denoted End(V)) is an associative algebra under composition of maps, since the composition of two linear maps is again a linear map, and the composition of maps is always associative. This case is discussed in more detail below. Given again the finite-dimensional case, if bases have been chosen, then the composition of linear maps corresponds to the matrix multiplication, the addition of linear maps corresponds to the matrix addition, and the multiplication of linear maps with scalars corresponds to the multiplication of matrices with scalars. Endomorphisms and automorphisms[] A linear transformation f: V → V is an endomorphism of V; the set of all such endomorphisms End(V) together with addition, composition and scalar multiplication as defined above forms an associative algebra with identity element over the field K (and in particular a ring). The multiplicative identity element of this algebra is the identity map id: V → V. An endomorphism of V that is also an isomorphism is called an automorphism of V. The composition of two automorphisms is again an automorphism, and the set of all automorphisms of V forms a group, the automorphism group of V which is denoted by Aut(V) or GL(V). Since the automorphisms are precisely those endomorphisms which possess inverses under composition, Aut(V) is the group of units in the ring End(V). If V has finite dimension n, then End(V) is isomorphic to the associative algebra of all n × n matrices with entries in K. The automorphism group of V is isomorphic to the general linear group GL(n, K) of all n × n invertible matrices with entries in K. Kernel, image and the rank–nullity theorem[] If f : V → W is linear, we define the kernel and the image or range of f by $\operatorname{\ker}(f)=\{\,x\in V:f(x)=0\,\}$ $\operatorname{im}(f)=\{\,w\in W:w=f(x),x\in V\,\}$ ker(f) is a subspace of V and im(f) is a subspace of W. The following dimension formula is known as the rank–nullity theorem: $\dim(\ker( f ))+ \dim(\operatorname{im}( f ))= \dim( V ).$ The number dim(im(f)) is also called the rank of f and written as rank(f), or sometimes, ρ(f); the number dim(ker(f)) is called the nullity of f and written as null(f) or ν(f). If V and W are finite-dimensional, bases have been chosen and f is represented by the matrix A, then the rank and nullity of f are equal to the rank and nullity of the matrix A, respectively. Cokernel[] Main article: Cokernel A subtler invariant of a linear transformation is the cokernel, which is defined as $\mathrm{coker}\,f := W/f(V) = W/\mathrm{im}(f).$ This is the dual notion to the kernel: just as the kernel is a subspace of the domain, the co-kernel is a quotient space of the target. Formally, one has the exact sequence $0 \to \ker f \to V \to W \to \mathrm{coker}\,f \to 0.$ These can be interpreted thus: given a linear equation f(v) = w to solve, • the kernel is the space of solutions to the homogeneous equation f(v) = 0, and its dimension is the number of degrees of freedom in a solution, if it exists; • the co-kernel is the space of constraints that must be satisfied if the equation is to have a solution, and its dimension is the number of constraints that must be satisfied for the equation to have a solution. The dimension of the co-kernel and the dimension of the image (the rank) add up to the dimension of the target space. For finite dimensions, this means that the dimension of the quotient space W/f(V) is the dimension of the target space minus the dimension of the image. As a simple example, consider the map f: R2 → R2, given by f(x, y) = (0, y). Then for an equation f(x, y) = (a, b) to have a solution, we must have a = 0 (one constraint), and in that case the solution space is (x, b) or equivalently stated, (0, b) + (x, 0), (one degree of freedom). The kernel may be expressed as the subspace (x, 0) < V: the value of x is the freedom in a solution – while the cokernel may be expressed via the map W → R, $(a,b) \mapsto (a):$ given a vector (a, b), the value of a is the obstruction to there being a solution. An example illustrating the infinite-dimensional case is afforded by the map f: R∞ → R∞, $\{a_n\} \mapsto \{b_n\}$ with b1 = 0 and bn + 1 = an for n > 0. Its image consists of all sequences with first element 0, and thus its cokernel consists of the classes of sequences with identical first element. Thus, whereas its kernel has dimension 0 (it maps only the zero sequence to the zero sequence), its co-kernel has dimension 1. Since the domain and the target space are the same, the rank and the dimension of the kernel add up to the same sum as the rank and the dimension of the co-kernel ( $\aleph_0 + 0 = \aleph_0 + 1$ ), but in the infinite-dimensional case it cannot be inferred that the kernel and the co-kernel of an endomorphism have the same dimension (0 ≠ 1). The reverse situation obtains for the map h: R∞ → R∞, $\{a_n\} \mapsto \{c_n\}$ with cn = an + 1. Its image is the entire target space, and hence its co-kernel has dimension 0, but since it maps all sequences in which only the first element is non-zero to the zero sequence, its kernel has dimension 1. Index[] For a linear operator with finite-dimensional kernel and co-kernel, one may define index as: $\mathrm{ind}\,f := \dim \ker f - \dim \mathrm{coker}\,f,$ namely the degrees of freedom minus the number of constraints. For a transformation between finite-dimensional vector spaces, this is just the difference dim(V) − dim(W), by rank–nullity. This gives an indication of how many solutions or how many constraints one has: if mapping from a larger space to a smaller one, the map may be onto, and thus will have degrees of freedom even without constraints. Conversely, if mapping from a smaller space to a larger one, the map cannot be onto, and thus one will have constraints even without degrees of freedom. The index comes of its own in infinite dimensions: it is how homology is defined, which is a central theory in algebra and algebraic topology; the index of an operator is precisely the Euler characteristic of the 2-term complex 0 → V → W → 0. In operator theory, the index of Fredholm operators is an object of study, with a major result being the Atiyah–Singer index theorem. Algebraic classifications of linear transformations[] No classification of linear maps could hope to be exhaustive. The following incomplete list enumerates some important classifications that do not require any additional structure on the vector space. Let V and W denote vector spaces over a field, F. Let T: V → W be a linear map. • T is said to be injective or a monomorphism if any of the following equivalent conditions are true: • T is one-to-one as a map of sets. • kerT = {0V} • T is monic or left-cancellable, which is to say, for any vector space U and any pair of linear maps R: U → V and S: U → V, the equation TR = TS implies R = S. • T is left-invertible, which is to say there exists a linear map S: W → V such that ST is the identity map on V. • T is said to be surjective or an epimorphism if any of the following equivalent conditions are true: • T is onto as a map of sets. • coker T = {0W} • T is epic or right-cancellable, which is to say, for any vector space U and any pair of linear maps R: W → U and S: W → U, the equation RT = ST implies R = S. • T is right-invertible, which is to say there exists a linear map S: W → V such that TS is the identity map on W. • T is said to be an isomorphism if it is both left- and right-invertible. This is equivalent to T being both one-to-one and onto (a bijection of sets) or also to T being both epic and monic, and so being a bimorphism. • If T: V → V is an endomorphism, then: • If, for some positive integer n, the n-th iterate of T, Tn, is identically zero, then T is said to be nilpotent. • If T2 = T, then T is said to be idempotent • If T = kI, where k is some scalar, then T is said to be a scaling transformation or scalar multiplication map; see scalar matrix. Change of basis[] Given a linear map whose matrix is A, in the basis B of the space it transforms vectors coordinates [u] as [v] = A[u]. As vectors change with the inverse of B, its inverse transformation is [v] = B[v']. Substituting this in the first expression $B[v'] = AB[u']$ hence $[v'] = B^{-1}AB[u'] = A'[u'].$ Therefore the matrix in the new basis is A′ = B−1AB, being B the matrix of the given basis. Therefore linear maps are said to be 1-co 1-contra -variant objects, or type (1, 1) tensors. Continuity[] Main article: Discontinuous linear map A linear transformation between topological vector spaces, for example normed spaces, may be continuous. If its domain and codomain are the same, it will then be a continuous linear operator. A linear operator on a normed linear space is continuous if and only if it is bounded, for example, when the domain is finite-dimensional. An infinite-dimensional domain may have discontinuous linear operators. An example of an unbounded, hence discontinuous, linear transformation is differentiation on the space of smooth functions equipped with the supremum norm (a function with small values can have a derivative with large values, while the derivative of 0 is 0). For a specific example, sin(nx)/n converges to 0, but its derivative cos(nx) does not, so differentiation is not continuous at 0 (and by a variation of this argument, it is not continuous anywhere). Applications[] A specific application of linear maps is for geometric transformations, such as those performed in computer graphics, where the translation, rotation and scaling of 2D or 3D objects is performed by the use of a transformation matrix. Another application of these transformations is in compiler optimizations of nested-loop code, and in parallelizing compiler techniques. References[] • Halmos, Paul R. (1974), Finite-dimensional vector spaces, New York: Springer-Verlag, ISBN 978-0-387-90093-3 [Amazon-US | Amazon-UK] • Lang, Serge (1987), Linear algebra, New York: Springer-Verlag, ISBN 978-0-387-96412-6 [Amazon-US | Amazon-UK] Content in this section is authored by an open community of volunteers and is not produced by, reviewed by, or in any way affiliated with MedLibrary.org. Licensed under the Creative Commons Attribution-ShareAlike 3.0 Unported License, using material from the Wikipedia article on "Linear map", available in its original form here: http://en.wikipedia.org/w/index.php?title=Linear_map • Finding More You are currently browsing the the MedLibrary.org general encyclopedia supplement. To return to our medication library, please select from the menu above or use our search box at the top of the page. In addition to our search facility, alphabetical listings and a date list can help you find every medication in our library. • Questions or Comments? If you have a question or comment about material specifically within the site’s encyclopedia supplement, we encourage you to read and follow the original source URL given near the bottom of each article. You may also get in touch directly with the original material provider. • About This site is provided for educational and informational purposes only and is not intended as a substitute for the advice of a medical doctor, nurse, nurse practitioner or other qualified health professional.

