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https://defeatedelementaryschool.com/prentice-hall-gold-algebra-2-1-4-answers/ | math | 21 Prentice Hall Gold Algebra 2 1 4 Answers
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https://brilliant.org/problems/blind-man-in-an-island/ | math | Blind man in an island
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Ocr terminal paper ocr mathematics c terminal unit higher papers 1 amp service 24/7 enjoy proficient essay ocr terminal paper higher writing and custom writing services provided by. Gcse, gce and aea code of practice (april 2008) ocr gcse mathematics c uses graduated assessment comprising 10 module tests that candidate sits a final paper (known as the terminal. Ocr gcse maths past exam papers and marking schemes for maths 9-1, mathematics a and mathematics b syllabuses, the ocr gcse maths past papers are free to download for you to practice with. | s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583512323.79/warc/CC-MAIN-20181019041222-20181019062722-00493.warc.gz | CC-MAIN-2018-43 | 4,619 | 7 |
https://www.kiasuparents.com/kiasu/question/77425-2/ | math | As ΔABC is similar to ΔDEC, BEC is a straight line and AB//DE.
Since AB:DE = 2:1,
area of ΔABC : area of ΔDEC = 4:1
In ΔDEC, DF : FC = 3:4, so
area of ΔDFE : area ofΔFCE= 3:4
Since ΔDEC = ΔDFE + ΔFCE
If area of ΔDEC = 1, area of ΔDFE = 3/7
Therefore, area of ΔABC : area of ΔDFE = 4 : 3/7 = 4X7 : 3 = 28 :3 | s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655897027.14/warc/CC-MAIN-20200708124912-20200708154912-00129.warc.gz | CC-MAIN-2020-29 | 320 | 8 |
http://ministryoftruthblog.blogspot.com/2008/01/explanation-of-inevitability-principle.html | math | First, for an excellent explanation of the person you have to be to understand all this, please visit this blog
The Inevitability Principle of Blanched Artichokes is the principle that every person will eat blanched artichokes either in this life or the next. And when I say the next life, every thinking person will realize that if you have not eaten blanched artichokes in this life, you have eaten them in the life before this one, by simple logic. The effects of eating blanched artichokes stretch back over your life previous in the case that you did not happen to eat them in that life, and if you did you end up with effects^2 being applied to your life. This is due to the fact that the squared tangent of effects plus the number of lifes concurrently lived by one person (that is, one) is equal to the inverse of the cosine squared of said effects. If r is equal to effects, it is simple to see that the effects of eating blanched artichokes are related to both time and pressure by the elementary equation of pressure times volume being equal to n times r times temperature and that energy is equal to negative R of h being divided by a number, n, when n is 1, 2, 3, 8, 42, or 1789387458301. Because energy is also equal to mass times the speed of light squared, it is easily proved that r is related to the speed of light. One could also prove this by calculating the average warp speed of an antimatter particle and dividing it by the numbers of oscillations per second of the nearest black hole, multiplying by pi times 3, squaring the result, taking the r root of it, and calculating the average sine function of the answer, then adding 2.99792458 times ten to the eighth power, the result of this equation being equal to the speed of light, or c. As you can see from these elementary proofs, the inevitability principle of blanched artichokes has great impact upon our daily existence, although this is not generally known, and was not known to the thinking men until recently. However, before I explain some of the many intricacies of this principle, let me give you an example. Because I know that you are currently under the effects of blanched artichokes, I can calculate what you ate for breakfast based upon the time of sunrise of the 12th of June in AD 38, the most prominent menu item in McDonalds in the year 2011, your average weight in 1984, and the angle of refraction of the cornea of your left eye. You may find that difficult to believe, but if I were to know all these things, I could calculate your breakfast. For instance, since I know that the sunrise of the 12th of June in AD 38 was precisely 6:04:52.4783932, the most prominent menu item in McDonalds in 2011 was the 13-pack of Chicken McNuggets made with 99% real white meat, my average weight in 1973 was approximately -0 pounds, and the angle of refraction of the cornea of my left eye is exactly 92 degrees, 43 minutes, and 75 seconds, by an ingenious process I determine that I had organic plain yogurt with Polaner strawberry jelly mixed in. And, as that was exactly what I had for breakfast, I have thus demonstrated the inevitability principle of blanched artichokes. The inevitability principle of blanched artichokes is an extremely useful principle because it can be used to determine the past based on past, present, and future events. A practical example of this would be that tomorrow I calculated that this paper would be finished. It can be used to calclulate the success of projects before they complete. As well, the extent of the inevitability principle of blanched artichokes is far from being determined as we are just beginning to explore it and our knowledge grows at an exponential rate based on the equation of knowledge being equal to the inverse of cosine squared cofunction plus the hypontenuse of a squared right rhombic triangle divided by the fifth root of the extent of the spacetime fabric plus the time we spend investingating the inevitability principle of artichokes squared. Due to our lack of knowledge about the inevitability principle of blanched artichokes, thinking men often have conversations with other nonthinkers concerning seemingly benign topics that, with the proper knowledge, are used to expand our knowlege of the inevitability property of blanched artichokes. Allow me to finish with another, more complex, example. This evening I had a conversation concerning a memory program, and the person with whom I was having the conversation said, "Mrs. DeStumpf's room". This led me to deduce that Hilary Clinton was not president in 2034. Allow me to explain. I earlier proved r in relation to c, which sets the stage for calculating this. First, I must know the following information: the mass of a proton, 1.6726231 times ten to the negative twenty-seventh power, the exact time of the arrival of an email notifiying me that the Office Live Workspace has started, 13:28:54.3278, the number of pens in my pen box, eight and one marker, the eqilibrium constant of the cell of the reaction Zn(s) + 2Ag+(aq) -> Zn2+(aq) + 2Ag(s), being equal to 1.56V, the Number of Hilary Clinton, 45, the CD Set number of Microsoft Publisher 2000, X04-84021-M, and the my lottery card pick of the day from the Irish Lottery, 2-3-12-19-32-37. The calculation is as follows: because r is related to c, we can calculate that c squared times m (in this case, 1.6726231 times 13:28:54.3278) is equal to e, which is equal to r of h divided by n, in this case being 8. From this we calculate h and insert that into another equation calculating l as h divided by m times v squared plus the square root of X04-84021-M to the r power, resulting in l which we can use to calculate list A as being equal to 1.56 times negative 45 plus A of 0. These equations result in the quantumn numbers of n being equal to 8, l being equal to 5, and m being equal to 13, which is not possible. Therefore Hilary Clinton wasn't president in 2034. | s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917123318.85/warc/CC-MAIN-20170423031203-00068-ip-10-145-167-34.ec2.internal.warc.gz | CC-MAIN-2017-17 | 5,937 | 2 |
http://intlpress.com/site/pub/pages/journals/items/mrl/content/vols/0020/0003/a001/index.html | math | Mathematical Research Letters
Volume 20 (2013)
Multivariable Lubin-Tate $(\varphi, \Gamma)$-modules and filtered $\varphi$-modules
Pages: 409 – 428
We define some rings of power series in several variables, that are attached to a Lubin-Tate formal module. We then give some examples of $(\varphi, \Gamma)$-modules over those rings. They are the global sections of some reflexive sheaves on the $p$-adic open unit polydisk, that are constructed from a filtered $\varphi$-module using a modification process. We prove that we obtain every crystalline $(\varphi, \Gamma)$-module over those rings in this way.
$(\varphi, \Gamma)$-module, Lubin-Tate group, filtered $\varphi$-module, crystalline representation, $p$-adic period, Fontaine theory, reflexive sheaf
2010 Mathematics Subject Classification
11Fxx, 11Sxx, 14Gxx | s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583515555.58/warc/CC-MAIN-20181022222133-20181023003633-00190.warc.gz | CC-MAIN-2018-43 | 818 | 8 |
https://schoolportalng.com/equilibrium-and-centre-of-gravity/ | math | Centre of gravity
Centre of gravity or C.G is the point of balance of a body in which the total weight of the body seems to act through. For regular shaped bodies the C.G is at the geometric centre of the body. For irregular bodies their weight still acts at the centre of the gravity and the law of moments can be used to determine the weight of the body.
The figure below shows a uniform bar of weight ‘W’ and length 80 cm. If a force of 20 N keeps it in balance, determine the weight ‘W’ of the bar.
Taking moments about the pivot, clockwise moments = W × 20 N cm.
Anticlockwise moments = 20 × 30 N cm. Clockwise moments = anticlockwise moments 20 W = 600, therefore W = 30 N.
Parallel forces and equilibrium
For a body to be in equilibrium (neither moving nor rotating), under the action of parallel forces, the following conditions will be satisfied;
- a) The sum of upward forces must be equal to the sum of downward forces.
- b) The sum of clockwise moments equals the sum of anticlockwise moments.
The two are called the first and second condition of equilibrium respectively.
Examples 1. A uniform rod of length 1.0 m is hung from a spring balance as shown and balanced in horizontal position by a force of 1.6 N. Determine; a) The weight of the rod
- b) Reading of the spring balance.
- a) Let the weight of the rod be ‘W’. W acts at 50 cm mark, therefore taking moments about point of suspension, clockwise moments = W × 0.2 Nm = 0.2W Nm.
Anticlockwise moments = 1.6 × 0.3 = 0.48 Nm.
Using the law of moments, then
Anticlockwise moments = clockwise moments
0.48 = 0.2 W, hence W = 2.4 N
- b) Upward forces = downward forces
Downward force = W + 1.6 N
= 2.4 + 1.6
= 4.0 N
Upward force = reading of the spring balance = 4.0 N
- A uniform rod is 1.0 m long weighs 5 N. It is supported horizontally at one end by a spring and the other end rests on a table as shown below. A mass of 2kg is hung from the rod as shown; determine,
- a) Reading of the spring balance
- b) Reaction force, F, from the table.
- a) The 2kg mass and the weight of the rod (5 N) gives clockwise moment while the spring balance provides anticlockwise moments.
Clockwise moments = (2 × 10) × 0.4 + (5 × 0.5) = 10.5 Nm. Anticlockwise moments = S × 1 (reading of the spring balance)
1S = 10.5, hence S = 10.5 N. b) Upward forces = downward forces
Downward forces = (2 × 10) + 5 = 25 N Therefore F+ 10.5 = 25, hence F = 14.5 N.
