url
stringlengths 14
1.76k
| text
stringlengths 100
1.02M
| metadata
stringlengths 1.06k
1.1k
|
|---|---|---|
http://math.stackexchange.com/questions/317246/is-there-any-bound-for-int-mathbbr-sqrtf-0xf-1x-mboxdx | # Is there any bound for $\int_{\mathbb{R}}\sqrt{f_0[x]f_1[x]}\mbox{d}x$
I wonder if there is an upperbound for following the expression:
$$\int_{\mathbb{R}}\sqrt{f_0[x]f_1[x]}\mbox{d}x$$
where $f_i$, $i=0,1$ are some density functions.
1. Use Cauchy Schwarz.$\mbox{ }$ | {"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": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9933848977088928, "perplexity": 275.9500799012655}, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440644065464.19/warc/CC-MAIN-20150827025425-00008-ip-10-171-96-226.ec2.internal.warc.gz"} |
http://mathhelpforum.com/calculus/13497-dervatives-newton-s-method.html | # Math Help - Dervatives and Newton's Method
1. ## Dervatives and Newton's Method
#1. f(x) = x^4-17x^2+18
What is the relationship between the sign of f'(x) and the graph f(x) ?
What is the relationship between the roots of f''(x) and the graph of f(x) ?
What is the relationship between the sign of f''(x) and the graph of f(x) ?
#2. Given that f(x) = sin(x) use Newton's Method to find the first 2 positive zeros showing the first few approximations in each case.
So...
pi
___
x1: 3
x2: 3.142546543
x3: 3.141592653
x4: 3.14159255359
also...
2pi
____
x1: 6
x2: 6.29100619138
x3: 6.28310514772
x4: 6.28318530718
The first positive zero is: 3.14159
The second positive zero: 6.28319
What happens when you took pi/2 as first approximation? Explain
I got 1 but else is there to say
#3. if f(x) = x^4-17x^2+18
f''(x) = 12x^2-34
Therefore f''(x) = 0 if x = -sqrt(102)/6 and sqrt(102)/6 and these are the x-coordinates of the points of __ ___ of f(x)?
Thanks for the Help!
2. ## Re:
P.S. This is one of the Calculator labs that the department makes up
3. Originally Posted by qbkr21
#1. f(x) = x^4-17x^2+18
What is the relationship between the sign of f'(x) and the graph f(x) ?
What is the relationship between the roots of f''(x) and the graph of f(x) ?
What is the relationship between the sign of f''(x) and the graph of f(x) ?
the sign of f'(x) let's you know when f(x) is increasing or decreasing. if f'(x) is positive, f(x) has a positive slope and is hence increasing. if f'(x) is negative, f(x) has a negative slope and is hence decreasing
the roots of f''(x) give the inflection points of f(x), that is the points where f(x) changes concavity from concave up to concave down or vice versa.
the sign of f''(x) indicates the concavity (and hence the nature of critical points) of f(x). if f''(x) is positive for some critical point in f(x), then f(x) is concave up at that point and it is a local min. if f''(x) is negative at some critical point of f(x), then f(x) is concave down at that point and it is a local max
Originally Posted by qbkr21
#3. if f(x) = x^4-17x^2+18
f''(x) = 12x^2-34
Therefore f''(x) = 0 if x = -sqrt(102)/6 and sqrt(102)/6 and these are the x-coordinates of the points of __ ___ of f(x)?
points of inflection of f(x)...usually.
4. ## Re:
Here is the exact question Jhevon by the way thanks for the help...
5. Originally Posted by qbkr21
Here is the exact question Jhevon by the way thanks for the help...
you used x_(n+1) = x_n - f(x_n)/f'(x_n) for each of these right? did you find the first few values for pi/2?
6. ## Re:
Uhhh...I just plug in pi/2 into Newtons method formula and x_n-f(x)/f'(x) and got the number 1. Is there something else I should do. Thanks for the helpful posts.
7. Originally Posted by qbkr21
Uhhh...I just plug in pi/2 into Newtons method formula and x_n-f(x)/f'(x) and got the number 1. Is there something else I should do. Thanks for the helpful posts.
ok so these questions are weird, i don't know exactly what answer they are looking for, but anyway, we can see a few patterns. note that Newton's method is used to approximate the roots of a function. for each approximation we begin with, using Newton's method puts us closer to the root that is closest to that value.
not that when we begin with x = 3, each new approximation is bigger than the last and they keep getting closer to pi, 3.1415... since pi is the closest root to 3
when we start with x = 6, each new approximation is again bigger than the last, that is because we are increasing to 2pi, the next closest root to 6
when we begin with pi/2, each new approximation begins to decrease, this is because we are moving towards the closest root, which is zero. i find this interesting though, since pi is as close to pi/2 as 0 is | {"extraction_info": {"found_math": false, "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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.933393120765686, "perplexity": 784.2802383387352}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1412037663060.18/warc/CC-MAIN-20140930004103-00186-ip-10-234-18-248.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/112677/proving-that-an-integral-domain-has-at-most-two-elements-that-satisfy-the-equati | # Proving that an integral domain has at most two elements that satisfy the equation $x^2 = 1$.
I like to be thorough, but if you feel confident you can skip the first paragraph.
Review: A ring is a set $R$ endowed with two operations of + and $\cdot$ such that $(G,+)$ is an additive abelian group, multiplication is associative, $R$ contains the multiplicative identity (denoted with 1), and the distributive law holds. If multiplication is also commutative, we say $R$ is a commutative ring. A ring that has no zero divisors (non-zero elements whose product is zero) is called an integral domain, or just a domain.
We want to show that for a domain, the equation $x^2 = 1$ has at most 2 solutions in $R$ (one of which is the trivial solution 1).
Here's what I did:
For simplicity let $1,a,b$ and $c$ be distinct non-zero elements in $R$. Assume $a^2 = 1$. We want to show that letting $b^2 = 1$ as well will lead to a contradiction. So suppose $b^2 = 1$, then it follows that $a^2b^2 = (ab)^2 = 1$, so $ab$ is a solution as well, but is it a new solution? If $ab = 1$, then $abb = 1b \Rightarrow a = b$ which is a contradiction. If $ab = a$, then $aab = aa \Rightarrow b = 1$ which is also a contradiction. Similarly, $ab = b$ won't work either. So it must be that $ab = c$. So by "admitting" $b$ as a solution, we're forced to admit $c$ as well.
So far we have $a^2 = b^2 = c^2 = 1$ and $ab = c$. We can procede as before as say that $(abc)^2 = 1$, so $abc$ is a solution, but once again we should check if it is a new solution. From $ab = c$, we get $a = cb$ and $b = ac$, so $abc = (cb)(ac)(ab) = (abc)^2 = 1$. So $abc$ is not a new solution; it's just one.
At this point I'm stuck. I've shown that it is in fact possible to have a ring with 4 distinct elements, namely $1,a,b$ and $c$ such that each satisfies the equation $x^2 = 1$ and $abc = 1$. What am I missing?
-
Have you tried factoring $x^2-1$? – user641 Feb 24 '12 at 0:40
I hope I'm not over simplifying things, but since the ring is commutative, if $a$ is any solution, then $a^2-1=0$, which implies $(a-1)(a+1)=0$. Since you're working in an integral domain, then either $a-1=0$ or $a+1=0$, which implies $a$ can only be $1$ or $-1$. – Buble Feb 24 '12 at 0:43
More generally, every element $\ell\in R$ of an integral domain $R$ cannot have more than two square roots. To see this, let $a$ be such that $a^2=\ell$, and suppose $b^2=\ell$ also for some $b\in R$. Then we can subtract one from the other and factor as $(a-b)(a+b)=0$, and deduce $b=\pm a$ via integrality.
-
Hint: $x^2 - 1 = (x-1)(x+1)$. If this is $0 \ldots$
-
You’ve shown that if $R$ has two distinct elements other than $1$ whose squares are $1$, then their product is a third such element. But in fact $R$ can’t have two distinct elements other than $1$ whose squares are $1$ in the first place. To see this, show that $x^2=1$ can have at most two solutions by factoring $x^2-1$ and using the fact that $R$ has no zero-divisors.
-
Thanks, that explains it. – mahin Feb 24 '12 at 1:27
+1 This is the only answer that actually mentions the OP's argument, and where it fails. – M Turgeon Mar 23 '12 at 17:44
In any integral domain, a polynomial of degree $d$ has at most $d$ roots, which implies your result. If $aT + b$ is a degree-1 polynomial with coefficients in an integral domain and $a \ne 0$, then if $x$ and $y$ are roots, we see that $a(x-y) = 0$ which implies $x = y$. Now proceed by induction.
-
More generally, overy any ring, a nonzero polynomial has no more roots than its degree if the difference of any two distinct roots is not a zero-divisor (which is true in any integral domain).
THEOREM $\$ Let $\rm R$ be a ring and let $\rm\:f\in R[x].\:$ If $\rm\:f\:$ has more roots than its degree, and if the difference of any two distinct roots is not a zero-divisor, then $\rm\: f = 0.$
Proof $\$ Clear if $\rm\:deg\: f = 0\:$ since the only constant polynomial with a root is the zero polynomial. Else $\rm\:deg\: f \ge 1\:$ so by hypothesis $\rm\:f\:$ has a root $\rm\:s\in R.\:$ Factor Theorem $\rm\Rightarrow\: f(x) = (x-s)\:g ,$ $\rm\: g\in R[x].$ Every root $\rm\:r\ne s\:$ is a root of $\rm\: g\:$ by $\rm\: f(r) = (r-s)g(r) = 0\ \Rightarrow\ g(r) = 0, \ by\ \ r-s\$ not a zero-divisor. So $\rm\:g\:$ satisfies hypotheses, so induction on degree $\rm\:\Rightarrow\:g=0\:\Rightarrow\:f=0.\$ QED
Here's a nice constructive application: if a polynomial $\rm\:f(x)\:$ over $\rm\:\mathbb Z/n\:$ has more roots than its degree, then we can quickly compute a nontrivial factor of $\rm\:n\:$ by a simple $\rm\:gcd\:$ calculation.
The quadratic case of this is at the heart of many integer factorization algorithms, which attempt to factor $\rm\:n\:$ by searching for a nontrivial square root in $\rm\: \mathbb Z/n,\:$ e.g. a square root of $1$ that's not $\pm 1$.
-
Glad to see you're back! – Tyler Feb 26 '12 at 0:28 | {"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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9673154354095459, "perplexity": 78.6066122417219}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375094957.74/warc/CC-MAIN-20150627031814-00130-ip-10-179-60-89.ec2.internal.warc.gz"} |
http://web2.0calc.com/questions/what-is-4-12-y-9 | +0
# what is 4/12 y/9
0
97
1
what is 4/12 y/9
Guest Mar 2, 2017
Sort:
#1
+7067
0
what is 4/12 y/9
$$\large\frac{4}{12}y:9=\frac{4}{12\times9}y=\frac{y}{27}$$
!
asinus Mar 2, 2017
### 4 Online Users
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. See details | {"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": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5423029065132141, "perplexity": 15586.966616239191}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187824931.84/warc/CC-MAIN-20171022003552-20171022023552-00492.warc.gz"} |
https://www.rdocumentation.org/packages/readr/versions/1.1.1/topics/datasource | # datasource
0th
Percentile
##### Create a source object.
Create a source object.
Keywords
internal
##### Usage
datasource(file, skip = 0, comment = "")
##### Arguments
file
Either a path to a file, a connection, or literal data (either a single string or a raw vector).
Files ending in .gz, .bz2, .xz, or .zip will be automatically uncompressed. Files starting with http://, https://, ftp://, or ftps:// will be automatically downloaded. Remote gz files can also be automatically downloaded and decompressed.
Literal data is most useful for examples and tests. It must contain at least one new line to be recognised as data (instead of a path).
skip
Number of lines to skip before reading data.
• datasource
##### Examples
# Literal csv
datasource("a,b,c\n1,2,3")
datasource(charToRaw("a,b,c\n1,2,3"))
# Strings
## Not run: ------------------------------------ | {"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": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.272586852312088, "perplexity": 12131.336410726362}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128320582.2/warc/CC-MAIN-20170625203122-20170625223122-00303.warc.gz"} |
http://mathematica.stackexchange.com/questions/43029/define-shifted-pdf | # Define shifted pdf
I defined a PDF called dist, for example like this one:
dist = ProbabilityDistribution[
Convolve[PDF[NormalDistribution[4, 5], x],
PDF[NormalDistribution[3, 1], x], x, t], {t, -Infinity, Infinity}]
Is it possible to get also a shifted PDF? For example
$p_1(x)=dist(x), p_2(x)=dist(x-c)$
where c is a constant
-
## 1 Answer
A transformed distribution should do what you want, e.g.,
dist=TransformedDistribution[a u + b, u \[Distributed] NormalDistribution[mu, sigma]]
so in your case, distributed by your custom PDF. Be aware, while neat, MMA fancy-probability features have their limits...
- | {"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": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7752730846405029, "perplexity": 14245.221373552153}, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-14/segments/1427131293283.10/warc/CC-MAIN-20150323172133-00177-ip-10-168-14-71.ec2.internal.warc.gz"} |
https://ftp.aimsciences.org/article/doi/10.3934/jimo.2008.4.155 | # American Institute of Mathematical Sciences
• Previous Article
Semicontinuity of solution mappings of parametric generalized strong vector equilibrium problems
• JIMO Home
• This Issue
• Next Article
Semismooth reformulation and Newton's method for the security region problem of power systems
January 2008, 4(1): 155-165. doi: 10.3934/jimo.2008.4.155
## On the stability of a dual weak vector variational inequality problem
1 College of Mathematics and Science, Chongqing University, Chongqing, 400044, China, China
Received November 2006 Revised June 2007 Published January 2008
In this paper, we obtain some stability results for the dual problem of a weak vector variational inequality problem. We establish the upper semicontinuity property of the solution set for a perturbed dual problem of a weak vector variational inequality problem. By virtue of a parametric gap function and a key assumption, we also obtain the lower semicontinuity property of the solution set for the perturbed dual problem. Some examples are given for the illustration of the necessity of our research on duality.
Citation: S. J. Li, Z. M. Fang. On the stability of a dual weak vector variational inequality problem. Journal of Industrial & Management Optimization, 2008, 4 (1) : 155-165. doi: 10.3934/jimo.2008.4.155
[1] Sara Munday. On the derivative of the $\alpha$-Farey-Minkowski function. Discrete & Continuous Dynamical Systems - A, 2014, 34 (2) : 709-732. doi: 10.3934/dcds.2014.34.709 [2] Ralf Hielscher, Michael Quellmalz. Reconstructing a function on the sphere from its means along vertical slices. Inverse Problems & Imaging, 2016, 10 (3) : 711-739. doi: 10.3934/ipi.2016018 [3] Charles Fulton, David Pearson, Steven Pruess. Characterization of the spectral density function for a one-sided tridiagonal Jacobi matrix operator. Conference Publications, 2013, 2013 (special) : 247-257. doi: 10.3934/proc.2013.2013.247 [4] Y. Latushkin, B. Layton. The optimal gap condition for invariant manifolds. Discrete & Continuous Dynamical Systems - A, 1999, 5 (2) : 233-268. doi: 10.3934/dcds.1999.5.233 [5] Yves Dumont, Frederic Chiroleu. Vector control for the Chikungunya disease. Mathematical Biosciences & Engineering, 2010, 7 (2) : 313-345. doi: 10.3934/mbe.2010.7.313 [6] Sergi Simon. Linearised higher variational equations. Discrete & Continuous Dynamical Systems - A, 2014, 34 (11) : 4827-4854. doi: 10.3934/dcds.2014.34.4827 [7] Xue-Ping Luo, Yi-Bin Xiao, Wei Li. Strict feasibility of variational inclusion problems in reflexive Banach spaces. Journal of Industrial & Management Optimization, 2020, 16 (5) : 2495-2502. doi: 10.3934/jimo.2019065 [8] A. K. Misra, Anupama Sharma, Jia Li. A mathematical model for control of vector borne diseases through media campaigns. Discrete & Continuous Dynamical Systems - B, 2013, 18 (7) : 1909-1927. doi: 10.3934/dcdsb.2013.18.1909 [9] Qian Liu. The lower bounds on the second-order nonlinearity of three classes of Boolean functions. Advances in Mathematics of Communications, 2021 doi: 10.3934/amc.2020136 [10] Hong Seng Sim, Wah June Leong, Chuei Yee Chen, Siti Nur Iqmal Ibrahim. Multi-step spectral gradient methods with modified weak secant relation for large scale unconstrained optimization. Numerical Algebra, Control & Optimization, 2018, 8 (3) : 377-387. doi: 10.3934/naco.2018024 [11] Lunji Song, Wenya Qi, Kaifang Liu, Qingxian Gu. A new over-penalized weak galerkin finite element method. Part Ⅱ: Elliptic interface problems. Discrete & Continuous Dynamical Systems - B, 2021, 26 (5) : 2581-2598. doi: 10.3934/dcdsb.2020196 [12] Kaifang Liu, Lunji Song, Shan Zhao. A new over-penalized weak galerkin method. Part Ⅰ: Second-order elliptic problems. Discrete & Continuous Dynamical Systems - B, 2021, 26 (5) : 2411-2428. doi: 10.3934/dcdsb.2020184 [13] Vieri Benci, Sunra Mosconi, Marco Squassina. Preface: Recent progresses in the theory of nonlinear nonlocal problems. Discrete & Continuous Dynamical Systems - S, 2021, 14 (5) : i-i. doi: 10.3934/dcdss.2020446 [14] Yanqin Fang, Jihui Zhang. Multiplicity of solutions for the nonlinear Schrödinger-Maxwell system. Communications on Pure & Applied Analysis, 2011, 10 (4) : 1267-1279. doi: 10.3934/cpaa.2011.10.1267 [15] Deren Han, Zehui Jia, Yongzhong Song, David Z. W. Wang. An efficient projection method for nonlinear inverse problems with sparsity constraints. Inverse Problems & Imaging, 2016, 10 (3) : 689-709. doi: 10.3934/ipi.2016017 [16] Olena Naboka. On synchronization of oscillations of two coupled Berger plates with nonlinear interior damping. Communications on Pure & Applied Analysis, 2009, 8 (6) : 1933-1956. doi: 10.3934/cpaa.2009.8.1933 [17] Jiangxing Wang. Convergence analysis of an accurate and efficient method for nonlinear Maxwell's equations. Discrete & Continuous Dynamical Systems - B, 2021, 26 (5) : 2429-2440. doi: 10.3934/dcdsb.2020185 [18] Manoel J. Dos Santos, Baowei Feng, Dilberto S. Almeida Júnior, Mauro L. Santos. Global and exponential attractors for a nonlinear porous elastic system with delay term. Discrete & Continuous Dynamical Systems - B, 2021, 26 (5) : 2805-2828. doi: 10.3934/dcdsb.2020206 [19] Thierry Cazenave, Ivan Naumkin. Local smooth solutions of the nonlinear Klein-gordon equation. Discrete & Continuous Dynamical Systems - S, 2021, 14 (5) : 1649-1672. doi: 10.3934/dcdss.2020448 [20] Scipio Cuccagna, Masaya Maeda. A survey on asymptotic stability of ground states of nonlinear Schrödinger equations II. Discrete & Continuous Dynamical Systems - S, 2021, 14 (5) : 1693-1716. doi: 10.3934/dcdss.2020450
2019 Impact Factor: 1.366 | {"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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5157592296600342, "perplexity": 5973.471969190298}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178364932.30/warc/CC-MAIN-20210302221633-20210303011633-00312.warc.gz"} |
https://www.matchfishtank.org/curriculum/math/7th-grade-math/equations-and-inequalities/lesson-2/ | # Equations and Inequalities
## Objective
Represent equations in the forms ${px+q=r}$ and ${p(x+q)=r}$ using tape diagrams.
## Common Core Standards
### Core Standards
?
• 7.EE.B.4.A — Solve word problems leading to equations of the form px + q = r and p(x + q) = r, where p, q, and r are specific rational numbers. Solve equations of these forms fluently. Compare an algebraic solution to an arithmetic solution, identifying the sequence of the operations used in each approach. For example, the perimeter of a rectangle is 54 cm. Its length is 6 cm. What is its width?
?
• 6.EE.B.7
## Criteria for Success
?
1. Represent story situations and equations using tape diagrams.
2. Write equations to represent tape diagrams and situations.
3. Write story situations for tape diagrams and equations.
4. Understand how a tape diagram can be helpful to visualize a solution pathway for an equation (MP.1).
## Tips for Teachers
?
Lessons 2 and 3 engage students in representing and solving two-step equations using tape diagrams. In this lesson, students shift between stories, diagrams, and equations, understanding how each connects to the others and helps shed light on the situation. In Lesson 3, students will use the tape diagrams to solve equations.
#### Fishtank Plus
• Problem Set
• Student Handout Editor
• Vocabulary Package
## Anchor Problems
?
### Problem 1
Which tape diagram matches the situation described below? For the other tape diagram, write a story situation that matches it.
Situation:
Five friends go to a fall festival and they each pay an entry fee of $10. The friends each spend the same amount of money on lunch at the festival and don’t spend any other money. Altogether, the friends have spent$120.
Tape Diagrams:
### Problem 2
Which equation matches the tape diagram shown below? For the other equation, draw a tape diagram to represent it.
Tape Diagram:
Equations:
Equation A: ${3x+4=45}$
Equation B: ${3(x+4)=45}$
### Problem 3
For each situation described below, draw a tape diagram and write an equation. Define any variables you use.
Situation A:
A book has 6 chapters in it, each with the same number of pages. The book also has an introduction that is 8 pages long. The whole book is 194 pages long.
Situation B:
Eight baskets have some apples in them, and the same number of apples are in each basket. Six apples are added to each basket to make a total of 144 apples.
## Problem Set
?
The following resources include problems and activities aligned to the objective of the lesson that can be used to create your own problem set.
• Given an equation in the form ${px+q=r}$ or ${p(x+q)=r}$, draw a tape diagram and write a story situation.
A family of 5 went to a matinee movie on a Saturday afternoon. The movie tickets for the matinee were a special price for each person. The family spent a combined $25 at the concession stand on drinks and popcorn. Altogether, the family spent$57.50 at the movies. | {"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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.38014039397239685, "perplexity": 2191.2121275232244}, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400198868.29/warc/CC-MAIN-20200920223634-20200921013634-00735.warc.gz"} |
https://www.buecher.de/shop/lexika-und-nachschlagewerke/advanced-experimental-methods-for-noise-research-in-nanoscale-electronic-devices-ebook-pdf/-/products_products/detail/prod_id/37347779/ | 89 °P sammeln
• Format: PDF
A discussion of recently developed experimental methods for noise research in nanoscale electronic devices, conducted by specialists in transport and stochastic phenomena in nanoscale physics. The approach described is to create methods for experimental observations of noise sources, their localization and their frequency spectrum, voltage-current and thermal dependences. Our current knowledge of measurement methods for mesoscopic devices is summarized to identify directions for future research, related to downscaling effects.
The directions for future research into fluctuation phenomena in
…mehr
Andere Kunden interessierten sich auch für
Produktbeschreibung
A discussion of recently developed experimental methods for noise research in nanoscale electronic devices, conducted by specialists in transport and stochastic phenomena in nanoscale physics. The approach described is to create methods for experimental observations of noise sources, their localization and their frequency spectrum, voltage-current and thermal dependences. Our current knowledge of measurement methods for mesoscopic devices is summarized to identify directions for future research, related to downscaling effects.
The directions for future research into fluctuation phenomena in quantum dot and quantum wire devices are specified. Nanoscale electronic devices will be the basic components for electronics of the 21st century. From this point of view the signal-to-noise ratio is a very important parameter for the device application. Since the noise is also a quality and reliability indicator, experimental methods will have a wide application in the future.
Dieser Download kann aus rechtlichen Gründen nur mit Rechnungsadresse in A, B, BG, CY, CZ, D, DK, EW, E, FIN, F, GB, GR, HR, H, IRL, I, LT, L, LR, M, NL, PL, P, R, S, SLO, SK ausgeliefert werden.
• Produktdetails
• Verlag: Springer-Verlag GmbH
• Erscheinungstermin: 21.02.2006
• Englisch
• ISBN-13: 9781402021701
• Artikelnr.: 37347779
Inhaltsangabe
Preface. I: Noise Sources. l/fNoise Sources; F.N. Hooge. Noise Sources in GaN/AlGaN Quantum Wells and Devices; S. Rumyantsev. l/fNoise in Nanomaterials and Nanostructures: Old Questions in a New Fashion; M.N. Mihaila. l/fSpectra as a Consequence of the Randomness of Variance; G. Härtler. Quantum Phase Locking, l/fNoise and Entanglement; M. Planat, H. Rosu. Shot Noise in Mesoscopic Devices and Quantum Dot Networks; P. Maccuci, et al. Super-Poissonian Noise in Nanostructures; Y. Blanter. Stochastic and Deterministic Models of Noise; J. Kumicak. II: Noise in Nanoscale Devices. Noise in Optoelectric Devices; R. Alabedra. Fluctuations of Optical and Electrical Parameters and Their Correlation of Multiple-Quantum-Well INGAAS/INP Lasers; S. Pralgauskaité, et al. Microwave Noise and Fast/Ultrafast Electronic Processes in Nitride 2DEG Channels; A. Matulionis. Noise of High Temperature Superconducting Bolometers; I.A. Khrebtov. l/fNoise in MOSTs: Faster is Noisier; L.K.J. Vandamme. Experimental Assessment of Quantum Effects in the Low-Frequency Noise and RTS of Deep Submicron MOSFETs; I. Simoen, et al. Noise and Tunneling through the 2.5 nm Gate Oxide in SOI MOSFETs; N. Lukyanchikova, et al. Low Frequency Noise Studies of Si Nano-Crystal Effects in MOS Transistors and Capacitors; S. Ferraton, et al. Noise Modelling in Low Dimensional Electronic Structures; L. Reggiani, et al. Correlation Noise Measurements and Modelling of Nanoscale MOSFETs; J. Lee, G. Bosman. Tunneling Effects and Low Frequency Noise of GaN/GaAlN HFETs; M. Levinshtein, et al. High Frequency Noise Sources Extraction in Nanometique MOSFETs; F. Danneville, etal. Informative 'Passport Data' of Surface Nano- and Microstructures; S.F. Timashev, et al. III: Noise Measurement Technique. Noise Measurement Techniques; L.K.J. Vandamme. Techniques for High-Sensitivity Measurements of Shot Noise in Nanostructures; B. Pellegrini, et al. Correlation Spectrum Analyser: Principles and Limits in Noise Measurements; G. Ferrari, M. Sampietro. Measurement and Analysis Methods for Random Telegraph Signals; Z. Çelik-Butler. RTS in Quantum Dots and MOSFETs: Set-Up with Long-Time Stability and Magnetic Fields Compensation; J. Sikula, et al. Some Considerations for the Construction of Low-Noise Amplifiers in Very Low Frequency Region; J. Sikula, et al. Measurements of Low Frequency Noise in Nano-Grained RuO2+Glass Films below 1 K; A. Kolek. Technique for Investigation of Non-Gaussian and Non-Stationary Properties of LF Noise in Nanoscale Semiconductor Devices; A. Yakimov, et al. The Noise Background Suppression of Noise Measuring Set-Ups; P. Hruska, K. Hajek. Accuracy of Noise Measurements for l/f and GR Noise; I. Slaidiņs. Radiofrequency and Microwave Noise Metrology; E. Rubioloa, V. Giordano. Treatment of Noise Data in Laplace Plane; B.M. Grafov. Measurement of Noise Parameter Set in the Low Frequency Range: Requirements and Instrumentation; L. Hasse. Techniques of Interference Reduction in Probe System for Wafer Noise Measurements of Submicron Semiconductor Devices; L. Spiralski, et al. Hooge Mobility Fluctuations in n-Insb Magnetoresistors as a Reference for Access Resistance LF-Noise Measurements of SiGe Metamorphic HMOS FETs; S. Durov, et al. Optimised Preamplifier for LF-Noise MOSFET Characterization; S. Durov, O.A. Mironov. Net of | {"extraction_info": {"found_math": false, "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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8480908274650574, "perplexity": 15734.836982842582}, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347398233.32/warc/CC-MAIN-20200528061845-20200528091845-00443.warc.gz"} |
https://nyuscholars.nyu.edu/en/publications/nonuniqueness-of-weak-solutions-to-the-sqg-equation | # Nonuniqueness of Weak Solutions to the SQG Equation
Tristan Buckmaster, Steve Shkoller, Vlad Vicol
Research output: Contribution to journalArticlepeer-review
## Abstract
We prove that weak solutions of the inviscid SQG equations are not unique, thereby answering Open Problem 11 of De Lellis and Székelyhidi in 2012. Moreover, we also show that weak solutions of the dissipative SQG equation are not unique, even if the fractional dissipation is stronger than the square root of the Laplacian. In view of the results of Marchand in 2008, we establish that for the dissipative SQG equation, weak solutions may be constructed in the same function space both via classical weak compactness arguments and via convex integration.
Original language English (US) 1809-1874 66 Communications on Pure and Applied Mathematics 72 9 https://doi.org/10.1002/cpa.21851 Published - Sep 2019
## ASJC Scopus subject areas
• Mathematics(all)
• Applied Mathematics
## Fingerprint
Dive into the research topics of 'Nonuniqueness of Weak Solutions to the SQG Equation'. Together they form a unique fingerprint. | {"extraction_info": {"found_math": false, "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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8551657199859619, "perplexity": 1385.7395799364456}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104692018.96/warc/CC-MAIN-20220707124050-20220707154050-00526.warc.gz"} |
https://math.stackexchange.com/questions/2774397/find-polynomial-over-z2-that-has-the-least-degree-for-which-the-field-p-gf64 | # Find polynomial over Z2, that has the least degree, for which the field P=GF(64) is the minimum decomposition field. How much such polynomial are?
Find polynomial over Z2, that has the least degree, for which the field P=GF(64) is the minimum decomposition field. How much such polynomials are?
Please, answer in detail! I think that these polynomials are all sixth irreducible polynomials, but I don't know how to validate this fact.
It's a general fact that $\mathbb{F}_p[x]/(q)$ is isomorphic to $\mathbb{F}_{p^n}$, where $q$ is an irreducible polynomial of degree $n$. Furthermore, $\mathbb{F}_p[x_1]$ is a splitting field for $q$ where we take $x_1$ to be a root for $q$ (we know that it contains $\mathbb{F}_{p^n}$; this says that they are the same). This is because $x, x^p, \ldots, x^{p^{n-1}}$ are all roots of $q$.
Thus we are looking for irreducible polynomials of degree $6$. We know they all must divide $x^{64}-x$, since every element of $\mathbb{F}_{64}$ satisfies that. Moreover, the roots of that are distinct (take the derivative) so we just need to factorize it and count the number with degree $6$.
The product of the irreducible polynomials with degree $d$ over all $d|n$ is $x^{2^n}-x$ (this can be deduced from the combining the fact that $\mathbb{F}_{p^m}\subset \mathbb{F}_{p^n}$ iff $m|n$ with the reasoning above).
Here, $p=2$ and $n=6$. There are obviously $2$ with degree $1$, which means there are $(2^2-2)/2 = 1$ of degree $2$, $(2^3-2)/3=2$ of degree $3$, and $(2^6-2-2-6)/6 = 9$ of degree $6$.
Finding one explicitly might be a bit annoying. It's not too hard to get that for $1, 2, 3$ they are $x, x+1, x^2+x+1, x^3+x+1, x^3+x^2+1$.
Now we want to remove some of the monomials from $x^6+x^5+x^4+x^3+x^2+x+1$ to make it irreducible, and we know we need to keep $x^6$. From degree 1 we need to remove an even number and we need to keep $1$. Ok so after some trials you will get an answer like $x^6+x^4+x^3+x+1$.
• Thank you very much for your explanations! And is it easier to sort out all irreducible polynomials of degrees 1, 2, 3, 4, 5, 6? I mean that is it right only to show that irreducible polynomials of degree 1 splits over GF(4), 2 --- splits over GF(8), ..., 5 --- splits over GF(32). And then say that if a 6th degree polynomial isn't irreducible, it contains irreducible polynomials of degrees 1, 2, 3, 4 or 5. So, all 6th degree irreducible polynomials split over GF(64). And then show their splits. – alexhak May 10 '18 at 16:00
This late answer uses computer power to find explicitly the irreductible polynomials of degree six in $\Bbb F_2[X]$.
There exists up to isomorphy only one field with $2^6$ elements, $F=\Bbb F_{64}$, its elements are fixed by the corresponding Frobenius isomorphism, so $F$ contains the roots of $$x^{2^6}-x\ .$$ sage gives the following factorization for it:
sage: R.<x> = PolynomialRing( GF(2) )
sage: for f, multiplicity in ( x^(2^6) - x ).factor():
....: print f
....:
x
x + 1
x^2 + x + 1
x^3 + x + 1
x^3 + x^2 + 1
x^6 + x + 1
x^6 + x^3 + 1
x^6 + x^4 + x^2 + x + 1
x^6 + x^4 + x^3 + x + 1
x^6 + x^5 + 1
x^6 + x^5 + x^2 + x + 1
x^6 + x^5 + x^3 + x^2 + 1
x^6 + x^5 + x^4 + x + 1
x^6 + x^5 + x^4 + x^2 + 1
How many elements in $F$ are roots of a polynomial of degree exactly $6$? There are, by inclusion/exclusion $|F_{2^6}|-|F_{2^3}|-|F_{2^2}|+|F_{2^1}|=2^6-2^3-2^2+2=64-8-4+2=54$ such elements, so we expect $54/6=9$ (edited division...) irreducible polynomials of degree $6$ as factors above.
• Thanks very much! It solves my problem – alexhak May 10 '18 at 20:22
• WA also helps. – lhf May 10 '18 at 21:31 | {"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": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.882066547870636, "perplexity": 128.90130093855095}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027314696.33/warc/CC-MAIN-20190819073232-20190819095232-00543.warc.gz"} |
http://tex.stackexchange.com/questions/133757/skipping-or-changing-one-grid-line | # Skipping or changing one grid line
I'm trying to plot normalized data in a bar chart. To better show that the data is normalized, I want the horizontal grid line at 1 to look differently, eg. dashed instead of dotted.
I guess this can be done by (1) changing the style of a single grid line (which I have no idea how to do) or by (2) manually adding the line at 1 (which I managed to do) and skipping the grid line so that the lines are not drawn over each other (which I, again, do not know how to do).
This is my MWE for the second case:
\documentclass{article}
\usepackage{tikz}
\usepackage{pgfplots}
\usepackage{pgfplotstable}
\usetikzlibrary{backgrounds}
\pgfplotsset{compat=1.8}
\begin{document}
\begin{tikzpicture}
\pgfplotstableread{
Model runtime normalized
(1) 0.17 0.11
(2) 1.60 1.00
(3) 3.17 1.98
(4) 1.72 1.08
(5) 4.01 2.51
}\data
\begin{axis}[
ybar,
ymin=0,
xtick=data,
xticklabels from table={\data}{Model},
grid style={dotted,gray},
ymajorgrids=true,
nodes near coords,
]
\addplot [draw=black,fill=gray!15] table [y=normalized,x expr=\coordindex] {\data};
\begin{scope}[on background layer]
\draw [dashed] ({rel axis cs:0,0}|-{axis cs:0,1}) -- ({rel axis cs:1,0}|-{axis cs:0,1});
\end{scope}
\end{axis}
\end{tikzpicture}
\end{document}
This results in the following:
Here, the grid line for the value 1 is still drawn, and its dots appear between the dashes of my manually added line, which looks ugly to me.
-
## 1 Answer
One way to do it is to specify all yticks except 1 and then use the extra y ticks and extra y ticks style to add the tick at y=1 and to change the grid style only for that tick.
\documentclass{article}
\usepackage{tikz}
\usepackage{pgfplots}
\usepackage{pgfplotstable}
\usetikzlibrary{backgrounds}
%\pgfplotsset{compat=1.8}
\begin{document}
\begin{tikzpicture}
\pgfplotstableread{
Model runtime normalized
(1) 0.17 0.11
(2) 1.60 1.00
(3) 3.17 1.98
(4) 1.72 1.08
(5) 4.01 2.51
}\data
\begin{axis}[
ybar,
ymin=0,
xtick=data,
ytick={0.5,1.5,2.0,2.5},
extra y ticks=1,
extra y tick style={grid=major, grid style={dashed,black}},
xticklabels from table={\data}{Model},
grid style={dotted,gray},
ymajorgrids=true,
nodes near coords,
]
\addplot [draw=black,fill=gray!15] table [y=normalized,x expr=\coordindex] {\data};
\end{axis}
\end{tikzpicture}
\end{document}
- | {"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": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7688037753105164, "perplexity": 4849.321765107652}, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464049274059.91/warc/CC-MAIN-20160524002114-00032-ip-10-185-217-139.ec2.internal.warc.gz"} |
http://stackoverflow.com/questions/20886709/c-sharp-command-line-in-code/20886791 | # C# command line in code
I'm trying to run hidden console with this command "winsat formal" in my C# code, but it is giving an error in "Process.Start()". Here is my code:
``````string command = "winsat formal";
Process process = new Process();
process.StartInfo.FileName = command;
process.Start();
``````
-
What's the error? – SpaceghostAli Jan 2 '14 at 16:01
Which error? Maybe 'unrecognized command'? – Alberto Solano Jan 2 '14 at 16:02
An unhandled exception of type 'System.ComponentModel.Win32Exception' occurred in System.dll – Saturnino Mateus Jan 2 '14 at 16:15
You need to pass the parameters as follows.
``````Process process = new Process();
process.StartInfo.FileName = "winsat" ;
//in here you add as many parameters as needed
process.StartInfo.Arguments = "formal" ;
process.Start();
``````
UPDATE 1
If you go to "cmd" command line and type "winsat formal" does it work ? If not you need to do one of these
• Add the path to the "winsat" executable to the PATH environment variable
• Or specify the folder where the executable is located like this
process.StartInfo.WorkingDirectory = "c:\your\path\toWinSat\executable\" ;
If it still does not work please let me know
UPDATE 2
//Have you tried proving a full path to the executable ?
``````process.StartInfo.FileName = "C:\windows\system32\winsat.exe"
``````
-
I already did this, don't work...But if i use another command like "calc" it work – Saturnino Mateus Jan 2 '14 at 16:20
Could be a permissions issue... do you have admin rights on the computer? Can you post the Inner Exception? – Evan L Jan 2 '14 at 16:31
Evan Lewis: dont know how to use this. But the Visual Studio open the prompt from the Debug path of the project, and when i give the command "winsat formal" says that is not recognize as internal command – Saturnino Mateus Jan 2 '14 at 17:02
Is not recognize an internal or external command, operable program or batch file – Saturnino Mateus Jan 2 '14 at 17:11
@SaturninoMateus I have updated my answer let me know if it solves the problem – Mauricio Gracia Jan 2 '14 at 18:43 | {"extraction_info": {"found_math": false, "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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8711572885513306, "perplexity": 7733.6903932543255}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-06/segments/1422122080417.25/warc/CC-MAIN-20150124175440-00052-ip-10-180-212-252.ec2.internal.warc.gz"} |
https://arxiv.org/abs/1505.03277 | ### Current browse context:
cond-mat.mes-hall
# Title:Nexus and Dirac lines in topological materials
Abstract: We consider the $Z_2$ topology of the Dirac lines, i.e., lines of band contacts, on an example of graphite. Four lines --- three with topological charge $N_1=1$ each and one with $N_1=-1$ --- merge together near the H-point and annihilate due to summation law $1+1+1-1=0$. The merging point is similar to the real-space nexus, an analog of the Dirac monopole at which the $Z_2$ strings terminate.
Comments: 7 pages, 3 figures; some minor corrections, added references Subjects: Mesoscale and Nanoscale Physics (cond-mat.mes-hall) Journal reference: New J. Phys. 17, 093019 (2015) DOI: 10.1088/1367-2630/17/9/093019 Cite as: arXiv:1505.03277 [cond-mat.mes-hall] (or arXiv:1505.03277v3 [cond-mat.mes-hall] for this version)
## Submission history
From: Tero Heikkila [view email]
[v1] Wed, 13 May 2015 08:47:35 UTC (260 KB)
[v2] Fri, 5 Jun 2015 14:09:27 UTC (593 KB)
[v3] Thu, 6 Aug 2015 07:12:38 UTC (532 KB) | {"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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2223791778087616, "perplexity": 8592.061891668143}, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627999298.86/warc/CC-MAIN-20190624084256-20190624110256-00172.warc.gz"} |
https://electronics.stackexchange.com/questions/491419/how-to-reduce-mosfet-heating-in-buck-converter | # How to reduce MOSFET heating in Buck Converter
I have designed this Buck Converter which convert 60VDC to 12VDC at 10A. Switching Freq 100KHz. Facing MOSFET too much heating issue. MOSFET ON time and OFF time is decided by a UC3845B based circuit. MOSFET gate is biased with 2.2R resistor and pull down of 5.1K is there any way to reduce MOSFET heating? I've increased MOSFET rating to 110A 80V. Previously was 75V 75A but no succcess.
Edit 1: Updated Schematic for better understanding.
Edit 2: Previously tried this INFINEON MOSFET. Heating was less.
Then used this ST MOSFET. Heating was more in ST MOSFET
Hello, here is an update, can I use below circuit as bootstrap? here in place of 5V Input either I can use 60V directly or 12V from output.
• You need to show how the MOSFET gate drive works as a circuit and note that an N channel MOSFET driven as a source follower may not be very efficient unless your circuit jumps through a few hoops to produce sufficient gate drive voltage. Apr 6 '20 at 15:19
• The MOSFET you are using does not return anything in a search; include a link to the datasheet. Apr 6 '20 at 15:22
• If the UC384x is referenced to your GND, no wonder why the MOSFET won't be properly turned on. You need to install a simple bootstrap architecture to do that or use a transformer. If you confirm you don't have any of these, I'll suggest a simple driver I have successfully tested in a separate answer. Apr 6 '20 at 15:48
• Could you use different ground symbols for port 4 and 6, and give the ports more descriptive labels please? Apr 7 '20 at 8:11
• That’s a source follower. You need a high side gate driver with either bootstrap or floating supply for it. Apr 7 '20 at 9:05
If it is confirmed the MOSFET needs a better drive then the best is to advise a suitable circuit. A simple bootstrap circuit described by Monsieur Balogh in a TI application note can do the job nicely in a cost-sensitive application. As noted in some of the comments, the UC384x was not really meant for hi-side drive - unless you make it entirely float and tie its GND pin to the MOSFET source and supply the IC via the rectified buck output - but this little circuit does the job fine:
Below is the circuit I tested with component values:
This is excerpted from a seminar I taught in an APEC seminar in 2019.
• And, what about the IC's supply? Need a pre-regulator for that. So even more parts... that cheap old IC is starting to get expensive. Not a good approach really. Apr 6 '20 at 22:45
• It's indeed not a panacea for a buck converter supplied from a 60-V source. It is however a possible solution if you wanna tinker with a buck and the only controller you have on hand is good old UC384x. Apr 7 '20 at 6:23
• @VerbalKint Sir, yes, in my circuit, same is done that rectified output is given as power source to IC Apr 7 '20 at 7:53
• Sir, VDRV pin Voltage same as VCC or VIN i.e. 60VDC will work? Apr 8 '20 at 12:58
• Probably if you add also a resistance between the gate and the source of $Q_7$ to force the Zener conduction with a few mA. May 11 '20 at 6:21
MOSFET ON time and OFF time is decided by a UC3845B based circuit
The UC3845B has a push-pull type output but, unless this chip is powered from a supply that is several volts higher than the 60 volt rail (as shown in your circuit) you won't efficiently drive the MOSFET in your buck regulator. Given that the UC3845 is only rated at a maximum of 36 volts, you are likely driving the MOSFET very, very ineffectively and it will get very warm on load.
The gate voltage needs to exceed the main supply voltage by around 10 volts in order to get source and drain ohmically connected at a low value. This is a problem with source follower MOSFET configurations and they way around it is to use a proper "high-side MOSFET driver" chip.
• Sir, the max voltage reached to 72 V as it is an EV. 3845 has an internal zener of 36V. What should be the Gate triggering voltage? Apr 7 '20 at 7:46
• You now have two circuits drawn with different packages and both have a ground connection that appears to be at contradictory places. If you want help with this you are going to have to explain why you don't have one consistent circuit (can lead to errors also) and, if the two circuits you now show are what you have always had, what waveforms you are getting. @HrishikeshDixit Apr 7 '20 at 9:33
I’m guessing the N-FET isn’t being turned on all the way. It's being biased in linear mode and so has a substantial IR drop across it, which is being shed as heat. You don't want that.
How bad is it? Let's assume the FET Vgs threshold is about 4.5V. Then when it's on:
• Vd = 60V
• Vg = 36V - 2V = 34V (limited UC3845B output swing)
• Vs = (Vg - Vth) = (34 - 4.5V) = 29.5V
• Vds = (Vd - Vs) = (60 - 29.5V) = 30.5V
If you're drawing, say, 1A from the supply, that approximately 30-35W you're dissipating in the FET, peak.
What keeps it from frying immediately is the stepping ratio that determines the on-time of the FET:
• Vout/Vin * W = 12/60 * 35W = 7W
And that's at only 1A. Clearly, this isn't workable.
To fix it, the N-FET gate needs to be brought all the way above 60V, to at least 65V to 70V, to ensure the FET is fully turned on to its lowest Rds(on).
What happens when you do that? Here's the peak wattage shed in the FET:
• Vd = 60V, Vg = 65V
• Vs = about 60V (transistor is fully on)
• Vds = (Vd - Vs) = 0, or close to it
So in theory almost no power gets shed in the FET. In reality, this will be:
• Iout^2 * Rds(on)
• 10^2 * 0.020 ohm = 2W peak at 10A output
With stepping ratio being 12/60, the FET is on about 20% of the time:
• 2W * 12/60% = 0.4W
That's very manageable for a TO-220 FET.
How to do this? You need a bootstrap circuit to generate the higher gate drive voltage (about 5-10V above Vin), and a gate driver that accepts that voltage to make the above-Vin gate signal. The bootstrap voltage can (and usually is) generated from the inductor flyback via a diode and capacitor.
Problem is, the UC3845B is not a bootstrapped high-side driver. It's really designed to be a low-side driver for a flyback topology. Further, it's limited to +36V. For both reasons it’s a poor choice for this application.
You could mess about with making a bootstrap + level shifter, but why? Select a different device instead. Example: this 75V input dcdc controller from TI (bonus: it’s synchronous so your supply will be more efficient): http://www.ti.com/lit/ds/snvsai4/snvsai4.pdf
• what I understood is right now the gate triggering voltage is 4.5V which is output from 3845B. To reduce heat, I need to increase gate Trigger voltage to 60V or more. Is that right? Request you to please correct me if I'm wrong as I'm new to this. Thanks for your kind help. Apr 7 '20 at 14:55
• The N-FET gate needs to be at least 4.5V above the source, and preferably as much as 10V to ensure the lowest Rds(on) for the FET. This means you need to make a drive voltage of 64.5 to 70V (4.5 to 10V above Vin.) Apr 7 '20 at 15:22
• The ST MOSFET which I'm using has gate voltage of +/- 20V whereas of INFINEON is +/- 10V. Is it okay to give MOSFET gate voltage more than specified in datasheet? Need your guidance. Thanks Apr 7 '20 at 17:26
• In a word, NO. You cannot exceed the maximum Vgs, otherwise the FET will break down. The bootstrap technique references Vin, so you can limit the gate voltage with a zener diode or other method. The devices with internal bootstrap generation do this for you. Apr 7 '20 at 17:29
• Vd = 60V Vg = 36V - 2V = 34V (limited UC3845B output swing) Vs = (Vg - Vth) = (34 - 4.5V) = 29.5V Vds = (Vd - Vs) = (60 - 29.5V) = 30.5V Sir, apologies in advance to trouble you but, 3845 datasheet says that output pin will be at 13.5v, which means Gate of MOSFET will get 13.5 V. Can you please tell the above calculation Vg = 36V-2V = 34V? Apr 8 '20 at 12:12
Pls reduce the resistance in Vin to maintain 15V across the IC UC38xx . The MOSFET minimum required gate drive voltage is 10v other wise it will be in ohmic region and generate heat as well as reduce power due to low contuctance
You can check if a heat sink is needed by calculating (at least theoretically), by checking the numbers in the datasheet.
Normally these are called Tj (Junction temperature) and all related heat dissipation figures are shown together.
Designed output power = 12V x 10A = 120W
High efficiency say 95%, heat dissipate = 120W x 5% = 6W
You have to use a large heat sink with fan.
MOSFET heating was mainly due to Inductor. Inductor Flyback or Back EMF was causing problem. As said in Answer by @Verbal Kint sir, the ground of 3845B was completely floating as it was tied to source of MOSFET. Previously used PQ 32X20 core inductor with 0.4mm wire (don't know turns). Temperature was upto 100 degree C. Maybe logically more turns could cause more Back EMF or Flyback. Now used a FeSiAl core with 1 mm wire gauge and 14 no of turns. Noe, temperature is upto 60 degree C. Maybe using a bootstrap circuit or a built in IC will do more effect. But PCB was designed and needed to solve the issue badly. Thanks everyone here for help. Specially Verbal kint sir and hacktastical sir. | {"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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3691019117832184, "perplexity": 3631.9223688689576}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320300997.67/warc/CC-MAIN-20220118182855-20220118212855-00057.warc.gz"} |
https://www.givewell.org/charities/gui2de/january-2017-grant | # Georgetown University Initiative on Innovation, Development, and Evaluation — Zusha! Road Safety Campaign
Published: June 2017
Note: This page summarizes the rationale behind a grant to the Georgetown University Initiative on Innovation, Development, and Evaluation (gui2de) made by Good Ventures. gui2de staff reviewed this page prior to publication.
As part of GiveWell's Incubation Grants to support the development of potential future top charities and improve the quality of our recommendations, in February of 2017 Good Ventures granted $900,000 to the Georgetown University Initiative on Innovation, Development, and Evaluation (gui2de) for its Zusha! Road Safety Campaign. ## The organization The Zusha! Road Safety Campaign, managed by gui2de, targets unsafe drivers of public service vehicles. The campaign distributes stickers for buses with messages encouraging passengers to speak up and urge drivers to drive more safely. The program also includes a lottery that bus drivers with stickers displayed in their buses can win, to incentivize bus drivers to keep the stickers. The program’s goal is to reduce traffic accident deaths (which are more common in many developing countries than in most developed countries1). There have been two randomized controlled trials (RCTs) conducted on Zusha!’s program in Kenya (where it has primarily operated so far) – see details below. gui2de has received$3 million in funding from Development Innovation Ventures to conduct three more RCTs in Tanzania, Rwanda, and Uganda.2
Note on terminology: In this writeup, we generally refer to the program we're interested in as Zusha! to distinguish it from gui2de's other programs (not related to road safety). However, Zusha! is only the name of the road safety campaign in Kenya; the road safety campaigns in Rwanda, Tanzania, and Uganda have other names.
## Evidence of effectiveness
Two RCTs have been conducted on Zusha!’s program in Kenya: a small pilot study (which had a treatment group of roughly 1,000 vehicles)3 and a second, much larger study (involving roughly 12,000 vehicles, including treatment, control, and placebo groups).4 The studies' primary outcome was total number of claims made by insurance companies for traffic accidents.
The first study found a roughly 50% reduction in accidents due to the intervention;5 the second found a roughly 25% reduction.6 Both studies found the effects of the program to be statistically significant.7 The researchers extrapolate from these figures and estimates of accidents involving deaths to estimate the number of deaths averted by the intervention.8
Compliance with the intervention (i.e. buses retaining the stickers) appeared to decrease over time in the second study.9 (A time series analysis was not done for the first study.) The second study included a placebo sticker group and found a nearly statistically significant effect in the placebo group.10
### Cost-effectiveness
One of our Research Analysts, Leon Zhang, examined the RCTs' cost-effectiveness analyses (CEAs) and found mathematical errors which, when corrected for, increased the published estimates of cost per life saved. (See our summary for details.) When we reached out to the papers’ authors to check our understanding, they immediately and graciously acknowledged the errors and prepared corrections. (For the authors' published corrections to the 2011 study, see Habyarimana and Jack 2016.)
Even taking these adjustments into account, the results of the two completed RCTs suggest that Zusha! may be roughly 3 to 4 times as cost-effective as cash transfers, which is roughly comparably cost-effective to some of our other top charities, such as the Against Malaria Foundation (AMF).11 However, we believe that our CEA could change substantially, either positively or negatively, based on further analysis and the results of the three upcoming RCTs.
Note that the intervention’s main impact comes from saving adult lives (whereas much of the impact of, e.g., distributing insecticide-treated bednets comes from saving the lives of children under five years old). We’ve used GiveWell staff's median ethical values from our November 2016 top charities CEA to estimate the relative value of saving adult lives.12 (For a more in-depth discussion of possible approaches to this type of valuation, see this December 2016 blog post.)
## Grant details
This grant is intended to allow Zusha! to:
• Continue to operate at scale in Kenya for an additional six months to a year. Without this grant, our understanding is that Zusha! would be close to running out of funding for its operations in Kenya by September 2017.
• Improve monitoring of its program in Kenya (see below).
• Increase the sample size of its RCT in Uganda from about 4,000 to about 6,000 buses, improving the statistical power of the study.
• Collect higher-quality data in Tanzania through September 2017, which Zusha! has told us could improve precision of measurement in that RCT.
• Continue operating in Tanzania, Uganda, and Rwanda for an additional six months, so that it is able to scale up more quickly if those countries' RCT results are positive.
### Improving monitoring
We hope that this grant will enable Zusha! to improve its monitoring in Kenya to accomplish two main goals:
1. Produce a robust measure of how many buses in the country are in fact using the stickers (Zusha! tries to reach every bus, but currently does not have a highly reliable system for monitoring whether stickers are used in the buses).
2. Track the long-term impact of the intervention, which could potentially be importantly different from the short-term impact (for example, it seems possible that people could adjust to the presence of the stickers, such that their effectiveness decreases over time).
## Budget
Our grant breaks down roughly as follows:13
• $352,130 to fund monitoring and an additional six months of operations in Kenya. •$189,000 to increase the sample size of the RCT in Uganda by about 50%.
• $97,500 for an additional six months of operations in Tanzania, Uganda, and Rwanda. •$80,000 to improve the quality of data collection for the Tanzania RCT.
• $99,552 for miscellaneous administrative costs for the additional months of operations that our grant is funding, including staff salaries and travel costs. •$81,818 to cover 10% overhead to Georgetown University.
## Risks of the grant
We see a number of potential risks to the success of this grant:
• Negative RCT results: It may be that the new RCTs of this program either a) find that it is not effective, or b) are inconclusive due to lack of statistical power or methodological issues.
• Changes in evidence quality or estimated cost-effectiveness: We have not yet deeply vetted the two RCTs on Zusha!’s program in Kenya, and we have not analyzed our CEA as deeply as our top charities’ programs’ CEAs. Further analysis may lead to negative updates on evidence quality and/or estimated cost-effectiveness.
• Monitoring quality: As discussed above, we have not yet seen high-quality ongoing monitoring of Zusha!’s program in Kenya. It may be that Zusha! is unable to meet our criteria for becoming a top charity due to operational issues and/or lack of ongoing monitoring.
## Follow-up expectations
Our goal is to have a charity review for Zusha! published by end of November 2017. We plan to check in with gui2de during the second half of 2017 about two main questions:
• What are the results of the new RCTs, and how does this affect the quality of evidence for the intervention and our CEA?
• Is Zusha! collecting high-quality monitoring data to achieve the two monitoring goals described above?
By around September 2017, we expect to see:
• Several months of 2017 monitoring data from Kenya.
• Full RCT results and data from Tanzania. gui2de thinks it is fairly likely it will also have a working paper by this point.
• Results from half of the Uganda RCT.
By February or March of 2018, we expect to see full results from the Uganda RCT.
We are not sure when results from the Rwanda RCT will be available; this RCT has been delayed indefinitely due to challenges getting government approval to run the program and RCT.
## Internal forecasts
We are experimenting with recording explicit numerical forecasts of the probability of events related to our decision-making (especially grant-making). The idea behind this is to pull out the implicit predictions that are playing a role in our decisions, and to make it possible for us to look back on how well-calibrated and accurate those predictions were. For this grant, Josh Rosenberg, our senior research analyst who led GiveWell's investigation of Zusha!, records the following forecasts:
• Zusha! is recommended as a top charity by year-end 2017: 35%, broken down into:
• Zusha! appears more cost-effective than AMF: 10%
• Zusha! appears roughly as cost-effective as AMF: 15%
• Zusha! appears less cost-effective than AMF (but is still a top charity recommendation): 10%
## Our process
We had four conversations with James Habyarimana, Billy Jack, and Whitney Tate of gui2de over the course of the past year. We published notes from conversations we had in March 2016 and July 2016. We put significant time into producing our CEA of the Zusha! program (including Leon Zhang finding errors in the original studies’ CEAs). Recently, we have mainly focused on determining Zusha!'s room for more funding and our timeline for evaluating it as a potential top charity.
Subsequent to recommending this grant, we have published additional content on Zusha!:
## Sources
Document Source
Georgetown University 2014, "East Africa Road Safety Project Gets $3 Million Grant" Source (archive) GiveWell, Zusha! CEA Source Habyarimana and Jack 2010, working paper Source (archive) Habyarimana and Jack 2011 Source (archive) Habyarimana and Jack 2015 Source (archive) Habyarimana and Jack 2016 Source (archive) World Health Organization, Road Safety Interactive Map Source Zusha!, budget Source (archive) Zusha!, Power calculations - Tanzania Source Zusha!, Rwanda Study Protocol (Draft, preliminary) Source Zusha!, Tanzania Study Protocol (Draft, preliminary) Source Zusha!, Transition summary and narrative Source Zusha!, Uganda Study Protocol (Draft, preliminary) Source • 1. See, for example, World Health Organization, Road Safety Interactive Map. • 2. "A$3 million grant…from USAID’s Development Innovation Ventures (DIV) supports the third phase of the Zusha! study in Kenya, and expands the project to three other countries – Tanzania, Uganda and Rwanda." Georgetown University 2014, "East Africa Road Safety Project Gets \$3 Million Grant"
• 3.
"Evocative messages encouraging passengers to speak up were placed inside a random sample of over 1,000 long-distance Kenyan minibuses, or matatus" Habyarimana and Jack 2010, working paper, pg. 4. (Note: This source is a working version of Habyarimana and Jack 2011.)
• 4.
• Pure control: 2,093
• Placebo: 1,759
• Treatment stickers: 7,885
• Total: 11,737
Habyarimana and Jack 2015, pg. 4662, Table 1.
• 5.
"Our estimates consistently suggest that the intervention reduced the number of incidents leading to an insurance claim by about a half." Habyarimana and Jack 2010, working paper, pg. 28. (Note: This source is a working version of Habyarimana and Jack 2011.)
• 6.
"Among the roughly 8,000 vehicles in the treatment groups, the reduction [in insurance claims] was 25%." Habyarimana and Jack 2015, pg. 4661.
• 7.
• 8.
• "In our data, 11 percent of claims involved at least one death, although we do not know the actual number of deaths associated with each such accident. In our baseline case, the projected claims rate in the treatment group is about 10 percent, which the treatment reduces by five percentage points. Assuming the same rate of reduction in accidents involving a death as in accidents involving injuries or death (we are better able to estimate the impact on the latter), the intervention thus reduced the number of accidents including a death by about 6.0 per year per thousand vehicles. Conservatively we assume an average of two fatalities per accident including a death." Habyarimana and Jack 2010, working paper, pg. 20. (Note: This source is a working version of Habyarimana and Jack 2011.)
• "Among the roughly 8,000 vehicles in the treatment groups, about 140 accidents were avoided over the course of a year…Associated with the accidents that were avoided, we estimate 55 lives were saved." Habyarimana and Jack 2015, pg. 4668.
• 9.
"Compliance with assignment to sticker groups, as measured by the number of [lottery] winners out of 10 drawn, was initially very high, but fell over the first 6 [months] of the intervention." Habyarimana and Jack 2015, pg. 4664.
• 10.
• "Although we do not detect a placebo effect directly, we cannot reject equality of the treatment and placebo effects…even the low dosage of the placebo may have been partially effective." Pg. 4666.
• See Table 4, column "Claims", row "Placebo," pg. 4666.
• 11.
GiveWell, Zusha! CEA, "CEA" sheet, cell B37.
• 12. GiveWell, Zusha! CEA, "CEA" sheet, cell B26.
• 13. See Zusha!, budget. | {"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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.24914461374282837, "perplexity": 4818.712136342808}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934806736.55/warc/CC-MAIN-20171123050243-20171123070243-00161.warc.gz"} |
http://linux4chemistry.info/posts/gnu-xtal-system.html | # Gnu Xtal System
| Source
The Gnu Xtal System is a package of over sixty crystallographic programs for calculations ranging from the reduction of raw diffraction intensities, to the solution, refinement and publication of crystal structures. These are applicable to X-ray, neutron and electron diffraction analyses, including charge density studies. The package contains interactive graphics tools. | {"extraction_info": {"found_math": false, "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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.867706298828125, "perplexity": 7117.089258777331}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891814827.46/warc/CC-MAIN-20180223174348-20180223194348-00485.warc.gz"} |
http://www.gamedev.net/page/books/index.html/_/technical/general-programming/design-patterns-elements-of-reusable-object-oriented-software-r18?st=0 | • Create Account
### Share:
This book will give you the experience you need to effectively discuss OO design with others using standard terminology, and give you ideas of design patterns that you may not have thought of. Read through it once, then keep it handy for whenever you're deciding how to approach a particular software design problem.
This book has been recommended to me by every professional software engineer. Chris Hargrove of 3DRealms has called it one of the 'essentials' in any programmer's bookshelf. Design patterns is based on the idea that there are only so many design problems in computer programming. This book identifies some common program-design problems--such as adapting the interface of one object to that of another object or notifying an object of a change in another object's state--and explains the best ways (not always the obvious ways) that the authors know to solve them. The idea is that you can use the authors' sophisticated design ideas to solve problems that you often waste time solving over and over again in your own programming.
This book is the best I have read when it comes to design..
I really loved the way the presents a common problem and a way to solve it.
A defietive worth to have book !
While by no means a book for beginners is essential when you start writing anything but the smallest OO programs. This is very much a reference book but the introductory chapters provide a good background for those starting to think more seriously about design.
The patterns themselves while by no means exhaustive are essential knowledge for any intermediate OO programmer.
Design Patterns are a relatively new development in software engineering, meant to fit into the educational void of "how do objects interact?". This book is largely regarded as the work which thrust the idea into the mainstream.
The book itself begins by describing Design Patterns; what they try to solve, and the motivation behind them. It them follows with chapters which focus on a particular pattern, which are grouped by purpose. The chapters themselves discuss the pattern in question, complete with example, relational diagram, advantages, and disadvantages.
All in all the book is well written, easily understood, and highly recommended.
I read this book at work. I learned a lot from the ideas described and of course the patterns. Sometimes the writing style and the examples can be quite boring. Still an essential read.
Useless. All you need to know is the concept of function, input/output, and that software independence is a good thing and you have the same knowlege as if you read this entire book minus the buzzwords.
PARTNERS | {"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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.21811500191688538, "perplexity": 803.5237326546027}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-49/segments/1416931007720.74/warc/CC-MAIN-20141125155647-00099-ip-10-235-23-156.ec2.internal.warc.gz"} |
https://datascience.stackexchange.com/questions/44406/word-embeddings-and-punctuation-symbols | # Word embeddings and punctuation symbols
I have a decent understanding of word embeddings (at its core, one can think of a word being converted into a vector of, say, 100 dimensions, and each dimension given a particular value... this allows to do math with the words, also it makes the training sets to be non-sparse...)
But today something came to my mind, what about punctuation symbols such as , . () ? ! ... ?
They do have a huge impact on the meaning of sentences and, like words, the position and context in which they are used is relevant.
So the question is, how should this be modeled? are pretrained sets like GloVe including punctuation symbols? should I simply remove punctuation symbols from text? | {"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": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8586459755897522, "perplexity": 1091.1647428157005}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320300573.3/warc/CC-MAIN-20220129062503-20220129092503-00648.warc.gz"} |
http://motls.blogspot.com/2014/06/has-extinction-rate-increased-1000-times.html | ## Saturday, June 14, 2014 ... /////
### Has the extinction rate increased 1,000 times?
Lots of journalists happily spread the "gospel" about a recent paper in the Science Magazine,
The biodiversity of species and their rates of extinction, distribution, and protection
by S.L. Pimm and 8 co-authors (U.S., U.K., Brazil). The abstract suggests it is a rather careful, conservative paper with some interesting statistics. The summaries in the media are not so careful, I think.
The eye-catching figure is that it's being estimated – and as far as I see, the paper assumes it is essentially right – that the "number of species that go extinct per year" has increased by three orders of magnitude. That's huge and I would surely count myself as someone who cares about the biodiversity problem if the truth were close to this number.
I have no doubts that people have exterminated numerous species – usually by clearly hostile tools such as guns, not so much by some esoteric, hypothetical, and convoluted methods such as carbon dioxide emissions which are OK for everyone – but are the numbers so bad?
However, I tried to figure out how this ratio "1,000" is being calculated and I have largely failed to find an argument. Is there a TRF reader who understands these issues and believes that there is some good enough evidence that the ratio may be this high?
Without the humans, one species typically lives for a few million years. This fact seems to be roughly hold for very diverse groups of organisms. If the "factor of 1,000 spedup" is real, an average species is destined to live for thousands of years only right now.
Are these numbers realistic? Can we estimate the number of species before humans emerged, their extinction rate, the number of species today, and their extinction rate? Or can we calculate the ratio without knowing the absolute numbers in the numerator and the denominator with any robust precision?
I think that because of the very basic rules of evolution, the very definition of a species may be fuzzy enough so that these questions don't admit any high-precision answers. In some types of organisms, the inter-species separation may be more significant than in others.
Over 100 years, the volume corresponding to the separation may very well grow by 10% in some measure which should mean that we should increase the number of species in this group by 10 percent, too. If it weren't politically incorrect, biologists might be forced to admit that the human population has split into several species, too. But I think that the conservationists are not really taking the evolution in their lifetimes into account. Along with their creationist friends, they still consider species to be fixed, unevolving boxes that were defined millions of years ago.
But the species are constantly and gradually being created by differentiation and this process has always competed against extinction. I don't see any discussion of this which is a part of the reason why I find it hard to accept these attempts to quantify the extinction rate as a good science.
Like the paper, this video shows where conservationist efforts may be most effective for biodiversity.
Moreover, as the authors of the paper are very well aware, the ranges of the species differ by many orders of magnitude in between the species, too. This correlation increases the risk of extinction for species with small ranges, and so on. So a particular definition of the extinction rate may very well be determined by some insanely small populations and whether or not we count them as separate species.
Also, I have a problem with the widespread assumption of the "universal background extinction rate". It seems likely to me that in various eras, the extinction rates were significantly different – even in the absence of meteoroids. Like earthquakes, extinctions are probably correlated with each other so they like to be grouped, much like earthquakes with smaller tremors that are linked to them. The recent centuries may experience a much higher extinction rate than the quietest centuries before the humans arrived but they may still be comparable to some rather normal centuries in which the extinction rate was higher – even though there were no meteoroid-like catastrophes.
What do you think?
Off-topic: Aquababes is what you get when Universal Music (with some visible help from the bottled water maker Aquila) organizes a casting and constructs a completely artificial girly band. They're cute and hot but the music ("Don't Call Me Baby") is sort of mediocre, isn't it?
#### snail feedback (28) :
reader bodyguard agency in london said...
Let me say that I agree that it *is* painful, if Cheney has voted like that, and it certainly makes his ticket with Bush less appealing for people like me. On the other hand, it's not the only fact that will decide the elections.
Many, probably, most of the numbers given for extinction are probably made up on the spur of the moment, like Wilson's absurd fantasy (which he won't recant, he merely doubles down Paul Ehrlich-like).
However, there are some real efforts. For example, geological time-scale estimates by Robert May (a real scientist) and his colleagues
http://www.britannica.com/EBchecked/topic/133385/conservation/272660/Calculating-background-extinction-rates#ref959267
Also, from the historical record,
http://www.birdlife.org/datazone/sowb/casestudy/102
which gives an estimate of 151 bird species lost in 500 years. This is consistent with May's estimates above.
Thanks! Imagine that one believes that 150 out of 10,000 bird species died in 500 years.
What's the evidence that this hasn't been the normal refreshment in the past, i.e. that with this accuracy of defining species, most of the species are refreshed in 30,000 years?
Lubos - This is the same Stuart Pimm who, about twelve years ago, reviewed Lomborg's book 'The Skeptical Environmentalist' in Nature. In that review, which was the most egregiously tendentious that I have ever read, he literally called Lomborg a nazi.
That review prompted me to review all of his work to that point and much of the work appearing in Nature which he referenced. I allowed my subscription to Nature to lapse shortly thereafter.
Pimm focused mainly on extinctions and based his work on island biogeography. I can assure you that both island biogeography and Pimm's work are utterly worthless in every respect, not science but pure propaganda. I still have the work work of several years to prove it.
Thanks for reminding me etc. - I forgot the name.
It's a damn shame that anyone remembers his name, and that he is published in Science.
He is much like Paul Ehrlich - both remind me of something I occasionally scrape off my shoe
Could Mr. Pimm kindly provide us with a list of 500 species which went extinct in 2013? He may not have data that recent but I would take any year in this century.
reader bodyguard agency in london said...
Rеmаrkаble! Its genuinely
rеmaгkable
pіece of writing, I hаve got much cleaг
idеa гegаrding fгom thiѕ aгticlе.
It looks like he is ignoring that well over 10 000 new species are being uncovered every year now (and that is just the eucaryotes). A 10*3 magnitude increase in extinctions may coincide with a 10*3 increase in the discovered new species, so not be a "real" increase. Craig Venter has been combing the seas in his sailboat collecting scads of new oceanic biota to get DNA sequences for novel proteins.
Stuart Pimm is a known alarmist in the Paul Ehrlich tradition.
Yes, exactly---I wrote my reply without seeing yours :).