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 30, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9001452326774597, "perplexity_flag": "head"}

http://math.stackexchange.com/questions/190656/geometric-proof-that-left-fraci-ziz-right1-for-z-in-mathbbh

Geometric proof that $\left|\frac{i-z}{i+z}\right|<1$ for $z\in \mathbb{H}$ I just read a proof in a complex analysis book that says there exists a biholomorphic (conformal) map between $\mathbb{H}$ and $\mathbb{D}$ and I don't understand one step in the proof, namely the norm of the map (in the title) is less than $1$, using a purely "geometric" argument (i.e. a picture was in the book). I don't understand "geometric division", so I don't immediately see the proof "geometrically", please I do NOT want an algebraic answer (this is fairly straight forward anyway). - 5 Answers Note that $|i-z|$ is the distance between $z$ and $i$, and that $|i+z|$ is the distance between $z$ and $-i$. The inequality says that $z$ lies closer to $i$ than to $-i$, which is true for points in the upper halfplane. To make this even more clear, write $\bigl|{i-z \over i+z}\bigr|={|i-z|\over |i+z|}<1$, and multiply through by the denominator $|i+z|$. - The real axis is in the middle between $i$ and $-i$, hence is the locus of all points having the same distance from $i$ and $-i$. The rest of $\mathbb C$ splits into those points closer to $i$ than to $-i$ (the upper half plane) and those closer to $-i$ (lower half plane). - Basically you want to prove that any point in $\mathbb{H}$ is closer to $i$ than to $-i$. Draw the triangle with vertices at $i, -i, z$ and observe that the angle at $-i$ is smaller than the angle at $i$. he easiest way to prove this is by extending the line through $i z$ till it intersects the real axix at $y$ and compare the angles in the triangle $i,-i,z$ with the angles in $i,-i,y$. - The norms of $i - z$ and $i+z$ are the squares of the distances between $z$ and $\pm i$. A point $z$ lies in the upper half plane if and only if it is strictly closer to $i$ than to $-i$, so that this ratio has norm strictly less than $1$ and so lies in the unit disk. - |z-i|=|z+i| is the set of points equidistant from $i$ and $-i$, the real line. |z-i|<|z+i| is the set of points closer to $i$ than $-i$, the upper half plane. -

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 46, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9620217084884644, "perplexity_flag": "head"}

http://mathematica.stackexchange.com/questions/5362/finding-a-fit-to-a-multi-dimensioned-function?answertab=oldest

# Finding a fit to a multi-dimensioned function I have a model function $f:\mathbb{R}^2\rightarrow\mathbb{R}^2$, and a bunch of data points for which I'd like Mathematica to fit for me. Unfortunately `FindFit` seems to only deal with functions $\mathbb{R}^n\rightarrow\mathbb{R}$. I guess I could make my own square difference from the data to the model and use `NMinimize` on that, but I wondering if there was an easier way? ### Edit - toy example with answers $g\left(x,y\right)=\left<abx+(2c+d)y,(2a+b)x+cdy\right>$ ````g[{x_, y_}, {a_, b_, c_, d_}] := {a b x + (2 c + d) y, (2 a + b) x + c d y}; points = Flatten[Table[{{x, y}, g[{x, y}, {1, 2, 3, 4}]}, {x, -1, 1, 0.5}, {y, -1, 1, 0.5}], 1]; ```` Finding a fit for each component separately doesn't work: ````data1 = points /. {{x_, y_}, {u_, v_}} -> {x, y, u}; data2 = points /. {{x_, y_}, {u_, v_}} -> {x, y, v}; FindFit[data1, g[{x, y}, {a, b, c, d}][[1]], {a, b, c, d}, {x, y}] FindFit[data2, g[{x, y}, {a, b, c, d}][[2]], {a, b, c, d}, {x, y}] ```` `{a -> 0.729723, b -> 2.74077, c -> 3.35659, d -> 3.28681}` `{a -> 1.48555, b -> 1.02891, c -> 2.81936, d -> 4.25629}` Jens suggestion works well: ````data3 = points /. {{x_, y_}, {u_, v_}} -> Sequence[{x, y, 0, u}, {x, y, 1, v}]; FindFit[data3, g[{x, y}, {a, b, c, d}].{1 - s, s}, {a, b, c, d}, {x, y, s}] ```` `{a -> 1., b -> 2., c -> 3., d -> 4.}` My original suggestion (which is similar to user840's links) ````error = Plus @@ (points /. {{x_, y_}, {u_, v_}} -> Norm[g[{x, y}, {a, b, c, d}] - {u, v}]^2); NMinimize[error, {a, b, c, d}] ```` Gives a different, but still correct, solution (or at least a close approximation thereof) `{1.41754*10^-13, {a -> 1.00004, b -> 1.99993, c -> 2., d -> 6.}}` - How about using NonlinearModelFit? Take a look into the documentation and the examples. You will see that under Scope the first example shows how one can fit a model of more than one variable. If you provide example data, it would be possible to help you more detailed. – partial81 May 9 '12 at 5:53 @partial81 More than one variable, not but not a vector value. That example is for $\mathbb{R}^2 \rightarrow \mathbb{R}$ while the OP wants $\mathbb{R}^2 \rightarrow \mathbb{R}^2$. `NonlinearModelFit` doesn't seem to support this. – Szabolcs May 9 '12 at 8:11 ok, I assume you are right. I would love to try it with `NonlinearModelFit`, but without example data, I do not want to spend more time on this problem. – partial81 May 9 '12 at 10:54 Why not two fits, one for each dependent variable? – Daniel Lichtblau May 9 '12 at 18:30 @DanielLichtblau Each variable depends on all the model parameters, so I end up with different parameter values for the different fits. – wxffles May 9 '12 at 21:55 show 2 more comments ## 3 Answers Without a specific example, I can only make general suggestions here. The first thing that comes to mind is that you could fit each of the two components of your vector field independently. I.e., if $\vec{f}: \mathbb{R}^2\to\mathbb{R}^2$ is split up into $\vec{f} = \{f_x, f_y\}$ with $f_{1,2}: \mathbb{R}^2\to\mathbb{R}$, then `FindFit` would work on each of these component functions. If you don't want the fits to be determined independently, it could still be possible to use `FindFit` by introducing an auxiliary variable $s$ that labels the two component functions above, $f_s$, and then provide the fitting data with this variable `s` included. Here, `s` can only have two discrete values (I borrowed the idea from spin-half quantum mechanics). Let's choose `s = 0` for the function $f_x$ and `s = 1` for $f_y$. So you'd have data in the form ````data = {{x1, y1, 0, fx1}, {x1, y1, 1, fy1}, ...} ```` where `fx1` is the first value of $f_x$ and `fy1` the first value of $f_y$, both at the point `x1, y1`. Your model could then look something like this: ````model[x_, y_, s_] := modely[x, y]*s + modelx[x, y]*(1 - s) ```` where `modely[x, y]` is the model for $f_y$ and `modelx[x, y]` is the model for $f_x$. The variables in the `FindFit` call would be `x, y, s`, and the parameters in the models `modelx`, `modely` could be the same (or different). There are probably many other reasonable ways to do such a fit, but these are some ideas. - Thanks. This is working well for my problem. `FindFit[data /. {{a_, b_}, {p_, q_}} -> Sequence[{a, b, 0, p}, {a, b, 1, q}], model.{1 - s, s}, ...]` – wxffles May 9 '12 at 21:33 Multi-layer perceptrons have this ability. The Neural network package from Wolfram offers this functionality. Neural Networks - Since that isn't part of the standard suite of packages, could you provide a link? – rcollyer May 9 '12 at 13:14 True, but now you have the additional problem of structuring the neural net appropriately (which is by no means trivial if you don't have any experience with neural nets). And after that you have to train it, which is equivalent to the original problem... – Oleksandr R. May 9 '12 at 13:46 Yes there is some effort involved, but it offers a different approach to the solution, which I thought might fill in parts of the landscape of the large selection of approximation techniques that might be deployed outside of the half a dozen or so that `FindFit` offers and MLPs have inherent support for output vectors in higher dimensions. – image_doctor May 9 '12 at 15:35 You can check Fitting Multiple-Response Data from Bruce Miller. He gives a step-by-step example of how to perform a simultaneous fit to multiple functions, and provides a fairly general function for performing such fits. Alternately, Multiple-Response Fitting notes from wolframThis package contains functions for simultaneously fitting multiple functions to data with a possibly shared set of parameters. It has flexible output options and the first part of the accompanying demonstration notebook can serve as an introduction to writing one's own fitting codes using FindMinimum. - lang-mma

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 16, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.905121386051178, "perplexity_flag": "middle"}

http://physics.stackexchange.com/questions/tagged/quantum-mechanics+research-level