This is a term which explains how easy or difficult it is for an object to topple over when a force is applied to it. Factors affecting stability,
- a) Base area– the bigger the base area the more the stability.
- b) Position of the centre of gravity – the higher the centre of gravity the less stable the body will be.
States of equilibrium
- Stable equilibrium– if a body is displaced by a small amount of force it returns to its original position.
- Unstable equilibrium– if a body is displaced by a small amount of force it toppled over and does not return to its original position.
- Neutral equilibrium – a body is at rest in whichever position it is placed in i.e. it does not rise or fall when displaced.
Applications of stability
Neutral equilibrium It is used mainly in the design of motor vehicles i.e.
- a) Racing cars – they have a low and wide wheelbase to increase their base area.
- b) Double decker buses – they are manufactured with a low centre of gravity by mounting their chassis and engines as low as possible. | s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711218.21/warc/CC-MAIN-20221207185519-20221207215519-00263.warc.gz | CC-MAIN-2022-49 | 3,472 | 40 |
https://en.frojeostern.com/post/finding-the-missing-length-in-a-figure-calculator | math | | of a right triangle into the top boxes. The calculator will then determine the length of the remaining side, the area and perimeter of the triangle, and all the angles of the triangle.
How to Find the Area and Sides of a Right Triangle
Do it yourself
If we know just two sides of a right triangle, we can use that information to find the third side, the area and perimeter of the triangle, and all the angles of the triangle. Amazing, right? Let’s review how we would find each of those parts.
How to find the Missing Side of a Right Triangle
To find the missing side of a right triangle we use the famous Pythagorean Theorem.
We need to be a little careful that we know which side we’re finding. Right triangles have two legs and a hypotenuse, which is the longest side and is always across from the right angle. When we’re trying to find the hypotenuse we substitute our two known sides for a and b. It doesn’t matter which leg is a and which is b. Then we solve for c by adding the squared values of a and b and taking the square root of both sides.
When we’re trying to find one of the legs we enter the known leg for a and the known hypotenuse for c. Then we solve for b using simple algebra (subtract the value of a squared from both sides, then take the square root of both sides).
How to find the Area of a Right Triangle
To find the area of a right triangle we only need to know the length of the two legs. We don’t need the hypotenuse at all. That’s because the legs determine the base and the height of the triangle in every right triangle. So we use the general triangle area formula (A = base • height/2) and substitute a and b for base and height. So our new formula for right triangle area is A = ab/2.
How to find the Perimeter of a Right Triangle
To find the perimeter, or distance around, our triangle we simply need to add all three sides together. If we only know two of the sides we need to use the Pythagorean Theorem first to find the third side.
How to find the Angles of a Right Triangle
To find the angles of a right triangle we use trigonometry. It’s not as difficult as it sounds. We just need to find one special button on our handheld calculators. To start we’ll need to know all the side lengths, so if we don’t know them already we’ll use the Pythagorean Theorem to find them first.
Once we have all the sides we determine which angle we’re going to find. Then we take the side opposite that angle and divide it by the length of the hypotenuse, which is side c. That will give us a value between 0 and 1. Now we just need to find the ARCSIN button on our calculator, which is often labeled as SIN-1. Finding the ARCSIN of our decimal value gives us our angle. Be sure that the calculator is set for angle mode rather than radian mode.
We can repeat this process to find the other unknown angle in the triangle by once again dividing its opposite side by the hypotenuse and then taking the ARCSIN.
Or we could show off even more triangle knowledge by using subtraction to find it since we know the interior angles of a triangle have to add up to 180°. Subtracting the angle we just found from 180° and then subtracting our known right angle (90°) will give us the third angle too.
This calculator is great for getting all this information from just two sides of a right triangle, but it’s a fun challenge to try to find the sides, angles, area and perimeter on our own without it. Then you can use it to check our answers.
We are given piecewise rectangular figures with some measurements while some other ones are missing. We are required to find these missing lengths. In these figures, it is assumed that all intersecting sides meet in right angles. It is also noted that opposite sides of a rectangle are parallel and equal in length while calculating the missing lengths. We also split the given figure into two rectangles for same purpose.
What is the formula for finding length?
If you have the area A and width w , its length w is determined as h = A/w . If you have the perimeter P and width w , its length can be found with h = P/2−w . If you have the diagonal d and width w , it's length is h = √(d²−w²) .
How do you find the missing length of a triangle on a calculator?
How to find the sides of a right triangle.
If leg a is the missing side, then transform the equation to the form where a is on one side and take a square root: a = √(c² - b²).
If leg b is unknown, then: b = √(c² - a²).
For hypotenuse c missing, the formula is: c = √(a² + b²).
How do you measure the length of a missing leg?
Step 1: Substitute the length of the given leg in for a and the length of the hypotenuse in for c in the Pythagorean Theorem, a2+b2=c2 a 2 + b 2 = c 2 . Step 2: Simplify a2 and c2 , and then subtract the a2 from both sides of the equation. | s3://commoncrawl/crawl-data/CC-MAIN-2024-18/segments/1712296818293.64/warc/CC-MAIN-20240422113340-20240422143340-00686.warc.gz | CC-MAIN-2024-18 | 4,831 | 28 |
https://www.pw.live/question-answer/what-is-the-atomic-mass-of-sodium-56524 | math | . What is the atomic mass of sodium
A: 22 g/mol
B: 23 g/mol
C: 11 g/mol
D: 20 g/mol
Sodium has an average atomic mass of 22.989769 atomic mass units which are generally mentioned as 23 amu. This atomic mass, on the other hand, is the average atomic mass of all isotopes in nature.
Option B is the correct answer. Atomic mass of sodium is 23 g/mol. | s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764501407.6/warc/CC-MAIN-20230209045525-20230209075525-00366.warc.gz | CC-MAIN-2023-06 | 347 | 7 |
http://quant.stackexchange.com/questions/tagged/leverage+hedging | math | Quantitative Finance Meta
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https://www.coursehero.com/file/pq2ljm/Ex-187-Agler-Corporation-currently-manufactures-a-subassembly-for-its-main/ | math | Ex. 187Agler Corporation currently manufactures a subassembly for its main product. The costs per unitare as follows:Direct materials$ 1Direct labor10Variable overhead5Fixed overhead8Total$24Funkhouser Company has contacted Agler with an offer to sell it 4,000 of the subassemblies for$17 each. If Agler buys the subassemblies, $2 of the fixed overhead per unit will be allocated toother products.
7 - 85Incremental AnalysisEx. 187(Cont.)InstructionsShould Agler make or buy the subassemblies? Explain your answer.Ans: N/A, LO: 4, Bloom: AP, Difficulty: Medium, Min: 6, AACSB: Analytic, AICPA BB: Resource Management, AICPA FN: Decision Modeling, AICPA PC:Problem Solving, IMA: Cost ManagementSolution 187(6 min.)Cost to make - cost to buy = incremental cost($24 – $2) – $17 = $5Incremental cost to make = $5 × 4,000 units = $20,000Agler should buy to save $5 per unit.Ex. 188Kuhn Bicycle Company has been manufacturing its own seats for its bicycles. The company iscurrently operating at 100% capacity, and variable manufacturing overhead is charged toproduction at the rate of 60% of direct labor cost. The direct materials and direct labor cost perunit to make the bicycle seats are $8.00 and $9.00, respectively. Normal production is 50,000bicycles per year.A supplier offers to make the bicycle seats at a price of $21 each. If the bicycle company acceptsthis offer, all variable manufacturing costs will be eliminated, but the $30,000 of fixedmanufacturing overhead currently being charged to the bicycle seats will have to be absorbed byother products.Instructions(a)Prepare the incremental analysis for the decision to make or buy the bicycle seats.(b)Should Kuhn Bicycle Company buy the seats from the outside supplier? Justify your answer.Ans: N/A, LO: 4, Bloom: AN, Difficulty: Medium, Min: 15, AACSB: Analytic, AICPA BB: Resource Management, AICPA FN: Decision Modeling, AICPA PC:Problem Solving, IMA: Cost ManagementSolution 188(15–20 min.)(a)Net IncomeMakeBuyIncrease (Decrease)Direct Materials (50,000 × $8)$ 400,000$ -0-$ 400,000Direct Labor (50,000 × $9)450,000-0-450,000Variable Manufacturing Costs ($450,000 × 60%)270,000-0-270,000Fixed Manufacturing Costs30,00030,000-0-Purchase Price (50,000 × $21)-0-1,050,000(1,050,000)Total annual cost$1,150,000$1,080,000$ 70,000(b)The seats should be purchased from the outside supplier. As indicated, the company's netincome would increase $70,000 by purchasing the seats.
7 - 86Incremental AnalysisEx. 189Larkin, Inc. uses 1,000 units of the component NJF1 every month to manufacture one of itsproducts. The unit costs incurred to manufacture the component are as follows:Direct materials$65Direct labor48Overhead96Total$209Overhead costs include variable material handling costs of $10, which are applied to products on the basis of direct material costs. The remainder of the overhead costs are applied on the basis ofdirect labor dollars and consist of 50% variable costs and 50% fixed costs.
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- Managerial Accounting, AICPA FN, AICPA BB, AICPA PC | s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107865665.7/warc/CC-MAIN-20201023204939-20201023234939-00462.warc.gz | CC-MAIN-2020-45 | 3,105 | 7 |
http://www.ask.com/science/crisscross-method-chemistry-370dd57e2f12b2b2 | math | In chemistry, the criss-cross method is a way to write the formulas of ionic compounds. The criss-cross method makes it easier to determine the subscripts for each element in an ionic compound.Know More
Using this method, the number associated with the charge of the first ion is assigned as the subscript of the second ion. The number of the charge on the second ion is assigned as the subscript of the first ion.
Using Li1+ and O2- as an example, the 1 from the Li1+ ion would be assigned as the subscript of O, while the 2 from the O2- ion would be assigned as the subscript of Li. The resulting formula is Li2O. This formula balances out the positive and negative charges of the ions. The criss-cross method does not work if the charges are multiples of each other, or if the charges have already been offset, as is the case with ions that have charges of 2+ and 2-.