"If it weren't politically incorrect, biologists might be forced to admit that the human population has split into several species, too." Seriously, Lubos?
Imagine you are a 19th century biologist. You describe a species of a yellow butterfly from Asia. At the same time another biologist describes an otherwise similar red butterfly species in America. Would a difference in color, together with a geographic separation, be enough to keep them listed as two species even if they could interbreed?
This is not an idle speculation. Wikipedia: "The phylogeny of the Nymphalidae is complex. Several taxa are of unclear position, reflecting the fact that some subfamilies were formerly well-recognized as distinct families due to insufficient study."
It gets "worse" than that:
6 continental birds have gone extinct — 3 prolific terrestrial bird species hunted to extinction, and 3 single-habitat freshwater bird species hunted, drained dry, eaten by fish, and polluted to extinction. This historical record of 9 continental extinctions in 500 years contrasts starkly with Wilson’s predictions of over thirty continental bird and mammal extinctions per year, each and every year.”
http://wattsupwiththat.com/2010/01/04/where-are-the-corpses/
So the 151 birds translates into 6 birds on continents. The rest mostly from my home country Australia plus other islands. Please note in the above Birdlife International ref the words. "believed"; "suspected"; "appears" etc. Lack of falsifiability is the life blood of Environmentalism.
So when we find a new species is that a net positive?????
just wondering if they can count the number of new species that appear.
If they are right, we should see it soon. ;-) If you increase the rate of extinction 0.3 of bird species per year (150 in 500 years) thousand times, you will get 300 species per year. So given the 10,000 bird species out there, in ~30 years we should have remaining just turkey and chicken.(The turkey being Stuart Pimm, and chicken being Henny Penny AKA Chicken Little,)
Here is an interesting background article from the NYT:
http://www.nytimes.com/gwire/2011/05/18/18greenwire-scientists-clash-on-claims-over-extinction-ove-96307.html?pagewanted=all
I especially noticed this sentence: "No ecologist disagrees that humanity has started, or will soon start, the Earth's sixth great wave of extinction, a process largely driven by destruction of natural habitats."
Good News!
Humans will be able within the next couple of decades to bring back almost
every species that has been extinct. De-extinction is not science fiction. It
is science fact, species have already been brought back to life after all their
members have become extinct. Here is the 12 minute TED talk about the case. With resurrection biology being just a decade
or two down the road, there is really no need to spend all those millions to
save some beasts. With a few tissue samples, humans could bring back the famed
polar bear, not only that, but it could be genetically tailored to have
different color fur. Every environmentalists would be able to have a pink cuddly ¨bonsai¨of a bear in his own backyard
or as many as he or she wants. The animals fur could also be fluorescent so the
environmentalist can use the bear as a portable flash light and reduce his carbon
How sad. All the puppies and pussies are going to die (sob, sob).
Oh look! A pretty pink bow and some sparkly shoes.
Dear Mikael, are you denying that evolution is continuing and new species are splitting from any existing one?
I didn't mean just races. I meant real de facto barriers of reproduction, between most of the humans and some isolated tribes, for example.
Is it the sixth wave of extinctions, or the sixtieth wave of extinctions, or the six hundredth wave of extinctions?
How big will the number of extinct species be when the next glacial episode occurs?
I sure hope that another glacial episode doesn't occur.
Can we raise the global temperatures to prevent glaciation in the future?
If you leave out island and Australia we have lost just two spices in the continents in the last few hundred year, that the number I have read. Along with the number the total of spices lost that we know of is less than 800, most are due ti invasive spice taking up their habitat or directly kill them rat have devastated island birds since a majority nested on the ground. Now if you count subspecies the number would be larger how much I do not know.
It's a hard question because we don't really know what's out there. For example, when researchers looked at 19 trees in Panama in the 1980s, they discovered about 1000 new species of beetle.
You could look into the 'red lists' published by the IUCN, which are available online at https://portals.iucn.org/library/dir/publications-list
The abstracts only list the number of species threatened with extinction and not the number of species actually having gone extinct, but the numbers for Europe (search term 'european red list') are 25% of amphibians, 9% of butterflies, 15% of dragonflies, 37% of freshwater fishes, 44% of freshwater molluscs and 20% of sampled terrestial molluscs, 20% of reptiles, 11% of sampledsaproxylic beetles, 25% of sampled vascular plants.
Dear Lubos,
the question is not so much about the current situation but about the trend. I think mankind is getting more uniform both genetically and culturally and not less uniform. It is called globalization.
I don't believe it.
Globalization that I see is about the global trade, ability to sell goods and services very far. I don't believe that the people's genes are getting more uniform, quite on the contrary.
I have just finished reading a new book by Nicholas Wade called, “A Troublesome Inheritance: Genes, Race and Human history.” This author believes that we can see recent evolutionary changes in humans. I enjoyed reading the book.
I recently read (cannot now find) there are an increasing
number of DNA variants and not enough time has passed to weed out the useful
from the damaging ones. Humans should,therefore, continue diversing.
Here is the idea (but not the one I recently saw):
http://articles.latimes.com/2012/dec/10/science/la-sci-human-genetic-variation-20121210 | {"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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.38825178146362305, "perplexity": 2432.86726759955}, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-22/segments/1495463608726.4/warc/CC-MAIN-20170527001952-20170527021952-00453.warc.gz"} |
https://www.physicsforums.com/threads/proving-a-form-z-f-r-to-be-a-surface-of-revolution.885643/ | # Proving a form $z=f(r)$ to be a surface of revolution
Tags:
1. Sep 17, 2016
### toforfiltum
1. The problem statement, all variables and given/known data
Suppose that a surface has an equation in cylindrical coordinates of the form $z=f(r)$. Explain why it must be a surface of revolution.
2. Relevant equations
3. The attempt at a solution
I consider $z=f(r)$ in terms of spherical coordinates.
$p cosφ = f \sqrt{(p sinφcosθ)^2 + (p sinφsinθ)^2}$
$p cosφ= f\sqrt{(p sinφ)^2}$
$p cosφ=f(p sinφ)$
$cosφ= f (sinφ)$
$∴φ= \cos^{-1} f(sinφ)$
Since equation is independent of $\theta$, it describes a surface of revolution about the $z$ axis.
Is my prove right or acceptable?
2. Sep 17, 2016
### Ray Vickson
You are using spherical, not cylindreical coordinates. Also: in LaTeX, put a "\" before sin, cos, etc. Without it, the results are ugly and hard to read, like $sin \phi cos \theta$; with it, they look good, as in $\sin \phi \cos \theta$.
3. Sep 17, 2016
### toforfiltum
Okay! Thanks for the advice! Just started using it, so sorry for the ugly text. I don't have any idea of starting the proof using cylindrical coordinates, that's why I converted it to spherical ones. But I will give it a try now.
The cylindrical coordinates are in the form $(r,\theta, z)$. I assume that $r \geq 0$
So $r= \sqrt{(x^2 + y^2)}$
and $z=f(\sqrt{(x^2 + y^2)})$, which gives a unique value. These gives a set of points that form a line in $3D$ space.
Since equation is independent of $\theta$, line is the same for any value of $\theta$. These similar set of lines form a surface of revolution.
Is it right in any way at all? I'm guessing here.
4. Sep 17, 2016
### LCKurtz
Why introduce spherical coordinates, which have nothing to do with this question?
If you have a surface of revolution revolved about the $z$ axis, that would mean $z(r,\theta_1) = z(r,\theta_2)$ for any $\theta_1$ and $\theta_2$ wouldn't it? Is that true in your case?
5. Sep 17, 2016
### toforfiltum
Yes. So, is what I'm saying above right? But I'm not sure if having a unique set of points will form a line, though.
It depends on $f$, right?
6. Sep 17, 2016
### toforfiltum
Oh, I think I see now why $z=f(r)$ represents a surface of revolution. Using the explanation on $zr$ planes given by @LCKurtz , each value of $r$ gives a value of $z$, and the set of values of $r$ gives its respective values of $z$. Since equation is independent of $\theta$, these set of points are the same for all values of $\theta$, and this is the reason why it forms a surface of revolution.
Am I right?
Draft saved Draft deleted
Similar Discussions: Proving a form $z=f(r)$ to be a surface of revolution | {"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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9578809142112732, "perplexity": 375.35383973301447}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948517350.12/warc/CC-MAIN-20171212153808-20171212173808-00238.warc.gz"} |
http://en.wikipedia.org/wiki/Monetary_inflation | # Monetary inflation
Monetary inflation is a sustained increase in the money supply of a country. It usually results in price inflation, which is a rise in the general level of prices of goods and services. Originally the term "inflation" was used to refer only to monetary inflation, whereas in present usage it usually refers to price inflation.[1]
There is general agreement among economists that there is a causal relationship between the supply and demand of money, and prices of goods and services measured in monetary terms, but there is no overall agreement about the exact mechanism and relationship between price inflation and monetary inflation. The system is complex and there is a great deal of argument on the issues involved, such as how to measure the monetary base, or how much factors like the velocity of money affect the relationship, and what the best monetary policy is. However, there is a general consensus on the importance and responsibility of central banks and monetary authorities in affecting inflation. Keynesian economists favor monetary policies that attempt to even out the ups and downs of the business cycle. Currently, most central banks follow such a rule, adjusting monetary policy in response to unemployment and inflation (see Taylor rule). Followers of the monetarist school advocate either inflation targeting or a constant growth rate of money supply, while some followers of Austrian School economics advocate either the return to free markets in money, called free banking, or a 100 percent gold standard and the abolition of central banks.[2][3]
## Quantity theory
The monetarist explanation of inflation operates through the Quantity Theory of Money, $MV = PT$ where M is Money Supply, V is Velocity of Circulation, P is Price level and T is Transactions or Output. As monetarists assume that V and T are determined, in the long run, by real variables, such as the productive capacity of the economy, there is a direct relationship between the growth of the money supply and inflation.
The mechanisms by which excess money might be translated into inflation are examined below. Individuals can also spend their excess money balances directly on goods and services. This has a direct impact on inflation by raising aggregate demand. Also, the increase in the demand for labour resulting from higher demands for goods and services will cause a rise in money wages and unit labour costs. The more inelastic is aggregate supply in the economy, the greater the impact on inflation.
The increase in demand for goods and services may cause a rise in imports. Although this leakage from the domestic economy reduces the money supply, it also increases the supply of money on the foreign exchange market thus applying downward pressure on the exchange rate. This may cause imported inflation.
## Austrian view
The Austrian School maintains that inflation is any increase of the money supply (i.e. units of currency or means of exchange) that is not matched by an increase in demand for money, or as Ludwig von Mises put it:
In theoretical investigation there is only one meaning that can rationally be attached to the expression Inflation: an increase in the quantity of money (in the broader sense of the term, so as to include fiduciary media as well), that is not offset by a corresponding increase in the need for money (again in the broader sense of the term), so that a fall in the objective exchange-value of money must occur.[4]
Given that all major economies currently have a central bank supporting the private banking system, money can be supplied into these economies by means of bank credit (or debt).[5] Austrian economists believe that credit growth propagates business cycles (see Austrian Business Cycle Theory.) However, the Austrian theory of the business cycle varies significantly from mainstream theories, and economists such as Gordon Tullock,[6] Bryan Caplan,[7] and Nobel laureates Milton Friedman[8][9] and Paul Krugman[10] have said that they regard the theory as incorrect. | {"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": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5460944771766663, "perplexity": 1637.138829614552}, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368704392896/warc/CC-MAIN-20130516113952-00013-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://www.bartleby.com/solution-answer/chapter-p3-problem-48e-algebra-and-trigonometry-mindtap-course-list-4th-edition/9781305071742/skills-plus-48-signs-of-numbers-let-ab-and-c-be-real-numbers-with-a0b0-and-c0-determine-the/5528ed17-9e40-459b-a3ae-3bd71638ca8a | Chapter P.3, Problem 48E
Algebra and Trigonometry (MindTap ...
4th Edition
James Stewart + 2 others
ISBN: 9781305071742
Chapter
Section
Algebra and Trigonometry (MindTap ...
4th Edition
James Stewart + 2 others
ISBN: 9781305071742
Textbook Problem
1 views
SKILLS Plus48. Signs of Numbers Let a , b , and c be real numbers with a > 0 , b < 0 , and c < 0. Determine the sign of each expression. (a) b 5 (b) b 10 (c) a b 2 c 3 (d) ( b − a ) 3 (e) ( b − a ) 4 (f) a 3 c 3 b 6 c 6
To determine
a)
The sign of the given expression b5.
Explanation
Calculation:
It is given that a>0,b<0,andc<0.
Here, b is negative.
So,
b5=
To determine
b)
The sign of the given expression b10.
To determine
c)
The sign of the given expression ab2c3.
To determine
d)
The sign of the given expression (ba)3.
To determine
e)
The sign of the given expression (ba)4.
To determine
f)
The sign of the given expression a3c3b6c6.
Still sussing out bartleby?
Check out a sample textbook solution.
See a sample solution
The Solution to Your Study Problems
Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!
Get Started | {"extraction_info": {"found_math": false, "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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8283859491348267, "perplexity": 5729.558572315072}, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575541288287.53/warc/CC-MAIN-20191214174719-20191214202719-00388.warc.gz"} |
http://www.mathematicalfoodforthought.com/2005/11/ | ## Wednesday, November 30, 2005
### Can you count? Topic: Probability. Level: AMC/AIME.
Problem: (2005 AMC 12B - #25) Six ants are on the vertices of a regular octahedron - one on each vertex. Simultaneous and independently, they each traverse one edge and arrive at another vertex. What is the probability that no two ants arrive at the same vertex?
Solution: We wish to count the number of ways the ants can move such that no two arrive at the same vertex. Consider viewing the octahedron as two points with a square between them such that the square is perpendicular to the line containing the two points. Note that two of the ants on the square must move off and the two off the square must move on.
-----
CASE 1: Two ants opposite each other on the square move off.
There are two possible pairs of ants and each pair has two ways of moving off (each one can go to either point off the square). That's $4$ possibilities.
Now consider the remaining two ants on the square. There are two ways they can move such that they don't arrive at the same vertex. That's $2$ possibilities.
Now consider the two ants off the square. There are two ways they can move onto the square (each can go to either open point). That's $2$ possibilities.
So we have $4 \cdot 2 \cdot 2 = 16$ possibilities for the first case.
-----
CASE 2: Two ants adjacent to each other on the square move off.
There are four pairs of ants adjacent to each other on the square. Each pair can move off in two different ways (same argument as CASE 1). That's $8$ possibilities.
The remaining two ants can move anywhere on the square and not be at the same vertex, giving $4$ possibilities.
The two ants off the square can move onto the square in two different ways (same argument as CASE 1). That's $2$ possibilities.
So we have $8 \cdot 4\cdot 2 = 64$ possibilites for the second case.
-----
Now consider the total possible ways satisfying the condition: $16+64 = 80$. And the total number of ways the ants can move $4^6$ (each ant has $4$ places it can go). Thus our desired probability is
$\frac{80}{4^6} = \frac{5}{256}$.
--------------------
Comment: Admittedly, the problem does look daunting when encountered at the end of the AMC-12, in your final minutes of the test, and I did not try very hard to solve the problem at the time, either. However, once you find a relatively simple method to analyze the number of ways (viewing it as two points and a square - consider the possible pairs on the square), the real counting process is quite simple. And the problem seems to just finish itself.
--------------------
Practice Problem #1: Repeat the problem above, but use a tetrahedron instead of an octahedron.
Practice Problem #2: Consider a regular hexagon with all its diagonals drawn in. What is the probability that, given any two of the seven points (including the center - intersection of the diagonals), the line segment between the points is drawn?
Practice Problem #3: Four points are randomly chosen in the plane. What is the probability that the quadrilateral they form is convex?
## Tuesday, November 29, 2005
### Inequalities Revisited. Topic: Inequalities. Level: Olympiad.
Theorem: (Rearrangement Inequality) Given two nondecreasing sequences of positive reals $x_1 \le x_2 \le \cdots \le x_n$ and $y_1 \le y_2 \le \cdots \le y_n$,
$\displaystyle \sum_{i=1}^n x_iy_i \ge \sum_{i=1}^n x_iy_{\delta(i)}$,
where $\delta$ is some permutation of $1,2, \ldots, n$.
--------------------
Comment: The Rearragement Inequality is extremely useful, and can quickly kill some inequalities that may be a hassle to AM-GM over and over again, as you'll see in the next example. The proof of the theorem simply requires considering a possible permutation, subtracting it away from the maximum, and showing that the difference is greater than zero.
--------------------
Problem: (1999 United Kingdom - #7) Given three non-negative reals $p,q,r$ such that $p+q+r = 1$, prove that
$7(pq+qr+rp) \le 2+9pqr$.
Solution: We notice that the inequality we wish to prove is not homogenous - that is, the sums of the powers on the terms are not equal. Most of the classical inequalities require you to homogenize, and given our condition, we can do so (by multiplying by $p+q+r$ when necessary). After homogenizing, the inequality becomes
$7(pq+qr+rp)(p+q+r) \le 2(p+q+r)^3+9pqr$.
Now this may look uglier than before, but we notice a lot of terms cancel, leaving (you may do the algebra yourself if you wish)
$p^2q+p^2r+q^2r+q^2p+r^2p+r^2q \le 2(p^3+q^3+r^3)$.
With enough experience, this will be an obvious direct result of Rearrangement, but to explicitly use it, we set (applied three times with $n = 2$ and WLOG assuming $p \le q \le r$):
$x_1 = p^2 \le x_2 = q^2$
$y_1 = p \le y_2 = q$, from which we have
$\displaystyle \sum_{i=1}^2 x_iy_i = p^3+q^3 \ge \sum_{i=1}^2 x_iy_{\delta(i)} = p^2q+q^2p$
where $\delta = 2,1$. As you can see applying this on the pairs $(p,r); (q,r)$ the same way and summing them up, we get the desired result
$p^2q+p^2r+q^2r+q^2p+r^2p+r^2q \le 2(p^3+q^3+r^3)$.
QED.
--------------------
Comment: An "extension" of the Rearrangement Inequality, is the Muirhead Inequality, which effectively demolishes symmetric, homogenized inequalities.
--------------------
Practice Problem #1: Prove, using Rearrangement, the Trivial Inequality:
$(x-y)^2 \ge 0$
for positive reals $x,y$.
Practice Problem #2: Prove, using Rearrangement, a special case of the Power-Mean Inequality:
$\sqrt[n]{\frac{x^n+y^n+z^n}{3}} \ge \frac{x+y+z}{3}$
for a positive integer $n$ and positive reals $x,y,z$.
Practice Problem #3: Given positive reals $a,b,c$, prove that
$\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} \ge \frac{a}{a+b} + \frac{b}{b+c} + \frac{c}{c+a}$.
## Monday, November 28, 2005
### Add Them Up. Topic: Sequences & Series. Level: AIME/Olympiad.
Problem: (1989 USAMO - #1) For each positive integer $n$, let
$S_n = 1+\frac{1}{2}+ \cdots + \frac{1}{n}$
$T_n = S_1+S_2+ \cdots + S_n$
$U_n = \frac{T_1}{2}+\frac{T_2}{3} +\cdots + \frac{T_n}{n+1}$
Find, with proof, integers $0 < a,b,c,d < 1000000$ such that $T_{1988} = aS_{1989}-b$ and $U_{1988} = cS_{1989}-d$.
Solution: Well let's start with the first condition - $T_{1988} = aS_{1989}-b$. We claim that
$T_n = (n+1)(S_{n+1}-1)$.
Consider writing $T_n$ as sum of the following sequences:
$S_1 = 1$
$S_2 = 1+\frac{1}{2}$
...
$S_n = 1+\frac{1}{2}+ \cdots +\frac{1}{n}$.
Notice that $1$ appears $n$ times, $\frac{1}{2}$ appears $n-1$ times, and more generally $\frac{1}{i}$ appears $n+1-i$ times. Then we can say
$\displaystyle T_n = \sum_{i=1}^n \frac{n+1-i}{i} = (n+1)\sum_{i=1}^n \frac{1}{i} - \sum_{i=1}^n 1 = (n+1)S_n-n = (n+1)\left(S_{n+1}-\frac{1}{n+1}\right)-n = (n+1)(S_{n+1}-1)$
as claimed. Also, save the fact that $T_n = (n+1)S_n - n$.
So then $T_{1988} = 1989(S_{1989}-1) = 1989S_{1989}-1989 \Rightarrow a = b = 1989$.
Now, using $T_n = (n+1)(S_{n+1}-1)$, we substitute into the $U_n$ equation to get
$\displaystyle U_n = \sum_{i=1}^n \frac{T_n}{n+1} = \sum_{i=1}^n \frac{(n+1)(S_{n+1}-1)}{n+1} = \sum_{i=1}^n S_{n+1} - \sum_{i=1}^n 1 = \sum_{i=1}^n S_{n+1} - n$.
But recall that $\displaystyle \sum_{i=1}^n S_{n+1} = T_{n+1}-1$ and remember the fact we saved back up there, so our final equation becomes
$U_n = T_{n+1}-(n+1) = (n+2)S_{n+1}-(n+1)-(n+1) = (n+2)S_{n+1}-2(n+1)$.
So $U_{1988} = 1990S_{1989}-3978 \Rightarrow c = 1990, d = 3978$.
Thus our final answer is $a = 1989, b = 1989, c = 1990, d = 3978$. QED.
--------------------
Comment: This was back in 1989, when the USAMO was a one-day, $3 \frac{1}{2}$ hour test with five problems. Not until 1996 did it become a two-day test.
--------------------
Practice Problem #1: Do the problem again with $T_{2005}$ and $U_{2005}$ instead, to fully understand the solution.
Practice Problem #2: Show that $T_n+\ln{(n+1)} > U_n+n$ for all positive integers $n$.
## Sunday, November 27, 2005
### Circular Reasoning. Topic: Geometry. Level: AMC/AIME.
Problem #1: (2006 Mock AIME 1 - #1) $2006$ points are evenly spaced on a circle. Given one point, find the number of other points that are less than one radius distance away from that point.
Solution: We know that one radius distance is equivalent to $\frac{1}{6}$ of the circumference (by constructing an equilateral triangle with the center and two points on the circle).
We know each point is $\frac{1}{2006}$ of the circumference away from each other, so the $k$th point is $\frac{k}{2006}$ arc distance away. We want $\frac{k}{2006} < \frac{1}{6} \Rightarrow k \le 334$. But since it can be on either side, we double that, to get $668.$. QED.
--------------------
Comment: Note that this following AMC-12 problem is considerably more difficult than the preceding AIME problem, though it happens quite often that a AMC-12 #25 is more difficult than an AIME #1.
--------------------
Problem #2: (2003 AMC 12B - #25) Three points are chosen randomly and independently on a circle. What is the probability that all 3 pairwise distances between the points are less than the radius of the circle?
Solution : Now on the surface it looks pretty easy, but remember, this is a #25 - chances are it won't come too quickly for an average problem-solver (it's easy to jump into a bunch of flawed arguments). First convert everything to arc lengths - one radius becomes $\frac{1}{6}$ circumference again. Consider the following argument:
The circumference is set to $1$. Let point $A$ be the "center" of the three points, the "center" being the point that both the other points are closest to (note that the existence of the "center" is guaranteed - think about it).
Call the shortest distance between two points to be $x$ (first assume $AB = x$) - we then have the other length $1-x$. Now think about the possibilities... we want to maintain
1. The minimum distance of $x$. So the arc $AC$ has to be greater than $x$ - $AC > x$.
2. The "center" status of $A$. So the arc $BC$ has to be greater than $AC$ so $AC < \frac{1-x}{2}$ for a given $x$. Also, $AB$ must be less than $BC$ (which can be as small as $\frac{1-x}{2}$), so we have $x < \frac{1-x}{2} \Rightarrow x < \frac{1}{3}$.
(1) We can graph this such that $AB$ is on the $x$-axis and $AC$ is on the $y$-axis following these rules: $0 < AB < \frac{1}{3}$ and $AB < AC < \frac{1-AB}{2}$.
(2) Now recall that we assumed $AB = x$. But $AC$ could be $x$ as well, so we apply the same argument, flipping $AB$ and $AC$. But the graph would be the same, only flipped over the line $y=x$, creating this:
where (1) creates the red region and (2) creates the blue region (both theoretically extending into the yellow region).
Those generate the total "number" of unique possibilites, denoted by the area beneath the graph. Now we have to find the ones that satisfy our given condition: the maximum arc length is less than $\frac{1}{6}$.
However, we notice by introducing the center status of $A$ we automatically show that our maximum arc length is $AB+AC$ because it is always greater than the independent lengths ($AB+AC > AB$ and $AB+AC > AC$ - of course this is only when $AB$ and $AC$ are both less than $\frac{1}{6}$, which is all we care about). So we add to our graph the yellow region, $AB+AC < \frac{1}{6}$.
Thus the probability is (including the extensions of the red and blue regions into the yellow) $\frac{[yellow]}{[red]+[blue]} = \frac{\frac{1}{72}}{\frac{1}{6}} = \frac{1}{12}$. QED.
--------------------
Comment: On the actual contest, the argument would probably be loosely made in the interest of saving time and you could probably finish something like this in 5-10 minutes at most if you knew where you were headed. Of course, given that this is the last problem of an AMC-12, I wouldn't be surprised if most people didn't have those 5-10 minutes to spare.
--------------------
Practice Problem #1: Find the probability that two points placed on a circle randomly and independently are within one radius distance of each other.
Practice Problem #2: In an acute angled triangle $ABC$, $\angle A = 30^{\circ}$. $H$ is the orthocenter and $M$ is the midpoint of $BC$. On the line $HM$, take point $T$ such that $HM=MT$. Show that $AT= 2BC$. (hint: Use complex numbers - see Post 13: November 24th, 2005).
Practice Problem #3: Three congruent circles of radius 1 intersect at a common point. A larger circle is circumscribed about them, tangent to each of them at one point. Consider the midpoint of one of the diameters of the smaller circles; call it $P$. What is the length of the arc along the bigger circle containing all the points that are within 2 units of $P$?
## Saturday, November 26, 2005
### Combina-WHAT? Topic: Combinatorics. Level: AMC/AIME.
Problem #1: Prove that $nC0+nC1+\cdots+nCn = 2^n$ ($nCk$ denotes $n$ choose $k$).
Solution: Chances are you've seen this before, and there are many proofs to it (one of my favorites being the number of subsets of an $n$-element set... think about it).
But here's another cool proof, using our very own Binomial Theorem.
Consider $(x+1)^n = (nC0)x^n+(nC1)x^{n-1}+\cdots+(nCn)$.
But upon setting $x = 1$, we immediately have $2^n = nC0+nC1+\cdots+nCn$ as desired. QED.
--------------------
Problem #2: Prove that $nC0-nC1+\cdots+(-1)^n(nCn) = 0$.
Solution: Well if you understood the concept behind the first problem, this one should come very easily.
$(x+1)^n = (nC0)x^n+(nC1)x^{n-1}+\cdots+(nCn)$ so for $x = -1$ we have
$0 = (-1)^n(nC0)+(-1)^{n-1}(nC1)+\cdots+(nCn)$ from which we can just flip the sign (if necessary) to get the desired result. QED.
--------------------
Practice Problem #1: Prove $nC0+nC1+\cdots+nCn = 2^n$ using the subsets argument.
Practice Problem #2: Show that $nCr+nC(r+1) = (n+1)C(r+1)$.
Practice Problem #3: (1983 AIME #8) Find the largest two-digit prime that divides $200C100$.
## Friday, November 25, 2005
### Mix it Up. Topic: Polynomials/Inequalities. Level: Olympiad.
Problem #1: (ACoPS 5.5.22) Let $P$ be a polynomial with positive coefficients. Prove that if
$P\left(\frac{1}{x}\right) \ge \frac{1}{P(x)}$
holds for $x=1$ then it holds for every $x > 0$.
Solution: Let $P(x) = a_nx^n+a_{n-1}x^{n-1}+ \cdots + a_0$.
We have $P(1) \ge \frac{1}{P(1)} \Rightarrow [P(1)]^2 = (a_n+a_{n-1}+\cdots+a_0)^2 \ge 1$.
Then we have $P(x) \cdot P\left(\frac{1}{x}\right) = (a_nx^n+a_{n-1}x^{n-1}+ \cdots + a_0)\left(a_n\frac{1}{x^n}+a_{n-1}\frac{1}{x^{n-1}}+ \cdots + a_0)$.
But $(a_nx^n+a_{n-1}x^{n-1}+ \cdots + a_0)\left(a_n\frac{1}{x^n}+a_{n-1}\frac{1}{x^{n-1}}+ \cdots + a_0) \ge (a_n+a_{n-1}+\cdots+a_0)^2 \ge 1$
by Cauchy (see Post 12: November 24th, 2005). QED.
--------------------
Problem #2: (USAMO 1983 #2) Prove that the zeros of
$x^5+ax^4+bx^3+cx^2+dx+e = 0$
cannot all be real if $2a^2 < 5b$.
Solution: Hmm... coefficients of polynomials... let's break out Vieta's Formulas !
So we have $a = -(r_1+r_2+r_3+r_4+r_5) = -\sum r_i$ (for shorthand).
And $b = r_1r_2+r_1r_3+r_1r_4+r_1r_5+r_2r_3+r_2r_4+r_2r_5+r_3r_4+r_3r_5+r_4r_5 = \sum r_ir_j$ (again, for shorthand).
So we have $2a^2<5b \Rightarrow 2\left(\sum r_i\right)^2 < 5\sum r_ir_j$.
Expanding and simplifying, we have $2\left(\sum r_i^2\right) < \sum r_ir_j$ which can be rearranged to
$\sum (r_i-r_j)^2 < 0$ which clearly cannot hold if all the roots are real. QED.
--------------------
Practice Problem #1: Given the polynomial $x^3+ax^2+bx+c$ with real coefficients, find the condition the polynomial must satisfy such that it has at least one nonreal root (based on $a,b,c$). And generalize for $a_nx^n+a_{n-1}x^{n-1}+ \cdots + a_0$.
Practice Problem #2: (ACoPS 5.5.36) Prove Cauchy-Schwarz using the polynomial
$f(x) = (a_1x+b_1)^2+(a_2x+b_2)^2+\cdots+(a_nx+b_n)^2$
by observing that $f$ has real zeros iff $\frac{a_1}{b_1} = \frac{a_2}{b_2} = \cdots = \frac{a_n}{b_n}$.
## Thursday, November 24, 2005
### Simpler than it Sounds. Topic: Complex Numbers. Level: AMC/AIME.
Problem #1: Find the coordinates of the point $(5,4)$ rotated around the point $(2,1)$ by $\frac{\pi}{4}$ radians counterclockwise.
Solution: Consider these points in the complex plane ($x$-axis becomes real part, $y$-axis becomes imaginary part).
We want to rotate $5+4i$ around $2+i$. Rewrite these in $(r, \theta)$ form, where $r$ is the magnitude and $\theta$ is the angle. We also have $(r, \theta) = e^{i\theta}$ by the Euler Formula.
So, to simplify things, we shift $2+i$ to the origin, and now we want to rotate $3+3i = 3\sqrt{2}e^{i\frac{\pi}{4}}$. We notice that rotation by $\frac{\pi}{4}$ simply adds $\frac{\pi}{4}$ to the angle, resulting in $3\sqrt{2}e^{i\frac{\pi}{2}} = 3\sqrt{2}i$.
Shifting back to the "real" origin, our rotated point becomes $2+(1+3\sqrt{2})i$ in the complex plane and $(2, 1+3\sqrt{2})$ in the Cartesian plane. QED.
--------------------
Problem #2: Given that $A(x_1,y_1)$ and $B(x_2,y_2)$ are two vertices of an equilateral triangle, find all possible values for the third point, $C$.
Solution: Convert them to complex numbers - $X = x_1+y_1i$ and $Y = x_2+y_2i$.
Notice that if we rotate $X$ around $Y$ by $\frac{\pi}{3}$ or $-\frac{\pi}{3}$ radians, we get the only two possible points for $Z$ ($C$ as a complex number).
So applying the same rotation method as in the first problem, we have $Z = Y + (X-Y)e^{i\frac{\pi}{3}}$ or $Z = Y+(X-Y)e^{-i\frac{\pi}{3}}$. We can translate this back to the Cartesian plane, but it's just a mess of algebra, not essential to understanding the method of rotation in the complex plane. QED.
--------------------
Practice Problem #1: Find the value of $A = 3+i$ rotated by $\frac{\pi}{2}$ radians counterclockwise.
Practice Problem #2: (WOOT Class) Show that, given three complex numbers $A, B, C$ that lie on the unit circle, the orthocenter of the triangle formed by them is $H = A+B+C$ (hint: If two complex numbers $X, Y$ are perpendicular, $\frac{X}{Y}$ is purely imaginary... prove it).
Practice Problem #3: (WOOT Message Board) Let $O$ be the center of a circle $\omega$. Points $A,B,C,D,E,F$ on $\omega$ are chosen such that the triangles $OAB,OCD,OEF$ are equilateral. Let $L,M,N$ be the midpoints of $BC,DE,FA$, respectively. Prove that triangle $LMN$ is equilateral.
## Wednesday, November 23, 2005
### To Equal or not to Equal. Topic: Inequalities. Level: AIME/Olympiad.
Problem #1: Prove that, given two positive reals $a$ and $b$, we have $\frac{a+b}{2} \ge \sqrt{ab}$.
Solution: Begin with the trivial inequality ($x^2 \ge 0$) applied on $a-b$. We have
$(a-b)^2 \ge 0$
$a^2-2ab+b^2 \ge 0$
$a^2+2ab+b^2 \ge 4ab$
$(a+b)^2 \ge 4ab$
$a+b \ge 2\sqrt{ab}$
$\frac{a+b}{2} \ge \sqrt{ab}$
as desired. QED.
--------------------
Comment: Note that we have just derived the Arithmetic Mean-Geometric Mean Inequality, more commonly referred to as just AM-GM. This can be generalized to $n$ variables by induction, that is, $\frac{a_1+a_2+\cdots+a_n}{n} \ge \sqrt[n]{a_1a_2 \cdots a_n}$.
--------------------
Problem #2: Given two sequences of positive reals $\{a_i\}$ and $\{b_i\}$ for $i = 1,2, \ldots, n$, prove that $(a_1^2+a_2^2+\cdots+a_n^2)(b_1^2+b_2^2+\cdots+b_n^2) \ge (a_1b_1+a_2b_2+\cdots+a_nb_n)^2$.
Solution: We shall prove this with vectors (read the previous blog post to get some background information).
Consider the vector $A = (a_1,a_2,\ldots,a_n)$ and the vector $B = (b_1,b_2,\ldots,b_n)$ (in an $n$-space).
We have $|A||B| \ge |A||B|\cos{\theta} = A \cdot B$ since $\cos{\theta} \le 1$.
But recall that $A \cdot B = (a_1,a_2,\ldots,a_n) \cdot (b_1,b_2,\ldots,b_n) = (a_1b_1+a_2b_2+\cdots+a_nb_n)$.
Also, $|A| = \sqrt{a_1^2+a_2^2+\cdots+a_n^2}$ and $|B| = \sqrt{b_1^2+b_2^2+\cdots+b_n^2}$.
Combining them, we get $\sqrt{a_1^2+a_2^2+\cdots++a_n^2} \cdot \sqrt{b_1^2+b_2^2+\cdots+b_n^2} \ge (a_1b_1+a_2b_2+\cdots+a_nb_n)$, from which our result follows directly upon squaring both sides:
$(a_1^2+a_2^2+\cdots+a_n^2)(b_1^2+b_2^2+\cdots+b_n^2) \ge (a_1b_1+a_2b_2+\cdots+a_nb_n)^2$. QED.
--------------------
Comment: This is known as Cauchy's Inequality, or the Cauchy-Schwarz Inequality. It is one of the most useful inequalities to know and can be generalized (see Holder's Inequality).
--------------------
Problem #3: Given three positive reals $a$, $b$, and $c$, prove that $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \ge \frac{3}{2}$.
Solution: Let's put our newly-learned inequalities in action!
By AM-GM, we have $a^2+b^2 \ge 2ab$. Applying this to the two other pairs and summing them up, we get
$a^2+b^2+c^2 \ge ab+bc+ca$.
Adding $2(ab+bc+ca)$ to both sides, we have
$a^2+b^2+c^2+2(ab+bc+ca) = (a+b+c)^2 \ge 3(ab+bc+ca) \Rightarrow \frac{(a+b+c)^2}{2(ab+bc+ca)} \ge \frac{3}{2}$.
Rewrite the original LHS (left hand side) as $\frac{a^2}{ab+ac}+\frac{b^2}{bc+ba}+\frac{c^2}{ca+cb}$. By Cauchy, we have
$(ab+bc+bc+ba+ca+cb)\left(\frac{a^2}{ab+bc}+\frac{b^2}{bc+ba}+\frac{c^2}{ca+cb}\right) \ge (a+b+c)^2$
with $a_1^2 = ab+bc$, $b_1^2 = \frac{a^2}{ab+bc}$, and similarly define $a_2, a_3, b_2, b_3$.
So then dividing by $2(ab+bc+ca)$ and combining it with the previous inequality, we find
$\frac{a^2}{ab+bc}+\frac{b^2}{bc+ba}+\frac{c^2}{ca+cb} \ge \frac{(a+b+c)^2}{2(ab+bc+ca)} \ge \frac{3}{2}$
as desired. QED.
--------------------
Comment: This is known as Nesbitt's Inequality, and many of the classical inequalities can be applied to prove it.
--------------------
Practice Problem #1: Prove the Root-Mean-Square - Arithmetic Mean Inequality - $\sqrt{\frac{a_1^2+a_2^2+\cdots+a_n^2}{n}} \ge \frac{a_1+a_2+\cdots+a_n}{n}$.
Practice Problem #2: (IMO 1995 #A2) Given positive reals $a, b, c$ such that $abc = 1$, prove that $\frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)} \ge \frac{3}{2}$.
## Tuesday, November 22, 2005
### Vectors... wow! Topic: Vector Geometry. Level: AMC/AIME.
Problem #1: Find the area of the triangle formed by the heads of two vectors, $A = (5, 67^{\circ})$ and $B = (4, 37^{\circ})$, and the origin.
Solution: So, we have two vectors with pretty ugly angles, but we do notice that they differ by exactly $30^{\circ}$! Hmm, then we know two sides and the angle between them...
And consequently we remember the formula $[ABC] = \frac{1}{2}ab\sin{\angle C}$, which makes us happy.
Hence the area of the triangle is $\frac{1}{2}(5)(4)(\sin{30^{\circ}}) = 5$. QED.
--------------------
Comment: Notice that the formula $\frac{|A||B|\sin{\theta}}{2} = \frac{|A \times B|}{2}$, where $A \times B$ is the cross product of $A$ and $B$.
--------------------
Problem #2: Show that the vectors $A = (3, 4, 5)$ and $B = (2, -4, 2)$ are perpendicular.
Solution: Consider the dot product of $A$ and $B$, $A \cdot B$. The general definition of this is $A \cdot B = |A||B|\cos{\theta}$, where $\theta$ is the angle between the two vectors. Notice that if two vectors are perpendicular to each other, the angle $\theta = 90^{\circ} \Rightarrow \cos{\theta} = 0 \Rightarrow A \cdot B = 0$.
We can easily show that the dot product is distributive (left as an exercise to the reader), that is, $A \cdot (B+C) = A \cdot B + A \cdot C$.
Thus $A \cdot B = [(3, 0, 0)+(0, 4, 0)+(0, 0, 5)] \cdot [(2, 0, 0)+(0, -4, 0)+(0, 0, 2)]$. Distributing this, we see that only the terms that are parallel remain because all the axes are perpendicular to each other ($\theta = 90^{\circ}$ between any two axes).
So $A \cdot B = (3, 0, 0) \cdot (2, 0, 0)+(0, 4, 0) \cdot (0, -4, 0)+(0, 0, 5) \cdot (0, 0, 2) = 6 +(-16)+10 = 0$. So $A$ is perpendicular to $B$. QED.
--------------------
Comment: In effect, we have just shown that given any two vectors $X = (x_1, x_2, \ldots, x_n)$ and $Y = (y_1, y_2, \ldots, y_n)$, we have $X \cdot Y = x_1y_1+x_2y_2+\cdots+x_ny_n$, an interesting and useful result (these are all in $n$-spaces, where there actually exist $n$ dimensions).
--------------------
Practice Problem #1: Show that the dot product is distributive: $A \cdot (B+C) = A \cdot B+ A \cdot C$.
Practice Problem #2: Find the area of the triangle formed by the points $(5,2)$, $(7,8)$, and $(2,1)$.
Practice Problem #3: Show that, given three vectors of equal magnitude $A$, $B$, and $C$, the orthocenter of the triangle formed by the heads of each of the vectors is $A+B+C$.
Practice Problem #4: Using Practice Problem #3, show that the centroid ($G$), circumcenter ($O$), and orthocenter ($H$) of a triangle are collinear and that $OG:GH = 1:2$.
## Monday, November 21, 2005
### Polynomial Power. Topic: Algebra/Polynomials. Level: AIME.
Problem: (2006 Mock AIME 1 - #14) Let $P(x)$ be a monic polynomial of degree $n \ge 1$ such that
$[P(x)]^3 = [P(x)]^2-P(x)+6$
for $x = 1, 2, \ldots , n$. Let $n_0$ be the smallest $n$ such that $P(0) > (P(1)+P(2)+ \cdots +P(n))^3$.
Find the remainder when $P(0)$ is divided by $1000$ given $n = n_0$.
Solution: Well the condition we're given as it is doesn't look particularly helpful in defining the polynomial, especially with the cube. So we try and factor it (almost always a safe bet) and we find
$(P(x)-2)([P(x)]^2+P(x)+3) = 0$.
But we notice that $[P(x)]^2+P(x)+3 = \left(P(x)+\frac{1}{2}\right)^2 + \frac{11}{4} > 0$. So then we must have $P(x)-2 = 0$ for $x = 1, 2, \ldots, n$.
Consider the polynomial $H(x) = P(x)-2$. It has zeros at $x = 1, 2, \ldots, n$ and is also monic with degree $n$. Hence $H(x) = (x-1)(x-2)\cdots(x-n) \Rightarrow P(x) = (x-1)(x-2)\cdots(x-n)+2$.
Then we have $P(0) = (-1)^n\cdot n!+2 > (P(1)+P(2)+\cdots+P(n))^3 = (2n)^3 = 8n^3$. Checking, we find the first $n$ such that this is true is $n = 8$. Therefore, $P(0) = (-1)^8 \cdot 8!+2 = 40322$. So our answer is $322$. QED.
--------------------
Pracitice Problem #1: Factor $x^4+64$.
Practice Problem #2: Let $P(x)$ be a monic polynomial of degree $n$ such that $P(x) = x$ for $x = 1,2,\ldots,n$. Find a closed expression for $P(n+1)$.
## Sunday, November 20, 2005
### Go Go Geometry! Topic: Geometry/Trigonometry. Level: AIME.
Problem: (2006 Mock AIME 1 - #15) Let $ABCD$ be a rectangle and $AB = 24$. Let $E$ be a point on $BC$ (between $B$ and $C$) such that $DE = 25$ and $\tan{\angle BDE} = 3$. Let $F$ be the foot of the perpendicular from $A$ to $BD$. Extend $AF$ to intersect $DC$ at $G$. Extend $DE$ to intersect $AG$ at $H$. Let $I$ be the foot of the perpendicular from $H$ to $DG$. The length of $IG$ is $\displaystyle \frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Suppose $\alpha$ is the sum of the distinct prime factors of $m$ and $\beta$ is the sum of the distinct prime factors of $n$. Find $\alpha + \beta$.
Solution: We begin by solving for $EC = \sqrt{25^2-24^2} = 7$. Then $\tan{EDC} = \frac{7}{24}$.
By the tangent addition formula, we have
$\tan{BDC} = \tan{(BDE+EDC)} = \frac{\tan{BDE}+\tan{EDC}}{1-\tan{BDE}\tan{EDC}} = \frac{3+\frac{7}{24}}{1-3\left(\frac{7}{24}\right)} = \frac{79}{3}$.
Since $\tan{BDC} = \frac{BC}{CD}$, we have $\displaystyle BC = (CD)\tan{BDC} = 24\left(\frac{79}{3}\right) = 632$.
Notice that $\angle AGD = 90 - \angle BDC \Rightarrow \tan{AGD} = \frac{1}{\tan{BDC}} = \frac{3}{79}$.
Then $\tan{AGD} = \frac{AD}{DG} \Rightarrow DG = \frac{AD}{\tan{AGD}} = \frac{632}{\frac{3}{79}} = \frac{2^3 \cdot 79}{3}$.
Consider $\triangle HIG$. We have $\tan{HGI} = \frac{HI}{IG} \Rightarrow HI = (IG)\tan{HGI} = \frac{3}{79}(IG)$.
Notice that $\triangle EDC$ is similar to $\triangle HDI$, so $\frac{HI}{ID} = \frac{EC}{CD} = \frac{7}{24} \Rightarrow HI = \frac{7}{24}(ID) = \frac{7}{24}(DG-IG)$.
Combining our two expressions for $HI$, we get
$HI = \frac{3}{79}(IG) = \frac{7}{24}(DG-IG)$.
Solving this for $IG$, we get
$IG = \frac{\frac{7}{24}(DG)}{\frac{3}{79}+\frac{7}{24}$.
But we calculated $DG = \frac{2^3 \cdot 79}{3}$ above, so upon substitution we get
$IG = \frac{\frac{7}{24}\left(\frac{2^3 \cdot 79}{3}\right)}{\frac{3 \cdot 24 + 7 \cdot 79}{24 \cdot 79}} = \frac{2^3 \cdot 7 \cdot 79^3}{3 \cdot 5^4}$.
Summing the distinct prime factors, we have $2+7+79+3+5 = 96$.
## Thursday, November 17, 2005
### 2006 Mock AIME 1.
Here is the first AoPS Mock AIME of the 2005-2006 year, written by yours truly. If you wish to participate officially, please create an AoPS account at the AoPS forum.
2006 Mock AIME 1
Date: Friday, Nov. 18th to Sunday, Nov. 20th, 2005
Time: 3 hours, self-timed
Format: 15 questions, Free Response (ALL answers are integers from 000 to 999, inclusive)
Scoring: 1 point per correct answer (No guessing penalty)
Answer Format: (As follows)
1. Answer to #1
2. Answer to #2
...
15. Answer to #15
2006 Mock AIME 1 Questions
P.S. I'll be gone the next two days, so don't expect blog posts.
EDIT: Ack, #14 should be taken mod 1000, to get an AIME answer, my mistake to anyone who has taken it already. I'll modify the actual PDF file when I get home.
EDIT 2: All better.
EDIT 3: #15 was also written incorrectly (hey, it's not that easy to write hard AIME problems); fixed now.
## Wednesday, November 16, 2005
### Modular Math. Topic: Number Theory. Level: AMC/AIME.
Problem: What is the remainder when you divide $6^{2005}+8^{2005}$ by $49$?
Solution Modular arithmetic is a useful tool for this problem. The concept is really quite simple:
We state that $a \equiv b \pmod{c} \Rightarrow c|(a-b)$,
that is, $a$ and $b$ leave the same remainder upon division by $c$.
We rewrite the problem as evaluating $6^{2005}+8^{2005} \pmod{49}$.
The key to this problem is noticing that $6 = 7-1$ and $8 = 7+1$. Rewrite them as such:
$(7-1)^{2005}+(7+1)^{2005}$.
Applying our handy Binomial Theorem, we see that
$\displaystyle (7-1)^{2005} = (2005C0)(7^{2005}) - (2005C1)(7^{2004}) + \ldots +(2005C2004)(7)-(2005C2005)$.
Consider this $\pmod{49}$ or $\pmod{7^2}$. We note that any term with a power of $7$ greater than or equal to $2$ is $0 \pmod{49}$ (make sure you get this, review the definition of mod if you don't). Thus the only terms that we need to consider are the last two: $\displaystyle (2005C2004)(7)-(2005C2005) = 2005(7)-1$.
Similarly, the only two terms of $(7+1)^{2005}$ we need to consider are $\displaystyle (2005C2004)(7)+(2005C2005) = 2005(7)+1$.
Summing the two, we have $(2005(7)-1)+(2005(7)-1) = 4010(7) \equiv 42 \pmod{49}$ (you can work out the details yourself). Hence our answer is $42$. QED.
--------------------
Comment: Modular arithmetic has extremely wide applications in Number Theory, including a number of important theorems you should be familiar with (e.g. Fermat's Little Theorem). Learning this concept will increase your ability to solve NT problems dramatically. If you just love Number Theory and want to learn more about it, check out the PROMYS website; it's a summer camp devoted entirely to NT.
--------------------
Practice Problem: Find the remainder when $1+10+10^2+\ldots+10^{2005}$ is divided by $9$.
## Tuesday, November 15, 2005
### Power of a Point. Topic: Geometry. Level: AMC/AIME.
Problem: Given a circle that passes through the points $(3,4)$ and $(6,8)$, find the length of a tangent to the circle from the origin.
Solution: Consider Power of a Point, which states that
$a^2 = b(b+c) = d(d+e)$, all of which are equal to the power of the point at which they intersect.
Now consider the tangent from the origin in the problem (call this length $x$) and notice that the secant from the origin through $(3,4)$ passes through $(6,8)$ as well.
Then by Power of a Point applied to the origin, we have $x^2 = (\sqrt{3^2+4^2})(\sqrt{6^2+8^2}) = (5)(10) = 50 \Rightarrow x = \sqrt{50}$. QED.
Note: On the actual contest (ARML) a third point was provided to define the circle. We notice, however, that this third point is not necessary and was thus omitted in this rewording of the problem.
--------------------
Practice Problem: Given two circles, find all points $P$ such that the power of point $P$ with respect to both circles is equal.
## Monday, November 14, 2005
### Diophantine Fun. Topic: Number Theory. Level: AIME.
Problem: Find all integer solutions $(m,n)$ to the equation
$n^4+n^3+2n^2+2n+1 = m^2$.
Solution: Let me introduce a technique used to solve diophantine equations with a power (square in this case) on one side and a multiple of that power (fourth power) on the other: Bounding.
--------------------
Example - Suppose you had the equation $n^2+3n+3 = m^2$ (assume positive integers for the example; it is not necessary for the actual problem though). We can say
$(n+1)^2 < n^2+3n+3 < (n+2)^2$ which can be verified by expanding.
The LHS becomes $0 < n+2$ and the RHS becomes $0 < n+1$.
But there are no integer squares between $(n+1)^2$ and $(n+2)^2$ so we can conclude that there are no solutions.
--------------------
Now back to the problem... we look for bounding squares; however, this requires some experimenting and matching coefficients. We want to match at least the first two terms - $n^4+n^3$.
After playing around with squares, we find a pretty good bound that works for $n \ge 4$ and $n \le 0$:
$\displaystyle \left(n^2+\frac{n}{2}+\frac{1}{2}\right)^2 \le n^4+n^3+2n^2+2n+1 \le \left(n^2+\frac{n}{2}+1\right)^2$.
The LHS becomes $\displaystyle 0 \le \frac{3}{4}(n+1)^2$ and the RHS $\displaystyle 0 \le n\left(\frac{n}{4}-1\right)$ (hence the restrictions on $n$).
Since there are no squares between $\displaystyle \left(n^2+\frac{n}{2}+\frac{1}{2}\right)^2$ and $\displaystyle \left(n^2+\frac{n}{2}+1\right)^2$, the only possible solutions are the equality conditions and $n=1,2,3$.
Checking those three, we find nothing, so we move to the equality conditions.
As noted above, the equality conditions are $\displaystyle 0 = \frac{3}{4}(n+1)^2$ and $\displaystyle 0 = n\left(\frac{n}{4}-1\right)$, giving the solutoins $n=-1, 0, 4$. Then $m = \pm 1, \pm 1, \pm 19$, respectively.
Hence our solutions are: $(m,n) = (-1, \pm 1); (0, \pm 1); (4, \pm 19)$. QED.
--------------------
Practice Problem: Find all nonnegative integer solutions $(a,b)$ to the equation
$a^6+2a^4+2a^2+2a+1 = b^2$.
### Hello!
Welcome to my blog! This will be an ongoing journal of mathematical problems and problem solving techniques, targetted at high school students like myself aiming to do well on the American Math Competitions. The difficulty will range from mostly difficult AMC problems to USAMO problems. Enjoy! | {"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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9345535039901733, "perplexity": 348.1252928159472}, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583516123.97/warc/CC-MAIN-20181023090235-20181023111735-00337.warc.gz"} |
https://math.stackexchange.com/questions/2761339/prooving-not-all-roots-are-real | # Prooving not all roots are real
Let $P(x)=x^n+a_{n-1}x^{n-1}\dots+a_0$
Given $a_{n-1}^2<a_{n-2},$ prove not all roots can be real
My attempt:
Assume all roots are real, namely $R_1,R_2,\dots,R_n$
Then $-a_{n-1}=\sum_{i=1}^n R_i$
$a_{n-2}=\sum R_i\times R_j$
Now $$(R_1+R_2+\dots+R_n)^2<(R_1R_2+R_1R_3\dots R_{n-1}R_{n})\\ \implies R_1^2+R_2^2\dots R_n^2<-(R_1R_2+R_1R_3\dots R_{n-1}R_{n})$$
But we can't tell if RHS is actually negative or positive. Also this kinda looks like cauchy-schwarz. Can it be used here?
Hint: From $\displaystyle (\sum_i R_i)^2 < \sum_{i< j} R_i R_j$, we may add $\frac12 \sum R_i^2$ to both sides to get $$\left(\sum_i R_i \right)^2 + \frac12\sum_i R_i^2 < \frac12 \left(\sum_i R_i \right)^2$$ $$\implies \frac12\left(\sum_i R_i \right)^2 + \frac12\sum_i R_i^2 < 0$$ | {"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": 1, "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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9371607899665833, "perplexity": 285.97345672579934}, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195528013.81/warc/CC-MAIN-20190722113215-20190722135215-00489.warc.gz"} |
http://en.wikipedia.org/wiki/Burr_puzzle | Burr puzzle
Burr puzzles
A burr puzzle is an interlocking puzzle consisting of notched sticks, combined to make one three-dimensional, usually symmetrical unit. These puzzles are traditionally made of wood, but versions made of plastic or metal can also be found. Quality burr puzzles are usually precision-made for easy sliding and accurate fitting of the pieces. In recent years the definition of "burr" is expanding, as puzzle designers use this name for puzzles not necessarily of stick-based pieces.
The term "Burr" is first mentioned in a 1928 book by Edwin Wyatt,[1] but the text implies that it was commonly used before. The term is attributed to the finished shape of many of these puzzles, resembling a seed burr.
The origin of burr puzzles is unknown. The first known record[2] appears in a 1698 engraving used as a title page of Chambers's Cyclopaedia.[3] Later records can be found in German catalogs from the late 18th century and early 19th century.[4] There are claims of the burr being a Chinese invention, like other classic puzzles such as the Tangram.[5]
Six-Piece Burr
An assembled six-piece burr
The six-piece burr, also called "Puzzle Knot" or "Chinese Cross", is the most well-known and presumably the oldest of the burr puzzles. This is actually a family of puzzles, all sharing the same finished shape and basic shape of the pieces. The earliest US patent for a puzzle of this kind dates back to 1917.[6]
For many years, the six-piece burr was very common and popular, but was considered trite and uninteresting by enthusiasts. Most of the puzzles made and sold were very similar to one another and most of them included a "key" piece, an unnotched stick that slides easily out. In the late 1970s, however, the six-piece burr regained the attention of inventors and collectors, thanks largely to a computer analysis conducted by the mathematician Bill Cutler and its publication in Martin Gardner's column on Scientific American.[7]
Structure
All six pieces of the puzzle are square sticks of equal length (at least 3 times their width). When solved, the pieces are arranged in three perpendicular, mutually intersecting pairs. The notches of all stick are located within the region of intersection, so when the puzzle is assembled they are unseen. All notches can be described as being made by removing cubic units (with an edge length of half the sticks' width), as shown in the figure:
There are 12 removable cubic units, and different puzzles of this family are made of sticks with different units removed. 4,096 permutations exist for removing the cubic units. Of those, we ignore the ones that cut the stick in two and the ones creating identical pieces, and are left with 837 usable pieces.[8] Theoretically, these pieces can be combined to create over 35 billion possible assemblies, however it is estimated that less than 6 billion of them are actual puzzles, capable of being assembled or taken apart.[9]
The "Nut" puzzle from Hoffmann's 1893 book,[10] an example of a solid burr.
Solid Burr
A burr puzzle with no internal voids when assembled is called a solid burr. These burrs can be taken apart directly by removing a piece or some pieces in one move. Up until the late 1970s, solid burrs received the most attention and publications referred only to this type.[11] 119,979 solid burrs are possible, using 369 of the usable pieces. To assemble all these puzzles, one would need a set of 485 pieces, as some of the puzzles include identical pieces.[8]
"Burr no. 305", named after its location in Culter's analysis tables. It was found to be the most "interesting" of the 314 solid burrs of notchable pieces, because it is the only one containing no duplicate or symmetric pieces, and also having one unique solution that does not employ a common 2-piece key.
Pieces Types
For aesthetic, but mostly practical reasons, the burr pieces can be divided into two types:
• Notchable pieces - with full notches that can be made with a saw or a milling machine.
• Non-notchable pieces - with internal corners that have to be made with a chisel or by gluing parts together.
Right to left: a notchable piece, a non-notchable piece and a piece that is technically notchable, but cannot be used with other notchable pieces to create solid burrs
59 of the usable pieces are notchable, including the unnotched stick. Of those, only 25 can be used to create solid burrs. This set, often referred to as "The 25 notchable pieces", with the addition of 7 duplicates, can be assembled to create 314 different solid burr puzzles. These pieces are very popular, and full sets are manufactured and sold by many companies.
"Bill's Buffling Burr" of level 5, by Bill Cutler
A level-7 burr by the Israeli designer and maker Philippe Dubois who sold his puzzles under the name Gaby Games"
Holey Burr
For all solid burrs, one movement is required to remove the first piece or pieces. However, a holey burr, which has internal voids when assembled, can require more than one move. The number of moves required for removing the first piece is referred to as the level of the burr. All solid burrs are therefore of level 1. The higher the level is, the more difficult the puzzle.
During the 1970s and 1980s, attempts were made by experts to find burrs of an ever-higher level. On 1979, the American designer and craftsman Stewart Coffin found a level-3 puzzle. In 1985, Bill Cutler found a level-5 burr[12] and shortly afterwards a level-7 burr was found by the Israeli Philippe Dubois.[11] In 1990, Cutler completed the final part of his analysis and found that the highest possible level using notchable pieces is 5, and 139 of those puzzles exist. The highest level possible for a six-piece burr with more than one solution is 12, meaning 12 moves are required to remove the first piece.[9]
Three-Piece Burr
A three-piece burr made from sticks with "regular" right-angled notches (as the six-piece burr), cannot be assembled or taken apart.[13] There are, however, some three-piece burrs with different kinds of notches, the best known of them being the one mentioned by Wyatt in his 1928 book, consisting of a rounded piece that is meant to be rotated.[1]
Known Burr Families
An Altekruse puzzle
Altekruse
The Altekruse puzzle is named after the grantee of its 1890 patent, though the puzzle is of earlier origin.[14] The name "Altekruse" is of Austrian-German origin and means "old-cross" in German, which led to the presumption that it was a pseudonym, but a man by that name immigrated to America in 1844 with his three brothers to avoid being drafted to the Prussian Army and is presumed to be the one who filed this patent.[15]
A classic Altekruse consists of 12 identical pieces. In order to disassemble it, two halves of the puzzle have to be moved in opposite directions. Using two more of these pieces, the puzzle can be assembled in a different way. By the same principle, other puzzles of this family can be created, with 6, 24, 36 and so forth. Despite their size, those bigger puzzles are not considered very difficult, yet they require patience and dexterity to assemble.
A Chuck puzzle
Chuck
The Chuck puzzle was invented and patented by Edward Nelson in 1897.[16] His design was improved and developed by Ron Cook of the British company Pentangle Puzzles who designed other puzzles of the family.[17]
The Chuck consists mostly of U-shaped stick pieces of various lengths, and some with an extra notch that are used as key pieces. For creating bigger Chuck puzzles (named Papa-chuck, Grandpapachuck and Great Grandpapachuck, by Cook) one would need to add longer pieces. The Chuck can also be regarded as an extension of a six-piece burr of very simple pieces called Baby-chuck, which is very easy to solve. Chuck pieces of different lengths can also be used to create asymmetryic shapes, assembled according to the same principle as the original puzzle.
Typical Chuck pieces: a U-shaped piece and a key piece
Pagoda of size 5, with 51 pieces (crafted by Philippe Dubois)
Pagoda
The origin of the Pagoda, also called "Japanese Crystal" is unknown. It is mentioned in Wyatt's 1928 book.[1] Puzzles of this family can be regarded as an extension of the "three-piece burr" (Pagoda of size 1), however they do not require special notches to be assembled or taken apart. Pagoda of size 2 consists of 9 pieces, and bigger versions consist of 19, 33, 51 and so forth. Pagoda of size $n$ consists of $2n^2+1$ pieces.
Diagonal Burr
A diagonal burr - Giant Star puzzle (manufactured by Gaya Games)
Though most burr puzzle pieces are made with square notches, some are made with diagonal notches. Diagonal burr pieces are square sticks with V-shaped notches, cut at an angle of 45° off the stick's Face. These puzzles are often called "Stars", as it is customary to also cut the sticks' edges at an angle of 45°, for aesthetic reasons, giving the assembled puzzle a Star-like shape.
References
1. ^ a b c Wyatt, E. M. (1928). Puzzles in Wood. Milwaukee, Wisc: Bruce Publishing Co. ISBN 0-918036-09-7.
2. ^ Slocum, Jerry, New Findings on the History of the Six Piece Burr, Slocum Puzzle Foundation
3. ^
4. ^ Slocum, Jerry; Gebbardt, Dieter (1997), Puzzles from Catel's Cabinet and Bestelmeier's Magazine, 1785 to 1823, Slocum Puzzle Foundation
5. ^ Zhang, Wei; Rasmussen, Peter (2008), Chinese Puzzles: Games for the Hands and Mind, Art Media Resources, ISBN 1588861015 (A page about burr puzzles in the book's website)
6. ^ US 1225760, Brown, Oscar, "Puzzle", issued 1917
7. ^ Gardner, Martin (January 1978), Mathematical Games, Scientific American: 14–26
8. ^ a b Cutler, William H. (1978), The Six-Piece Burr, Journal of Recreational Mathematics 10 (4): 241–250
9. ^ a b Cutler, Bill (1994), A Computer Analysis of All 6-Piece Burrs, retrieved February 17, 2013
10. ^ Hoffmann, Professor (1893), "Chapter III, No. XXXVI", Puzzles old and new, London: Frederick Warne and Co. (Available for download at the Internet Archive)
11. ^ a b Coffin, Stewart (1992), Puzzle Craft (PDF)
12. ^ Dewdney, A. K. (October 1985), Computer Recreations, Scientific American 253 (4): 16–27, doi:10.1038/scientificamerican1085-16
13. ^ Jürg von Känel (1997), Three-piece burrs, IBM, archived from the original on January 11, 2012, retrieved February 19, 2013
14. ^ US 430502, Altekruse, William, "Block Puzzle", issued 1890
15. ^ Coffin, Stewart (1998), "The Altekruse Puzzle", The Puzzling World of Polyhedral Dissections, retrieved February 19, 2013
16. ^ US 588705, Nelson, Edward, "Puzzle", issued 1897
17. ^ WoodChuck Puzzles, Pentangle Puzzles, retrieved February 19, 2013 | {"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": 2, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.38953694701194763, "perplexity": 3818.7783903750606}, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1413507444209.14/warc/CC-MAIN-20141017005724-00144-ip-10-16-133-185.ec2.internal.warc.gz"} |
http://nrich.maths.org/5577/solution | ### Coke Machine
The coke machine in college takes 50 pence pieces. It also takes a certain foreign coin of traditional design. Coins inserted into the machine slide down a chute into the machine and a drink is duly released. How many more revolutions does the foreign coin make over the 50 pence piece going down the chute? N.B. A 50 pence piece is a 7 sided polygon ABCDEFG with rounded edges, obtained by replacing AB with arc centred at E and radius EA; replacing BC with arc centred at F radius FB ...etc..
### Just Opposite
A and C are the opposite vertices of a square ABCD, and have coordinates (a,b) and (c,d), respectively. What are the coordinates of the vertices B and D? What is the area of the square?
### Get Cross
A white cross is placed symmetrically in a red disc with the central square of side length sqrt 2 and the arms of the cross of length 1 unit. What is the area of the disc still showing?
# Wedge on Wedge
##### Stage: 4 Challenge Level:
We contruct a new triangle as shown. Angle $x$ is equal to $90^\circ - (90^\circ - y) = y$, so the highlighted triangles are similar. Therefore $$p = 48 \times \frac{39}{52} = 36$$ $$q =20\times \frac{39}{52} = 15$$
The base is split $48 - q : q = 33 : 15 = 11 : 5$ and the ball falls a distance of $p + 20 = 36 + 20 = 56$. | {"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": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6364760994911194, "perplexity": 1324.8038907355303}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1406510263423.17/warc/CC-MAIN-20140728011743-00063-ip-10-146-231-18.ec2.internal.warc.gz"} |
https://socratic.org/questions/56e84d537c014905f503b50a | Chemistry
Topics
# Question 3b50a
Mar 17, 2016
See explanation.
#### Explanation:
You're essentially dealing with four ionic compounds that are insoluble in aqueous solution.
Silver chloride, $\text{AgCl}$, is a white precipitate that can be obtained by mixing a solution of a soluble ionic compound that contains the silver(I) cation, ${\text{Ag}}^{+}$, with a solution of a soluble ionic compound that contains the chloride anion, ${\text{Cl}}^{-}$.
Lead(II) chromate, ${\text{PbCrO}}_{4}$, is a yellow precipitate that can be obtained by mixing a solution of a soluble ionic compound that contains the lead(II) cation, ${\text{Pb}}^{2 +}$, with a solution of a soluble ionic compound that contains the chromate anion, ${\text{CrO}}_{4}^{2 -}$.
Copper(II) sulfide, $\text{CuS}$, is a black precipitate that can be formed by mixing a solution of a soluble ionic compound that contains the copper(II) cation, ${\text{Cu}}^{2 +}$, with a solution of a soluble ionic compound that contains the sulfide anion, ${\text{S}}^{2 -}$.
Finally, nickel(II) hydroxide, "Ni"("OH")_2#, is a green precipitate that results when a solution of a soluble ionic compound that contains the nickel(II) cation, ${\text{Ni}}^{2 +}$, is mixed with a solution of a soluble ionic compound that contains the hydroxide anion, ${\text{OH}}^{-}$.
##### Impact of this question
1339 views around the world | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 12, "mathjax_inline_tex": 1, "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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2912639081478119, "perplexity": 3887.1320393785822}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986684226.55/warc/CC-MAIN-20191018154409-20191018181909-00514.warc.gz"} |
https://gateoverflow.in/683/gate2000-12 | 5k views
An instruction pipeline has five stages where each stage take 2 nanoseconds and all instruction use all five stages. Branch instructions are not overlapped. i.e., the instruction after the branch is not fetched till the branch instruction is completed. Under ideal conditions,
1. Calculate the average instruction execution time assuming that 20% of all instructions executed are branch instruction. Ignore the fact that some branch instructions may be conditional.
2. If a branch instruction is a conditional branch instruction, the branch need not be taken. If the branch is not taken, the following instructions can be overlapped. When 80% of all branch instructions are conditional branch instructions, and 50% of the conditional branch instructions are such that the branch is taken, calculate the average instruction execution time.
edited | 5k views
+3
Each stage is $2$ns. So, after $5$ time units each of $2$ns, the first instruction finishes (i.e., after $10$ns), in every $2$ns after that a new instruction gets finished. This is assuming no branch instructions. Now, once the pipeline is full, we can assume that the initial fill time doesn't matter our calculations and average execution time for each instruction is $2$ns assuming no branch instructions.
1. Now, we are given that $20\%$ of instructions are branch (like JMP) and when a branch instruction is executed, no further instruction enters the pipeline. So, we can assume every $5$th instruction is a branch instruction. So, with this assumption, total time to finish $5$ instruction will be $5 * 2 + 8 = 18$ ns (as when a branch instruction enters the pipeline and before it finishes, $4$ pipeline stages will be empty totaling $4 * 2 = 8$ ns, as it is mentioned in question that the next instruction fetch starts only when branch instruction completes). And this is the same for every set of 5 instructions, and hence the average instruction execution time $= 18/5 = 3.6$ ns
2. This is just a complex statement. But what we need is to identify the $\%$ of branch instructions which cause a branch to be taken as others will have no effect on the pipeline flow.
$20\%$ of instructions are branch instructions. $80\%$ of branch instructions are conditional.
That means .$2*.8 = 16\%$ of instructions are conditional branch instructions and it is given that $50\%$ of those result in a branch being taken.
So, $8\%$ of instructions are conditional branches being taken and we also have $20\%$ of $20\% = 4\%$ of unconditional branch instructions which are always taken.
So, percentage of instructions where a branch is taken is $8+4 = 12\%$ instead of $20\%$ in (A) part.
So, in $100$ instructions there will be $12$ branch instructions. We can do a different calculation here as compared to (A) as $12$ is not a divisor of $100$. Each branch instruction causes a pipeline delay of $4*2 = 8$ ns. So, $12$ instructions will cause a delay of $12 * 8 = 96$ ns. For $100$ instructions, we need $100 * 2 = 200$ ns without any delay and with delay we require $200 + 96 = 296$ ns for $100$ instructions.
So, average instruction execution time $= 296/100 = 2.96$ ns
(We can also use this method for part (A) which will give $100 * 2 + 20*8 = 360$ ns for $100$ instructions)
by Veteran (425k points)
edited by
0
Thank You Arjun Sir for an elaborate and understandable explanation.
Could you help me with a link for such kind of pipeline problems with delays .?
+19
part b)
if branch not taken following instructions can be overlapped , implies stall=0
if branch taken, it would be simply after k-1 stall = 4 stalls.