# Tagged Questions 1answer 32 views ### Name of a state with $d-1$ excitations, distributed uniformly among $n$ qudits Is there a particular name for a quantum state of the form (up to the normalization): $$\sum_{i_1+\ldots+i_n = d-1} |i_1\rangle |i_2\rangle \ldots |i_n\rangle$$ or was it studied is some papers? ... 0answers 101 views ### Looking for modern results in semiclassical physics and relevant references What are some important approximations, especially those that are state-of-the-art, used to approximate the many-body dynamics of atoms and molecules in the semiclassical regime? To be clear, I'm not ... 2answers 100 views ### How is the energy/eigenvalue gap plot drawn for adiabatic quantum computation? I was going through arXiv:quant-ph/0001106v1, the first paper by Farhi on adiabatic quantum computation. Equation 2.24 says, $$\tilde{H}(s) = (1-s)H_B + sH_P$$ which means the adiabatic evolution ... 0answers 110 views ### Local explanation of the Aharonov-Bohm effect in terms of force fields Here is an interesting paper for the Physics SE community: On the role of potentials in the Aharonov-Bohm effect, Lev Vaidman, published in PHYSICAL REVIEW A 86, 040101(R) (2012). You should check it ... 1answer 164 views ### Are Born-Oppenheimer energies analytic functions of nuclear positions? I am looking for references to bibliography that explores the smoothness and analyticity of eigenvalues and eigenfunctions (and matrix elements in general) of a hamiltonian that depends on some ... 1answer 331 views ### Schwinger representation of operators for n-particle 2-mode symmetric states A bosonic (i.e. permutation-symmetric) state of $n$ particles in $2$ modes can be written as a homogenous polynomial in the creation operators, that is \left(c_0 \hat{a}^{\dagger n} + c_1 ... 1answer 563 views ### Is resonating valence bond (RVB) states long-range entangled? Quantum liquid is at the core of condensed matter theory study, examples include superfluid in Bose Hubbard model, quantum spin liquid around the RK point of a quantum dimer model, string-net ... 10answers 1k views ### What is spontaneous symmetry breaking in QUANTUM systems? Most descriptions of spontaneous symmetry breaking, even for spontaneous symmetry breaking in quantum systems, actually only give a classical picture. According to the classical picture, spontaneous ... 0answers 109 views ### What is a Hilbert space filter? In a recent paper, Side-Channel-Free Quantum Key Distribution, by Samuel L. Braunstein and Stefano Pirandola. Phys. Rev. Lett. 108, 130502 (2012). doi:10.1103/PhysRevLett.108.130502, ... 1answer 158 views ### partial trace with sparse matrices Let $\rho_{ABCD}$ be a sparse matrix of 4 systems each in a $d$-dimensional Hilbert space. For $d<7$ in a reasonable time (few seconds) I able to perform the partial trace $\rho_{AD}$ using the ... 3answers 99 views ### What is the physical difference between states and unital completely positive maps? Mathematically, completely positive maps on C*-algebras generalize positive linear functionals in that every positive linear functional on a C*-algebra $A$ is a completely positive map of $A$ into ... 2answers 56 views ### Heuristics for definitions of open and closed quantum dynamics I've been reading some of the literature on "open quantum systems" and it looks like the following physical interpretations are made: Reversible dynamics of a closed quantum system are represented ... 6answers 105 views ### Multiqubit state tomography by performing measurement in the same basis For a $n$-qubit state $\rho$ we perform all projective measurement consisting of one-particle measurements in the same basis, that is, p_{i_1i_2\ldots i_n}(\theta,\varphi) = \text{Tr}\left \{ \rho ... 1answer 109 views ### Monte Carlo integration over space of quantum states I am currently facing the problem of calculating integrals that take the general form $\int_{R} P(\sigma)d\sigma$ where $P(\sigma)$ is a probability density over the space of mixed quantum states, ... 1answer 452 views ### Entanglement in time Quantum entanglement links particles through time, according to this study that received some publicity last year: New Type Of Entanglement Allows 'Teleportation in Time,' Say Physicists at The ... 1answer 34 views ### Tip of a spreading wave-packet: asymptotics beyond all orders of a saddle point expansion This is a technical question coming from mapping of an unrelated problem onto dynamics of a non-relativistic massive particle in 1+1 dimensions. This issue is with asymptotics dominated by a term ... 1answer 46 views ### Unknown quantum state with promise of classical data I am trying to solve a problem in the measurement and identification of quantum states with a promise as to what states it could be. Here is the problem. Imagine a system that produces qubits in ... 3answers 68 views ### Constructing a Hamiltonian (as a polynomial of $q_i$ and $p_i$) from its spectrum For a countable sequence of positive numbers $S=\{\lambda_i\}_{i\in N}$ is there a construction producing a Hamiltonian with spectrum $S$ (or at least having the same eigenvalues for $i\leq s$ for ... 0answers 41 views ### Spectrum of a quantum relativistic “distance squared” operator This question disusses the same concepts as that question (this time in quantum context). Consider a relativistic system in spacetime dimension $D$. Poincare symmetry yields the conserved charges $M$ ... 1answer 86 views ### random matrix ensembles from BMN model My friends working on Thermalization of Black Holes explained solutions to their matrix-valued differential equations (from numerical implementation of the Berenstein-Maldacena-Nastase matrix model) ... 1answer 73 views ### Many body quantum states analyzed as probabilistic sequences Measurements of consecutive sites in a many body qudit system (e.q. a spin chain) can be interpreted as generating a probabilistic sequence of numbers $X_1 X_2 X_3 \ldots$, where \$X_i\in ... 3answers 130 views ### POVMs that do not require enlargement of the Hilbert space The usual justification for regarding POVMs as fundamental measurements is via Neumark's theorem, i.e., by showing that they can always be realized by a projective measurement in a larger Hilbert ... 1answer 37 views ### Optimality of the CHSH strategy The maximum achievable probability of the Clauser-Horne-Shimony-Holt game is $\cos^2(\pi/8)\approx85.355\%,$ which can be proved with Tsirelson's inequality. But I don't imagine that this remained ... 1answer 48 views ### Spekkens Toy Model, Internal Comonoids I have been thinking about Spekkens Toy model in terms of interfaces. The Spekkens paper concerns a physics based on only being able to receive answers to half the number of questions necessary to ... 1answer 35 views ### Convexity — reference request I've been reading a few papers on generalized probabilistic theories, and have been struggling through proofs of some results that involve use of convexity and group theory, e.g. this paper on bit ... 3answers 134 views ### Hilbert-Schmidt basis for many qubits - reference Every density matrix of $n$ qubits can be written in the following way \hat{\rho}=\frac{1}{2^n}\sum_{i_1,i_2,\ldots,i_n=0}^3 t_{i_1i_2\ldots i_n} ... 1answer 65 views ### How does one geometrically quantize the Bloch equations? I've just now rated David Bar Moshe's post (below) as an "answer", for which appreciation and thanks are given. Nonetheless there's more to be said, and in hopes of stimulating further posts, I've ... 3answers 364 views ### Rigorous proof of Bohr-Sommerfeld quantization Bohr-Sommerfeld quantization provides an approximate recipe for recovering the spectrum of a quantum integrable system. Is there a mathematically rigorous explanation why this recipe works? In ... 0answers 424 views ### Experimental test of the non-statisticality theorem? Context: The recent paper The quantum state cannot be interpreted statistically by Pusey, Barrett and Rudolph shows under suitable assumptions that the quantum state cannot be interpreted as a ... 1answer 45 views ### Accurate quantum state estimation via “Keeping the experimentalist honest” Bob has a black-box, with the label "V-Wade", which he has been promised prepares a qubit which he would like to know the state of. He asks Alice, who happens also to be an experimental physicist, to ... 1answer 81 views ### Quantum mechanical gravitational bound states The quantum mechanics of Coloumb-force bound states of atomic nuclei and electrons lead to the extremely rich theory of molecules. In particular, I think the richness of the theory is related to the ... 1answer 69 views ### Quantum mechanics as a Markov process I am currently involved in some understanding on this matter with a colleague of mine. I know all the literature about but I do not know the state of art. Please, could you provide some relevant ... 1answer 42 views ### Allowed states vis-a-vis allowed dynamics in generalized probabilistic theories (GPTs) In his work on information processing in GPTs http://arxiv.org/abs/quant-ph/0508211 Barrett speculates that the trade-off between allowed states and the allowed dynamics in a GPT is optimal in quantum ... 2answers 123 views ### Geometric quantization of identical particles Background: It is well known that the quantum mechanics of $n$ identical particles living on $\mathbb{R}^3$ can be obtained from the geometric quantization of the cotangent bundle of the manifold ... 3answers 91 views ### Which are the simplest known contextual inequalities? It is well-known that quantum mechanics does not admit a noncontextual ontological model, and there are countless different proofs of it. I'm interested in the simplest proofs that can be cast as an ... 1answer 111 views ### Majorana-like representation for mixed symmetric states? Is there a generalization of the Majorana representation of pure symmetric $n$-qubit states to mixed states (made of pure symmetric $n$-qubit)? By Majorana representation I mean the decomposition of ... 1answer 129 views ### What proof techniques have failed for solving the SIC-POVM problem and what new insights have been gleaned from them? The SIC-POVM problem is remarkably easy to state given that it has not yet been solved. It goes like this. With dim($\mathcal H$) $=d$, find states $|\psi_k\rangle\in\mathcal H$, $k=1,\ldots,d^2$ ... 1answer 55 views ### Explicit construction for unitary extensions of CPTP maps? Given a completely positive and trace preserving map $\Phi : \textrm{L}(\mathcal{H})\to\textrm{L}(\mathcal{G})$, it is clear by the Kraus representation theorem that there exist \$A_k \in ... 1answer 81 views ### What is the Holevo-Schumacher-Westmoreland capacity of a Pauli channel? Suppose you are given an $n$-qubit quantum channel defined as $\mathcal{E}(\rho) = \sum_{i} p_i X_i \rho X_i^\dagger$, where $X_i$ denotes an $n$-fold tensor product of Pauli matrices and $\{p_i\}$ is ... 2answers 117 views ### Relevance of SIC-POVMs to quantum information What is the real relevance of SIC-POVMs (symmetric informationally complete POVMs) to concrete tasks in quantum information theory? A lot of work has been put into giving explicit constructions, and ... 1answer 31 views ### Connections of iterative solvers for large systems of equation in Physics? I am trying to find the domains in physics where solving large systems of equations is computationally expensive. The sparse systems are of my particular interest, where the input matrix A is in GBs ... 4answers 208 views ### direct sum of anyons? In the topological phase of a fractional quantum Hall fluid, the excitations of the ground state (quasiparticles) are anyons, at least conjecturally. There is then supposed to be a braided fusion ... 11answers 827 views ### Negative probabilities in quantum physics Negative probabilities are naturally found in the Wigner function (both the original one and its discrete variants), the Klein paradox (where it is an artifact of using a one-particle theory) and the ... 6answers 252 views ### Is there a theorem that says that QFT reduces to QM in a suitable limit? A theorem similar to Ehrenfest's theorem? Is there a theorem that says that QFT reduces to QM in a suitable limit? Of course, it should be, as QFT is relativisitc quantum mechanics. But, is there a more manifest one? such as Ehrenfest's ... 0answers 174 views ### Orbits of maximally entangled mixed states It is well known (Please, see for example Geometry of quantum states by Bengtsson and Życzkowski ) that the set of $N-$dimensional density matrices is stratified by the adjoint action of $U(N)$, where ... 1answer 49 views ### Principle behind fidelity balance in quantum cloning If we do optimal state estimation on an unknown qubit, we can recreate a state with fidelity $F_c=2/3$ with respect to the original. Let us define the "quantum information content" $I_q=1-2/3=1/3$ as ... 1answer 61 views ### Quasiparticles in Bohmian mechanics My questions are about de Broglie-Bohm "pilot wave" interpretation of quantum mechanics (a.k.a. Bohmian mechanics). Do quasiparticles have any meaning in Bohmian mechanics, or not? Specifically, is ... 2answers 50 views ### Counting complete sets of mutually unbiased bases composed of stabilizer states Consider $N$ qubits. There are many complete sets of $2^N+1$ mutually unbiased bases formed exclusively of stabilizer states. How many? Each complete set can be constructed as follows: partition the ... 1answer 23 views ### Sub and super multiplicativity of norms for understanding non-locality In relation to various problems in understanding entanglement and non-locality, I have come across the following mathematical problem. It is most concise by far to state in its most mathematical form ... 1answer 51 views ### Stabilizer formalism for symmetric spin-states? This question developed out of conversation between myself and Joe Fitzsimons. Is there a succinct stabilizer representation for symmetric states, on systems of n spin-1/2 or (more generally) n higher ...