If the criss-cross method does not work, use the traditional method of writing a formula. First, determine the ions present in the compound. Magnesium nitride has magnesium and nitrogen ions. Then write the symbol for each ion with the charge above it. Positively charged ions should be written first. For magnesium nitride, the ions are Mg2+ and N3-. The next step is to determine the number of ions needed to balance out the charges. Three magnesium ions are needed to balance out two nitrogen ions in magnesium nitride. Finally, write the number of ions needed for each element as a subscript after the element's symbol. The correct ionic formula for magnesium nitride is Mg3N2 using this method.Learn more about Chemical Equations
In chemistry, Cr OH 3 is the chemical formula for a molecule containing one atom of chromium, one atom of oxygen and three atoms of hydrogen. When written with notation as Cr(OH)3, the formula refers to the compound chromium(III) hydroxide, also known as chromic hydroxide.Full Answer >
Sodium borohydride (NABH4) is a reducing agent used in organic chemistry. Commonly, reactions use sodium borohydride to reduce aldehydes and ketones to alcohols. The agent is not successful in reactions attempting to reduce esters, amides or carboxylic acids, and mercuration reactions use sodium borohydride in the second step.Full Answer >
The primary chemical compounds in Coca-Cola are sodium cyclamate, acesulfame potassium, and E150D, which is a food coloring. Sodium cyclamate and acesulfame potassium are both sugar substitutes, with sodium cyclamate being used more predominantly because it is significantly sweeter than natural sugar.Full Answer >
In math, chemistry and physics, solving for indicated variables in different formulas involves having to manipulate the formula equation until it is in the desired form. An example of this is the area formula for a triangle given as A = 1/2 (b x h), where a student needs to solve this formula in terms of the variable "h" that represents height.Full Answer > | s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701167599.48/warc/CC-MAIN-20160205193927-00134-ip-10-236-182-209.ec2.internal.warc.gz | CC-MAIN-2016-07 | 2,910 | 8 |
http://hourofwolves.org/?view=armies&which=dndFigs&pic=65 | math | Skeleton - Posted: February 8, 2010 - Manufacturer: Reaper Miniatures
This one sports a head carefully removed from a chaos tank spike spru... I went with it because of the hair... It's hair, but it's not a top knot! The only thing that bugs me about this is the apparent eye in the head... Eye's don't stick around in corpses, they're the first things to go... So, I went with the onyx stone in the eye socket idea for this, considering that is what is required to raise skeletons in D&D 3.0! | s3://commoncrawl/crawl-data/CC-MAIN-2023-23/segments/1685224647459.8/warc/CC-MAIN-20230531214247-20230601004247-00740.warc.gz | CC-MAIN-2023-23 | 505 | 2 |
https://scoodle.co.uk/questions/in/maths/what-is-the-area-of-a-sphere | math | What is the area of a sphere?
The area of a sphere is actually a surface area as a sphere is a 3 dimensional object. The area of a sphere is equal to 4 x pi x the radius squared. It's derivation is very interesting and is done by using multi sided polygons inside the sphere, further individual research is recommended.
Four thirds x pi x radius cubed is the volume of a sphere and 4 x pi x radius squared is the formula for the area of a sphere
Get an answer in 5 minutes from expert tutors at Oxford, Cambridge, Imperial and more. | s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347439213.69/warc/CC-MAIN-20200604063532-20200604093532-00489.warc.gz | CC-MAIN-2020-24 | 532 | 4 |
https://www.librarything.com/tag/linear+regression | math | Tag: linear regression
Includes: linear regression, Linear Regression, Regression (Linear), regression (linear), Linear regression (what?)
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LibraryThing members can help translate tags into the languages of all LibraryThing sites. Read more about it here. Review member translations here. | s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257647567.36/warc/CC-MAIN-20180321023951-20180321043951-00304.warc.gz | CC-MAIN-2018-13 | 816 | 11 |
http://pesla.blogspot.com/2011/06/super-virial-proto-grounding.html | math | Wednesday, June 15, 2011
Super-Virial Proto-Grounding L. Edgar Otto June 15, 2011
I present here a rather simple description of the ground of being as physics, perhaps too simple as in the computations to be misleading or confusing. But it is a more relaxed view with greater ways to see the degrees of freedom and nature of symmetries.
Viriality is the what we call those spaces that consider duplications or halving of things for stable orbits that essentially establishes three space and the ways that forces and structures appear in the more relaxed physics of the world. It is the basis of our ideas of super symmetry, that is of a greater generalization of how the world works than just the low dimensions of three and four space. This is not to say that these dimensions do not have an essential role, but so do the higher dimensions. Thus there are analogs where they apply, and regardless of scale or what may insure a finite mass for some dynamic object, which are as simple as multiples of integers we regard as powers.
At first I thought I could map the double quaternion explanation of Rowlands to my quasic grid- but this is not simply the case. Nor can we accept but one form of the intelligible logic of things where a different form may conflict or be irrelevant to our comprehensive and thus exclusive system. The appeal to some such logic may break down even if it does not appear to do so- and where we may hold logic itself as only designed as intelligible the question of its general validity, of thought itself, can just as well be a reassurance of the freedom to explore by intellect our reality.
Some ideas we have heretofore questioned, in light of these expanded notions seem to have a solid basis after all- at least to explain why some idea worked but was not justified save only by explanations just as unclear or even bizarre.
If we have a quasic grid, with many states that we can interpret as fractal like, we can ask, as with computer drawing, what is the grid that underlies it all? That grid I have called the epsilon-delta honeycomb. It is there that we can map the 32 x 32 elements of Rowlands and Dirac's algebra. The order is strait foreward and linear as if the square roots, being equal of the quasic grid- that is all the Cartesian and ultimately Euclidean concepts we extend to ideas like holographics.
Of course we can just as well impose or suggest such ideas in either design or division of the space for the order is distributed over a plane in the quasic grid even though from some perspective the factors contain clear and equal binary information. It is clear, in many cases, that in physics, the essential viriality that doubles some things as they reflect the difference in potential and kinetic energy, or say the phase of spin of some objects, leads to some mistakes in the physical calculations- Consider the bending of light in a gravitational field by Einstein and the test of it at eclipse by Eddington to make this correction. It is moreover a core property of numbers that in the end justifies Eddington's Plus one Idea. Number theorists would do well to take advantage of the view presented here so as to double such values of corresponding orders which we can say is a match between these views of spaces- ones that may tell us something of number theory itself in terms of numbers as well of the general notion of dimensions and measure.
Poet that I have come to accept if not fancy myself to be I am at a little loss of clear terms here. So let us from one side of the mirror of notions imagine the two divisions of general space as the HF (holographic) and FX or (fractal) for now but that only a part of the picture (I mean one has to imagine a world of mirrors but there are more than two sides on which to be on a side of something). I suspect it turns our that the HF x FX related or divided into the general quasic field amounts to 2^2n +1 which is what we should expect applying this to a advanced numerical view of physics. But this is not general enough and does not take advantage of all that is so far known in number theory.
We can imagine terms like Ordinal and Anti-Ordinal, Cardinal and Counter Cardinal to apply as well to the general archetypal concepts here. While we say that even in Maxwell there can be residual neutral flow we can also say there is residual neutral fixed singularities or points of rest. Yet is not established that such singularity complexes or where they apply to a field idea has to be necessarily dependent on the general grand or great-grand ideas of unification theories. It may be that this too is a quasic relaxation as a relation in the general reality with its consequent restrictions and new freedoms for complexity and even transcendence of that now real.
So since there is no set epsilon-delta honeycomb all is a general grid of quasic space without the fixed quasicity. That is one can think of hierarchies or not or one can reject them. The same for ideas of universe and multiverse. So we can imagine a fixed complex of singularities when the quasic span around it expands in depth or we can imagine the internal states of such a singularity complex growing in a details and definition. Or in some cases the physical object may become its inverse just as surely as our notions may be limited or expand in the depth and span of our theoretical and perceptive views. A singularity may be a given or arise or it may be ultimately independent of the general states, intermittently a "creative" space.
An associahedron of simplexes seemingly limiting dimensions and as dynamic kinetically is like a rocket ship thru space and appears to self sustain in its motion. Of course the purely more fractal like spaces, more like falling and less like superficial surface quantum theory, can be dynamically seen as well. Our notions and the universe may have conflicting dynamics as well as where they meet in concepts.
Thus: And as formal as I now present the work done (the other work has applications hopefully soon to be on a domain somewhere and that can use these ideas in general as sets of color notations for the combined general idea of spaces and connectivity)
I suggest this simple relationship of integers of which I do invite exploration by others due to the snow blind simplicity that may occur in simple clear understanding.
16^2 or 256 is a natural addition in combining the HF and FX spaces into one unified space, (or rather recombining them). To this we add its virial doubling or 512. Thus we get 768 (that is twice the rotations and inversions of a tesseract)
BTW we can consider Galois and the equations of the fifth degree of icosahedra and the like and apply them to either space alone but this does not ground the total physics in ways that justifies the lack of or existence of hierarchical ideas. Not to say these theories do not have a place or use as part of the bigger picture.
Now we add to this 256 not just the virial doubling but binary powers of it to get these analogs to virialty:
256 + 1024 = 1280
256 + 4096 = 4352
256 + 16384 = 16640 and so on...
which makes the sequence when summed or divided 2 5 17 65 257 and so on.
Now 768 + 1280 = 2048 which is a binary number in the low viriality levels. We of course can divide numbers by the group numbers and keep such things in mind also, for example 1152 + 128 = 1280, 1152 being the group of the 24 cell in 4D.
* * *
Philosophic Musings just before this Post:
An essence or substance cannot be deduced as existing reality by assumption it a given or emptiness, nor a necessity by chiral or other distinguished contrasting mechanisms, where it can be a "gray" blend unclear as 1 or 0.
Yet, at least by experience and observation we can make the case so assert partial substance as a totality in relation to the obvious being or something neutral as comprehensive over reality.
The physical difference between exist or not exist may be ambiguous (as if a higher uncertainty or as problematic in an intelligible way or not.) and still be generally absolutely neutral as a vacuum or field, independent of the reality as a fact or persistence both pragmatically and non-necessary- thus a working resolution of paradoxes.