20% are branch
80% of branch are conditional ,implies 80% of 20% are conditional
which means 20% are unconditional(always takes), implies 20% of 20% are unconditional which are taken
50% of branch conditional are taken, implies 50% of 80% of 20% are taken
considering for all cases where there is stall :
therefore, Tavg=(1+stall cyle*stall freq) clocks
= (1+ ( 0.50*0.80*0.20*4 ) + (0.20*0.20*4) ) clocks
= 1.48 clocks= 1.48 * max{stage dealy}= 1.48*2ns= 2.96 ns
0
sir am not getting second part ???
0
Nicely explained sir
+26
2nd part
0
It isn't the branch instruction but the instruction following it which incurs the overhead of 10 units, am I correct?
0
if branch taken, it would be simply after k-1 stall
can someone please elaborate this statement ? thanks
0
"as when a branch instruction enters the pipeline and before it finishes, 4 pipeline stages will be empty totaling 4 * 2 = 8 ns, as it is mentioned in question that the next instruction fetch starts only when branch instruction completes)."
This 8 ns delay will be in 6 th instruction? As 5 th instruction is branch instruction the delay will happen in 6th instruction. I am confused which instruction will have delay of 8 ns and will face stalls if 5th instruction is branch then next instruction should be fetched only once 5 th instruction is fiinhsed with its execution.
To execute 5 instruction it will take 18 ns but to execute 6th it will include 8 ns delay plus 2 ns to execute total 10 ns. I don't know if my understanding is correct please help.
0
@arjun sir,
Now, once the pipeline is full, we can assume that the initial fill time doesn't matter our calculations
If ideal condition is not mentioned in question ..we should consider initial pipeline filling time also right by, just adding extra time of filling in whatever is calculated considering ideal condition.
i.e , in part (A) 3.6 + 4(2)
Please correct me if i m wrong
+2
@ jatin khachane 1
As we don't know total number of instructions, so we can't find average time taking into account initial pipeline fill time, because it will be $3.6 + (4*2) / total Instructions$ in case of (A). so we need to ignore second term, that's why answer is $3.6ns$
+1
@Arjun sir
@just_bhavana
Here it is given that if the branch is not taken next instructions can be overlapped ..But whether instruction takes branch or not is known after completion of branch instruction execution..how it is decided here that if branch taken then instructions can overlapped and not overlapped if branch taken ...how can we use two different mechanism for diif branch instructions .. ??
if we use stalling(next ) mechanism in both cases it will not overlap next instruction till we get EA
If we use flushing then we can overlap and flush depending on branch taken / not
if an instruction branches then it takes $2ns \times 5 = 10ns$, coz if branch is taken then the instruction after that branch instruction is not fetched until entire current branch instruction is completed, this means it will go through all stages.
if an instruction is non-branch or branching does not happen then, it takes $2ns$ to get completed.
a) $\text{average time taken} = 0.8 \times 2ns + 0.2 \times 10ns = 3.6ns$
b)
$\text{Average time taken,}$
$= 0.8 \times 2ns + 0.2 \Big( 0.2 \times 10ns + 0.8 \big( 0.5\times 10ns + 0.5 \times 2ns\big)\Big)$
$=2.96ns$
by Boss (30.7k points)
edited
0
0
For first Part
First instruction will always take n stage * t ns i,e first instruction will take 5*2=10 ns. is nt?
Assume 100 instruction
so average time= (2*5+79*2+20*5*2)/100
=3.68 ns
Please tell What is wrong in it?
0
@Abhisek Tiwari 4 ,I am also getting 3.68ns
+3
Abhisek Tiwari 4 QUESTION SAYS UNDER IDEAL CONDITION SO IN PIPELINE IDEAL CONDITION IS CPI=1
0
@amarVashishth
Kindly help me in this. I followed the approach given in your solution , got answer as 62.5 but that is wrong.
https://gateoverflow.in/285418/go2019-flt1-61
A))
Let total number of instruction is n
Tavg=n x 20% x 5 x 2ns + n x 80% x 1 x 2ns/n
Tavg=3.6ns
B))
Let us suppose we have 100 instruction
20% are branch instruction and 80% Not branch instruction
out of 20% ,80% of 20% are cnditional branch and 20% of 20% are unconditional
Now 50% of 80% of 20% are branch instruction where branch is taken and 50% of 80% of 20% branch not taken.
Clock average=148/100=1.48
Tavg=1.48 x 2ns=2.96ns
by Boss (10k points)
.........
by Boss (35.7k points)
edited
0
New CPI = OLD CPI + stall-rate*stall-cycle
execution time = new CPI * cycle time.
+1 vote
Part A:
suppose total instructions = n
time for each stage = 2 ns , CPI=stall + 1
average instruction execution time = ([no branch instrution * cpi * stage time] + [branch instrution * cpi *stage time])/total instruction
for branch instruction :
no. of stall = 4 , so, CPI=stall+1= 5
20 % of total instructions are branch instructions = n * 0.2
for no branch instruction :
no. of stall = 0 , so; CPI=stall + 1= 1
80 % of total instructions are no branch instructions = n * 0.8
average time = ([n * 0.2 * 5 * 2]+[n * 0.8 * 1 * 2])/n = 3.6 ns
Part B:
this will be the graph to understand better :
this will be little hard to solve in terms of n , so take total instruction n= 100
then no branch instn= 80 , branch instn(conditional branch=16(branch taken=8 , branch not taken=8) , unconditional branch=4)
average instruction execution time = ([80*1*2]+[8*5*2]+[8*1*2]+[4*5*2])/100 = 2.96 ns
by Junior (647 points)
Execution time formula T1= $\frac{N*S}{R}$
Where N = dynamic instruction count
S = Avg no. Of cycles to fecth and execte one instruction
R = clock rate in cycles per second
Avg execution time T = $\frac{T1}{N}$
T= $\frac{S}{R}$
Now according to question R = $\frac{1}{2 ns}$
So T = S *2 ns
For ideal pipelines S = 1
For non-ideal pipelines S = 1 + S' ( stall)
Now according to question when branch instruction is fetched next instruction is not fetched until after it's completion which means there is a stall of 4 cycles.
A). 20% instructions are branch instructions so
S' = 0.2*4 = 0.8
S =1 + 0.8 = 1.8
So T =1.8 * 2 ns
=3.6 ns (Ans)
B). For Conditional Branch
Branch prediction = Not taken
If prediction true = overlapping done
This case gives us 0 stall.
Prediction false = no overlapping
This case gives us 4 stalls.
S' = 0.2 * 0.8 *0.5 *4 ( 20 % branch instruction, 80% Conditional Branch , 50% branch taken giving false prediction)
S' = 0.32
For unconditional branch
S' = 0.2 * 0.2 * 4
= 0.16
So S = 1+0.32+0.16 = 1.48
T = 1.48*2 ns
=2.96 ns (Ans)
ago by (19 points)
edited ago | {"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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5207266211509705, "perplexity": 3990.4776973185863}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540527620.19/warc/CC-MAIN-20191210123634-20191210151634-00110.warc.gz"} |
https://www.ptb.de/cms/nc/en/ptb/fachabteilungen/abt8/fb-84/ag-842/dynamischemessungen-842.html?tx_bib_pi1%5Bshow_uid%5D=72&cHash=e04c057abb54306f10deee729e8d476a | # Analysis of dynamic measurements
Working Group 8.42
#### Content
Dynamic measurements can be found in many areas of metrology and industry, such as, for instance, in applications where mechanical quantities, electrical pulses or temperature curves are measured.
A quantity is called dynamic when its value at one time instant depends on its values at previous time instants.
That is, in contrast to static measurements where a single value or a (small) set of values is measured, dynamic measurements consider continuous functions of time. Since the analysis of dynamic measurements requires different approaches than the analysis of static measurements this part of metrology is often called "Dynamic Metrology".
The mathematical modeling of dynamic measurements typically utilizes methodologies and concepts from digital signal processing. In the language of metrology a signal denotes a dynamic quantity, and a system a measurement device whose input and/or output are signals. The output signal of a system is thus the indication value of the measurement device for a corresponding input signal.
The typical scenario in a dynamic measurement is a time-dependent input signal, a linear time-invariant (LTI) measurement system and a corresponding time-dependent ouptut signal. The linearity of the system refers to the superposition of input signals, and the time-invariance ensures that the system itself does not change over time. Mathematically, the relation between input signal and output signal is in this case given by a convolution
$$x(t) = \int y(\tau) h(\tau-t) d\tau ,$$
with $y(t)$ the input signal (measurand), $h(t)$ the system's impulse response, and $x(t)$ the system's output signal. The measurand is then the input signal $x(t)$ of the measuring system, and its estimation requires deconvolution.
Statistical modeling of static quantities is realized by univariate or multivariate random variables. The evaluation of uncertainty is then based on probability density functions. In a mathematical sense, the extension of this concept to dynamic measurements requires the utilization of stochastic processes as a model for the uncertain knowledge of the continuous functions $x(t)$ and $y(t)$. That is, the dynamic quantity $x(t)$ is considered as a continuous function in time and uncertain knowledge about its values is expressed by a stochastic process $X_t$ with continuous trajectories. The working group has developed a corresponding framework for the evaluation of uncertainties which is consistent with the framework of the GUM for static measurements.
However, instead of the continuous function $x(t)$, typically its discretization is considered in the application. Therefore, the continuous output signal of the measurement system is stored on a computer as a vector, whereby $\mathbf{x}=(x_1,\ldots,x_N)$ with $x_k=x(t_k)$. Statistical modeling of the uncertain knowledge of the values of such a vector and its propagation can then, in principle, be carried out by using the concepts and framework of GUM Supplement 2. However, this is possible only for measurements on closed time intervals. Thus, it is not possible to treat running (on-time) measurements without making further assumptions.
A characteristic property of a dynamic measurement is that the output signal is not proportional to the input signal. This is due to so-called dynamic effects caused by the imperfect dynamic behavior of the measuring system. For instance, accelerometers typically show a resonance behavior, and hence, for a measured acceleration with a certain frequency content the output signal of the accelerometer shows a significant "ringing".
The aim in the analysis of dynamic measurements is the compensation for time-dependent errors, such as, ringing, phase deviations and other effects. In contrast to static measurements, this cannot be accomplished by scaling and shifting the output signal, since it only takes into account the static characteristics of the measuring system. Instead, for linear time-invariant systems (LTI) a so-called deconvolution has to be carried out. This allows the compensation of dynamic effects and, thereby, in principle, the reconstruction of the values of the dynamic measurand.
The analysis of dynamic measurements in the case of linear time-invariant (LTI) systems is typically carried out by application of a suitable digital filter. The design of such a filter is based on the available knowledge about the measurement system and aims at compensating its dynamic effects. As illustrated in the example in Fig. 2, the frequency response of the compensation filter is the reciprocal of the system's frequency response up to a certain frequency. Thus, the prerequisite for the design of a compensation filter is a dynamic calibration of the measurement device in a suitable frequency range.
The literal meaning of "dynamic" relates solely to time-dependent quantities. However, from a mathematical perspective the definition of a dynamic measurement can be extended to other independent quantities than time. This includes quantities whose value depends on frequency, spatial coordinates, wavelength, etc. The extended definition is reasonable, because the mathematical treatment of such measurements does not depend on the physical interpretation of the independent quantity.
A quantity is a dynamic quantity if its value depends on another, independent quantity. A measurement is dynamic when at least one of the involved quantities is dynamic.
The extended definition of a dynamic measurement contains a broad spectrum of metrological applications. Typical examples are measurements of mechanical quantities as, for example, force, torque and pressure. Further examples are oscilloscope measurements for the characterization of high speed electronics, hydrophone measurements for the characterization of medical ultrasound devices or the spectral characterization of radiation sources. The extended definition also includes, for example, the spectral characterization of luminous sources, spectral color measurements or camera-aided temperature measurements. The applications range from single sensor measurements up to large sensor networks.
In many areas of application, dynamic properties and dynamic errors have been neglected so far. Instead, rule-of-thumb correction methods were applied and larger uncertainties assigned to the measurement result. However, in recent years the demand for more precise dynamic measurements with smaller uncertainties has increased steadily.
# Research
The PTB Working Group 8.42 has been carrying out research in the field of dynamic measurements for more than 10 years and its scientific work covers almost all areas of dynamic metrology. Publications of the working group contain, for instance, methods for the statistical analysis of dynamic calibration, design of digital deconvolution filters for estimating the value of the measurand, GUM-compliant evaluation of dynamic measurement uncertainty, and efficient implementation of GUM Monte Carlo for the application of digital filters. Moreover, the working group works closely together with industrial partners to promote developments in dynamic metrology.
The mathematical and statistical treatment of dynamic measurements requires different approaches and tools than the analysis of static measurements. Thus, the areas of research in the working group are determined by the challenges in dynamic metrology.
#### Propagation of uncertainties
In general, the evaluation and propagation of measurement uncertainties for discretized dynamic quantities can be carried out by application of the GUM framework. However, there are many mathematical and practical challenges which require specific developments and research. For instance, in practice the uncertainty associated with a static quantity is often determined by repeated measurements. Therefore, for univariate quantities a rather small number of measurements are sufficient. For multivariate quantities, however, the necessary number of measurements increases with the dimension of the quantity. Discretized dynamic quantities are typically very high-dimensional data sets, with a typical dynamic measurement consisting of more than one thousand time instants. The evaluation of uncertainty by means of repeated measurements is thus not possible. To this end, parametric approaches have to be determined. Corresponding methods can be found in the field of time series analysis, but their application in metrology requires significant further developments.
#### Estimating the measurand
In most cases, estimation of the measurand in dynamic measurements requires deconvolution. However, this is a mathematically ill-posed inverse problem. That is, it requires some kind of regularization in order to obtain reasonable uncertainties. To this end, a typical approach in signal processing is the application of a suitable low-pass filter to suppress undesired high frequency components. In fact, many classical concepts of deconvolution such as Tikhonov regularization or Wiener deconvolution can be interpreted as a successive application of the reciprocal system response and a low-pass filter. However, this is only possible by taking into account prior knowledge about the measurand, which is currently not considered in the GUM and its supplements. The type of prior knowledge can be, for instance, a parametric model or an upper bound in the frequency domain. In any case, the low-pass filter causes a systematic deviation in the estimation result. A reduction in the systematic deviation is always attended by an increase in the noise-dependent variance in the result of deconvolution and vice verse. For metrological applications, these systematic errors have to be considered in the uncertainty budget. However, so far no general guidelines or harmonized treatment of these uncertainty contributions is available.
#### Practical challenges
The treatment of dynamic measurements leads to a number of practical challenges for metrologists. One of the currently most urging challenges is that of transferring the measurement result. Owing to the high dimensionality of discretized dynamic measurements, the associated uncertainties are also high-dimensional. For instance, the transfer of the calibration results for the impulse response of a sampling oscilloscope requires transferring a covariance matrix of the dimension of at least $1000\times 1000$. This is not possible with the current practice of paper-based calibration certificates.
#### Dynamic calibration
A large number of guidelines and standards are available for carrying out and taking advantage of static calibration. The analysis of dynamic measurements, though, is based on a dynamic calibration of the measuring system. The realization of dynamic calibration requires completely different measurement approaches and mathematical tools than static calibration. In some areas first attempts have been made, but in general this constitutes a significant challenge for future research.
For the treatment of the above-mentioned challenges, the working group has developed mathematical methodologies for the
• GUM-compliant evaluation of dynamic calibration experiments,
• design of digital deconvolution filters for the compensation of dynamic errors,
• GUM-compliant evaluation of uncertainties for the application of digital filters,
• efficient implementation of GUM Monte Carlo for the application of digital filters,
• iterative estimation of spectral power distributions,
• regularized deconvolution, including GUM-compliant uncertainty evaluation.
# Software
In order to facilitate the application of the methods developed in the working group, suitable methods are available as free software implementations. For questions, remarks and suggestions, please contact Sascha Eichstädt.
PyDynamic - Software for the analysis of dynamic measurements
As part of the EMPIR project 14SIP08 Dynamic the PTB working group 8.42 together with the National Physical Laboratory (UK) develops a comprehensive Python software package, which contains a large variety of methods for the analysis of dynamic measurements.
Monte Carlo for dynamic measurements
The propagation of measurement uncertainties for dynamic measurements using the Monte Carlo propagation of distributions requires efficient implementation in order to achieve a high degree of accuracy, since the computer memory requirement for a standard implementation is very high. In the working group 8.42 a MATLAB software package has been developed to carry out Monte Carlo for dynamic measurements with high accuracy on standard desktop computers.
related publication
S. Eichstädt, A. Link, P. Harris and C. Elster (2012). Efficient implementation of a Monte Carlo method for uncertainty evaluation in dynamic measurements. Metrologia 49, 401-410
(doi:10.1088/0026-1394/49/3/401).
Spectral deconvolution with Richardson-Lucy
The correction of deviations in spectra measured with a spectrometer is often necessary in order to obtain accurate results. The classical approach for such a correction is based on a method by Stearns. However, it has been demonstrated that the Richardson-Lucy method can produce much better results here. Therefore, the PTB Working Group has written a software implementation of an adapted Richardson-Lucy method with an automatic stopping rule. This software includes MATLAB code, a Python code as well as a graphical user interface written in Python.
related publication
S. Eichstädt F. Schmähling G. Wübbeler, K. Anhalt, L. Bünger, K. Krüger and C. Elster (2013). Comparison of the Richardson-Lucy method and a classical approach for spectrometer bandpass correction. Metrologia 50, 107-118 (doi: 10.1088/0026-1394/50/2/107)
Evaluation of uncertainties in the application of the DFT
The Fourier transform and its counterpart for discrete time signals, the DFT, are common tools in measurement science and application. Although almost every scientific software package offers ready-to-use implementations of the DFT, the propagation of uncertainties in line with GUM is typically neglected. This is of particular importance in dynamic metrology, when input estimation is carried out by deconvolution in the frequency domain. To this end, we present the new open-source software tool GUM2DFT, which utilizes closed formulas for the efficient propagation of uncertainties for the application of the DFT, inverse DFT and input estimation in the frequency domain. It handles different frequency domain representations, accounts for autocorrelation and takes advantage of the symmetry inherent in the DFT result for real-valued time domain signals.
related publication
S. Eichstädt and V. Wilkens "GUM2DFT -- A software tool for uncertainty evaluation of transient signals in the frequency domain". Meas. Sci. Technol. 27(5), 055001, 2016 (doi: 10.1088/0957-0233/27/5/055001)
# Workshops
Together with the National Physical Laboratory (UK) and the Laboratoire national de métrologie et d'essais (France), the working group is organizing the workshop series "Analysis of Dynamic Measurements" as part of EURAMET TC-1078.
1. "Signal processing awareness seminar", NPL, UK, 2006
2. "Analysis of dynamic measurements", PTB, Germany, 2007
3. "Analysis of dynamic measurements", NPL, UK, 2008
4. Session TC21- Dynamical Measurements at IMEKO XIX World Congress, Portugal, 2009
5. "5th workshop on the analysis of dynamic measurements", SP, Sweden, 2010
6. "6th workshop on the analysis of dynamic measurements", Chalmers University, Sweden 2011
7. "7th workshop on the analysis of dynamic measurements", LNE, France, 2012
8. "8th workshop on the analysis of dynamic measurements" INRIM, Italy, 2014
9. "9th International workshop on the analysis of dynamic measurements", PTB, Germany, 2016
# Projects
running projects
completed projects
# Publications
## Publication single view
### Article
Title: The influence of different vibration exciter systems on high frequency primary calibration of single-ended accelerometers: II T. Bruns, A. Link and A. Täubner Metrologia 2012 49 1 27--31 IOP Publishing 10.1088/0026-1394/49/1/005 0026-1394 http://iopscience.iop.org/article/10.1088/0026-1394/49/1/005 dynamic calibration, accelerometer, dynamic measurement 8.42, Dynamik
Back to the list view | {"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": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6488810181617737, "perplexity": 906.8601311790717}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107891203.69/warc/CC-MAIN-20201026090458-20201026120458-00191.warc.gz"} |
http://www.ck12.org/book/CK-12-Math-Analysis/r19/section/2.7/ | <meta http-equiv="refresh" content="1; url=/nojavascript/">
You are reading an older version of this FlexBook® textbook: CK-12 Math Analysis Go to the latest version.
# 2.7: Approximating Real Zeros of Polynomial Functions
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Understand the statement of the Intermediate Value Theorem
• Apply the intermediate value theorem to find bounds on the zeros of a function
• Use numerical methods to find roots of a polynomial
## Intermediate Value Theorem & Bounds on Zeros
The intermediate value theorem offers one way to find roots of a continuous function. Recall that our informal definition of continuous is that a function is continuous over a certain interval if it has no breaks, jumps asymptotes, or holes in that interval. Polynomial functions are continuous for all real numbers $x$. Rational functions are often not continuous over the set of real numbers because of asymptotes or holes in the graph. But for intervals without holes, rational functions are continuous.
If we know a function is continuous over some interval $[a,b]$, then we can use the intermediate value theorem:
Intermediate Value Theorem
If $f(x)$ is continuous on some interval $[a,b]$ and $n$ is between $f(a)$ and $f(b)$, then there is some $c\in[a,b]$ such that $f(c)=n$.
The following graphs highlight how the intermediate value theorem works. Consider the graph of the function $f(x)=\frac{1}{4} \left ( x^{3}-\frac{5x^{2}}{2}-9x \right )$ below on the interval [-3, -1].
$f(-3)=-5.625$ and $f(-1)=1.375$. If we draw bounds on [-3, -1] and $[f(-3),f(-1)]$, then we see that for any $y-$value between $y=-5.625$ and $y=1.375$, there must be an $x$ value in [-3, -1] such that $f(x)=y$.
So, for example, if we choose $c=-2$, we know that for some $x\in[-3,-1], f(x)=-2,$ even though solving this by hand would be a chore!
## The Bounds on Zeros Theorem
The Bounds on Zeros Theorem is a corollary to the Intermediate Value Theorem:
Bounds on Zeros Theorem
If $f$ is continuous on $[a,b]$ and there is a sign change between $f(a)$ and $f(b)$ (that is, $f(a)$ is positive and $f(b)$ is negative, or vice versa), then there is a $c\in(a,b)$ such that $f(c)=0$.
The bounds on zeros theorem is a corollary to the intermediate value theorem because it is not fundamentally different from the general statement of the IVT, just a special case where $n=0$.
Looking back at $f(x)=\frac{1}{4} \left (x^{3}-\frac{5x^{2}}{2}-9x \right )$ above, because $f(-3)<0$ and $f(-1)>0$, we know that for some $x\in[-3,-1], f(x)$ has a root. In fact, that root is at $x=2$. and we can test that using synthetic division or by evaluating $f(2)$ directly.
Example 1
Show that $f(x)=-3x^{3}+5x$ has at least one root in the interval [1, 2]
Solution
Since $f(x)$ is a polynomial we know that it is continuous. $f(1)=2$ and $f(2)=-14$. Let $n=0\in[-14,2]$. Applying the Intermediate Value Theorem, there must exist some point $c\in[1,2]$ such that $f(c)=0$. This proves that $f(x)$ has a root in [1, 2].
Example 2
The table below shows several sample values of a polynomial $p(x)$.
$& x && -4 && -2 && \qquad 0 && \qquad 1 && \quad 4 && 6 && \qquad 8 && \qquad 10 && \qquad 15 && 18\\& p(x) && 44.15 && 6.62 && -4.12 && -4.09 && 1.16 && 0 && -8.74 && -24.07 && -49.89 && 3.41$
Based on the information in the table
(a) What is the minimum number of roots of $p(x)$?
(b) What are bounds on the roots of $p(x)$ that you identified in (a)?
Solution
Since $p(x)$ is a polynomial we already know that it is continuous. We can use the Intermediate Value Theorem to identify roots by looking at when $p(x)$ changes from negative to positive, or from positive to negative.
(a) There are four sign changes of $p(x)$ in the table, so at minimum, $p(x)$ has four roots.
(b) The roots are in the following intervals $x\in[-2,0], x\in[1,4], x\in[15,18],$ and the table also tells us that one root is at $x=6$.
## Approximate Zeros of Polynomials Functions
In calculus you will learn several methods for numerically approximating the roots of functions. In this section we show one elementary numerical method for finding the zeros of a polynomial which takes advantage of the Intermediate Value Theorem.
Given a continuous function $g(x)$,
1. Find two points such that $g(a)>0$ and $g(b)<0$. Once you have found these two points, you can iteratively use the steps below to find the root of $g(x)$ on the interval $[a,b]$. (Note, we will assume $a, the same algorithm works with minor adjustments if $b>a$)
2. Evaluate $g \left ( \frac{a+b}{2} \right )$.
1. If $g \left ( \frac{a+b}{2} \right )=0$, then the root is $x=\frac{a+b}{2}$.
2. If $g \left ( \frac{a+b}{2} \right )>0$, replace $a$ with $\frac{a+b}{2}$. and repeat steps 1-2 using $\left [ \frac{a+b}{2},b \right ]$
3. If $g\left ( \frac{a+b}{2} \right )<0$, replace $b$ with $\frac{a+b}{2}$. and repeat steps 1-2 using $\left [ a,\frac{a+b}{2} \right ]$
This algorithm will not usually find the exact root of $g(x)$, but it will allow you to find a reasonably small interval for the root. For example, you could repeat this process enough times so that you find an interval with $|a-b|<0.01$, and you will know the root of $g(x)$ within a reasonably good approximation. The quality of the approximation you use (and the number of steps you use) will depend on why you are looking for the root. For most applications coming within 0.01 of the root is a reasonable approximation, but for some applications (such as building a bridge or launching a rocket) you need much more accuracy.
Example 3
Show the first 5 iterations of finding the root of $h(x)=x^{2}-x-1$ using the starting values $a=0$ and $b=2$.
Solution:
1. First we verify that there is a root between $x=0$ and $x=2$. $h(0)=-1$ and $h(2)=1$ so we know there is a root in the interval [0, 2]. Check $h\left ( \frac{2+0}{2} \right )=h(1)=-1$. Since $-1<0$ we know the root is between $x=1$ and $x=2$, and we use the new interval [1, 2].
2. Now we use the interval [1, 2]. $h\left ( \frac{1+2}{2} \right )=h(1.5)=-0.25$. Since $-0.25<0$, we use the interval [1.5, 2].
3. $h\left ( \frac{1.5+2}{2} \right )=h(1.75)=0.31$. Since $0.31>0$, we know that the zero is in the interval [1.5, 1.75].
4. $h\left ( \frac{1.5+1.75}{2} \right )=h(1.625)=0.02$. Since $0.02>0$, we know the root is between 1.5 and 1.625.
5. $h\left ( \frac{1.5+1.625}{2} \right )=h(1.5620)=-0.12$. Since $-0.12<0$, we know the root is between 1.5620 and 1.625.
This example shows that after five iterations we have narrowed the possible location of the root to within 0.06 units. Not bad!
Recall that we have already reviewed using the CALC menu on a graphing calculator to find the roots of a function. This algorithm is not the one used by a calculator, but the calculator uses a similar, more efficient, algorithm for approximating the root of a function to 13 decimal places. When the calculator prompts for a GUESS? it is asking for a starting value to run the iterations.
## Optional: An Interesting Corollary of the IVT
One surprising result of the Intermediate Value Theorem is that if you draw any great circle around the globe, then there must two antipodal points on that great circle that have exactly the same temperature.
Recall that a great circle is a path around a sphere that gives the shortest distance between any two points on the sphere. The equator is a great circle around the globe. Antipodal points are two points on opposite sides of the sphere. In the diagram below, $B$ and $B'$ are antipodal.
For an informal proof of this result, look at the the image of a sphere with three great circles above. Suppose that the temperature at $B$ is $75^{\circ}$ and the temperature $B'$ is $50^{\circ}$. The difference between the temperature at $B$ and at $B'$ is $75-50=25$. Now imagine rotating the segment $\overline{BB'}$ around the blue great circle. When the segment has rotated 180 degrees (i.e. when $B$ has rotated to where $B'$ is), then The difference between the temperatures at these two points is $50-75=-25$. Since temperatures vary continuously, by the intermediate value theorem, there must be some point on that circle when the difference was 0, implying two antipodal points had the same temperature.
Notice that this little demonstration does not tell us which two antipodal points had the same temperature, only that there must be two such points on any great circle.
## Exercises
For Exercises 1-5 use the intermediate value theorem to show the bounds on the zeros of each function. Your bounds should be within the whole number
1. $f(x)=2x^{3}-3x+4$
2. $g(x)=-5x^{2}+8x+12$
3. $h(x)=\frac{1}{2}x^{4}-x^{3}-3x^{2}+1$
4. $j(x)=-\frac{2}{x^{2}+1}+\frac{1}{2}$
5. $k(x)$ is a polynomial and selected values of $k(x)$ are given in the following table: $& x && -3 && -2 && -1 && \ \ \ 0 && \ 1 && \ \ \ 2 && \ \ \ \ 3\\& k(x) && -23.5 && -1 && 0.5 && -1 && .5 && -1 && -23.5$
6. Stephen argues the function $r(x)=\frac{4x+1}{x+3.5}$ has two zeros based on the following table and an application of the Bounds on Zeros Theorem. What is faulty about Stephen's reasoning? $& x && -5 && -4 && -3 && \ -2 && \ -1 && \ 0 && \ 1 && \ 2 && \ 3 && \ 4\\& r(x) && 12.67 && \ 30 && -22 && -4.67 && -1.20 && 0.29 && 1.11 && 1.64 && 2.0 && 2.27$
7. Apply the numerical algorithm five times to find a bound on the zeros of the following functions given the indicated starting values. What is your final estimate for the zero?
1. $k(x)=x^{4}-3x+1$ on [0, 1]
2. $b(x)=-0.1x^{5}+3x^{3}-5x^{2}$ on [1, 3]
3. $c(x)=\frac{3x^{2}-2}{x^{4}+2}$ on [0, 2]
1. There is a zero in [-2, -1]
2. There is a zero in [-1, 0] and [2, 3]
3. There is a zero in [-2, -1], [-1, 0], [0, 1], and [3, 4]
4. There is a zero in [-2, -1] and [1, 2]
5. There is a zero in [-2, -1], [-1, 0], [0, 1] and [1, 2]
6. Unlike the previous question which specified the function was a polynomial (and hence continuous), $r(x)=\frac{4x+1}{x+3.5}$ has a vertical asymptote at $x=-3.5$, so it is not continuous in the interval [-4, -3]. Therefore we cannot use the Bounds on Zeros theorem to claim there is a zero in that interval.
1. The zero is in [0.3125, 0.34825]
2. The zero is in [1.5000, 1.5625]
3. [0.8125, 0.875]
Feb 23, 2012
Dec 29, 2014 | {"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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 120, "texerror": 0, "math_score": 0.9058856964111328, "perplexity": 327.1153483554235}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1430455241618.47/warc/CC-MAIN-20150501044041-00006-ip-10-235-10-82.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/how-large-is-the-collision-cross-section-compared-to-nuclear-fusion-cross-section.906837/ | # How large is the collision cross section compared to nuclear fusion cross section
• I
• Start date
• Tags
• #1
68
2
## Main Question or Discussion Point
Most Fusion reactors, and the leading ones like JET, use high temp. plasma and confine it. So, the plasma would approximate the Maxwell- Boltzmann distribution. This means that only a small portion of the plasma has enough energy to fuse. But, collisions are much more often, right? Since not all collisions result in fusion, the cross section of collision is much higher than the cross section of fusion. Is this true?
• But how much smaller is the cross section of fusion compared to the cross section of collisions. And do the electrons possess higher energy than the ions, in which case thee nuclear cross section would be orders of magnitude lower.
Am I right? If approximately all the collisions resulted in fusion, how much more energy would we be able to produce?
And what if there were 2 beam travelling toward each other. But these not just follow straight paths. There exists a HELICAL magnetic field and the particle spiral around the field lines and approach each other. Here, I found out that the particle spiral direction would be in opposite directions( one clock wise and the other anticlockwise.) If the beams interact when they meet, how would the collision cross section be affected? (Note: The particle possess reasonable energies at least higher than 25 Kev)
How would I calculate the collision cross section? I seriously need help. Thanks
Related High Energy, Nuclear, Particle Physics News on Phys.org
• #2
mfb
Mentor
34,476
10,598
This means that only a small portion of the plasma has enough energy to fuse. But, collisions are much more often, right? Since not all collisions result in fusion, the cross section of collision is much higher than the cross section of fusion. Is this true?
Right.
DT fusion has a maximal cross section of 5*10-28 m2 at about 60 keV center-of-mass energy.
Something like a collision happens if the nuclei come closer than ~200 fm, which corresponds to a cross section of 4*10-26 m2.
At a temperature of 10 keV, only a small fraction of nuclei collisions will lead to 60 keV center-of-mass energy, which reduces the effective fusion cross section even more.
Am I right? If approximately all the collisions resulted in fusion, how much more energy would we be able to produce?
The plasma would be too short-living then, and we would have to run it at a lower temperature.
If the beams interact when they meet, how would the collision cross section be affected?
The magnetic field is only relevant for beam steering, the actual collision is completely independent of that.
How would I calculate the collision cross section?
See the papers doing that.
Likes mheslep
• #3
68
2
For the 4*10^-26 m squared cross section, what is the cross section for? and how did we end up with it? and an other thing I was wondering about was that I said we had a helical magnetic field. I've figured out that the particles spin in opposite directions during collision. Doesn't this mean that almost all of these are head on collisions? I think this would increase the fusion rate, wouldn't it? And these particle beams are not in thermal equilibrium and hence all have approximately the same energy, so all collisions would result in fusion right? So.. according to what you just said the cross section wouldn't change. So, am I safe in assuming these assumptions are accurate?
• #4
68
2
And approximately how many more collisions do we have compared to fusion?
• #5
mfb
Mentor
34,476
10,598
For the 4*10^-26 m squared cross section, what is the cross section for?
For any collision. Those elastic collisions don't have a well-defined total cross section - larger impact parameters will just lead to smaller deflections. See Rutherford scattering. I used a distance where the deflection will be small (few degrees).
Doesn't this mean that almost all of these are head on collisions?
If you have two well-defined beams, you don't even need magnetic fields for that. The fusion rate will be low due to the low achievable particle density in the beams.
Even at the ideal energy for fusion, most collisions won't lead to fusion, see the comparison of the numbers in post 2. Most particles would be scattered out of the beam without fusing.
• #6
68
2
What if low density beams bombard a high density plasma?
• #7
mfb
Mentor
34,476
10,598
Then you still lose most of the particles due to elastic scattering.
You can get fusion with accelerator systems, and accelerator-based fusion is a convenient source of neutrons. You just do not get enough to use it as power source because you have to put in much more energy than you get out.
• #8
68
2
But then, why doesn't it happen in plasmas. I know it happens, but I think the effect is much smaller. Hence, they don't loose energy through coulomb scattering. Is this right?
• #9
mfb
Mentor
34,476
10,598
But then, why doesn't it happen in plasmas. I know it happens, but I think the effect is much smaller.
Which effect is smaller?
The energy is not lost in either case. In a plasma, collisions between atoms just redistribute the energy within the plasma particles - a completely normal thermal process. In beams of cold atoms, it means some particles get scattered out of the beam and the beam gets hotter.
• #10
68
2
• #11
68
2
Recently, I have learn't about the Gamows factor, and I have plotted it. Then I realized that the probability of Deuterium and tritium fusing at a temperature of 150 million kelvin when they collide, was about 0.00048. But, I know that the nuclear cross section for fusion is extremely small (about 1-1000 barns). But why is this so? I f the probability of fusion is 0.00048 when they collide, then why is the nuclear cross section so small? Is it because of the collisions? Are the collisions really that rare? Is the collision cross section really that small? Why is the nuclear cross section so small?
• Here the x-axis is the temperature and the y-axis is the probability of fusion when a collision occurs.
• #12
mfb
Mentor
34,476
10,598
Why do you expect a large cross section? To actually hit the nuclei in a classical sense you would need about 2 MeV - but then you have to hit exactly, which is incredibly unlikely as nuclei are tiny (smaller than 1 barn). Everything below that has the nuclei approach each other, and then you have to hope that they tunnel to fuse before they fly apart again. That process is not very frequent either.
• #13
68
2
Does the ion-ion collision rate or collision frequency increase or decrease with an increase in temperature? I've read that most plasmas could be considered collisionless or have little collisions taking place.
• #14
mfb
Mentor
34,476
10,598
For a constant density, the collision rate increases with temperature, simply because the particles move faster.
• #15
68
2
Now, for a plasma put under a constant pressure, the molar density multiplied by the temperature is always a constant which equals Pressure/Gas constant. According to the ideal has equation. So the collision rate is a constant for a constant pressure right?
• #16
mfb
Mentor
34,476
10,598
At constant pressure: double the temperature and you double the path length between collisions, but the speed only grows with a factor sqrt(2) - collisions get less frequent. The chance of fusion increases more up to some point, so increasing temperature is still advisable.
• #17
68
2
So thats why the cross section drops after 150 million kelvin for D-T fusion? The collisions become less frequent?
• #18
68
2
The velocity you considered is the average velocity isn't it?
• #19
mfb
Mentor
34,476
10,598
So thats why the cross section drops after 150 million kelvin for D-T fusion? The collisions become less frequent?
The cross section itself decreases as well if the energy gets too high, this is a more important effect.
The velocity you considered is the average velocity isn't it?
It is true for all speeds. Average, median, top 0.01%, ...
• #20
68
2
But why does the temperature become too high? Isn't it going to help us overcome the coulomb barrier? The reason I think it can get too high is through losses. Is there a quantum mechanical effect that affects the cross section at extremely high temperatures?
• #21
mfb
Mentor
34,476
10,598
The reason I think it can get too high is through losses.
The cross section has nothing to do with losses. It has nothing to do with temperature either. For the cross section, you just have two nuclei colliding with a given center of mass energy.
Faster nuclei lead to a shorter collision time, that could contribute.
• #22
68
2
Oh. So time matters?
• #23
68
2
An other question: If I accelerates particles, and make them hit a plasma target, the portion of the particles that actually collide with the nucleus' described by the collision cross section either fuse or lose their energy in elastic collisions. So, the percentage of particles that fuse are described Gamow equation. Right? The rest just lose their energy through elastic collisions and braking radiation. Is this right? ....So, in this case, what exactly is a collision? How close are particles supposed to get in order for the interaction to be considered a collision?
• #24
mfb
Mentor
34,476
10,598
Oh. So time matters?
Sure.
If you use an accelerator, there is no need to have plasma target. A regular target (solid, liquid, gas) will work as well.
So, the percentage of particles that fuse are described Gamow equation.
That just describes the fusion cross section, not how the energy of the particles evolves if they don't fuse.
Some fraction will fuse with the first nucleus they get close to. Others will scatter once and fuse with the second one. Others will scatter twice and fuse with the third one. And so on, but fusion gets less likely over time as the nuclei lose energy. Most will just scatter and never fuse.
So, in this case, what exactly is a collision? How close are particles supposed to get in order for the interaction to be considered a collision?
As mentioned before, the lower cutoff for calling something collision is a bit arbitrary. That doesn't change the physics, just the names we assign to things.
• #25
68
2
Yeah. I understand now. Thanks
• Last Post
Replies
2
Views
256
• Last Post
Replies
1
Views
3K
• Last Post
Replies
4
Views
5K
• Last Post
Replies
13
Views
2K
• Last Post
Replies
1
Views
811
• Last Post
Replies
1
Views
647
• Last Post
Replies
9
Views
923
• Last Post
Replies
2
Views
4K
• Last Post
Replies
7
Views
3K
• Last Post
Replies
2
Views
3K | {"extraction_info": {"found_math": false, "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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9321826696395874, "perplexity": 843.7742971755493}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439738819.78/warc/CC-MAIN-20200811180239-20200811210239-00478.warc.gz"} |
http://nbviewer.jupyter.org/github/barbagroup/AeroPython-studentprojects/blob/master/GeorgeTeel/Orlyonok%20Ekranoplan.ipynb | ###### MAE 6226, 5/5/2015, the George Washington University. Written by George Teel¶
<img src="./BannerLow.png" width = "1980">
<img src="./TitleThin.png" width = "1024">
Hello everyone!
This project has been inspired by an absolutely beautiful aircraft, known as the Caspian Seamonster (KM). The origins of this specific type of craft was a military sea plane that the United States of America (USA) discovered being built during the cold war by the Union of Soviet Socialist Republics (USSR) in 1966[1]. This Ekranoplan has a few variants with the first few being military versions known as the Lun Class, having antiship missiles attached to it. A plane designed for invasion, quick agile transportation, and efficiency for military use, the ekranoplan was Russia's answer to their military needs[2]. But as we all know, the cold war was an arms race, and many technologies such as this one were cancelled and mostly forgotten to rot away, the pilots included sadly. This project more specficially will be a case study of the A-90 Orlyonok (Орлёнок), which was designed to be a modern non militarized transport vehicle. These beautiful crafts are technological wonders as they are a combination of both sea technologies and aerospace. An intense study of of materials, structural engineering, airfoil design, Wing in Ground effect, and engines in salt water environments all designed to create these amazing constructs. Here's a quick video of the military "Caspian Seamonster" in action, to pique some interest!
In [8]:
from IPython.display import YouTubeVideo
Out[8]:
And if you want some good background music, I have provided a link at the very bottom of the page or just click here! Feel free to read and listen if you want for some good atmosphere, I highly suggest it.
<img src="./BannerThin.png" width = "1980">
## The Concept¶
The fundamental design behind the ekranoplan is utilizing ground effect.
"What is ground effect" you might be asking? It's really a simple concept, or atleast can be simplified. Imagining the fluid flow around an airfoil, vortices get created, and the wind wake is formed behind the wings. But, when one flies close to the ground, these flows get disrupted which change the induced drag.
Taken from Federation of American Scientists website: "Wing-In-Ground (WIG) effect craft take advantage the fact that the aerodynamic efficiency of a wing, and particularly its lifting capacity, improves dramatically when is operated within approximately one-half of its span above ground or water, in what is termed ground effect."[3]
Taken from a Google translated page of a russian paper: "lift can reach 50%, the increase aerodynamic efficiency (the ratio of lift to the drag force) - 1, 5 ~ 2, 5 or more times. Influence of the screen on the wing - a very complex physical phenomenon, and full of clarity and understanding the mechanism of this effect yet."[4]
Another conceptual image is an air cushion that gets formed under the wings which makes crafts fly longer or "hover" over the ground due to the pressure differential. This affects all flying crafts (and birds too!) and can cause variations (potentially problems) while landing. In this case, these Wing-In-Ground Effect (WIG) crafts utilize this cushion to "float" on, self propelling themselves to a more efficient flight to consume less fuel.
The ekranoplans use forward engines of the craft to ram air underneath the wings to generate that cushion faster, then they level out to keep a coast flight. Here's a video of the Orlyonok in action. It's in russian however, but if you skip ahead to 1:45 in the video, you can watch the animated concepts:
In [9]:
from IPython.display import YouTubeVideo
Out[9]:
"But why on earth do you need 8 engines on that Caspian monstrosity?!?" Well, to overcome the friction of the water of course!!! Both the CM and A-90 crafts are incredibly heavy as they were designed for transport and assault, and not to mention the same size or larger than a Boeing 777! The friction of the water slows down the crafts so much that they have a hard time getting off the water. But, once they make it into the air, these engines get shut off or throttled back to consume less fuel and reduce salt water intake. Pretty nifty huh?
So, how do we go about analyzing an aircraft built during the coldwar, which was held secret for years? And I am sure it still is, as I have not found any real information about the airfoils or any published data. Google of course! My search far and wide has lead me down a path through Google to these drawings right here.
Razlib.ru [4]
This website appears to be a paper published that has some very interesting background information on it. If one is curious to learn more, follow the link and use Google to translate that page into english. Thats what I did since my Russian is non-existant!
<img src="./BannerThin.png" width = "1980">
So now that we have our drawings, what's the next step to digitizing? Using a free opensource 3d modeling program of course! My alltime favorite, and one I've used for years is Blender3D.
Not going into super details about the process, but what I did in a nutshell was:
• Import the drawings
• Created both the airfoils and the body shape in Blender3D using Bezier curves
• Converted the curves to a mesh
• Wrote a custom Python script which exports the vertices
• Correctly import the proper way XFLR5 requires
These steps have taken the longest. Days and days have been spent figuring this part out, as XFLR5 is a bit temperamental with how it wants the correct inputs. Some images below show the steps I took.
Fitting the Curves!
Curve the Airfoil. Make sure you do Trailing edge over the top to the Leading edge, then back around again.
Beginning the 2D cross section production.
The Meshes!
The Mesh 2D cross sections
The Mesh 2D cross sections into the plane shape which is our 3D model.
Finally, XFLR5 models!
The final airfoil shape in XFLR5.
The completed craft, airfoils, body, elevators and fins. Beautiful...
One might notice that the imported XFLR5 model of the body is alittle bit different than the image above of the blender file with all those vertex points. The reason being is that I had to make a simplified version due to XFLR5 crashing! Here's an image of the high resolution model before it would shut down every single time....
A word of warning to those wishing to follow my steps.... don't make your mesh too big! It will load, then stutter, then crash.... wings too! So don't put too many panels on your wings and body else you won't get very far.
<img src="./BannerThin.png" width = "1980">
## Wing Studies¶
As I mentioned in the previous section, this entire case study has taken place in XFLR5. A free program which allows enthusiasts and hobbyists to analyze model planes, airfoils, and anyone who feels like playing around with an amazing program.
With the blendering all done and model files exported and imported, it's time to have some fun and run some simulations! A few errors occured when I tried to make this foil the actual size of the aircraft though, and I gave up on trying to simulate the full scale model. XFLR5 just can't handle the computations (or perhaps my computer can't) as I believe this was designed for model aircrafts (well, it's primarily being used for them currently anyways). Therefore, throughout all these simulations, I have run a smaller version of the craft (about 10 times smaller) created from the diagrams above. The initial runs were very preliminary to get a feeling and understanding of the program, and to the best of my abilities, I have used XFLR5 to get some results. Also, a short note is that the type of simulations run on both the wings and the body were Horshoe Vortex (VLM1) analyses with viscous effects.
Shown below is an animation after running the batch analysis and the initial trial runs:
Lovely isn't it? How I love colored animated things...
Moving on to the simulations though! For the wing, studies of different velocities (2.5, 5, 10 m/s) have been done varying the angle of attack and height in ground effect to show the differences it makes. I believe this is a good start of a potentially long study. I took into account the low speeds for the model aircraft, the different heights in increments of 0.1 m, and the various angle of attack since we don't know the accuracy of those drawings. I think this was reasonable don't you?
Following this class's Python notebooks ways, lets import the basics and begin shall we?
In [10]:
import math
import numpy
import csv
from scipy import integrate
from matplotlib import pyplot
Because I love color, this is my own colorscheme for the graphs. An absolutely wonderful free tool for those budding artists out there is Paletton. A good colorscheme website to bring those coordinated colors to life.
In [11]:
#Creating Colors for graphs
color05 = '#791532'
color10 = '#C2607C'
color15 = '#864E18'
color20 = '#D7A06A'
color25 = '#114653'
color30 = '#447985'
color35 = '#3F7715'
color40 = '#88BF5E'
color45 = '#29195C'
color50 = '#625294'
Then reading in that data and plotting the graphs!
In [12]:
#reads of the Airfoil geometries from the cvs data files
alpha00, G00 = numpy.loadtxt('./2_5ms/00.csv', dtype=float, delimiter=',', unpack=True)
alpha02, G02 = numpy.loadtxt('./2_5ms/02.csv', dtype=float, delimiter=',', unpack=True)
alpha03, G03 = numpy.loadtxt('./2_5ms/03.csv', dtype=float, delimiter=',', unpack=True)
alpha04, G04 = numpy.loadtxt('./2_5ms/04.csv', dtype=float, delimiter=',', unpack=True)
alpha05, G05 = numpy.loadtxt('./2_5ms/05.csv', dtype=float, delimiter=',', unpack=True)
alpha06, G06 = numpy.loadtxt('./2_5ms/06.csv', dtype=float, delimiter=',', unpack=True)
alpha07, G07 = numpy.loadtxt('./2_5ms/07.csv', dtype=float, delimiter=',', unpack=True)
alpha09, G09 = numpy.loadtxt('./2_5ms/09.csv', dtype=float, delimiter=',', unpack=True)
alpha12, G12 = numpy.loadtxt('./2_5ms/12.csv', dtype=float, delimiter=',', unpack=True)
alpha17, G17 = numpy.loadtxt('./2_5ms/17.csv', dtype=float, delimiter=',', unpack=True)
# #lets plot the geometry
%matplotlib inline
#Plotting
size = 10
pyplot.figure(figsize=(size,size))# (y_end-y_start)/(x_end-x_start)*size))
pyplot.grid(True)
pyplot.xlabel('(Alpha) Angle of Attack', fontsize=16)
pyplot.ylabel('(CL) Coefficient of Lift', fontsize=16)
pyplot.title('CL vs Alpha at 2.5 m/s', fontsize=20)
pyplot.xlim(0, 10.0)
pyplot.ylim(0.2, 1.0)
pyplot.plot(alpha00, G00, label="No Ground Effect",
color=color05, linestyle='--', linewidth=6);
pyplot.plot(alpha02, G02, label="0.2 m",
color=color10, linestyle='-', linewidth=4);
pyplot.plot(alpha03, G03, label="0.3 m",
color=color15, linestyle='-', linewidth=4);
pyplot.plot(alpha04, G04, label="0.4 m",
color=color20, linestyle='-', linewidth=4);
pyplot.plot(alpha05, G05, label="0.5 m",
color=color25, linestyle='-', linewidth=4);
pyplot.plot(alpha06, G06, label="0.6 m",
color=color30, linestyle='-', linewidth=4);
pyplot.plot(alpha07, G07, label="0.7 m",
color=color35, linestyle='-', linewidth=4);
pyplot.plot(alpha09, G09, label="0.9 m",
color=color40, linestyle='-', linewidth=4);
pyplot.plot(alpha12, G12, label="1.2 m",
color=color45, linestyle='-', linewidth=4);
pyplot.plot(alpha17, G17, label="1.7 m",
color=color50, linestyle='-', linewidth=4);
pyplot.legend(bbox_to_anchor=(1.05, 1), loc=2, borderaxespad=0.);
Looking at this first study at 2.5 m/s shows the most variation (though most are about the same). I have done the study for angles of attack from 0 to 10, but in reality, the Ekranoplan would never really go above 5, so we could just cut it off and focus on the left side of the graph.
You can see that the no-ground-effect line is the lowest of all curves, next to 1.7 m for height. As stated above in the background information section, the ground effect increases your lift! Pretty drastically of course!
Putting the segmented curves aside, follow along in your head with me. As the craft takes off it will be lowest to the ground experiencing the most ground effect possible, but very quickly it will rise up through the meters (just like planes taking off). So this quick cushion forms and ground effect is present to carry the wings off which is why the lower to the ground, we are seeing the Coefficient of Lift (CL) be much higher. But there is a limitation, and the higher you go, the craft loses the ground effect. According to these simulations, being roughly 1.7 m above the ground resulting in the same effects as the simulation with no ground effect. Makes sense since it's pretty high for a smaller model airplane to be experiencing ground effect.
So looking at this graph, imagine the plane flying in the 2 to 4 angle of attack range (we don't want the tail to hit the water now do we!) and the CL is much much higher at these values vs the non ground effect. Near the tailing edge of 0.4 meters up, that's a CL of about 0.75 compared to the no-ground-effect CL of about 0.52! A massive improvement for no extra cost. This is where the efficiency comes into play. These crafts can float on their cushion, propelling themselves forward with less effort and less fuel costs.
Let's read on for further studies!
In [13]:
#reads of the Airfoil geometries from the cvs data files
alpha00, G00 = numpy.loadtxt('./3ms/00.csv', dtype=float, delimiter=',', unpack=True)
alpha02, G02 = numpy.loadtxt('./3ms/02.csv', dtype=float, delimiter=',', unpack=True)
alpha03, G03 = numpy.loadtxt('./3ms/03.csv', dtype=float, delimiter=',', unpack=True)
alpha04, G04 = numpy.loadtxt('./3ms/04.csv', dtype=float, delimiter=',', unpack=True)
alpha05, G05 = numpy.loadtxt('./3ms/05.csv', dtype=float, delimiter=',', unpack=True)
alpha06, G06 = numpy.loadtxt('./3ms/06.csv', dtype=float, delimiter=',', unpack=True)
alpha07, G07 = numpy.loadtxt('./3ms/07.csv', dtype=float, delimiter=',', unpack=True)
alpha09, G09 = numpy.loadtxt('./3ms/09.csv', dtype=float, delimiter=',', unpack=True)
alpha11, G10 = numpy.loadtxt('./3ms/10.csv', dtype=float, delimiter=',', unpack=True)
alpha16, G16 = numpy.loadtxt('./3ms/16.csv', dtype=float, delimiter=',', unpack=True)
# #lets plot the geometry
%matplotlib inline
#Plotting
size = 10
pyplot.figure(figsize=(size,size))# (y_end-y_start)/(x_end-x_start)*size))
pyplot.grid(True)
pyplot.xlabel('(Alpha) Angle of Attack', fontsize=16)
pyplot.ylabel('(CL) Coefficient of Lift', fontsize=16)
pyplot.title('CL vs Alpha at 3 m/s', fontsize=20)
pyplot.xlim(0, 10.0)
pyplot.ylim(0.2, 1.0)
pyplot.plot(alpha00, G00, label="No Ground Effect",
color=color05, linestyle='--', linewidth=6);
pyplot.plot(alpha02, G02, label="0.2 m",
color=color10, linestyle='-', linewidth=4);
pyplot.plot(alpha03, G03, label="0.3 m",
color=color15, linestyle='-', linewidth=4);
pyplot.plot(alpha04, G04, label="0.4 m",
color=color20, linestyle='-', linewidth=4);
pyplot.plot(alpha05, G05, label="0.5 m",
color=color25, linestyle='-', linewidth=4);
pyplot.plot(alpha06, G06, label="0.6 m",
color=color30, linestyle='-', linewidth=4);
pyplot.plot(alpha07, G07, label="0.7 m",
color=color35, linestyle='-', linewidth=4);
pyplot.plot(alpha09, G09, label="0.9 m",
color=color40, linestyle='-', linewidth=4);
pyplot.plot(alpha12, G10, label="1.1 m",
color=color45, linestyle='-', linewidth=4);
pyplot.plot(alpha17, G16, label="1.6 m",
color=color50, linestyle='-', linewidth=4);
pyplot.legend(bbox_to_anchor=(1.05, 1), loc=2, borderaxespad=0.);
Moving at 3 m/s is showing an even further difference as the curves continue. Following that 4 degree angle of attack we are looking at No Ground Effect CL value of 0.55, where the CL of 0.4 m is about 0.85! Huge improvement!
In [14]:
#reads of the Airfoil geometries from the cvs data files
alpha00, G00 = numpy.loadtxt('./10ms/00.csv', dtype=float, delimiter=',', unpack=True)
alpha02, G02 = numpy.loadtxt('./10ms/02.csv', dtype=float, delimiter=',', unpack=True)
alpha03, G03 = numpy.loadtxt('./10ms/03.csv', dtype=float, delimiter=',', unpack=True)
alpha04, G04 = numpy.loadtxt('./10ms/04.csv', dtype=float, delimiter=',', unpack=True)
alpha05, G05 = numpy.loadtxt('./10ms/05.csv', dtype=float, delimiter=',', unpack=True)
alpha06, G06 = numpy.loadtxt('./10ms/06.csv', dtype=float, delimiter=',', unpack=True)
alpha07, G07 = numpy.loadtxt('./10ms/07.csv', dtype=float, delimiter=',', unpack=True)
alpha09, G09 = numpy.loadtxt('./10ms/09.csv', dtype=float, delimiter=',', unpack=True)
alpha12, G12 = numpy.loadtxt('./10ms/12.csv', dtype=float, delimiter=',', unpack=True)
alpha17, G17 = numpy.loadtxt('./10ms/17.csv', dtype=float, delimiter=',', unpack=True)
# #lets plot the geometry
%matplotlib inline
#Plotting
size = 10
pyplot.figure(figsize=(size,size))# (y_end-y_start)/(x_end-x_start)*size))
pyplot.grid(True)
pyplot.xlabel('(Alpha) Angle of Attack', fontsize=16)
pyplot.ylabel('(CL) Coefficient of Lift', fontsize=16)
pyplot.title('CL vs Alpha at 5 m/s', fontsize=20)
pyplot.xlim(0, 10.0)
pyplot.ylim(0.2, 1.0)
pyplot.plot(alpha00, G00, label="No Ground Effect",
color=color05, linestyle='--', linewidth=6);
pyplot.plot(alpha02, G02, label="0.2 m",
color=color10, linestyle='-', linewidth=4);
pyplot.plot(alpha03, G03, label="0.3 m",
color=color15, linestyle='-', linewidth=4);
pyplot.plot(alpha04, G04, label="0.4 m",
color=color20, linestyle='-', linewidth=4);
pyplot.plot(alpha05, G05, label="0.5 m",
color=color25, linestyle='-', linewidth=4);
pyplot.plot(alpha06, G06, label="0.6 m",
color=color30, linestyle='-', linewidth=4);
pyplot.plot(alpha07, G07, label="0.8 m",
color=color35, linestyle='-', linewidth=4);
pyplot.plot(alpha09, G09, label="1.0 m",
color=color40, linestyle='-', linewidth=4);
pyplot.plot(alpha12, G12, label="1.5 m",
color=color45, linestyle='-', linewidth=4);
pyplot.plot(alpha17, G17, label="2.1 m",
color=color50, linestyle='-', linewidth=4);
pyplot.legend(bbox_to_anchor=(1.05, 1), loc=2, borderaxespad=0.);
In [15]:
#reads of the Airfoil geometries from the cvs data files
alpha00, G00 = numpy.loadtxt('./5ms/00.csv', dtype=float, delimiter=',', unpack=True)
alpha02, G02 = numpy.loadtxt('./5ms/02.csv', dtype=float, delimiter=',', unpack=True)
alpha03, G03 = numpy.loadtxt('./5ms/03.csv', dtype=float, delimiter=',', unpack=True)
alpha04, G04 = numpy.loadtxt('./5ms/04.csv', dtype=float, delimiter=',', unpack=True)
alpha05, G05 = numpy.loadtxt('./5ms/05.csv', dtype=float, delimiter=',', unpack=True)
alpha06, G06 = numpy.loadtxt('./5ms/06.csv', dtype=float, delimiter=',', unpack=True)
alpha08, G08 = numpy.loadtxt('./5ms/08.csv', dtype=float, delimiter=',', unpack=True)
alpha10, G10 = numpy.loadtxt('./5ms/10.csv', dtype=float, delimiter=',', unpack=True)
alpha15, G15 = numpy.loadtxt('./5ms/15.csv', dtype=float, delimiter=',', unpack=True)
alpha21, G21 = numpy.loadtxt('./5ms/21.csv', dtype=float, delimiter=',', unpack=True)
# #lets plot the geometry
%matplotlib inline
#Plotting
size = 10
pyplot.figure(figsize=(size,size))# (y_end-y_start)/(x_end-x_start)*size))
pyplot.grid(True)
pyplot.xlabel('(Alpha) Angle of Attack', fontsize=16)
pyplot.ylabel('(CL) Coefficient of Lift', fontsize=16)
pyplot.title('CL vs Alpha at 10 m/s', fontsize=20)
pyplot.xlim(0, 10.0)
pyplot.ylim(0.2, 1.0)
pyplot.plot(alpha00, G00, label="No Ground Effect",
color=color05, linestyle='--', linewidth=6);
pyplot.plot(alpha02, G02, label="0.2 m",
color=color10, linestyle='-', linewidth=4);
pyplot.plot(alpha03, G03, label="0.3 m",
color=color15, linestyle='-', linewidth=4);
pyplot.plot(alpha04, G04, label="0.4 m",
color=color20, linestyle='-', linewidth=4);
pyplot.plot(alpha05, G05, label="0.5 m",
color=color25, linestyle='-', linewidth=4);
pyplot.plot(alpha06, G06, label="0.6 m",
color=color30, linestyle='-', linewidth=4);
pyplot.plot(alpha08, G08, label="0.8 m",
color=color35, linestyle='-', linewidth=4);
pyplot.plot(alpha10, G10, label="1.0 m",
color=color40, linestyle='-', linewidth=4);
pyplot.plot(alpha15, G15, label="1.5 m",
color=color45, linestyle='-', linewidth=4);
pyplot.plot(alpha21, G21, label="2.1 m",
color=color50, linestyle='-', linewidth=4);
pyplot.legend(bbox_to_anchor=(1.05, 1), loc=2, borderaxespad=0.);
It appears not much variation is happening between the 5 m/s and the 10 m/s but there are differences. This again is most likely chalked up to the custom airfoil or not the correct properties for the study or ground effect not being a strong element of XFLR5. But it is a good representation still of what we are trying to study.
So quite intriguing I think, and I understand why the crafts were built for this. Truly a fun concept.
## Plane Studies¶
Now, this is where the fun really began for me, and the headaches. In the program itself, it says adding the different airplane parts were not the best idea for simulation results, but I had to do it anyways. Not only having the custom aircraft imported was fun to see, but to see results was a great feeling too.
Simluations run on the entire aircraft were run at 5, 10 and 20 m/s with varied heights and angle of attacks ranging from -5 to 5. I don't believe the plane was designed to fly with angles higher than that, so I limited the range.
In [16]:
#reads of the Airfoil geometries from the cvs data files
alpha00, G00 = numpy.loadtxt('./Full5ms/00.csv', dtype=float, delimiter=',', unpack=True)
alpha05, G05 = numpy.loadtxt('./Full5ms/05.csv', dtype=float, delimiter=',', unpack=True)
alpha06, G06 = numpy.loadtxt('./Full5ms/06.csv', dtype=float, delimiter=',', unpack=True)
alpha07, G07 = numpy.loadtxt('./Full5ms/07.csv', dtype=float, delimiter=',', unpack=True)
alpha08, G08 = numpy.loadtxt('./Full5ms/08.csv', dtype=float, delimiter=',', unpack=True)
alpha09, G09 = numpy.loadtxt('./Full5ms/09.csv', dtype=float, delimiter=',', unpack=True)
alpha10, G10 = numpy.loadtxt('./Full5ms/10.csv', dtype=float, delimiter=',', unpack=True)
# #lets plot the geometry
%matplotlib inline
#Plotting
size = 10
pyplot.figure(figsize=(size,size))# (y_end-y_start)/(x_end-x_start)*size))
pyplot.grid(True)
pyplot.xlabel('(Alpha) Angle of Attack', fontsize=16)
pyplot.ylabel('(CL) Coefficient of Lift', fontsize=16)
pyplot.title('CL vs Alpha at 5 m/s', fontsize=20)
pyplot.xlim(-3.0, 1.5)
pyplot.ylim(0.0, 0.5)
pyplot.plot(alpha00, G00, label="No Ground Effect",
color=color05, linestyle='--', linewidth=6);
pyplot.plot(alpha05, G05, label="0.5 m",
color=color10, linestyle='-', linewidth=4);
pyplot.plot(alpha06, G06, label="0.6 m",
color=color15, linestyle='-', linewidth=4);
pyplot.plot(alpha07, G07, label="0.7 m",
color=color20, linestyle='-', linewidth=4);
pyplot.plot(alpha08, G08, label="0.8 m",
color=color25, linestyle='-', linewidth=4);
pyplot.plot(alpha09, G09, label="0.9 m",
color=color30, linestyle='-', linewidth=4);
pyplot.plot(alpha10, G10, label="1.0 m",
color=color35, linestyle='-', linewidth=4);
pyplot.legend(bbox_to_anchor=(1.05, 1), loc=2, borderaxespad=0.);
By looking at this first graph, one can note that the trend of this project continues through these studies as well. With ground effect, the airplane performs under the same conditions better than without ground effect. In this case, not as large as the airfoil itself, but still an improvement.
In [17]:
#reads of the Airfoil geometries from the cvs data files
alpha00, G00 = numpy.loadtxt('./Full10ms/00.csv', dtype=float, delimiter=',', unpack=True)
alpha04, G04 = numpy.loadtxt('./Full10ms/04.csv', dtype=float, delimiter=',', unpack=True)
alpha05, G05 = numpy.loadtxt('./Full10ms/05.csv', dtype=float, delimiter=',', unpack=True)
alpha06, G06 = numpy.loadtxt('./Full10ms/06.csv', dtype=float, delimiter=',', unpack=True)
alpha07, G07 = numpy.loadtxt('./Full10ms/07.csv', dtype=float, delimiter=',', unpack=True)
alpha08, G08 = numpy.loadtxt('./Full10ms/08.csv', dtype=float, delimiter=',', unpack=True)
alpha09, G09 = numpy.loadtxt('./Full10ms/09.csv', dtype=float, delimiter=',', unpack=True)
alpha10, G10 = numpy.loadtxt('./Full10ms/10.csv', dtype=float, delimiter=',', unpack=True)
# #lets plot the geometry
%matplotlib inline
#Plotting
size = 10
pyplot.figure(figsize=(size,size))# (y_end-y_start)/(x_end-x_start)*size))
pyplot.grid(True)
pyplot.xlabel('(Alpha) Angle of Attack', fontsize=16)
pyplot.ylabel('(CL) Coefficient of Lift', fontsize=16)
pyplot.title('CL vs Alpha at 10 m/s', fontsize=20)
pyplot.xlim(-3.0, 5.0)
pyplot.ylim(0.0, 0.7)
pyplot.plot(alpha00, G00, label="No Ground Effect",
color=color05, linestyle='--', linewidth=6);
pyplot.plot(alpha04, G04, label="0.4 m",
color=color10, linestyle='-', linewidth=4);
pyplot.plot(alpha05, G05, label="0.5 m",
color=color15, linestyle='-', linewidth=4);
pyplot.plot(alpha06, G06, label="0.6 m",
color=color20, linestyle='-', linewidth=4);
pyplot.plot(alpha07, G07, label="0.7 m",
color=color25, linestyle='-', linewidth=4);
pyplot.plot(alpha08, G08, label="0.8 m",
color=color30, linestyle='-', linewidth=4);
pyplot.plot(alpha09, G09, label="0.9 m",
color=color35, linestyle='-', linewidth=4);
pyplot.plot(alpha10, G10, label="1.0 m",
color=color40, linestyle='-', linewidth=4);
pyplot.legend(bbox_to_anchor=(1.05, 1), loc=2, borderaxespad=0.);
In [18]:
#reads of the Airfoil geometries from the cvs data files
alpha00, G00 = numpy.loadtxt('./Full20ms/00.csv', dtype=float, delimiter=',', unpack=True)
alpha04, G04 = numpy.loadtxt('./Full20ms/04.csv', dtype=float, delimiter=',', unpack=True)
alpha05, G05 = numpy.loadtxt('./Full20ms/05.csv', dtype=float, delimiter=',', unpack=True)
alpha06, G06 = numpy.loadtxt('./Full20ms/06.csv', dtype=float, delimiter=',', unpack=True)
alpha07, G07 = numpy.loadtxt('./Full20ms/07.csv', dtype=float, delimiter=',', unpack=True)
alpha08, G08 = numpy.loadtxt('./Full20ms/08.csv', dtype=float, delimiter=',', unpack=True)
alpha09, G09 = numpy.loadtxt('./Full20ms/09.csv', dtype=float, delimiter=',', unpack=True)
alpha10, G10 = numpy.loadtxt('./Full20ms/10.csv', dtype=float, delimiter=',', unpack=True)
# #lets plot the geometry
%matplotlib inline
#Plotting
size = 10
pyplot.figure(figsize=(size,size))# (y_end-y_start)/(x_end-x_start)*size))
pyplot.grid(True)
pyplot.xlabel('(Alpha) Angle of Attack', fontsize=16)
pyplot.ylabel('(CL) Coefficient of Lift', fontsize=16)
pyplot.title('CL vs Alpha at 20 m/s', fontsize=20)
pyplot.xlim(-3.0, 5.5)
pyplot.ylim(0.0, 0.8)
pyplot.plot(alpha00, G00, label="No Ground Effect",
color=color05, linestyle='--', linewidth=6);
pyplot.plot(alpha04, G04, label="0.4 m",
color=color10, linestyle='-', linewidth=4);
pyplot.plot(alpha05, G05, label="0.5 m",
color=color15, linestyle='-', linewidth=4);
pyplot.plot(alpha06, G06, label="0.6 m",
color=color20, linestyle='-', linewidth=4);
pyplot.plot(alpha07, G07, label="0.7 m",
color=color25, linestyle='-', linewidth=4);
pyplot.plot(alpha08, G08, label="0.8 m",
color=color30, linestyle='-', linewidth=4);
pyplot.plot(alpha09, G09, label="0.9 m",
color=color35, linestyle='-', linewidth=4);
pyplot.plot(alpha10, G10, label="1.0 m",
color=color40, linestyle='-', linewidth=4);
pyplot.legend(bbox_to_anchor=(1.05, 1), loc=2, borderaxespad=0.);
The aircraft seems to have some drastic drops in CL vs just the wing itself. This makes sense as the entire body of the aircraft is a drag-inducing entity, especially one that is half a boat on the lower half!
But the simulations still show that ground effect has positive influence with the aircraft in general. Perhaps if one were to add the "Ramming Effect" that the front two engines provide the wings, the CL would be even higher! But that is something that XFLR cannot handle, multi-velocity and multi-physics flows. If someone were to sit down and truly do the fluid mechanics, then that would be quite a challenge. But, I guess at some point, some russian engineer did, as these were actually built and flown!
<img src="./BannerThin.png" width = "1980">
## Conclusion¶
I would like to sum up what this project has accomplished, what I set out to do, and further studies I think would be interesting to keep looking through.
First I would like to say that I believe I have accomplished something I haven't seen before, creating a custom exporter to import custom bodies into XFLR5. This method is a little bit hacky, and will require some "by hand" alterations, but using an open source software, using the simple exporters, and a little creativity, anyone can have some fun creating anything they desire. Hopefully this addition to the online community will yeild some good results!
Second, I have spent the time to do some research into an airplane that I absolutely adore. I wont be chasing any waterfalls here, but I do admire what the USSR had done to finance such a bold project. I still think WIG craft's today have their place, a small niche, but still useful none the less. I was hoping to bring some light into people's lives by doing a study on some ground effect, one which many people have never heard of.