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 47, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9068542718887329, "perplexity_flag": "middle"}

http://mathoverflow.net/revisions/31637/list

## Return to Answer 3 pedantic TeX correction Since I could use the reputation... letting $\psi: G \to Diff(M)$ \operatorname{Diff}(M)$be the action then$X' = \psi_* X$, where we identify$Lie(Diff(M))$\operatorname{Lie}(\operatorname{Diff}(M))$ with vector fields on $M$. This is because they both correspond to the one-parameter subgroup $t\mapsto \psi(\exp tX)$ of $Diff(M)$. \operatorname{Diff}(M)$. Then$[X', Y'] = [X,Y]'\$ is immediate. 2 added 107 characters in body Since I could use the reputation... letting $\psi: G \to Diff(M)$ be the action then $X' = \psi_* X$, where we identify $Lie(Diff(M))$ with vector fields on $M$. This is because they both correspond to the one-parameter subgroup $t\mapsto \psi(\exp tX)$ of $Diff(M)$. Then $[X', Y'] = [X,Y]'$ is immediate. 1 Since I could use the reputation... letting $\psi: G \to Diff(M)$ be the action then $X' = \psi_* X$, where we identify $Lie(Diff(M))$ with vector fields on $M$. Then $[X', Y'] = [X,Y]'$ is immediate.

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 20, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8569443821907043, "perplexity_flag": "head"}

http://mathoverflow.net/revisions/107938/list

## Return to Question 2 added 8 characters in body A number of sources concerning Speiser's 1934 result state that the Riemann Hypothesis (RH) implies $\zeta'(s)\neq 0$ for all $0<\text{Re}(s)<1/2$. But I have seen some (possibly less reliable) sources without proof suggesting this is an if and only if relationship, i.e. RH$\Leftrightarrow\zeta'(s)\neq RH$\Longleftrightarrow\zeta'(s)\neq 0$. However, those (perhaps more reliable) sources state only forward implication, i.e. RH$\Rightarrow\zeta'(s)\neq RH$\Longrightarrow\zeta'(s)\neq 0$. My question is this: is Speiser's result an if and only if relationship or not? 1 # A question about Speiser's 1934 result on the Riemann hypothesis A number of sources concerning Speiser's 1934 result state that the Riemann Hypothesis (RH) implies $\zeta'(s)\neq 0$ for all $0<\text{Re}(s)<1/2$. But I have seen some (possibly less reliable) sources without proof suggesting this is an if and only if relationship, i.e. RH$\Leftrightarrow\zeta'(s)\neq 0$. However, those (perhaps more reliable) sources state only forward implication, i.e. RH$\Rightarrow\zeta'(s)\neq 0$. My question is this: is Speiser's result an if and only if relationship or not?

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 9, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9003283977508545, "perplexity_flag": "middle"}