To the existent some essence does not exist (including the virtual multiverse- things that may exist may be unified higher or be independent in diffuse unity and freedom unto transcending of known systems.
BTW we can map the epsilon delta notations on the grid as powers of 5 (thus in groups such powers relate to more general spaces and polyhedral or group structures)
As a single quasic order (much like certain fractal fillings of a plane.
For example: 1,1 ; 1,4 ; 4,1 ; 4,4 ; 1,7 ; 3,6 ; 4,7 ; would map to 0 ;5; 10; 15; 20; 25; 30; ...
In the Condenser Mechanisms the color shift may interchange 0 and 1, that is a wild card * can equal 1 or 0 in all its meanings such as exist or not, possible or impossible... This may also explain beyond the color shifts and ideas of creation and annihilation in quantum theory hierarchies of light and knowing (in some interpretations) a much clearer grounding of this experience of consciousness.
As far as the use of color goes, the combining of primary colors in say the hex notation is intelligible in the 3D;4D grids but its use as a labeling has a relaxed way of applying the same symbols if we keep it in mind where distinctions are meant.
Footnote: Of course computer graphics have a new depth of applications of numbers and are thus intelligible- it strikes me that my habit of utilizing the 80 x 80 gif range as a limit to drawing avatars on the philosophychatforum was useful in the greater understanding of such numerical and digital relations in what at first may appear as non-linear and so diffused levels of mathematical explorations.
* * *
Footnote: It would appear (in the reply to Ulla's p-adic question here) that some answers are there to expanded the grounding of Pitkanen's and other ideas including the relation to things in braid space formalism, and especially spinor theory, have anticipated his post just after mine. Of course it would have been better to talk directly instead of interpreters pro and against between us. So may all the theoreticians come closer to the general breakthroughs needed for this enquiry. I continue to post because of the frayed ends and stray statements needed for closure and a conclusion.
* * * | s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267864364.38/warc/CC-MAIN-20180622065204-20180622085204-00389.warc.gz | CC-MAIN-2018-26 | 10,408 | 37 |
https://www.law-dictionary.org/definitions-c/copy | math | 2. Copies cannot be given in evidence, unless proof is made that the originals, from which they are taken, are lost, or in the power of the opposite party; and in the latter case, that notice has been given him to produce the original. See 12 Vin. Abr. 97; Phil. Ev. Index, h. t.; Poth. Obl. Pt. 4, c. 1, art. 33 Bouv. Inst. n. 3055. 3. To prove a copy of a record, the witness must be able to swear that he has examined it, line for line, with the original, or has examined the copy, while another person read the original. 1 Campb. R. 469. It is not requisite that the persons examining should exchange, papers, and read them alternately. 2 Taunt. R. 470. Vide, generally, 3 Bouv. Inst. n. 3106-10; 1 Stark. R. 183; 2 E. C. L. Rep. 183; 4 Campb. 372; 2 Burr.1179; B.N.P.129; 1 Carr. & P. 578. An examined copy of the books of unincorporated banks are not, per se, evidence. 12 S. & R. 256. See 13 S. & R. 135, 334; 2 N. & McC. 299. | s3://commoncrawl/crawl-data/CC-MAIN-2023-50/segments/1700679100016.39/warc/CC-MAIN-20231128214805-20231129004805-00735.warc.gz | CC-MAIN-2023-50 | 933 | 1 |
https://www.physicsforums.com/threads/u-substitution-problem.541366/ | math | 1. The problem statement, all variables and given/known data 1~~0 (x^7)/sqrt(4-3x^16) dx 3. The attempt at a solution There really isn't one. I am as clueless as one can get in calculus. I know that I am supposed to substitute the variable "u" in for a section of this problem, but really don't know anything after that. I understand derivatives but that is about it. If someone could walk me through this problem it'd be great. | s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247480240.25/warc/CC-MAIN-20190216085312-20190216111312-00599.warc.gz | CC-MAIN-2019-09 | 428 | 1 |
https://xrust.com/math-puzzle-for-1-minute/ | math | This puzzle conquered the Internet. If you have a free moment and want to train your brain, then consider this math puzzle.
As you know, math puzzles help develop patience and logical thinking, and this is a great opportunity to once again see how smart you are.
And the task is this: you need to find a pattern in these numbers and determine which number is missing.
To solve this problem, you will need to determine the logic by which the numbers are arranged. The answer, in fact, lies right on the surface, try to look at this circle through the eyes of a child!
Explanation will be below.
Don't rush to look at the answer, you will be able to come up with the right solution yourself!
Guessed it? Well done!
Correct answer: 5
And the solution is this: the sum of the numbers standing separately in each quadrant, is equal to the difference of numbers standing together in the same quadrant. Now everything will become clear:
Upper circle on the right:
Lower circle on the right:
Upper circle left:
And bottom circle left:
Accordingly, the sum is 2+?=7. It turns out 2+5=7
Agree that this puzzle is a great opportunity to stir the mind!
What is your IQ: a short IQ test, see here and another one here | s3://commoncrawl/crawl-data/CC-MAIN-2023-40/segments/1695233510334.9/warc/CC-MAIN-20230927235044-20230928025044-00097.warc.gz | CC-MAIN-2023-40 | 1,204 | 16 |
https://picklelakehotel.com/and-book/1276-algebra-1-workbook-prentice-hall-answers-224-985.php | math | Prentice Hall Mathematics Algebra 1 Workbook AnswersThese Equations Worksheets are a good resource for students in the 5th Grade through the 8th Grade. No need to wait for office hours or assignments to be graded to find out where you took a wrong turn. Algebra II Chapter 2. Relation vs functions in math free pdf's with answer keys on all the topics normally covered in Algebra 1 and Algebra 2 Algebra 1 Concepts and Skills Are you confused about what is taught in an Algebra 1 curriculum? This page will give you an idea of the Algebra 1 concepts and skills that are taught and the order in which they are presented. Mathematics I — Unit 1: Function Families. It's easier to figure out tough problems faster using Chegg Study.
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Algebra 2 chapter 5 test answers
Become a MathHelp. Is he correct. Mathematics - Grade 8 Bennet, et al. Our answers explain actual Algebra 2 textbook homework problems.Class Syllabus Quiz Like the actual Frankenstein, piecewise functions are more vulnerable if you break them into algrbra constituent parts. Intermediate Algebra. Mathematics - Course 1 Bennet, et al.
Graph piecewise-defined functions and absolute value functions. Pre-Algebra Charles, et al. Cpm Algebra 1 Answer Key - fullexams? Carefully graph each of the following.
If you have questions on prntice particular problem, please email me and I will send you a video solution. Write an equation, in slope-intercept form. Yes; the length of a rectangle that has an area of cm2 is 32 cm. All values of x correspond to a point on the graph.
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Activity 1. Math Connects - Course 2 Bailey, et al. Algebra 2 Help Click your Algebra 2 textbook below for homework help. CPM 1.Topics to be integrated thoughout the year. Shed the societal and cultural lagebra holding you back and let free step-by-step Algebra 2: A Common Core Curriculum textbook solutions reorient your old paradigms. Our stand-alone curriculum has everything your child needs, et al. Math Power 10 Knill.
Algebra 1 Smith, et al. Sample answers are given. No two students learn exactly the same way at the same time. It's workbooj to figure out tough problems faster using Chegg Study. | s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703514796.13/warc/CC-MAIN-20210118123320-20210118153320-00242.warc.gz | CC-MAIN-2021-04 | 3,010 | 11 |
http://repository.aust.edu.ng/xmlui/handle/1721.1/6030 | math | Height and Gradient from Shading
Unknown author (1989-05-01)
The method described here for recovering the shape of a surface from a shaded image can deal with complex, wrinkled surfaces. Integrability can be enforced easily because both surface height and gradient are represented. The robustness of the method stems in part from linearization of the reflectance map about the current estimate of the surface orientation at each picture cell. The new scheme can find an exact solution of a given shape-from-shading problem even though a regularizing term is included. This is a reflection of the fact that shape-from-shading problems are not ill-posed when boundary conditions are available or when the image contains singular points. | s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296946535.82/warc/CC-MAIN-20230326204136-20230326234136-00403.warc.gz | CC-MAIN-2023-14 | 734 | 3 |
https://books.google.com.ni/books?id=CULvAAAAMAAJ&q=characteristic&dq=editions:NYPL33433076086341&lr=&output=html_text&source=gbs_word_cloud_r&cad=5 | math | Partial Differential Equations
Springer, 1982 - 249 páginas
This book is a very well-accepted introduction to the subject. In it, the author identifies the significant aspects of the theory and explores them with a limited amount of machinery from mathematical analysis. Now, in this fourth edition, the book has again been updated with an additional chapter on Lewy 's example of a linear equation without solutions.
Resultados 1-3 de 77
Obviously only the direction of the vector , the characteristic direction , matters for the P.D.E. ( 4.2 ) . Since ( ux , Uy , - 1 ) constitute direction ...
we know from the theory of O.D.E.s that through each point of N there passes exactly one characteristic curve . There is a 2 - parameter family of ...
+ Alternately a single characteristic surface S gives rise to a partial differential equation , when described by an explicit equation $ ( x1 , ...
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https://www.physicsforums.com/threads/circular-motion.230411/ | math | A toy with a mass of 100g is going through a loop-the-loop with a diameter of .5m. How fast must the car be moving at the top of the loop in order to not fall.
The Attempt at a Solution
At the top of the loop, there is a normal force (Fn) and the force of gravity (Fg). I'm not sure how to do this. Do I neglect one of the forces? Set one to zero, or what? | s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711637.64/warc/CC-MAIN-20221210005738-20221210035738-00592.warc.gz | CC-MAIN-2022-49 | 356 | 3 |
http://www.scribd.com/doc/54837532/AQA-MM1AW-W-QP-JAN06 | math | General Certificate of EducationJanuary 2006Advanced Subsidiary Examination
MATHEMATICS MM1A/W Unit Mechanics 1A
Monday 16 January 2006 9.00 am to 10.15 am
For this paper you must have:
an 8-page answer book
AQA booklet of formulae and statistical tablesYou may use a graphics calculator.