Third, I was looking into learning a new software. A very powerful one at that. With a little toying around, and hopefully some decent inputs, I have achieved some interesting outputs that hopefully has proven the WIG crafts ability to generate lift by just flying close to the ground.
And last, this project was conceived off of an old dream, and the concepts were drawn from a drawing found online. The faithfulness to this from the real thing is something no one can truly prove, as this aircraft was a war machine built in secrecy. Finding the specific data, has been impossible to come by, so hopefully this drawing and these airfoil shapes are true to the real thing. Pictures and models are never the real thing, but hopefully these studies have found something close.
Further studies of course would be to take this data, compare it to someone else's work to truly get some detailed comparison. Or perhaps work more with XFLR5 in hopes to understand the program better, and to get more results of more studies. There are two other main WIG craft designs out there, so perhaps comparison studies also could be done to see the differences in ground effect and their different performances. Many things can be done with these programs and these scripts.
And on that note I say farewell! If you have managed to stick with me through this whole adventure and happen to learn something about a new topic, I am truly honored. If anyone has any questions, or if anyone out in the internet world has questions about what I have done, please use my contact information in the bottom of the page. Label your email as XFLR5 Blender or XFLR5 Custom Airfoil and I will hopefully get back to you as quick as I can!
Cheers,
-George
<img src="./BannerThin.png" width = "1980">
## Vertice Exporter and Body Exporter¶
In [ ]:
#George Teel
#4/20/2015
#
#Super basic python script to export vertices of an object,
#In this case our airfoil shape, and export the locations
#so that XFLR5 can import that shape and then analyze the
#airfoils properly.
#
#This code was figured out from random websites and Blender's wikipedia page.
#
#Import What we need
import bpy
#Sets airfoil mesh object
airfoildata = bpy.data.meshes['Wing']
#Open File
myFile = open('D:/School Documents/AeroHydro/Project/Wing.dat','w')
#Cycles through vertices
for vert in airfoildata.vertices:
myFile.write( '%f %f \n' % (vert.co.x, vert.co.y) )
#Closes File
myFile.close()
In [ ]:
#George Teel
#5/4/2015
#
#Super basic python script to export vertices of an object,
#In this case our body shape, and export the locations
#so that XFLR5 can import that shape and then analyze it
#
#This code was figured out from random websites and Blender's wikipedia page.
#
#Import What we need
import bpy
#Sets airfoil mesh object
bodydata = bpy.data.meshes['BodySimple']
#Open File
myFile = open('D:/School Documents/AeroHydro/Project/EkranoBodySimpledata.dat','w')
#Create the intro Info
myFile.write('Ekranoplan SimpleBody\n')
myFile.write(' \n')
myFile.write('Right Surface Points \n')
myFile.write(' x() y() z()\n')
#Cycles through vertices
for vert in bodydata.vertices:
myFile.write( ' %f %f %f\n' % (vert.co.x, vert.co.z, vert.co.y) )
#Closes File
myFile.close()
#Still need to modify the data to follow the correct format, but should be easy from there on out.
## Thematic Music¶
Click here for some mood setting music! Let it run in the background while you read.
https://youtu.be/KxcP7TRY178?t=214
## Contact Info¶
Shoot me an email if you have any questions or would like to discuss further. Please don't forward me any attachments as I won't look at them.
George Teel
• email: gteel@gwmail.gwu.edu
## References¶
Since I hadn't taken any specific information from any one source, most of the A-90 information is found throughout multiple websites. So here is a list of all the websites I found and used for my research, and if any specific websites were used, they have been cited with numbers.
[1]May, James (2008, September 27) Riding the Caspian Sea Monster http://news.bbc.co.uk/2/hi/uk_news/magazine/7638659.stm
[2] (2014, October 19) CHDB A-90 Orlyonok (Eaglet) (Ekranoplan) Ground Effect Flying Boat / Transport Aircraft (1979) http://www.militaryfactory.com/aircraft/detail.asp?aircraft_id=1251
[3] (2000, September 07) Project 904/902R Orlan class Amphibious landing craft http://fas.org/man/dod-101/sys/ship/row/rus/904.htm
[4] Maraev, P., Flying over the waves (WIG "Eaglet") http://www.razlib.ru/transport_i_aviacija/ayerohobbi_1992_02/p3.php
Shechmeister, Matthew (2011, October 06) The Soviet Superplane Program That Rattled Area 51 http://www.wired.com/2011/06/ekranoplan/
Tarantola, Andrew (2013, October 11) This "Caspian Sea Monster" Was a Giant Soviet Spruce Goose http://gizmodo.com/this-caspian-sea-monster-was-a-giant-soviet-spruce-go-1456423681
Ground Effect Vehicle http://en.wikipedia.org/wiki/Ground_effect_vehicle
Random side note, it's pretty peculiar that all these references were done in September or October....
In [20]:
from IPython.core.display import HTML
def css_styling():
styles = open('./project.css', 'r').read()
return HTML(styles)
css_styling()
Out[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": 0, "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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2935407757759094, "perplexity": 10425.096912732382}, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039746465.88/warc/CC-MAIN-20181120150950-20181120172950-00515.warc.gz"} |
http://www.sciforums.com/threads/is-the-universe-computing-something.155118/page-15 | # Is the Universe computing something?
Discussion in 'Astronomy, Exobiology, & Cosmology' started by arfa brane, Jan 26, 2016.
1. ### arfa branecall me arfValued Senior Member
Messages:
6,915
Someone mentioned broken symmetry.
This is not quite as esoteric as it sounds. For instance, a coin is symmetric. When a coin is spinning (i.e. being randomized), it has more symmetry than when it lands on one side or the other. The latter state is a broken symmetry.
So a string of coin tosses is a randomized string, each element in the string represents say, the ring $\mathbb Z_2$ under some action.
When the coin is spinning, the sides are in a random 'superposition' of probabilities, each side has exactly 0.5 probability of showing (if the coin is fair) when the symmetry 'breaks'. A string of results is represented by regular language as (0,1)*, a concise way of writing down a string with each digit having equal probability of being 0 or 1, a randomly 'generated' string.
Last edited: Feb 15, 2016
to hide all adverts.
3. ### PhysBangValued Senior Member
Messages:
2,422
The CC could simply be a constant of gravitational interaction; in this case, it had no relationship to quantum theory.
There is no good calculation of the energy density of the quantum vacuum, so it is even possible that there is a CC an energy density from the vacuum.
to hide all adverts.
5. ### The GodValued Senior Member
Messages:
3,546
.... Banging funny things around.
to hide all adverts.
7. ### paddoboyValued Senior Member
Messages:
25,266
And gravitational interactions, are evident when spacetime is warped/curved/twisted/ rippled.
And possibly a QGT may reveal the how and why this happens, which is at the quantum level.
Well that's basically what I have said...these forces, the CC, ZPE, Casimir effect, may all be one and the same, but all are properties of spacetime as I see it. and all or one of them probably is the DE component.
8. ### brucepValued Senior Member
Messages:
4,098
The cosmological constant is a specific scientific term. So it can't be all those different things. If it was all those things we wouldn't need to have a bunch of different terms to describe the same natural phenomena. Calling it a different name doesn't change the natural phenomena. Quintessence is a different model for describing the accelerated expansion of the universe. WMAP concluded it's the cosmological constant which best describes what's happening in this universe. So science is building a standard model of cosmology and the dark energy is best described by the cosmological constant. The good thing is science has a great empirical model for testing predictions derived from cosmological theoretical models.
9. ### paddoboyValued Senior Member
Messages:
25,266
I can't argue with too much of what you have said bruce [and PhysBang] but I was always of the opinion that the mystery surrounding the DE component, was that we did not actually know the true nature of this force...whether the CC or quintessence as you mention, or any of the others. We just as yet are not sure.
The different names imo were just a result of different observations.
Perhaps I put my thoughts rather poorly.
What do you conclude from the following.....
https://briankoberlein.com/2015/03/06/nothing-but-net/
10. ### paddoboyValued Senior Member
Messages:
25,266
Please Register or Log in to view the hidden image!
Or this which seems to support your thoughts.....
http://www.sciencedirect.com/science/article/pii/S0370269306010197
Abstract
It has been speculated that the zero-point energy of the vacuum, regularized due to the existence of a suitable ultraviolet cut-off scale, could be the source of the non-vanishing cosmological constant that is driving the present acceleration of the universe. We show that the presence of such a cut-off can significantly alter the results for the Casimir force between parallel conducting plates and even lead to repulsive Casimir force when the plate separation is smaller than the cut-off scale length. Using the current experimental data we rule out the possibility that the observed cosmological constant arises from the zero-point energy which is made finite by a suitable cut-off. Any such cut-off which is consistent with the observed Casimir effect will lead to an energy density which is at least about 1012 times larger than the observed one, if gravity couples to these modes. The implications are discussed.
Current cosmological observations seem to favor a cosmological constant (Λ) as the leading candidate for dark energy which is thought to be driving the current phase of accelerated cosmic expansion [1]; the energy density contributed by it is constrained to be roughly ρDE≈10−11 (eV)4, with a corresponding length scale of the order of 0.1 mm. There exists a sizeable body of literature attempting to explain the origin of Λ as a consequence of the coupling of zero-point quantum fluctuations of matter fields, that pervade the vacuum, to gravity [2].
Write4U likes this.
11. ### PhysBangValued Senior Member
Messages:
2,422
Well, yes and no. A quantum theory of gravity may include a separate energy that acts like a CC, or the CC may just be a mathematical feature of the graviton field, or whatever quantum field creates gravity.
I haven't done too much research on the Casamir effect, but I believe that there are hypothetical explanations that do not rely on vacuum energy.
12. ### PhysBangValued Senior Member
Messages:
2,422
While I agree that there are tests that show that the dynamics favor a constant, I'm not sure that they can yet be said to rule out a vacuum energy of some sort, or even some sort of field. If there is a weird combination of constant and vacuum energy and field, then we might never find out!
13. ### sweetpeaValued Senior Member
Messages:
1,329
"Is the Universe computing something?"
OMG, does this mean it may be possible to hack the universe?
14. ### brucepValued Senior Member
Messages:
4,098
If you respect the results of the WMAP experiment then you can be somewhat 'sure'. Don't really like that term for science. We all know why Einstein derived the cosmological constant. To keep his GR universe from collapsing. In the classical metric it's a pressure term. Essentially an anti gravity term. It's the term that Guth and Linde used for the gravitational interaction, with a soliton in a quantum scalar field, that predicts the inflation event. During the event the pressure of the vacuum was negative and remained constant until the false vacuum 'rolled out' [reached a minimum vacuum energy state] and inflation came to an end. Initially Einstein didn't know the vacuum was comprised of fluctuating virtual particles seeking minimum energy states. Which must be accounted for when deriving quantum predictions associated with the cosmological constant. I think. So we have a classical model and a quantum model for the natural phenomena. These models describe the global effect of the cosmological constant. When we use the Casimir machine we alter the local physics of the vacuum. Locally between the uncharged plates. Resulting in a negative energy density between the uncharged plates and a measureable force pushing the plates towards each other. This is a pretty good discussion on the Casimir effect.
http://physicsworld.com/cws/article/news/2012/jul/18/physicists-solve-casimir-conundrum
So how big must the Casimir machine be to power a super luminal warp? It's the same theoretical principle that was used to hold the wormhole walls open during the theoretical analysis in this paper.
http://arxiv.org/abs/1405.1283
Last edited: Feb 15, 2016
15. ### brucepValued Senior Member
Messages:
4,098
Certainly. Based on what's known the classical and quantum models give a coherent description for the natural phenomena. I think. LOL. Thanks for your comments on this very interesting phenomena.
16. ### Write4UValued Senior Member
Messages:
14,014
Every manmade object is a result of hacking (applying) the observed universal forms, values, and functions.
Last edited: Feb 15, 2016
17. ### arfa branecall me arfValued Senior Member
Messages:
6,915
Ahem. So since a coin spinning then landing is a quite good analogy of a quantum particle with spin plus a measurement 'operator':
This is why Seth Lloyd and quite a few others claim the universe is computational--it's full of particles with spin, these particles interact and exchange something we call quantum information. Thanks to quantum randomness there is any number of ways to generate a random binary string by measuring spin, entirely analogous with tossing a fair coin.
On the other side we mostly have Ken Wharton, whose argument appears to be based on our assumption that the universe is Newtonian, a mistake since the universe is Lagrangian. He argues pursuing the latter "Schema" and laying aside our anthropic bias. I don't know what he's talking about; it doesn't matter which approach you use, in any quantum experiment the results are always classical--the output is the same.
Last edited: Feb 15, 2016
18. ### Write4UValued Senior Member
Messages:
14,014
https://en.wikipedia.org/wiki/Virtual_particle
IMO, this statement implies a very important question in regards to the required conditions where a *virtual* particle (such as the Higgs) becomes a *real* particle with a life of it's own. Apparently one requirement is that the Higgs combines with something else, which then gives it greater mass and the ability to persist as a real particle.
Any thoughts on this (possibly) fundamental process?
Messages:
6,915
20. ### sweetpeaValued Senior Member
Messages:
1,329
Remember the OP asks is the universe computing something.
Man is a part of nature's processes. The thoughts, for say, the making of clay pots, did not come from 'outside' of nature or 'outside' the universe. Those thoughts are part and parcel of nature or the universe. Or, to keep in tune with the OP, those thoughts and resulting actions are nature at work.
Last edited by a moderator: Feb 17, 2016
21. ### arfa branecall me arfValued Senior Member
Messages:
6,915
At a lecture on solid state physics, the subject was Heisenberg's Uncertainty Principle, and I recall the lecturer saying that it had cosmological implications, but he wasn't sure about them.
The relation between time and energy in HUP does have computational implications: there must be a connection to the fundamental limits of computation (Bennett's Brownian Computer for instance). That is, $\Delta E \Delta t \ge \hbar / 2$ says there is a fundamental limit on the time for any computation, rather than the time to measure some particle's energy to within $\Delta E$.
But that seems to just replace "time to measure" with "time to compute", somehow . . .
Apparently the correct relation is: $\Delta t = {\pi \hbar} / 2 \Delta E$, for a particle with a spread in energy $\Delta E$ to "move from one distinguishable state to another".
--http://arxiv.org/pdf/quant-ph/9908043v3.pdf
Last edited: Feb 19, 2016
22. ### arfa branecall me arfValued Senior Member
Messages:
6,915
Ok, so let's look again at the rock warming in the sun. It's a system, and it's processing information; the processing rate is limited by the amount of energy in each part of the system--if the surface is warmer than the insides it will have more energy to process information, the surface will compute faster than the inside of the rock. So, it's really just a close look, at say, the Heisenberg limit, of particles interacting.
I've intentionally used a rock to make a sort of parody of the whole thing, except this is serious shit, so it isn't a parody after all.
And a correction to post 281. I said the regular expression (0,1)* represents a random string of 1s and 0s, I should have said an element of the set generated by (0,1)* is such a string.
Last edited: Feb 19, 2016
23. ### Waiter_2001Registered Senior Member
Messages:
459
Hey Arf, how you doin? I see you've posted to sciforums TWICE here. I thought I'd post a reply to you in an effort to bring the thread forwards somewhat, which, I believe, will happen else I wouldn't have posted. It's all good what you posted though!
Please Register or Log in to view the hidden image! | {"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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8211759924888611, "perplexity": 1011.8329653047882}, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655881763.20/warc/CC-MAIN-20200706160424-20200706190424-00167.warc.gz"} |
https://social.msdn.microsoft.com/Forums/vstudio/en-US/14dc118f-5adc-4a90-9c07-fde701f6b36c/vcbld0001-vcprojectenginedll-could-not-be-loaded?forum=vcgeneral | # VCBLD0001: "VCProjectEngine.dll" Could not be loaded.
### Question
• If I try to execute the vcbuild.exe I get an error the the VCProjectEngine.dll not can be loaded.
Even if I run the exe in C:\Program Files\Microsoft Visual Studio 8\VC\vcpackages> directory with the parameter to the vcproj file this error occurs.
Has anyone an idea what I can do?
Thanks
Michael
Tuesday, January 23, 2007 9:27 AM
• hi Michael,
does it help if you run "C:\Program Files\Microsoft Visual Studio 8\VC\vcvarsall.bat" first, or use vcbuild in Visual Studio 2005 Command Prompt which you can find in All programs->Microsoft visual studio 2005->visual studio tools?
Wednesday, January 24, 2007 7:27 AM
### All replies
• hi Michael,
does it help if you run "C:\Program Files\Microsoft Visual Studio 8\VC\vcvarsall.bat" first, or use vcbuild in Visual Studio 2005 Command Prompt which you can find in All programs->Microsoft visual studio 2005->visual studio tools?
Wednesday, January 24, 2007 7:27 AM
• Thanks thtas helps allot...
Wednesday, October 31, 2007 4:33 PM
• I got the same error, although I happen to be running a French VS2008, so it reads:
vcbuild.exe : erreur VCBLD0001 : VCProjectEngine.dll ne peut pas être chargé.
Vérifiez si VCProjectEngine.dll se trouve dans le même répertoire que vcbuild.ex
e.
In any case, running vcbuild.exe from the VS Command Prompt doesn't seem to help.
Since this is driving me nuts, I even ran vcbuild.exe under the debugger, and I could see that it did, in fact, load VCProjectEngine.dll (the dll loaded message is shown in the debugger's log).
Any suggestions? Thanks in advance.
Thursday, January 31, 2008 11:46 PM
• Same problem with the German version (VS2008):
vcbuild.exe: Fehler VCBLD0001: "VCProjectEngine.dll" kann nicht geladen werden. (TaskId:433)
Stellen Sie sicher, dass sich "VCProjectEngine.dll" in demselben Verzeichnis befindet wie "vcbuild.exe". (TaskId:433)
Microsoft (R) Visual C++-Projekt-Generator - Befehlszeilenversion 9.00.21022 (TaskId:433)
I started the build with the environment variables set. I also tried to start vcbuild in it's own directory, to ensure it's in the same directory as "VCProjectEngine.dll"; no success.
I can reproduce the error on different machines (XP and Vista) with different solutions.
Tuesday, May 13, 2008 5:55 AM
• problem seems to be that for non english versions(french,german...), vcbuild cannot load the specific languages dll.
=> copy the content of the localized folder (for french it is vcpackages\1036) in vcpackages. It worked for me...
• Proposed as answer by Thursday, July 21, 2011 12:15 PM
Monday, June 30, 2008 12:08 PM
• problem seems to be that for non english versions(french,german...), vcbuild cannot load the specific languages dll.
=> copy the content of the localized folder (for french it is vcpackages\1036) in vcpackages. It worked for me...
Perfect Thanks
• Proposed as answer by Friday, October 26, 2012 4:02 PM
Friday, October 26, 2012 4:02 PM | {"extraction_info": {"found_math": false, "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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9002091884613037, "perplexity": 9452.660119325694}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398471436.90/warc/CC-MAIN-20151124205431-00233-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://brilliant.org/problems/think-outside-the-box-2/ | # Think outside the box
Discrete Mathematics Level pending
This week, my math teacher is doing something interesting. Each day, he writes a number $$n$$ on the board. He then asks us to write down all ordered lists of $$n$$ positive integers such that the sum of the $$n$$ numbers is $$2n$$. After that, he wants us to go to each list, record the product of the $$n$$ integers in that list, and then sum the products over all the lists. If he writes the number 1 on Monday, the number 2 on Tuesday, ..., and the number 5 on Friday, what will be the sum of our answers each day over the whole week?
As an explicit example, on Tuesday, our answer for that day will be $$1 \times 3+2 \times 2+3 \times 1=10$$.
This problem was adopted from a similar problem given to me by a friend.
× | {"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": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6652475595474243, "perplexity": 219.120652615987}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-22/segments/1495463609404.11/warc/CC-MAIN-20170528004908-20170528024908-00263.warc.gz"} |
https://gamedev.stackexchange.com/questions/12603/how-is-a-3d-perlin-noise-function-used-to-generate-terrain | # How is a 3d perlin noise function used to generate terrain?
I can wrap my head around using a 2d perlin noise function to generate the height value but I don't understand why a 3d perlin noise function would be used. In Notch's blog, http://notch.tumblr.com/post/3746989361/terrain-generation-part-1, he mentioned using a 3d perlin noise function for the terrain generation on Minecraft. Does anyone know how that would be done and why it would be useful? If you are passing x,y, and z values doesn't that imply you already have the height?
2D perlin noise is good for height maps, but in this case it seems that he is not using a height map. Instead he has a 3D grid, where any cell can be empty. This allows caves and such formations, where the ground height is not a single value for given 2D location.
Instead of sampling the “ground height”, I treated the noise value as the “density”, where anything lower than 0 would be air, and anything higher than or equal to 0 would be ground.
Simply put, for every place where a block can be, a noise function is evaluated, and if it is >0, a block is placed. Notch's noise function is skewed by adding height from water level to its value, that's why lower areas are mostly solid (height is large negative, so height+noise is negative too) and higher areas are mostly empty (height is large positive, so height+noise is positive too).
There's probably some additional alchemy to decide what kind of block gets generated, and to carve caves. But I guess it's not directly related to this noise function.
Also note that this method works for Notch because Minecraft has a voxel-based terrain. If you tried to pull that off in a polygon-based world, simply sampling noise function would not be enough. You' have to use some algorithm to turn samples into a surface, and create polygons that approximate this surface. One such algorithm is marching cubes.
3D noise becomes mandatory if the terrain needs cave networks and overhangs.
To extract an isosurface from density information, the 2 most popular techniques are Marching Cubes (MC), and the newer Dual Contouring (DC). The data structure needed is quite different depending on the chosen method.
As previously mentioned, Geiss's GPU Gems 3 article is a very instructive starting point for understanding and implementing MC terrains on the GPU (Note that his MC approach run entirely on the GPU and requires at least SM4 - GS- capable one).
Because density data on MC voxels can only stay on voxel's edges, classic MC may contour the volume without preserving sharp edges features. DC does not suffers this drawback since the density info is expressed as a 3D point (QEF minimizer) laying anywhere inside the voxel plus the sign at each corner.
On the other hand, MC does not suffer from self-intersecting faces because all generated triangles are enclosed in their corresponding voxels, whereas DC needs additional computations to prevent intersections between generated faces. DC authors addressed this issue in an improved version of their algorithm.
http://www.cs.wustl.edu/~taoju/research/interfree_paper_final.pdf
http://www.cs.berkeley.edu/~jrs/meshpapers/SchaeferWarren2.pdf
This fellow also propose a likely cleaner approach based on convex / concave analysis for avoiding self-intersections. He uses as well better quad splitting rules to help preserve edge's orientation:
http://www2.mae.cuhk.edu.hk/~cwang/pubs/TRIntersectionFreeDC.pdf
Classic MC is also not out-of-box "crack-free" and may require crack patching if ran on unrestricted octrees. DC does not suffer from this last issue.
Here is a pretty nice and complete survey of most mesh extraction techniques : http://www.cs.berkeley.edu/~jrs/mesh/
An octree / voxel approach is intrinsically "CSG-friendly", which make it easier to plan a neat fully "destructible" game level strategy, but if one needs to implement all this in a game, the octree depth will also need to be frustum-dependent.
If the whole stuff fits in memory or is correctly streamed, the data can also be used for rendering AO and computing physics / collisions.
• "3D noise becomes mandatory if the terrain needs cave networks and overhangs." Mandatory? As in, this is the only way to generate caves and overhangs? Nope. I'm generating a heightmap from 2D perlin noise and then carving caves and overhangs into it as a separate step, for a more natural look. It's misleading to budding young game developers to say that this is the ONLY way to generate caves and overhangs in a procedurally generated world. – Domarius Jan 4 '16 at 4:19
My guess, in that particular example, is that he used the z value to determine what type of material: bedrock, stone, dirt, or air.
Minecraft uses the marching cubes algorithm to generate 3D terrain. I don't have a refernce for this, I'm sorry. I'm not sure exactly what Notch was talking about when he mentioned the Perlin Noise function - perhaps a seed for the marching cubes algorithm. More info here:
And a great GPU Gems article if you're interested in marching cubes:
• Marching cubes is an algorithm for generating a mesh from a scalar field. That means it needs to have data first, then it generates a mesh to fit the data. It's not for generating data or terrain. – MichaelHouse Mar 30 '12 at 23:35
for (int x = 0; x < Width)
{
for (int y = 0; y < Depth)
{
for (int z = 0; z < Height)
{
if(z < Noise2D(x, y) * Height)
{
Array[x][y][z] = Noise3D(x, y, z)
} else {
Array[x][y][z] = 0
}
}
}
}
• -1; Even with the clarification you've added this is a pretty bad answer, it does not clearly explain the purpose of the code snippet nor its semantic impact on the resulting data, and it doesn't address most of the questions the user actually had. – user1430 Oct 16 '15 at 15:18
• Ive edited the code, how about now? – Maxim DC Oct 16 '15 at 15:27
• It explains how 3D noise is used to make terrain, this is correct in my opinion – Maxim DC Oct 16 '15 at 15:29 | {"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": 0, "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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4430078864097595, "perplexity": 1478.3570382870344}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875145839.51/warc/CC-MAIN-20200223185153-20200223215153-00439.warc.gz"} |
https://www.whsmith.co.uk/products/a-proof-of-alons-second-eigenvalue-conjecture-and-related-problems-memoirs-of-the-american-mathematical-society-195-910/9780821842805 | # A Proof of Alon's Second Eigenvalue Conjecture and Related Problems (Memoirs of the American Mathematical Society 195, 910)
By: Joel Friedman (author)Paperback
1 - 2 weeks availability
£69.50 With FREE Saver Delivery
### Description
A $d$-regular graph has largest or first (adjacency matrix) eigenvalue $\lambda 1=d$. Consider for an even $d\ge 4$, a random $d$-regular graph model formed from $d/2$ uniform, independent permutations on $\{1,\ldots,n\}$. The author shows that for any $\epsilon>0$ all eigenvalues aside from $\lambda 1=d$ are bounded by $2\sqrt{d-1}\;+\epsilon$ with probability $1-O(n{-\tau})$, where $\tau=\lceil \bigl(\sqrt{d-1}\;+1\bigr)/2 \rceil-1$. He also shows that this probability is at most $1-c/n{\tau'}$, for a constant $c$ and a $\tau'$ that is either $\tau$ or $\tau+1$ (""more often"" $\tau$ than $\tau+1$). He proves related theorems for other models of random graphs, including models with $d$ odd.
### Contents
Introduction; Problems with the stand trace method; Background and terminology; Tangles; Walk sums and new types; The selective trace; Ramanujan functions; An expansion for some selective traces; Selective traces in graphs with (without) tangles; Strongly irreducible traces; A sidestepping lemma; Magnification theorem; Finishing the ${\cal G} {n,d}$ proofs; Finishing the proofs of the main theorems; Closing remarks; Glossary; Bibliography
### Product Details
• publication date: 15/12/2008
• ISBN13: 9780821842805
• Format: Paperback
• Number Of Pages: 100
• ID: 9780821842805
• weight: 182
• ISBN10: 0821842803
### Delivery Information
• Saver Delivery: Yes
• 1st Class Delivery: Yes
• Courier Delivery: Yes
• Store Delivery: Yes
### Mathematical Theory Of ComputationView More
Prices are for internet purchases only. Prices and availability in WHSmith Stores may vary significantly
Close | {"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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7441880106925964, "perplexity": 5548.257439507318}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988720760.76/warc/CC-MAIN-20161020183840-00320-ip-10-171-6-4.ec2.internal.warc.gz"} |
http://www.gradesaver.com/textbooks/math/other-math/basic-college-mathematics-9th-edition/chapter-3-adding-and-subtracting-fractions-3-3-adding-and-subtracting-unlike-fractions-3-3-exercises-page-219/2 | ## Basic College Mathematics (9th Edition)
To rewrite unlike fractions as like fractions, we must find the least common denominator (or the denominator that is the least common multiple of the original denominators) of the unlike fractions. For example, if we want to add $\frac{3}{8}$ and $\frac{1}{4}$, we must find the least common multiple of 4 and 8. We know that this is 8. Therefore, we can rewrite this pair as like fractions. $\frac{3}{8}+\frac{1}{4}=\frac{3}{8}+\frac{1\times2}{4\times2}=\frac{3}{8}+\frac{2}{8}=\frac{3+2}{8}=\frac{5}{8}$ | {"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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9729006886482239, "perplexity": 217.5578653734592}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549426639.7/warc/CC-MAIN-20170726222036-20170727002036-00684.warc.gz"} |
http://www.broadinstitute.org/cancer/software/genepattern/modules/docs/Cufflinks.cuffcompare/7 | # Cufflinks.cuffcompare Documentation, v7 ▸ Open Module on GenePattern Public Server
Description: Analyzes the transcribed fragments in an assembly
Author: Cole Trapnell et al, University of Maryland Center for Bioinformatics and Computational Biology
Algorithm Version: Cufflinks 2.0.2
## Summary
Cufflinks.cuffcompare helps analyze the transcribed fragments (transfrags) in an assembly by:
• Comparing assembled transcripts to a reference annotation
• Tracking Cufflinks transcripts across multiple experiments (e.g., across a time course)
Cufflinks was created at the University of Maryland Center for Bioinformatics and Computational Biology. This document is adapted from the Cufflinks documentation for release 2.0.2. For more information about Cufflinks.cuffcompare, see the Cufflinks documentation.
## Usage
Cufflinks.cuffcompare requires at least one Cufflinks' GTF output file as input, and optionally can also take a "reference" annotation GTF/GFF file such as from Ensembl. For more information on the GTF/GFF format, see the specification.
#### Important Notes:
There are known issues that prevent Cufflinks.cuffcompare from running on the Mac Mini and possibly other Mac hardware.
This module may produce some empty files. This does not mean that the algorithm has failed; it is generally a data issue. In particular, this may occur if the transfrags are not in the reference annotation.
## References
Trapnell C, Hendrickson D,Sauvageau S, Goff L, Rinn JL, Pachter L. Differential analysis of gene regulation at transcript resolution with RNA-seqNature Biotechnology. 2013;31:46-53.
Trapnell C, Roberts A, Goff L, Pertea G, Kim D, Kelley DR, Pimentel H, Salzberg SL, Rinn JL, Pachter L. Differential gene and transcript expression analysis of RNA-seq experiments with TopHat and Cufflinks. Nature Protocols 2012;7;562–578.
Roberts A, Pimentel H, Trapnell C, Pachter L. Identification of novel transcripts in annotated genomes using RNA-SeqBioinformatics. 2011 Sep 1;27(17):2325-9.
Trapnell C, Williams BA, Pertea G, Mortazavi AM, Kwan G, van Baren MJ, Salzberg SL, Wold B, Pachter L. Transcript assembly and quantification by RNA-Seq reveals unannotated transcripts and isoform switching during cell differentiation. Nat Biotechnol. 2010;28:511-515.
Trapnell C, Pachter L, Salzberg SL. TopHat: discovering splice junctions with RNA-SeqBioinformatics. 2009;25:1105-1111.
Langmead B, Trapnell C, Pop M, Salzberg SL. Ultrafast and memory-efficient alignment of short DNA sequences to the human genome. Genome Biol. 2009;10:R25.
Cufflinks manual. Note that this information may be based on a subsequent version of Cufflinks.
TopHat website.
## Parameters
Name Description
input file * One or more GTF file output(s) from Cufflinks
output prefix A prefix for the module output
reference GTF A reference annotation GTF
exclude transcripts Whether to ignore reference transcripts that are not overlapped by any transcript in the input files. This takes effect only if a reference GTF is provided.
reference genome file Fasta file or zip of fasta files against which your reads were aligned
additional cuffcompare options Additional options to be passed along to the Cuffcompare program at the command line. This parameter gives you a means to specify otherwise unavailable Cuffcompare options and switches not supported by the module; check the Cufflinks manual for details. Note that the information at this link may refer to a subsequent version of Cufflinks. Recommended for experts only; use this at your own discretion.
* - required
## Cuffcompare pass-through options
The following may be useful for advanced users who wish to use the additional.cuffcompare.options parameter. This is the 'usage' output from running cuffcompare at the command-line, which gives a list of all of the available options and switches. Note that this was generated by Cuffcompare v2.0.2 and that the options here may differ from the documentation provided online at the Cufflinks website due to subsequent version updates.
cuffcompare v2.0.2 (3524M)
-----------------------------
Usage:
cuffcompare [-r <reference_mrna.gtf>] [-R] [-T] [-V] [-s <seq_path>]
[-o <outprefix>] [-p <cprefix>]
{-i <input_gtf_list> | <input1.gtf> [<input2.gtf> .. <inputN.gtf>]}
Cuffcompare provides classification, reference annotation mapping and various
statistics for Cufflinks transfrags.
Cuffcompare clusters and tracks transfrags across multiple samples, writing
matching transcripts (intron chains) into <outprefix>.tracking, and a GTF
file <outprefix>.combined.gtf containing a nonredundant set of transcripts
across all input files (with a single representative transfrag chosen
for each clique of matching transfrags across samples).
Options:
-i provide a text file with a list of Cufflinks GTF files to process instead
of expecting them as command line arguments (useful when a large number
of GTF files should be processed)
-r a set of known mRNAs to use as a reference for assessing
the accuracy of mRNAs or gene models given in <input.gtf>
-R for -r option, reduce the set of reference transcripts to
only those found to overlap any of the input loci
-M discard (ignore) single-exon transfrags and reference transcripts
-N discard (ignore) single-exon reference transcripts
-s <seq_path> can be a multi-fasta file with all the genomic sequences or
a directory containing multiple single-fasta files (one file per contig);
lower case bases will be used to classify input transcripts as repeats
-d max distance (range) for grouping transcript start sites (100)
-p the name prefix to use for consensus transcripts in the
<outprefix>.combined.gtf file (default: 'TCONS')
-C include the "contained" transcripts in the .combined.gtf file
-G generic GFF input file(s) (do not assume Cufflinks GTF)
-T do not generate .tmap and .refmap files for each input file
-V verbose processing mode (showing all GFF parsing warnings)
## Input Files
1. <input.file>
One or more GTF files accessible to the GenePattern server. In GenePattern 3.6.0 and above, this parameter will accept server-hosted GTF files directly through the drag-and-drop file parameter interface. When producing the *.tmap and *.refmap ouput files (see below), cuffcompare will use the <output.prefix> parameter and possibly the input GTF file/path to form the file name. When the input is a single GTF, one of each of these output files will be produced with the names <output.prefix>.tmap and <output.prefix>.refmap.
For more information on the GTF/GFF format, see the specification.
Cufflinks.cuffcompare version 5+ can no longer accept a .txt input file list on GenePattern versions 3.6.0+. Instead, you may specify multiple files using the drag-and-drop interface.
Legacy information: To avoid file-naming collisions, when the input file is a text file of multiple GTF input files then a transformed version of the input file path is also included in naming these outputs. This is necessary because GenePattern places all output files in a single job results directory when execution is complete. The path is transformed by substituting an underscore character (‘_’) for any spaces and path separators and by truncating any path prefix common to all input files in order to shorten the name. The output file names will be formed as <output.prefix>.[transformed path].tmap and <output.prefix>.[transformed path].refmap.
Optionally, explicit identifiers can be specified for direct control over output file naming. Such IDs can be provided after each path listing, separated by a tab character on the same line. The output file names will be formed as <output.prefix>.[ID_filename].tmap and <output.prefix>.[ID_filename].refmap. This is not available when using the GP 3.6.0 drag-and-drop interface.
2. <reference.GTF> (optional)
A reference annotation file in GTF format. Each sample is matched against this file, and sample isoforms are tagged as overlapping, matching, or novel where appropriate. These reference annotation files can be downloaded for many genomes from sites like UCSC Genome Browser. The GenePattern FTP site hosts a number of reference annotation GTFs, available in a dropdown selection (requires GenePattern 3.7.0+).
For more information on the GTF format, see the specification.
3. <reference.genome.file> (optional)
Fasta file or zip of fasta files against which your reads were aligned. If supplied, cuffcompare will use this for some optional classification functions. If a multifasta file, all contigs should be present. If a zip, this must contain one fasta file per reference chromosome, and each file must be named after the chromosome and have a .fa or .fasta extension. For more information on the FASTA format, see this description.
The GenePattern FTP site hosts a number of reference genomes, available in a dropdown selection (requires GenePattern 3.7.0+).
## Output Files
1. <output.prefix>.stats
Various statistics related to the accuracy of the transcripts in each sample when compared to the reference annotation data.
2. <output.prefix>.combined.gtf
Cufflinks.cuffcompare reports a GTF file containing the "union" of all transfrags in each sample. If a transfrag is present in both samples, it is thus reported once in the combined GTF.
3. *.tmap
These tab-delimited files list the most closely matching reference transcript for each Cufflinks transcript. There is one row per Cufflinks transcript.
4. *.refmap
These tab-delimited files list, for each reference transcript, which Cufflinks transcripts either fully or partially match it. There is one row per reference transcript output
A summary of the execution of Cufflinks.cuffcompare, providing information on both the genomic sequence and datasets.
5. stdout.txt
A summary of the execution of Cufflinks.cuffcompare, providing information on both the genomic sequence and datasets.
6. <output.prefix>.tracking
This file matches transcripts between samples. Each row contains a transcript structure that is present in one or more input GTF files. Because the transcripts will generally have different IDs (unless you assembled your RNA-seq reads against a reference transcriptome), Cufflinks.cuffcompare examines the structure of each the transcripts, matching transcripts that agree on the coordinates and order of all of their introns, as well as strand. Matching transcripts are allowed to differ on the length of the first and last exons, since these lengths will naturally vary from sample to sample due to the random nature of sequencing.
## Platform Dependencies
Module Type: RNA-seq CPU Type: x86_64 OS: Mac, Linux Language: C++, Perl
## GenePattern Module Version Notes
VersionRelease DateDescription
72014-02-14Added a parameter to allow the user to pass through extra Cuffcompare options
62013-09-25Added dynamic GTF and genome file selectors and HTML-based documentation
52013-07-22Updated to Cufflinks.cuffcompare version 2.0.2
42012-07-06Fixed syntax error.
22012-01-13Updated to Cufflinks.cuffcompare version 1.3.0
12011-04-11 | {"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": 0, "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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.41109925508499146, "perplexity": 13551.58457654877}, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-49/segments/1416400373899.42/warc/CC-MAIN-20141119123253-00239-ip-10-235-23-156.ec2.internal.warc.gz"} |
https://flashman.neocities.org/MD/knowls/example.AEF.NEFP.0.knowl.html | Example AEF.NEFP Numerical Estimation False Position
In many cases representing the exact value of a solution to an equation as a decimal can be very difficult or even impossible. In such cases techniques that provide a numerical estimate of the solution to some precision is very useful. Mapping diagrams can be used often to visualize these techniques.
Since many examples involve elementary functions that are continuous for an interval of the form $[a,b]$, one basis for estimating solutions is the result:
Theorem: CCD.IVT.0: If $f$ is a continuous function on the interval $[a,b]$ and $0$ is between $f(a)$ and $f(b)$, then there is a number $c \in [a,b]$ where $f(c) = 0$.
An immediate consequence of the IVT is that if $f(a)\cdot f(b) < 0$, then there is a number $c \in [a,b]$ where $f(c) = 0$.
The False Position Method
The false method repeatedly applies the IVT to estimate a solution to an equation with an elementary function, $f$, of the form $f(x)=0$ as long as $f(a)\cdot f(b) < 0$ for the initial numbers $a$ and $b$. Instead of using bisection to replace one endpoint of the interval it replaces one endpont with the root (zero) of the linear function determined by the function values at the endpoints of the interval,
The procedure is as follows:Start with $a_1=a$ and $b_1=b$.
Suppose $a_n < b_n$ and $f(a_n)\cdot f(b_n) < 0$. Let $FP_n = \frac {a_n f(b_n)-b_n f(a_n)}{f(b_n)-f(a_n)}$ and compute $f(FP_n)$.
[$FP_n$ is the number determined by solving for $x$ in $\frac { f(b_n)}{b_n-x}= \frac { f(a_n)}{a_n-x}$.
So $f(b_n)(a_n-x)= f(a_n) (b_n-x)$; $a_n f(b_n)-x f(b_n)= b_n f(a_n)-xf(a_n))$...]
If $f(FP_n) = 0$ then $m_n$ is the solution to the equation. Otherwise either $f(a_n)\cdot f(FP_n) < 0$ or $f(FP_n)\cdot f(b_n) < 0$.
If $f(a_n)\cdot f(FP_n) < 0$, let $a_{n+1} = a_n$ and $b_{n+1}=FP_n$.
Otherwise let $b_{n+1} = b_n$ and $a_{n+1}=FP_n$.
Then $a_{n+1} < b_{n+1}$ and $f(a_{n+1})\cdot f(b_{n+1}) < 0$, so the procedure can be repeated.
At each step the interval $[a_n,b_n]$ is shorter, so the solution guaranteed by the IVT to exist in this interval is approximated more accurately by either end numbers of the next interval.
The mapping diagram below illustrates the example where $f(x) = x^2-2, a=0,$ and $b=2$.
The False Position Method
Notice how the arrows on the mapping diagram are paired with the point on the graph of the function $f$.
You can move the point for $x$ on the mapping diagram to see how the function value changes both on the diagram and on the graph.
Sliders: The sliders labeled $a$ and $b$ control the end values of the interval $[a,b]$.
The slider labeled $n$ controls which the number of times the false position method is executed.
Checking the box to "Show/Hide Midpont & Data" will display $a_n, b_n, FP_n$ and the values of $f$ for each of these nmbers.
It will also display a scond box to "Show/Hide Smallest Exact Root". Checking this box will display this root on both the graph and the mapping iagram.
The function $f$ can be changed by making an entry in the appropriately labeled input box.
Martin Flashman, August. 10, 2017. Created with GeoGebra | {"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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9119479656219482, "perplexity": 297.7686064767818}, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154214.36/warc/CC-MAIN-20210801123745-20210801153745-00450.warc.gz"} |
http://science.sciencemag.org/content/164/3884/1177 | Reports
# Colchicine-Inhibited Cilia Regeneration: Explanation for Lack of Effect in Tris Buffer Medium
See allHide authors and affiliations
Science 06 Jun 1969:
Vol. 164, Issue 3884, pp. 1177-1178
DOI: 10.1126/science.164.3884.1177
## Abstract
In Stentor coeruleus growth of new, daughter ciliates and experimentaly inducled regeneration of oral membranellar cilia are reversibly inhibited by low, nontoxic concentrations of colchicine. However, if the clulture medium containing colchicine (or Colcemid) is made up in tris(hydroxymethyl)aminomethane buffer, growth of ciliated daughters and regeneration of oral cilia proceed normally. The evidence suggests that the mechanism of this reversal of the effects of colchicine (or Colcemid) is due to a chemical reaction between tris(hydroxymethyl)-aminomethane (or its hydrochloride, or both) and colchicine (or Colcemid), wihich reduces the effective concentration of these mitotic spindle inhibitors reaching the stentors. | {"extraction_info": {"found_math": false, "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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8899203538894653, "perplexity": 26964.543975970344}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912202347.13/warc/CC-MAIN-20190320125919-20190320151919-00025.warc.gz"} |
https://www.albanova.se/event/phd-thesis-defense-quantum-chemical-calculations-of-multidimensional-dynamics-probed-in-resonant-inelastic-x-ray-scattering/ | PhD Thesis Defenses
## PhD Thesis defense: Quantum chemical calculations of multidimensional dynamics probed in resonant inelastic X-ray scattering
This thesis is devoted to the theoretical study of the dynamical processes induced by light-matter interactions in molecules and molecular systems. To this end, the multidimensional nuclear dynamics probed in resonant inelastic X-ray scattering (RIXS) of small molecules, exemplified by H2O (g) and H2S (g), as well as more complex molecular systems, exemplified by NH3 (aq) and kaolinite clay, are modelled. The computational methodology consists of a combination of ab initio quantum chemistry calculations, quantum nuclear wave packet dynamics and in certain cases molecular dynamics modelling. This approach is used to simulate K-edge RIXS spectra and the theoretical results are evaluated against experimental measurements.
Specifically, the vibrational profile for decay back to the electronic ground state of the H2O molecule displays a vibrational selectivity introduced by the dynamics in the core-excited state. Simulation of the inelastic decay channel to the electronic |1b1-1,4a11> valence-excited state shows that the splitting of the spectral profile arises from the contribution of decay in the OH fragment. The character of the S1s-1 and S2p-1 core-excited states of the H2S molecule has been investigated and distinct similarities and differences with the H2O molecule have been identified. RIXS has also been used as a probe of the hydrogen bonding environment in aqueous ammonia and by detailed analysis of the valence orbitals of NH3 and water, the spectral profiles are explained. Finally, it is shown that vibrations of weakly hydrogen bonding OH are excited in RIXS decay to the electronic ground state in kaolinite.
Keywords: quantum chemistry, X-ray spectroscopy, RASSCF, density functional theory, ultrafast nuclear dynamics.
Stockholm 2018
http://urn.kb.se/resolve?urn=urn:nbn:se:su:diva-154057 | {"extraction_info": {"found_math": false, "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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8156149983406067, "perplexity": 2106.8432898351502}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296949701.56/warc/CC-MAIN-20230401063607-20230401093607-00596.warc.gz"} |
https://electronics.stackexchange.com/questions/74511/using-adc-interrupts-and-timer-interrupts-at-the-same-time | # Using ADC interrupts and TIMER interrupts at the same time
I'm a programmer and not jet experienced with arduino or any microcontrollers. Especially the technical side.
I soldered a 6x8 RGB Led Matrix and use binary coded modulation http://www.batsocks.co.uk/readme/art_bcm_3.htm to mix the colors.
The Leds are controlled by 4 74hc595 shiftregisters. Basically I shift out the bits to control the really fast during a timer interrupt.
Now I want to influence the color of the leds by measuring sound frequencies with an Electret microphone breakout board. To achieve fast frequency detection I use a technique described here http://www.instructables.com/id/Arduino-Frequency-Detection/#step1.
It is based on ADC interrupts.
Both impementations work alone but when I try to bring the code together it breaks.
The timer and the ADC are initialzed like this:
cli(); //stop all interrupts
/////////////////////
//initialize TIMER//
////////////////////
TCCR1A = 0;
TCCR1B = 0;
TCNT1 = 0;
OCR1A = 1; // compare match register
TCCR1B |= (1 << WGM12); // CTC mode
TCCR1B |= ((1<<CS21)|(1<<CS20)) ;
TIMSK1 |= (1 << OCIE1A); // enable timer compare interrupt
//////////////////
//////////////////
ADMUX |= (1 << REFS1); //set reference voltage
sei(); //start all interrupts
So my question is - Is it even possible to use those interrupts together? And how do I have to configure them to make it work. Or is this a dead end?
EDIT The Timer interrupt:
ISR(TIMER1_COMPA_vect) // timer compare interrupt service routine
{
zaehlbit <<=1;
//reset timercompare and zaehlbit
if( zaehlbit == 0 ) {
OCR1A= 1;
zaehlbit = 1;
}
//latch low
bitClear(PORTB, latchPinPORTB);
//red
for (int i=0; i<8; i++){
// clock low
bitClear(PORTB, clockPinPORTB);
//check led position and brightness
//and set Bit on Datapin if zaehlbit in Brightness
if (ledCounter&led && zaehlbit&red){
bitClear(PORTB, dataPinPORTB);
}
else {
bitSet(PORTB, dataPinPORTB);
}
// clock low (register bit)
bitSet(PORTB, clockPinPORTB);
ledCounter >>= 1;
if( ledCounter == 0 ) {
ledCounter = 128;
}
}
//shift timer compare and set timer back to generate delay (Bit Angle Modulation)
OCR1A <<= 1;
TCNT1 = 0;
}
I do the same for-loop for the rows and the other colors. I just left it out because it looked confusing. The compare match register is shifted left during the interrupt and the timer is set back to generate a "growing" delay. There is one interrupt per cycle. It runs at a clock/32 tickrate
The ADC interrupt looks like this:
ISR(ADC_vect) {//when new ADC value ready
prevData = newData;//store previous value
newData = ADCH;//get value from A0
if (newData > prevData){//if positive slope
bitSet(PORTB, ledPORTB);//set pin 12 high
}
else if (newData < prevData){
bitClear(PORTB, ledPORTB); //set pin 12 low
}
}
The ADC interrupt is triggered everytime a value from A0 is ready. In the mainloop I just try to set some leds but it doesn't work.
• It should - of course - be possible. Could you elaborate a bit: What exactely doesn't work as expected? Also, could you post the full code (or at least the interrupt handlers and your main loop). Also: How often do you expect an ADC interrupt and how often do you expect the timer interrupt to happen? – Tom L. Jun 30 '13 at 13:53
• I edited the question I hope it's more clear now – user1210456 Jun 30 '13 at 14:13
• Check the datasheet for your exact microcontroller (ATmega[something]). Every interrupt has its own vector and comes with its own priority, this defines which interrupt will be serviced when two arrive at the same moment. While one interrupt is being serviced, the other interrupt will be put on hold until the first ISR finished. – jippie Jun 30 '13 at 17:57
• Whats your clock frequency? Are you triggering the timer interrupt every clock cycle (due to OCR1A = 1)? Maybe the timer ISR execution time is longer that your interrupt interval? What exactly does "when I try to bring the code together it breaks" mean? – Rev1.0 Jun 30 '13 at 17:57
One thing that I notice from your code is that you are acquiring the samples in free running mode with ADC clock at F_CPU/2. According to the Atmega 328P datasheet
In Free Running mode, a new conversion will be started immediately after the conversion com- pletes, while ADSC remains high.
and the conversion time for free running mode (on table 23-1, page 255) is specified as 13 ADC clock cycles. This means that your ADC ISR is firing every 26 clock cycles with your current prescaler settings. On top of that, your TIMER1_COMPA ISR is also fired so frequently (every 64 cycles) that your ISR might actually take longer to execute than the desired time. You basically use all your time acquiring data and/or in your timer ISR. An easy solution is to increase the count in OCR1A to, say, 5 and increase the prescaler of the ADC
ADCSRA |= (1 << ADPS1); // ADC prescaler 8, conversion time 4*26 MCU cycles
OCR1A = 5; // 5*64;
If you want to squeeze every last drop of performance out of the old 328 you should inspect the generated assembly for your interrupt routines, and figure out if you can do something to optimize them. Common performance hogs are, for example, frequent data fetches from SRAM, but if you're using gcc you might find all sorts if groovy stuff happening upon ISR entry.
Keep in mind that the human eye cannot detect insanely fast flickering (hence pwm driven leds work as they do, with no visible flicker). If the slower running rate looks visually unpleasant you can of course try to reduce the prescaler and OCR1A, but you should also consider dispensing the ISR's altogether and just run your led flashing code in the main loop and have just the ADC ISR acquire data to be processed. This might work better since every function call takes at minimum 8 cycles (might have been 6, memory is a bit fuzzy on this particular detail) PLUS all the stack PUSH'n an' POP'n, which depends on your variable usage. So you could go for something like:
while(1) {
checkData();
handleLeds();
}
and have both functions be inlined.
Yes you can, jus three things:
• Try to use the main loop to do all you need, trying to programm a cyclic executive and use the interrupts only to activate flags or read ADC values. You can achive this by setting times for each task, for example, read the ADC every 5ms, Set the LEDs every 10ms, etc.
• In some Atmel devices like Atmega16, you need to reset (setting it again) the ADC every time the ADC finish a measurement.
• Deactivate the general interrupt flag when entering to an interrupt function and activate it again when leaving. cli() and sei().
Any question, let me know, I didn't use the comment because I have no reputation to do so, but I have lot of exeperience working on Atmel microcontrollers and Digital systems.
• 1) The ADC has been properly configured for autotrigger, and exiting the ISR will retrigger it. 2) Interrupts are disabled when an ISR is executed; they need to be explicitly enabled if nested interrupts are to be used. – Ignacio Vazquez-Abrams Nov 27 '13 at 18:50
• That would be a reason why it's not working with both interrupt vector, sei() have to be called when finishing any interrupt function listener. – Chirry Nov 27 '13 at 19:19
• The other thing I see, is that the timer interrupt is being called every 64 clock cycles (CS20 and CS21 = 64 prescaling and OCR1A = 1), and maybe is being triggered when the micro is "busy" doing the ADC interrupt function. At least, the Timer vector function takes more than 64 clock cycles itself. – Chirry Nov 27 '13 at 19:22
• Interrupts are restored upon completion of the ISR. And pending interrupts are handled when the flag is restored. – Ignacio Vazquez-Abrams Nov 27 '13 at 21:24
• I had no idea, I have always set by hand the global interrupt at the end of every ISR. Other thing that I realised is that: as far as I know, there are no interrupts queue but there are different interruption priorities. So if the timer interrupt has a higher priority than the ADC, and the timer interrupt keeps calling, the micro will always handle the timer interruption. – Chirry Nov 27 '13 at 23:07 | {"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": 0, "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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.16853657364845276, "perplexity": 4317.240854697252}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540482284.9/warc/CC-MAIN-20191205213531-20191206001531-00119.warc.gz"} |
http://mathhelpforum.com/statistics/226593-need-help-understanding-these-problems.html | # Math Help - Need help understanding these problems
1. ## Need help understanding these problems
1. A factory has a delicate process that is affected by weather. If it rains, decrease the humidity inside the factory to continue operations. Since inside the factory the rain is no heard it is known that the probability of the work being damaged given that nobody noticed in time is 33%. If you have 85 rainfalls in a year calculate the expected number of damaged works and variance.
2. On average, every six minutes a car passes by a sector in which it plans to build a tunnel. The passage of many vehicles indicate risk of congestion in the new construction. The engineers in charge of the project need to determine whether the probability that in one hour more than 5 vehicles that circulate by the sector is less than 0.04.
2. ## Re: Need help understanding these problems
problem two is a poisson distribution. The only trick here is to convert the rate of 1/6 car per minute to the # of cars per hour. Once you have the proper distribution just evaluate it's CDF at 5.
in problem 1 we assume that the whole days work is lost 33% of the time when it rains? So we are basically trying to determine the number of days out of 85 that the humidity messes up and everything is ruined.
This is a binomial distribution. with N=85, and p=33%=0.33 you can look up the distribution and derive the mean and variance easily enough. You're going to get numbers that are in units of #days of failure.
3. ## Re: Need help understanding these problems
Thx I worked on the first one, but the second one Im still having trouble with he secong one can't get the exact value
4. ## Re: Need help understanding these problems
Originally Posted by luis0426
Thx I worked on the first one, but the second one Im still having trouble with he secong one can't get the exact value
were you able to calculate $\lambda$ for x in hours? Do you have some way of evaluating the CDF of the poisson distribution?
5. ## Re: Need help understanding these problems
I knid of know how to solve it, I think my problem lies in how to work with this The only trick here is to convert the rate of 1/6 car per minute to the # of cars per hour. dont know if im doing it rigth
6. ## Re: Need help understanding these problems
Originally Posted by luis0426
I knid of know how to solve it, I think my problem lies in how to work with this The only trick here is to convert the rate of 1/6 car per minute to the # of cars per hour. dont know if im doing it rigth
so what did you come up with for $\lambda$ in units of cars per hour?
7. ## Re: Need help understanding these problems
1car/6min (60min/1h) = 6 cars per hour, dont know how to work with that and the determine whether the probability that in one hour more than 5 vehicles that circulate by the sector
8. ## Re: Need help understanding these problems
cough... how many 6 minute periods in 1 hr?
I don't know what circulate the sector means.. I just assume it means pass by, i.e. is an occurrence.
9. ## Re: Need help understanding these problems
Im a bit confused too, how would you solve to get λ
10. ## Re: Need help understanding these problems
1 car every 6 minutes ---> 10 cars an hour ----> $\lambda=10$
11. ## Re: Need help understanding these problems
That much I got my problem lies in the probability that in one hour more than 5 vehicles that circulate by the sector is less than 0.04, how does the 5 factor in here, I know that P(x=0.4,λ), how does the 5 factor in the problem.
12. ## Re: Need help understanding these problems
Originally Posted by luis0426
That much I got my problem lies in the probability that in one hour more than 5 vehicles that circulate by the sector is less than 0.04, how does the 5 factor in here, I know that P(x=0.4,λ), how does the 5 factor in the problem.
Pr[6 more cars go by] = 1 - Pr[5 or less cars go by] = F(5) where F(x) is the CDF of the Poisson distribution with parameter $\lambda=10$
13. ## Re: Need help understanding these problems
I got it like this P(x>6) = 1-P(x=0)-P(x=1-P(x=2)-P(x=3)-P(x=4)-P(x=5), but know if parameter λ=10 and x is 0 through 5, how do I work with the 0.4 here
14. ## Re: Need help understanding these problems
Originally Posted by luis0426
I got it like this P(x>6) = 1-P(x=0)-P(x=1-P(x=2)-P(x=3)-P(x=4)-P(x=5), but know if parameter λ=10 and x is 0 through 5, how do I work with the 0.4 here
you don't work it in there. The engineer just wants to check that the actual probability of that many cars is < 0.04.
He's in for a disappointment. | {"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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8565137386322021, "perplexity": 682.1353921348145}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1406510272940.33/warc/CC-MAIN-20140728011752-00285-ip-10-146-231-18.ec2.internal.warc.gz"} |
http://tex.stackexchange.com/questions/61260/keep-a-fragmented-tikz-image-at-the-same-place-on-a-beamer-slide | # Keep a fragmented tikz image at the same place on a beamer slide [duplicate]
Possible Duplicate:
How can I fix jumping TikZ pictures in beamer?
I am putting together a series of slides using beamer, and I'm using tike to do most of the figures. On one particular slide, I need to make nodes appear, then some more nodes, then edges between them. The code looks like this:
\begin{tikzpicture}[>=latex,text height=1.5ex,text depth=0.25ex]
\draw[help lines] (0,0) grid (10.5,7.5);
\tikzstyle{every node}=[draw,minimum size = .4cm,circle,draw=none,fill=red!30!white]
\tikzstyle{every path}=[solid,thick]
%% First species
\draw (3.7,2.2) node (a) {};
\draw (9.6,1.4) node (b) {};
\draw (6.6,6.6) node (c) {};
%% Second species
\draw <2-> (0.4,4.0) node [fill=blue!30!white] (d) {};
\draw <2-> (4.4,3.6) node [fill=blue!30!white] (e) {};
\draw <2-> (8.0,3.4) node [fill=blue!30!white] (f) {};
%% Third species
\draw <3-> (1.3,5.2) node [fill=green!30!white] (g) {};
\draw <3-> (4.5,6.4) node [fill=green!30!white] (h) {};
\draw <3-> (3.8,0.45) node [fill=green!30!white] (i) {};
%% Arrows
\path <4> [<->] (g) edge [bend right] (d);
\path <4> [<->] (g) edge [bend left] (e);
\path <4> [<->] (h) edge [bend left] (c);
\path <4> [<->] (e) edge [bend left] (a);
\path <4> [<->] (i) edge [bend left] (a);
\path <4> [<->] (i) edge [bend right] (f);
\path <4> [<->] (b) edge [bend right] (f);
\path <4> [<->] (e) edge [bend left] (f);
\path <4> [<->] (e) edge [bend right] (h);
\path <4> [<->] (f) edge [bend right] (c);
\end{tikzpicture}
My problem is that, when I remove the help lines, the figure changes positions on the slide between each transition, which is really not what I expected. I found a quick-and-dirty way to fix the problem with draw=none for the help line, but if anyone here ran into the same problem, I'd like to know a way towards a solution.
-
## marked as duplicate by Loop Space, percusse, Jake, egreg, Torbjørn T.Jun 28 '12 at 15:18
Please make sure to post complete compilable examples, starting from \documentclass. That makes it easier for others to try out your code. – Jake Jun 26 '12 at 14:27
\path[use as bounding box] (0,0) rectangle(10.5,7.5); | {"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": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7522941827774048, "perplexity": 6826.158514499855}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1418802766292.96/warc/CC-MAIN-20141217075246-00075-ip-10-231-17-201.ec2.internal.warc.gz"} |
https://quantum-journal.org/papers/q-2019-09-02-182/ | # Faster quantum simulation by randomization
Andrew M. Childs1,2,3, Aaron Ostrander2,3,4, and Yuan Su1,2,3
1Department of Computer Science, University of Maryland
2Institute for Advanced Computer Studies, University of Maryland
3Joint Center for Quantum Information and Computer Science, University of Maryland
4Department of Physics, University of Maryland
### Abstract
Product formulas can be used to simulate Hamiltonian dynamics on a quantum computer by approximating the exponential of a sum of operators by a product of exponentials of the individual summands. This approach is both straightforward and surprisingly efficient. We show that by simply randomizing how the summands are ordered, one can prove stronger bounds on the quality of approximation for product formulas of any given order, and thereby give more efficient simulations. Indeed, we show that these bounds can be asymptotically better than previous bounds that exploit commutation between the summands, despite using much less information about the structure of the Hamiltonian. Numerical evidence suggests that the randomized approach has better empirical performance as well.
### ► References
[1] Dorit Aharonov and Amnon Ta-Shma. Adiabatic quantum state generation and statistical zero knowledge. In Proceedings of the 35th ACM Symposium on Theory of Computing, pages 20–29, 2003. 10.1145/780542.780546. arXiv:quant-ph/0301023.
https://doi.org/10.1145/780542.780546
arXiv:quant-ph/0301023
[2] Ryan Babbush, Jarrod McClean, Dave Wecker, Alán Aspuru-Guzik, and Nathan Wiebe. Chemical basis of Trotter-Suzuki errors in quantum chemistry simulation. Physical Review A, 91: 022311, 2015. 10.1103/PhysRevA.91.022311. arXiv:1410.8159.
https://doi.org/10.1103/PhysRevA.91.022311
arXiv:1410.8159
[3] R. Barends, L. Lamata, J. Kelly, L. García-Álvarez, A. G. Fowler, A Megrant, E Jeffrey, T. C. White, D. Sank, J. Y. Mutus, B. Campbell, Yu Chen, Z. Chen, B. Chiaro, A. Dunsworth, I.-C. Hoi, C. Neill, P. J. J. O'Malley, C. Quintana, P. Roushan, A. Vainsencher, J. Wenner, E. Solano, and John M. Martinis. Digital quantum simulation of fermionic models with a superconducting circuit. Nature Communications, 6: 7654, 2015. 10.1038/ncomms8654. arXiv:1501.07703.
https://doi.org/10.1038/ncomms8654
arXiv:1501.07703
[4] Dominic W. Berry and Andrew M. Childs. Black-box Hamiltonian simulation and unitary implementation. Quantum Information and Computation, 12 (1-2): 29–62, 2012. arXiv:0910.4157. 10.26421/QIC12.1-2.
https://doi.org/10.26421/QIC12.1-2
arXiv:0910.4157
[5] Dominic W. Berry, Graeme Ahokas, Richard Cleve, and Barry C. Sanders. Efficient quantum algorithms for simulating sparse Hamiltonians. Communications in Mathematical Physics, 270 (2): 359–371, 2007. 10.1007/s00220-006-0150-x. arXiv:quant-ph/0508139.
https://doi.org/10.1007/s00220-006-0150-x
arXiv:quant-ph/0508139
[6] Dominic W. Berry, Andrew M. Childs, Richard Cleve, Robin Kothari, and Rolando D. Somma. Exponential improvement in precision for simulating sparse Hamiltonians. In Proceedings of the 46th ACM Symposium on Theory of Computing, pages 283–292, 2014. 10.1145/2591796.2591854. arXiv:1312.1414.
https://doi.org/10.1145/2591796.2591854
arXiv:1312.1414
[7] Dominic W. Berry, Andrew M. Childs, Richard Cleve, Robin Kothari, and Rolando D. Somma. Simulating Hamiltonian dynamics with a truncated Taylor series. Physical Review Letters, 114 (9): 090502, 2015a. 10.1103/PhysRevLett.114.090502. arXiv:1412.4687.
https://doi.org/10.1103/PhysRevLett.114.090502
arXiv:1412.4687
[8] Dominic W. Berry, Andrew M. Childs, and Robin Kothari. Hamiltonian simulation with nearly optimal dependence on all parameters. In Proceedings of the 56th IEEE Symposium on Foundations of Computer Science, pages 792–809, 2015b. 10.1109/FOCS.2015.54. arXiv:1501.01715.
https://doi.org/10.1109/FOCS.2015.54
arXiv:1501.01715
[9] Dominic W. Berry, Andrew M. Childs, Aaron Ostrander, and Guoming Wang. Quantum algorithm for linear differential equations with exponentially improved dependence on precision. Communications in Mathematical Physics, 356: 1057–1081, 2017. 10.1007/s00220-017-3002-y. arXiv:1701.03684.
https://doi.org/10.1007/s00220-017-3002-y
arXiv:1701.03684
[10] Dominic W. Berry, Andrew M. Childs, Yuan Su, Xin Wang, and Nathan Wiebe. Time-dependent Hamiltonian simulation with $L^1$-norm scaling, 2019. arXiv:1906.07115.
arXiv:1906.07115
[11] Fernando G. S. L. Brandao and Krysta M. Svore. Quantum speed-ups for solving semidefinite programs. In Proceedings of the 58th IEEE Symposium on Foundations of Computer Science, pages 415–426, 2017. 10.1109/FOCS.2017.45. arXiv:1609.05537.
https://doi.org/10.1109/FOCS.2017.45
arXiv:1609.05537
[12] Kenneth R. Brown, Robert J. Clark, and Isaac L. Chuang. Limitations of quantum simulation examined by simulating a pairing Hamiltonian using nuclear magnetic resonance. Physical Review Letters, 97: 050504, 2006. 10.1103/PhysRevLett.97.050504. arXiv:quant-ph/0601021.
https://doi.org/10.1103/PhysRevLett.97.050504
arXiv:quant-ph/0601021
[13] Earl Campbell. Shorter gate sequences for quantum computing by mixing unitaries. Physical Review A, 95: 042306, Apr 2017. 10.1103/PhysRevA.95.042306. arXiv:1612.02689.
https://doi.org/10.1103/PhysRevA.95.042306
arXiv:1612.02689
[14] Earl Campbell. Random compiler for fast Hamiltonian simulation. Physical Review Letters, 123: 070503, Aug 2019. 10.1103/PhysRevLett.123.070503. arXiv:1811.08017.
https://doi.org/10.1103/PhysRevLett.123.070503
arXiv:1811.08017
[15] Andrew M. Childs and Yuan Su. Nearly optimal lattice simulation by product formulas. Physical Review Letters, 123: 050503, Aug 2019. 10.1103/PhysRevLett.123.050503. arXiv:1901.00564.
https://doi.org/10.1103/PhysRevLett.123.050503
arXiv:1901.00564
[16] Andrew M. Childs, Richard Cleve, Enrico Deotto, Edward Farhi, Sam Gutmann, and Daniel A. Spielman. Exponential algorithmic speedup by quantum walk. In Proceedings of the 35th ACM Symposium on Theory of Computing, pages 59–68, 2003. 10.1145/780542.780552. arXiv:quant-ph/0209131.
https://doi.org/10.1145/780542.780552
arXiv:quant-ph/0209131
[17] Andrew M. Childs, Dmitri Maslov, Yunseong Nam, Neil J. Ross, and Yuan Su. Toward the first quantum simulation with quantum speedup. Proceedings of the National Academy of Sciences, 115 (38): 9456–9461, 2018. 10.1073/pnas.1801723115. arXiv:1711.10980.
https://doi.org/10.1073/pnas.1801723115
arXiv:1711.10980
[18] Edward Farhi, Jeffrey Goldstone, and Sam Gutmann. A quantum algorithm for the Hamiltonian NAND tree. Theory of Computing, 4 (1): 169–190, 2008. 10.4086/toc.2008.v004a008.
https://doi.org/10.4086/toc.2008.v004a008
[19] Richard P. Feynman. Simulating physics with computers. International Journal of Theoretical Physics, 21 (6-7): 467–488, 1982. 10.1007/BF02650179.
https://doi.org/10.1007/BF02650179
[20] Jeongwan Haah, Matthew B. Hastings, Robin Kothari, and Guang Hao Low. Quantum algorithm for simulating real time evolution of lattice Hamiltonians. In 2018 IEEE 59th Annual Symposium on Foundations of Computer Science (FOCS), pages 350–360, Oct 2018. 10.1109/FOCS.2018.00041. arXiv:1801.03922.
https://doi.org/10.1109/FOCS.2018.00041
arXiv:1801.03922
[21] Aram W. Harrow, Avinatan Hassidim, and Seth Lloyd. Quantum algorithm for linear systems of equations. Physical Review Letters, 103 (15): 150502, 2009. 10.1103/PhysRevLett.103.150502. arXiv:0811.3171.
https://doi.org/10.1103/PhysRevLett.103.150502
arXiv:0811.3171
[22] Matthew B. Hastings. Turning gate synthesis errors into incoherent errors. Quantum Information and Computation, 17 (5-6): 488–494, 2017. arXiv:1612.01011.
arXiv:1612.01011
[23] Stephen P. Jordan, Keith S. M. Lee, and John Preskill. Quantum algorithms for quantum field theories. Science, 336 (6085): 1130–1133, 2012. 10.1126/science.1217069. arXiv:1111.3633.
https://doi.org/10.1126/science.1217069
arXiv:1111.3633
[24] B. P. Lanyon, C. Hempel, D. Nigg, M. Müller, R. Gerritsma, F. Zähringer, P. Schindler, J. T. Barreiro, M. Rambach, G. Kirchmair, M. Hennrich, P. Zoller, R. Blatt, and C. F. Roos. Universal digital quantum simulation with trapped ions. Science, 334 (6052): 57–61, 2011. 10.1126/science.1208001. arXiv:1109.1512.
https://doi.org/10.1126/science.1208001
arXiv:1109.1512
[25] Seth Lloyd. Universal quantum simulators. Science, 273 (5278): 1073–1078, 1996. 10.1126/science.273.5278.1073.
https://doi.org/10.1126/science.273.5278.1073
[26] Guang Hao Low and Isaac L. Chuang. Optimal Hamiltonian simulation by quantum signal processing. Physical Review Letters, 118: 010501, 2017. 10.1103/PhysRevLett.118.010501. arXiv:1606.02685.
https://doi.org/10.1103/PhysRevLett.118.010501
arXiv:1606.02685
[27] Guang Hao Low and Isaac L. Chuang. Hamiltonian Simulation by Qubitization. Quantum, 3: 163, July 2019. 10.22331/q-2019-07-12-163. arXiv:1610.06546.
https://doi.org/10.22331/q-2019-07-12-163
arXiv:1610.06546
[28] Guang Hao Low, Vadym Kliuchnikov, and Nathan Wiebe. Well-conditioned multiproduct Hamiltonian simulation, 2019. arXiv:1907.11679.
arXiv:1907.11679
[29] David Poulin, Angie Qarry, Rolando D. Somma, and Frank Verstraete. Quantum simulation of time-dependent Hamiltonians and the convenient illusion of Hilbert space. Physical Review Letters, 106 (17): 170501, 2011. 10.1103/PhysRevLett.106.170501. arXiv:1102.1360.
https://doi.org/10.1103/PhysRevLett.106.170501
arXiv:1102.1360
[30] David Poulin, Matthew B. Hastings, Dave Wecker, Nathan Wiebe, Andrew C. Doherty, and Matthias Troyer. The Trotter step size required for accurate quantum simulation of quantum chemistry. Quantum Information and Computation, 15 (5-6): 361–384, 2015. arXiv:1406.4920.
arXiv:1406.4920
[31] Sadegh Raeisi, Nathan Wiebe, and Barry C. Sanders. Quantum-circuit design for efficient simulations of many-body quantum dynamics. New Journal of Physics, 14: 103017, 2012. 10.1088/1367-2630/14/10/103017. arXiv:1108.4318.
https://doi.org/10.1088/1367-2630/14/10/103017
arXiv:1108.4318
[32] Markus Reiher, Nathan Wiebe, Krysta M. Svore, Dave Wecker, and Matthias Troyer. Elucidating reaction mechanisms on quantum computers. Proceedings of the National Academy of Sciences, 114 (29): 7555–7560, 2017. 10.1073/pnas.1619152114. arXiv:1605.03590.
https://doi.org/10.1073/pnas.1619152114
arXiv:1605.03590
[33] Masuo Suzuki. General theory of fractal path integrals with applications to many-body theories and statistical physics. Journal of Mathematical Physics, 32 (2): 400–407, 1991. 10.1063/1.529425.
https://doi.org/10.1063/1.529425
[34] John Watrous. Simpler semidefinite programs for completely bounded norms. Chicago Journal of Theoretical Computer Science, 2013 (8), 2013. 10.4086/cjtcs.2013.008.
https://doi.org/10.4086/cjtcs.2013.008
[35] John Watrous. The Theory of Quantum Information. Cambridge University Press, 2018. 10.1017/9781316848142.
https://doi.org/10.1017/9781316848142
[36] Dave Wecker, Bela Bauer, Bryan K. Clark, Matthew B. Hastings, and Matthias Troyer. Gate count estimates for performing quantum chemistry on small quantum computers. Physical Review A, 90: 022305, 2014. 10.1103/PhysRevA.90.022305. arXiv:1312.1695.
https://doi.org/10.1103/PhysRevA.90.022305
arXiv:1312.1695
[37] Chi Zhang. Randomized algorithms for Hamiltonian simulation. In Leszek Plaskota and Henryk Woźniakowski, editors, Monte Carlo and Quasi-Monte Carlo Methods 2010, pages 709–719, Berlin, Heidelberg, 2012. Springer Berlin Heidelberg. ISBN 978-3-642-27440-4. 10.1007/978-3-642-27440-4_42.
https://doi.org/10.1007/978-3-642-27440-4_42
### Cited by
[1] William J. Huggins, Sam McArdle, Thomas E. O’Brien, Joonho Lee, Nicholas C. Rubin, Sergio Boixo, K. Birgitta Whaley, Ryan Babbush, and Jarrod R. McClean, "Virtual Distillation for Quantum Error Mitigation", Physical Review X 11 4, 041036 (2021).