http://math.stackexchange.com/questions/tagged/convergence+calculus

# Tagged Questions 2answers 33 views ### Checking $\displaystyle \int_{0}^{\infty} {\frac{\sin^2x}{\sqrt[3]{x^7 + 1}} dx}$ for convergence Given $\displaystyle\int_{0}^{\infty} {\frac{\sin^2x}{\sqrt[3]{x^7 + 1}} dx}$, prove that it converges. So of course, I said: We have to calculate \$\displaystyle \lim_{b \to \infty} ... 2answers 18 views ### Develop the next function:$f(x)=\frac{4x+53}{x^2-x-30}$ into power series, Find the radius on convergence and find $f^{(20)}(0)$ Develop the next function:$\displaystyle f(x)=\frac{4x+53}{x^2-x-30}$ into power series, Find the radius on convergence and find $f^{(20)}(0).$ For the first part: \$\displaystyle\frac ... 7answers 124 views ### How to prove that $1/n!$ is less than $1/n^2$? I want to prove $$\sum_{n=0}^{\infty} \frac{1}{n!}$$ is a converging series. So I want to compare it with $\sum_{n=0}^{\infty} \frac{1}{n^2}$. I want to do direct comparison test. How to prove ... 1answer 16 views ### Absolute convergence. Determine if absolutely convergent or not; Justify. $$\sum_{n=1}^\infty (-1)^n n^2 3^{1-n} x^n \text{ s.t }|x|<3$$ if we take the abs value of $(-1)^n$ we are left with $n^{2} 3^{1-n} x^{n}$ now ... 3answers 28 views ### behavior of a sequence Imaging a sequence $a_{k} \in \Omega$ with $\Omega \subset \Bbb{R}$ closed, $\lim\limits_{k \to \infty} \| a_{k+1} - a_{k} \| = 0$. My Professor said that because of this the sequence would ... 2answers 32 views ### What's the radius of convergence of the next sum: $\sum_{n=0}^\infty (\int_o^n\frac{\sin^2t}{\sqrt[3]{t^7+1}}dt)x^n$ What's the radius of convergence of the next sum: $$\sum_{n=0}^\infty \left(\int_0^n\frac{\sin^2t}{\sqrt[3]{t^7+1}}dt\right)x^n$$ I know that $$\int_0^\infty\frac{\sin^2t}{\sqrt[3]{t^7+1}}dt$$ does ... 1answer 49 views ### Conditions for taking a limit into an infinite sum Suppose $f\left(x\right)={\displaystyle \sum_{n=0}^{\infty}g_{n}\left(x\right)}$ under what conditions is it true that: \lim_{x\to c}f\left(x\right)={\displaystyle \sum_{n=0}^{\infty}\lim_{x\to ... 1answer 36 views ### Radius of convergence of a power series with Bernoulli numbers Say, we use the definition: Bernoulli numbers arise in Taylor series in the expansion $$\frac{x}{e^x-1}=\sum_{k=0}^\infty B_k \frac{x^k}{k!}$$ and then derive power series representations of the ... 1answer 30 views ### Calculate (& determine if converges): $\sum_{n=1}^\infty (-1)^n\frac 1n(0.5)^n$ Calculate (& determine if converges): $\sum_{n=1}^\infty (-1)^n\frac 1n(0.5)^n$ The above is a specific equation for $x_0=0.5$, but from here i'm pretty much stuck. 1answer 67 views ### Radius of convergence of: $\sum_{n=1}^\infty\frac {x^{n!}}{n!}$? What the Radius of convergence of: $\sum\limits_{n=1}^\infty\frac {x^{n!}}{n!}$? So far I tried finding the ... 2answers 63 views ### A question on the convergence of a Taylor series of some prominent function The function $f:\mathbb{R}\to\mathbb{R}$ defined by $$f(x)=\begin{cases}e^{-\frac{1}{x^2}} &if &x\neq 0\\ 0 & else \end{cases}$$ is a prominent example of a function whose Taylor series ... 3answers 91 views ### How can I prove that $\int_1^\infty \left\lvert\frac{\sin x}{x}\right\rvert dx$ diverges? I know a start could be to try and prove that $\int_1^\infty \frac{\sin^2x}{x} dx$ diverges since $\frac{\sin^2x}{x} \le \left\lvert\frac{\sin x}{x}\right\rvert$ in this interval, but I wouldn't know ... 1answer 45 views ### Radius of convergence of $\sum_{n=1}^{\infty} { (n \sin{\frac{1}{n}})^{n} x^n }$ We need to calculate the radius of convergence $R$ of: $$\sum_{n=1}^{\infty} {\left(n \sin{\frac{1}{n}}\right)^{n} x^n }.$$ Here's what I did: \lim_{n\to\infty} { \left| ... 3answers 78 views ### Radius of convergence for $\sum_{n=1}^{\infty} { \frac{1}{n!} \cdot x^{n!}}$ How do we find the radius of convergence $R$ of this power sum? $$\sum_{n=1}^{\infty} { \frac{1}{n!} \cdot x^{n!}}$$ How do we handle the $n!$ as the power of $x$? 2answers 43 views ### Finding the radius of convergence and what does it mean We have just started learning this and I do not really understand it fully. Given: $$\sum_{n=1}^{\infty} {(\pi^n + n + 1) \cdot x^n}$$ We should check what is the radius of convergence $R$ and what ... 3answers 68 views ### Show that $(x_n-y_n)$ converges to $x-y$. Given $(x_n)$ and $(y_n)$ are sequences of real number which converge to $x$ and $y$ respectively. Show that $(x_n-y_n)$ converges to $x-y$. If it's asking about $(x_n+y_n)$. I know that I can ... 1answer 29 views ### Study the convergence of this sequence of functions I have the following sequence of function: $$f_n(\lambda)=\bigg[\alpha-i\bigg(\lambda+\frac{1}{n}\bigg)\bigg]^{-1}-\bigg[\alpha-i\lambda\bigg]^{-1},\,\,\,\alpha\neq 0$$ and I have to study its ... 1answer 57 views ### Series of Functions - Pointwise and Uniform Convergence. I was hoping for some help for the following questions. Prove that the series $\sum_{n=1}^\infty x^n(1-x)$ converges pointwise but not uniformly on $[0,1]$. Prove that the series \$\sum_{n=1}^\infty ... 3answers 66 views ### Show that the sequence $(x_n)=\left( \sum_{i=1}^n\frac 1 i\right)$ diverge by epsilon delta definition. Show that the sequence $\displaystyle (x_n)=\left( \sum_{i=1}^n\frac 1 i\right)$ diverge by epsilon delta definition. I'm not familiar with proving divergent sequence. Do anyone have any des? ... 1answer 49 views ### Prove convergence of improper integral using change of variable. This may be trivial, but I could use some help... Consider a real function $f: (0,1) \rightarrow \mathbb{R}$, continuous, positive, but not necessarily bounded. Let $g: [0,1] \rightarrow [0,1]$ be a ... 2answers 48 views ### Prove or disprove a result for a double sequence. Suppose that a double sequence $\{a_{n,k}\}=\left\{\frac{1}{n^{\frac{k-1}{k}}}\right\}$. Prove or disprove $\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{k}a_{n,k}=0$. 2answers 88 views ### What is $\lim_{n\to\infty}\frac{1}{e^n}\Bigl(1+\frac1n\Bigr)^{n^2}$? How to solve the following limit question? $$\lim_{n\to\infty}\frac{1}{e^n}\Bigl(1+\frac1n\Bigr)^{n^2}$$ Thanks a lot. 1answer 62 views ### When does $\lim\limits_{n\to\infty}\int_{b}^{a_n}f_n(x)dx=\lim\limits_{n\to\infty}\int_b^\infty f_n(x)dx$ hold? Let $\{a_n\}\subset \mathbb{R}$ be sequence and $$f_n:[b,\infty)\longrightarrow \mathbb{R}, \qquad n=1,2,\dots .$$ Assume that $$\lim_{n\longrightarrow\infty}a_n=+\infty.$$ Obviously, from the ... 3answers 125 views ### Evaluating $\lim \limits_{n\to \infty} \left( n \int_{0}^{\frac \pi 2} 1-\sqrt [n]{\sin x} \,\mathrm dx \right)$ Evaluate the following limit: $$\lim \limits_{n\to \infty} \;\; n \int_{0}^{\frac \pi 2} \left(1-\sqrt [n]{\sin x} \right)\,\mathrm dx$$ I have done the problem . How I solved is First I ... 2answers 85 views ### Interval of convergence of $\sum_{n=4}^\infty x^n/n^5$ Find the interval of convergence $$\sum_{n=4}^\infty x^n/n^5$$ I'm lost here. My intuition was to use the ratio test. $$\lim_{n \to \infty} \frac{x^{n+1}}{(n+1)^5} \times \frac{n^5}{x^n}$$ ... 3answers 97 views ### How to solve $\sum_{k=2}^\infty {\frac{1}{k^2-1}}$ I'm using the integral test to determine if this series converges. From what I have so far it seems that it diverges, but according to wolfram alpha it converges. Where is my mistake? ... 5answers 135 views ### Convergence of a sequence $c_n$ Suppose that $(a_n)$ and $(b_n)$ be sequences such that $\lim (a_n)=0$ and $\displaystyle \lim \left( \sum_{i=1}^n b_i \right)$ exists. Define $c_n = a_1 b_n + a_2 b_{n-1} + \dots + a_n b_1$. Prove ... 1answer 189 views ### binomial expansion formula proof, bases on Lagrange form of Taylor series remainder Another exercise from Bartle/Sherbert Introduction to Real Analysis book (this one is exercise 9.4.14): Use the Lagrange form of the remainder to justify the general Binomial Expansion ... 0answers 36 views ### Is this pointwise convergence sequence also uniform convergence? $f_{n}$ and $f$ are continuous functions and $f_{n}\rightarrow f$ pointwise. Which of the following are correct? $\int _{0}^{x}F_{n}\left( t\right) dt\rightarrow\int _{0}^{x}F\left( t\right) dt$ ... 1answer 47 views ### convergence of complex series. For what values of $(a,b\in \mathbb{C})$ does this series converge or diverge? $\sum\frac{(k-a)^2}{(k-b)^3}$ if $a,b\in\mathbb{R}$ by the Limit comparison test (with $\sum\frac{1}{k}$ ) I know ... 3answers 67 views ### convergence of $\sum\frac{a_{n}}{n}$ if $\sum_{k=1}^{n}a_{k}\le M*n^{r}$ where $r<1$ Show that if the partial sums $s_{n}$ of the series $\sum_{k=1}^{\infty}a_{k}$ satisfy $|s_{n}|\le M*n^{r}$ for some $r<1$, then the series $\sum_{n=1}^{\infty}\frac{a_{n}}{n}$ converges. 2answers 30 views ### Radius of convergence in a series. Ratio test. I am having a hard time with this question. $$\sum_{k=0}^{\infty} \frac{-(1)^k (4^k -3)x^{2k}}{k^4+3}$$ I used the ratio test and got stuck here: x^2 \lim_{k\to\infty} \frac ... 1answer 100 views ### Proof that $\sum\limits_{k=1}^n\frac{\sin(kx)}{k^2}$ convergences uniformly using the Cauchy criterion I'd like to use the Cauchy criterion to show that $$f_n(x)=\sum\limits_{k=1}^n\frac{\sin(kx)}{k^2} \mbox{convergences uniformly}$$ Here is what I did: We want to show that \$\forall \, ... 3answers 89 views ### Does this series converge or diverge? I have a series here, and I'm supposed to determine whether it converges or diverges. I've tried the different tests, but I can't quite get the answer. ... 1answer 49 views ### Composition Taylor Series Is there any theorem that specifies when we are allowed to compose the taylor series of two functions? Does it have a name? Thanks. 1answer 31 views ### Two quick questions about convergence (in the context of pointwise vs. uniform convergence) I found this example online: Let $\{f_{n}\}$ be the sequence of functions on $(0,\infty)$ defined by $f_{n}(x)=\frac{nx}{1+n^{2}x^{2}}$ .This function converges pointwise to zero. ... 3answers 64 views ### Representing Functions as Power Series Rewrite $$f(x)=(1+x)/(1-x)^2$$ as a power series. Work thus far: I separated it into two parts: $$1/(1-x)^2 + x/(1-x)^2$$ I realize that the first expression is the derivative of $1/(1-x)$ and ... 1answer 29 views ### Specify conditions for $\alpha$ so that the iteration $x_{n+1} = x_n - \alpha f(x_n)$ converges to root of f. Specify conditions on $\alpha$ so that the iterative process $x_{n+1} = x_n - \alpha f(x_n)$ converges to root of f if started with $x_0$ close to the root. It is suggested that the proof should ... 1answer 56 views ### Value of limsup i? This is a part of my question. $\lim \sup \cos(n\pi/12)$ as n goes to infinity What is the value of this limit? 1answer 81 views ### Cauchy but not fast cauchy As the title indicates, I am trying to find a Cauchy sequence that is not fast (or rapidly) Cauchy. Could anyone suggest something? A sequence $\{a_n\}_{n \in \Bbb N}$is termed fast (or rapidly) ... 3answers 55 views ### Convergence of definite integral I have to find out the convergence of the next integral: $$\int^{\pi/2}_0{\frac{\ln(\sin(x))}{\sqrt{x}}}dx$$ Any help? Thanks 2answers 29 views ### What does $\sum_{n=0}^{\infty} (-1)^n .\frac{x^{2n+1}}{2n+1}$ converge to at x= 1 and x = -1 This is what I did. $\sum_{n=0}^{\infty} (-1)^n .\frac{1^{2n+1}}{2n+1}=\sum_{n=0}^{\infty}(-1)^n .\frac{1}{2n+1}$ Now I broke it up to positive and negative. \$\sum_{n=0}^{\infty}(-1)^n ... 1answer 89 views ### A question on uniform convergence Let $f,g:[0,1]\rightarrow \mathbb{R}$ be continuous functions. Define $$\displaystyle x_{n}(t)=f(t)+ \int_{0}^{t}x_{n-1}(s)ds\quad 0\leq t\leq 1,\;n=1,2,\cdots ,$$ where $x_{0}(t)=g(t),0\leq t\leq 1$ ... 2answers 68 views ### Checking convergence of an improper integral I did a quick search here but couldn't find a similar problem (it's probably out there somewhere...) I'm stuck with this rather simple improper integral: $\int_{1}^{\infty} \frac{1}{x^{\alpha}-1}dx$ ... 2answers 85 views ### Question about convergence of series if $\{n a_n\} \to 0$ This is part of Rudin's PMA Exercise 3.14 (d). If I understand correctly, it would be helpful to prove the following: Let $a_n$ be some sequence. Assume that $\lim_{n\to\infty} na_n = 0$. Prove ... 3answers 87 views ### Uniform convergence of a sequence of functions Let $(u_n(x))$ be a sequence of functions in $(0,\infty)$ such that: $u_1(x)=x$, $u_{n+1}=\frac12\left(u_n(x)+\frac1{u_n(x)}\right)$ for $n\in\mathbb{N}$. Check if $u_n(x)$ converge unfiormly in ... 0answers 42 views ### Convergence of integrals over divergent parts I'm wondering if it is possible for an integral which diverges in the limits $1$ to $\infty$ to converge in the limits from $0$ to $\infty$. And if so: how could I find this out? For example ... 2answers 121 views ### $\sum\limits_{n=1}^\infty |a_n|$ converges implies $\sum\limits_{n=1}^\infty |a_n|^2$ converges? [duplicate] Possible Duplicate: Prove that $\sum_{n=1}^{\infty}\ a_n^2$ is convergent if $\sum_{n=1}^{\infty}\ a_n$ is absolutely convergent If $\sum\limits_{n=1}^\infty |a_n|$ converges, the ... 1answer 38 views ### Convergence integral causal function I have an exercise where there is the following given: $f$ is a causal function. $f$ is Laplace transformable:$\int_{0}^{\infty} f(t)e^{-zt} \, dt = L(z)$ with $Real(z)> -1$ I have to ... 1answer 59 views ### Why one comparison test and not the other I'm studying integral comparison tests and I come to this one. $$f(x)=\int_{1}^{\infty}\frac{\sqrt{x}}{x^5+\sqrt[3]{x}}dx$$ The solution provided is to do \$\displaystyle ...