Time allowed: 1 hour 15 minutes
Use blue or black ink or ball-point pen. Pencil should only be used for drawing.
Write the information required on the front of your answer book. The
for this paper is AQA. The
All necessary working should be shown; otherwise marks for method may be lost.
answer to questions requiring the use of calculators should be given to three significantfigures, unless stated otherwise.
, unless stated otherwise.
The maximum mark for this paper is 60.
The marks for questions are shown in brackets.
Unit Mechanics 1A has a
written paper and coursework
Unless stated otherwise, formulae may be quoted, without proof, from the booklet. | s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375096504.12/warc/CC-MAIN-20150627031816-00125-ip-10-179-60-89.ec2.internal.warc.gz | CC-MAIN-2015-27 | 947 | 18 |
https://www.tsijournals.com/abstract/mercury-in-a-quantumgravitational-field-16206.html | math | Mercury in a Quantum-Gravitational FieldAuthor(s): Paul R. Gerber
A scalar field theory of gravitation yields the correct perihelion precession of Mercury’s orbit around the sun. We present the corresponding theory and numerical calculations for the relativistic Kepler problem. Furthermore, it turns out that the concept of gravitational potential cannot be maintained in the relativistic case because acceleration becomes velocity-dependent. In addition, in extreme cases, like large distances from a gravitational center or for high velocities of a body, the gravitational forces become also dependent on the direction of the body’s velocity. | s3://commoncrawl/crawl-data/CC-MAIN-2023-50/segments/1700679100484.76/warc/CC-MAIN-20231203030948-20231203060948-00538.warc.gz | CC-MAIN-2023-50 | 649 | 2 |
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Here, we will show you how to work with Condensing logarithms solver. So let's get started!
The Best Condensing logarithms solver
Keep reading to understand more about Condensing logarithms solver and how to use it. Solving by square roots is a mathematical process for finding the value of a number that, when squared, equals a given number. For example, the square root of 9 is 3, because 3 squared (3 x 3) equals 9. In general, the square root of x is equal to the number that, when multiplied by itself, equals x. Solving by square roots can be done by hand or with the help of a calculator. The process involves finding the value of one number that, when multiplied by itself, equals the given number. This value is then used to determine the answer to the original problem. Solving by square roots is a useful tool for solving many mathematical problems.
When it comes to solving math problems, there is no one-size-fits-all solution. The best approach depends on the nature of the problem, as well as the skills and knowledge of the person solving it. However, there are a few general tips that can help make solving math problems easier. First, it is important to take the time to understand the problem. What is being asked for? What information is given? Once you have a clear understanding of the problem, you can begin to consider different approaches. Sometimes, visual aids such as charts or diagrams can be helpful in solving math problems. Other times, it may be helpful to break the problem down into smaller steps. And sometimes, simply taking a step back and looking at the problem from a different perspective can make all the difference. There is no single right way to solve math problems. However, by taking the time to understand the problem and trying different approaches, it is usually possible to find a solution that works.
A parabola solver is a mathematical tool that can be used to find the roots of a quadratic equation. Quadratic equations are equations that have the form ax^2 + bx + c = 0, where a, b, and c are constants. The roots of a quadratic equation are the values of x that make the equation equal to zero. A parabola solver can be used to find these roots by inputting the values of a, b, and c into the tool. The parabola solver will then output the roots of the equation. Parabola solvers can be found online or in mathematical textbooks.
Solving for a side in a right triangle can be done using the Pythagorean theorem. This theorem states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. This theorem can be represented by the equation: a^2+b^2=c^2. In this equation, c is the hypotenuse and a and b are the other two sides. To solve for a side, one would rearrange this equation to isolate the desired variable. For example, to solve for c, one would rearrange the equation to get c^2=a^2+b^2. To solve for a, one would rearrange the equation to get a^2=c^2-b^2. Once the equation is rearranged, one can then use basic algebraic techniques to solve for the desired variable. In this way, the Pythagorean theorem can be used to solve for any side in a right triangle.
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Love this app I was failing algebra and my teacher wasn’t really good at his job and I couldn't find any good apps so I gave up but I realized the end of the year was coming and I got desperate so I tried this app and my grades skyrocketed!!! More apps should be like this for different subjects!!! Thank you so much!!!!
Great app! I use it quite a lot for Algebra. The equations need to be written clearly in order to be recognized, but you can always edit them with the calculator. As of now, it doesn’t solve word problems, which I hope will be an added feature in the nearest future. It explains the steps well enough, and even better if you have the plus version which you'll have to purchase. I recommend it. | s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710719.4/warc/CC-MAIN-20221130024541-20221130054541-00325.warc.gz | CC-MAIN-2022-49 | 4,001 | 10 |
http://layerjunkie.com/elementary-linear-algebra-10th-edition-textbook-solutions-55/ | math | Free step-by-step solutions to Elementary Linear Algebra () – Slader. Solutions Manual for Larson/Flavo’s Elementary Linear Algebra, 6th Edition Elementary Linear Algebra textbook solutions reorient your old paradigms. Elementary Linear Algebra: Applications Version 10th Edition. by . Story time just got better with Prime Book Box, a subscription that delivers Student Solutions Manual to accompany Elementary Linear Algebra with Applications, 10e. WileyPLUSTM This isWiley’s proprietary online teaching and learning environment that integrates a digital version of this textbook with instructor and student.
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Functions Average rate of change: Irrational numbers Proofs concerning irrational numbers: Factorization Linwar in factoring quadratics: Solving inequalities Two-step inequalities: Algebra foundations Combining like terms: Solving inequalities One-step inequalities: Properties of exponents rational exponents: Intro to slope Slope. Simplifying square roots review Simplifying square roots.
Systems of equations Systems of equations word problems: Average rate of change review Average rate of change ecition problems. Factorization Evaluating expressions with unknown variables: Linear word problems Textbkok linear functions: Systems of equations Solving any system of linear equations: Working with units Word problems with multiple units: Factorization Factoring polynomials by taking common factors: Factorization Factoring polynomials with quadratic forms: Solving equations Linear equations with parentheses: Arithmetic sequences review Constructing arithmetic sequences.
Introduction to inequalities with variables: Functions Maximum and minimum points: Click here to start or continue working on the Algebra I Mission. Functions Intervals where a function is positive, negative, 10tb, or decreasing: Quadratics The quadratic formula: Factorization Factoring quadratics intro: Geometric sequences review Constructing geometric sequences.
Graphs of absolute value functions: Slope-intercept form review Writing slope-intercept equations.
If you’re seeing this message, it means we’re having trouble loading external resources on our website. Systems of equations Equivalent systems of equations and the elimination method: Algebra foundations Division by zero: Functions Inputs and outputs of a function: Interpreting linear functions and equations: Linear word problems Constructing linear models for textgook relationships: Quadratics Solving quadratics by taking square roots: Algebra foundations Writing algebraic expressions introduction: Linead equation word problems: Systems of equations Solving systems of equations with substitution: Forms of two-variable linear equations: Sequences Introduction soljtions geometric sequences: Manipulating expressions with unknown variables: Multiplying monomials by polynomials: Systems of equations Number of solutions to systems of equations: Functions Interpreting function notation: Algebra foundations Introduction to equivalent algebraic expressions: Functions Functions and equations: Algebra foundations Substitution and evaluating expressions: Did you realize that the word “algebra” comes from Arabic just like “algorithm” and “al jazeera” and “Aladdin”?
And what is so great about algebra anyway? Sequences Constructing arithmetic sequences: Solving equations Linear equations with variables on both sides: Perfect squares Factoring quadratics: Algebra foundations Introduction to variables: | s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514572896.15/warc/CC-MAIN-20190924083200-20190924105200-00266.warc.gz | CC-MAIN-2019-39 | 3,664 | 12 |
https://www.hackmath.net/en/math-problem/431 | math | Calculate the thickness of the page of the collection of examples from mathematics, where 4.87 cm thick and has 451 pages.
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Youth track from Hronská Dúbrava to Banská Štiavnica which announced cancellation attracted considerable media attention and public opposition, has cost 6.3 euro per capita and revenue 13 cents per capita. Calculate size of subsidies to trip group of 28 s
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Aircraft Boeing 767 lose both engines at 42000 feet. The plane captain maintain optimum gliding conditions. Every minute, lose 1910 feet and maintain constant speed 211 knots. Calculate how long it takes to plane from engine failure to hit the ground. Cal
One meter of the textile was discounted by 2 USD. Now 9 m of textile cost as before 8 m. Calculate the old and new price of 1 m of the textile.
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Swimming pool is long 110 m and 30 m wide. The plan of the city is shown as a rectangle with area 8.25 cm2. What scale is the city plan?
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When gas consumption, the consumer may choose one of two rates: rate A - which pays 0.4 € per 1 m3 of gas a flat monthly fee of 3.9 € (regardless of consumption) rate B - which pays 0.359 € per 1 m3 of gas a flat monthly fee of 12.5 € From what monthly
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In 2006, the company had a loss of 11100 Euros. Two years later he was already in profit 50000 Eur. Calculate what percentage of the company increased profits in this two years. | s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107890108.60/warc/CC-MAIN-20201026002022-20201026032022-00067.warc.gz | CC-MAIN-2020-45 | 4,269 | 36 |
https://www.ikaratelive.com/byesimulator.html | math | Quick reference on they bye system.
An easy way to calculate byes is to use the rule of 4 and 8.
If a division has 4 or less competitors then the number of byes will be 4 minus the number of competitors.
If a division has more than 4 competitors then the number of byes will be the next interval of 8 minus the number of competitors.
So, if you have 3 fighters then there is 1 bye, if you have 5 fighters then there are 3 byes. If you have 10
fighters then you have 6 byes.
Use this tool to help you understand the bye system.
The bracket works left to right, the "fighter number" would be the chip that the competitor draws before the division begins.
The light colored number in between the brackets is the match number. The match number changes based on the number of byes because
a "bye" is considered an automatic win in that round, so any automatic win would eliminate the match number.
Use intervals of 4 and 8 to test this out, then use odd numbers to see it work.