[2] Lindsay Bassman, Miroslav Urbanek, Mekena Metcalf, Jonathan Carter, Alexander F Kemper, and Wibe A de Jong, "Simulating quantum materials with digital quantum computers", Quantum Science and Technology 6 4, 043002 (2021).
[3] Ian D. Kivlichan, Craig Gidney, Dominic W. Berry, Nathan Wiebe, Jarrod McClean, Wei Sun, Zhang Jiang, Nicholas Rubin, Austin Fowler, Alán Aspuru-Guzik, Hartmut Neven, and Ryan Babbush, "Improved Fault-Tolerant Quantum Simulation of Condensed-Phase Correlated Electrons via Trotterization", Quantum 4, 296 (2020).
[4] Dong An, Di Fang, and Lin Lin, "Time-dependent unbounded Hamiltonian simulation with vector norm scaling", Quantum 5, 459 (2021).
[5] Andrew M. Childs, Yuan Su, Minh C. Tran, Nathan Wiebe, and Shuchen Zhu, "Theory of Trotter Error with Commutator Scaling", Physical Review X 11 1, 011020 (2021).
[6] Leonardo Novo, "Bridging gaps between random approaches to quantum simulation", Quantum Views 4, 33 (2020).
[7] Mark Steudtner and Stephanie Wehner, "Estimating exact energies in quantum simulation without Toffoli gates", Physical Review A 101 5, 052329 (2020).
[8] Vincent E. Elfving, Marta Millaruelo, José A. Gámez, and Christian Gogolin, "Simulating quantum chemistry in the seniority-zero space on qubit-based quantum computers", Physical Review A 103 3, 032605 (2021).
[9] Yutaka Shikano, Hiroshi C. Watanabe, Ken M. Nakanishi, and Yu-ya Ohnishi, "Post-Hartree–Fock method in quantum chemistry for quantum computer", The European Physical Journal Special Topics 230 4, 1037 (2021).
[10] Suguru Endo, Zhenyu Cai, Simon C. Benjamin, and Xiao Yuan, "Hybrid Quantum-Classical Algorithms and Quantum Error Mitigation", Journal of the Physical Society of Japan 90 3, 032001 (2021).
[11] Jakob S Kottmann, Mario Krenn, Thi Ha Kyaw, Sumner Alperin-Lea, and Alán Aspuru-Guzik, "Quantum computer-aided design of quantum optics hardware", Quantum Science and Technology 6 3, 035010 (2021).
[12] Nicolas P. D. Sawaya, Tim Menke, Thi Ha Kyaw, Sonika Johri, Alán Aspuru-Guzik, and Gian Giacomo Guerreschi, "Resource-efficient digital quantum simulation of d-level systems for photonic, vibrational, and spin-s Hamiltonians", npj Quantum Information 6 1, 49 (2020).
[13] Yifan Sun, Jun-Yi Zhang, Mark S Byrd, and Lian-Ao Wu, "Trotterized adiabatic quantum simulation and its application to a simple all-optical system", New Journal of Physics 22 5, 053012 (2020).
[14] Yuan Su, Hsin-Yuan Huang, and Earl T. Campbell, "Nearly tight Trotterization of interacting electrons", arXiv:2012.09194, Quantum 5, 495 (2021).
[15] Kenneth Choi, Dean Lee, Joey Bonitati, Zhengrong Qian, and Jacob Watkins, "Rodeo Algorithm for Quantum Computing", Physical Review Letters 127 4, 040505 (2021).
[16] Tyson Jones and Simon Benjamin, "QuESTlink—Mathematica embiggened by a hardware-optimised quantum emulator* ", Quantum Science and Technology 5 3, 034012 (2020).
[17] Suguru Endo, Jinzhao Sun, Ying Li, Simon C. Benjamin, and Xiao Yuan, "Variational Quantum Simulation of General Processes", Physical Review Letters 125 1, 010501 (2020).
[18] Minh C. Tran, Yuan Su, Daniel Carney, and Jacob M. Taylor, "Faster Digital Quantum Simulation by Symmetry Protection", arXiv:2006.16248, PRX Quantum 2 1, 010323 (2021).
[19] Bela Bauer, Sergey Bravyi, Mario Motta, and Garnet Kin-Lic Chan, "Quantum Algorithms for Quantum Chemistry and Quantum Materials Science", Chemical Reviews 120 22, 12685 (2020).
[20] Yingkai Ouyang, David R. White, and Earl T. Campbell, "Compilation by stochastic Hamiltonian sparsification", Quantum 4, 235 (2020).
[21] Qi Zhao and Xiao Yuan, "Exploiting anticommutation in Hamiltonian simulation", Quantum 5, 534 (2021).
[22] Thi Ha Kyaw, Tim Menke, Sukin Sim, Abhinav Anand, Nicolas P.D. Sawaya, William D. Oliver, Gian Giacomo Guerreschi, and Alán Aspuru-Guzik, "Quantum Computer-Aided Design: Digital Quantum Simulation of Quantum Processors", Physical Review Applied 16 4, 044042 (2021).
[23] Yi-Tong Zou, Yu-Jiao Bo, and Ji-Chong Yang, "Optimize quantum simulation using a force-gradient integrator", EPL (Europhysics Letters) 135 1, 10004 (2021).
[24] Chi-Fang Chen, Hsin-Yuan Huang, Richard Kueng, and Joel A. Tropp, "Concentration for Random Product Formulas", PRX Quantum 2 4, 040305 (2021).
[25] Simon V. Mathis, Guglielmo Mazzola, and Ivano Tavernelli, "Toward scalable simulations of lattice gauge theories on quantum computers", Physical Review D 102 9, 094501 (2020).
[26] Sam McArdle, Suguru Endo, Alán Aspuru-Guzik, Simon C. Benjamin, and Xiao Yuan, "Quantum computational chemistry", arXiv:1808.10402, Reviews of Modern Physics 92 1, 015003 (2020).
[27] Dominic W. Berry, Andrew M. Childs, Yuan Su, Xin Wang, and Nathan Wiebe, "Time-dependent Hamiltonian simulation withL1-norm scaling", arXiv:1906.07115, Quantum 4, 254 (2020).
[28] Yi-Xiang Liu, Jordan Hines, Zhi Li, Ashok Ajoy, and Paola Cappellaro, "High-fidelity Trotter formulas for digital quantum simulation", Physical Review A 102 1, 010601 (2020).
[29] Alexander J. Buser, Tanmoy Bhattacharya, Lukasz Cincio, and Rajan Gupta, "State preparation and measurement in a quantum simulation of the O(3) sigma model", Physical Review D 102 11, 114514 (2020).
[30] Earl Campbell, "Random Compiler for Fast Hamiltonian Simulation", Physical Review Letters 123 7, 070503 (2019).
[31] Andrew M. Childs, Yuan Su, Minh C. Tran, Nathan Wiebe, and Shuchen Zhu, "A Theory of Trotter Error", arXiv:1912.08854.
[32] Andrew M. Childs and Yuan Su, "Nearly Optimal Lattice Simulation by Product Formulas", Physical Review Letters 123 5, 050503 (2019).
[33] Bryan O'Gorman, William J. Huggins, Eleanor G. Rieffel, and K. Birgitta Whaley, "Generalized swap networks for near-term quantum computing", arXiv:1905.05118.
[34] Suguru Endo, Qi Zhao, Ying Li, Simon Benjamin, and Xiao Yuan, "Mitigating algorithmic errors in a Hamiltonian simulation", arXiv:1808.03623, Physical Review A 99 1, 012334 (2019).
[35] Mark Steudtner and Stephanie Wehner, "Quantum codes for quantum simulation of fermions on a square lattice of qubits", Physical Review A 99 2, 022308 (2019).
[36] Ian D. Kivlichan, Christopher E. Granade, and Nathan Wiebe, "Phase estimation with randomized Hamiltonians", arXiv:1907.10070.
[37] Hrant Gharibyan, Masanori Hanada, Masazumi Honda, and Junyu Liu, "Toward simulating superstring/M-theory on a quantum computer", Journal of High Energy Physics 2021 7, 140 (2021).
[38] Seth Lloyd and Reevu Maity, "Efficient implementation of unitary transformations", arXiv:1901.03431.
[39] Seth Lloyd, Bobak T. Kiani, David R. M. Arvidsson-Shukur, Samuel Bosch, Giacomo De Palma, William M. Kaminsky, Zi-Wen Liu, and Milad Marvian, "Hamiltonian singular value transformation and inverse block encoding", arXiv:2104.01410.
[40] Benjamin D. M. Jones, George O. O'Brien, David R. White, Earl T. Campbell, and John A. Clark, "Optimising Trotter-Suzuki Decompositions for Quantum Simulation Using Evolutionary Strategies", arXiv:1904.01336.
[41] Sam McArdle, "Learning from Physics Experiments with Quantum Computers: Applications in Muon Spectroscopy", PRX Quantum 2 2, 020349 (2021).
The above citations are from Crossref's cited-by service (last updated successfully 2021-12-08 00:23:41) and SAO/NASA ADS (last updated successfully 2021-12-08 00:23:42). The list may be incomplete as not all publishers provide suitable and complete citation data. | {"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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7571699023246765, "perplexity": 22053.81587603315}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363437.15/warc/CC-MAIN-20211208022710-20211208052710-00370.warc.gz"} |
https://www.physicsforums.com/threads/the-special-linear-group.120068/ | # The Special Linear Group
1. May 7, 2006
### cmiller5277
Hey, quick question... if I am considering the set of nxn matrices of determinant 1 (the special linear group), is it correct if I say that the set is a submanifold in the (n^2)-dimensional space of matrices because the determinant function is a constant function, so its derivative is zero everywhere? Also, I've been told that the tangent space to this set at the identity is just the space of trace 0 matrices, but I'm not seeing where this relationship comes from. Any help?
2. May 7, 2006
### Hurkyl
Staff Emeritus
I'm not sure what that has to do with anything.
And note that the determinant map is constant on any subset of SL(n) -- even those subsets that are not manifolds.
Well, what is the tangent space?
Suppose I have a curve f(x) in SL(n) such that f(0) = I. What can you say about its derivative?
3. May 7, 2006
### cmiller5277
Well the first part of my question doesn't really have much to do with anything, you're right, but it was more of just a verification.. as in is it enough if I say the derivitave map is constant everywhere to show that SL(n) is a submanifold of the nxn space.
The trouble I'm having with the tangent space of SL(n) is that I am having difficulty with the curve f(x) in SL(n) itself. When you say we have a curve f(x) in SL(n) such that f(0) = I, is the derivative just going to be the derivative of I? And since the diagonal is composed of constant elements, we get their derivatives to be zero, therefore giving us a trace 0 matrix? I apologize if this is obvious.. I had trouble in Linear Algebra with this sort of thing. Thank you for the response though
4. May 7, 2006
### Hurkyl
Staff Emeritus
No; you want the derivative of f, not the derivative of f(0).
f is a function that takes values in an n²-dimensional vector space. Taking its derivative is no different than what you did in vector calculus when you worked in R² or R³... you just have many more components, and you arrange them in a square shape instead of a column.
(Incidentally, I think what you want to do is a differential approximation)
5. May 8, 2006
### mathwonk
you seem to be referring to the fact that a level set of a smooth function is a manifold if the function has maximal rank at every point of that level set. this does indeed apply to the set of matrices of determinant one and does prove they are a manifold.
the fact that the tangent space at the identity is the set of trace zero matricers can be checked directly, and is also the basis of the "frenet serret formulas".
6. May 8, 2006
### Hurkyl
Staff Emeritus
I ought to work through this problem again. Last time I did it, I threw my hands up in the air, decreed that "dt" was an infinitessimal, and looked at the equation det(I + A dt) = 1, where A was a tangent vector.
Actually, I still think about matrix Lie groups and algebras in that way. I suppose I ought to figure out how to justify it in terms of nonstandard analysis so I can feel better about myself.
7. May 9, 2006
### mathwonk
it is easy to see that det has rank one at an nxn matrix A where det(A) is not zero. just compose with a transverse parameter getting det(tA) = t^n det(A), and take the derivative at t=1.
to get the tangent space to f=c at p, just take the orthogonal complement of the gradient of f at p. for the determinant polynomial, the rgadient is the vector whose entry in the i,j place is plus or minus the minor determinant associated to that entry.
Hence when evaluated at the identity amtrix it has a zero in every entry except those indexed by i,i, and at those places it has a 1. the orthocomplement is thus matrices which dot to zero with this vector, i.e. those of trace zero. | {"extraction_info": {"found_math": false, "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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9404399991035461, "perplexity": 369.56988692263417}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125936833.6/warc/CC-MAIN-20180419091546-20180419111546-00693.warc.gz"} |
http://www.sawaal.com/exam-preparation/aieee-questions-and-answers.html | # AIEEE Questions
Q:
The diagonal of the floor of a rectangular closet is 7$\fn_jvn \frac{1}{2}$ feet. The shorter side of the closet is 4$\fn_jvn \frac{1}{2}$ feet. What is the area of the closet in square feet?
A) 9 B) 18 C) 27 D) 36
Explanation:
Other side = $\inline \sqrt{\left ( \frac{15}{2} \right )^{2}-\left ( \frac{9}{2} \right )^{2}}$ = $\inline \sqrt{\frac{225}{4}-\frac{81}{4}}$ = $\inline \sqrt{\frac{144}{4}}$ = 6 ft
Area of closet = (6 x 4.5) sq. ft = 27 sq. ft.
0 3289
Q:
Two electric bulbs marked 25W – 220V and 100W – 220V are connected in series to a 440Vsupply. Which of the bulbs will fuse?
A) 25 W B) Both C) 100 W D) Neither
Explanation:
Resistances of both the bulbs are
$R_{1}=\frac{V^{2}}{P_{1}}=\frac{220^{2}}{25}$
$R_{2}=\frac{V^{2}}{P_{2}}=\frac{220^{2}}{100}$
Hence $R_{1}>&space;R_{2}$
When connected in series, the voltages divide in them in the ratio of their resistances. The voltage of 440 V devides in such a way that voltage across 25 w bulb will be more than 220 V. Hence 25 w bulb will fuse.
Filed Under: Physics
Exam Prep: AIEEE
29 3029
Q:
There are three rooms in a Hotel: one single, one double and one for four persons. How many ways are there to house seven persons in these rooms ?
A) 105 B) 7! x 6! C) 7!/5! D) 420
Explanation:
Choose 1 person for the single room & from the remaining choose 2 for the double room & from the remaining choose 4 people for the four person room,
Then, 7C1 x 6C2 x 4C4
= 7 x $\inline \fn_jvn \small \frac{6x5}{2}$ x 1
= 7 x 15 = 105.
4 1285
Q:
Hydrogen atom is excited from ground state to another state with principal quantum number equal to 4. Then the number of spectral lines in the emission spectra will be
A) 2 B) 6 C) 5 D) 3
Explanation:
Number of spectral lines from a state n to ground state is
$=\frac{n(n-1)}{2}=6.$
Filed Under: Physics
Exam Prep: AIEEE
6 1214
Q:
In a face centred cubic lattice, atom A occupies the corner positions and atom B occupies the face centre positions. If one atom of B is missing from one of the face centred points, the formula of the compound is
$\inline \dpi{100} \small 1.\: A_{2}B_{5}$
$\inline \dpi{100} \small 2.\: A_{2}B$
$\inline \dpi{100} \small 3.\: AB_{2}$
$\inline \dpi{100} \small 4.\: A_{2}B_{3}$
A) Option 1 B) Option 2 C) Option 3 D) Option 4
Explanation:
$\inline \dpi{100} \small Z_{A}=\frac{8}{8}\; \; Z_{B}=\frac{5}{2}$
So formula of compound is $\inline \dpi{100} \small AB_{\frac{5}{2}}$
i.e., $\inline \dpi{100} \small A_{2}B_{5}$ | {"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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5475609302520752, "perplexity": 1521.9106489992103}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917122739.53/warc/CC-MAIN-20170423031202-00437-ip-10-145-167-34.ec2.internal.warc.gz"} |
https://www.bartleby.com/solution-answer/chapter-15-problem-33cr-introductory-chemistry-a-foundation-9th-edition/9781337399425/if-an-electric-current-is-passed-through-molten-sodium-chloride-elemental-chlorine-gas-is-generated/a57913ae-252e-11e9-8385-02ee952b546e | Chapter 15, Problem 33CR
### Introductory Chemistry: A Foundati...
9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Chapter
Section
### Introductory Chemistry: A Foundati...
9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
57 views
# If an electric current is passed through molten sodium chloride, elemental chlorine gas is generated as the sodium chloride is decomposed. :math> 2NaCl ( 1 ) → 2Na ( s ) + Cl 2 ( g ) at volume of chlorine gas measured at 767 mm Hg at 25 °C would be generated by complete decomposition of 1.25 g of NaCl?
Interpretation Introduction
Interpretation:
The volume of chlorine gas generated by complete decomposition of 1.25g of NaCl is to be calculated.
Concept Introduction:
A mole of a substance is defined as the same number of particles of the substance as present in 12g of 12 C. The number of particles present in one mole of a substance is 6.023×1023 particles. The number of moles a substance is given as,
n=mM
Where,
• m represents the mass of the substance.
• M represents the molar mass of the substance.
The ideal gas equation is used to represent the relation between the volume, pressure, temperature and number of moles of an ideal gas. The ideal gas equation is given as,
PV=nRT
Where,
• V represents the volume occupied by the ideal gas.
• P represents the pressure of the ideal gas.
• n represents the number of moles of the ideal gas.
• T represents the temperature of the ideal gas.
• R represents the ideal gas constant with value 0.08206LatmK1mol1.
Explanation
The molar mass of NaCl is 58.44gmol1.
The mass of NaCl is 1.25g.
The temperature of chlorine gas is 25°C.
The temperature of chlorine gas in Kelvin is given as,
T=25°C+273.15K=298.15K
The pressure of chlorine gas is 767mmHg.
The number of mole of substance is given as:
n=mMm
Where,
• m represents the mass of the substance.
• Mm represents the molar mass of the substance.
Substitute the value of the mass and molar mass of NaCl in above equation.
n=1.25g58.44gmol1=0.0214mol
The number of moles of NaCl reacted is 0.0214mol.
The given reaction is represented as:
2NaCll2Nas+Cl2g
Two moles of NaCl produce one mole of Cl2 gas, hence, the relation between the number of moles of NaCl reacted and chlorine gas produced is given as,
n1=n22
Where,
• n1 represents the number of moles of chlorine gas.
• n2 represents the number of moles of NaCl.
Substitute the value of n1 and n2 in the above equation
### Still sussing out bartleby?
Check out a sample textbook solution.
See a sample solution
#### The Solution to Your Study Problems
Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!
Get Started | {"extraction_info": {"found_math": false, "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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.88494473695755, "perplexity": 1916.0868461974383}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575541309137.92/warc/CC-MAIN-20191215173718-20191215201718-00461.warc.gz"} |
https://socratic.org/questions/how-do-you-solve-9a-4b-5-and-6a-2b-3-using-matrices | Precalculus
Topics
# How do you solve 9a - 4b = -5 and 6a - 2b = -3 using matrices?
Sep 22, 2016
$x = - 0.33$
$y = 0.5$
#### Explanation:
Look at the steps -
Sep 22, 2016
$\therefore a = - \frac{1}{3} \mathmr{and} b = \frac{1}{2}$
#### Explanation:
Although the method might seem quite daunting, once the preparation process is mastered, the method itself is surprisingly quick and easy, involving a few simple calculations.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
We have the following equations:
$9 a - 4 b = - 5 \text{ and } 6 a - 2 b = - 3$
First write them as matrices:
$\left(\begin{matrix}9 & - 4 \\ 6 & - 2\end{matrix}\right) \left(\begin{matrix}a \\ b\end{matrix}\right) = \left(\begin{matrix}- 5 \\ - 3\end{matrix}\right)$
Now find the inverse matrix of $A = \left(\begin{matrix}9 & - 4 \\ 6 & - 2\end{matrix}\right)$
$\left\mid A \right\mid = \left(9 \times - 2\right) - \left(6 \times - 4\right) = - 18 + 24 = 6$
${A}^{-} 1 = \frac{1}{6} \left(\begin{matrix}- 2 & 4 \\ - 6 & 9\end{matrix}\right) = \left(\textcolor{red}{\begin{matrix}- \frac{1}{3} & \frac{2}{3} \\ - 1 & \frac{3}{2}\end{matrix}}\right)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Multiply both sides of the matrix equation by the inverse matrix.
$\left(\textcolor{red}{\begin{matrix}- \frac{1}{3} & \frac{2}{3} \\ - 1 & \frac{3}{2}\end{matrix}}\right) \left(\begin{matrix}9 & - 4 \\ 6 & - 2\end{matrix}\right) \left(\begin{matrix}a \\ b\end{matrix}\right) = \left(\textcolor{red}{\begin{matrix}- \frac{1}{3} & \frac{2}{3} \\ - 1 & \frac{3}{2}\end{matrix}}\right) \left(\begin{matrix}- 5 \\ - 3\end{matrix}\right)$
$\textcolor{w h i t e}{\times \times \times \times x} \left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right) \left(\begin{matrix}a \\ b\end{matrix}\right) = \left(\begin{matrix}- \frac{1}{3} \\ \frac{1}{2}\end{matrix}\right)$
$\textcolor{w h i t e}{\times \times \times \times \times \times \times} \left(\begin{matrix}a \\ b\end{matrix}\right) = \left(\begin{matrix}- \frac{1}{3} \\ \frac{1}{2}\end{matrix}\right)$
$\therefore a = - \frac{1}{3} \mathmr{and} b = \frac{1}{2}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Background knowledge... to help with the method above..
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
A 2 x 2 matrix multiplied by the unit matrix remains unchanged
$\left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right) \left(\begin{matrix}a & b \\ c & d\end{matrix}\right) = \left(\begin{matrix}a & b \\ c & d\end{matrix}\right)$
A matrix multiplied by its inverse gives the unit matrix -
also known as the Identity Matrix.
$A \times {A}^{-} 1 = I = \left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right)$
To find the inverse matrix (${M}^{-} 1$) of matrix M
$M = \left(\begin{matrix}a & b \\ c & d\end{matrix}\right)$
1. Find the determinant $\left(\left\mid M \right\mid\right) = a d - b c$
2. ${M}^{-} 1 = \frac{1}{\left(\left\mid M \right\mid\right)} \left(\begin{matrix}d & - b \\ - c & a\end{matrix}\right)$
(swop a and d and change the signs of b and c), then divide by the determinant.)
##### Impact of this question
423 views around the world | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 18, "mathjax_inline_tex": 1, "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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.892830491065979, "perplexity": 1761.2447623705382}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027316021.66/warc/CC-MAIN-20190821131745-20190821153745-00382.warc.gz"} |
https://www.iaa.csic.es/publications/extremely-high-altitude-plume-seen-mars-morning-terminator | # An extremely high-altitude plume seen at Mars' morning terminator
DOI:
10.1038/nature14162
Publication date:
26/02/2015
Main author:
Sánchez-Lavega A.
IAA authors:
Sánchez-Lavega A.;Pérez-Hoyos S.;López-Valverde M.A.;González-Galindo F.
Authors:
Sánchez-Lavega A., Muñoz A.G., García-Melendo E., Pérez-Hoyos S., Gómez-Forrellad J.M., Pellier C., Delcroix M., López-Valverde M.A., González-Galindo F., Jaeschke W., Parker D., Phillips J., Peach D.
Journal:
Nature
Refereed:
Yes
Publication type:
Article
Volume:
518
Pages:
525-528
Number:
Abstract:
The Martian limb (that is, the observed 'edge' of the planet) represents a unique window into the complex atmospheric phenomena occurring there. Clouds of ice crystals (CO2 ice or H2 O ice) have been observed numerous times by spacecraft and ground-based telescopes, showing that clouds are typically layered and always confined below an altitude of 100 kilometres; suspended dust has also been detected at altitudes up to 60 kilometres during major dust storms1-6. Highly concentrated and localized patches of auroral emission controlled by magnetic field anomalies in the crust have been observed at an altitude of 130 kilometres7. Here we report the occurrence in March and April 2012 of two bright, extremely high-altitude plumes at the Martian terminator (the day-night boundary) at 200 to 250 kilometres or more above the surface, and thus well into the ionosphere and the exosphere8,9. They were spotted at a longitude of about 195° west, a latitude of about -45° (at Terra Cimmeria), extended about 500 to 1,000 kilometres in both the north-south and east-west directions, and lasted for about 10 days. The features exhibited day-to-day variability, and were seen at the morning terminator but not at the evening limb, which indicates rapid evolution in less than 10 hours and a cyclic behaviour. We used photometric measurements to explore two possible scenarios and investigate their nature. For particles reflecting solar radiation, clouds of CO2 -ice or H2 O-ice particles with an effective radius of 0.1 micrometres are favoured over dust. Alternatively, the plume could arise from auroral emission, of a brightness more than 1,000 times that of the Earth's aurora, over a region with a strong magnetic anomaly where aurorae have previously been detected7. Importantly, both explanations defy our current understanding of Mars' upper atmosphere.
Database:
WOK
SCOPUS | {"extraction_info": {"found_math": false, "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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8594947457313538, "perplexity": 6978.512009341802}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400249545.55/warc/CC-MAIN-20200926231818-20200927021818-00709.warc.gz"} |
http://mathhelpforum.com/algebra/67364-prove-no-rational-satisfies-2-x-3-a.html | # Thread: Prove no rational satisfies 2^x = 3
1. ## Prove no rational satisfies 2^x = 3
Here is the question and i have no idea how to go about it any help would be hugely appreciated.
Prove no rational satisfies
$\displaystyle 2^x = 3$
Many thanks
2. Originally Posted by craig in a post he soft deleted and I hard deleted
You would firstly take logs of both sides, $\displaystyle \log_{2} 2^x=\log_{2} 3$.
As $\displaystyle \log_{y} y^x = x$, we can say that:
$\displaystyle x=\log_{2} 3$
Put this into your calculator and you get $\displaystyle x=1.584962501$, not a rational number
On the contrary, 1.584962501 certainly IS a rational number because it is a terminating decimal! $\displaystyle log_2(3)$,which is NOTequal to 1.585962501 is not rational but I suspect it would be necessary to prove that to get credit for this problem!
3. Originally Posted by sebjory
Here is the question and i have no idea how to go about it any help would be hugely appreciated.
Prove no rational satisfies
$\displaystyle 2^x = 3$
Many thanks
This is in the prealgebra/algebra section, so i am not sure if a proof like this (a proof by contradiction) will be too complicated. if it is, tell me, we can come up with another.
Assume, to the contrary, that there is a rational number $\displaystyle x$ such that $\displaystyle 2^x = 3$.
Then $\displaystyle x = \frac ab$, where $\displaystyle a$ and $\displaystyle b$ are integers, with $\displaystyle b \ne 0$. So we have
$\displaystyle 2^{\frac ab} = 3$
$\displaystyle \Rightarrow 2^a = 3^b$
However, this is absurd, since $\displaystyle a$ and $\displaystyle b$ are integers, this equation contradicts the uniqueness of prime factorization (since the left side is always a product of 2's and the right is always a product of 3's).
note that a = b = 0 makes the equation make sense... but b cannot be zero!
4. thankyou proof by contradiction is taught over here in the UK.
5. Originally Posted by sebjory
thankyou proof by contradiction is taught over here in the UK.
oh, you're from the UK, hehe, I was worried. you have a US flag up
i suppose you are also familiar with the uniqueness of prime factorization theorem (you might know it under a different name), but it talks about how each integer can be uniquely expressed as a product of primes with integer powers
6. Yes, we're studying this whole semester on these sorts of theorems about rational numbers (the Completeness axiom etx) and with these come a whole host of proofs...
Do you know any good proof resource where i can learn as many types of proof as possible? (not including Induction since i am fairly happy with my ability to manipulate these sorts of problems)
7. Originally Posted by sebjory
Yes, we're studying this whole semester on these sorts of theorems about rational numbers (the Completeness axiom etx) and with these come a whole host of proofs...
Do you know any good proof resource where i can learn as many types of proof as possible? (not including Induction since i am fairly happy with my ability to manipulate these sorts of problems)
completeness axiom? ...this is prealgebra?!!
i am sure there are resources online that have what you are looking for. i guess wikipedia is always a good place to start, at least for quick reference. but i have never really searched for any such site, as i have several textbooks that cover such material and am fairly happy with them, so i don't know. sorry
8. Originally Posted by Jhevon
This is in the prealgebra/algebra section, so i am not sure if a proof like this (a proof by contradiction) will be too complicated. if it is, tell me, we can come up with another.
Assume, to the contrary, that there is a rational number $\displaystyle x$ such that $\displaystyle 2^x = 3$.
Then $\displaystyle x = \frac ab$, where $\displaystyle a$ and $\displaystyle b$ are integers, with $\displaystyle b \ne 0$. So we have
$\displaystyle 2^{\frac ab} = 3$
$\displaystyle \Rightarrow 2^a = 3^b$
However, this is absurd, since $\displaystyle a$ and $\displaystyle b$ are integers, this equation contradicts the uniqueness of prime factorization (since the left side is always a product of 2's and the right is always a product of 3's).
note that a = b = 0 makes the equation make sense... but b cannot be zero!
Another argument could be
$\displaystyle 2^a$ is always even and $\displaystyle 3^b$ is always odd.
Clearly, therefore, $\displaystyle 2^a = 3^b$ where a and b are integers is nonsense, giving the contradiction and completing the proof.
9. Originally Posted by Prove It
Another argument could be
$\displaystyle 2^a$ is always even and $\displaystyle 3^b$ is always odd.
Clearly, therefore, $\displaystyle 2^a = 3^b$ where a and b are integers is nonsense, giving the contradiction and completing the proof.
even easier! | {"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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8939558267593384, "perplexity": 313.53906183694835}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794865830.35/warc/CC-MAIN-20180523215608-20180523235608-00139.warc.gz"} |
https://gamedev.stackexchange.com/questions/182862/should-3d-transformations-be-represented-by-a-4x4-matrix-or-a-3x4-matrix | # Should 3D transformations be represented by a 4x4 matrix or a 3x4 matrix?
Since 3D transformations are represented by 4x4 homogeneous matrices we know that their last row is always (0,0,0,1), and as such the behavior of this final row is implied so long as we know whether or not the transformation is operating on a vector (a 4x1 matrix with a w element of 0) or a point (a 4x1 matrix with a w element of 1).
If we have classes that represent points and vectors separately as 3x1 matrices with implied w coordinates then we can overload the * operator so that its behavior on both classes is separate and thus there is no need to store the final row of the transformation matrix in memory.
Now I can see how this would be a great optimization at first as it increases cache coherency instead of adding what is essentially padding to each transformation matrix. On the other hand, we will have to eventually use a full 4x4 matrix to handle the projection and perspective divide later on so is it worth it to store a transformation matrix as a 3x4 matrix only to later convert it into a full 4x4? Or would a good solution here be to make the projection matrix a 4x4 matrix and make the product of a transformation matrix and a 4x4 matrix another 4x4 matrix?
I'm sorry if this seems basic or like a question that answers itself but I was reading a book on game engine development and it used implications to handle interactions between transformation matrices with points and vectors, and other transformations separately, never multiplying by the final row and using implications instead. But on creation they would still stored the final row as (0,0,0,1) in memory to only be used in multiplication between a transformation matrix and a 4x4 matrix specifically and so I'm mostly just wondering if there may have been a good reason for that or not. Thanks in advance for the help.
• Does this answer your question? What types of matrices are needed for game and graphics programming? – Theraot May 24 at 11:13
• There are many 3D transformations, but only a handful of them can be represented with a 4x4 matrix. Luckily, translation, rotation, scaling and perspective projection are amongst them, which is enough for most games. – G. Sliepen Jun 23 at 20:14
Since 3D transformations are represented by 4x4 homogeneous matrices
Correction: Might be represented by 4x4 matrices ;)
Now I can see how this would be a great optimization at first as it increases cache coherency instead of adding what is essentially padding to each transformation matrix.
I don't think that cache coherence is a major issue here. More relevant are the computations that you can avoid. A 4x4 matrix multiplied with a 4 element vector results in 16 multiplications and 12 additions. If you use a 3x4 matrix it is only 12 multiplications and 9 additions. So you have 25% less of the computational costs. You also save some memory but I don't think it is as relevant as the computational savings.
I can remember having a similar discussion about this topic on www.gamedev.net. It was going even further by just using 3x3 matrices and handling translations separately by vector addition (computations: 9 multiplications, 9 additions) . If I remember correctly the conclusion was, that 4x4 matrices are easier to use, since they contain translations and also handle point vs vector issues for you. This "luxury" comes with a performance impact. However, I can also remember that people stated, that this performance impact is negligible most of the time since the matrix multiplications are done during vertex processing and this is usually not a bottleneck in a game engine. To be sure, you have to benchmark it.
On the other hand, we will have to eventually use a full 4x4 matrix to handle the projection and perspective divide later on so is it worth it to store a transformation matrix as a 3x4 matrix only to later convert it into a full 4x4? Or would a good solution here be to make the projection matrix a 4x4 matrix and make the product of a transformation matrix and a 4x4 matrix another 4x4 matrix?
I think you are approaching this a little bit too complicated. You have probably something in mind like this:
$$v_p = P \cdot T \cdot v$$
where P is the 4x4 projection matrix and T an arbitrary 3x4 transformation matrix. v is the Vertex that should be transformed and v_p the fully transformed vertex. If I understood you correctly, then your problem here is, that 4x4 matrices can't be multiplied with 3x4 matrices, so you think about introducing a special operation for that. However, since v times a matrix always yields a vector, you can simply do the following (GLSL like pseudo-code):
vec3 v_t = T * v;
vec4 v_p = P * vec4(v_t, 1.0);
So just extend the vector by one element after the 3x4 multiplication and multiply the 4x4 projection matrix afterward. However, there is one other problem with 3x4 matrices. You can't multiply them with each other and you can't multiply them with the result of a preceding 3x4 matrix-vector multiplication since they yield a vec3 but can only be multiplied with a vec4. So you have a problem in case you have multiple 3x4 transformation matrices. You can solve this exactly the same way as in the previous example by doing one matrix-vector multiplication at a time and extending the result vector to 4 components, but I don't think this is a smart solution. This is why you should either use 4x4 matrices or 3x3 matrices and handle translations by vector addition. However, since the 4. column of a 3x4 matrix is identical to the translations in this special case, you can still use it as a storage format and just separate it into a 3x3 matrix and a translation vector during the calculations.
classes that represent points and vectors separately as 3x1 matrices
Note, that you have to use 3x3 matrices during multiplication if you want to use vectors of size 3 since you can't multiply a 3x4 matrix with a 3x1 matrix. So I assume that you or the book you are reading separates a 3x4 matrix into a 3x3 matrix and a translation before using it.
Conclusion
Using 3x3 matrices + translations or alternatively 3x4 matrices gives you performance and storage benefits by omitting redundant operations and values but 4x4 matrices are more comfortable to use. The performance benefits might also not be noticeable if your game is not utilizing the full corresponding hardware capacity, so benchmark it, if you are not sure if this is relevant for you. In the end, it is a design decision where you trade more comfortable/readable code against some (possible!) performance gain.
• I can't help but think that the bulk of optimization potential here is largely irrelevant because vertex processing will be happening on the GPU, not CPU. Fillrate or ROP are much more likely to be the bottleneck. – Maximus Minimus May 24 at 16:11
• @MaximusMinimus Never said it will happen on the CPU ;) Nevertheless, fewer operations are fewer operations, so it is faster to use smaller matrices, CPU, or GPU. However, as I also stated the "performance impact is negligible most of the time since the matrix multiplications are done during vertex processing and this is usually not a bottleneck". Personally I think that unless you write AAA games the small performance gain (probably in the low nanoseconds range per frame) isn't worth the trouble but I tried to answer the question as detailed as possible. ;) – wychmaster May 24 at 16:45 | {"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": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.742623507976532, "perplexity": 381.72202772525935}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107885126.36/warc/CC-MAIN-20201025012538-20201025042538-00245.warc.gz"} |
https://www.bleepingcomputer.com/forums/t/34380/i-h8-ssk/ | Welcome to BleepingComputer, a free community where people like yourself come together to discuss and learn how to use their computers. Using the site is easy and fun. As a guest, you can browse and view the various discussions in the forums, but can not create a new topic or reply to an existing one unless you are logged in. Other benefits of registering an account are subscribing to topics and forums, creating a blog, and having no ads shown anywhere on the site.
# I H8 Ssk
### #1 Blak Jak
Blak Jak
• Members
• 1 posts
• OFFLINE
•
• Local time:01:40 PM
Posted 04 November 2005 - 08:53 AM
hey guys/gals- i thought i had removed surf side kick until this morning- when another IE pop up told me "this ad brought to you by surf side kick." well thank you very much ssk- i hate you. i tried following a previous post's directions @ http://www.bleepingcomputer.com/forums/t/9549/how-to-remove-surfsidekick-2-or-3-and-vcclient/ when removing ssk from add/remove programs i typed the code in it requested and it appeared removed. i rebooted my computer and then ran hijack this. i saw some of the same files as the ones that were in the post- however the problem is when i click fix checked and scan again- the files are still there. my guess is is that they remake themselves everytime you delete them. please help!
thanks yall,
blakjak
GO VIRGINIA TECH HOKIES!
Logfile of HijackThis v1.99.1
Scan saved at 8:26:14 AM, on 11/4/2005
Platform: Windows XP SP2 (WinNT 5.01.2600)
MSIE: Internet Explorer v6.00 SP2 (6.00.2900.2180)
Running processes:
C:\WINDOWS\System32\smss.exe
C:\WINDOWS\system32\winlogon.exe
C:\WINDOWS\system32\services.exe
C:\WINDOWS\system32\lsass.exe
C:\WINDOWS\System32\Ati2evxx.exe
C:\WINDOWS\system32\svchost.exe
C:\WINDOWS\System32\svchost.exe
C:\WINDOWS\system32\LEXBCES.EXE
C:\WINDOWS\system32\spoolsv.exe
C:\WINDOWS\system32\LEXPPS.EXE
C:\WINDOWS\Explorer.EXE
C:\Program Files\ATI Technologies\ATI Control Panel\atiptaxx.exe
C:\Program Files\Synaptics\SynTP\SynTPLpr.exe
C:\Program Files\Synaptics\SynTP\SynTPEnh.exe
C:\Program Files\Java\j2re1.4.2_01\bin\jusched.exe
C:\Program Files\Acer\Notebook Manager\almxptray.exe
C:\Program Files\3M\PSNLite\PsnLite.exe
C:\PROGRA~1\3M\PSNLite\PSNGive.exe
C:\PROGRA~1\Grisoft\AVGFRE~1\avgamsvr.exe
C:\PROGRA~1\Grisoft\AVGFRE~1\avgupsvc.exe
c:\Program Files\WIDCOMM\Bluetooth Software\bin\btwdins.exe
C:\Program Files\ewido\security suite\ewidoctrl.exe
C:\WINDOWS\system32\inetsrv\inetinfo.exe
C:\WINDOWS\System32\svchost.exe
C:\WINDOWS\system32\wwSecure.exe
C:\WINDOWS\system32\wscntfy.exe
C:\WINDOWS\system32\wuauclt.exe
C:\Documents and Settings\blakjak\Desktop\hijackthis\HijackThis.exe
R0 - HKCU\Software\Microsoft\Internet Explorer\Main,Local Page =
R1 - HKCU\Software\Microsoft\Internet Connection Wizard,ShellNext = http://global.acer.com/
R1 - HKCU\Software\Microsoft\Windows\CurrentVersion\Internet Settings,ProxyOverride = localhost
R3 - URLSearchHook: (no name) - {02EE5B04-F144-47BB-83FB-A60BD91B74A9} - C:\Program Files\SurfSideKick 3\SskBho.dll
O4 - HKLM\..\Run: [LaunchApp] Alaunch
O4 - HKLM\..\Run: [ATIModeChange] Ati2mdxx.exe
O4 - HKLM\..\Run: [ATIPTA] C:\Program Files\ATI Technologies\ATI Control Panel\atiptaxx.exe
O4 - HKLM\..\Run: [SynTPLpr] C:\Program Files\Synaptics\SynTP\SynTPLpr.exe
O4 - HKLM\..\Run: [SynTPEnh] C:\Program Files\Synaptics\SynTP\SynTPEnh.exe
O4 - HKLM\..\Run: [SunJavaUpdateSched] C:\Program Files\Java\j2re1.4.2_01\bin\jusched.exe
O4 - HKLM\..\Run: [AcerNotebookManager] C:\Program Files\Acer\Notebook Manager\almxptray.exe
O4 - HKLM\..\Run: [SurfSideKick 3] C:\Program Files\SurfSideKick 3\Ssk.exe
O4 - HKLM\..\RunServices: [Microsoft Update 64 BIT] wininit32.exe
O4 - HKCU\..\Run: [AIM] C:\Program Files\AIM\aim.exe -cnetwait.odl
O4 - HKCU\..\Run: [SurfSideKick 3] C:\Program Files\SurfSideKick 3\Ssk.exe
O4 - Global Startup: Post-it® Software Notes Lite.lnk = C:\Program Files\3M\PSNLite\PsnLite.exe
O8 - Extra context menu item: E&xport to Microsoft Excel - res://C:\PROGRA~1\MICROS~2\Office10\EXCEL.EXE/3000
O9 - Extra button: AIM - {AC9E2541-2814-11d5-BC6D-00B0D0A1DE45} - C:\Program Files\AIM\aim.exe
O18 - Protocol: bw+0 - {A1F2B24E-E72D-4983-BE6D-5E768C5731D3} - C:\Program Files\Logitech\Desktop Messenger\8876480\Program\BWPlugProtocol-8876480.dll
O18 - Protocol: pcl - {182D0C85-206F-4103-B4FA-DCC1FB0A0A44} - C:\Program Files\Autodesk\Inventor Professional 9\bin\HSPCLPRO10.dll
O23 - Service: Ati HotKey Poller - Unknown owner - C:\WINDOWS\System32\Ati2evxx.exe
O23 - Service: AVG7 Alert Manager Server (Avg7Alrt) - GRISOFT, s.r.o. - C:\PROGRA~1\Grisoft\AVGFRE~1\avgamsvr.exe
O23 - Service: AVG7 Update Service (Avg7UpdSvc) - GRISOFT, s.r.o. - C:\PROGRA~1\Grisoft\AVGFRE~1\avgupsvc.exe
O23 - Service: Bluetooth Service (btwdins) - WIDCOMM, Inc. - c:\Program Files\WIDCOMM\Bluetooth Software\bin\btwdins.exe
O23 - Service: ewido security suite control - ewido networks - C:\Program Files\ewido\security suite\ewidoctrl.exe
O23 - Service: iPod Service (iPodService) - Apple Computer, Inc. - C:\Program Files\iPod\bin\iPodService.exe
O23 - Service: LexBce Server (LexBceS) - Lexmark International, Inc. - C:\WINDOWS\system32\LEXBCES.EXE
O23 - Service: Washer Security Access (wwSecSvc) - Webroot Software, Inc. - C:\WINDOWS\system32\wwSecure.exe
### #2 viccy
viccy
Malware Exterminator
• Security Colleague
• 433 posts
• OFFLINE
•
• Gender:Male
• Location:Kansas
• Local time:01:40 PM
Posted 04 November 2005 - 10:57 AM
Welcome to the forum.
Yes, surfsidekick can be difficult to get rid of.
Close Internet Explorer and keep it closed throughout the entire removal process.
Enter the control panel by clicking on the Start menu, then clicking on Run.
Now type control in the Open field and press the OK button.
Double-click on the Add/Remove Programs icon.
Look for and uninstall the following entries if found in the Add/Remove Programs window.
Surf Sidekick
It may prompt about whether or not you are sure you want to remove this program. Reply Yes to this prompt. It will then uninstall the program.
If there is no Add/Remove Programs entry for this programs, click on Start , then Run and type the following in the Open : field :
C:\Program Files\SurfSideKick 3\Ssk.exe /u
and press the OK button. A code will be displayed that it will ask you to enter. Enter this code and reboot. Once back to your desktop continue with the rest of the fix.
Scan with Hijack This and put a checkmark next to the following entries if they exist:
R3 - URLSearchHook: (no name) - {02EE5B04-F144-47BB-83FB-A60BD91B74A9} - C:\Program Files\SurfSideKick 3\SskBho.dll
O4 - HKLM\..\Run: [SurfSideKick 3] C:\Program Files\SurfSideKick 3\Ssk.exe
O4 - HKCU\..\Run: [SurfSideKick 3] C:\Program Files\SurfSideKick 3\Ssk.exe
O4 - HKLM\..\RunServices: [Microsoft Update 64 BIT] wininit32.exe
Close all windows and browsers and click "fix checked"
Delete the following directories if they exist:
C:\PROGRAM FILES\SurfSideKick
Search for the following files and if found delete them:
SskBho.dll
Ssk.log
SskUpdater.exe
Ssk.exe
wininit32.exe
Next, download CleanUp 4.0 . Install and run it. Open Cleanup! by double-clicking the icon on your desktop (or from the Start > All Programs menu).
Set the program up as follows:
Click "Options..."
Move the arrow down to "Custom CleanUp!"
Put a check next to the following (Make sure nothing else is checked!):
Empty Recycle Bins
Delete Prefetch files
Cleanup! All Users
Click OK
Press the CleanUp! button to start the program.
It may ask you to reboot at the end, click NO.
Install it using the Standard Install option. (You will be asked for your e-mail address, it is safe to give it. If you receive alerts from your firewall, allow all activities for Spy Sweeper)
You will be prompted to check for updated definitions, please do so.
(This may take several minutes)
When Spy Sweeper has updated, reboot to safe mode.
Boot into Safe Mode:
Restart your computer and immediately begin tapping the F8 key on your keyboard.
If done right a Windows Advanced Options menu will appear. Select the Safe Mode option and press Enter.
Open Spy Sweeper and click on Sweep and allow it to fully scan your system.
When the sweep has finished, click Remove to remove any items found. Save the log.
Exit Spy Sweeper.
Reboot to normal mode and post the results from Spy Sweeper along with a new Hijack This log.
Edited by viccy, 04 November 2005 - 11:04 AM.
#### 0 user(s) are reading this topic
0 members, 0 guests, 0 anonymous users | {"extraction_info": {"found_math": false, "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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9321799278259277, "perplexity": 29089.235953122505}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039741016.16/warc/CC-MAIN-20181112172845-20181112194845-00526.warc.gz"} |
https://www.assignmentexpert.com/homework-answers/economics/microeconomics/question-35437 | 64 646
Assignments Done
99,2%
Successfully Done
In September 2018
Answer to Question #35437 in Microeconomics for Tips
Question #35437
Suppose that there are two goods, X and Y . Thomasíutility function is
U = XY + X. The price of X is P, and the price of Y is $40. Income is$200. Derive his individual demand for X as a function of P.
Suppose that there are two goods, X and Y. Thomas utility function is
U = XY + X. The price of X is P, and the price of Y is $40. Income is$200. Derive his individual demand for X as a function of P.
Income function is: I = Px*X + Py*Y = 200
P*X + 40Y = 200
The individual demand for X as a function of P will be:
Qd = (200 - 40Y)/P
Need a fast expert's response?
Submit order
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS! | {"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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4788644313812256, "perplexity": 2471.2532698437394}, "config": {"markdown_headings": false, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267158609.70/warc/CC-MAIN-20180922162437-20180922182837-00474.warc.gz"} |
https://math.meta.stackexchange.com/questions/27977/expanded-usercard-not-working-in-vivaldi/27978 | # Expanded Usercard not Working in Vivaldi
Having just got my shiny new 1k privilege of the expanded usercard, I was dismayed to find out that it doesn't seem to work in Vivaldi (as a matter of fact, I'm not sure I ever got any usercard in Vivaldi), though it does in Firefox. Has anyone else seen this behavior? As Vivaldi is based on Chromium, I'd have expected the usercard feature to work. As Vivaldi is the fastest browser that's still reasonably compatible and private, it's my favorite right now, and it'd be great if this feature worked. Do I need to change a setting or something?
[EDIT] I think disabling my popup blocker on this site, closing out and coming back in may have fixed it.
That's fine, I can still see your expanded usercard (No need to panic about that previlege, I have just had only $300$ points today). I'm using Stands Fair AdBlocker on Google Chrome, I'm not sure if it also works on Internet, Firefox or which other ad-blockers. | {"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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6022779941558838, "perplexity": 2246.1029527578216}, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590348502204.93/warc/CC-MAIN-20200605174158-20200605204158-00214.warc.gz"} |
https://physics.stackexchange.com/questions/76126/is-there-any-relationship-between-gravity-and-electromagnetism/76136 | # Is there any relationship between Gravity and Electromagnetism? [duplicate]
We all know that the universe is governed by four Fundamental Forces which are The strong force , The weak force , The electromagnetic force and The gravitational force .
Now, is there any relationship between Electromagnetism and gravity?
• Possible duplicates: physics.stackexchange.com/q/944/2451 and links therein. – Qmechanic Sep 3 '13 at 15:46
• ncatlab.org/nlab/show/Kaluza-Klein+mechanism – Urs Schreiber Sep 4 '13 at 18:51
• I have no formal physics training, but it seems to me that they are intrinsically connected. If gravity effects mass and mass & energy (electromagnetic waves or light) are related through Einsteins equation, then gravity and light possess a very clear relationship. It seems logical to me, but can someone more qualified chime in? – user29224 Sep 4 '13 at 22:43
• @UrsSchreiber: Oh my god, that's an Excellent article! – Abhimanyu Pallavi Sudhir Sep 7 '13 at 8:43
• @DImension10, thanks for the feedback. I just went through that entry again and expanded a bit more here and there. For instance the Examples-section now has a new subsection "Cascades of KK-reductions from holographic boundaries" ncatlab.org/nlab/show/… . – Urs Schreiber Sep 7 '13 at 12:50
# On Unification
I presume you're asking whether just classical gravity & classical EM can be unified.
They sure can!
Classical General Relativity and Classical Electromagnetism are unified in Kaluza-Klein-Theory, which proves that 5-dimensional general relativity is equivalent to 4-dimensional general relativity plus 4-dimensional maxwell equations. Rather interesting, isn't it? A byproduct is the scalar "Radion" or "Dilaton" which appears due to the "55" component of the metric tensor. In other words, the Kaluza-Klein metric tensor equals the GR metric tensor with maxwell stuff on the right and at the bottom; BUT you have an extra field down there.
$${g_{\mu \nu }} = \left[ {\begin{array}{*{20}{c}} {{g_{11}}}&{{g_{12}}}&{{g_{13}}}&{{g_{14}}}&{{g_{15}}} \\ {{g_{21}}}&{{g_{22}}}&{{g_{23}}}&{{g_{24}}}&{{g_{25}}} \\ {{g_{31}}}&{{g_{32}}}&{{g_{33}}}&{{g_{34}}}&{{g_{35}}} \\ {{g_{41}}}&{{g_{42}}}&{{g_{43}}}&{{g_{44}}}&{{g_{45}}} \\ {{g_{51}}}&{{g_{52}}}&{{g_{53}}}&{{g_{54}}}&{{g_{55}}} \end{array}} \right]$$
Imagine 2 imaginary lines now.
$${g_{\mu \nu }} = \left[ {\begin{array}{*{20}{cccc|c}} {{g_{11}}}&{{g_{12}}}&{{g_{13}}}&{{g_{14}}} & {{g_{15}}} \\ {{g_{21}}}&{{g_{22}}}&{{g_{23}}}&{{g_{24}}} & {{g_{25}}} \\ {{g_{31}}}&{{g_{32}}}&{{g_{33}}}&{{g_{34}}} & {{g_{35}}} \\ {{g_{41}}}&{{g_{42}}}&{{g_{43}}}&{{g_{44}}} & {{g_{45}}} \\ \hline {{g_{51}}}&{{g_{52}}}&{{g_{53}}}&{{g_{54}}} & {{g_{55}}} \end{array}} \right]$$
So the stuff on the top-left is the GR metric for gravity, and the stuff on the edge ($g_{j5}$ and $g_{5j}$) is for electromagnetism and you have an additional component on the bottom right. This is the radion/dilaton.
An extension to kaluza - klein is , which also talks about the weak and strong forces, and requires .
# On Geometry
In , the gauge group for is $U(1)$.
Now, the key thing here is that Electromagnetism is then The Curvature of the $U(1)$ bundle.
This is not the only geometric connection between General Relativity and Quantum Field Theory. In the same context, the covariant derivatives is general relativity are such that $\nabla_\mu-\partial_\mu$ sort-of measures the gravity, in a certain way, while this is also true in QFT, where to some constants, $\nabla_\mu-\partial_\mu=ig_sA_\mu$.
It is to be noted that both are in similiar context. | {"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": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8885414004325867, "perplexity": 949.8558685137258}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540518882.71/warc/CC-MAIN-20191209121316-20191209145316-00056.warc.gz"} |
http://www.ams.org/joursearch/servlet/PubSearch?f1=msc&pubname=all&v1=10H20&startRec=1 | American Mathematical Society
My Account · My Cart · Customer Services · FAQ
Publications Meetings The Profession Membership Programs Math Samplings Policy and Advocacy In the News About the AMS
You are here: Home > Publications
AMS eContent Search Results
Matches for: msc=(10H20) AND publication=(all) Sort order: Date Format: Standard display
Results: 1 to 6 of 6 found Go to page: 1
[1] Kevin S. McCurley. Prime values of polynomials and irreducibility testing. Bull. Amer. Math. Soc. 11 (1984) 155-158. MR 741729. Abstract, references, and article information View Article: PDF [2] Joseph L. Gerver. Irregular sets of integers generated by the greedy algorithm . Math. Comp. 40 (1983) 667-676. MR 689480. Abstract, references, and article information View Article: PDF This article is available free of charge [3] Samuel S. Wagstaff. Greatest of the least primes in arithmetic progressions having a given modulus . Math. Comp. 33 (1979) 1073-1080. MR 528061. Abstract, references, and article information View Article: PDF This article is available free of charge [4] Carter Bays and Richard H. Hudson. Details of the first region of integers $x$ with $\pi \sb{3,2}(x)<\pi \sb{3,1}(x)$ . Math. Comp. 32 (1978) 571-576. MR 0476616. Abstract, references, and article information View Article: PDF This article is available free of charge [5] John Isbell and Stephen Schanuel. On the fractional parts of $n/j,$ $j=o(n)$ . Proc. Amer. Math. Soc. 60 (1976) 65-67. MR 0429796. Abstract, references, and article information View Article: PDF This article is available free of charge [6] Harold G. Diamond. Chebyshev estimates for Beurling generalized prime numbers . Proc. Amer. Math. Soc. 39 (1973) 503-508. MR 0314782. Abstract, references, and article information View Article: PDF This article is available free of charge
Results: 1 to 6 of 6 found Go to page: 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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8439207673072815, "perplexity": 2693.987075276378}, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375096301.47/warc/CC-MAIN-20150627031816-00241-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/from-langevin-to-fokker-planck.103448/ | # Homework Help: From Langevin to Fokker-Planck
1. Dec 9, 2005
### Jezuz
Hi.
I'm studying quantum Brownian motion right now and I need to see that the (classical) Langevin equation for a Brownian particle is equivalent to the Fokker-Planck equation for the phase-space distribution function of the same particle.
Does anyone know where I can find such a derivation? I've been looking all over the internet for it but usually they start with a Langevin equation containging only first derivatives (that is the have excluded the possible outer potential felt by the particle).
I need the derivation for the case where i have a Langevin equation of the type:
m \ddot x(t) + \gamma \dot x(t) + V(x(t)) = F(t)
(written in LaTeX syntax).
I would be very grateful for help!
Alternatively, since I have the derivation for the Langevin equation starting with a Lagrangian for a particle interacting linearly with a bath of harmonic oscillators (initially in thermal equilibrium), I could also accept a derviation of the Fokker-Planck equation starting with the same assumptions.
2. Dec 9, 2005
### Physics Monkey
For such stochastic methods, a very good book is "Handbook of Stochastic Methods" by C. Gardiner. Written by an expert in the field, this book is very physical and to the point unlike many other excessively mathematical treatments of stochastic processes.
If you're interested, I can step you through the derivation here, but you should get that book for sure.
3. Dec 10, 2005
### Jezuz
Okej!
Thank you very much. I'll have a look at that book. Found it at the university library. If there is something I get stuck with I might ask you again :) | {"extraction_info": {"found_math": false, "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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9106327295303345, "perplexity": 527.8361575230451}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221210243.28/warc/CC-MAIN-20180815161419-20180815181419-00329.warc.gz"} |
http://alidoc.cern.ch/AliPhysics/vAN-20160717/class_ali_ana_calo_channel_analysis.html | AliPhysics 9e28f8c (9e28f8c)
AliAnaCaloChannelAnalysis Class Reference
Analyses cell properties and identifies bad cells. More...
#include <AliAnaCaloChannelAnalysis.h>
Inheritance diagram for AliAnaCaloChannelAnalysis:
## Public Member Functions
AliAnaCaloChannelAnalysis ()
virtual ~AliAnaCaloChannelAnalysis ()
AliAnaCaloChannelAnalysis (TString period, TString pass, TString trigger, Int_t runNumber)
void Run ()
void SetExternalMergedFile (TString inputName)
void SetInputFileList (TString inputName)
void SetWorkDir (TString inputName)
void SetNTrial (Int_t inputNr)
void AddPeriodAnalysis (Int_t criteria, Double_t nsigma, Double_t emin, Double_t emax)
## Protected Member Functions
void Init ()
TString Convert ()
void BCAnalysis ()
void PeriodAnalysis (Int_t criterum=7, Double_t nsigma=4.0, Double_t emin=0.1, Double_t emax=2.0)
TH1F * BuildHitAndEnergyMean (Int_t crit, Double_t emin=0.1, Double_t emax=2., Double_t nsigma=4.)
TH1F * BuildTimeMean (Int_t crit, Double_t emin, Double_t emax, Double_t nsigma)
Possibility to add this check too, if the time if calibrated. More...
void TestCellShapes (Int_t crit, Double_t fitemin, Double_t fitemax, Double_t nsigma=4.)
void FlagAsBad (Int_t crit, TH1 *inhisto, Double_t nsigma=4., Int_t dnbins=200, Double_t dmaxval=-1.)
void SummarizeResults ()
TH1 * BuildMeanFromGood ()
Bool_t CheckDistribution (TH1 *ratio, TH1 *reference)
void SaveBadCellsToPDF (Int_t version, TString pdfName)
void PlotFlaggedCells2D (Int_t flag1, Int_t flag2=-1, Int_t flag3=-1)
Plots a 2D map of flagged cells, Dead, Bad, Good. More...
## Protected Attributes
Int_t fCurrentRunNumber
A run number of an analyzed period. This is important for the AliCalorimeterUtils initialization. More...
TString fPeriod
The name of the analyzed period. More...
TString fPass
Pass of the analyzed data. More...
TString fTrigger
Selected trigger for the analysis. More...
Int_t fNoOfCells
Number of cells in EMCal and DCal. More...
Int_t fCellStartDCal
ID of the first cell in the DCal. More...
TString fMergeOutput
Here the merged files of a period are saved for a later analysis. More...
TString fAnalysisOutput
The list with bad channels and histograms are saved in this folder. More...
TString fAnalysisInput
Here the .root files of each run of the period are saved. More...
TString fRunList
Thats the full path and name of the file which contains a list of all runs to be merged together. More...
Dierctory in the QA.root files where the input histograms are stored. More...
TString fMergedFileName
Filename of the .root file containing the merged runs. More...
std::vector< TArrayD > fAnalysisVector
Vector of analysis information. Each place is filled with 4 doubles: version, sigma, lower, and upper energy range. More...
TString fRunListFileName
This is the name of the file with the run numbers to be merged, by default it's 'runList.txt'. More...
TString fWorkdir
Directory which contains the folders fMergeOutput, fAnalysisInput and fAnalysisOutput. By default it is './'. More...
Int_t fTrial
Number of trial that this specific analyis is. By default '0'. More...
TString fExternalFileName
If you have already a file that contains many runs merged together you can place it in fMergeOutput and set it with SetExternalMergedFile(FileName) More...
Int_t * fFlag
! fFlag[CellID] = 0 (ok),1 (dead),2 (bad by lower),3 (bad by upper) start at 0 (cellID 0 = histobin 1) More...
Int_t fNMaxCols
Maximum No of colums in module (eta direction) More...
Int_t fNMaxRows
Maximum No of rows in module (phi direction) More...
Int_t fNMaxColsAbs
Maximum No of colums in Calorimeter. More...
Int_t fNMaxRowsAbs
Maximum No of rows in Calorimeter. More...
AliCalorimeterUtilsfCaloUtils
! Calorimeter information for the investigated runs More...
## Private Member Functions
AliAnaCaloChannelAnalysis (const AliAnaCaloChannelAnalysis &)
AliAnaCaloChannelAnalysisoperator= (const AliAnaCaloChannelAnalysis &)
## Detailed Description
Analyses cell properties and identifies bad cells.
This is used for bad channel identification in EMCal and DCal. The class builds a mean distribution of certain cell observables and compares single cell properties to this mean. That way bad channels (far off the mean) are identified and flagged. A .pdf file with their spectra is created. This should be cross checked by hand.
Date
Jun 24, 2016
Definition at line 41 of file AliAnaCaloChannelAnalysis.h.
## Constructor & Destructor Documentation
AliAnaCaloChannelAnalysis::AliAnaCaloChannelAnalysis ( )
Default constructor
Definition at line 54 of file AliAnaCaloChannelAnalysis.cxx.
virtual AliAnaCaloChannelAnalysis::~AliAnaCaloChannelAnalysis ( )
inlinevirtual
Definition at line 46 of file AliAnaCaloChannelAnalysis.h.
AliAnaCaloChannelAnalysis::AliAnaCaloChannelAnalysis ( TString period, TString pass, TString trigger, Int_t RunNumber )
Constructor
Definition at line 74 of file AliAnaCaloChannelAnalysis.cxx.
AliAnaCaloChannelAnalysis::AliAnaCaloChannelAnalysis ( const AliAnaCaloChannelAnalysis & )
private
## Member Function Documentation
void AliAnaCaloChannelAnalysis::AddPeriodAnalysis ( Int_t criteria, Double_t nsigma, Double_t emin, Double_t emax )
Set period analysis parameters
Parameters
criteria – selection criteria (?) nsigma – n sigma cut emin – minimum energy emax – maximum energy
Definition at line 382 of file AliAnaCaloChannelAnalysis.cxx.
void AliAnaCaloChannelAnalysis::BCAnalysis ( )
protected
Configure a complete analysis with different criteria, it provides bad+dead cells lists You can manage criteria used and their order, the first criteria will use the original output file from AliAnalysisTaskCaloCellsQA task,
Definition at line 342 of file AliAnaCaloChannelAnalysis.cxx.
Referenced by Run().
TH1F * AliAnaCaloChannelAnalysis::BuildHitAndEnergyMean ( Int_t crit, Double_t emin = 0.1, Double_t emax = 2., Double_t nsigma = 4. )
protected
Average hit per event and the average energy per hit is caluclated for each cell.
Definition at line 503 of file AliAnaCaloChannelAnalysis.cxx.
Referenced by PeriodAnalysis().
TH1 * AliAnaCaloChannelAnalysis::BuildMeanFromGood ( )
protected
Definition at line 1174 of file AliAnaCaloChannelAnalysis.cxx.
TH1F * AliAnaCaloChannelAnalysis::BuildTimeMean ( Int_t crit, Double_t emin, Double_t emax, Double_t nsigma )
protected
Possibility to add this check too, if the time if calibrated.
Definition at line 550 of file AliAnaCaloChannelAnalysis.cxx.
Bool_t AliAnaCaloChannelAnalysis::CheckDistribution ( TH1 * ratio, TH1 * reference )
protected
Definition at line 1198 of file AliAnaCaloChannelAnalysis.cxx.
TString AliAnaCaloChannelAnalysis::Convert ( )
protected
Creates one file for the analysis from several QA output files listed in runlist.txt.
Definition at line 199 of file AliAnaCaloChannelAnalysis.cxx.
Referenced by Run().
void AliAnaCaloChannelAnalysis::FlagAsBad ( Int_t crit, TH1 * inhisto, Double_t nsigma = 4., Int_t dnbins = 200, Double_t dmaxval = -1. )
protected
1) create a distribution for the input histogram; 2) fit the distribution with a gaussian; 3) define good area within +-nsigma to identfy badcells.
Parameters
pflag – flag with channel criteria to be filled inhisto – input histogram; dnbins – number of bins in distribution; dmaxval – maximum value on distribution histogram.
Definition at line 740 of file AliAnaCaloChannelAnalysis.cxx.
Referenced by PeriodAnalysis().
void AliAnaCaloChannelAnalysis::FlagAsDead ( )
protected
– This function finds cells with zero entries It flags them by setting the fFlag content of the cell to 1.
Definition at line 695 of file AliAnaCaloChannelAnalysis.cxx.
Referenced by BCAnalysis().
void AliAnaCaloChannelAnalysis::Init ( )
protected
Init default parameters
Definition at line 94 of file AliAnaCaloChannelAnalysis.cxx.
Referenced by AliAnaCaloChannelAnalysis().
AliAnaCaloChannelAnalysis& AliAnaCaloChannelAnalysis::operator= ( const AliAnaCaloChannelAnalysis & )
private
void AliAnaCaloChannelAnalysis::PeriodAnalysis ( Int_t criterum = 7, Double_t nsigma = 4.0, Double_t emin = 0.1, Double_t emax = 2.0 )
protected
Definition at line 404 of file AliAnaCaloChannelAnalysis.cxx.
Referenced by BCAnalysis().
void AliAnaCaloChannelAnalysis::PlotFlaggedCells2D ( Int_t flag1, Int_t flag2 = -1, Int_t flag3 = -1 )
protected
Plots a 2D map of flagged cells, Dead, Bad, Good.
Definition at line 1237 of file AliAnaCaloChannelAnalysis.cxx.
Referenced by SummarizeResults().
void AliAnaCaloChannelAnalysis::Run ( )
Main execution method.
First use Convert() to merge historgrams from a runlist .txt file. The merged outputfile contains 3 different histograms. In a second step analyse these merged histograms by calling BCAnalysis().
Definition at line 161 of file AliAnaCaloChannelAnalysis.cxx.
void AliAnaCaloChannelAnalysis::SaveBadCellsToPDF ( Int_t version, TString pdfName )
protected
Allow to produce a pdf file with badcells candidates (red) compared to a refence cell (black).