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 120, "mathjax_display_tex": 23, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9022445678710938, "perplexity_flag": "head"}

http://cs.stackexchange.com/questions/tagged/linear-algebra

# Tagged Questions The linear-algebra tag has no wiki summary. 0answers 38 views ### how to represent Sparse Matrices I have been using Harwell Boeing format, also known as Compressed Column Strorage (CCS) in order to store Sparse Matrices. Could you please suggest me some other way to store/represent sparse ... 2answers 62 views ### Finding the required value of an algebric expression I have an expression $$Ax+By+Cz.$$ where $A$, $B$ and $C$ are positive constants $\ge1$. The variables $x$, $y$ and $z$ are non-negative integers. I am also given a number $T$. I want to find the ... 3answers 30 views ### Solving system of linear inequalities I am trying to solve a system of inequalities in the following form: $\ x_i - x_j \leq w$ I know these inequalities can be solved using Bellman-Ford algorithm. ... 2answers 78 views ### Can you complete a basis in polynomial time? Here is the problem: we are given vectors $v_1, \ldots, v_k$ lying in $\mathbb{R}^n$ which are orthogonal. We assume that the entries of $v_i$ are rational, with numerator and denominator taking $K$ ... 1answer 41 views ### Counting solutions to system of linear equations modulo prime I have implemented Gaussian elimination for solving system of linear equations in the field of modulo prime remainders. If there is a pivot equal to zero I assume the system has no solution but how to ... 2answers 95 views ### What is the complexity of this matrix transposition? I'm working on some exercises regarding graph theory and complexity. Now I'm asked to give an algorithm that computes a transposed graph of $G$, $G^T$ given the adjacency matrix of $G$. So basically ... 0answers 28 views ### multigrid method to solve PDE [closed] I need explanation of the Multigrid Method or some literature. I am familiar with iterational methods including BiCGStab,CG,GS,Jacobi and preconditioning, but I am a beginner with multigrid method. ... 1answer 46 views ### LU decomposition with pivoting I have to solve system of linear algebraic equations $AX=B$, where $A$ is a two-dimensional matrix with all elements of main diagonal equal to zero. How to solve this problem? Iterational methods are ... 1answer 121 views ### Machine Learning: how to correctly calculate gradient descent for simple linear problem So, I was trying to learn machine learning, and, after watching a couple of Andrew Ng's lectures decided to try and write a simple piece of code to determine what someone's salary would be based on ... 0answers 22 views ### Maximum feasible subsystem problem (MaxFS) in 2 variables Topic: The maximum feasible subsystem problem, which is generally NP-hard [1]. Question: Are there special algorithms in case of only 2 variables (2D linear constraints)? The problem seems to be a ... 0answers 103 views ### Alternatives to SVD for rank factorization I have rank-deficient matrix $M \in \mathbb{R}^{n\times m}$ with $\text{rank}(M) = k$ and I want to find a rank factorization $M = PQ$ with $P \in \mathbb{R}^{n \times k}$ and \$Q \in \mathbb{R}^{k ... 0answers 40 views ### Time - Complexity Convex Optimization and Eigen Decomposition Say I had the choice of choosing one out of the following two optimization problems which I could use to solve my problem. Which choice is the fastest? How much of a trade-off would it be? Is the ... 1answer 117 views ### Probabilistic test of matrix multiplication with one-sided error Given three matrices $A, B,C \in \mathbb{Z}^{n \times n}$ we want to test whether $AB \neq C$. Assume that the arithmetic operations $+$ and $-$ take constant time when applied to numbers from ... 1answer 352 views ### Complexity of checking whether linear equations have a positive solution Consider a system of linear equations $Ax=0$, where $A$ is a $n\times n$ matrix with rational entries. Assume that the rank of $A$ is $<n$. What is the complexiy to check whether it has a solution ...

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 31, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.920265793800354, "perplexity_flag": "head"}