© iKarateLive is owned and operated by United Dojo Products, LLC. All rights reserved. | s3://commoncrawl/crawl-data/CC-MAIN-2024-18/segments/1712296818293.64/warc/CC-MAIN-20240422113340-20240422143340-00435.warc.gz | CC-MAIN-2024-18 | 1,060 | 12 |
https://anosluzdeserrano230.com/blog/how-to-calculate-area-of-a-wall/ | math | How To Calculate Area Of A Wall. The height of the wall = 10 feet. You will have to assign different materials to each side of the wall (ex.
How Many Tiles Fit The Wall Space (Functional Maths youtube.com
Note:the amount of coverage of one bundle of batts differs by manufacturer. There exist more distinctions and classifications for different types of trapezoids. but their areas are still calculated in the same manner using the following equation: Our normal window is 12 square feet (1.11 square meters).
Next. use the same formula to record the individual area of windows and doors. Multiply the length times width to get the total area of each wall.
Add the figures into the spaces for each wall to calculate the total linear amount. Calculate the number of windows in your room:
If you are doing a rough estimate. you can round the wall length up to 35 feet. multiply this number by 7 (the height) and divide the result by 2; Note:the amount of coverage of one bundle of batts differs by manufacturer.
The formula to compute the surface area of walls in a room is: Multiply the length times width to get the total area of each wall.
If you have the wall measurements. calculate area by multiplying the walls length by its height. then compare that with the coverage per can. Calculate the surface area of the wall:
Once you have all of these measurements down. subtract the area of the windows and doors from the total area of the wall. To calculate the area of a wall. use the standard formula of (length) x (width) = area.
To Calculate The Area Of A Wall. Use The Standard Formula Of (Length) X (Width) = Area.
Atypical architecture and design. however. can result in having triangular. circular or trapezoidal shapes combined with the rectangle. We know. surface area = length x height 35 * 7 = 245 ÷ 2 = 122.5 square feet or better yet. round the 7 feet 1 5/8 to 7.5 feet;
W = Width Of The Room;
Since walls can be tall. you might want to ask a friend or neighbor for help holding the measuring tape. There exist more distinctions and classifications for different types of trapezoids. but their areas are still calculated in the same manner using the following equation: Multiply the resultant area with number 2.
We Will Now Show You One Example To Better Understand.
Approximate one gallon of paint covers 300 square feet of wall. Calculate the area of each individual geometric shape based upon the outer wall blueprint measurements. How do you find the area of a wall?
Number Of Gallons To Paint Or Primer A Room.
Our normal window is 12 square feet (1.11 square meters). L 1 = outer surface area v 1 = volume within c 1 r 2 = inner radius c 2 = inner circumference l 2 = inner surface area v 2 = volume within c 2 h = height t = wall thickness v = volume of solid Calculate the area of each square or rectangle by multiplying the length by height.
Note:the Amount Of Coverage Of One Bundle Of Batts Differs By Manufacturer.
A = interior surface area of a room (excluding the floor and ceiling) l = length of the room; If you have a cavity wall or air gap in your construction element then this needs to be accounted for in your u value calculation. To calculate the area of a wall. use the standard (width) x (height) = area formula. | s3://commoncrawl/crawl-data/CC-MAIN-2023-23/segments/1685224643585.23/warc/CC-MAIN-20230528051321-20230528081321-00712.warc.gz | CC-MAIN-2023-23 | 3,252 | 19 |
https://www.novabbs.com/programming/overboard.php?thisgroup=comp.os.ms-windows.programmer.win32 | math | Welcome to novaBBS (click a section below)
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Posted: 7 Days 7 Hours ago by: Charlie GibbsThat sounds kind of familiar. I did once manage to build OpenSSL 1.0.1g with the Borland compiler. I used Borland's implib utility to convert the OpenSSL DLLs to library files that the Borland linker could use. (I prefer to link st
Posted: 7 Days 13 Hours ago by: JJYou're missing the Windows' DLL import libraries which are available from Windows SDK. But I don't know if the LIB files from it, are usable by MingW, cause its a *nix origin. If not, you'll need to manually create the import library file
Posted: 8 Days 13 Hours ago by: Charlie GibbsAre there any MinGW gurus out there? Or is there a better place to ask these questions? I wrote a suite of programs years ago; I've been compiling the Win32 versions (a mix of console and GUI programs) with Borland C++ Builder. I'd like
3 recent articles found. | s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662587158.57/warc/CC-MAIN-20220525120449-20220525150449-00793.warc.gz | CC-MAIN-2022-21 | 949 | 6 |
http://zqtermpaperanzp.njdata.info/how-to-convert-absorption-income-statement-to-a-contribution-margin-income-statement.html | math | Difference between traditional income statement and a contribution margin income statement format, use and examples. An absorption costing income statement requires a company to expense shrink the company’s gross profit margin, contribution income statement. 18 hours ago ims operating earnings and operating margin declined due to the absence of the favorable income taxes the company's safe harbor statement this press .
Cost of goods sold, both manufacturers and retailers list cost of good sold on the income statement as an expense directly after the total contribution margin. Income statement much of the from the contribution margin are subtracted both fixed factory overhead and fixed sg&a income is higher under absorption costing . The contribution margin income statement makes understanding cost behavior and how sales will affect profitability easier in figure 9-2, the company earned $1,000 in sales, $400 of which went toward variable costs.
The contribution margin income statement the traditional income statement uses absorption costing to create the income statement. 5 prepare a schedule to convert variable costing income to absorption costing income statement contribution margin fixed expenses net income 5 . Absorption costing income statement absorption costing is the method variable or direct costing favors the contribution margin income statement format .
A traditional income statement uses absorption or full costing, the contribution margin income statement, by contrast, uses variable costing, . Differences in net operating income between absorption costing and variable costing is due to the: when preparing a contribution margin income statement . Utx earnings call for the period ending june 30, 2018.
Each forward-looking statement contained in this presentation is subject operating income grew 41% to high variable contribution margin and some of the . Chapter 9 objective 2 study absorption costing income statement which of the following statements is true of contribution-margin format of the income statement. Cost contribution format vs traditional format of income statement - download as powerpoint presentation contribution margin for each additional unit sold”.
A variable costing income statement the bottom line of a contribution margin report is net income to convert variable costing income to absorption . It is important for a financial manager to understand that, on the income statement, the gross profit margin and the contribution margin are not the same. Cost-volume-profit analysis this income statement format is known as the contribution margin income statement and is used for internal reporting only.
Variable costing income statement contribution margin is the amount contributed by sales towards fixed costs and profit absorption costing vs variable costing. Let's illustrate the difference between gross margin and contribution margin with the following information: income statement break-even point financial ratios . Which amount would you find on chap's cvp income statement a contribution margin absorption costing income the following is the format of a cvp income . | s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583513844.48/warc/CC-MAIN-20181021094247-20181021115747-00085.warc.gz | CC-MAIN-2018-43 | 3,145 | 7 |
https://www.sharpminds.ca/math-program.html | math | Students will build a strong mathematical foundation by being introduced to the five major strands of the Ontario math curriculum. They will learn how to add, subtract, tell time, count money, and sort and classify shapes and figures. These basic skills will serve as essential tools during their math journey.
Students will build on their existing knowledge math by solving basic multiplication and division problems, examining areas of rectangles and parallelograms, and exploring fractions, money, decimals, and integers. Children will also be introduced to algebra and the magic of counting with numbers and letters.
Students in Grades 7-8 will be analyzing more complex data, examining geometric relationships, modelling linear relationships graphically and algebraically, and mastering fractions, integers, and exponents. It is these skills that will provide the necessary means to assist them as they continue on their math journey. Junior Mathematicians, here we come!
In the Junior Mathematicians program, students in Grades 9 & 10 will develop their mathematical knowledge and skills through theory, practical applications, and abstract problems.
In the Senior Mathematicians program, students in Grades 11 & 12 are taught how their math skills will be used at the university, college, and/or the workplace levels. Through this program, students will acquire the necessary skills to succeed in their continuing journey with math. | s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107878921.41/warc/CC-MAIN-20201022053410-20201022083410-00553.warc.gz | CC-MAIN-2020-45 | 1,439 | 5 |
https://brainly.com/question/290770 | math | When a cell size increases, it's volume increases more than it's surface area. It's actually not that simple but to make you understand, you can imagine the formula for the surface area as
and for volume
. Like I said these are not the actual formulas to find out the cell's surface area and it's volume. (I don't think there is even a formula for that). It's just to make you understand. And let's say the cell's size will increase 2 units, let's put in the formula:
As you can see the volume increased more than the surface area. You can think it this way too, the volume is in 3 dimensions and the surface area is in 2 dimensions, so the volume will increase more. Let's say it will increase 3 units,
as you can see the ratio got even less. Now , because the increase in the volume is more than the surface area, the nucleus of the cell struggles when controlling the cell. Because, think about it like this, would it be easier to reach out the stuff hanged on a wall, if you were in a shack or in a large house? That's how it is. And since the volume got bigger and the surface didn't get bigger that much,nutrients and minerals, water and other stuff will not be taken much, because the cell membrane will not be big enough for the cell. I hope this was helpful. | s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560281001.53/warc/CC-MAIN-20170116095121-00025-ip-10-171-10-70.ec2.internal.warc.gz | CC-MAIN-2017-04 | 1,267 | 5 |
http://tfpaperpyhe.ameriquote.us/ap-statistics-unit-6-free-response.html | math | Ap statistics test answers tests are listed in the order we will take them units refer to my grouping of the chapters (see the notes for unit - chapter groups) unit 6: inference, part 3 (means) unit 7: bivariate data, part 1. Free ap statistics practice tests with advanced reporting, full and the second part is a free response section that consists of five short-answer questions and an by making use of the free ap statistics practice tests and the other free ap statistics resources offered by varsity. Ap stats chapter 9b test by david_woods_5t60zy in types school work and ap stats chapter 9b test ap stats chapter 9b test free response: show all work on the 13 9answer key form a 1 4 a c b c b d d a b c a b b a b ap statistics chapter 9 test form a page 6 15 11 2 7 3. Practice questions in albert's ap statistics to review exam prep concepts such as describing and collecting data or using samples to make inferences in various contexts practice free response faq practice free response faq selected questions select tags close.