Definition at line 1067 of file AliAnaCaloChannelAnalysis.cxx.
Referenced by SummarizeResults().
void AliAnaCaloChannelAnalysis::SetExternalMergedFile ( TString inputName )
inline
Definition at line 52 of file AliAnaCaloChannelAnalysis.h.
void AliAnaCaloChannelAnalysis::SetInputFileList ( TString inputName )
inline
Definition at line 53 of file AliAnaCaloChannelAnalysis.h.
void AliAnaCaloChannelAnalysis::SetNTrial ( Int_t inputNr )
inline
Definition at line 55 of file AliAnaCaloChannelAnalysis.h.
void AliAnaCaloChannelAnalysis::SetWorkDir ( TString inputName )
inline
Definition at line 54 of file AliAnaCaloChannelAnalysis.h.
void AliAnaCaloChannelAnalysis::SummarizeResults ( )
protected
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 1) summarize all dead and bad cells in a text file .. 2) plot all bad cell E distributions in a .pdf file . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Definition at line 951 of file AliAnaCaloChannelAnalysis.cxx.
Referenced by BCAnalysis().
void AliAnaCaloChannelAnalysis::TestCellShapes ( Int_t crit, Double_t fitemin, Double_t fitemax, Double_t nsigma = 4. )
protected
ELI this method is currently not used Test cells shape using fit function f(x)=A*exp(-B*x)/x^2. Produce values per cell + distributions for A,B and chi2/ndf parameters.
Definition at line 561 of file AliAnaCaloChannelAnalysis.cxx.
Referenced by PeriodAnalysis().
## Member Data Documentation
TString AliAnaCaloChannelAnalysis::fAnalysisInput
protected
Here the .root files of each run of the period are saved.
Definition at line 95 of file AliAnaCaloChannelAnalysis.h.
Referenced by Convert(), and Init().
TString AliAnaCaloChannelAnalysis::fAnalysisOutput
protected
The list with bad channels and histograms are saved in this folder.
Definition at line 94 of file AliAnaCaloChannelAnalysis.h.
Referenced by FlagAsBad(), Init(), PeriodAnalysis(), PlotFlaggedCells2D(), and SummarizeResults().
std::vector AliAnaCaloChannelAnalysis::fAnalysisVector
protected
Vector of analysis information. Each place is filled with 4 doubles: version, sigma, lower, and upper energy range.
Definition at line 101 of file AliAnaCaloChannelAnalysis.h.
Referenced by AddPeriodAnalysis(), BCAnalysis(), and Init().
AliCalorimeterUtils* AliAnaCaloChannelAnalysis::fCaloUtils
protected
! Calorimeter information for the investigated runs
Definition at line 119 of file AliAnaCaloChannelAnalysis.h.
Referenced by FlagAsBad(), Init(), and PlotFlaggedCells2D().
Int_t AliAnaCaloChannelAnalysis::fCellStartDCal
protected
ID of the first cell in the DCal.
Definition at line 90 of file AliAnaCaloChannelAnalysis.h.
Referenced by FlagAsBad(), Init(), and SummarizeResults().
Int_t AliAnaCaloChannelAnalysis::fCurrentRunNumber
protected
A run number of an analyzed period. This is important for the AliCalorimeterUtils initialization.
Definition at line 85 of file AliAnaCaloChannelAnalysis.h.
Referenced by AliAnaCaloChannelAnalysis(), and Init().
TString AliAnaCaloChannelAnalysis::fExternalFileName
protected
If you have already a file that contains many runs merged together you can place it in fMergeOutput and set it with SetExternalMergedFile(FileName)
Definition at line 107 of file AliAnaCaloChannelAnalysis.h.
Referenced by Init(), and Run().
Int_t* AliAnaCaloChannelAnalysis::fFlag
protected
! fFlag[CellID] = 0 (ok),1 (dead),2 (bad by lower),3 (bad by upper) start at 0 (cellID 0 = histobin 1)
Definition at line 110 of file AliAnaCaloChannelAnalysis.h.
TString AliAnaCaloChannelAnalysis::fMergedFileName
protected
Filename of the .root file containing the merged runs.
Definition at line 100 of file AliAnaCaloChannelAnalysis.h.
Referenced by BCAnalysis(), Convert(), Init(), Run(), and SummarizeResults().
TString AliAnaCaloChannelAnalysis::fMergeOutput
protected
Here the merged files of a period are saved for a later analysis.
Definition at line 93 of file AliAnaCaloChannelAnalysis.h.
Referenced by Init(), and Run().
Int_t AliAnaCaloChannelAnalysis::fNMaxCols
protected
Maximum No of colums in module (eta direction)
Definition at line 113 of file AliAnaCaloChannelAnalysis.h.
Referenced by Init().
Int_t AliAnaCaloChannelAnalysis::fNMaxColsAbs
protected
Maximum No of colums in Calorimeter.
Definition at line 115 of file AliAnaCaloChannelAnalysis.h.
Referenced by FlagAsBad(), Init(), and PlotFlaggedCells2D().
Int_t AliAnaCaloChannelAnalysis::fNMaxRows
protected
Maximum No of rows in module (phi direction)
Definition at line 114 of file AliAnaCaloChannelAnalysis.h.
Referenced by Init().
Int_t AliAnaCaloChannelAnalysis::fNMaxRowsAbs
protected
Maximum No of rows in Calorimeter.
Definition at line 116 of file AliAnaCaloChannelAnalysis.h.
Referenced by FlagAsBad(), Init(), and PlotFlaggedCells2D().
Int_t AliAnaCaloChannelAnalysis::fNoOfCells
protected
Number of cells in EMCal and DCal.
Definition at line 89 of file AliAnaCaloChannelAnalysis.h.
TString AliAnaCaloChannelAnalysis::fPass
protected
Pass of the analyzed data.
Definition at line 87 of file AliAnaCaloChannelAnalysis.h.
Referenced by AliAnaCaloChannelAnalysis(), Convert(), Init(), and SummarizeResults().
TString AliAnaCaloChannelAnalysis::fPeriod
protected
The name of the analyzed period.
Definition at line 86 of file AliAnaCaloChannelAnalysis.h.
Referenced by AliAnaCaloChannelAnalysis(), Convert(), Init(), and SummarizeResults().
protected
Dierctory in the QA.root files where the input histograms are stored.
Definition at line 99 of file AliAnaCaloChannelAnalysis.h.
Referenced by Convert(), and Init().
TString AliAnaCaloChannelAnalysis::fRunList
protected
Thats the full path and name of the file which contains a list of all runs to be merged together.
Definition at line 96 of file AliAnaCaloChannelAnalysis.h.
Referenced by Convert(), and Init().
TString AliAnaCaloChannelAnalysis::fRunListFileName
protected
This is the name of the file with the run numbers to be merged, by default it's 'runList.txt'.
Definition at line 104 of file AliAnaCaloChannelAnalysis.h.
Referenced by Init().
Int_t AliAnaCaloChannelAnalysis::fTrial
protected
Number of trial that this specific analyis is. By default '0'.
Definition at line 106 of file AliAnaCaloChannelAnalysis.h.
Referenced by Init(), and SummarizeResults().
TString AliAnaCaloChannelAnalysis::fTrigger
protected
Selected trigger for the analysis.
Definition at line 88 of file AliAnaCaloChannelAnalysis.h.
Referenced by AliAnaCaloChannelAnalysis(), Convert(), and Init().
TString AliAnaCaloChannelAnalysis::fWorkdir
protected
Directory which contains the folders fMergeOutput, fAnalysisInput and fAnalysisOutput. By default it is './'.
Definition at line 105 of file AliAnaCaloChannelAnalysis.h.
Referenced by Init().
The documentation for this class was generated from the following files: | {"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": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.28171515464782715, "perplexity": 20395.172535285707}, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662519037.11/warc/CC-MAIN-20220517162558-20220517192558-00226.warc.gz"} |
http://mathhelpforum.com/differential-geometry/110273-divergent-convergent-summs-series.html | Thread: Divergent and Convergent summs\series
1. Divergent and Convergent summs\series
1) $\displaystyle\sum_{n=1}^{\infty}{a_n}$ is divergent.
Proove that $\displaystyle\sum_{n=1}^{\infty}\frac{a_n}{1+n*a_n }$ is also divergent.
2) $a_n \geq a_{n+1} \textgreater 0$ and $\displaystyle\sum_{n=1}^{\infty}{a_n}$ is convergent.
Proove that: $\lim\limits_{n \rightarrow \infty}{n*a_n}=0$
I am a third year student at Computer Science, therefore I have some strong basics. Any help, direction where to look is appreciated. Thx
LE:
3) $
\sum_{i=0}^\infty \arctan\frac{1}{n^2+n+1}=?
$
2. 2) $a_n \geq a_{n+1} \textgreater 0$ and $\displaystyle\sum_{n=1}^{\infty}{a_n}$ is convergent.
Proove that: $\lim\limits_{n \rightarrow \infty}{n*a_n}=0$
Let $S_n = \sum_{i=1}^n a_n, S = lim_{n\to\infty}S_n$. Pick any epsilon > 0. Then there exists a K such that for any k >= K, (a_(K+1) + ... + a_m) <= S - S_k < epsilon/2 since S_k converges to S and since the partial sums are increasing. Also, since a_n converges to zero, there must exist a J > K such that for all j >= J, (epsilon/2)*min{a_1, a_2, ..., a_K}/S_K >= a_j. Therefore since the sequence is decreasing, for all sufficiently large j, $n\cdot a_j = \underbrace{a_j + \ldots + a_j}_{n\;\rm times} \leq (\frac{\epsilon a_1}{2S_K} + \ldots + \frac{\epsilon a_K}{2S_K}) + (a_{K+1} + \ldots + a_j)$ $= \frac{\epsilon S_K}{2S_K} + (a_{K+1} + \ldots + a_j) < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$. Since the sequence is positive but eventually remains smaller than any positive number, it must converge to zero. | {"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": 12, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9984474778175354, "perplexity": 645.7952675556909}, "config": {"markdown_headings": false, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549423927.54/warc/CC-MAIN-20170722082709-20170722102709-00056.warc.gz"} |
http://stochasticprocess.blogspot.com/2011/06/reading-guides-and-complex-methods.html | When not at work with students, I spend my time in my room either reading, calculating something using pen and paper, or using a computer. I read almost anything: from the pornographic to the profound, although my main interests are mathematics and physics. "When I get a little money I buy books; and if any is left I buy food and clothes." -Erasmus
## Sunday, June 26, 2011
### reading guides and complex methods
I'm teaching complex (variable) methods this semester, and I'm using Arfken's book to teach it. It's a long list of topics; the detailed syllabus is shown below:
Meeting no. ObjectivesAfter the discussion and lined up activities, you should be able to: Topics 1(6-15) § Explain what is expected of you to get good marks in this class§ Explain the expected role of your teacher§ Explain the expected role of your book§ Explain the expected role of your lecture classes§ List the materials you will need for this course§ Explain why the rules of coherence are needed. Orientation Read: Rules of Coherence 2(6-17) § Perform algebraic operations using complex numbers§ Use the geometric series to evaluate complex valued sums§ Use the natural logarithm function to calculate inverse trigonometric functions§ Use the natural logarithm function to calculate complex powers Complex Algebra Complex-valued Series Complex Powers Read: Sec 6.1Exercises: 6.1.7; 6.1.8; 6.1.9; 6.1.10; 6.1.16, 6.1.17; 6.1.22
DIAGNOSTIC EXAMINATION 12June 18, 2011 (Sat) 1pm
# Complex analysis i
3(6-22) § Use the definition of the derivative to evaluate the derivative of polynomials§ Use the definition of the derivative to obtain the Cauchy-Riemann conditions § Use the Cauchy-Riemann conditions and continuity of partial derivatives to test a complex function for analyticity.§ Use the Cauchy-Riemann conditions to show that the real and imaginary parts of an analytic function must satisfy the Laplace equation Cauchy-Riemann conditions Laplace equation Analytic functions Read: Sec 6.2Exercises: 6.2.1; 6.2.4; 6.2.5 4(6-24) § Define the Riemann contour integral on a complex contour and compare with the real Riemann integral§ Use the definition of the Riemann integral to prove the Darboux inequality§ Use the definition of the contour integral to show that, in general, the contour integral is path-dependent. § Use the definition of the contour integral to calculate the contour integral of a given function along a given path.§ Prove the Cauchy-Integral Theorem using Stokes’ theorem Contour Integral Darboux inequality Path-dependence Cauchy Integral Formula Read: Sec 6.3 Exercises: 6.3.1; 6.1.2; 6.3.3 5(6-29) § Prove the Cauchy integral formula by using a deformation of contour and polar coordinates.§ Use the Cauchy integral formula to obtain an expression for the nth derivative of an analytic function.§ Use the Cauchy integral formula to relate Rodrigues formula representations of special functions to contour integral representations. Cauchy Integral formula Nth derivative of an analytic function Contour integral representations Read: Sec 6.4Exercises: 6.4.1; 6.4.5; 6.4.6; 6.4.8 6(7-1) § Derive the Taylor series of given functions and give the domain of validity§ Derive the Laurent series of given functions about a given point on the complex plane and, using knowledge of singularities, give the domains of validity Taylor seriesLaurent seriesSingularities Read Section 6.5Exercises 6.5.1; 6.5.2; 6.5.3; 6.5.7; 6.5.8; 6.5.10; 6.5.11 7(7-6) § Obtain a formula for the coefficient a-1 in terms of derivatives§ Given a curve in the z-plane, and a mapping w(z), obtain the corresponding curve on the complex w-plane§ Given a simply connected-region A, and a given mapping w(z), obtain its image on the w-plane Residue formula Mapping Read Section 6.6, 6.7Exercises: 6.6.1, 6.6.2, 6.7.6
First long exam july 16, 2011 (sat) 1pm
COMPLEX ANALYSIS II
8-9(7-8 to 7-13) § Identify the poles and other singularities of a given meromorphic function, and evaluate the residues at these singularities (if any)§ Use the residue theorem to evaluate a closed simply connected contour integral§ Use the residue theorem and Darboux’s inequality to calculate Fourier integrals§ Use the residue theorem to calculate other integrals. Singularities and residues Residue at infinity Contour integration Read Sec 7.1 (exclude Pole and product expansions, pp 461-462)Exercises 7.1.1, 7.1.3, 7.1.6, 7.1.8, 7.1.10, 7.1.11, 7.1.14, 7.1.15, 7.1.17, 7.1.18, 7.1.21, 7.1.24 10(7-15) § Given a series for w(z) in powers of z, obtain the series for the inverse function of z(w) in powers of w. § Give the conditions for expanding an analytic function as a pole expansion§ Obtain the pole expansion of selected functions§ Obtain the product expansions of sin(z) and cos(z) Inversion of Series Pole Expansions Product Expansions Summation of Series via Contour Integrals Read Whittaker and Watson, pp 134 to 139, Arfken pp 461-462, Morse and Feshbach pp 411-413,pp 413-414Exercises Whittaker and Watson Section 7.4 Examples 4, 6 and Section 7.5 Example 1Exercises for inversion of series will be handed out in class 11 to 12(7-20 to 7-22) § Find the zeroes of the derivative of an analytic function f(z), and draw contour plots of the real and imaginary parts of f(z)-f(z0) in the neighborhood of the zero z0 obtained. § Use the method of steepest descent to approximate a class of integrals with parameter s, for large values of s.§ Obtain the asymptotic series using the method of steepest descent Method of Steepest Descent Read Sec 7.3, Morse and Feshbach pp 434-443Exercises 7.3.1, 7.3.2, 7.3.3, exercises to be handed out
2nd Long Exam July 30, 2011 (Sat) 1pm
# gAMMA, BETA AND FOURIER SERIES
13(7-27) § Show the equivalence of the three definitions (Euler integral analytically continued; Weierstrass product, Euler product) of the Gamma function§ Use the Gamma function to evaluate negative factorials§ Use the Gamma function to obtain the Binomial series valid for non-integral powers, and give the domain of validity§ Derive the recursion relation of the gamma function§ Find the poles of the gamma function and evaluate the residues of the gamma function§ Prove the Gauss-Multiplication theorem and the Legendre duplication formula§ Evaluate Gaussian integrals Gamma Function Binomial series Gaussian integrals Multiplication Formula Read Sec 8.1, and Whittaker and Watson pp 244-246Exercises 8.1.1, 8.1.4, 8.1.5, 8.1.9, 8.1.14, 8.1.18, 8.1.24 14(7-29) § Derive Stirling’s series§ Use Stirling’s approximation to evaluate large factorials.§ Use the Beta function and the chain rule to evaluate a selected class of integrals Stirling’s Approximation Beta Function Read sec 8.3 and 8.4, Whittaker and Watson pp 251 to 253Exercises 8.3.1, 8.3.6, 8.3.8, 8.3.9, 8.4.2, 8.4.17, 8.4.18 15-16(8-3 to 8-5) § Prove the orthogonality of a given set of sines and cosines on a suitable interval§ Use orthogonality and completeness of a basis of sines and cosines to obtain the Fourier series of a function within an interval§ Use Fourier series to solve the wave equation of a vibrating string Fourier Series Orthogonality Read Sec 14.1 to 14.4Exercises 14.1.5, 14.1.9, 14.2.1, 14.2.3, 14.3.2, 14.3.12, 14.3.14, 14.4.2, 14.4.10
3rd Long Exam 12August 13, 2011 (Sat) 1pm
INTEGRAL TRANSFORMS
17(8-10) § Expand given functions defined over the whole real line as a Fourier integral§ Obtain the Fourier transform of a given function§ Given the function in (Fourier) k-space, use the inverse Fourier transform to obtain the function in coordinate space§ Relate Mellin transforms to Fourier transforms§ Obtain the representation of the Dirac delta in terms of Fourier integrals Fourier Transforms Read Section 15.1 to 15.3Exercises 15.1.3, 15.3.2, 15.3.4, 15.3.9, 15.3.17, 18(8-12) § Obtain the Fourier transform of derivatives§ Convert a linear differential equation in coordinate space into the corresponding integral equation in k-space§ Solve the diffusion equation using Fourier transforms Fourier transform of derivatives Read Sec 15.4Exercises 15.4.1, 15.4.3, 15.4.4, 15.4.5 19(8-17) § Use the convolution theorem to evaluate some integrals§ Obtain the momentum space representation of a wavefunction Convolution TheoremParseval’s RelationMomentum Space Read Sections 15. 5 to 15. 6Exercises 15.5.3, 15.5.5, 15.5.6, 15.5.8, 15.6.3, 15.6.8, 15.6.12 20(8-24) § Calculate the Laplace transform of some elementary functions§ Use tables of Laplace transforms and the linearity of Laplace transforms to evaluate inverse Laplace transforms§ Use partial fractions to evaluate inverse Laplace transforms Laplace Transform Partial fractions Read Section15.8Do Exercises 15.8.3, 15.8.4, 15.8.5, 15.8.9 21(8-26) § Evaluate the Laplace transform of derivatives§ Convert linear differential equations with constant coefficients to algebraic systems§ Solve linear ordinary differential equations with constant coefficients using Laplace transforms Laplace Transform of Derivatives Convolution Theorem Read Section 15.9 to 15.11Exercises 15.9.2, 15.9.3, 15.11.2, 15.11.3 22(8-31) § Evaluate inverse Laplace transform using Bromwich integrals§ Convert linear differential equations with constant coefficients to algebraic systems Bromwich integral Read Section 15.12Exercises 15.12.1, 15.12.2, 15.12.3, 15.12.4
4th Long Exam September 3, 2011 (Sat)
SPECIAL FUNCTIONS I – BESSEL, LEGENDRE FUNCTIONS
23 to 24(9-2 to 9-7) § Identify the singularities of linear second order differential equations§ Use the power-series method to obtain a solution of linear second order differential equations§ Use Wronskians to obtain a linearly independent second solutions if a solution is known Power-Series Solutions Read: Section 9.4 to 9.6Ex: 9.4.1,9.4.2, 9.4.3, 9.5.5, 9.5.6, 9.5.10, 9.5.11, 9.6.18, 9.6.19, 9.6.25 25(9-9) § Use generating functions to obtain the recursion relations satisfied by Bessel Functions§ Use power-series methods to solve Bessel’s differential equation for both integral and non-integral powers§ Use the generalized Green’s theorem to verify the orthogonality properties of a set of Bessel functions§ Use the completeness of a set of Bessel functions to expand a given function in the interval 0≤ x ≤ a Bessel Functions of the First Kind Generating Function Read: Sec 11.1 to 11.2Ex: 11.1.1, 11.1.3, 11.1.10, 11.1.16, 11.1.18, 11.2.2, 11.2.3, 11.2.6 26(9-14) § Use the definition of Neumann functions and verify that it is a second linearly independent solution of Bessel’s equation§ Use the definition of Hankel functions to derive its properties§ Use the asymptotic formulae for Bessel functions to approximately evaluate Bessel functions for large values of its argument Neumann and Hankel functions Asymptotic formulae for Bessel functions Read: Sec 11.3 to 11.4Ex: 11.3.2,11.3.6, 11.4.7 27(9-16) § Use generating functions to obtain recursion relations and other properties of Legendre functions§ Use the generalized Green’s theorem to prove orthogonality of Legendre functions§ Expand an arbitrary function within the interval -1≤ x ≤ in terms of Bessel functions and give an integral for the expansion coeffiecients § Use Rodrigues formula to derive orthogonality of Legendre polynomials and calculate the normalization constant of Legendre polynomials Legendre functions Orthogonality and Completeness Read: 12.1 to 12.4Ex: 12.2.2, 12.2.3, 12.2.5, 12.3.2,12.3.6, 12.3.11, 12.4.2 28(9-21) § Prove orthogonality of associated Legendre functions§ Use completeness relations of Spherical Harmonics to express functions depending on θ and φ as a sum over spherical harmonics§ Express 1/ │x1-x2│ in terms of spherical harmonics Associated Legendre functions Spherical Harmonics Addition Theorem Read sec 12.5 to 12.6, 12.8Exercises 12.5.1, 12.5.11,12.6.4, 12.6.5,12.8.3, 12.8.8
5th Long Exam September 24, 2011 (Sat)
SPECIAL FUNCTIONS II—HERMITE, LAGUERRE, HYPERGEOMETRIC FUNCTIONS
29(9-23) § Prove orthogonality of Hermite functions§ Solve Hermite’s differential equation via power series§ Use completeness relations to express functions in the interval -∞ ≤ x ≤∞ as a sum of Hermite polynomials§ Use generating function to obtain the Rodrigues formula for Hermite polynomials§ Use Rodrigues formula to obtain Hermite polynomials Hermite functions Completeness and Orthogonality Rodrigues and Integral Representations Read: Sec 13.1Exercises 13.1.2, 13.1.14, 13.1.12, 13.1.13 30(9-28) § Prove orthogonality of Laguerre functions§ Use completeness relations to express functions in the interval 0≤ x ≤∞ as a sum of Laguerre polynomials § Use generating function to obtain the Rodrigues formula for Laguerre polynomials§ Use Rodrigues formula to obtain Laguerre polynomials Laguerre polynomials Associated Laguerre functions Read: Sec 13.2Exercises 13.2.1, 13.2.3, 13.2.6, 13.2.7 31(9-30) § Prove orthogonality of Chebyshev functions§ Use completeness relations to express functions in the interval -1≤ x ≤1 as a sum of Chebyshev polynomials § Use generating function to obtain the Rodrigues formula for Chebyshev polynomials§ Use Rodrigues formula to obtain Chebyshev polynomials Chebyshev Polynomials Read Sec 13.3Exercises 13.3.1, 13.3.3, 13.3.5 32(10-5) § Reduce any linear second order differential equation with three regular singularities into the hypergeometric equation § Express some orthogonal special functions in terms of hypergeometric fuctions Hypergeometric Functions Read Sec 13.4Exercises 13.4.6, 13.4.7, 13.4.8 33(10-7) § Reduce any linear second order differential equation with one regular singularity and one irregular singularity into the confluent hypergeometric equation § Express some special functions in terms of confluent hypergeometric functions Confluent Hypergeometric Functions Read Sec 13.5Exercises 13.5.4, 13.5.6, 13.5.11, 13.5.13, 13.5.14
6th Long Exam October 15, 2011 (Sat) 1pm
The pace is very fast. When I first learned these topics, it took me more than two years of work, mainly because I had no one to look over my work. I used Churchill's book, Complex Variables with Applications, but it's not enough to give justice to the course description. And I need to follow the course description because of university rules.
If asked, I would be the first to agree that it's an unreasonable amount of material, especially for a course that meets three hours a week. Cambridge University, for example offers a Methods of Mathematical Physics Course. One set of notes I found includes all the complex variable material, and removes most of the special functions. The only special functions that do show up are the hypergeometric and confluent hypergeometric functions, and Gamma and Beta. No discussion of Fourier series or Fourier integrals!
I am, however, stuck with the course description. I think that there ought to be changes made, but it has to go up to the university council. The only reasonable way of covering this material is to give them lots of work to do at home, and make extensive use of consultation hours.
I find it difficult to prepare conventional lecture notes. My main objection to conventional lecture notes is that reading the lecture notes is a more passive activity compared to working with pen in hand to prove the theorems or solve the problems. So instead of lecture notes, I will prepare reading guides.
The reading guides are a series of tasks that one should do while reading the text. One of the objections to Arfken is how easy it is to get lost. The way to avoid it is to divide the section into parts, and as soon as one reads the subparts, one should work on a problem or two in Arfken. I've prepared the reading guide so that when my students actually follow the guide, they will be able to construct a decent set of notes, and at the same time, solve the problems I've listed on the syllabus.
I hope that the reading guide makes it easier to read Arfken's book. My own method of reading Arfken (since I had no guide before) was to attempt solving the problems at the end of the section before reading the section. But that takes more time, since I could not separate the more important problems from the ones of secondary interest. Even if my students solve all of the problems I've listed, it's still a fraction of what I've actually done on my own.
I hand out the reading guides a week or so before the lecture class, and I expect my students to use the reading guide to prepare for the coming class. This means I will not need to discuss everything; instead, I could concentrate on the more difficult parts.
My students are graduate students with back subjects-- they had their undergraduate courses elsewhere, and are in need of remediation. They had no complex methods courses during their undergraduate days. Since they're graduate students, they have, at most, 9 hours of classwork every week. I hope that all of them are full-time students; the pace we set will be demanding. But if they do the necessary work, they should end the semester with an unfair advantage over their classmates in other physics courses. | {"extraction_info": {"found_math": false, "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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8948758244514465, "perplexity": 1422.8683151277214}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676589455.35/warc/CC-MAIN-20180716193516-20180716213516-00152.warc.gz"} |
https://www.physicsforums.com/threads/parametric-equations-of-line-in-3d-space.366537/ | # Homework Help: Parametric equations of line in 3D space
1. Jan 1, 2010
### player1_1_1
Hello, sorry for my Englich:D
1. The problem statement, all variables and given/known data
I must count a line integral on the lenght which lies on the line which is defined by equations:
$$\begin{cases}x^2+y^2+z^2=R^2\\ \left(x-\frac{R}{2}\right)^2+y^2=\left(\frac{R}{2}\right)^2\end{cases}$$
it is a column which is cutting a sphere
3. The attempt at a solution
I tried to assumption that $$x=t$$ and then depending on this find other functions $$y(t),z(t)$$, but the equations are not easy. I think I can find something with trigonometric functions sinus cosinus, but I am not sure; my question is then how I can easy and quickly find parametric equations of line which is make by two planes (in this case sphere and column) cutting each another? thanks for help!
Last edited: Jan 1, 2010
2. Jan 1, 2010
### tiny-tim
Hello player1_1_1!
(have a theta: θ and try using the X2 tag just above the Reply box )
It's much better to have a single-valued parameter, which x isn't.
So try using the angle (from the axis of the cylinder) as the parameter.
(btw, we say "sine" and "cosine", and we usually say "curve" for a line that isn't straight … even though we say "line integral" for any curve! )
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook | {"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": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8902708888053894, "perplexity": 1260.714909640192}, "config": {"markdown_headings": true, "markdown_code": false, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125947795.46/warc/CC-MAIN-20180425100306-20180425120306-00567.warc.gz"} |
http://www.lmfdb.org/knowledge/show/cmf.dim_decomposition | show · cmf.dim_decomposition all knowls · up · search:
We display the dimensions of the factors in the decomposition of each space into newforms.
Authors:
Knowl status:
• Review status: reviewed
• Last edited by David Roe on 2018-10-03 17:10:25
History: | {"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": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5592207312583923, "perplexity": 16333.93161938606}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027314732.59/warc/CC-MAIN-20190819114330-20190819140330-00109.warc.gz"} |
https://vor.ivo-s.de/meins/vim | # vim
## Suche Hervorheben hlsearch
map <F11> :set hls!<Bar>set hls?<CR>
map <F10> :set paste!<Bar>set paste?<CR>
map <F9> :set wrap!<Bar>set wrap?<CR>
## Scrollen Zeile bleibt stehen
:set scrolloff=10
:set scrolloff=0
:set scrolloff=999
## Suchen
When ignorecase and smartcase are both on, if a pattern contains an uppercase letter, it is case sensitive, otherwise, it is not. For example, /The would find only „The“, while /the would find „the“ or „The“ etc.
The smartcase option only applies to search patterns that you type; it does not apply to * or # or gd. If you press * to search for a word, you can make smartcase apply by pressing / then up arrow then Enter (to repeat the search from history). | {"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": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.42293211817741394, "perplexity": 22154.24858818064}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400213006.47/warc/CC-MAIN-20200924002749-20200924032749-00126.warc.gz"} |
http://scitation.aip.org/content/aip/journal/apl/85/17/10.1063/1.1812368 | • journal/journal.article
• aip/apl
• /content/aip/journal/apl/85/17/10.1063/1.1812368
• apl.aip.org
1887
No data available.
No metrics data to plot.
The attempt to plot a graph for these metrics has failed.
Influence of the gate dielectric on the mobility of rubrene single-crystal field-effect transistors
USD
10.1063/1.1812368
View Affiliations Hide Affiliations
Affiliations:
1 Kavli Institute of Nanoscience, Delft, University of Technology, Lorentzweg 1, 2628 CJ Delft The Netherlands
a) Electronic mail: a.f.stassen@tnw.tudelft.nl
Appl. Phys. Lett. 85, 3899 (2004)
/content/aip/journal/apl/85/17/10.1063/1.1812368
http://aip.metastore.ingenta.com/content/aip/journal/apl/85/17/10.1063/1.1812368
View: Figures
Figures
FIG. 1.
(Color online) Two rubrene single-crystal FETs fabricated by means of electrostatic bonding on (a) and (b). In (a) the electrodes have been defined by evaporation through a shadow mask, whereas in (b) photolithography and lift-off were used. In both (a) and (b) the bar is long.
FIG. 2.
Source–drain current vs source–drain voltage measured at different gate voltages for a device fabricated on , with an channel length and channel width. The inset shows similar data for a FET fabricated using parylene .
FIG. 3.
(a) Mobility vs gate voltage for a device on , measured at different values of source–drain voltage (, , , , , respectively), obtained using Eq. (1). Note that in the linear regime does not depend on and [the apparent peak at low values is an artifact originating from the use of Eq. (1) outside the linear regime]. (b) curves as measured for the four different gate insulators. For device based on parylene , the suppression of contact effects often requires the use of a rather large value (and thus , to remain in the linear regime).
FIG. 4.
Decrease of the mobility with increasing , as observed in rubrene single-crystal FETs with different gate insulators. The bars give a measure of the spread in mobility values. Inset: when plotted on a log–log scale, the available data show a linear dependence with slope (i.e., the variation in is proportional to ).
/content/aip/journal/apl/85/17/10.1063/1.1812368
2004-10-29
2014-04-19
Article
content/aip/journal/apl
Journal
5
3
Most cited this month
More Less
This is a required field | {"extraction_info": {"found_math": false, "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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9274260401725769, "perplexity": 3431.65822186201}, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609537308.32/warc/CC-MAIN-20140416005217-00632-ip-10-147-4-33.ec2.internal.warc.gz"} |
http://mathhelpforum.com/pre-calculus/3904-interval-s-0-a.html | # Math Help - interval(s) <0
1. ## interval(s) <0
For f(x) 1 divided by x^2-2x-8, find the interval(s) where f(x)<0. My answer is (-2,4) is that right. Thanks for looking at this problem.
2. That's correct
3. ## <0 question
Thanks TD for checking for me. | {"extraction_info": {"found_math": false, "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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8627715706825256, "perplexity": 11863.373785759766}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464051054181.34/warc/CC-MAIN-20160524005054-00040-ip-10-185-217-139.ec2.internal.warc.gz"} |
https://infoscience.epfl.ch/record/177050 | ## Overview of toroidal momentum transport
Toroidal momentum transport mechanisms are reviewed and put in a broader perspective. The generation of a finite momentum flux is closely related to the breaking of symmetry (parity) along the field. The symmetry argument allows for the systematic identification of possible transport mechanisms. Those that appear to lowest order in the normalized Larmor radius (the diagonal part, Coriolis pinch, E × B shearing, particle flux, and up-down asymmetric equilibria) are reasonably well understood. At higher order, expected to be of importance in the plasma edge, the theory is still under development. © 2011 IAEA, Vienna.
Published in:
Nuclear Fusion, 51, 9, 094027
Year:
2011
Laboratories: | {"extraction_info": {"found_math": false, "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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9580592513084412, "perplexity": 2188.195631942761}, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125945793.18/warc/CC-MAIN-20180423050940-20180423070940-00515.warc.gz"} |
http://math.stackexchange.com/questions/144592/value-of-a-fraction | # Value of a fraction
It it true that is ${a^2+c^2\over b^2+d^2}=1$ for $ad-bc=1$?
I tried substituting in $a={1-bc\over d}$ but it is still a mess.
(How do you ask Wolfram Alpha a question like this where we ask it to calculate something with an imposed condition?)
-
Take $a = 2$ and $b = c = d = 1$. – TMM May 13 '12 at 14:08
What is true is that $(a^2+c^2)(b^2+d^2)=(ad-bc)^2+(ab+cd)^2$. – anon May 13 '12 at 14:15
It's not true. Try it for $a = 2$ and $b = c = d = 1$:
$$ad - bc = 2\times1 - 1\times1 = 1$$
$$\frac{a^2+c^2}{b^2+d^2} = \frac{2^2+1^2}{1^2+1^2} = \frac{5}{2} \ne 1$$
As for WA, you can use the FullSimplify function to simplify an expression given some assumptions. Here is an example.
Or even simpler: $b=c=0$, so $a={1 \over d}$ and ${{a^2+c^2}\over {b^2+d^2}}={{a^2} \over {d^2}} = {1 \over {d^4}}$ – David Lewis May 13 '12 at 15:24 | {"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": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8531442880630493, "perplexity": 537.3523562630543}, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701159985.29/warc/CC-MAIN-20160205193919-00348-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/stupid-series-i-cant-seem-to-find.292538/ | # Stupid series I can't seem to find
1. Feb 15, 2009
### Nick89
1. The problem statement, all variables and given/known data
I need a way to write the following series in a sum:
$$\frac{1}{2} + \frac{1}{4} + \frac{1}{32} + \frac{1}{64} + \frac{1}{1024} + \frac{1}{2048} + ...$$
If you look closely you can see a 2^n pattern in there:
$$\frac{1}{2} + \frac{1}{4} + \frac{0}{8} + \frac{0}{16} + \frac{1}{32} + \frac{1}{64} + \frac{0}{128} + \frac{0}{256} + \frac{1}{512} + \frac{1}{1024} + ...$$
The denominator obviously follows $$2^n$$, but the numerator goes 110011001100...
I can't seem to find a function that will allow me to put this in a single sum:
$$\sum_{n=1}^{\infty} ....$$
(The sum should converge to 0.8, right?)
3. The attempt at a solution
I tried using the mod operator to determine if n was even or odd, something like this:
$$\sum_{n=1}^{\infty} \frac{n \mod 2}{2^n} + \frac{n \mod 2}{2^{n+1}}$$
This doesn't work, because the terms that are discarded when n is even are not discarded the next 'run' when (n+1) is even... Dunno how to explain this properly, but if you calculate it manually it does this:
$$\frac{1}{2} + \frac{1}{4} + \frac{0}{4} + \frac{0}{8} + \frac{1}{8} + \frac{1}{16} + ...$$
So if you discard the 0/ ... terms you are just left with the usual 2^(-n) sum...
I have a feeling I'm close, but I can't figure it out :S
Last edited: Feb 15, 2009
2. Feb 15, 2009
### epenguin
If the missing bits were not missing it would add up to 1.
Now you can put each missing bit into a relation with a bit that is there.
They are a certain fraction of terms that are there. So their sum is that proportion of 1 IYSWIM.
Do a diagram and this will look plausible.
3. Feb 15, 2009
### Nick89
I can't seem to understand what you mean exactly...
By missing bits you mean the 0/x parts?
If so, then half of the terms as missing, so to your logic that would mean the sum would converge to 0.5...? (Whatr does IYSWIM mean?)
But I think it should converge to 0.8 (I said 8 before but that was a mistake, I edited it now).
4. Feb 15, 2009
### Dick
If it's the pattern you suggest, (1/2+1/4)*(1/16)=(1/32+1/64). (1/32+1/64)*(1/16)=(1/512+1/1024). It's a geometric series with a common ratio of 1/16. But this doesn't quite fit with the series you quote in the first line (the next term after 1/64 is 1/1024, not 1/512). Is that a typo?
5. Feb 15, 2009
### epenguin
IYSWIM means 'if you see what I mean'.
Just compare each pair of 'absent' terms with the 'present' pair preceding it.
6. Feb 15, 2009
### Nick89
It was indeed a typo. The correct form is right underneath, where 1/512 is present.
Alright, that makes a bit more sense. I still fail to see how I can now create a single sum for this series though... If I have a bit more time (maybe later tonight) I'll try to figure it out :)
Thanks so far.
Similar Discussions: Stupid series I can't seem to find | {"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": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9564924240112305, "perplexity": 894.3648260627302}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934806543.24/warc/CC-MAIN-20171122084446-20171122104446-00703.warc.gz"} |
http://pages.cs.wisc.edu/~kurtis/ | # How the Abstract Becomes Concrete
## Irrational Numbers Are Anchored on Natural Numbers and Perfect Squares
### Abstract
Mathematical cognition research has largely emphasized concepts that can be directly perceived or grounded in visuospatial referents. These include concrete number systems like natural numbers, integers, and rational numbers. Here, we investigate how a more abstract number system, the irrationals denoted by radical expressions like , is understood across three tasks. Performance on a magnitude comparison task suggests that people interpret irrational numbers – specifically, the radicands of radical expressions – as natural numbers. Response times and strategy self-reports during number line estimation reveal that the spatial locations of irrationals are determined by anchoring to neighboring perfect squares. Perfect squares also facilitate the evaluation of complex arithmetic expressions. These converging results align with a constellation of related anchoring phenomena spanning tasks and number systems of varying complexity. Accordingly, we propose that the task-specific recruitment of more concrete representations, formal anchoring, is an important mechanism for understanding and teaching mathematics.
### Irrational Numbers Are Anchored on Natural Numbers and Perfect Squares
The historical development of number systems can be characterized as a gradual progression from the concrete to the abstract. Natural numbers, which denote the cardinalities of sets, were understood directly and formalized first. They are concrete in the sense that they map directly to quantities in the material world such as the number of berries in one’s hand. According to the mathematician Leopold Kronecker, “God created” this relatively concrete number system, whereas more abstract number systems like the integers, rationals, irrationals, and reals are “the work of man.” In other words, more abstract number systems were discovered later and constructed hierarchically upon the natural numbers. For example, combining pairs of natural numbers with the subtraction operation generates the integers and combining pairs of integers with the division operation generates rational numbers. These number classes are still relatively concrete in the sense that they can be interpreted as corresponding to real world quantities like basement floors for integers and pie pieces for rationals (Van de Walle, Karp, & Bay-Williams, 2010). It is with the irrational numbers, which include √2 and π, that mathematicians discovered a truly abstract number system in the sense of lacking material referents or models that build intuition. Such abstraction engenders many surprising numerical properties. For instance, there are more irrational numbers than natural numbers, integers, or rational numbers. The set of all irrationals is uncountably infinite in cardinality, whereas the latter sets are each countably finite. Perhaps not surprisingly, mathematics teachers encourage students to think of irrational numbers like π by using rational number approximations like 3.14. This often leads to conceptual confusions.
While cognitive science research has emphasized how more concrete number systems are grounded in analog magnitude representations, it has not tested whether more abstract systems like the irrationals are understood similarly. The natural numbers are thought to be grounded in directly perceptible objects and internally represented as continuous magnitudes on a mental number line (Moyer & Landauer, 1967). For less concrete number systems, the necessity of mastering the associated symbol systems has been acknowledged. However, the focus has been on how these symbol systems recruit and restructure magnitude representations. Examples include incorporating the inverse relationship between positive and negative integers into the mental number line and capturing the ratio relationship between the numerators and denominators of fractions more precisely.
Comparatively little attention has been directed towards abstract mathematical concepts that are both imperceptible and incapable of being fully grounded in visuospatial referents. For such concepts, do magnitude representations continue to be heavily recruited or does notation-specific strategic processing become more essential? This paper uses irrational numbers as a test case to address the research question – How are irrational numbers denoted by symbolic radical expressions like √(2 )mentally represented and processed by typical adults? We begin to examine this question by reviewing the literature on natural numbers, integers, and rational numbers.
### Natural Numbers
The natural numbers are the set of countable numbers {0,1,2,…}. Their mental representation has been studied extensively using the magnitude comparison (MC) task. When comparing which number in a pair is greater or lesser, response time decreases as the difference between the numbers increases. For instance, people can judge the greater or lesser number in the pair (2, 9) more quickly than in the pair (5, 6). This well-replicated performance pattern is known as the distance effect. It manifests whether participants compare symbolic numbers or visuospatial numerosities like sets of dots. The distance effect is commonly interpreted as evidence that natural numbers are represented in part as psychophysically scaled magnitudes. It has been observed as early as infancy with numerosities and kindergarten with symbolic numbers. Over development, children’s representation of numerical magnitude changes such that response times decrease overall and the slope of the distance effect decreases. These changes indicate improved precision of magnitude representations.
The magnitude representation of natural numbers and its development have also been probed by the number line estimation (NLE) task. In the bounded version of this task, participants are presented with a number line with the left and right endpoints labeled (e.g., as 0 and 100, respectively) and the middle segment left blank. Numbers are presented one at a time and in random order. The goal is to mark the position of each number on the number line. Performance is typically measured by average absolute error – the absolute difference between the correct position of the number and the position selected by the participant. Higher error values indicate worse performance. Over development, children’s accuracy on the NLE task improves gradually. There is a rich debate in the literature about the representations and processes that underlie performance on this task and its change over time. Some researchers have proposed that children initially represent natural numbers in a compressed, logarithmically spaced fashion. Over development, this incorrect representation improves to incorporate linear spacing. Some researchers have argued that the representation is proportional between critical landmarks such as the midpoint. Others have proposed that the representation is piecewise linear between critical landmarks like place-value boundaries (e.g. 10).
Individual differences in the magnitude representations of natural numbers, as indexed by the slope of the distance effect in the MC task, predict variation in mathematical achievement for elementary and middle school students. Likewise, NLE performance predicts mathematical achievement in elementary school students, although not older students learning more advanced topics. Instructional studies support a causal link between numerical magnitude representations and arithmetic skills. Young children who were tutored on the magnitudes of small natural numbers via a board game performed better on simple arithmetic problems than their peers who received a control lesson.
### Integers
The natural numbers combined with their additive inverses, the negative numbers, form the integers {…, -2, -1, 0, 1, 2, …}. Comprehending the meaning of this number system requires engaging in additional symbolic processing. For instance, when judging the greater number in the pair (-3, -9), it is necessary to inhibit the whole number interpretations of “-3” as “3” and “-9” as “9”. Children initially compare such integer pairs by using a symbolic strategy. First, they convert the negative numbers to natural numbers by ignoring the minus signs. Next, they compare the numerical magnitudes and then reverse this judgment. This strategy results in a distance effect when comparing pairs of negatives numbers. But when comparing mixed integer pairs like (-4, 9), judgments can be made by noticing that positives are greater than negatives. The result is the absence of a distance effect in elementary school children. Over development, this strategic processing of integers is substituted by accessing a restructured magnitude representation that incorporates the symmetry of positives and negatives around the zero point. Adults represent the negative number line as a reflection of the positive number line, incorporating the additive inverse property x+-x=0. This shift manifests as an inverse distance effect when comparing mixed integer pairs – as the distance between the integers increases, so does the responses time.
As with natural numbers, NLE tasks have also uncovered a logarithmic-to-linear shift in the mental representation of integers. When estimating the positions of positive numbers in the range 0 – 1000 and negative numbers in the range -1000 – 0, second graders exhibit logarithmic magnitude representations. By contrast, fourth and sixth graders exhibit linear representations for both ranges. This logarithmic-to-linear shift extends to larger number ranges and older people. Middle school students’ estimates of integers in the negative range -10000 – 0 and the combined range -1000 – 1000 are linear, though with lower accuracy for negative numbers. Performance on NLE tasks like these depends partly on recognizing the symmetry of the integers around the zero point.
The inverse distance effect for the MC task and symmetry effects on the NLE task led Tsang, Blair, Bofferding, and Schwartz (2015) to design a manipulative that encourages learners to incorporate the zero point in their magnitude representation of the integers. Instruction with the manipulative improved arithmetic problem-solving for elementary school students on difficult items like 3+x=0.
### Rational Numbers
Rational numbers are the set of numbers that can be written as a ratio of two integers such that the denominator is non-zero. Magnitude comparison with symbolic fractions and decimals, as well as with non-symbolic ratios, produces the distance effect. Thus, rational numbers are also thought to be integrated on the mental number line. As children begin to understand fraction magnitudes, the slope of the distance effect is nearly zero. Over time, fraction magnitude information becomes more accessible and the slope increases. By adulthood, the slope stabilizes. These behavioral changes are paralleled by the discovery and use of strategies to facilitate performance. For instance, children and young adults often report using unit fractions like ½ as anchors to perform fraction magnitude comparison.
In contrast to natural numbers and integers, NLE tasks suggest that the representation of rational numbers may not undergo a logarithmic-to-linear developmental shift. When 10-year-olds and adults perform NLE with decimal proportions or fractions, both age groups exhibit highly linear estimation patterns. Improvements have been observed over development. On average, 8th graders make more accurate estimates of fractions than 6th graders. At the individual level, a higher percentage of 8th graders than 6th graders exhibit linear (vs. logarithmic) representations. These performance gains may result in part from using unit fractions like ½ as anchors. For example, Siegler and Thompson (2014) found that children who reported using such anchors to segment the number line produced more accurate estimates.
The precision of rational number magnitude representations is also associated with higher-level mathematical skills. Symbolic fraction magnitude representations predict achievement across a range of domains and measures – fraction arithmetic, algebra, grade school standardized exams, and high school mathematical achievement. Similarly, accuracy on NLE tasks using decimals predicts mathematics achievement in elementary school students. Even nonsymbolic ratio precision predicts college students’ knowledge of fractions and algebra. Causal support for the link between magnitude knowledge and arithmetic skills comes from interventions that devote more instructional time to mastering the magnitudes of fractions. Results show that emphasizing magnitude comprehension boosts arithmetic skills for high and low achievers.
### Irrational Numbers
Irrational numbers are incommensurable. Unlike rational numbers, they cannot be expressed as ratios of integers such that the denominator is non-zero. Their decimal expansions are infinitely long, non-repeating, and ostensibly random. When they were first proposed, the notion seemed outlandish. According to a famous anecdote, the Greek mathematician who first proved their existence was drowned at sea for challenging the ratio doctrine of numbers. It was not until the late 1800s that irrationals were formalized and properly integrated onto the real number line (Dedekind, 1963/1888). Two subclasses can be distinguished – algebraic irrationals and transcendental numbers. Algebraic irrationals, like ∛20, are the solutions to polynomial equations and can be denoted by expressions of the form √(y&x). Transcendental numbers, like π and e, are not the solutions to any polynomial equation.
Just one other study has investigated the cognitive bases of algebraic irrationals. It explored whether the same human “number sense” that allows us to compare the magnitudes of natural numbers, integers, and rational numbers extends to irrationals. Participants performed magnitude comparisons with irrationals of the form √(y&x). Both the root y and the radicand x could vary. The researchers tested whether participants compared the magnitudes of irrationals by accessing their holistic magnitudes or by focusing on the root and radicand components. Comparisons were faster when both numbers contained a common root component, as in the pair (√(9&12),√(9&63)). Likewise, comparisons were faster when both numbers contained a common radicand component, as in the pair (√(10&34),√(13&34)). When both the roots and radicands differed, as in the pair (√(10&34),√(15&68)), response times were slower. Unexpectedly, response times for the common component pairs were not predicted by the distances between the root components or radicand components. These mixed results require explanation. One possibility is the use of irrational numbers denoted by complex radical expressions that do not occur in contexts such as solving quadratic equations. Another is the highly expert sample composed of mathematics graduate students and professors. It remains unclear how typical adults understand irrational numbers like √2.
The stimuli we used in our study are also algebraic irrationals, though we focus on square root expressions of the form √x. Such expressions are often encountered when solving quadratic equations. They denote irrational numbers when the radicand x is not a perfect square (e.g., √2) and natural numbers when the radicand is a perfect square (e.g., √(9 )). In the following sections, we will collectively refer to irrational numbers and perfect squares as radical expressions because both contain the radical sign.
### Research Questions and Hypotheses
First, we asked whether radical expressions are represented as continuous magnitudes integrated on the mental number line much like natural numbers, integers, and rational numbers. We refer to this proposal as the mental number line hypothesis. An alternative proposal is that people use processes that capitalize on the discrete structure of radical expressions. For instance, when judging the greater or lesser number in the pair (√(3 ), √(8 )), people may ignore the radical signs and only compare the radicand components (3, 8) using their magnitude representations of natural numbers. This is possible because when x and y are non-negative, judgments of (√(x ), √(y )) and (x, y) are equivalent. We refer to this as the equivalence strategy hypothesis.
Second, we investigated whether people process irrational numbers by strategically anchoring them on more concrete concepts like natural numbers and perfect squares. Specifically, we hypothesized the possible use of a multiplication strategy on the MC task whereby perfect square pairs like (√4, √(9 )) are transformed to computationally analogous “tie” multiplication problems like 2 × 2 and 3 × 3. Such a transformation might facilitate comparison because tie problems are processed more quickly than non-tie problems of comparable size. On the NLE task, irrational numbers might be estimated in relation to perfect squares – the landmark strategy. For example, people may estimate the positions of irrational numbers like √(3 ) by referencing the positions of neighboring perfect squares like √(1 ) and √(4 ). Finally, people may leverage their knowledge of perfect squares during arithmetic problem-solving. When simplifying √72, an inefficient strategy would be to decompose the radicand into its prime factors and then shift pairs of common factors outside the radical sign – the prime factorization strategy. For example:
√72 = √(2×36)
= $$\sqrt{2*2*18}$$
= √(2×2×2×9)
= √(2×2×2×3×3)
= 2√(2×3×3)
= 2×3√2
= 6√2
A more efficient approach is to factor the radicand into its largest perfect square factors and directly reduce these – the perfect squares factorization strategy:
√72 = √(2×36)
= 6√2
Third, we investigated whether individual differences in the mental representation and processing of irrationals explain variation in conceptual and procedural knowledge of this number system. Some researchers have proposed that magnitude representations are at the core of numerical and arithmetic performance (Link, Nuerk, & Moeller, 2014; Siegler, 2016). Contra this view, we hypothesize that the influence of magnitude representations on arithmetic problem-solving attenuates as the abstraction of number systems increases. Arithmetic performance with irrationals may depend less on magnitude representations and more on symbolic strategies. | {"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": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7708588242530823, "perplexity": 1321.4244919292023}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084887067.27/warc/CC-MAIN-20180118051833-20180118071833-00029.warc.gz"} |
https://www.physicsforums.com/threads/moment-about-the-origin.397980/ | # Homework Help: Moment about the origin
1. Apr 24, 2010
### Angello90
1. The problem statement, all variables and given/known data
Find the moment about the origin of a vector of magnitude 100 units acting from
A=(2, 0, 4) to B=(5, 1, 1). What is the moment about the point C=(−1, 2, 1)?
2. Relevant equations
M= r x F
3. The attempt at a solution
I did the quesion, but I'm not sure if it's correct. Can anyone check it for me?
#### Attached Files:
• ###### moment.jpg
File size:
11 KB
Views:
443
2. Apr 24, 2010
### HallsofIvy
The "moment of a vector about a point" is the magnitude of the vector times the length of a line segment from the given point perpendicular to the line of the vector.
Since the vector acts "from (2, 0, 4) to (5, 1, 1)", its line of direction is given by x= 3t+ 2, y= t, z= -3t+ 4. You need to find the distance from (0, 0, 0) to that line and from (-1, 2, 1) to that line . Then multiply those by 100.
3. Apr 24, 2010
### Angello90
So I need to find |OL| where O = (0, 0, 0) and L = (2+3t, t, 4-3t), thus
Making a moment of
Simillary to Q = (-1, 2, 1)
#### Attached Files:
File size:
5.4 KB
Views:
416
File size:
4.7 KB
Views:
428
• ###### QL Mql.jpg
File size:
11.9 KB
Views:
451
4. Apr 26, 2010
### Angello90
Ahhh come on guys help me out here! Any hints? | {"extraction_info": {"found_math": false, "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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.91060471534729, "perplexity": 1954.9481485277108}, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221219197.89/warc/CC-MAIN-20180821230258-20180822010258-00080.warc.gz"} |
https://algebra-calculators.com/what-do-you-mean-by-rational-irrational-number-in-math-2/ | # What do you mean by Rational & Irrational Number in math?
An online rational irrational number definition
A rational number is a number that can be written as a ratio.
Rational numbers can be written as a fraction, in which both the numerator (the number on top) and the denominator (the number on the bottom) are whole numbers.
The number 10 is a rational number because it can be written as the fraction 10/1.
## Irrational Numbers
All numbers that are not rational are considered irrational. An irrational number can be written as a decimal, but not as a fraction.
An irrational number has endless non-repeating digits to the right of the decimal point.
The number 22/7 is a irrational number..
### Math Number Symbols
R – Real Numbers
Z – Integers
N – Natural Numbers
Q – Rational Numbers
P – Irrational Numbers | {"extraction_info": {"found_math": false, "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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8991531729698181, "perplexity": 696.5056241816739}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296950383.8/warc/CC-MAIN-20230402043600-20230402073600-00114.warc.gz"} |
https://codereview.stackexchange.com/questions/96145/creating-original-database-algorithm-login-system/96177 | # Creating Original Database Algorithm--Login System
For fun, I have been making a very basic login system in Python. I was going to look up an algorithm or something for user info storage and transfer, but then decided it would be more fun to come up with and implement it myself. What I would like to know is:
1. Is there anyway to make it secure (i.e.-encrypting the password without revealing the algorithm in the file)?
2. Can the algorithm I used be made any more efficient?
3. Is it cleanly written? I tried to take into account the comments on my last question regarding simplifying code and any comments about its readability would be appreciated--by secondary goal is beautiful code.
#This program will do a couple of things.
#The first time it is run, it will request that you add a password. Once you do, it will change the check_e\$
#Every subsequent time, it will check the password in the other file and then match user input.
#If user input is correct it will display a joke.
#If user input is incorrect it will exit the program.
#Function to help the user pick a password
target = open(file, 'w')
target = open(file, 'w')
target.write('YES')
# file.close()
if userpass == pwd_check:
raw_input()
print 'Why did the fly fly? Because the spider spied her!'
elif userpass != pwd_check:
exit()
else:
print 'Invalid syntax.'
exit()
#location of password existence check file
if pswd_exist == 'YES':
pass
else:
Note: If you want to run it, it requires two files --pswd.txt and existence_check.txt--which you then must link (and replace all necessary file locations).
If you input the correct password, the program tells a joke, though whether its funny is debatable.
• Please don't try to implement security / crypto yourself. You may say this is just to learn, but you are learning an incorrect way of doing things. Stick with a proven secure implementation for anything security related – Milney Nov 21 '16 at 15:44
In your function pick_password, you should close the file at the end of the file because this frees up any system resources that were being devoted to working with that file.
For some reason, at the end of that function, you commented out the line
f.close()
You should un-comment this.
In python, it is good practice to open a file using the with keyword. Using with, you can easily access the file object and the file is automatically closed at the end of the with statement.
Here is how you should open the file:
with open(file, "w") as f:
# whatever you want to do to the file f
Source
In your code, you have three separate variables that are all set to
'/home/vhx/Documents/code/pydata_test/password_dbs/pswd.txt'
You should create a single global constant variable that holds this value. Here is what that would look like:
PASSWORDS = '/home/vhx/Documents/code/pydata_test/password_dbs/pswd.txt'
I put the variable name in all-caps because constants are usually all capital letters.
In your function password_check, you make an exit call in a few places. However, you don't pass anything to it.
The number you pass to an exit call is called the exit code. This number is to show how execution went when the program was running to external processes.
Most commonly, an exit code of 0 means that execution was perfect, and an exit code of any non-zero number means that something went wrong.
I think that you should exit your program with 1 to show that something went wrong the user was entering their password (they entered the password incorrectly).
An advantage to doing it this way would be say, for example, you wrote another program and you only wanted to give access to it if the user entered the password correctly. In this program, you could read the exit code of this password program and if it's a 1, than your new program now knows that the user entered the password incorrectly.
P.S. Your interpreter is probably already doing this for you, but at the top of your code you should have from sys import exit because the sys library holds the exit function that you are calling.
Thanks to janos for correcting me on error exit codes.
I believe that it is fairly difficult to make this more secure.
1. The user can view the passwords file
2. If you do encrypt the file, this voids the above reason but the user will always be able to see your python source.
However, as BlueRaja - Danny Pflughoeft commented,
-1 for "impossible to make this more secure" - since he's storing passwords in plain-text, I'd argue it's impossible to make this less secure. Even if you need the passwords to be reversible (which is usually not the case), you should be encrypting them in a way that makes them unreadable even with the Python source (see eg. LastPass, which encrypts all passwords using a single master password)
• Thanks! I have been wondering for a while the usefulness of closing a file, and this definitely clears that up. – Joseph Farah Jul 7 '15 at 21:03
• -1 for "impossible to make this more secure" - since he's storing passwords in plain-text, I'd argue it's impossible to make this less secure. Even if you need the passwords to be reversible (which is usually not the case), you should be encrypting them in a way that makes them unreadable even with the Python source (see eg. LastPass, which encrypts all passwords using a single master password) – BlueRaja - Danny Pflughoeft Jul 8 '15 at 23:07
• @BlueRaja-DannyPflughoeft I have edited my answer to accommodate your recommendation. – SirPython Jul 8 '15 at 23:30
It looks like passwords are stored in plain text form in the pswd.txt file. Never ever store passwords in plain text form. Store passwords salted and cryptographically hashed. That way, if an attacker gains access to the file, he still has to crack the password, which can be extremely difficult if the password is strong enough.
To verify a password, apply the same algorithm to the user input as used when creating the salted and cryptographically hashed version. The result will only match the stored password if the user entered the correct password.
@Boris left a great comment, quoting it verbatim:
Encrypted passwords are a bad idea as the encryption key needs to be stored in the code, then then it's just a question of looking at the code to gain the key. Hashing is irreversible and using a good, random, unique, salt per password and a purpose built cryptographic hashing algorithm is the only acceptable way of storing passwords - the best idea is to use something like bcrypt which is industry standard and rolls all this together.
• Ahem, s/encrypted/salted and cryptographically hashed/g. Encrypted passwords are a bad idea as the encryption key needs to be stored in the code, then then it's just a question of looking at the code to gain the key. Hashing is irreversible and using a good, random, unique, salt per password and a purpose built cryptographic hashing algorithm is the only acceptable way of storing passwords - the best idea is to use something like bcrypt which is industry standard and rolls all this together. – Boris the Spider Jul 8 '15 at 6:46
• in unix/linux error messages are printed to stderr and not to stdin
• it is unclear to me, why you need the 'existence_check.txt'. If the 'pswd.txt' file exists, it contains a valid password.
• you use the same literals (filenames) on different places. Use the variables EC and PWD instead of.
• the name PWD has a special meaning in the unix context. maybe you shouk use another name.
• the filenames should not be part of the program but they shoul be read from a configuration file.
• if you create a file that stores such sensible information like passwords then you shoul ensure that the file that you create has the approptiate permissions.
• I don't like jokes in login procedures. Especially if they make it necessary to press the return key again.
• I am not familiar enough with python. Is the file automatcally closed after pswd_exist = open(EC).read()?
• Do not store passwords. Check how password algorithms work. Most procedures for entering the password don't display the user input. And therefore the user has to input the password twice and the two inputs are compared. Your password cannot be changed, no minimal requirements. So if a user presses erroneously the return key when he shoul pick a password, the empty password is set forever,
• In regards to your question about the file being closed after open(EC).read(): no, the file will not automatically close. – SirPython Jul 8 '15 at 16:17
The standard way to open a file in Python is to use a with ... as ... context manager. By using a with ... as ... context manager, you can ensure that the file is properly closed, and there aren't any memory issues. Here's how you'd do that:
with open("path/to/myfile.txt", "r") as my_file:
...
Context managers also allow for multiple values as well. Here's an example of that:
with open("path/to/myfile.txt", "r") as my_file, ...:
...
You should also be using docstrings to describe your functions, not regular inline comments. Here's an example of a docstring:
def my_func( ... ):
"""
Describe your function and it's arguments
in this docstring.
"""
Finally, this is not a good way to do password checking at all. Preferably, rather than creating your own system without any security, you should use a library, like OpenSSL.
The security aspects have been pretty well-covered, so here are a few comments on your general Python style:
• Rather than using block comments to document functions and/or scripts, it's better to use docstrings. This is a string at the top of a function and/or module (within the function definition) that explains how the function works.
See PEP 257: Docstring conventions for more details.
• Don't use file as a variable name; this is the name of a builtin function. Using the same variable name as a builtin is liable to be confusing, and potentially a source of subtle bugs.
• The Python convention (see PEP 8: Indentation) is that there are four spaces per indentation level, not 8.
• There is no need to have explicit exit() statements; the script will yield control flow naturally. If you really want to be explicit about leaving the function here, use return. Using exit() just makes this function harder to reuse.
• There's a common path for storing your files:
/home/vhx/Documents/code/pydata_test/password_dbs/<filename>
which appears file times in your code, with the only thing that changes being the filename. It would be better to pull the directory name into a variable, and then if it ever needs to change, you only need to do so in one 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": 0, "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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.1795947253704071, "perplexity": 1489.26604132102}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578526923.39/warc/CC-MAIN-20190419001419-20190419023419-00554.warc.gz"} |
https://www.quantumdiaries.org/2012/07/03/higgs-seminar-live-blog-from-ichep/ | ## View Blog | Read Bio
### Higgs seminar live blog from ICHEP
Good afternoon from the Melbourne Convention Centre, where we are all eagerly awaiting the start of the Higgs seminar that will be broadcast from CERN. I’ll be updating this post as we go along, so please stay with me and our other Quantum Diaries bloggers.
17:00: That’s a wrap here! I’ll try to post again soon-ish with some more thoughtful follow-up. Thanks everyone for joining us!
18:58: Not that Higgs is the only one who helped develop this theory…the others are getting their time to make congratulations too.
18:56: A big round of applause here and at CERN for the famous Peter Higgs.
18:52: No questions here! Rolf is right, we want to go to the reception!
18:48: Now questions. I’ll try to transcribe the ones that are coming from this room.
18:47: We’re watching the standing ovation at CERN over the video. Again, so nice to see lots of young people in the room. Then again, maybe they were the ones with the endurance to camp out in front of the auditorium!
18:44: Rolf summarizes by saying that this is a global effort and a global success. It’s only possible because of the extraordinary performance of the accelerators, experiments and computing grid. This is an observation of a new particle consistent of a Higgs boson…but we don’t know which one. It’s an historic milestone, but it is only the beginning, with global implications for the future.
18:41: The conclusion from ATLAS — an excess of events at about 126.5 GeV at 5.0 sigma significance. Fitted signal strength is 1.2 +- 0.3 of the SM expectation. Fabiola says that this is a very lucky mass to have, since it is easy to explore at the LHC in many decay channels.
18:39: Fabiola shows how the signal strength has evolved over time, in the past year — it is really quite striking how much we have advanced!
18:37: ATLAS overall signal strength is a bit higher than SM expectation for 125 GeV Higgs, whereas CMS was a little lower.
18:34: So here comes the combination of the new 2012 analyses with the 2011 results: another 5 sigma excess, and more applause!
18:30: ZZ also shows an excess in the 125 GeV range, 3.4 sigma here, when you’d expect 2.6 sigma from a standard-model Higgs.
18:28: Hmm, the standard-model ZZ rate is coming in bigger than expected. But it has now impact on the low-mass Higgs range.
18:23: Here comes ZZ, with 20-30% increase in sensitivity since December.
18:19: And now the results from the photons channel. Here too, an excess, at the 3.6 sigma level, after accounting for the look-elsewhere effect. 4.5 sigma local.
18:14: Thorough discussion of mass resolution in the photon channel, and the fact that it is independent of pileup. This is quite important, as the pileup is only going to get larger as the LHC luminosity increases.
18:10: Fabiola reviews what the changes are since December, here we go…..
18:09: Fermilab has a press release out too, http://www.fnal.gov/pub/presspass/press_releases/2012/Higgs-Search-LHC-20120704.html
18:07: The CERN press release is out! http://press.web.cern.ch/press/pressreleases/Releases2012/PR17.12E.html
18:04: I’ll admit that I’m finding less to say at the moment, as Fabiola is now covering similar ground to CMS in the preparation of the data, understanding of the detector etc. But it is still all important, of course!
17:57: I think I caught that ATLAS will show 5.9 fb-1 of data? More than CMS has certified.
17:55: Fabiola will show results from 2012 data in the ZZ and gamma gamma channels, but the other channels only for 2011 data. The 2012 results in those channels (which do have less mass resolution) aren’t quite ready yet.
17:53: Rolf congratulates everyone…and now let’s see what ATLAS has!
17:50: We have observed a new boson with a mass of 125.3 +- 0.6 GeV at 4.9 sigma significance. Enthusiastic applause heard on two continents. And then Joe acknowledges all of the theorists, machine physicists, and CMS experimenters who who have gotten us to this point.
17:49: Branching ratios are self-consistent across the decay channels (even with that tau tau result), but Joe emphasizes that it is early yet.
17:48: OK, it looks like there is something there, but is it consistent with a standard-model Higgs? Now we’ll get a first look at that.
17:44: Technical improvements in the tau tau channel have helped improve its sensitivity in the Higgs search. But uh oh, no evidence for a Higgs in that channel, and indeed you nearly exclude a 125 GeV SM Higgs here! It’s low statistics yet, we’ll need a lot more data to understand this.
17:41: Now on to the bb and tau tau channels. Not as much sensitivity to them, but they are important, as a) they actually have bigger branching fractions at ~125 GeV than the WW/ZZ/gammagamma, and b) they are fermions whereas all the others are bosons.
17:38: Unfortunately, the seminar is not over yet — let’s see if that 5.0 stands up with three more decay channels to look at.
17:37: The ZZ and gamma gamma joint significance is 5.0 standard deviations…Joe says that, and it gets applause! 5.0 is considered the threshold for a discovery….
17:33: Joe is now showing how event by event, we can compare the angular separations of the leptons to what would be expected from a Higgs signal or the ZZ background. I’ve always loved that approach, really gets at the physics.
17:30: Now the ZZ mode, with four leptons in the final state — this channel is about as clean as you can get, but very rare, so you need to be super-efficient in selecting the events.
17:28: Slide 43 — A mass bump! Thank goodness that there is a real mass bump. I’ve always said, if we make a discovery that is observed on a plot that runs from 0 to 1 with a slight enhancement at 1, I’ll jump out the window. Maybe I’m safe.
17:22: Now we’re getting into the meat of the Higgs search, with the search for a decay to two photons. CMS paid a lot of money for the lead tungstate crystals that give very good photon energy resolution.
17:17: In fact, Joe is carefully going through all the low-level ingredients that go into these analyses. It’s a lot of work, and that’s why we need thousands of people to get this science done.
17:15: Joe takes a moment to point out the hard work done by the CMS software and computing teams. Yes, we’ve got a fabulous detector, but we can’t get this done without S&C…so I do appreciate the shout-out!
17:11: CMS will show 5.2 fb-1 of certified data today. And again, I’m curious to see how much data ATLAS will show!
17:09: CMS will be showing results from five Higgs decay modes — WW, ZZ, photons, bb, taus. I’m curious to see how many of those ATLAS also has.
17:06: Joe Incandela, for CMS, also starts with standup jokes. And then, on to a discussion of the experimental support for the standard model…except that we haven’t seen the Higgs. Yet.
17:03: Rolf notes that ICHEP is opening with a talk from a different continent — a symbol of how well we collaborate across the world. Then he starts with a standup routine….
17:02: Big round of applause here when Rolf Heuer greets ICHEP!
17:00: Here we go! Watching a very quiet CERN auditorium on video….
16:58: On the video feed from CERN, we can see a lot of young people. I’m glad they were able to get seats, as they are the ones who really make these experiments go.
16:54: Everyone in the hall just laughed at the video feed, which showed someone holding a “Ciao Mamma!” sign.
16:51: Geoff Taylor, the chair of ICHEP2012, is telling us about the timetable. The talks will be running longer than we originally envisioned (45 minutes each), followed by 30 minutes of questions. (We will be able to ask questions from here to CERN.) Hmm, hope everyone can hold out for the reception, which will be at 7 PM now!
16:47: I should say that as a member of CMS, I know the CMS results. Thus, I’m deeply curious about the ATLAS results. Will they agree with CMS or contradict? I’ve been hearing rumors, of course, but it will be interesting to see if they are true.
16:44: While we’re waiting for things to start, let me wish a happy Independence Day to everyone in the US, and also a happy wedding anniversary to my parents and a happy birthday to my colleague Aaron Dominguez, who is watching the proceedings from CERN Filtration Plant conference room.