http://en.wikisource.org/wiki/On_General_Relativity

# On General Relativity From Wikisource On General Relativity (1918) by Harry Bateman Philosophical Magazine 6, 37 (218), 219-229, Online On General Relativity Harry Bateman 1918 On General Relativity. To the Editors of the Philosophical Magazine. Gentlemen,— THE appearance of Dr. Silberstein's recent article[1] on "General Relativity without the Equivalence Hypothesis" encourages me to restate my own views on the subject. I am perhaps entitled to do this as my work on the subject of General Relativity was published before that of Einstein and Kottler, and appears to have been overlooked by recent writers. In 1909 I proposed a scheme of electromagnetic equations[2] which are covariant for all transformations of co-ordinates which are biuniform in the domain we are interested in. These equations were similar to Maxwell's equations, except that the familiar relations B = μH, D = kE of Maxwell's theory were replaced by more general equations, which implied that two fundamental integral forms were reciprocals with regard to a quadratic differential form $\sum\sum g_{m,n}dx_{m}dx_{n}$ which was assumed to be invariant for all transformations of co-ordinates. The coefficients of the quadratic form were regarded as characteristics of the medium supporting the electromagnetic field and of the motion of the medium and its parts. The vanishing of the quadratic form was regarded as the condition that two neighbouring particles should be in positions such that a disturbance starting from one at the associated time should arrive at the other at its associated time[3]. The idea that the coefficients of the quadratic form might be considered as characteristics of the mind interpreting the phenomena was also entertained[4], and it was suggested that a correspondence or transformation of co-ordinates might be considered as a crude mathematical symbol for a mind. The phenomena here considered were those occurring in the brain and body; and although the correspondence by which the universe is reconstructed, so to speak, may be totally different[5] from the type contemplated here, yet it was thought that some of the general conclusions might still be valid if a transformation of co-ordinates was adopted as a working model of the correspondence. It was thought, for instance, that there was an analogy between the relativity principle that the earth's motion in space cannot be detected from experiments with terrestrial objects: and the interesting fact that we are unaware of the flow of blood and other processes taking place in our own bodies so long as they take place in the normal way. It was thought that the correspondence by which the elementary processes in the brain are interpreted may be adjusted in such a way that some of the changes are obscured. Again, if we assume that the nature of an electromagnetic field depends on the type of fundamental quadratic form which determines the constitutive relations, and thus depends indirectly on a transformation which filters the coefficients of this quadratic form, this dependence may be a symbol for the relation between physical and mental phenomena instead of giving the influence of gravitation on light as in Einstein's theory. Einstein and others have attempted to formulate a set of equations of motion which will cover all physical phenomena; but the present writer does not feel inclined to accept them as final, because in his opinion the true equations of motion should be capable of accounting for the phenomena of life, which after all are the most important physical phenomena. To make my position more definite, let us consider one of the methods by which the equations of motion of an electron are obtained in the usual electromagnetic theory. The principle is adopted that at each instant the integral over the electron of the total force on each element must be zero. Now before this principle can be used to write down the equations of motion we must know the design of the electron, and we must know the way in which the motions of the different elements are co-ordinated. This co-ordination or organization of the motions of the elements may be represented mathematically by a sequence of infinitesimal transformations, by which some of the features of the design are preserved. The design of the electron and the co-ordinated motion of its parts may, perhaps, be specified by a quadratic differential form in four variables, which determines a mapping of the interior of the electron on the interior of a stationary sphere; but I doubt if this is sufficiently general. A knowledge of this quadratic differential form is necessary then before we can write down the equations of motion of the electron as a whole. What we usually regard as the equations of motion of matter need then to be supplemented by geometrical conditions which specify the design and organization of each elementary portion of matter. Furthermore, when this design and organization is assumed to be known, the ordinary equations of motion may be regarded as a consequence of the electromagnetic laws and the above-mentioned principle. It must be confessed, however, that this principle does not seem satisfactory for a fundamental principle, and is probably a consequence of some deep underlying principles which are the true equations of motion. These new principles should indicate the reason for a similarity of design of the different electrons. One of the fundamental facts of life is that a good design is copied, and that there is a certain characteristic of the design of an object and its surrounding medium, depending perhaps on the closeness of fit of an imperfect correspondence, which determines the extent to which the design of the object is copied and preserved in the surrounding medium. This may be called the value of the design in relation to the medium, and it is a quantity which I feel must be taken into account in the true equations of motion, and a number assigned to it at each instant. As an example of standardization, the Ford motor-car is not in it with the electron; and, according to the above view, we must regard the design of the electron as one of very great value in relation to the surrounding medium. Returning to our generalized scheme of electromagnetic equations, and looking at matters from the point of view of physical optics, it may be remarked that the scheme of constitutive relations mentioned above is not sufficiently general to cover the case of a doubly-refracting crystalline medium[6]. To remedy this defect we may use a biquadratic integral form instead of a quadratic differential form to specify the constitutive relations. The vanishing of the biquadratic integral form may perhaps be regarded as the condition for action of a moving curve on a particle, a type of condition that seems natural if we regard moving Faraday tubes as fundamental. With this generalized theory it is possible for the elementary wave surface in a medium to be a general Kummer surface, a surface of which Fresnel's wave surface is a particular case. It is doubtful whether this generalized theory is sufficiently general for all purposes, and the above example is given just to emphasize that the absolute calculus of Ricci and Levi Civita can be used to develop a theory of generalized relativity on many lines in addition to that adopted by Einstein. Going back to the case in which a quadratic form is sufficient to determine the optical properties of a medium, we may remark that if Einstein's idea of the gravitational equations is accepted, it is still by no means certain that his quadratic form from which the gravitational equations are derived is the same as the quadratic form which determines the optical properties of the medium. Indeed, the example which I considered on p. 262 of my first paper would seem to indicate that this was not the case. It should be mentioned that in the first seven equations in this example there is a misprint, $\epsilon\mu$ should be replaced by $(\epsilon\mu)^{-1}$. On the above view Einstein's idea of an influence of gravitation on light is simply an hypothesis, but a very interesting and reasonable one. It may be remarked, however, that in the theory of surfaces there are two fundamental quadratic forms, and we may perhaps expect something similar in general relativity. With regard to possible extensions of the idea of relativity it may be worth while to consider transformations analogous to the contact transformations of dynamics in which the co-ordinates x, y, z, t and the component velocities u, v, w correspond to a new set $\left(x_{1}y_{1}z_{1}t_{1}u_{1}v_{1}w_{1}\right)$ in such a way that the differential equations $\frac{dx_{1}}{u_{1}}=\frac{dy_{1}}{v_{1}}=\frac{dz_{1}}{w_{1}}=dt_{1}$ are a consequence of the equations $\frac{dx}{u}=\frac{dy}{v}=\frac{dz}{w}=dt$ This may be secured by making a single quadratic form, such as $\begin{array}{cc} \left(dx^{2}+dy^{2}+dz^{2}-c^{2}dt^{2}\right)\left(c^{2}-u^{2}-v^{2}-w^{2}\right)\\ +\left(c^{2}dt-udx-vdy-wdz\right)^{2}, & \left(c^{2}>u^{2}+v^{2}+w^{2}\right)\end{array}$ an invariant[7]. Various other quadratic forms consisting of sums of squares may, of course, be adopted instead. H. Bateman. Throop College. Pasandena, Cal. Aug. 10th, 1918. 1. Phil. Mag. July 1918 2. Loc. cit. p. 225. See also Amer. Journ. of Math. vol. xxxiv. p. 340 (1912). 3. The term correspondence is used here in a very general sense, and is by no means restricted to the familiar one to one correspondence of entities of the same type, such as points. We should say, for instance, that there is a correspondence between the disturbance running along a telephone-wire and the sound-waves which produce it, because we can pass from one to the other by mathematical equations of a definite type, or rather by solving the equations and the boundary conditions. A correspondence is, moreover, regarded as an entity which may have real existence and be capable of growth and variation. 4. Proc. London Math. Soc. ser. 2, vol. viii. p. 375. See also p. 261 of my first paper. 5. This is a positive definite quadratic form in the variables dx - udt, dy - vdt, dz - wdt, and so can only vanish when all these quantities are zero. This work is in the public domain in the United States because it was published before January 1, 1923. The author died in 1946, so this work is also in the public domain in countries and areas where the copyright term is the author's life plus 60 years or less. This work may also be in the public domain in countries and areas with longer native copyright terms that apply the rule of the shorter term to foreign works.

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 7, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9534021019935608, "perplexity_flag": "head"}

http://math.stackexchange.com/questions/177125/holomorphic-function-on-the-complex-plane?answertab=votes

# holomorphic function on the complex plane let $f(z) \in Hol(\mathbb{C})$ holomorphic function such that for each $z_0 \in \mathbb{C}$ there exists $N(z_0)$ such that $f^{(N(z_0))}(z_0) = 0$ Prove: that f is a polynom - ## 2 Answers Modifying Davide's argument but avoiding Baire's category theorem: Let $\overline{\mathbb E}$ be the closed unit ball and $$F_n := \{ z \in \overline{\mathbb E} : f^{(n)}(z) = 0 \}.$$ Then, by assumption, $\overline{\mathbb E} = \bigcup_n F_n$. Since $\overline{\mathbb E}$ is uncountable, not all $F_n$ can be finite. Let $n \in \mathbb N$ s.t. $F_n$ is infinite. Since $\overline{\mathbb E}$ is compact, $F_n$ has a limit point in $\overline{\mathbb E}$, i.e. there is a $z_0 \in \overline{\mathbb E}$ and a sequence $(z_k)$ in $F_n \setminus \{z_0\}$ that converges to $z_0$. We have $f^{(n)}(z_k) = 0$ for all $k$, hence $f^{(n)} = 0$ by the identity theorem, so $f$ is a polynomial. - +1 Damn, that was beautiful! – DonAntonio Aug 2 '12 at 21:06 $f^n=0$ means $f$ is a polynomial? how? – Taxi Driver Aug 3 '12 at 4:15 $f^{(n)}$ means $f$ power n or what? – Taxi Driver Aug 3 '12 at 4:27 $f^{(n)}$ denotes the $n$-th derivative of $f$. I assumed that you know that since you used that notation in your question. To see that $f^{(n)} = 0$ implies that $f$ is a polynomial just look at the Taylor expansion of $f$. – marlu Aug 3 '12 at 15:19 Let $F_n:=\{z\in\Bbb C, f^{(n)}(z)=0\}$. Since $f^{(n)}$ is continuous, $F_n$ is closed and by hypothesis, $\Bbb C=\bigcup_{n\in\Bbb N}F_n$. By Baire's theorem, there exists a $n_0$ such that $F_{n_0}$ has a non-empty interior. Hence $f^{(n_0)}$ vanishes on a ball, and it's necessarily the null function. This proves that $f$ is a polynomial. Note that there is a similar result for $C^{\infty}$ functions from $\Bbb R$ to $\Bbb R$, but it's harder to show. It has been discussed here. - is there a proof without baire theorem? i didnt learn that in the complex analysis course – bar Jul 31 '12 at 11:33 And what did you learn? In which chapter such an exercise appear? Actually, Baire is more a theorem from topology than complex analysis. – Davide Giraudo Jul 31 '12 at 11:38 i think the relevant theorems for this exercise are: uniqeness theorem: if $f\in Hol(D)$ D is a domain and there is U a subset of D such that U has accumulation point and $\forall z \in U f(z)=0$ then $\forall z\in D f(z) = 0$ – bar Jul 31 '12 at 11:43

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 45, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9520666003227234, "perplexity_flag": "head"}

http://quant.stackexchange.com/questions/2148/how-to-detect-regime-change-when-estimating-asset-correlation-from-historical-ti?answertab=votes

# How to detect regime change when estimating asset correlation from historical time series? Suppose I have two asset time series, $X_t$ and $Y_t$, and I'm estimating their correlation from historical data. I'd like to apply some systematic criterion to estimate what time window I should use to estimate the correlation reliably, and also to spot "regime changes" (when the correlation jumps suddenly) after which I should discard old data completely (as opposed to rolling the window in a continuous manner). Can you recommend some approaches which have a decent theoretical background? - ## 3 Answers You can use changepoint analysis to identify regime change. You can also look at large angle differences in the eigenvectors between your most up-to-date/recent covariance matrix and the covariance matrix from the prior window. Another way to identify regime change is using a factor model. If the returns on a particular set of factors is X standard deviations from its usual terrain for a sustained period then you can call this regime change. I do not believe you will find a single time window that is best. Regime duration is variable. Key here is identifying an estimation procedure for a covariance matrix that produces reasonable out-of-time forecasts. You will need to do some empirical testing, or develop a rule to re-estimate your model based on the how you identify regimes, or use Garch (or other dynamic model) as Patrick suggests. Technical side note: You probably don't want to discard old data completely but instead weight more recent data with exponential weighting, or re-scale the covariance matrix to reflect current volatility. The eigenvectors of the correlation matrix (after the 1st eigenvector which is the market factor) will correspond to sector and industry groups. These correlations will persist. When market flip from bull to bear market (let's call this a 1st order approximation of regime - as opposed to style and industry changes) what is happening is that the variance explained by the largest eigenvector has increased substantially. - Do you have any references about changepoint analysis? – quant_dev Oct 14 '11 at 6:50 1 – strimp099 Oct 14 '11 at 15:32 Yep - those links look great – Quant Guy Oct 14 '11 at 17:47 @strimp099 Are there any resources in these search results you find particularly instructive and interesting? Introductions, surveys, papers, books? – vanguard2k Dec 27 '12 at 15:38 @vanguard2k I found the first two links interesting but in fair disclosure have done very little in the field area. I have not seen any papers either... – strimp099 Jan 7 at 18:23 I would suggest a multivariate garch model as a possibility. We aren't exactly overrun with wonderful software for that, but with just bivariate data I would think that the in-sample correlation estimates would be reasonably robust over models and estimation. It would be good to try two or three ways of doing it to make sure I'm right about that. You may find that the garch route is good enough to use as your solution if you aren't forecasting very far ahead. - Could I use R for that? – quant_dev Oct 12 '11 at 15:14 1 of course, garch is implemented in several R packages – RockScience Oct 13 '11 at 7:05 One approach would be Engle (2002) dynamic conditional correlations. Taking your $Y_t$ and $X_t$, I will make the simplifying assumption that the mean equation of these is: $$\boxed{Y_t = \mu_y + \varepsilon_{y,t}}$$ $$\boxed{X_t = \mu_x + \varepsilon_{x,t}}$$ with $\varepsilon_{y,t} = z_{y,t} \sigma_{y,t} \sim N(0,\sigma_{y,t})$, $\varepsilon_{x,t} = z_{x,t} \sigma_{x,t} \sim N(0,\sigma_{x,t})$. In practice you might want to specify a GARCH-M, or a GARCH with exogenous variables inside the mean equation. For example if $Y_t$ and $X_t$ are individual stocks in the same market, you might want to include $R_{M,t}$ so that you don't detect any correlation due to this shared factor. If you're looking across borders GARCH-M terms may become important if you want to control for portfolio rebalancing due to changing relative risks or something. The mean equations can be written in vector form as: $$\boxed{ Z_t = \mathbf{\mu} + \mathbf{\varepsilon}_t}$$ with $Z_t := [Y_t,X_t]'$, $\mathbf{\varepsilon} \sim N(0,H_t)$, $H_t := D_t R_t D_t$. Here, $R_t$ is the (possibly) time-varying correlation matrix and $D_t$ is simply $\text{diag}(\sigma_{y,t},\sigma_{x,t})$ when we assume no news or variance spillovers in the variance equation. The DCC estimator in the `rmgarch` package will provide you with the dynamics of $R_t$ with no effort on your part. You can then visually inspect for a break in the correlation. However, for an objective approach withing the DCC framework, take a look at this paper for the ability to hypothesis test structural breaks in the correlation. -