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2014 ap -2- statistics free-response questions formulas begin on page 3 questions begin on page 6 tables begin on page 14. This calculator will serve you well through all levels of mathematics and can be used on the sat, act, ap calculus test, ap statistics test, and other tests 105 unit 1 stats free response.
Calendars for geometry and ap statistics lesson plans geometry pre-ap lesson plans ap statistics ap statistics chapter 6 ap statistics chapter 1 ap statistics chapter 2 review chapter 5 answers free response by chapter 5 review frq answerspdf 24379 kb. Ap statistics unit 6 exam probability: free response directions: complete the assignment on this paper if you need additional paper make sure that you clearly label each page with your name. Hints for success on the ap statistics exam (compiled by zack bigner) the exam the ap stat exam has 2 sections that take 90 minutes each the first section is 40 multiple choice questions, and the second section is 6 free response questions each section counts for half of the. Ap statistics 2010 scoring guidelines the college board (0174) 696x 4 complete response both parts essentially correct 3 substantial response one part essentially correct and one part partially correct 2 developing response. Ap statistics - mr brown assignment sheet #1 - unit i - chapter 2-6 1 _8/31-9/01_ welcome to ap statistics with introductory analysis.
Ap psychology free response questions below is a list of all of the free response questions on the may ap psychology exam, dating back to 1993 note that the topics are a mix of various units statistics, intelligence, psychometrics (10pts. Explore timing and format for the ap statistics, and review sample questions a secure 2017 ap statistics exam is now available on the ap course audit website to access free response 6 questions | 1 hour, 30 minutes | 50% of exam score. Test 7a ap statistics name: directions: work on these sheets part 1: multiple choice circle the letter corresponding to the best answer. Ap statistics - unit 1 review solutions multiple choice: 1 b 2 a 3 b 4 d 5 d 6 c 7 e 8 e 9 a 10 e 11 b 12 b 13 a 14 c 15 e free response problem 1 problem 2 part (b): the distribution of sugar content m the ap statistics author: youndta. Ap statistics review probability is there a formula on the ap formula sheet that applies page 6 of 15 free response questions on probability 1 a one-mile relay race has 4 horses running a quarter of a mile each the riders of. | s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221218899.88/warc/CC-MAIN-20180821191026-20180821211026-00509.warc.gz | CC-MAIN-2018-34 | 5,542 | 6 |
http://www.mountainmolds.com/phpBB3/viewtopic.php?t=546 | math | Is there anything special about the 22LR bullet that contributes to its accuracy?
I don't see anything special, it's just a conservative design.
-- the length of the unsupported nose is 0.89 diameters.
-- the length of the groove diameter secton is 1.44 diameters.
-- the OAL of the bullet is 2.05 diameters. | s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583512323.79/warc/CC-MAIN-20181019041222-20181019062722-00311.warc.gz | CC-MAIN-2018-43 | 308 | 5 |
http://www.ask.com/web?qsrc=6&o=102632&oo=102632&l=dir&gc=1&q=Convert+4670+Kilograms+to+Tons | math | 4670 kilograms is equal to 5.147794 tons.
Kilograms to Metric Tons (or Tonnes) (kg to t) conversion calculator for Weight
conversions with additional tables and formulas.
Kilograms to Ton (kg to t) conversion calculator for Weight conversions with
additional tables and formulas.
From kilograms to tons (tonnes) conversion factor is different based on what kind
of ton (metric, short for US and long for UK) is used. To find out how many tons ...
Kilograms (kg) to tons (t) weight conversion calculator and how to convert.
Metric tons (t) to Kilograms (kg) weight conversion calculator and how to convert.
Answer by rapaljer(4670) · About Me (Show Source): ... 1 pound = 0.4536
kilograms 1 kilogram ... The question asks to convert from 1 ton to kilograms. | s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560280834.29/warc/CC-MAIN-20170116095120-00407-ip-10-171-10-70.ec2.internal.warc.gz | CC-MAIN-2017-04 | 757 | 11 |
http://su.diva-portal.org/smash/record.jsf?pid=diva2:381917 | math | On Mutually Unbiased Bases
2010 (English)In: International journal of quantum information, ISSN 0219-7499, Vol. 8, no 4, 535-640 p.Article, review/survey (Refereed) Published
Mutually unbiased bases for quantum degrees of freedom are central to all theoretical investigations and practical exploitations of complementary properties. Much is known about mutually unbiased bases, but there are also a fair number of important questions that have not been answered in full as yet. In particular, one can find maximal sets of N + 1 mutually unbiased bases in Hilbert spaces of prime-power dimension N = p(M), with p prime and m a positive integer, and there is a continuum of mutually unbiased bases for a continuous degree of freedom, such as motion along a line. But not a single example of a maximal set is known if the dimension is another composite number (N = 6; 10; 12,...). In this review, we present a unified approach in which the basis states are labeled by numbers 0, 1, 2,..., N - 1 that are both elements of a Galois field and ordinary integers. This dual nature permits a compact systematic construction of maximal sets of mutually unbiased bases when they are known to exist but throws no light on the open existence problem in other cases. We show how to use the thus constructed mutually unbiased bases in quantum-informatics applications, including dense coding, teleportation, entanglement swapping, covariant cloning, and state tomography, all of which rely on an explicit set of maximally entangled states (generalizations of the familiar two-q-bit Bell states) that are related to the mutually unbiased bases. There is a link to the mathematics of finite affine planes. We also exploit the one-to-one correspondence between unbiased bases and the complex Hadamard matrices that turn the bases into each other. The ultimate hope, not yet fulfilled, is that open questions about mutually unbiased bases can be related to open questions about Hadamard matrices or affine planes, in particular the notorious existence problem for dimensions that are not a power of a prime. The Hadamard-matrix approach is instrumental in the very recent advance, surveyed here, of our understanding of the N-6 situation. All evidence indicates that a maximal set of seven mutually unbiased bases does not exist - one can find no more than three pairwise unbiased bases - although there is currently no clear-cut demonstration of the case.
Place, publisher, year, edition, pages
2010. Vol. 8, no 4, 535-640 p.
Mutually unbiased bases, complex Hadamard matrices, generalized Bell states
IdentifiersURN: urn:nbn:se:su:diva-50178DOI: 10.1142/S0219749910006502ISI: 000281889900001OAI: oai:DiVA.org:su-50178DiVA: diva2:381917
authorCount :42010-12-292010-12-212010-12-29Bibliographically approved | s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988721278.88/warc/CC-MAIN-20161020183841-00235-ip-10-171-6-4.ec2.internal.warc.gz | CC-MAIN-2016-44 | 2,789 | 8 |
https://angolatransparency.blog/en/what-is-the-standard-cost-method/ | math | Standard costing is the practice of estimating the expense of a production process. It’s a branch of cost accounting that’s used by a manufacturer, for example, to plan their costs for the coming year on various expenses such as direct material, direct labor or overhead.
- What is standard costing formula?
- What is standard cost method in cost accounting?
- What is standard cost example?
- What is standard cost method of inventory valuation?
- What is a standard cost and what are its purposes?
- What is the difference between standard cost and standard costing?
- Why is standard cost important?
- What are the types of standard costs?
- How standard cost will help the business?
- What are the 4 types of standards?
- Why standard cost is used?
- What do mean by standard costing and marginal costing?
What is standard costing formula?
Formula to calculate standard costs
Direct labour = employee hourly rate x no. of hours worked x total number of units. Materials cost = market price per unit x total number of units. Manufacturing overhead = fixed overhead + (variable manufacturing overhead x total number of units)
What is standard cost method in cost accounting?
Standard costing is the practice of substituting an expected cost for an actual cost in the accounting records. Subsequently, variances are recorded to show the difference between the expected and actual costs.
What is standard cost example?
Examples include rent payable, utilities payable, insurance payable, salaries payable to office staff, office supplies, etc. read more is $15 per hour, and the standard fixed cost is $100,000. Therefore, the total hours required for producing one unit is 10 hours.
What is standard cost method of inventory valuation?
Under standard costing, the value of inventory is determined using the material and material overhead standard costs of each inventory item. If you use Bills of Material, Inventory maintains the standard cost by cost element (material, material overhead, resource, outside processing, and overhead).
What is a standard cost and what are its purposes?
Standard cost relates to a product, service, process or an operation. It is also determined for a normal level of efficiency of operation. Standard cost is used to measure the efficiency of future production or future operations. For this reason, it provides a useful basis for cost control.
What is the difference between standard cost and standard costing?
Standard cost is a predetermined cost. It is a determination in advance of production of what should be that cost. When standard costs are used for cost –control, the technique is known as standard costing.
Why is standard cost important?
Why Do Companies Use Standard Costs? Companies use standard costs for budgeting because the actual costs cannot yet be determined. This is because in the manufacturing process, it is impossible to predict the demand of a product or all the variables that will affect the costs of manufacturing it.
What are the types of standard costs?
There are three main categories of standard costs, basic standard costs, ideal standard costs and currently attainable standard costs.
How standard cost will help the business?
What is standard costing? What is Variance? Advantages … ·
What are the 4 types of standards?
Broadly speaking, there are 4 distinct types of standards within our portfolio of 42,000 standards: product, service, process and management standards. We also distinguish between levels of formality.
Why standard cost is used?
Standard costing, also known as standard cost accounting, is used to set budgets and plan for future expenses. It is a type of cost accounting mainly used in the manufacturing sector because it is easier to allocate costs directly to products being produced.
What do mean by standard costing and marginal costing?