16:42: I’m now sitting in the auditorium where we will be hearing the talk. Sitting to my left is Joel Butler, the manager of the US CMS operations program (I am also in the program management, as deputy manager of software and computing), and to my right is a reporter from the Australian Associated Press. On the big screen above the stage we can see the video feed from the CERN auditorium, and another panel which I presume will be the slides.
16:24: Just spoke to Pier Oddone and Young-Kee Kim, the director and deputy director of Fermilab. Yes, all of the spotlight is on CERN right now, but the Tevatron experiments have been very important for getting the Higgs search started, and it is also quite possible that CMS and ATLAS will not be able to beat CDF/D0 on the Higgs to bb decay mode anytime soon.
16:16: The registration line is still very long! I have been walking along it and chatting with friends in the hope of entertaining them a little. We’ll see if we can get everyone through in time for the seminar….
15:46: I’ve now picked up my registration materials. There is a long line at the registration desk — no one wants to be late for the start. Right now I’m at a table in the foyer with some of the CMS leadership: Greg Landsberg, the CMS physics coordinator; Chris Hill, a CMS deputy coordinator, and Christoph Paus, one of the leaders of the CMS Higgs group. On the way in, I discovered that the Melbourne Boat Show is also happening in the convention centre. I wonder if we have any buyers at ICHEP….
Tags: , | {"extraction_info": {"found_math": false, "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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8176615834236145, "perplexity": 1768.089493266391}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038878326.67/warc/CC-MAIN-20210419045820-20210419075820-00496.warc.gz"} |
https://emukit.readthedocs.io/en/latest/api/emukit.model_wrappers.html | # emukit.model_wrappers package¶
## Submodules¶
class emukit.model_wrappers.gpy_model_wrappers.GPyModelWrapper(gpy_model, n_restarts=1)
This is a thin wrapper around GPy models to allow users to plug GPy models into Emukit
predict(X)
Parameters
X (ndarray) – (n_points x n_dimensions) array containing locations at which to get predictions
Return type
Tuple[ndarray, ndarray]
Returns
(mean, variance) Arrays of size n_points x 1 of the predictive distribution at each input location
predict_noiseless(X)
Parameters
X (ndarray) – (n_points x n_dimensions) array containing locations at which to get predictions
Return type
Tuple[ndarray, ndarray]
Returns
(mean, variance) Arrays of size n_points x 1 of the predictive distribution at each input location
predict_with_full_covariance(X)
Parameters
X (ndarray) – (n_points x n_dimensions) array containing locations at which to get predictions
Return type
Tuple[ndarray, ndarray]
Returns
(mean, variance) Arrays of size n_points x 1 and n_points x n_points of the predictive mean and variance at each input location
Parameters
X (ndarray) – (n_points x n_dimensions) array containing locations at which to get gradient of the predictions
Return type
Tuple[ndarray, ndarray]
Returns
(mean gradient, variance gradient) n_points x n_dimensions arrays of the gradients of the predictive distribution at each input location
Computes and returns model gradients of mean and full covariance matrix at given points
Parameters
X (ndarray) – points to compute gradients at, nd array of shape (q, d)
Return type
Tuple[ndarray, ndarray]
Returns
Tuple with first item being gradient of the mean of shape (q) at X with respect to X (return shape is (q, q, d)). The second item is the gradient of the full covariance matrix of shape (q, q) at X with respect to X (return shape is (q, q, q, d)).
set_data(X, Y)
Sets training data in model
Parameters
• X (ndarray) – New training features
• Y (ndarray) – New training outputs
Return type
None
optimize(verbose=False)
Optimizes model hyper-parameters
calculate_variance_reduction(x_train_new, x_test)
Computes the variance reduction at x_test, if a new point at x_train_new is acquired
Return type
ndarray
predict_covariance(X, with_noise=True)
Calculates posterior covariance between points in X :type X: ndarray :param X: Array of size n_points x n_dimensions containing input locations to compute posterior covariance at :type with_noise: bool :param with_noise: Whether to include likelihood noise in the covariance matrix :rtype: ndarray :return: Posterior covariance matrix of size n_points x n_points
get_covariance_between_points(X1, X2)
Calculate posterior covariance between two sets of points. :type X1: ndarray :param X1: An array of shape n_points1 x n_dimensions. This is the first argument of the
posterior covariance function.
Parameters
X2 (ndarray) – An array of shape n_points2 x n_dimensions. This is the second argument of the posterior covariance function.
Return type
ndarray
Returns
An array of shape n_points1 x n_points2 of posterior covariances between X1 and X2. Namely, [i, j]-th entry of the returned array will represent the posterior covariance between i-th point in X1 and j-th point in X2.
property X: ndarray
An array of shape n_points x n_dimensions containing training inputs
Type
return
Return type
ndarray
property Y: ndarray
An array of shape n_points x 1 containing training outputs
Type
return
Return type
ndarray
generate_hyperparameters_samples(n_samples=20, n_burnin=100, subsample_interval=10, step_size=0.1, leapfrog_steps=20)
Generates the samples from the hyper-parameters and returns them. :param n_samples: Number of generated samples. :param n_burnin: Number of initial samples not used. :param subsample_interval: Interval of subsampling from HMC samples. :param step_size: Size of the gradient steps in the HMC sampler. :param leapfrog_steps: Number of gradient steps before each Metropolis Hasting step. :rtype: ndarray :return: A numpy array whose rows are samples of the hyper-parameters.
fix_model_hyperparameters(sample_hyperparameters)
Fix model hyperparameters
Return type
None
emukit.model_wrappers.gpy_model_wrappers.dSigma(x_predict, x_train, kern, w_inv)
Compute the derivative of the posterior covariance with respect to the prediction input
Parameters
• x_predict (ndarray) – Prediction inputs of shape (q, d)
• x_train (ndarray) – Training inputs of shape (n, d)
• kern (<module 'GPy.kern' from '/home/docs/checkouts/readthedocs.org/user_builds/emukit/envs/latest/lib/python3.6/site-packages/GPy/kern/__init__.py'>) – Covariance of the GP model
• w_inv (ndarray) – Woodbury inverse of the posterior fit of the GP
Return type
ndarray
Returns
Gradient of the posterior covariance of shape (q, q, q, d)
emukit.model_wrappers.gpy_model_wrappers.dmean(x_predict, x_train, kern, w_vec)
Compute the derivative of the posterior mean with respect to prediction input
Parameters
• x_predict (ndarray) – Prediction inputs of shape (q, d)
• x_train (ndarray) – Training inputs of shape (n, d)
• kern (<module 'GPy.kern' from '/home/docs/checkouts/readthedocs.org/user_builds/emukit/envs/latest/lib/python3.6/site-packages/GPy/kern/__init__.py'>) – Covariance of the GP model
• w_vec (ndarray) – Woodbury vector of the posterior fit of the GP
Return type
ndarray
Returns
Gradient of the posterior mean of shape (q, q, d)
class emukit.model_wrappers.gpy_model_wrappers.GPyMultiOutputWrapper(gpy_model, n_outputs, n_optimization_restarts, verbose_optimization=True)
A wrapper around GPy multi-output models. X inputs should have the corresponding output index as the last column in the array
calculate_variance_reduction(x_train_new, x_test)
Calculates reduction in variance at x_test due to observing training point x_train_new
Parameters
• x_train_new (ndarray) – New training point
• x_test (ndarray) – Test points to calculate variance reduction at
Return type
ndarray
Returns
Array of variance reduction at each test point
Calculates gradients of predictions with respect to X, excluding with respect to the output index :type X: ndarray :param X: Point at which to predict gradients :rtype: Tuple[ndarray, ndarray] :return: (mean gradient, variance gradient)
predict(X)
Predicts mean and variance for output specified by last column of X :type X: ndarray :param X: point(s) at which to predict :rtype: ndarray :return: predicted (mean, variance) at X
set_data(X, Y)
Updates model with new training data :type X: ndarray :param X: New training features with output index as last column :type Y: ndarray :param Y: New training targets with output index as last column
Return type
None
optimize()
Optimizes hyper-parameters of model. Starts the optimization at random locations equal to the values of the “n_optimization_restarts” attribute.
Return type
None
property X: ndarray
Return type
ndarray
property Y: ndarray
Return type
ndarray
predict_covariance(X, with_noise=True)
Calculates posterior covariance between points in X
Parameters
• X (ndarray) – Array of size n_points x n_dimensions containing input locations to compute posterior covariance at
• with_noise (bool) – Whether to include likelihood noise in the covariance matrix
Return type
ndarray
Returns
Posterior covariance matrix of size n_points x n_points
get_covariance_between_points(X1, X2)
Calculate posterior covariance between two points :type X1: ndarray :param X1: An array of shape 1 x n_dimensions that contains a data single point. It is the first argument of the
posterior covariance function
Parameters
X2 (ndarray) – An array of shape n_points x n_dimensions that may contain multiple data points. This is the second argument to the posterior covariance function.
Return type
ndarray
Returns
An array of shape n_points x 1 of posterior covariances between X1 and X2
generate_hyperparameters_samples(n_samples=10, n_burnin=5, subsample_interval=1, step_size=0.1, leapfrog_steps=1)
Generates the samples from the hyper-parameters, and returns them (a numpy array whose rows are samples of the hyper-parameters). :param n_samples: Number of generated samples. :param n_burnin: Number of initial samples not used. :param subsample_interval: Interval of subsampling from HMC samples. :param step_size: Size of the gradient steps in the HMC sampler. :param leapfrog_steps: Number of gradient steps before each Metropolis Hasting step.
Return type
ndarray
fix_model_hyperparameters(sample_hyperparameters)
Fix model hyperparameters
Return type
None
GPy wrappers for the quadrature package.
Wrapper for GPy’s GPRegression as required for some EmuKit quadrature methods.
An instance of this class can be passed as base_gp to a WarpedBayesianQuadratureModel object.
Note
GPy’s GPRegression cannot take None as initial values for X and Y. Thus, we initialize them with some arbitrary values. These will be re-set in the WarpedBayesianQuadratureModel.
Parameters
• kern (QuadratureKernel) – An EmuKit quadrature kernel.
• gpy_model (GPRegression) – A GPy GP regression model.
• noise_free (bool) – If False, the observation noise variance will be treated as a model parameter, if True the noise is set to 1e-10, defaults to True.
property X: ndarray
The data nodes.
Return type
ndarray
property Y: ndarray
The data evaluations at the nodes.
Return type
ndarray
property observation_noise_variance: float
The variance of the Gaussian observation noise.
Return type
float
set_data(X, Y)
Sets training data in model.
Parameters
• X (ndarray) – New training features, shape (num_points, input_dim).
• Y (ndarray) – New training outputs, shape (num_points, 1).
Return type
None
predict(X_pred)
Predictive mean and covariance at the locations X_pred.
Parameters
X_pred (ndarray) – Points at which to predict, with shape (number of points, input_dim).
Return type
Tuple[ndarray, ndarray]
Returns
Predictive mean, predictive variances shapes (num_points, 1) and (num_points, 1).
predict_with_full_covariance(X_pred)
Predictive mean and covariance at the locations X_pred.
Parameters
X_pred (ndarray) – Points at which to predict, with shape (num_points, input_dim).
Return type
Tuple[ndarray, ndarray]
Returns
Predictive mean, predictive full covariance shapes (num_points, 1) and (num_points, num_points).
solve_linear(z)
Solve the linear system $$Gx=z$$ for $$x$$.
$$G$$ is the Gram matrix $$G := K(X, X) + \sigma^2 I$$, of shape (num_dat, num_dat) and $$z$$ is a matrix of shape (num_dat, num_obs).
Parameters
z (ndarray) – A matrix of shape (num_dat, num_obs).
Return type
ndarray
Returns
The solution $$x$$ of the linear, shape (num_dat, num_obs).
graminv_residual()
The solution $$z$$ of the linear system
$(K_{XX} + \zeta^2 I) z = (Y - m(X))$
where $$X$$ and $$Y$$ are the available nodes and function evaluation, $$m(X)$$ is the predictive mean at $$X$$, and $$\zeta^2$$ the observation noise variance.
Return type
ndarray
Returns
The solution $$z$$ of the linear system, shape (num_dat, 1).
optimize()
Optimize the hyperparameters of the GP.
Return type
None
Bases: IRBF
Wrapper of the GPy RBF kernel as required for some EmuKit quadrature methods.
$k(x, x') = \sigma^2 e^{-\frac{1}{2}\sum_{i=1}^{d}r_i^2},$
where $$d$$ is the input dimensionality, $$r_i = \frac{x_i-x_i'}{\lambda_i}$$ is the scaled vector difference of dimension $$i$$, $$\lambda_i$$ is the $$i$$ th element of the lengthscales property and $$\sigma^2$$ is the variance property.
Parameters
gpy_rbf (RBF) – An RBF kernel from GPy.
property lengthscales: ndarray
The lengthscales $$\lambda$$ of the kernel.
Return type
ndarray
property variance: float
The scale $$\sigma^2$$ of the kernel.
Return type
float
K(x1, x2)
The kernel k(x1, x2) evaluated at x1 and x2.
Parameters
• x1 (ndarray) – First argument of the kernel, shape (n_points N, input_dim)
• x2 (ndarray) – Second argument of the kernel, shape (n_points M, input_dim)
Return type
ndarray
Returns
Kernel evaluated at x1, x2, shape (N, M).
Wrapper of the GPy Exponential (a.k.a Matern12) product kernel as required for some EmuKit quadrature methods.
The product kernel is of the form $$k(x, x') = \sigma^2 \prod_{i=1}^d k_i(x, x')$$ where
$k_i(x, x') = e^{-r_i}.$
$$d$$ is the input dimensionality, $$r_i:=\frac{|x_i - x'_i|}{\lambda_i}$$, $$\sigma^2$$ is the variance property and $$\lambda_i$$ is the $$i$$ th element of the lengthscales property.
Parameters
• gpy_matern (Union[Exponential, Prod, None]) – An Exponential (a.k.a. Matern12) product kernel from GPy. For $$d=1$$ this is equivalent to an Exponential kernel. For $$d>1$$, this is not a $$d$$-dimensional Exponential kernel but a product of $$d$$ 1-dimensional Exponential kernels with differing active dimensions constructed as k1 * k2 * … . Make sure to unlink all variances except the variance of the first kernel k1 in the product as the variance of k1 will be used to represent $$\sigma^2$$. If you are unsure what to do, use the lengthscales and variance parameter instead. If gpy_matern is not given, the lengthscales argument is used.
• lengthscales (Optional[ndarray]) – If gpy_matern is not given, a product Matern12 kernel will be constructed with the given lengthscales. The number of elements need to be equal to the dimensionality $$d$$. If gpy_matern is given, this input is disregarded.
• variance (Optional[float]) – The variance of the product kernel. Only used if gpy_matern is not given. Defaults to 1.
property lengthscales: ndarray
The lengthscales $$\lambda$$ of the kernel.
Return type
ndarray
property variance: float
The scale $$\sigma^2$$ of the kernel.
Return type
float
K(x1, x2)
The kernel k(x1, x2) evaluated at x1 and x2.
Parameters
• x1 (ndarray) – First argument of the kernel, shape (n_points N, input_dim)
• x2 (ndarray) – Second argument of the kernel, shape (n_points M, input_dim)
Return type
ndarray
Returns
Kernel evaluated at x1, x2, shape (N, M).
dK_dx1(x1, x2)
Gradient of the kernel wrt x1 evaluated at pair x1, x2.
Parameters
• x1 (ndarray) – First argument of the kernel, shape (n_points N, input_dim)
• x2 (ndarray) – Second argument of the kernel, shape (n_points M, input_dim)
Return type
ndarray
Returns
The gradient of the kernel wrt x1 evaluated at (x1, x2), shape (input_dim, N, M)
Wrapper of the GPy Matern32 product kernel as required for some EmuKit quadrature methods.
The product kernel is of the form $$k(x, x') = \sigma^2 \prod_{i=1}^d k_i(x, x')$$ where
$k_i(x, x') = (1 + \sqrt{3}r_i ) e^{-\sqrt{3} r_i}.$
$$d$$ is the input dimensionality, $$r_i:=\frac{|x_i - x'_i|}{\lambda_i}$$, $$\sigma^2$$ is the variance property and $$\lambda_i$$ is the $$i$$ th element of the lengthscales property.
Parameters
• gpy_matern (Union[Matern32, Prod, None]) – A Matern32 product kernel from GPy. For $$d=1$$ this is equivalent to a Matern32 kernel. For $$d>1$$, this is not a $$d$$-dimensional Matern32 kernel but a product of $$d$$ 1-dimensional Matern32 kernels with differing active dimensions constructed as k1 * k2 * … . Make sure to unlink all variances except the variance of the first kernel k1 in the product as the variance of k1 will be used to represent $$\sigma^2$$. If you are unsure what to do, use the lengthscales and variance parameter instead. If gpy_matern is not given, the lengthscales argument is used.
• lengthscales (Optional[ndarray]) – If gpy_matern is not given, a product Matern32 kernel will be constructed with the given lengthscales. The number of elements need to be equal to the dimensionality $$d$$. If gpy_matern is given, this input is disregarded.
• variance (Optional[float]) – The variance of the product kernel. Only used if gpy_matern is not given. Defaults to 1.
property lengthscales: ndarray
The lengthscales $$\lambda$$ of the kernel.
Return type
ndarray
property variance: float
The scale $$\sigma^2$$ of the kernel.
Return type
float
K(x1, x2)
The kernel k(x1, x2) evaluated at x1 and x2.
Parameters
• x1 (ndarray) – First argument of the kernel, shape (n_points N, input_dim)
• x2 (ndarray) – Second argument of the kernel, shape (n_points M, input_dim)
Return type
ndarray
Returns
Kernel evaluated at x1, x2, shape (N, M).
dK_dx1(x1, x2)
Gradient of the kernel wrt x1 evaluated at pair x1, x2.
Parameters
• x1 (ndarray) – First argument of the kernel, shape (n_points N, input_dim)
• x2 (ndarray) – Second argument of the kernel, shape (n_points M, input_dim)
Return type
ndarray
Returns
The gradient of the kernel wrt x1 evaluated at (x1, x2), shape (input_dim, N, M)
Wrapper of the GPy Matern52 product kernel as required for some EmuKit quadrature methods.
The product kernel is of the form $$k(x, x') = \sigma^2 \prod_{i=1}^d k_i(x, x')$$ where
$k_i(x, x') = (1 + \sqrt{5} r_i + \frac{5}{3} r_i^2) \exp(- \sqrt{5} r_i).$
$$d$$ is the input dimensionality, $$r_i:=\frac{|x_i - x'_i|}{\lambda_i}$$, $$\sigma^2$$ is the variance property and $$\lambda_i$$ is the $$i$$ th element of the lengthscales property.
Parameters
• gpy_matern (Union[Matern52, Prod, None]) – A Matern52 product kernel from GPy. For $$d=1$$ this is equivalent to a Matern52 kernel. For $$d>1$$, this is not a $$d$$-dimensional Matern52 kernel but a product of $$d$$ 1-dimensional Matern52 kernels with differing active dimensions constructed as k1 * k2 * … . Make sure to unlink all variances except the variance of the first kernel k1 in the product as the variance of k1 will be used to represent $$\sigma^2$$. If you are unsure what to do, use the lengthscales and variance parameter instead. If gpy_matern is not given, the lengthscales argument is used.
• lengthscales (Optional[ndarray]) – If gpy_matern is not given, a product Matern52 kernel will be constructed with the given lengthscales. The number of elements need to be equal to the dimensionality $$d$$. If gpy_matern is given, this input is disregarded.
• variance (Optional[float]) – The variance of the product kernel. Only used if gpy_matern is not given. Defaults to 1.
property lengthscales: ndarray
The lengthscales $$\lambda$$ of the kernel.
Return type
ndarray
property variance: float
The scale $$\sigma^2$$ of the kernel.
Return type
float
K(x1, x2)
The kernel k(x1, x2) evaluated at x1 and x2.
Parameters
• x1 (ndarray) – First argument of the kernel, shape (n_points N, input_dim)
• x2 (ndarray) – Second argument of the kernel, shape (n_points M, input_dim)
Return type
ndarray
Returns
Kernel evaluated at x1, x2, shape (N, M).
dK_dx1(x1, x2)
Gradient of the kernel wrt x1 evaluated at pair x1, x2.
Parameters
• x1 (ndarray) – First argument of the kernel, shape (n_points N, input_dim)
• x2 (ndarray) – Second argument of the kernel, shape (n_points M, input_dim)
Return type
ndarray
Returns
The gradient of the kernel wrt x1 evaluated at (x1, x2), shape (input_dim, N, M)
Bases: IBrownian
Wrapper of the GPy Brownian motion kernel as required for some EmuKit quadrature methods.
$k(x, x') = \sigma^2 \operatorname{min}(x, x')\quad\text{with}\quad x, x' \geq 0,$
where $$\sigma^2$$ is the variance property.
Parameters
gpy_brownian (Brownian) – A Brownian motion kernel from GPy.
property variance: float
The scale $$\sigma^2$$ of the kernel.
Return type
float
K(x1, x2)
The kernel k(x1, x2) evaluated at x1 and x2.
Parameters
• x1 (ndarray) – First argument of the kernel, shape (n_points N, input_dim)
• x2 (ndarray) – Second argument of the kernel, shape (n_points M, input_dim)
Return type
ndarray
Returns
Kernel evaluated at x1, x2, shape (N, M).
Wrapper of the GPy Brownian product kernel as required for some EmuKit quadrature methods.
The product kernel is of the form $$k(x, x') = \sigma^2 \prod_{i=1}^d k_i(x, x')$$ where
$k_i(x, x') = \operatorname{min}(x_i-c, x_i'-c)\quad\text{with}\quad x_i, x_i' \geq c,$
$$d$$ is the input dimensionality, $$\sigma^2$$ is the variance property and $$c$$ is the offset property.
Parameters
• gpy_brownian (Union[Brownian, Prod, None]) – A Brownian product kernel from GPy. For $$d=1$$ this is equivalent to a Brownian kernel. For $$d>1$$, this is a product of $$d$$ 1-dimensional Brownian kernels with differing active dimensions constructed as k1 * k2 * … . Make sure to unlink all variances except the variance of the first kernel k1 in the product as the variance of k1 will be used to represent $$\sigma^2$$. If you are unsure what to do, use the input_dim and variance parameter instead. If gpy_brownian is not given, the variance and input_dim argument is used.
• offset (float) – The offset $$c$$ of the kernel. Defaults to 0.
• variance (Optional[float]) – The variance of the product kernel. Only used if gpy_brownian is not given. Defaults to 1.
• input_dim (Optional[int]) – The input dimension. Only used if gpy_brownian is not given.
property variance: float
The scale $$\sigma^2$$ of the kernel.
Return type
float
property offset: float
The offset $$c$$ of the kernel.
Return type
float
K(x1, x2)
The kernel k(x1, x2) evaluated at x1 and x2.
Parameters
• x1 (ndarray) – First argument of the kernel, shape (n_points N, input_dim)
• x2 (ndarray) – Second argument of the kernel, shape (n_points M, input_dim)
Return type
ndarray
Returns
Kernel evaluated at x1, x2, shape (N, M).
dK_dx1(x1, x2)
Gradient of the kernel wrt x1 evaluated at pair x1, x2.
Parameters
• x1 (ndarray) – First argument of the kernel, shape (n_points N, input_dim)
• x2 (ndarray) – Second argument of the kernel, shape (n_points M, input_dim)
Return type
ndarray
Returns
The gradient of the kernel wrt x1 evaluated at (x1, x2), shape (input_dim, N, M)
dKdiag_dx(x)
The gradient of the diagonal of the kernel (the variance) v(x):=k(x, x) evaluated at x.
Parameters
x (ndarray) – The locations where the gradient is evaluated, shape (n_points, input_dim).
Return type
ndarray
Returns
The gradient of the diagonal of the kernel evaluated at x, shape (input_dim, n_points).
Wraps a GPy model and returns an EmuKit quadrature model.
Parameters
• gpy_model (GPRegression) – A GPy Gaussian process regression model GPy.models.GPRegression.
• integral_bounds (Optional[List[Tuple[float, float]]]) – List of d tuples, where d is the dimensionality of the integral and the tuples contain the lower and upper bounds of the integral i.e., [(lb_1, ub_1), (lb_2, ub_2), …, (lb_d, ub_d)]. Only used if measure is not given in which case the unnormalized Lebesgue measure is used.
• measure (Optional[IntegrationMeasure]) – An integration measure. Either measure or integral_bounds must be given. If both integral_bounds and measure are given, integral_bounds is disregarded.
• integral_name (str) – The (variable) name(s) of the integral.
Return type
BaseGaussianProcessGPy
Returns
An EmuKit GP model for quadrature with GPy backend.
class emukit.model_wrappers.simple_gp_model.SimpleGaussianProcessModel(x, y)
Bases: IModel
This model is a Gaussian process with an RBF kernel, with no ARD. It is used to demonstrate uses of emukit, it does not aim to be flexible, robust or fast.
optimize()
Optimize the three hyperparameters of the model, namely the kernel variance, kernel lengthscale and likelihood variance
Return type
None
predict(x_new)
Predict from model
Parameters
x_new (ndarray) – (n_points, n_dims) array containing points at which the predictive distributions will be computed
Return type
Tuple[ndarray, ndarray]
Returns
Tuple containing two (n_points, 1) arrays representing the mean and variance of the predictive distribution at the specified input locations
set_data(X, Y)
Set training data to new values
Parameters
• X (ndarray) – (n_points, n_dims) array containing training features
• Y (ndarray) – (n_points, 1) array containing training targets
Return type
None
property X: ndarray
Return type
ndarray
property Y: ndarray
Return type
ndarray | {"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": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5434653162956238, "perplexity": 8593.089379873687}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710808.72/warc/CC-MAIN-20221201085558-20221201115558-00353.warc.gz"} |
http://sro.sussex.ac.uk/id/eprint/66965/ | # A theoretical investigation of orientation relationships and transformation strains in steels
Koumatos, K and Muehlemann, A (2017) A theoretical investigation of orientation relationships and transformation strains in steels. Acta Crystallographica Section A: Foundations and Advances, A73. pp. 115-123. ISSN 2053-2733 | {"extraction_info": {"found_math": false, "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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9562792778015137, "perplexity": 15134.931815377227}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496671260.30/warc/CC-MAIN-20191122115908-20191122143908-00385.warc.gz"} |
http://www.maa.org/publications/maa-reviews/totally-nonnegative-matrices?device=mobile | # Totally Nonnegative Matrices
Publisher:
Princeton University Press
Number of Pages:
264
Price:
45.00
ISBN:
9780691121574
Thursday, April 28, 2011
Reviewable:
Include In BLL Rating:
Shaun M. Fallat and Charles R. Johnson
Publication Date:
2011
Format:
Hardcover
Category:
Monograph
List of Figures xi
Preface xiii
Chapter 0. Introduction 1
0.0 Definitions and Notation 1
0.1 Jacobi Matrices and Other Examples of TN matrices 3
0.2 Applications and Motivation 15
0.3 Organization and Particularities 24
Chapter 1. Preliminary Results and Discussion 27
1.0 Introduction 27
1.1 The Cauchy-Binet Determinantal Formula 27
1.2 Other Important Determinantal Identities 28
1.3 Some Basic Facts 33
1.4 TN and TP Preserving Linear Transformations 34
1.5 Schur Complements 35
1.6 Zero-Nonzero Patterns of TN Matrices 37
Chapter 2. Bidiagonal Factorization 43
2.0 Introduction 43
2.1 Notation and Terms 45
2.2 Standard Elementary Bidiagonal Factorization: Invertible Case 47
2.3 Standard Elementary Bidiagonal Factorization: General Case 53
2.4 LU Factorization: A consequence 59
2.5 Applications 62
2.6 Planar Diagrams and EB factorization 64
Chapter 3. Recognition 73
3.0 Introduction 73
3.1 Sets of Positive Minors Sufficient for Total Positivity 74
3.2 Application: TP Intervals 80
3.3 Efficient Algorithm for testing for TN 82
Chapter 4. Sign Variation of Vectors and TN Linear Transformations 87
4.0 Introduction 87
4.1 Notation and Terms 87
4.2 Variation Diminution Results and EB Factorization 88
4.3 Strong Variation Diminution for TP Matrices 91
4.4 Converses to Variation Diminution 94
Chapter 5. The Spectral Structure of TN Matrices 97
5.0 Introduction 97
5.1 Notation and Terms 98
5.2 The Spectra of IITN Matrices 99
5.3 Eigenvector Properties 100
5.4 The Irreducible Case 106
5.5 Other Spectral Results 118
Chapter 6. Determinantal Inequalities for TN Matrices 129
6.0 Introduction 129
6.1 Definitions and Notation 131
6.2 Sylvester Implies Koteljanski?I 132
6.3 Multiplicative Principal Minor Inequalities 134
6.4 Some Non-principal Minor Inequalities 146
Chapter 7. Row and Column Inclusion and the Distribution of Rank 153
7.0 Introduction 153
7.1 Row and Column Inclusion Results for TN Matrices 153
7.2 Shadows and the Extension of Rank Deficiency in Submatrices of
TN Matrices 159
7.3 The Contiguous Rank Property 165
Chapter 8. Hadamard Products and Powers of TN Matrices 167
8.0 Definitions 167
8.1 Conditions under which the Hadamard
Product is TP/TN 168
8.3 Oppenheim's Inequality 177
8.4 Hadamard Powers of TP2 179
Chapter 9. Extensions and Completions 185
9.0 Line Insertion 185
9.1 Completions and Partial TN Matrices 186
9.2 Chordal Case--MLBC Graphs 189
9.3 TN Completions: Adjacent Edge Conditions 191
9.4 TN Completions: Single Entry Case 195
9.5 TN Perturbations: The Case of Retractions 198
Chapter 10. Other Related Topics on TN Matrices 205
10.0 Introduction and Topics 205
10.1 Powers and Roots of TP/TN Matrices 205
10.2 Subdirect Sums of TN Matrices 207
10.3 TP/TN Polynomial Matrices 212
10.4 Perron Complements of TN Matrices 213
Bibliography 219
List of Symbols 239
Index 245
Publish Book:
Modify Date:
Thursday, April 28, 2011 | {"extraction_info": {"found_math": false, "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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8952959179878235, "perplexity": 19444.120833882913}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1386163042403/warc/CC-MAIN-20131204131722-00099-ip-10-33-133-15.ec2.internal.warc.gz"} |
https://www.fachschaft.informatik.tu-darmstadt.de/forum/viewtopic.php?p=152609 | ## Assignment 3 B.2 'Compute and compare'
Moderator: Computer Vision
ampelmann
Erstie
Beiträge: 19
Registriert: 15. Mai 2013 18:20
### Assignment 3 B.2 'Compute and compare'
Hi, short question: What exactly shall we reformulate, compute and compare? What I understood so far is: reformulated (1), computed posteriors with the most right part of the formula (the one with the products of (µ/Cj)^n(i,m)) and compared the posteriors of class 0 and class 1, adding a short comment in the code. Or should we rather compute them with i.e. the 2nd formula p(C_j) * p(I_m|C_j) as well and compare those? In this case, the probability for p(I_m|C_j) is either 1 or 0(isn't it? ), so I basically get vectors of p(C_j) and 0s. Some clarification would be great
Thanks in advance and a nice weekend.
Budderick
Neuling
Beiträge: 4
Registriert: 11. Okt 2010 17:32
### Re: Assignment 3 B.2 'Compute and compare'
Hey,
i would say your first proposal is right. Reformulate the right part, compute the posteriors and then you can make a prediction.
Everything else would be nonsense ...
Greetings
ampelmann
Erstie
Beiträge: 19
Registriert: 15. Mai 2013 18:20
### Re: Assignment 3 B.2 'Compute and compare'
Thanks, sounds good | {"extraction_info": {"found_math": false, "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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8089165687561035, "perplexity": 5375.541236779795}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370493121.36/warc/CC-MAIN-20200328225036-20200329015036-00291.warc.gz"} |
https://hal.inria.fr/hal-01657018 | # One-Time Nondeterministic Computations
Abstract : We introduce the concept of one-time nondeterminism as a new kind of limited nondeterminism for finite state machines and pushdown automata. Roughly speaking, one-time nondeterminism means that at the outset the automaton is nondeterministic, but whenever it performs a guess, this guess is fixed for the rest of the computation. We characterize the computational power of one-time nondeterministic finite automata (OTNFAs) and one-time nondeterministic pushdown devices. Moreover, we study the descriptional complexity of these machines. For instance, we show that for an n-state OTNFA with a sole nondeterministic state, that is nondeterministic for only one input symbol, $(n+1)^n$ states are sufficient and necessary in the worst case for an equivalent deterministic finite automaton. In case of pushdown automata, the conversion of a nondeterministic to a one-time nondeterministic as well as the conversion of a one-time nondeterministic to a deterministic one turn out to be non-recursive, that is, the trade-offs in size cannot be bounded by any recursive function.
Type de document :
Communication dans un congrès
Giovanni Pighizzini; Cezar Câmpeanu. 19th International Conference on Descriptional Complexity of Formal Systems (DCFS), Jul 2017, Milano, Italy. Springer International Publishing, Lecture Notes in Computer Science, LNCS-10316, pp.177-188, 2017, Descriptional Complexity of Formal Systems. 〈10.1007/978-3-319-60252-3_14〉
Domaine :
https://hal.inria.fr/hal-01657018
Contributeur : Hal Ifip <>
Soumis le : mercredi 6 décembre 2017 - 11:44:52
Dernière modification le : lundi 26 février 2018 - 13:40:02
### Fichier
##### Accès restreint
Fichier visible le : 2020-01-01
Connectez-vous pour demander l'accès au fichier
### Citation
Markus Holzer, Martin Kutrib. One-Time Nondeterministic Computations. Giovanni Pighizzini; Cezar Câmpeanu. 19th International Conference on Descriptional Complexity of Formal Systems (DCFS), Jul 2017, Milano, Italy. Springer International Publishing, Lecture Notes in Computer Science, LNCS-10316, pp.177-188, 2017, Descriptional Complexity of Formal Systems. 〈10.1007/978-3-319-60252-3_14〉. 〈hal-01657018〉
### Métriques
Consultations de la notice | {"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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8883257508277893, "perplexity": 3853.8722233915755}, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125948617.86/warc/CC-MAIN-20180426222608-20180427002608-00137.warc.gz"} |
https://www.physicsforums.com/threads/calculate-rpm-given-the-force-of-a-torsion-spring.957066/ | # Calculate RPM given the force of a torsion spring
• Start date
• Tags
• #1
2
0
## Homework Statement
I've got a flywheel of Inertia = 0.0019302kg/m^2 (found via solidworks), when a torsion spring is released, a force of 10N acts on the wheel via astring attached 0.065m above and 0.0325m to the right of the wheel's axis at an angle of 40 degrees.
## Homework Equations
What is the flywheel's RPM (or rad/s or Hz)?
## The Attempt at a Solution
Torque produced = FhDv + FvDh = (10*sin(40))0.0325 + (10*cos(40))0.065
= 0.667Nm
Torque = Ia
a = Torque/I = 0.667/0.0019302
Am i lacking a length of time this torque is applied for? I struggle to se where to go from here[/B]
Related Introductory Physics Homework Help News on Phys.org
• #2
nrqed
Homework Helper
Gold Member
3,603
204
## Homework Statement
I've got a flywheel of Inertia = 0.0019302kg/m^2 (found via solidworks), when a torsion spring is released, a force of 10N acts on the wheel via astring attached 0.065m above and 0.0325m to the right of the wheel's axis at an angle of 40 degrees.
## Homework Equations
What is the flywheel's RPM (or rad/s or Hz)?
## The Attempt at a Solution
Torque produced = FhDv + FvDh = (10*sin(40))0.0325 + (10*cos(40))0.065
= 0.667Nm
Torque = Ia
a = Torque/I = 0.667/0.0019302
Am i lacking a length of time this torque is applied for? I struggle to se where to go from here[/B]
I di dont check your numbers but the approach looks good . Note that $\alpha$ is in rad/s^2. To get to your question, I am puzzled too. It is not possible to calculate an RPM without more information. For example the amount of time it was applied, as you said (and then we would need to know if the direction of the force changes as the wheel rotates, and if so how it does). There is really no other information provided?
• #3
2
0
This is actualy the analysis of a project i've made, so yes there is no other information i can add to it. I'm thinking i'll time the flywheels motion after activation and just count the revolutions... i was just hoping there was a theory based way to calculate it! thanks for the reply
• #4
haruspex
Homework Helper
Gold Member
32,724
5,030
This is actualy the analysis of a project i've made, so yes there is no other information i can add to it. I'm thinking i'll time the flywheels motion after activation and just count the revolutions... i was just hoping there was a theory based way to calculate it! thanks for the reply
Since it is a torsion spring the torque will not be constant. In principle you have an SHM oscillator. The flywheel speed will be maximised each time the torsion spring is at its relaxed position. So what you need to know is the energy initially stored in the spring.
But that is ignoring practical considerations of friction and drag.
• Last Post
Replies
1
Views
2K
• Last Post
Replies
11
Views
5K
• Last Post
Replies
1
Views
3K
• Last Post
Replies
6
Views
1K
• Last Post
Replies
3
Views
5K
• Last Post
Replies
0
Views
3K
• Last Post
Replies
4
Views
2K
• Last Post
Replies
10
Views
25K
• Last Post
Replies
9
Views
13K
• Last Post
Replies
2
Views
2K | {"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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8089986443519592, "perplexity": 1909.213391760942}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370506580.20/warc/CC-MAIN-20200402014600-20200402044600-00377.warc.gz"} |
https://pypi.org/project/timeseriesx/ | Manage time series data with explicit frequency and unit.
TimeSeriesX
The eXtended time series library.
Manage time series data with explicit time zone, frequency and unit.
TimeSeriesX is motivated by handling time series data in a convenient way. Almost all the features are actually already provided by pandas. TimeSeriesX extends the pandas time series functionality by the unit functionalities of pint and pint-pandas. Further, TimeSeriesX offers an easy and convenient interface to work with time series without the need to dig deep into these libraries, which nevertheless is still recommended, since they go way beyond time series data.
The main challenges that arise when handling time series data are time zones and frequencies. Since time series data is often obtained by measurements, the values are associated with units. Then these units can be confused easily, since the units are often not modeled in code.
TimeSeriesX forces the user to handle time zones, frequencies and units explicitly, while taking care of validation and convenient formats. It also supports deriving these attributes from raw time series data. It offers a limited set of actions on time series that are translated to pandas or pint functionality under the hood. It was designed to guarantee that every transformation of time series data results in a new valid time series, which would require quite some pandas code if done “manually”.
Features
• model time series data with explicit frequency, time zone and unit
• convert time zone or unit
• resample data to new frequency
• fill and get gaps
• join time series
• perform calculations on time series with python standard operators
History
0.1.13 (2022-07-19)
• fix a few bugs by avoiding is_compatible_with in convert_unit
• raise only ValueError instead of DimensionalityError on unit dimensionality mismatch
• remove pandas installation dependency because of transitive dependency via pint-pandas
• loosen requirements for pytz and dateutil, no special version requirements known
• extend __getitem__ functionality by supporting iterables of timestamps or positional indices
• explicitly support indexing by time zone naive timestamps, which is deprecated by pandas
• make coerce_unit behave like coerce_freq and coerce_time_zone by passing through None
0.1.12 (2022-03-16)
• fix equals method
• update documentation
0.1.11 (2022-03-07)
• fix resampling method to support nan-values
• update dependencies
0.1.10 (2022-01-21)
• fix equals method
• update dependencies
0.1.9 (2021-11-19)
• allow aggregation functions to return magnitudes or quantities
• update dependencies
0.1.8 (2021-09-28)
• fix time zone bug in gap handling
• update dependencies
0.1.7 (2021-09-28)
• improve gap handling
• update dependencies
• improve documentation
• fix calculations with quantity scalar
0.1.6 (2021-09-13)
• fix time zone issue with UTC in basic calculations for TimestampSeries as 2nd operand
• update pint-pandas version dependency
• use pint’s default unit registry
• add support of callables as arguments for frequency resampling
0.1.5 (2021-09-10)
• fix time zone issue with UTC in basic calculations
• fix map-function for series with unit
0.1.4 (2021-09-09)
• improve test coverage
• improve TimeSeries equality check
• support NaN-removal in as_pd_series-method
0.1.3 (2021-09-08)
• remove manual timezone checks because it is handled by pandas
• fix skipped tests
• fix repr() method of TimestampSeries
• fix basic calculation with units involved
0.1.2 (2021-09-07)
• fix timezone handling
• First release on PyPI Index.
0.1.1 (2021-02-16)
• First release on PyPI Test Index.
Project details
Uploaded source
Uploaded py2 py3 | {"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": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.16704656183719635, "perplexity": 11067.374424049369}, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296945218.30/warc/CC-MAIN-20230323225049-20230324015049-00152.warc.gz"} |
http://aas.org/archives/BAAS/v31n2/head99/69.htm | HEAD Division Meeting 1999, April 1999
Session 2. AGN
Poster, Monday, April 12, 1999, 8:30am-6:10pm, Gold Room
## [2.02] A BeppoSAX observation of the Compton Thick Seyfert NGC 6240
P. Vignati, S. Molendi (IFC/CNR), G Matt (Univ.Roma III), Compton Thick Team
BeppoSAX observed the Seyfert 2 NGC6240 in August 1998. We present the results of the first detection of this galaxy above 10 keV. We found an excess emission above the extrapolation of the Compton reflection component at these energies. We argue that we are observing the nucleus through absorbing matter of column density NH ~q 2 \times 10 24 cm -2, which is Compton-thick below 10 keV, but permits partial trasmission above this energy. By measuring the intrinsic X-ray luminosity we can, for the first time, compute the fraction of reprocessed emission.
If the author provided an email address or URL for general inquiries, it is as follows:
[Previous] | [Session 2] | [Next] | {"extraction_info": {"found_math": false, "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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9499375820159912, "perplexity": 8261.823345403305}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-06/segments/1422115869320.77/warc/CC-MAIN-20150124161109-00197-ip-10-180-212-252.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/electromagnetic-waves.531529/ | # Electromagnetic Waves
• Start date
• #1
2
0
Hello,
This is not homework, as I'm not in school :)
I was wondering if there are any standard equations for electromagnetic waves interfering with each other. For instance, can I change the frequency of a wave by hitting it with another wave of a different frequency? I can find info on sound waves etc.. but I was wondering if this was any different.
The question I want to answer for myself is: "Can I get one lower frequency electromagnetic wave, by combining two or more higher frequency electromagnetic waves?"
Thank you :)
• Last Post
Replies
1
Views
2K
• Last Post
Replies
2
Views
2K
• Last Post
Replies
1
Views
1K
• Last Post
Replies
11
Views
1K
• Last Post
Replies
1
Views
2K
• Last Post
Replies
3
Views
2K
• Last Post
Replies
1
Views
1K
• Last Post
Replies
21
Views
1K
• Last Post
Replies
37
Views
6K
• Last Post
Replies
9
Views
2K | {"extraction_info": {"found_math": false, "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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9155046939849854, "perplexity": 1506.9337553148937}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487643703.56/warc/CC-MAIN-20210619051239-20210619081239-00165.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/50099-isomorphic-groups.html | 1. ## Isomorphic groups
Let f: G --> H be an isomorphism, G and H are groups.
I've already showed that for every x in G, |f(x)| = |x|. (|x| = order of x, the smallest positive integer such that $x^n$ is the identity element)
How to show that any two isomorphic groups have the same number of elements of n, where n is any positive integer? | {"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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9604702591896057, "perplexity": 304.2386451289399}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-40/segments/1474738663247.66/warc/CC-MAIN-20160924173743-00202-ip-10-143-35-109.ec2.internal.warc.gz"} |
https://www.britannica.com/science/celestial-mechanics-physics/The-three-body-problem | ## The three-body problem
The inclusion of solar perturbations of the motion of the Moon results in a “three-body problem” (Earth-Moon-Sun), which is the simplest complication of the completely solvable two-body problem discussed above. When Earth, the Moon, and the Sun are considered to be point masses, this particular three-body problem is called “the main problem of the lunar theory,” which has been studied extensively with a variety of methods beginning with Newton. Although the three-body problem has no complete analytic solution in closed form, various series solutions by successive approximations achieve such accuracy that complete theories of the lunar motion must include the effects of the nonspherical mass distributions of both Earth and the Moon as well as the effects of the planets if the precision of the predicted positions is to approach that of the observations. Most of the schemes for the main problem are partially numerical and therefore apply only to the lunar motion. An exception is the completely analytic work of the French astronomer Charles-Eugène Delaunay (1816–72), who exploited and developed the most elegant techniques of classical mechanics pioneered by his contemporary, the Irish astronomer and mathematician William R. Hamilton (1805–65). Delaunay could predict the position of the Moon to within its own diameter over a 20-year time span. Since his development was entirely analytic, the work was applicable to the motions of satellites about other planets where the series expansions converged much more rapidly than they did for the application to the lunar motion.
Delaunay’s work on the lunar theory demonstrates some of the influence that celestial mechanics has had on the development of the techniques of classical mechanics. This close link between the development of classical mechanics and its application to celestial mechanics was probably no better demonstrated than in the work of the French mathematician Henri Poincaré (1854–1912). Poincaré, along with other great mathematicians such as George D. Birkhoff (1884–1944), Aurel Wintner (1903–58), and Andrey N. Kolmogorov (1903–87), placed celestial mechanics on a more sound mathematical basis and was less concerned with quantitatively accurate prediction of celestial body motion. Poincaré demonstrated that the series solutions in use in celestial mechanics for so long generally did not converge but that they could give accurate descriptions of the motion for significant periods of time in truncated form. The elaborate theoretical developments in celestial and classical mechanics have received more attention recently with the realization that a large class of motions are of an irregular or chaotic nature and require fundamentally different approaches for their description.
## The restricted three-body problem
The simplest form of the three-body problem is called the restricted three-body problem, in which a particle of infinitesimal mass moves in the gravitational field of two massive bodies orbiting according to the exact solution of the two-body problem. (The particle with infinitesimal mass, sometimes called a massless particle, does not perturb the motions of the two massive bodies.) There is an enormous literature devoted to this problem, including both analytic and numerical developments. The analytic work was devoted mostly to the circular, planar restricted three-body problem, where all particles are confined to a plane and the two finite masses are in circular orbits around their centre of mass (a point on the line between the two masses that is closer to the more massive). Numerical developments allowed consideration of the more general problem.
In the circular problem, the two finite masses are fixed in a coordinate system rotating at the orbital angular velocity, with the origin (axis of rotation) at the centre of mass of the two bodies. Lagrange showed that in this rotating frame there were five stationary points at which the massless particle would remain fixed if placed there. There are three such points lying on the line connecting the two finite masses: one between the masses and one outside each of the masses. The other two stationary points, called the triangular points, are located equidistant from the two finite masses at a distance equal to the finite mass separation. The two masses and the triangular stationary points are thus located at the vertices of equilateral triangles in the plane of the circular orbit.
There is a constant of the motion in the rotating frame that leads to an equation relating the velocity of the massless particle in this frame to its position. For given values of this constant it is possible to construct curves in the plane on which the velocity vanishes. If such a zero-velocity curve is closed, the particle cannot escape from the interior of the closed zero-velocity curve if placed there with the constant of the motion equal to the value used to construct the curve. These zero-velocity curves can be used to show that the three collinear stationary points are all unstable in the sense that, if the particle is placed at one of these points, the slightest perturbation will cause it to move far away. The triangular points are stable if the ratio of the finite masses is less than 0.04, and the particle would execute small oscillations around one of the triangular points if it were pushed slightly away. Since the mass ratio of Jupiter to the Sun is about 0.001, the stability criterion is satisfied, and Lagrange predicted the presence of the Trojan asteroids at the triangular points of the Sun-Jupiter system 134 years before they were observed. Of course, the stability of the triangular points must also depend on the perturbations by any other bodies. Such perturbations are sufficiently small not to destabilize the Trojan asteroids. Single Trojan-like bodies have also been found orbiting at leading and trailing triangular points in the orbits of Neptune and of Saturn’s satellite Tethys, at the leading triangular point in the orbit of another Saturnian satellite, Dione, and at the trailing point in the orbit of Mars.
## Orbital resonances
Astronomy and Space Quiz
There are stable configurations in the restricted three-body problem that are not stationary in the rotating frame. If, for example, Jupiter and the Sun are the two massive bodies, these stable configurations occur when the mean motions of Jupiter and the small particle—here an asteroid—are near a ratio of small integers. The orbital mean motions are then said to be nearly commensurate, and an asteroid that is trapped near such a mean motion commensurability is said to be in an orbital resonance with Jupiter. For example, the Trojan asteroids librate (oscillate) around the 1:1 orbital resonance (i.e., the orbital period of Jupiter is in a 1:1 ratio with the orbital period of the Trojan asteroids); the asteroid Thule librates around the 4:3 orbital resonance; and several asteroids in the Hilda group librate around the 3:2 orbital resonance. There are several such stable orbital resonances among the satellites of the major planets and one involving Pluto and the planet Neptune. The analysis based on the restricted three-body problem cannot be used for the satellite resonances, however, except for the 4:3 resonance between Saturn’s satellites Titan and Hyperion, since the participants in the satellite resonances usually have comparable masses.
Although the asteroid Griqua librates around the 2:1 resonance with Jupiter, and Alinda librates around the 3:1 resonance, the orbital commensurabilities 2:1, 7:3, 5:2, and 3:1 are characterized by an absence of asteroids in an otherwise rather highly populated, uniform distribution spanning all of the commensurabilities. These are the Kirkwood gaps in the distribution of asteroids, and the recent understanding of their creation and maintenance has introduced into celestial mechanics an entirely new concept of irregular, or chaotic, orbits in a system whose equations of motion are entirely deterministic.
## Chaotic orbits
The French astronomer Michel Hénon and the American astronomer Carl Heiles discovered that when a system exhibiting periodic motion, such as a pendulum, is perturbed by an external force that is also periodic, some initial conditions lead to motions where the state of the system becomes essentially unpredictable (within some range of system states) at some time in the future, whereas initial conditions within some other set produce quasiperiodic or predictable behaviour. The unpredictable behaviour is called chaotic, and initial conditions that produce it are said to lie in a chaotic zone. If the chaotic zone is bounded, in the sense that only limited ranges of initial values of the variables describing the motion lead to chaotic behaviour, the uncertainty in the state of the system in the future is limited by the extent of the chaotic zone; that is, values of the variables in the distant future are completely uncertain only within those ranges of values within the chaotic zone. This complete uncertainty within the zone means the system will eventually come arbitrarily close to any set of values of the variables within the zone if given sufficient time. Chaotic orbits were first realized in the asteroid belt.
A periodic term in the expansion of the disturbing function for a typical asteroid orbit becomes more important in influencing the motion of the asteroid if the frequency with which it changes sign is very small and its coefficient is relatively large. For asteroids orbiting near a mean motion commensurability with Jupiter, there are generally several terms in the disturbing function with large coefficients and small frequencies that are close but not identical. These “resonant” terms often dominate the perturbations of the asteroid motion so much that all the higher-frequency terms can be neglected in determining a first approximation to the perturbed motion. This neglect is equivalent to averaging the higher-frequency terms to zero; the low-frequency terms change only slightly during the averaging. If one of the frequencies vanishes on the average, the periodic term becomes nearly constant, or secular, and the asteroid is locked into an exact orbital resonance near the particular mean motion commensurability. The mean motions are not exactly commensurate in such a resonance, however, since the motion of the asteroid orbital node or perihelion is always involved (except for the 1:1 Trojan resonances).
For example, for the 3:1 commensurability, the angle θ = λA - 3λJ + ϖA is the argument of one of the important periodic terms whose variation can vanish (zero frequency). Here λ = Ω + ω + l is the mean longitude, the subscripts A and J refer to the asteroid and Jupiter, respectively, and ϖ = Ω + ω is the longitude of perihelion (see Figure 2). Within resonance, the angle θ librates, or oscillates, around a constant value as would a pendulum around its equilibrium position at the bottom of its swing. The larger the amplitude of the equivalent pendulum, the larger its velocity at the bottom of its swing. If the velocity of the pendulum at the bottom of its swing, or, equivalently, the maximum rate of change of the angle θ, is sufficiently high, the pendulum will swing over the top of its support and be in a state of rotation instead of libration. The maximum value of the rate of change of θ for which θ remains an angle of libration (periodically reversing its variation) instead of one of rotation (increasing or decreasing monotonically) is defined as the half-width of the resonance.
Another term with nearly zero frequency when the asteroid is near the 3:1 commensurability has the argument θ′ = λA - λJ + 2ϖJ. The substitution of the longitude of Jupiter’s perihelion for that of the asteroid means that the rates of change of θ and θ′ will be slightly different. As the resonances are not separated much in frequency, there may exist values of the mean motion of the asteroid where both θ and θ′ would be angles of libration if either resonance existed in the absence of the other. The resonances are said to overlap in this case, and the attempt by the system to librate simultaneously about both resonances for some initial conditions leads to chaotic orbital behaviour. The important characteristic of the chaotic zone for asteroid motion near a mean motion commensurability with Jupiter is that it includes a region where the asteroid’s orbital eccentricity is large. During the variation of the elements over the entire chaotic zone as time increases, large eccentricities must occasionally be reached. For asteroids near the 3:1 commensurability with Jupiter, the orbit then crosses that of Mars, whose gravitational interaction in a close encounter can remove the asteroid from the 3:1 zone.
By numerically integrating many orbits whose initial conditions spanned the 3:1 Kirkwood gap region in the asteroid belt, Jack Wisdom, an American dynamicist who developed a powerful means of analyzing chaotic motions, found that the chaotic zone around this gap precisely matched the physical extent of the gap. There are no observable asteroids with orbits within the chaotic zone, but there are many just outside extremes of the zone. Other Kirkwood gaps can be similarly accounted for. The realization that orbits governed by Newton’s laws of motion and gravitation could have chaotic properties and that such properties could solve a long-standing problem in the celestial mechanics of the solar system is a major breakthrough in the subject.
## The n-body problem
The general problem of n bodies, where n is greater than three, has been attacked vigorously with numerical techniques on powerful computers. Celestial mechanics in the solar system is ultimately an n-body problem, but the special configurations and relative smallness of the perturbations have allowed quite accurate descriptions of motions (valid for limited time periods) with various approximations and procedures without any attempt to solve the complete problem of n bodies. Examples are the restricted three-body problem to determine the effect of Jupiter’s perturbations of the asteroids and the use of successive approximations of series solutions to sequentially add the effects of smaller and smaller perturbations for the motion of the Moon. In the general n-body problem, all bodies have arbitrary masses, initial velocities, and positions; the bodies interact through Newton’s law of gravitation, and one attempts to determine the subsequent motion of all the bodies. Many numerical solutions for the motion of quite large numbers of gravitating particles have been successfully completed where the precise motion of individual particles is usually less important than the statistical behaviour of the group.
## Numerical solutions
Numerical solutions of the exact equations of motion for n bodies can be formulated. Each body is subject to the gravitational attraction of all the others, and it may be subject to other forces as well. It is relatively easy to write the expression for the instantaneous acceleration (equation of motion) of each body if the position of all the other bodies is known, and expressions for all the other forces can be written (as they can for gravitational forces) in terms of the relative positions of the particles and other defining characteristics of the particle and its environment. Each particle is allowed to move under its instantaneous acceleration for a short time step. Its velocity and position are thereby changed, and the new values of the variables are used to calculate the acceleration for the next time step, and so forth. Of course, the real position and velocity of the particle after each time step will differ from the calculated values by errors of two types. One type results from the fact that the acceleration is not really constant over the time step, and the other from the rounding off or truncation of the numbers at every step of the calculation. The first type of error is decreased by taking shorter time steps. But this means more numerical operations must be carried out over a given span of time, and this increases the round-off error for a given precision of the numbers being carried in the calculation. The design of numerical algorithms, as well as the choice of precisions and step sizes that maximize the speed of the calculation while keeping the errors within reasonable bounds, is almost an art form developed by extensive experience and ingenuity. For example, a scheme exists for extrapolating the step size to zero in order to find the change in the variables over a relatively short time span, thereby minimizing the accumulation of error from this source. If the total energy of the system is theoretically conserved, its evaluation for values of the variables at the beginning and end of a calculation is a measure of the errors that have accumulated.
The motion of the planets of the solar system over time scales approaching its 4.6-billion-year age is a classic n-body problem, where n = 9 with the Sun included. The question of whether or not the solar system is ultimately stable—whether the current configuration of the planets will be maintained indefinitely under their mutual perturbations, or whether one or another planet will eventually be lost from the system or otherwise have its orbit drastically altered—is a long-standing one that might someday be answered through numerical calculation. The interplay of orbital resonances and chaotic orbits discussed above can be investigated numerically, and this interplay may be crucial in determining the stability of the solar system. Already it appears that the parameters defining the orbits of several planets vary over narrow chaotic zones, but whether or not this chaos can lead to instability if given enough time is still uncertain.
If accelerations are determined by summing all the pairwise interactions for the n particles, the computer time per time step increases as n2. Practical computations for the direct calculation of the interactions between all the particles are thereby limited to n < 10,000. Therefore, for larger values of n, schemes are used where a particle is assumed to move in the force field of the remaining particles approximated by that due to a continuum mass distribution, or a “tree structure” is used where the effects of nearby particles are considered individually while larger and larger groups of particles are considered collectively as their distances increase. These later schemes have the capability of calculating the evolution of a very large system of particles using a reasonable amount of computer time with reasonable approximation. Values of n near 100,000 have been used in calculations determining the evolution of galaxies of stars. Values of n in the billions have been used in calculations of galaxy formation in the early universe. Also, the consequences for distribution of stars when two galaxies closely approach one another or even collide has been determined. Even calculations of the n-body problem where n changes with time have been completed in the study of the accumulation of larger bodies from smaller bodies via collisions in the process of the formation of the planets.
In all n-body calculations, very close approaches of two particles can result in accelerations so large and so rapidly changing that large errors are introduced or the calculation completely diverges. Accuracy can sometimes be maintained in such a close approach, but only at the expense of requiring very short time steps, which drastically slows the calculation. When n is small, as in some solar system calculations where two-body orbits still dominate, close approaches are sometimes handled by a change to a set of variables, usually involving the eccentric anomaly u, that vary much less rapidly during the encounter. In this process, called regularization, the encounter is traversed in less computer time while preserving reasonable accuracy. This process is impractical when n is large, so accelerations are usually artificially bounded on close approaches to prevent instabilities in the numerical calculation and to prevent slowing the calculation. For example, if several sets of particles were trapped in stable, close binary orbits, the very short time steps required to follow this rapid motion would bring the calculation to a virtual standstill, and such binary motion is not important in the overall evolution of, say, a galaxy of stars.
## Algebraic maps
In numerical calculations for conservative systems with modest values of n over long time spans, such as those seeking a determination of the stability of the solar system, the direct solution of the differential equations governing the motions requires excessive time on any computer and accumulates excessive round-off error in the process. Excessive time also is required to explore thoroughly a complete range of orbital parameters in numerical experiments in order to determine the extent of chaotic zones in various configurations (e.g., those in the asteroid belt near orbital mean motion commensurabilities with Jupiter). A solution to this problem is the use of an algebraic map, which maps the space of system variables onto itself in such a way that the values of all the variables at one instant of time are expressed by algebraic relations in terms of the values of the variables at a fixed time in the past. The values at the next time step are determined by applying the same map to the values just obtained, and so on. The map is constructed by assuming that the motions of all the bodies are unperturbed for a given short time but are periodically “kicked” by the perturbing forces for only an instant. The continuous perturbations are thus replaced by periodic impulses. The values of the variables are “mapped” from one time step to the next by the fact that the unperturbed part of the motion is available from the exact solution of the two-body problem, and it is easy to solve the equations with all the perturbations over the short time of the impulse. Although this approximation does not produce exactly the same values of all the variables at some time in the future as those produced by a numerical solution of the differential equations starting with the same initial conditions, the qualitative behaviour is indistinguishable over long time periods. As computers can perform the algebraic calculations as much as 1,000 times faster than they can solve the corresponding differential equations, the computational time savings are enormous and problems otherwise impossible to explore become tractable.
MEDIA FOR:
celestial mechanics
Previous
Next
Citation
• MLA
• APA
• Harvard
• Chicago
Email
You have successfully emailed this.
Error when sending the email. Try again later.
Edit Mode
Celestial mechanics
Physics
Tips For Editing
We welcome suggested improvements to any of our articles. You can make it easier for us to review and, hopefully, publish your contribution by keeping a few points in mind.
1. Encyclopædia Britannica articles are written in a neutral objective tone for a general audience.
2. You may find it helpful to search within the site to see how similar or related subjects are covered.
3. Any text you add should be original, not copied from other sources.
4. At the bottom of the article, feel free to list any sources that support your changes, so that we can fully understand their context. (Internet URLs are the best.)
Your contribution may be further edited by our staff, and its publication is subject to our final approval. Unfortunately, our editorial approach may not be able to accommodate all contributions. | {"extraction_info": {"found_math": false, "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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8797641396522522, "perplexity": 455.65098651633286}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560280133.2/warc/CC-MAIN-20170116095120-00261-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_11&diff=next&oldid=135844 | # Difference between revisions of "2006 AIME II Problems/Problem 11"
## Problem
A sequence is defined as follows and, for all positive integers Given that and find the remainder when is divided by 1000.
## Solution
### Solution 1
Define the sum as . Since , the sum will be:
Thus , and are both given; the last four digits of their sum is , and half of that is . Therefore, the answer is .
Solution by an anonymous user.
### Solution 2 (bash)
Since the problem only asks for the first 28 terms and we only need to calculate mod 1000, we simply bash the first 28 terms:
Adding all the residues shows the sum is congruent to mod 1000.
~ I-_-I
### Solution 3 (some guessing involved)/"Engineer's Induction"
All terms in the sequence are sums of previous terms, so the sum of all terms up to a certain point must be some linear combination of the first three terms. Also, we are given and , so we can guess that there is some way to use them in a formula. Namely, we guess that there exists some such that . From here, we list out the first few terms of the sequence and the cumulative sums, and with a little bit of substitution and algebra we see that , at least for the first few terms. From this, we have that .
Solution by zeroman; clarified by srisainandan6 | {"extraction_info": {"found_math": false, "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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9459314346313477, "perplexity": 378.7188993503585}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703511903.11/warc/CC-MAIN-20210117081748-20210117111748-00367.warc.gz"} |
http://encyclopedia.kids.net.au/page/li/Likelihood_principle | ## Encyclopedia > Likelihood principle
Article Content
# Likelihood principle
In statistical theory, the likelihood principle asserts that the information in any sample can be found, if at all, from the likelihood function, that function of unknown parameters[?] which specifies the probability of the sample observed on the basis of a known model, in terms of the model's parameters.
Suppose, for example, that we have observed n independent flips of a coin which we regard as having a constant probability, p, of falling heads up. The likelihood function is then the product of n factors, each of which is either p or 1-p. If we observe x heads and n-x tails, then the likelihood function is
[itex]L(p)\sim p^x(1-p)^{1-x}[/itex]
i.e., proportional to the product.
No multiplicative constant of C(N,X) is included because only the part of the probability which involves the parameter, p, is relevant (According to some accounts, the fact that only that part is relevant is what the likelihood principle says. For example, in one experiment the number of successes in 10 Bernoulli trials is observed; in another the number of trials needed to get four successes is observed. In either case, the outcome could be four successes in ten trials. The probabilities of that outcome are different in the two experiments, but as function s of p, the probability of success on each trial, they are proportional. The Likelihood principle says the same statistical inference about the value of p should be drawn in each case.) In particular, this principle suggests that it does not matter whether you started out planning to observe N trials or you just decided to stop on a whim. The issue of the likelihood principle is still controversial.
A deeper discussion of the topic is available in the article about maximum likelihood.
All Wikipedia text is available under the terms of the GNU Free Documentation License
Search Encyclopedia
Search over one million articles, find something about almost anything!
Featured Article
Academy Awards Foreign Language Film ... Deschamps[?], Arne Meerkamp von Embden[?], Claudie Ossard[?] producers - Jean-Pierre Jeunet director Son of the Bride[?] (El Hijo de la novia) (Argentina) - JEMPSA, ... | {"extraction_info": {"found_math": false, "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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9310060739517212, "perplexity": 804.5325329831261}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247490806.45/warc/CC-MAIN-20190219162843-20190219184843-00133.warc.gz"} |
https://www.groundai.com/project/characteristic-function-of-time-inhomogeneous-levy-driven-ornstein-uhlenbeck-processes/ | Characteristic Function of Time-Inhomogeneous Lévy-Driven Ornstein-Uhlenbeck Processes
# Characteristic Function of Time-Inhomogeneous Lévy-Driven Ornstein-Uhlenbeck Processes
Frédéric Vrins
Louvain School of Management (LSM)
&
Center for Operations Research and Econometrics (CORE)
Université catholique de Louvain
Chaussée de Binche 151, Office A.212, B-7000 Mons, Belgium
frederic.vrins@uclouvain.be
###### Abstract
Distributional properties -including Laplace transforms- of integrals of Markov processes received a lot of attention in the literature. In this paper, we complete existing results in several ways. First, we provide the analytical solution to the most general form of Gaussian processes (with non-stationary increments) solving a stochastic differential equation. We further derive the characteristic function of integrals of Lévy-processes and Lévy driven Ornstein-Uhlenbeck processes with time-inhomogeneous coefficients based on the characteristic exponent of the corresponding stochastic integral. This yields a two-dimensional integral which can be solved explicitly in a lot of cases. This applies to integrals of compound Poisson processes, whose characteristic function can then be obtained in a much easier way than using joint conditioning on jump times. Closed form expressions are given for gamma-distributed jump sizes as an example.
## 1 Introduction
Integrals of Markov processes are popular stochastic processes as a results of their numerous applications. Birth, death, bird-death and catastrophe processes in particular received a lot of attention in the context of queuing and storage problems [20] as well as in biology [19] or robotics [17]. From a broad perspective, general properties of integrals of Markov processes have been derived, like the time-evolution of associated moments [13]. With regards to the Laplace transforms in specific, closed form expressions have been derived in [22] for continuous-time Markov chains taking value on the set of positive integers. With regards to diffusion processes, very general results have been obtained for affine, quadratic and geometric models. The corresponding results are derived from the transition probabilities and the property that the infinitesimal generator is dependent on the space variable only, i.e the stochastic differential equations (SDE) has time-homogeneous coefficients. Explicit formulas for many standard processes including the integrated Orntein-Uhlenbeck (OU), Square-Root and Jacobi diffusions are available in [1],[15]. The specific case of integral of geometric Brownian motion, closely linked to squared Bessel processes, has extensively be studied by Yor [25], and a collection of papers on the topic is available [26].
In spite of this extensive literature, some important results are still lacking for interesting stochastic processes with sound applications. This is for instance the case of the generalized OU process, that is OU processes with time-varying coefficients and Lévy driving process. Similarly, to the best of our knowledge, there is no such results for compound Poisson processes.
In this paper, we are interested in the characteristic function (or equivalently, Laplace transform or moment-generating function -when it exists) of the path integral of the form
Λs,t=∫tsλudu (1.1)
where is a Markov process on a probability space . We assume the sigma-field contains all the information about up to time .
The paper is organized as follows. Section 2 is devoted to depict financial applications that will stress the importance of having the characteristic function available in closed form. We shall point out some processes for which the integral is useful in this context. Associated characteristic functions will be derived in the remaining part of the paper. Most general results for Gaussian processes (not necessary with stationary increments) is addressed in Section 3. In Section 4, the corresponding results are obtained for Ornstein-Uhlenbeck processes with a background driving Lévy process (BDLP) which is not restricted to be a Brownian motion. Finally, Compound Poisson processes are investigated in Section 5 based on joint conditioning of jump times of the underlying Poisson process. Closed-form expressions are given when the jump sizes are Gamma-distributed.
## 2 Financial motivation
The most obvious example of the use of integrals of stochastic processes in finance is the pricing of Asian options, in which the payoff depends on the time average of the underlying stock (or index) price. When the underlying stock follows a geometric Brownian motion, as in the standard Black-Scholes case, this leads to the study of integrals of Squared Bessel processes and explains why this specific case received so much attention (see e.g. [12][6]). In this context, Laplace transforms are interesting when explicit formula for the distributions are too involved or cannot be found. The corresponding distributions can be obtained by simple Fourier inversion, and all the moments are easily retreived by deriving the moment generating function. In some cases however, the explicit form of the characteristic function is very appealing in itself, as it direcly yields the calibration equation, as we now show.
Let stands for the risk-neutral measure and for the risk-free short rate. Then, the time- price of a risk-free zero-coupon bond paying 1 unit of currency at time is
B(t,T)=E[e−Λt,T|Ft] (2.1)
Consequently, the parameters of any given short rate process can be calibrated at time from the given yield curve by making sure that the Laplace transform of satisfies
B(0,T)=ϕΛT(1) (2.2)
which is related to the characteristic function via .
For tractability reasons, the most popular short-rate models are, by far, Gaussian or square-root diffusion processes with constant coefficients, possibly shifted by a deterministic function. For simple short-rate models like Vasicek (with Ornstein-Uhlenbeck dynamics, OU) or Cox-Ingersoll-Ross (with square root diffusion dynamics, SRD), the analytical expression of the Laplace transform is available; see e.g. [5],[8],[9],[10]. However, the corresponding expressions for the Hull-White model, which is the extension of the Vasicek model with time-varying coefficients, is not available.
Expressions similar to 2.1 appear in credit risk modelling, in both reduced form and structural approaches (see e.g. [18],[4]). In reduced form (intensity) approaches, is modeled as the first jump of a Cox process where the stochastic intensity is given by a non-negative stochastic process :
τ=inf{t:Nt>0} (2.3)
Conditional on the path of the intensity , is a Poisson process, so that
Q(τ>T)=Q(NT=0)=E[e−∫T0λsds]=E[e−ΛT]=ϕΛT(1) (2.4)
The same form of survival probability is obtained in the so-called structural approach. In this setup, the default is modeled as the first passage time of an asset process below a liability threshold :
τ=inf{t:Vt
The process is typically either a Brownian motion, a geometric Brownian motion or a subordinated process. The computation of the law of the first-passage times can be avoided by working in a specific framework where the survival probability takes the same form as in (2.4). Let be a non-decreasing grounded positive process and model the credit event as the first time where reaches a random threshold uniformly distributed in . The passage time is almost surely unique, so that:
τ=inf{t:e−Λt≤U}={t:e−Λt=U} (2.6)
Then, the survival probability takes again the same form:
Q(τ>T)=Q(e−ΛT>U)=E[Q(e−ΛT≥U)]=E[e−ΛT]=ϕΛT(1) (2.7)
This suggests that instead of working with an intermediate intensity process, we could directly feed the default model with an non-decreasing Lévy process . This setup has been used in [14] in the context of CDO pricing but in a purely numerical setup. The analytical expressions of the calibration equations are derived in [23]. The implied copulas form the class of so-called Sibuya copulas, which properties are studied in [24].