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 16, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9419596791267395, "perplexity_flag": "middle"}

http://www.physicsforums.com/showthread.php?t=514850

Physics Forums ## Diagonalization 1. The problem statement, all variables and given/known data I think my teacher made a mistake in his homework answer. I need to verify this for practice. The answer I got is below. The answer the teacher has is in the pdf. 2. Relevant equations 3. The attempt at a solution So there is two eigenvalues= 4 and 2 but the eigenvalue 2 has 2 eigenvectors [-1 1 0]T and [0 0 1]T but my teacher has only one [-1 1 0]T. That's why he says A is not diagonalizable. Do you think it's correct? Attached Thumbnails PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug I get the same result as your teacher for $$\lambda = 2$$ $$A= \begin{bmatrix} 3 & 1 & 0\\ 0 & 2 & 1\\ 1 & 1 & 3 \end{bmatrix}$$ so for lamda = 2, $$(A-2I)\vec{v}=\vec{0}$$ $$\begin{bmatrix} 1 & 1 & 0\\ 0 & 0 & 1\\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} v_1\\ v_2\\ v_3 \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\\ \end{bmatrix}$$ I've personally always found it easier to not do row operations here and just jump straight in. From that you get $$v_3 = 0$$ and $$v_1 = -v_2$$ so therefore $$\vec{v} = \begin{bmatrix} 1\\ -1\\ 0 \end{bmatrix}$$ Since it is a 3x3 matrix, it needs 3 eigenvectors to be diagonalizable. Hope this helps. Hi, thanks for replying. Attached is how I got my vectors, do you think my steps are correct. Attached Thumbnails Mentor ## Diagonalization No. Why do you think (0, 0, 1) is an eigenvector? You seem to just pull that out of thin air. You seem to have made a mistake in the step $$(A-2I)\vec{v}=\vec{0}$$ Why is your second row 1 0 1 instead of 0 0 1 ? Have you said $$v_2= \delta$$ ? I'm not sure if I have read that correctly. If it is a delta, you can't say that unless the whole row equals zero. i.e. $$\begin{bmatrix} 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} v_1 \end{bmatrix} = \begin{bmatrix} 0 \end{bmatrix}$$ Tags algebra, diagonalization, eigenvalue, linear Thread Tools | | | | |--------------------------------------|----------------------------|---------| | Similar Threads for: Diagonalization | | | | Thread | Forum | Replies | | | Calculus & Beyond Homework | 0 | | | Calculus & Beyond Homework | 3 | | | Calculus & Beyond Homework | 5 | | | Quantum Physics | 1 | | | Quantum Physics | 2 |

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 10, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9221944808959961, "perplexity_flag": "middle"}

http://math.stackexchange.com/questions/292533/convergent-or-divergent

# Convergent or Divergent Determine if the series converges or diverges: $$\sum_{n=1}^\infty \frac{\ln(n)}{n \left|\sin(n)\right|} =$$ We know that $$\sum_{n=1}^\infty \frac{1}i = \text{Divergent}$$ $$\sum_{n=1}^1 \ln(i )= 0$$ How to solve this? - 1 you need to remove 1 from right hand side of first eq. The other two can be reworded too. – Maesumi Feb 2 at 3:26 ## 2 Answers Another way: also note $\frac{\log n }{n|\sin n|} \geq \frac{\log n}{n}$. Now look at $\sum_{n=1}^{\infty}\frac{\log n }{n}$. By intergral test if the integral diverges, then the sum diverges too. $\int_{1}^{\infty} \frac{\log x dx}{x} = \left. \frac{ \log^{2} x}{2} |^{\infty}_{1}\right.=\infty$. The integral diverges, hence the sum diverges too. Since the original sum is lower-bounded by this one, by comparison test it diverges too. - Hint: Since $|\sin n|\le 1$, and $\ln n \gt 1$ for $n\ge 3$, the $n$-th term of our sequence is greater than $\frac{1}{n}$ if $n\ge 3$. Now use the Comparison Test. - Got it thanks! That seems easier now – user1730308 Feb 2 at 3:18

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 9, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8597381114959717, "perplexity_flag": "middle"}

http://math.stackexchange.com/questions/139549/how-to-show-path-connectedness

# How to show path-connectedness Well, I am not getting any hint how to show $GL_n(\mathbb{C})$ is path connected. So far I have thought that let $A$ be any invertible complex matrix and $I$ be the idenity matrix, I was trying to show a path from $A$ to $I$ then define $f(t)=At+(1-t)I$ for $t\in[0,1]$ which is possible continous except where the $\operatorname{det}{f(t)}=0$ i.e. which has $n$ roots and I can choose a path in $\mathbb{C}\setminus\{t_1,\dots,t_n\}$ where $t_1,\dots,t_n$ are roots of $\operatorname{det}{f(t)}=0$, is my thinking was correct? Could anyone tell me the solution? - This path can fail in liying on $GL_n(\mathbb{C})$. For example, if you take $A=-I$, at the middle point $t=1/2$ you will get $f(t)=0$ – matgaio May 1 '12 at 19:45 +1 Looks good to me. @matgaio, the idea seems to be that you go around that troublesome $t=1/2$ - there is room for that, when you let $t$ be a complex number! It might have been clearer to define $f(t)$ for all complex $t$, exclude the finite set of points, and then select a path from $0$ to $1$. – Jyrki Lahtonen May 1 '12 at 19:47 Uhm, ok. I was understanding he was trying to go straight from $A$ to $I$. – matgaio May 1 '12 at 19:50 2 – Antonio Vargas May 1 '12 at 20:02 ## 3 Answers • If $P$ is a polynomial of degree $n$, the set $\{\lambda,P(\lambda)\neq 0\}$ is path connected (because its complement is finite, so you can pick a polygonal path). • Let $P(t):=\det(A+t(I-A))$. We have that $P(0)=\det A\neq 0$, and $P(1)=\det I=1\neq 0$, so we can find a path $\gamma\colon[0,1]\to\mathbb C$ such that $\gamma(0)=0$, $\gamma(1)=1$, and $P(\gamma(t))\neq 0$ for all $t$. Finally, put $\Gamma(t):=A+\gamma(t)(I-A)$. • If $B_1$ and $B_2$ are two invertible matrices, consider $\gamma(t):=B_2\cdot\gamma(t)$, where we chose $\gamma$ for $A:=B_2^{-1}B_1$. - what is $f(\gamma(t))$, will it be $P(\gamma(t))$? – Taxi Driver Feb 28 at 11:19 Since any matrix $A\in GL_n(\mathbb C)$ has only finitely many eigenvalues, and 0 isn't one of them, there is a point $z\in S^1$ such that the line through the origin containing $z$ doesn't intersect any of the eigenvalues of $A$. Now, consider the path $f(t)=At+z(1-t)I$. This has determinant 0 iff $z(t-1)$ is an eigenvalue of $At$, which happens iff $z(1-1/t)$ is an eigenvalue of $A$ (this doesn't work when $t=0$, but then it is clear that the determinant is non-zero). By construction, it isn't for any $t\in[0,1]$ so this defines a path form $A$ to $zI$. now there is a path not passing through 0 from $z$ to 1, and this gives rise to a path from $zI$ to $I$, and so concatenating the two paths, we get a path from $A$ to $I$, showing that $GL_n(\mathbb C)$ is path connected. - Use $\Gamma(t) = e^{t \log A + (1-t) \log B}$. This is well defined since $A,B$ are invertible. $\Gamma(t)$ is clearly invertible for all $t$, $\Gamma(0) = B$, $\Gamma(1) = A$. - This idea is not mine, I just can't remember where I saw it, it was in a control theory context. – copper.hat May 1 '12 at 20:20

{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 68, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9570044279098511, "perplexity_flag": "head"}