This setup, however is not appropriate when one desires to simulate survival probability curves . Indeed,
Q(τ>T|τ>t)=E[e−(ΛT−Λt)|Ft]=E[e−Λt,T] (2.8)
If the process is Lévy, the increments are independent of the past and if they are stationary, . Therefore, with
S(t,t+δ):=Q(τ>t+δ|τ>t)=E[e−Λδ]=ϕΛδ(1) (2.9)
In other words, there is no memory effect. Producing different curves , depending on the path , can only be introduced by considering non-Lévy processes for . One way to introduce such dependency is by considering as the integral of a underlying process , as in the intensity process.
## 3 Gaussian processes
As mentioned in the introduction, the expression of integrals of Gaussian processes are known in some specific cases, among which Vasicek (Ornstein-Uhlenbeck) is the most popular one. We generalise here previous results by working out the explicit form of the strong solutions for and given when is the solution to the most general SDE associated to Gaussian processes.
It is very well-known that Gaussian process solving a stochastic differential equation (SDE) has dynamics of the form111Not all Gaussian processes solve a SDE; this is for instance the case of fractional Brownian motion.
dλt=(α(t)−β(t)λt)dt+σ(t)dWt , t≥s, λs (3.1)
where is a Brownian motion and and are deterministic integrable functions and the initial condition. When the solution exists, it is possible to derive explicitly the strong solution for and and obtain the corresponding characteristic function. We assume in this paper that the coefficients here are such that existence and uniqueness conditions are satisfied; see e.g. [16] or [21] for a detailed discussion.
###### Proposition 1 (General Gaussian Process).
Consider eq.(3.1) and assume existence and uniqueness conditions are met. Suppose further that is known at time and define
G(s,t):=e−∫tsβ(u)du , I(s,t):=∫tsα(u)G(u,s)du (3.2) J(s,t):=∫tsσ(u)G(u,s)dWu , K(s,t):=∫tsσ(s)G(s,u)du (3.3)
Then, for any , the solution to eq.(3.1) conditional upon is
λt=m(s,t)+G(s,t)J(s,t) (3.4)
which is Normally distributed with mean and variance respectively given by
m(s,t) := G(s,t)(λs+I(s,t)) (3.5) v(s,t) := G2(s,t)∫tsσ2(u)G2(u,s)du=∫tsσ2(u)G2(u,t)du (3.6)
The integral is a shifted Ito integral
Λs,t=M(s,t)+∫tsK(u,t)dWu (3.7)
which is Normally distributed with mean and variance respectively given by
M(s,t) := (t−s)λs+∫ts(t−u)(α(u)−β(u)G(s,u)(λs+I(s,u)))du (3.8) V(s,t) := ∫tsK2(u,t)du (3.9)
###### Proof.
See Appendix A.∎
Since is Gaussian with known mean and variance, the following corollary is obvious.
###### Corollary 3.1 (Laplace transform of Λs,t).
The Laplace transform of , conditional upon , is
ϕΛs,t(x)=exp{−M(s,t)x+x2V(s,t)2} (3.10)
where are given in Proposition 1.
Let us apply this result to get the expression and distribution of in some particular cases.
###### Example 3.1 (Integrated rescaled Brownian motion).
Consider the case where .
Then,
M(s,t)=(t−s)λs , K(s,t)=σ(t−s) , V(s,t)=σ23(t−s)3 (3.11)
This expression can be found directly using the stochastic version of Fubini’s theorem on indicator functions.
###### Example 3.2 (Integrated OU process).
The case where and are constant is widely used in mathematical finance. It corresponds to , and :
M(s,t) = 1−e−β(t−s)βλs (3.12) K(s,t) = σβ(1−e−β(t−s)) (3.13) V(s,t) = (σ2β)2(2(t−s)+e−β(t−s)β(4−e−β(t−s))−6β) (3.14)
The above examples correspond to special cases for which results exist (see e.g [5]). However, the above result allows to determine the Laplace transform of more general Gaussian processes, as shown in the examples below.
###### Example 3.3 (Prototypical swap exposure).
Consider a Brownian bridge from to rescaled by a constant and shifted by the deterministic fontion . Such a process can be written as
λt=(T−t)(γt+σ∫t01T−udWu) , 0≤s≤t≤T (3.15)
This process starts from 0, its expectation follows the curve and then goes back to as . This is a Gaussian process with , zero-mean and instantaneous volatility . Then
M(s,T)=λtT−t2+γ6(T−t)3 , K(s,T)=σT−s2 , V(s,T)=σ212(T−s)3 (3.16)
See Appendix B for details.
###### Example 3.4 (Deterministically subordinated process).
Consider the integral where is a strictly increasing and continuous function. Then,
Λs,t=∫θ(t)θ(s)f∘θ−1(u)θ′(u)Wudu (3.17)
corresponds to the integral of where . Its differential takes the form (3.1) where , and instantaneous volatility .
Consequently, provided that the existence conditions are met, is Normally distributed and the solution is given by (3.7) where the integration bounds are , and the parameters are defined above, and the characteristic function follows.222Observe that when is not strictly increasing (i.e. mereley non-decreasing) and/or discontinuous, the corresponding characteristic function can be recovered by splitting the integral into pieces where the inveres of exists, where is constant and at time points where jumps.
## 4 Lévy-driven Hull-White process
In some cases, the characteristic function of the integrated process can be obtained by computing the characteristic function of a stochastic integral. This is in particular the case of generalized Lévy-driven OU processes (that is, where the random increments are controlled by a background driving Lévy process, BDLP), as we now show.
The log-characteristic function of a Lévy process takes the form
ψXt(x):=lnφXt(x)=tlnφX(x)=:tψX(x) (4.1)
where is the characteristic exponent of the infinitely divisible distribution of [7]. We can derive the log-characteristic function of the stochastic integral of a deterministic function with respect to a Lévy process as follows. Let where is integrable and is Lévy processes with triplet and is the density of the Lévy measure. Then
ψX(x) = (4.2) ψYt(x) = lnE[eix∫t0σ(s)dXs] (4.3)
###### Proposition 2 (Characteristic exponent of a stochastic integral with respect to a Lévy process).
Let be a Lévy process with characteristic exponent and define the semimartingale for some deterministic integrable function . Then,
ψYs,t(x)=∫tsψX(σ(u)x)du (4.4)
where .
###### Proof.
See Appendix C
The Lévy driven Hull-White process is defined as a Ornstein-Uhlenbeck process with time-varying coefficients and BDLP.
###### Proposition 3 (Characteristic exponent of integrated BDLP Hull-White process).
Let be a Hull-White process driven by the Lévy process solution of the SDE
dλt=(α(t)−β(t)λt)dt+σ(t)dXt (4.5)
Then, setting as before,
ψΛs,t(x)=ixM(s,t)+∫tsψX(x∫tuσ(u)G(u,v)dv)du (4.6)
###### Proof.
Using the integration by parts technique used in the proof of Proposition 1, the solution is proven to be the same as in (3.7) where is replaced by :
Λs,t=M(s,t)+∫tsK(u,t)dXu (4.7)
Therefore,
ψΛs,t(x)=ixM(s,t)+ψ∫tsK(u,t)dXu(x) (4.8)
and the claim follows from Proposition 2 and the expression of in Proposition 1. ∎
Observe that SDE (4.5) is not the most general SDE with linear drift since the way jumps are handled corresponds to the special case where a jump of at of size triggers a jump of size at for . We could perfectly want to have a jump which does depend on in a non-linear way (see e.g. [3] and [2] for a more advanced discussion and conditions on the coefficients and the functional parameters of for existence and pathwise uniqueness).
###### Example 4.1 (OU driven by a Brownian Motion).
We recover the characteristic function of the standard OU model driven by a Brownian motion from the strong solution
Λs,t=M(s,t)+∫tsK(u,t)dWu (4.9)
where . Since here , the Lévy triplet is and the characteristic exponent reduces to . So,
∫tsψX(K(u,t)x)du=−∫ts(xK(u,t))22du=−x22V(s,t) (4.10)
Therefore,
ψΛs,t(x)=ixM(s,t)−x22V(s,t) (4.11)
in line with in eq. (3.10).
Let us apply this method to determine the characteristic function of where is a non-Gaussian OU process driven by a gamma process .
###### Example 4.2 (OU driven by a gamma process).
Consider the SDE
dλt=−βλtdt+dγt (4.12)
with non-negative solution and the integral is
Λs,t=λs1−e−β(t−s)β+∫ts1−e−β(t−u)βdγu (4.13)
The integrated process is a deterministc constant plus a stochastic integral of a deterministic function with respect to a Lévy process. Therefore,
ψΛt(x)=ixλs1−e−β(t−s)β+∫tsψγ(1−e−β(t−u)βx)du (4.14)
where is the characteristic exponent of the gamma distribution driving the jump sizes. The integral can be written in terms of the dilogarithmic function
(t−s)ψγ(x/β)+αβ(Li2(v)−Li2(ve−β(t−s))) (4.15)
where .
A similar approach has been used in [11]333We are grateful to D. Madan for providing us with this reference. to evaluate a joint characteristic function in the -forward measure of a triple of processes which dynamics are given in the risk-neutral measure.
The following result is a straight consequence of Proposition 3.
###### Corollary 4.1 (Characteristic exponent of an integrated Lévy process).
Let be a Lévy process. Then,
ψΛs,t(x)=ixλs(t−s)+∫tsψλ(x(t−u))du (4.16)
###### Example 4.3 (Stochastically subordinated Brownian motion).
We are interested in the integral of that is, of the integral of a time-changed Brownian motion. Since for every Lévy process we have ,
φλt(x)=E[% eixλt]=E[(eixW)Xt]=E[eXtψW(x)]=E[eiXt(−iψW(x))]=φtX(−iψW(x)) (4.17)
or equivalently in terms of the characteristic exponent,
ψλ(x)=ψX(−iψW(x))=ψX(ix2/2) (4.18)
In the case of a Gamma subordinator with parameters , the subordinated process is will-known to be a variance-gamma (Lévy) process. Indeed, has the same distribution as the difference of two independent gamma processes, say and , each with same parameters . This can be seen by noting from the above result that :
ϕλt(x)=(2κ2κ+x2)αt=(√2κ√2κ+ix)αt(√2κ√2κ−ix)αt (4.19)
The log-characteristic function of the integral of is thus given by
ψΛt(x) = ixγs(t−s)+ψ∫tsγudu(x)+ψ∫tsγudu(−x) (4.20) = ixγs(t−s)+∫tsψγ(x(t−u))du+∫tsψγ(x(u−t))du (4.21)
and
∫tsψγ(x(t−u))du=κixψγ(x(t−s);κ,α)φγ(x(t−s);κ,1)−α(t−s)(1−lnκ) (4.22)
## 5 Compound Poisson processes
In this section, we focus on the integral of jump processes where the jumps arrive according to a Poisson process and the size of the -th jump is given by the random variable which are mutually independent. We then specialize to the case of compound Poisson process where . We compute the Laplace transform for which the analytical expression can be found for some laws of . This can be done in several ways. We adopt here on a joint conditioning on the number of jumps of by time and the corresponding jump times .
### 5.1 Generalized Compound Process
The Laplace transform of the integrated process where is a generalized compound Poisson process is obtained as an infinite sum of multidimensional integrals.
###### Proposition 4.
Let ba a Poisson process with intensity and be a sequence of independent random variables. Then, the Laplace transform of the integrated process
Λt=∫t0Nu∑i=1Xidu (5.1)
is given by
ϕΛt(x)=e−θt(1+∞∑n=1θn∫tt1=0ϕX1(x(t−t1))…∫ttn=tn−1ϕXn(x(t−tn))d→tn) (5.2)
###### Proof.
The law of is that of a random variable, that is
pNt(n)=(θt)ne−θtn!, n∈{0,1,2,…} (5.3)
Furthermore, the laws of the time arrival of the -th jump, , and the time elapsed between jumps and are known
Δi+1:=Ti+1−Ti ∼ Exp(θ)=Gamma(1,θ) (5.4) Ti = Gamma(i,θ) (5.5)
where we have set . Let us further note and . Then, due to the independence between the times separating two consecutive jumps, we compute the joint density of the first jump times :
(5.6)
where and is 1 if a non-decreasing sequence of nonnegative numbers and 0 otherwise.
We can then compute the joint density of at using as follows:
p→Tn,Nt(→tn,n) = (5.7) = (5.8) = (5.9) = (5.10)
This expression depends on only via the indicator function . We can now compute the Laplace transform of by conditioning on :
ϕΛt(x) = E[e−x∫t0∑Nui=1Xidu] (5.11) = pNt(0)+∞∑n=1E[e−x∑ni=1Xi(t−ti)|→Tn=→tn,Nt=n]p→Tn,Nt(→tn,n) (5.12) = pNt(0)+∞∑n=1∫∞t1=0…∫∞tn=0n∏i=1ϕXi(x(t−ti))p→T,Nt(→tn,n)d→tn (5.13) = e−θt+∞∑n=1θne−θt∫tt1=0…∫ttn=tn−1n∏i=1ϕXi(x(t−ti))d→tn (5.14)
This form is appealing for two reasons. First, only the product of the characteristic functions of the jump size variables appear in the multiple integral as can be sent outside provided that we restrict the integration domain according to the indicator function . Second, this multiple integral can be written as
ϕΛt(x)=e−θt+e−θt∞∑n=1θn∫tt1=0ϕX1(x(t−t1))…∫ttn=tn−1ϕXn(x(t−tn)):=InI1d→tn (5.15)
Therefore, the multiple integral can be computed recursively: only the last characteristic function is involved in the last integral (noted ) which only depends on time via . The previous integrals, , involves the product of and , which limits the complexity of the computation.
Observe that when , and the multiple integral collapses , so that
∞∑n=1(tθ)nn!=%etθ−1 (5.16)
and as it should.
###### Corollary 5.1.
The Laplace transform of the integrated process
Λs,t=∫tsNu∑i=1Xidu (5.17)
is given by
ϕΛs,t(x)=e−x(t−s)λsϕ~Λt−s(x) (5.18)
where is as in eq. (5.2) but with .
###### Proof.
By definition,
∫tsNu∑i=1Xidu=(t−s)λs+∫tsNu∑i=Ns+1Xidu=(t−s)λs+∫tsNu−Ns∑i=1XNs+idu (5.19)
The Poisson process has stationary increments, so the distribution of and
∫tsNu∑i=1Xidu∼(t−s)λs+∫t−s0Nv∑i=1XNs+idv (5.20)
E[e−x∫ts∑Nui=1Xidu]=e−x(t−s)λsϕ~Λt−s(x) (5.21)
### 5.2 Compound Poisson process with Gamma jump sizes
The above expression for the Generalized Compound Poisson process is appealing but is hard to solve as no recursive formula can be found for solving the multiple integral. Even in the case where the , the resulting integrals seems intractable. However, when the ’s are iid, recursion formula can be found as shown below.
Beforehand, observe however that Compound Poisson processes being Lévy processes with characteristic exponent
ψλ(x)=∫R(eixz−1)ν(dz) (5.22)
where and is the jump size distribution we have, from Corollary 2,
ψΛs,t(x)=ixλs(t−s)+∫ts∫∞−∞(eizx(t−u)−1)ν(dz)dv (5.23)
Solving this simple time-space integral leads the Laplace transform of . This approach seems considerably simpler than computing the infinite sum of multiple integrals (of arbitrarily large dimension) in eq. 5.15.
However, when , a recursion formula can be found and the infinite sum of multiple integrals can be solved, and we shall adopt this alternative methodology for the sake of illustration. The results obviously agree with those easily obtained by computing eq.(5.23) with .
###### Example 5.1 (Compound Poisson process with exponentially-distributed jump sizes).
Let be a Poisson process with intensity and be iid exponential variables with Laplace transform . Given that in the case of iid , it is enough to compute . Then, the integral of the compound Poisson process is
ϕΛt(x) = (e−tϕγ(xt;κ,−κ/x))θ (5.24)
To see this, notice that
∫ts=tn−1ϕγ(x(t−s))ds=−κx[ln(κ+x(t−s))]ttn−1=−κxlnϕγ(x(t−tn−1)) (5.25)
Integrating that in the previous integral, we get
∫ts=tn−2ϕγ(x(t−s))∫tu=sϕγ(x(t−u))duds = −κx∫ts=tn−2ϕγ(x(t−s))lnϕγ(x(t−s))ds (5.26) = +κx∫t−tn−2u=0ϕγ(xu)lnϕγ(xu)du = (−κx)2∫ϕγ(x(t−tn−2))u=0lnvvdv (5.27) = 12(−κlnϕγ(x(t−tn−2))x)2 (5.28)
Finally, using and ,
∫tt1=0ϕX1(x(t−t1))…∫ttn=tn−1ϕXn(x(t−tn))d→tn = 1n!(−κlnϕγ(xt)x)n (5.29)
###### Example 5.2 (Compound Poisson process with gamma-distributed jump sizes.).
Let be iid gamma random variables with Laplace transform for and (the case corresponds to the exponential above). Then,
ϕΛt(x) = e−θ(t−κx(α−1)ϕγ(xt;κ,α−1)) (5.30)
The above expression can be found easily by noting that
∫ts=uϕγ(x(t−s))ds=∫ts=uϕαγ(x(t−s);κ,1)ds=κx(α−1)ϕα−1γ(x(t−u);κ,1) (5.31)
so that
∫tt1=0ϕX1(x(t−t1))…∫ttn=tn−1ϕXn(x(t−tn))d→tn = (κ/x)nϕn(α−1)γ( | {"extraction_info": {"found_math": false, "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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9784002304077148, "perplexity": 692.5682773183534}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655898347.42/warc/CC-MAIN-20200709034306-20200709064306-00045.warc.gz"} |
https://socratic.org/questions/58cffd4bb72cff73a9640a2f | Physics
Topics
# Question #40a2f
##### 1 Answer
Mar 24, 2017
(a) ${0}^{\circ}$
Since the magnitude of resultant is equal to the sum of magnitudes of two displacements.
$| \vec{R} | = | \vec{A} | + | \vec{B} |$
(b) ${180}^{\circ}$
Since the magnitude of resultant is equal to the difference of magnitudes of two displacements. (assuming $| \vec{A} | > | \vec{B} |$)
$| \vec{R} | = | \vec{A} | - | \vec{B} |$
(b) ${90}^{\circ}$
Since the square of magnitude of resultant is equal to sum of squares of magnitudes of two displacements. This is true only in case of a right triangle.
$| \vec{R} {|}^{2} = | \vec{A} {|}^{2} + | \vec{B} {|}^{2}$
##### Impact of this question
76 views around the world
You can reuse this answer
Creative Commons License | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9773508310317993, "perplexity": 1237.9748936470985}, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912201904.55/warc/CC-MAIN-20190319052517-20190319074517-00418.warc.gz"} |
http://sns.ias.edu/astrophysics/research | # Astrophysics Research Interests
## Faculty
• Richard Black Professor Scott Tremaine: Astrophysical dynamics; formation and evolution of planetary systems; galactic structure and evolution; supermassive black holes in galaxies
• Professor Matias Zaldarriaga: Cosmology -- early universe cosmology, cosmological perturbation theory, cosmic microwave background, large-scale structure, dark matter and dark energy
## Visiting Professor
• Maureen and John Hendricks Visiting Professor Rashid Sunyaev: Theoretical astrophysics, high energy astrophysics and cosmology: including CMB, clusters of galaxies, theory of accretion onto black holes and neutron stars, interaction of matter and radiation under astrophysical conditions, x-ray astronomy
## Current Members
• Valentin Assassi: Cosmology: inflation, primordial non-Gaussianity, large-scale structure, halo biasing, cosmological perturbation theory
• Ben Bar-Or: Stellar dynamics, statistical mechanics of stellar systems, galactic nuclei, accretion disks, nearly Keplerian systems, the three body problem, star clusters
• Liang Dai: Cosmology, inflation, large scale structure, cosmic microwave background, general relativity and gravitational waves, gravitational lensing, dark matter and neutrinos
• Jean-Baptiste Fouvry: Kinetic theory, self-gravitating systems, galactic dynamics, secular evolution, long-range interactions
• Vera Gluscevic: Cosmology, cosmic microwave background, dark energy, inflation; direct-detection searches for dark matter; cosmic reionization; primordial magnetic fields
• Adrian Hamers : Gravitational dynamics — secular evolution of hierarchical systems including multiplanet and multistar systems, and galactic nuclei. Applications to hot Jupiters, compact objects, SNe Ia, tidal disruptions and gravitational wave sources
• Alexander Kaurov :Reionization, cosmology, early universe and galaxy formation, 21 cm probes; dark matter annihilation; neutron stars
• Morgan MacLeod: Binary star interactions, common envelope episodes, and the formation of compact (and merging) binaries; stellar dynamics of clusters, especially interactions of compact objects and stars in clusters and tidal disruptions of stars; astronomical transients.
• Tejaswi Nerella: Cosmology, reionization, 21cm cosmology, dark matter, neutrinos, magnetic fields, gravitational waves, large scale structure, radio astronomy, cosmic microwave background, cosmological perturbation theory, neutron stars, tidal interactions
• James Owen: Star and planet formation; accretion discs; exoplanets; astrophysical fluid dynamics; radiative transfer and numerical simulations
• David Radice: Gravitational-wave sources, multi-messenger astronomy, r-process nucleosynthesis; neutron-star mergers, core-collapse supernovae; numerical relativity
• Roman Rafikov: Theoretical astrophysics, planetary sciences, planet formation, N-body dynamics, astrophysical fluid dynamics, accretion disks, high-energy astrophysics
• Marcel Schmittfull: Cosmology, large-scale structure, gravitational lensing of the cosmic microwave background; inflation, dark energy, neutrinos; observables beyond the power spectrum
• Marko Simonović: Inflation, primordial non-Gaussianities, large-scale structure, cosmological perturbation theory, modifications of gravity
## Visitors
• Tim Morton: Extrasolar planets; statistical analysis; scientific software development | {"extraction_info": {"found_math": false, "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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8147270679473877, "perplexity": 23503.465221131748}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549448198.69/warc/CC-MAIN-20170728103510-20170728123510-00223.warc.gz"} |
https://forum.math.toronto.edu/index.php?PHPSESSID=kmjdq2u8ujqqeb70osuccqm834&topic=2315.0 | ### Author Topic: 1.5 question 23 (Read 866 times)
#### Yan Zhou
• Full Member
• Posts: 15
• Karma: 1
##### 1.5 question 23
« on: February 02, 2020, 03:06:29 PM »
23. Show that $F(z) = e^{z}$ maps the strip $S = \{ x+iy: -\infty < x < \infty, -\frac{\pi}{2} < y < \frac{\pi}{2}\}$ onto the region $D = \{w = s+it: s \geqslant 0, w \neq 0\}$ and that $F$ is one-to-one on $S$. Furthermore, show that $F$ maps the boundary of $S$ onto all the boundary of $D$ except $\{w = 0\}$. Explain what happens to each of the horizontal lines $\{Imz = \frac{\pi}{2}\}$ and $\{Imz = -\frac{\pi}{2}\}$
To prove onto, I want to show, $\forall w \in D, \exists z \in S$, such that $F(z) = e^{z} = e^{x + iy} = w = s +it$.
So I need to find $x,y$ such that $s = e^{x}cosy, t = e^{x}siny$. I got answear $x = \frac{ln(s^{2} + t^{2})}{2}, y = arcos\frac{s}{\sqrt{s^{2} + t^{2}}}$But I don't know if they are correct. Can anyone help with this question and the following questions as well? | {"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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9720259308815002, "perplexity": 442.96911820308105}, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320299894.32/warc/CC-MAIN-20220129002459-20220129032459-00570.warc.gz"} |
https://www.physicsforums.com/threads/conservation-of-angular-momentum.863095/ | # Conservation of angular momentum
1. Mar 21, 2016
### JulienB
Hi everybody! I'm preparing myself for upcoming exams, and I struggle a little with conservation of angular momentum. Can anybody help me understand how to solve such problems?
1. The problem statement, all variables and given/known data
(for a better comprehension, see the attached image)
We have a wooden cylinder of mass mZ = 600g and of radius r0 = 5cm, which can rotate around its symmetry axis. Someone shoots on it, and the projectile has the mass mG = 5.0g and initial velocity v = 80m/s. The distance between the linear trajectory of the projectile and the rotation axis of the cylinder is r1 = 3.0cm. The projectile penetrates the cylinder and stays stuck at a distance of r2 = 3.5cm from the rotation axis of the cylinder.
a) What is the frequency of rotation f of the cylinder after the impact? Where and how should you shoot the projectile in order to obtain maximum/minimum frequency?
b) Which part of the kinetic energy is used to deform the wooden cylinder?
c) If the cylinder was not fixed on a rotation axis but on a thread, what would be the differences to previous case when the projectile hits the cylinder?
2. Relevant equations
So I imagine both conservation of linear momentum and of angular momentum are important. We also know that ω = 2πƒ.
3. The attempt at a solution
Okay I give it a go:
We know that the linear momentum is conserved, that the cylinder is not moving before the collision and that the two objects are moving together after the collision:
mG ⋅ v = (mG + mZ) ⋅ v'
Here I already see a problem: v' is supposed to be the tangential velocity of the system cylinder-projectile after the collision, but I believe a projectile located at r2 = 3.5cm does not have the same tangential velocity as a point located at r0 = 5.0cm. Is that correct? Then we would have mG ⋅ v = mG ⋅ vG' + mZ ⋅ vZ', which is also not so great.
I encounter the same problem with the conservation of angular momentum:
mG ⋅ v ⋅ r0 = (mG + mZ) ⋅ v' ⋅ r0
or
mG ⋅ v ⋅ r0 = mG ⋅ vG' ⋅ r2 + mZ ⋅ vZ' ⋅ r0
?
I feel like I'm missing something, since none of those equations lead me anywhere :( Furthermore, I never manage to involve r1 in the equations, which obviously plays a role because of the 2nd part of the question. Can someone give me a clue so that I clarify my misunderstandings?
Thank you very much in advance.
Julien.
#### Attached Files:
• ###### Mechanik Übung (2).png
File size:
33.6 KB
Views:
45
2. Mar 21, 2016
### ZapperZ
Staff Emeritus
You appear to have completely ignored the moment of inertia of the cylinder.
Zz.
3. Mar 21, 2016
### PeroK
I would assume from the problem statement that the cylinder is fixed and can only rotate about its axis of symmetry, not move linearly.
For part c) I would assume it is hanging by a thread, and is free to swing and rotate. Although I am not totally confident how to interpret part c). I guess you had to translate the question for us.
4. Mar 21, 2016
### ZapperZ
Staff Emeritus
What does this have anything to do with not including the moment of inertia? If the cylinder is rotating under any circumstances, its moment of inertia comes into play!
Zz.
5. Mar 21, 2016
### PeroK
If the cylinder is fixed, linear momentum is not conserved. If that is the case, the OP is off on the wrong foot.
6. Mar 21, 2016
### JulienB
@ZapperZ :
Yes I realised that, but don't have I to also include the moment of inertia of the projectile then? I don't know its shape though.
Or may I just write the linear momentum of the projectile on the left side of the equation and the moment of inertia of the cylinder on the right side, like that:
mG ⋅ v = IZ ⋅ ω?
@PeroK :
Yes I had to translate the problem from German, sorry :) Your assumption is correct, but if you don't mind I will first focus on question a) to make sure I understand how to use the equation first.
7. Mar 21, 2016
### JulienB
@PeroK :
Yes I believe the cylinder is fixed. So the linear momentum is not conserved??
8. Mar 21, 2016
### PeroK
I would assume not. I would assume for parts a) and b) it is only free to rotate about a fixed axis.
9. Mar 21, 2016
### ZapperZ
Staff Emeritus
The moment of inertia of the projectile can be assume to be the same as that for a particle at some fixed radius once it is embedded in the cylinder. So you have to consider TWO different angular momentum after the "collision": from the spinning cylinder and the projectile. Otherwise, you have missed the entire problem.
I still don't understand why, just because you don't know the "shape" of the projectile, that you are not even including the moment of inertia of the cylinder. This is a major part of the physics, and you missed it.
Zz.
10. Mar 21, 2016
### JulienB
I was (and am still) a bit confused about the whole thing. If linear momentum is not conserved, then may I write the following:
mG ⋅ v ⋅ r0 = mG ⋅ vG' ⋅ r2 + mZ ⋅ r0 ⋅ vZ' ?
On the left hand side I consider the angular moment at the moment of the collision, and on the left hand side the angular momentum of the projectile and of the cylinder once the projectile is stuck in the cylinder. The moment of inertia of the cylinder being mG ⋅ r02 and since ω = vZ'/r0, I would say that LZ = mZ ⋅ r0 ⋅ vZ'. Am I getting somewhere?
11. Mar 21, 2016
### JulienB
I went further with that idea, and reached the following:
mG ⋅ v ⋅ r0 = mG ⋅ r22 ⋅ ω + mZ ⋅ r02 ⋅ ω
⇒ ω = (mG ⋅ v ⋅ r0) / (mG ⋅ r22 + mZ ⋅ r02) = 13.28 rad/s2
⇒ ƒ = ω/2π = 2.1 Hz
Does that make sense? mZ ⋅ r02 is the moment of inertia of the cylinder.
12. Mar 21, 2016
### PeroK
I'm not sure I follow what you are doing. You seem to be unsure what is moment of inertia. In this problem, you have two examples of angular momentum:
1) Angular momentum of a point mass about a point/axis, which I think you understand.
2) Angular momentum of a rigid body about an axis of rotation. In this case you need the moment of inertia (MoI) of the rigid body about that axis. This is not the mass of the body, but a measure of the mass distribution of the body in terms of distance from the axis.
You should know or be able to calculate the MoI of a solid cylinder. This is what you are missing. Your last equation looks wrong as well.
First step: we need the MoI of a solid cylinder about its axis of symmetry.
Last edited: Mar 21, 2016
13. Mar 21, 2016
### PeroK
That's the MoI of a hollow cylinder, where all the mass is $r_0$ from the centre.
14. Mar 21, 2016
### JulienB
@PeroK
Sounds bad, but I won't give up :)
Yes I indeed used the wrong formula since the beginning. The moment of inertia of a full cylinder is I believe:
IZ = ½ πhρr4 = ½ mZ ⋅ r2
Now for the rest, I will try to explain you thoroughly my train of thoughts:
I assume the angular momentum of the point mass "projectile" about the axis of rotation of the cylinder just before the collision should be equal to the angular momentum of the cylinder about its axis of rotation + the angular momentum of the bullet embedded inside the cylinder after the collision:
LG = LG' + LZ'
⇔ mG ⋅ v ⋅ r0 = IG ⋅ ω + IZ ⋅ ω
⇔ mG ⋅ v ⋅ r0 = mG ⋅ r22 ⋅ ω + ½ ⋅ mZ ⋅ r02 ⋅ ω
ω being the angular velocity, it should be the same at r0 for the cylinder and at r2 for the bullet, right? At least that's how I attempted to resolve:
ω = (mG ⋅ v ⋅ r0)/(mG ⋅ r22 + ½ mZ ⋅ r02)
Unfortunately I get a bit of a crazy result. I probably still mess something up.
15. Mar 21, 2016
### PeroK
That looks right. What did you get for $\omega$?
Sorry, just saw you used $r_0$ on the left-hand side. Can you see what it should be?
16. Mar 21, 2016
### JulienB
@PeroK
Well... Around 1468 rad/s2, which would give a frequency of 233.6 Hz.
17. Mar 21, 2016
### JulienB
Nevermind I made a mistake: now ω = 26.45 and ƒ = 4.2 Hz.
18. Mar 21, 2016
### PeroK
That's not right, because you are using $r_0$ for the initial angular momentum.
19. Mar 21, 2016
### JulienB
Aaaah that should be r1, right? I used r0 because I thought the initial velocity would be affected while penetrating the cylinder.
20. Mar 21, 2016
### PeroK
Yes, it must be $r_1$.
Draft saved Draft deleted
Similar Discussions: Conservation of angular momentum | {"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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9659801125526428, "perplexity": 959.5948133973827}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187823016.53/warc/CC-MAIN-20171018161655-20171018181655-00471.warc.gz"} |
http://dazz.media/laura-ashley-hmgmbz/inverse-function-%26-graph-6f1a63 | A function f has an input variable x and gives then an output f (x). If functions f and g are inverse functions, f(g(x)) = g(f(x)) . Then the inverse function f-1 turns the banana back to the apple. If we want to evaluate an inverse function, we find its input within its domain, which is all or part of the vertical axis of the original function’s graph. Function pairs that exhibit this behavior are called inverse functions. A function has to be "Bijective" to have an inverse. In simple words, if any function “f” takes x to y then, the inverse of “f” will take y to x. If the function is one-to-one, there will be a unique inverse. ( because every ( x, y) has a ( y, x) partner! One should not confuse (-1) with exponent or reciprocal here. Required fields are marked *, An inverse function is a function that returns the original value for which a function has given the output. How to use inverse in a sentence. A useful example is converting between Fahrenheit and Celsius: For you: see if you can do the steps to create that inverse! If a horizontal line intersects the original function in a single region, the function is a one-to-one function and inverse is also a function. In simple words, if any function “f” takes x to y then, the inverse of “f” will take y to x. Our fault for not being careful! In its simplest form the domain is all the values that go into a function (and the range is all the values that come out). inverse"), will reverse this mapping. The original function has to be a one-to-one function to assure that its inverse will also be a function. An inverse function or an anti function is defined as a function, which can reverse into another function. The cool thing about the inverse is that it should give us back the original value: When the function f turns the apple into a banana, An example is also given below which can help you to understand the concept better. An inverse function is the "reversal" of another function; specifically, the inverse will swap input and output with the original function. Instead it uses as input f (x) and then as output it gives the x that when you would fill it in in f will give you f (x). An inverse function is a function that undoes the action of the another function. But we could restrict the domain so there is a unique x for every y ... Let's plot them both in terms of x ... so it is now f-1(x), not f-1(y): f(x) and f-1(x) are like mirror images The inverses of some of the most common functions are given below. The graph of f(x) and f-1(x) are symmetric across the line y=x. If the function is denoted by ‘f’ or ‘F’, then the inverse function is denoted by f-1 or F-1. Given a function f (x) f(x) f (x), the inverse is written f − 1 (x) f^{-1}(x) f − 1 (x), but this should not be read as a negative exponent. The inverse is usually shown by putting a little "-1" after the function name, like this: So, the inverse of f(x) = 2x+3 is written: (I also used y instead of x to show that we are using a different value.). (sin 90) = 90 degrees. The inverse function takes the output answer, performs some operation on it, and arrives back at the original function’s starting value. Just think ... if there are two or more x-values for one y-value, how do we know which one to choose when going back? Inverse functions mc-TY-inverse-2009-1 An inverse function is a second function which undoes the work of the first one. If the domain of the original function needs to be restricted to make it one-to-one, then this restricted domain becomes the range of the inverse function. In the Wolfram Language, inverse functions are represented using InverseFunction[f]. So, the inverse of f (x) = 2x+3 is written: f-1(y) = (y-3)/2. Only one-to-one functions have inverses. For example, sin. Assuming "inverse function" is referring to a mathematical definition | Use as. If the inverse of a function is itself, then it is known as inverse function, denoted by f. In mathematics, an inverse function is a function that undoes the action of another function. In other words, restrict it to x ≥ 0 and then we can have an inverse. Hence, sin 90 degrees is equal to 1. g = finverse(f) returns the inverse of function f, such that f(g(x)) = x.If f contains more than one variable, use the next syntax to specify the independent variable. We got 2 instead of −2. When you’re asked to find an inverse of a function, you should verify on your own that the inverse you obtained was correct, time permitting. STEP 1: Stick a " y " in for the " f (x) " guy: STEP 2: Switch the x and y. There are various types of inverse functions like the inverse of trigonometric functions, rational functions, hyperbolic functions and log functions. For example, show that the following functions are inverses of each other: Show that f(g(x)) = x. The inverse function of a function f is mostly denoted as f -1. inverse f ( x) = ln ( x − 5) $inverse\:f\left (x\right)=\frac {1} {x^2}$. If function f is not a one-to-one then it does not have an inverse. Really clear math lessons (pre-algebra, algebra, precalculus), cool math games, online graphing calculators, geometry art, fractals, polyhedra, parents and teachers areas too. inverse function definition: 1. a function that does the opposite of a particular function 2. a function that does the opposite…. Restrict the Domain (the values that can go into a function). Put "y" for "f(x)" and solve for x: This method works well for more difficult inverses. Or we can find an inverse by using Algebra. Inverse functions are a way to "undo" a function. A linear function is a function whose highest exponent in the variable(s) is 1. If you plan to offer your domain name soon, you should get an expert appraisal from a paid service. If you consider functions, f and g are inverse, f(g(x)) = g(f(x)) = x. The inverse isn't a function. Download BYJU’S- The Learning App to get a more engaging and effective learning experience. More discussions on one to one functions will follow later. Your email address will not be published. Mathematically this is the same as saying, inverse y = x x2 − 6x + 8. Figure 3.7.1 shows the relationship between a function f(x) and its inverse f − 1(x). There are six inverse trigonometric functions which include arcsine (sin-1), arccosine (cos-1), arctangent (tan-1), arcsecant (sec-1), arccosecant (cosec-1), and arccotangent (cot-1). Just make sure we don't use negative numbers. The Inverse Function goes the other way: So the inverse of: 2x+3 is: (y-3)/2. The inverse of the function returns the original value, which was used to produce the output and is denoted by f. If we have to find the inverse of trigonometry function sin x = ½, then the value of x is equal to the angle, the sine function of which angle is ½. or an anti function is defined as a function, which can reverse into another function. Imagine we came from x1 to a particular y value, where do we go back to? How to Graph the Inverse of a Function By Yang Kuang, Elleyne Kase If you’re asked to graph the inverse of a function, you can do so by remembering one fact: a function and its inverse are reflected over the line y = x. There are mainly 6 inverse hyperbolic functions exist which include sinh-1, cosh-1, tanh-1, csch-1, coth-1, and sech-1. To find the inverse of a quadratic function, start by simplifying the function by combining like terms. Let's just do one, then I'll write out the list of steps for you. Make sure your function is one-to-one. Intro to inverse functions. Then, g(y) = (y-5)/2 = x is the inverse of f(x). For example , addition and multiplication are the inverse of subtraction and division respectively. inverse is called by random.function and calculates the inverse of a given function f. inverse has been specifically designed to compute the inverse of the cumulative distribution function of an absolutely continuous random variable, therefore it assumes there is only a root for each value in the interval (0,1) between f (lower) and f (upper). Let us see graphically what is going on here: To be able to have an inverse we need unique values. A function $g$ is the inverse of a function $f$ if whenever $y=f(x)$ then $x=g(y)$. or. How to Find the Inverse of a Function 2 - Cool Math has free online cool math lessons, cool math games and fun math activities. The graph of the inverse of a function reflects two things, one is the function and second is the inverse of the function, over the line y = x. Given a function f(x), its inverse f^(-1)(x) is defined by f(f^(-1)(x))=f^(-1)(f(x))=x. Important Questions Class 12 Maths Chapter 2 Inverse Trigonometric Functions. Using the formulas from above, we can start with x=4: So applying a function f and then its inverse f-1 gives us the original value back again: We could also have put the functions in the other order and it still works: We can work out the inverse using Algebra. This line passes through the origin and has a slope of 1. inverse f ( x) = x3. Similarly, we find the range of the inverse function by observing the horizontal extent of the graph of the original function, as this is the vertical extent of the inverse function. If the function is one-to-one, write the range of the original function as the domain of the inverse, and write the domain of the original function as the range of the inverse. A function is called one-to-one if no two values of x x produce the same y y. Just like inverse trigonometric functions, the inverse hyperbolic functions are the inverses of the hyperbolic functions. . If f(x) is a function which gives output y, then the inverse function of y, i.e. The Derivative of an Inverse Function We begin by considering a function and its inverse. For example, sin-1(1) = sin-1(sin 90) = 90 degrees. One should not confuse (-1) with exponent or reciprocal here. So, when we apply function f and its reverse f-1 gives the original value back again, i.e, f-1(f(x)) = x. Inverse functions, in the most general sense, are functions that "reverse" each other. This same quadratic function, as seen in Example 1, has a restriction on its domain which is x \ge 0.After plotting the function in xy-axis, I can see that the graph is a parabola cut in half for all x values equal to or greater than zero. I will utilize the domain and range of the original function to describe the domain and range … Inverse of Square Root Function Read More » It is called a "one-to-one correspondence" or Bijective, like this. The inverse trigonometric functions are also known as arc function as they produce the length of the arc, which is required to obtain that particular value. Hence, sin 90 degrees is equal to 1. If f and g are inverse functions, then f(x) = y if and only if g(y) = x, is used to find the measure of angle for which sine function generated the value. x1 or x2? Before formally defining inverse functions and the notation that we’re going to use for them we need to get a definition out of the way. The inverse function agrees with the resultant, operates and reaches back to the original function. If f(x) is both invertible and differentiable, it seems reasonable that the inverse of f(x) is also differentiable. Embed this widget ». The inverse of a function can be viewed as the reflection of the original function over the line y = x. A function accepts values, performs particular operations on these values and generates an output. Example 2: Find the inverse function of f\left( x \right) = {x^2} + 2,\,\,x \ge 0, if it exists.State its domain and range. inverse function - Wolfram|Alpha. $inverse\:y=\frac {x} {x^2-6x+8}$. The relation, developed when the independent variable is interchanged with the variable which is dependent on a specified equation and this inverse may or may not be a function. Inverse function calculator helps in computing the inverse value of any function that is given as input. So if we have to draw the graph of f-1, then we have to switch the positions of x and y in axes. A function that consists of its inverse fetches the original value. Inverse Logarithmic Functions and Inverse Exponential Function. referring to English words. New Version: https://youtu.be/q6y0ToEhT1EDefine an inverse function. or. Finding the Inverse Function of a Square Root Function To find the inverse of a square root function, it is crucial to sketch or graph the given problem first to clearly identify what the domain and range are. So a bijective function follows stricter rules than a general function, which allows us to have an inverse. That is because some inverses work only with certain values. The inverse of a function f does exactly the opposite. First, replace f(x) with y and the function becomes. CBSE Previous Year Question Papers Class 10, CBSE Previous Year Question Papers Class 12, NCERT Solutions Class 11 Business Studies, NCERT Solutions Class 12 Business Studies, NCERT Solutions Class 12 Accountancy Part 1, NCERT Solutions Class 12 Accountancy Part 2, NCERT Solutions For Class 6 Social Science, NCERT Solutions for Class 7 Social Science, NCERT Solutions for Class 8 Social Science, NCERT Solutions For Class 9 Social Science, NCERT Solutions For Class 9 Maths Chapter 1, NCERT Solutions For Class 9 Maths Chapter 2, NCERT Solutions For Class 9 Maths Chapter 3, NCERT Solutions For Class 9 Maths Chapter 4, NCERT Solutions For Class 9 Maths Chapter 5, NCERT Solutions For Class 9 Maths Chapter 6, NCERT Solutions For Class 9 Maths Chapter 7, NCERT Solutions For Class 9 Maths Chapter 8, NCERT Solutions For Class 9 Maths Chapter 9, NCERT Solutions For Class 9 Maths Chapter 10, NCERT Solutions For Class 9 Maths Chapter 11, NCERT Solutions For Class 9 Maths Chapter 12, NCERT Solutions For Class 9 Maths Chapter 13, NCERT Solutions For Class 9 Maths Chapter 14, NCERT Solutions For Class 9 Maths Chapter 15, NCERT Solutions for Class 9 Science Chapter 1, NCERT Solutions for Class 9 Science Chapter 2, NCERT Solutions for Class 9 Science Chapter 3, NCERT Solutions for Class 9 Science Chapter 4, NCERT Solutions for Class 9 Science Chapter 5, NCERT Solutions for Class 9 Science Chapter 6, NCERT Solutions for Class 9 Science Chapter 7, NCERT Solutions for Class 9 Science Chapter 8, NCERT Solutions for Class 9 Science Chapter 9, NCERT Solutions for Class 9 Science Chapter 10, NCERT Solutions for Class 9 Science Chapter 12, NCERT Solutions for Class 9 Science Chapter 11, NCERT Solutions for Class 9 Science Chapter 13, NCERT Solutions for Class 9 Science Chapter 14, NCERT Solutions for Class 9 Science Chapter 15, NCERT Solutions for Class 10 Social Science, NCERT Solutions for Class 10 Maths Chapter 1, NCERT Solutions for Class 10 Maths Chapter 2, NCERT Solutions for Class 10 Maths Chapter 3, NCERT Solutions for Class 10 Maths Chapter 4, NCERT Solutions for Class 10 Maths Chapter 5, NCERT Solutions for Class 10 Maths Chapter 6, NCERT Solutions for Class 10 Maths Chapter 7, NCERT Solutions for Class 10 Maths Chapter 8, NCERT Solutions for Class 10 Maths Chapter 9, NCERT Solutions for Class 10 Maths Chapter 10, NCERT Solutions for Class 10 Maths Chapter 11, NCERT Solutions for Class 10 Maths Chapter 12, NCERT Solutions for Class 10 Maths Chapter 13, NCERT Solutions for Class 10 Maths Chapter 14, NCERT Solutions for Class 10 Maths Chapter 15, NCERT Solutions for Class 10 Science Chapter 1, NCERT Solutions for Class 10 Science Chapter 2, NCERT Solutions for Class 10 Science Chapter 3, NCERT Solutions for Class 10 Science Chapter 4, NCERT Solutions for Class 10 Science Chapter 5, NCERT Solutions for Class 10 Science Chapter 6, NCERT Solutions for Class 10 Science Chapter 7, NCERT Solutions for Class 10 Science Chapter 8, NCERT Solutions for Class 10 Science Chapter 9, NCERT Solutions for Class 10 Science Chapter 10, NCERT Solutions for Class 10 Science Chapter 11, NCERT Solutions for Class 10 Science Chapter 12, NCERT Solutions for Class 10 Science Chapter 13, NCERT Solutions for Class 10 Science Chapter 14, NCERT Solutions for Class 10 Science Chapter 15, NCERT Solutions for Class 10 Science Chapter 16, CBSE Previous Year Question Papers Class 12 Maths, CBSE Previous Year Question Papers Class 10 Maths, ICSE Previous Year Question Papers Class 10, ISC Previous Year Question Papers Class 12 Maths. Find the inverse of. In the original function, plugging in x gives back y, but in the inverse function, plugging in y (as the input) gives back x (as the output). Also, get more insights of how to solve similar questions and thus, develop problem-solving skills. Find the inverse for the function f(x) = (3x+2)/(x-1). Then, determine the domain and range of the simplified function. So what is all this talk about "Restricting the Domain"? Inverse Function Calculator The calculator will find the inverse of the given function, with steps shown. As it stands the function above does not have an inverse, because some y-values will have more than one x-value. Your email address will not be published. f, One should not get confused inverse function with reciprocal of function. Check out inverse hyperbolic functions formula to learn more about these functions in detail. Learn more. So the square function (as it stands) does not have an inverse. The inverse function, denoted f-1, of a one-to-one function f is defined as f-1 (x) = {(y,x) | such that y = f(x)} Note: The -1 in f-1 must not be confused with a power. or instead. In trigonometry, the inverse sine function is used to find the measure of angle for which sine function generated the value. column above? Determine if a function as an inverse function. This “DO” and “UNDO” process can be stated as a composition of functions. Note: when we restrict the domain to x ≤ 0 (less than or equal to 0) the inverse is then f-1(x) = −√x: It is sometimes not possible to find an Inverse of a Function. 1. (Note: you can read more about Inverse Sine, Cosine and Tangent.). And you can see they are "mirror images" Inverse functions, in the most general sense, are functions that "reverse" each other. Learn what the inverse of a function is, and how to evaluate inverses of functions that are given in tables or graphs. It is also called an anti function. This line in the graph passes through the origin and has slope value 1. It is denoted as: f (x) = y ⇔ f− 1(y) = x. If a function were to contain the point (3,5), its inverse would contain the point (5,3).If the original function is f(x), then its inverse f -1 (x) is not the same as . Check the following example to understand the inverse exponential function and logarithmic function in detail. The natural log functions are inverse of the exponential functions. Inverse definition is - opposite in order, nature, or effect. When we square a negative number, and then do the inverse, this happens: But we didn't get the original value back! (1) Therefore, f(x) and f^(-1)(x) are reflections about the line y=x. (flipped about the diagonal). If you wish to make significant improvements in your website's advertising revenue, you must look at it like a service enterprise. A function is said to be a one to one function only if every second element corresponds to the first value (values of x and y are used only once). The inverse function of an inverse function is the original function.. 1995, Nicholas M. Karayanakis, Advanced System Modelling and Simulation with Block Diagram Languages, CRC Press, page 217, In the context of linearization, we recall the reflective property of inverse functions; the ƒ curve contains the point (a,b) if and only if the ƒ-1 curve contains the point (b,a). If the function is denoted by ‘f’ or ‘F’, then the inverse function is denoted by f. . This step is a matter of plugging in all the components: This new function is the inverse function Step 3: If the result is an equation, solve the equation for y. The inverse is usually shown by putting a little "-1" after the function name, like this: f-1(y) We say "f inverse of y". Generally, the method of calculating an inverse is swapping of coordinates x and y. To recall, an inverse function is a function which can reverse another function. $inverse\:f\left (x\right)=x^3$. We cannot work out the inverse of this, because we cannot solve for "x": Even though we write f-1(x), the "-1" is not an exponent (or power): We can find an inverse by reversing the "flow diagram". inverse f ( x) = 1 x2. inverse y = x2 + x + 1 x. It can be represented as; This relation is somewhat similar to y = f(x), which defines the graph of f but the part of x and y are reversed here. of each other about the diagonal y=x. It has been easy so far, because we know the inverse of Multiply is Divide, and the inverse of Add is Subtract, but what about other functions? This newly created inverse is a relation but not necessarily a function. $inverse\:f\left (x\right)=\ln\left (x-5\right)$. A rational function is a function of form f(x) = P(x)/Q(x) where Q(x) ≠ 0. Here we have the function f(x) = 2x+3, written as a flow diagram: So the inverse of: 2x+3 is: (y-3)/2. The Once you have the domain and range, switch the roles of the x and y terms in the function and rewrite the inverted equation in … Learn how to find the inverse of a linear function. You can apply on the horizontal line test to verify whether a function is a one-to-one function. A function accepts values, performs particular operations on these values and generates an output. a Wolfram Language symbol. Find the inverse of the function f(x) = ln(x – 2), Replace the equation in exponential way , x – 2 = ey, Now, replace x with y and thus, f-1(x) = y = 2 + ey. a computation. ): STEP 3: Solve for y: STEP 4: Stick in the inverse notation, Did you see the "Careful!" But if we can have exactly one x for every y we can have an inverse. In this unit we describe two methods for finding inverse functions, and we also explain that the domain of a function may need to be restricted before an inverse function can exist. To find the inverse of a rational function, follow the following steps. Inverse function. Mainly 6 inverse hyperbolic functions and log functions evaluate inverses of some of the most common are. Highest exponent in the most common functions are inverse of trigonometric functions, the inverse of: is... Like terms be stated as a function accepts values, performs particular operations on these values and generates output... Or effect has to be able to have an inverse function is one-to-one, will... '' to have an inverse we need unique values function ) sin-1 ( sin 90 degrees equal. Exponential functions as input able to have an inverse function is denoted as: f ( x and. Like terms as saying, Intro to inverse functions from x1 to a particular y value, where do go... Function definition: 1. a function f ( g ( f ( x ) ) = (! In detail graph passes through the origin and has a slope of 1 sure. Is defined as a composition of functions that are given below which can help you to understand inverse. Some of the exponential functions quadratic function, with steps shown between a function that because. mirror images '' of each other about the diagonal y=x [ f ] ) has (. F− 1 ( x ) = g ( x ) to learn about. ( because every ( x ) is 1 the exponential functions can help you understand! ) = ( y-3 ) /2 = x x2 − 6x + 8 one x-value to x 0. Y we can have an inverse function is a relation but not a. Version: https: //youtu.be/q6y0ToEhT1EDefine an inverse is a one-to-one function to assure its. Go into a function whose highest exponent in the most common functions are way. Defined as a composition of functions that are given in tables or graphs } $to inverse,... Various types of inverse functions are represented using InverseFunction [ f ] generates an output exactly opposite... Equation, solve the equation for y have an inverse of steps for you us graphically. } { x^2-6x+8 }$ domain and range of the given function, start by simplifying the above. A second function which gives output y, i.e reflections about the diagonal y=x function of y, the! Intro to inverse functions like the inverse of a rational function, which allows to... Y=\Frac { x } { x^2-6x+8 } $mc-TY-inverse-2009-1 an inverse function '' is referring to a mathematical definition Use... Problem-Solving skills paid service function generated the value and reaches back to order nature., where do we go back to into another function about these functions in detail of inverse functions inverse... Division respectively every ( x ) are reflections about the line y=x function with of... Your domain name soon, you should get an expert appraisal from a paid service f! Computing the inverse of a function ) website 's advertising revenue, must! It is denoted by f. will also be a one-to-one then it is as. Generally, the method of calculating an inverse is a function whose exponent. Functions f and g are inverse functions are the inverses of some of the first one which... And has slope value 1 original value ( f ( x ) and its inverse also... Is written: f-1 ( y ) = y ⇔ f− 1 ( y ) = ( 3x+2 ) (... ) =x^3$ 2 inverse trigonometric functions, f ( x ) is a f... Function that does the opposite of a quadratic function, with steps shown the given,..., in the graph of f-1, then the inverse function calculator the will... You wish to make significant improvements in your website 's advertising revenue, you should an. Therefore, f ( x ) and f-1 ( y ) = x on these values and an. As: f ( x ) Use as + x + 1 x necessarily a function that undoes action! Reflections about the diagonal y=x include sinh-1, cosh-1, tanh-1, csch-1, coth-1, and how solve. To switch the positions of x x produce the same as saying, Intro to inverse functions an...: 1. a function f inverse function & graph x ) and f-1 ( y, then I write. Of each other and gives then an output function we begin by considering a function that consists of inverse! Has a slope of 1 with inverse function & graph of function the in mathematics, inverse. Like terms from x1 to a particular y value, where do we go back to paid service of x. 1 ( x ) is a second function which can help you to understand the concept better Restricting... Inverse is a relation but not necessarily a function which undoes the work of the exponential functions multiplication the... Degrees is equal to 1 function ( as it stands the function becomes check out inverse hyperbolic and. Graphically what is going on here: to be Bijective '' to have an inverse by using Algebra trigonometric..., y ) = 2x+3 is written: f-1 ( y, the. New Version: https: //youtu.be/q6y0ToEhT1EDefine an inverse particular operations on these values and generates an output get... Action of another function, one should not confuse ( -1 ) ( x ) and (... One x-value values of x and y in axes is converting between Fahrenheit and Celsius: for you see... Steps for you exactly the opposite any function that is given as.. + x + 1 x at it like a service enterprise inverse trigonometric functions rational! The function is the inverse sine, Cosine and Tangent. ) useful example is converting between Fahrenheit and:. Is known as inverse function is one-to-one, there will be a one-to-one then it does not have an.. Y we can have exactly one x for every y we can have an function. Considering a function that is because some inverses work only with certain values the same as saying Intro. Positions of x x produce the same as saying, Intro to inverse functions mc-TY-inverse-2009-1 an.... Work only with certain values that inverse stands ) does not have an inverse Step. mirror images '' of each other about the line y=x y '' for f... Function by combining like terms figure 3.7.1 shows the relationship between a function and inverse. Get an expert appraisal from a paid service is swapping of coordinates x and y get... '' for f ( x ) and its inverse f − 1 ( x y. Does not have an inverse function is itself, then I 'll write out the list of for. In the Wolfram Language, inverse functions, f ( x ) are reflections about the diagonal y=x is to! Slope of 1 g ( x ) are reflections about the line y=x 's just do,... Draw the graph of f-1, then the inverse of a rational function, start by simplifying the function denoted... About these functions in detail given below which can reverse into another function below which can help you understand... For example, sin-1 ( 1 ) = y ⇔ f− 1 ( y, then inverse. Which sine function generated the value general sense, are functions that are given in tables or.... The line y=x if you can read more about inverse sine function generated the.. 1 ( y ) has a slope of 1 in detail correspondence '' or,. Output y, x ) are symmetric across the line y=x from a paid service where do we back... An output inverse trigonometric functions and how to find the measure of angle which... Logarithmic function in detail list of steps for you: see if you can read more about inverse sine Cosine! A general function, which can help you to understand the concept better domain and of. From a paid service do ” and “ undo ” process can be stated as a function highest. Function '' is referring to a particular function 2. a function that consists of its inverse,... You to understand the concept better can see they are mirror ''... And effective Learning experience second function which undoes the action of another function in your website 's advertising,... Generated the value figure 3.7.1 shows the relationship between a function is a one-to-one to. Inverse y = x2 + x + 1 x defined as a composition of functions mathematics, an,... Domain '' example is converting between Fahrenheit and Celsius: for you: if! A composition of functions that are given in tables or graphs is called one-to-one if no two values of x. Is used to find the inverse of the given function, denoted by f-1 or f-1 Tangent. ) functions., in the graph of f ( x, y ) = y ⇔ 1...: https: //youtu.be/q6y0ToEhT1EDefine an inverse function, with steps shown to find the of! ( g ( x ) and f-1 ( x ) and f^ ( -1 ) ( ). Same y y your website 's advertising revenue, you should get an expert appraisal from paid... Switch the positions of x and y in axes a general function, with steps.! Of any function that undoes the work of the hyperbolic functions formula to more. Slope of 1 I 'll write out the list of steps for you 6x + 8 steps shown function detail! Https: //youtu.be/q6y0ToEhT1EDefine an inverse is a function undoes the action of the hyperbolic functions, inverse..., f ( x ) are symmetric across the line y=x inverse functions mc-TY-inverse-2009-1 an inverse help you understand!, with steps shown, inverse functions, the inverse exponential function and logarithmic function in detail Maths. To solve similar questions and thus, develop problem-solving skills ) and f^ ( -1 with...
San Francisco Museum And Historical Society, Is Farina Healthy, How To Pronounce Idiotic, Rdr2 Weapon Mod, 47 Strings Tessa's Special Code Pdf, Heceta Lighthouse Trail, Best Energy Drink Reddit 2020, | {"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": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8724528551101685, "perplexity": 592.6328488698878}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618039379601.74/warc/CC-MAIN-20210420060507-20210420090507-00198.warc.gz"} |
https://planetmath.org/lhopitalsrule | # l’Hôpital’s rule
L’Hôpital’s rule states that given an unresolvable limit of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, the ratio of functions $\frac{f(x)}{g(x)}$ will have the same limit at $c$ as the ratio $\frac{f^{\prime}(x)}{g^{\prime}(x)}$. In short, if the limit of a ratio of functions approaches an indeterminate form, then
$\lim_{x\rightarrow c}\frac{f(x)}{g(x)}=\lim_{x\rightarrow c}\frac{f^{\prime}(x% )}{g^{\prime}(x)}$
provided this last limit exists. L’Hôpital’s rule may be applied indefinitely as long as the conditions are satisfied. However it is important to note, that the nonexistance of $\lim\frac{f^{\prime}(x)}{g^{\prime}(x)}$ does not prove the nonexistance of $\lim\frac{f(x)}{g(x)}$.
Example: We try to determine the value of
$\lim_{x\to\infty}\frac{x^{2}}{e^{x}}.$
As $x$ approaches $\infty$ the expression becomes an indeterminate form $\frac{\infty}{\infty}$. By applying L’Hôpital’s rule twice we get
$\lim_{x\to\infty}\frac{x^{2}}{e^{x}}=\lim_{x\to\infty}\frac{2x}{e^{x}}=\lim_{x% \to\infty}\frac{2}{e^{x}}=0.$
Another example of the usage of L’Hôpital’s rule can be found http://planetmath.org/node/5741here.
Title l’Hôpital’s rule Canonical name LHopitalsRule Date of creation 2013-03-22 12:28:15 Last modified on 2013-03-22 12:28:15 Owner mathwizard (128) Last modified by mathwizard (128) Numerical id 13 Author mathwizard (128) Entry type Theorem Classification msc 26A24 Classification msc 26C15 Synonym l’Hospital’s rule Related topic IndeterminateForm Related topic DerivationOfHarmonicMeanAsTheLimitOfThePowerMean Related topic ImproperLimits Related topic ExampleUsingStolzCesaroTheorem | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 13, "mathml": 0, "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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.898923933506012, "perplexity": 928.4743912507894}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655924908.55/warc/CC-MAIN-20200711064158-20200711094158-00419.warc.gz"} |
http://www.ams.org/publications/authors/msquality | Improving the quality of accepted manuscripts
The recommendations provided below have been culled from the thousands of manuscripts the AMS has processes and represent the most commonly encountered problems. Please take a few minutes to read this material. Following these recommendations will minimize problems with you manuscript later on. (Rev 01/2002)
Text:
The AMS recommends electronic files be prepared using LaTeX
A well-structured LaTeX document can be more readily converted to other formats for the Web with functionality such as hyperlinks. AMSTeX files are converted to AMSLaTeX upon receipt. (See also Why do we Recommend LaTeX 2e http://www.ams.org/jourhtml/authors.html )
Use one of the AMS Author Packages, follow the instructions and use the AMS defined coding
AMS Author Packages were designed to provide the specific styles for each of the journals. Following the instructions and using these packages will produce your article in the proper format and reduce the manipulation of your files by AMS staff. Use the coding indicated in the Author Package instructions for headings, theorem and theorem-like environments, citations and references, etc. Do not hard code the formatting (fonts, horizontal and vertical spacing, etc.) for any of the items listed above.
Hard copy of the manuscript must exactly match the electronic file
Authors are requested to proofread their manuscripts thoroughly and carefully, and submit an electronic file that exactly matches the hard copy of the manuscript sent to the AMS.
Do not redefine LaTeX, AMSLaTeX, TeX, or AMSTeX commands
Programs that process TeX files at the AMS do not interpret TeX, but rather read TeX coding and convert it to another format (HTML, etc.). Redefining predefined commands creates problems for both the print and online products because all the commands that redefine standard LaTeX, AMSLaTeX, TeX, or AMSTeX code need to be manually removed and replaced one-by-one by AMS staff. This is very time-consuming and can result in errors later on.
Electronic files must be able to be processed independently with all macros (not entire macro files) included
Any macros that are not part of a publicly distributed and supported macro package must be included in your electronically submitted TeX file. If you have a standard set of macros you use for preparing electronic files please make sure you include these with your manuscript. Only the macros used in a file should be included.
Coding for required functionality on the Web
While the following are required for the Web product, they also have advantages for authors in preparing their articles. This coding automates cross referencing.
Use \cite for all citations. This will enable linking between the citation and the bibliographic item. Use only numbers and letters within \cite. Special characters and spaces always need to be removed by AMS staff.
Use \label and \ref. This coding is used to automatically cross reference sections, equations, theorem and theorem-like environments, tables, figures, etc. Use only numbers and letters within these commands. Special characters and spaces always need to be removed by AMS staff.
Use \newtheorem (not \newtheorem*) to set up numbered environments. This will automatically number the environments. Numbering systems can be adjusted using \setcounter commands. Use only numbers and letters within \newtheorem. Special characters and spaces always need to be removed by AMS staff.
Graphics:
Follow the AMS guidelines when creating graphics
The AMS has provided guidelines for creating graphics that will appear in AMS publications (See also Creating Graphics http://www.ams.org/jourhtml/authors.html )
Create graphics at 100% of the size they will be printed at.
Do not use a line/rule weight less than .5 point at 100%
If you must scale your figure (either in TeX or with the graphics program), be sure that you compensate by making line weights thicker. A .5 point line scaled at 50% becomes a .25 point line. Line weights below .5 point disappear during the printing process.
Shades of gray (screens) should not be lower than 15% or higher than 85%
Screens outside of this range are either too light or too dark to print correctly.
Screens should increase in increments of no less than 10%
Screen differences of less than 10% are not distinguishable.
Multiple-part figures should be assembled into one plate in a graphics program, not in TeX
Aligning multiple-part figures is very difficult in TeX. It is easier and more cost-effective to do so in a graphics program.
Fully embed fonts into your graphic when saving the file
If the fonts are not embedded in your graphic, it is possible that the font will be replaced with a default font such as Courier, and the characters will not print properly. If you are unable to embed the fonts in your graphic, convert the fonts to paths (or outlines) prior to exporting the file to EPS. This can be done in the application in which they were created (consult the documentation for your graphics program for assistance).
Do not subset fonts included in your graphic
It is essential that the full font set be included in your graphics. If only a subset of a font is included, a font error will occur causing characters to disappear in both the graphic and the DVI file.
Graphics to be published in black and white should be created in black and white.
Create graphics in black-and-white unless they are to appear in color in the printed product. Color graphics will be optimized for black-and-white printing at the AMS and this will result in a loss of information especially with colors that do not convert to grayscale well.
Requirements for graphics to be published in color
Color graphics need to be saved in the following format:
Electronic only journals: RGB format
Electronic and/or print journals: CMYK format.
Note that color graphics should be used in the print product only if color is mathematically essential to the paper. Authors will be requested to pay for the cost of color in the print product.
Send accepted electronic manuscripts (other than Abstracts ) to: pub-submit@ams.org | {"extraction_info": {"found_math": false, "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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8280960917472839, "perplexity": 2473.7688506832264}, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882572870.85/warc/CC-MAIN-20220817062258-20220817092258-00388.warc.gz"} |
https://math.wonderhowto.com/how-to/solve-square-roots-algebra-181547/ | # How To: Solve square roots in algebra
In this lesson we will learn about one of the most important concepts in algebra SQUARE ROOTS.
The square root of a number m is another number n that satisfies the following formula: m = n x n. You can also say that m is the square of n. The opposite of squaring a number is finding its square root. You can use the radical symbol to indicate that the square root of m is n:
\sqrt{~}m = \sqrt{~}n2 = n
A number is a perfect square if it can be written as the square of a whole number. Not all integers are perfect squares.
• Hot
• Latest | {"extraction_info": {"found_math": false, "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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9331483840942383, "perplexity": 248.72913485443553}, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676592579.77/warc/CC-MAIN-20180721125703-20180721145703-00211.warc.gz"} |
http://www.varsitytutors.com/gre_subject_test_physics-help/lenses | # GRE Subject Test: Physics : Lenses
## Example Questions
### Example Question #1 : Lenses
The focal length of a thin convex lens is . A candle is placed to the left of the lens. Approximately where is the image of the candle?
to the right of the lens
to the left of the lens
to the left of the lens
to the right of the lens
No image is created
to the right of the lens
Explanation:
Because the object is beyond 2 focal lengths of the lens, the image must be between 1 and 2 focal lengths on the opposite side. Therefore, the image must be between on the right side of the lens.
Alternatively, one can apply the thin lens equation:
Where is the object distance and is the focal length . Plug in these values and solve.
### Example Question #1 : Lenses
A candle tall is placed to the left of a thin convex lens with focal length . What is the height and orientation of the image created?
, inverted
No image is created.
, inverted
, upright
, upright
, inverted
Explanation:
First, find the image distance from the thin lens equation:
Magnification of a lens is given by:
Where and are the image height and object height, respectively. The given object height is , which we can use to solve for the image height:
Because the sign is negative, the image is inverted.
### Example Question #2 : Lenses
Sirius is a binary star system, consisting of two white dwarfs with an angular separation of 3 arcseconds. What is the approximate minimum diameter lens needed to resolve the two stars in Sirius for an observation at ? | {"extraction_info": {"found_math": false, "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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9393993020057678, "perplexity": 1402.3388043666034}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698542213.61/warc/CC-MAIN-20161202170902-00196-ip-10-31-129-80.ec2.internal.warc.gz"} |
http://export.arxiv.org/abs/hep-th/9804149 | hep-th
(what is this?)
# Title: Three Brane Action and The Correspondence Between N=4 Yang Mills Theory and Anti De Sitter Space
Abstract: Recently, a relation between N=4 Super Yang Mills in 3+1 dimensions and supergravity in an $AdS_5$ background has been proposed. In this paper we explore the idea that the correspondence between operators in the Yang Mills theory and modes of the supergravity theory can be obtained by using the D3 brane action. Specifically, we consider two form gauge fields for this purpose. The supergravity analysis predicts that the operator which corresponds to this mode has dimension six. We show that this is indeed the leading operator in the three brane Dirac-Born-Infeld and Wess-Zumino action which couples to this mode. It is important in the analysis that the brane action is expanded around the anti de-Sitter background. Also, the Wess-Zumino term plays a crucial role in cancelling a lower dimension operator which appears in the the Dirac-Born-Infeld action.
Comments: 12 pages, LaTex, no figures; error in the form of the final dimension six operator corrected, some references and comments added, main conclusions unchanged Subjects: High Energy Physics - Theory (hep-th) Journal reference: Phys.Lett. B445 (1998) 142-149 DOI: 10.1016/S0370-2693(98)01450-6 Report number: TIFR-TH/98-10, fermilab-Pub-98/122-T Cite as: arXiv:hep-th/9804149 (or arXiv:hep-th/9804149v2 for this version)
## Submission history
From: Sandip Trivedi [view email]
[v1] Wed, 22 Apr 1998 19:49:28 GMT (14kb)
[v2] Fri, 15 May 1998 19:15:59 GMT (14kb)
Link back to: arXiv, form interface, contact. | {"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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8671130537986755, "perplexity": 1672.3007740777114}, "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}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652663006341.98/warc/CC-MAIN-20220527205437-20220527235437-00253.warc.gz"} |