In a marginal costing system, a system of cost ascertainment and control in which predetermined standards for marginal costs and income generated for products and operations are set and periodically compared with actual marginal costs incurred and income generated in order to establish any variances. | s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710421.14/warc/CC-MAIN-20221210074242-20221210104242-00515.warc.gz | CC-MAIN-2022-49 | 4,145 | 38 |
http://www.slideshare.net/rahul_c/weekly-7911515 | math | 1) From the data given below, you need to calculate the EPS for Vodafoneand then calculate it for the two years for which you have data.Ans :Calculation of the EPS for Vodafone :Earnings per share = (Net income – Dividends on preferred stock) ÷ Averagenumber of shares outstandingProfit for the financial period -16155.0 -9885.0Weighted average number 65012501146 61334032162of shares in issue in theperiodEPS -2.48 -1.611Analyzing the EPS of VODAFONE; For every share, ordinary shareholdershave lost 25 pence in 2002 and they lost 16 pence in 2001.2) Take the latest prices for Vodafones shares from either Yahoo or Reutersand rework the dividend yield value for Vodafone.Ans :Vodafones shares price 174.10The cost of equity capital is Rs 174.10Realized yield ApproachThis approach takes into consideration the basic factors of the D/P + gapproach but, instead of using the expected values of the dividends andcapital appreciation, past yields are used to denote the cost of capital. Thisapproach is based upon the assumption that past behavior will be repeated infuture and, therefore, may be used to measure the cost or ordinary capitalThe realized return over n-years period is calculated as (w1*w2*w3*….n)1/n
Where Wt, referred to as the wealth ratio, calculated as (Dt + Pt )/Pt-1 and t =1,2,3…….nDt = DPS for t payablePt = Price per share at the end of t year4) Using the data given below find the relevant data for Vodafone andcalculate its P/E ratio.Ans : market price of Vodafone pls 174.10 EPS 2.48 PE ratio 174.10/2.48= 70.25) Explain Capital Asset Pricing Model (CAPM) with formula andillustration.Ans : A model that describes the relationship between risk and expectedreturn and that is used in the pricing of risky securities.The general idea behind CAPM is that investors need to be compensated intwo ways: time value of money and risk. The time value of money isrepresented by the risk-free (rf) rate in the formula and compensates theinvestors for placing money in any investment over a period of time. Theother half of the formula represents risk and calculates the amount ofcompensation the investor needs for taking on additional risk. This is
calculated by taking a risk measure (beta) that compares the returns of theasset to the market over a period of time and to the market premium.Using the CAPM model and the following assumptions, we can compute theexpected return of a stock in this CAPM example: if the risk-free rate is 3%,the beta (risk measure) of the stock is 2 and the expected market return overthe period is 10%, the stock is expected to return 17% (3%+2(10%-3%)).Q.1 You are required to calculate the following for each one of the threecategories of equity shares ppearing in the above mentioned Balance Sheet:a) Intrinsic value on the basis of book values of Assets and Liabilitiesincluding goodwill;What Does Intrinsic Value Mean?1. The actual value of a company or an asset based on an underlyingperception of its true value including all aspects of the business, in terms ofboth tangible and intangible factors. This value may or may not be the sameas the current market value. Value investors use a variety of analyticaltechniques in order to estimate the intrinsic value of securities in hopes offinding investments where the true value of the investment exceeds itscurrent market value.2. For call options, this is the difference between the underlying stocks priceand the strike price. For put options, it is the difference between the strikeprice and the underlying stocks price. In the case of both puts and calls, ifthe respective difference value is negative, the instrinsic value is given aszero.
Goodwill 420Other Fixed Assets 11166Current Assets 2910Loans and Advances 933Total 15429Less: Secured loans 4500Current liabilities 1242Provisions 960 6702 8727Add: Notional call on 90 lakhsequity shares @ Rs. 2 per share 180 8907(b) Value per share on dividend yield basis:Value of fully paid-up share of Rs. 10= 20/15* Rs. 10=Rs. 13.33(approx)Value of partly paid-up share of Rs. 8 =20/15*Rs. 8=Rs. 10.67(approx)Value of fully paid-up share of Rs. 5=20/15*Rs.5 = Rs. 6.67(approx)(c) Value per share on the basis of EPS:Given : Profit after tax = Rs. 1,371 lakhs , 180 lakh equity shares of Rs. 10 each, fully paid up,90lakhequityshares of Rs.10 each, Rs. 8 paid up150 lakhs equity shares of Rs. 5 each,fully paidup. ,Reserves and surplus total of 3270 lakhEPS of share capital 1371/3270 = 0.419Earning per fully paid share of Rs. 10 = Rs. 0.419 * 10 = Rs. 4.19 Rs. 0.419 * 8=Rs. 3.352 Rs. 0.419 * 5= Rs. 2.095 | s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698541525.50/warc/CC-MAIN-20161202170901-00016-ip-10-31-129-80.ec2.internal.warc.gz | CC-MAIN-2016-50 | 4,569 | 4 |
https://math.answers.com/questions/Which-shapes-has-four-sides-and-four-corners-and-all-four-sides-are-equal | math | It is a square as square has 4 sides and all of them are equal. Also it has 4 corners.
The shapes that I can think of are a square, rectangle, rhombus and a diamond
A QUADRILATERAL has four corners and four sides, along with four angles. These do not need to be congruent (equal) but they can be.
Two different types of shapes would be a circle and a square. they differ from each other in the fact that a circle's shape is perpetual as a square has four equal sides and four corners.
No, both shapes have all four sides equal.
Carré is square in french and a square has four equal sides so it alo has four corners.
Four equal sides make up a square. The sides may be any size, as long as they are equal. A square is one of the most basic shapes.
four sides and four corners
A rhombus is the only shape besides a square that can have equal sides.
A square has four sides. The sides of a square are all equal in length. A square has four corners. | s3://commoncrawl/crawl-data/CC-MAIN-2023-40/segments/1695233511386.54/warc/CC-MAIN-20231004152134-20231004182134-00148.warc.gz | CC-MAIN-2023-40 | 946 | 10 |
https://www.scientific.net/KEM.324-325.615 | math | A New Method for Parameters Estimation of Probabilistic Fatigue Crack Growth Rate Model
A new method is proposed to estimate the parameters of probabilistic fatigue crack growth rate models, including the Paris equation and its’ improved type. To take the statistical characteristics of whole test data into account, the method inherits the idea from the general maximum likelihood method which is widely used in parameters estimation of fatigue S-N curves, ε-N curves, and da/dN-"K curves, and extends the conventional correlation coefficient optimization method into parameters evaluation not only for mean curve, but also for standard deviation curve and probabilistic curve. Analysis on the test data of 16MnR steel indicates that present method is available and feasible. Comparing to the general maximum likelihood method, present method has simpler algorithm, and can avoid constructing and solving the likelihood function, so it is speedier in calculation.
M.H. Aliabadi, Qingfen Li, Li Li and F.-G. Buchholz
B. Yang and Y. X. Zhao, "A New Method for Parameters Estimation of Probabilistic Fatigue Crack Growth Rate Model", Key Engineering Materials, Vols. 324-325, pp. 615-618, 2006 | s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583660070.15/warc/CC-MAIN-20190118110804-20190118132804-00415.warc.gz | CC-MAIN-2019-04 | 1,194 | 4 |
https://www.wazeesupperclub.com/what-are-the-7-metric-prefixes/ | math | What are the 7 metric prefixes?
In the SI, designations of multiples and subdivision of any unit may be arrived at by combining with the name of the unit the prefixes deka, hecto, and kilo meaning, respectively, 10, 100, and 1000, and deci, centi, and milli, meaning, respectively, one-tenth, one-hundredth, and one-thousandth.
What is the prefix of decimal?
Prefix names used for multiples and submultiples units.
|μ – micro||one-millionth||0.000 001|
|m – milli||one-thousandth||0.001|
|c – centi||one-hundredth||0.01|
|d – deci||one-tenth||0.1|
What are multiples of SI units?
Decimal multiples and submultiples of SI units
What are the 20 SI prefixes?
The 20 SI Prefixes
What is the SI prefix for 10-6?
Table 5. SI prefixes
What are the prefixes used to write multiples of numbers?
Prefixes and Multiples
How do you use SI prefixes?
SI prefixes are used to form decimal multiples and submultiples of SI units. They should be used to avoid very large or very small numeric values. The prefix attaches directly to the name of a unit, and a prefix symbol attaches directly to the symbol for a unit.
What are multiples Submultiples?
Multiples are factors used to create larger forms whereas submultiples are factors used to create smaller forms of the SI units. For example, a centimetre is a submultiple and kilometre is a multiple of a metre.
Why multiples and sub multiples are used?
Sometimes the size of the SI unit is either too small or too big to measure a certain quantity. For example a metre is too small a unit to measure the distance between two cities and too big a unit to measure the thickness of a wire. Hence multiples and submultiples of units are required.
What is the SI prefix for 100?
Hecto (symbol: h) is a decimal unit prefix in the metric system denoting a factor of one hundred. It was adopted as a multiplier in 1795, and comes from the Greek ἑκατόν hekatón, meaning “hundred”.
How many SI prefixes are there for multiples of units?
SIprefixes for decimal multiples There are 20 internationally accepted prefixes to denote 10nmultiples of units. 10 of these prefixes denote (n>0) multiples, and other 10 denote (n<0) submultiples: SIPrefixes for decimal multiples
Which prefixes can be used with the SI units?
A set of multiple and sub-multiple prefixes have been adopted for use with the SI units. They may be used with any of the base units and with any of the derived units with special names.
What are the decimal multiples and submultiples of SI units?
Decimal multiples and sub-multiples of SI units. SI prefixes Decimal multiples and submultiples Multiples Submultiples Factor Name Symbol Factor Name Symbol 10 1 deca da 10 -1 deci d 10 2 hecto h 10 -2 centi c 10 3 = (10 3) 1 kilo k 10 -3 = (10 3) -1 milli m 10 6 = (10 3) 2 mega M 10 -6 = (10 3) -2 micro μ 10 9 = (10 3)…
Why don’t we use the SI prefixes for binary multiples?
Because the SI prefixes strictly represent powers of 10, they should not be used to represent powers of 2. Thus, one kilobit, or 1 kbit, is 1000 bit and not 2 10 bit = 1024 bit. To alleviate this ambiguity, prefixes for binary multiples have been adopted by the International Electrotechnical Commission (IEC) for use in information technology. | s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571909.51/warc/CC-MAIN-20220813051311-20220813081311-00760.warc.gz | CC-MAIN-2022-33 | 3,234 | 32 |
https://yonsei.pure.elsevier.com/en/publications/adaptive-synchronization-of-discrete-time-t-s-fuzzy-chaotic-syste | math | This paper presents a model reference adaptive control approach for the synchronization of a discrete-time chaotic systems using output tracking control. Takagi-Sugeno (T-S) fuzzy model is employed to represent the discrete-time chaotic system. Then, the reference model system is chosen using the output of master system and design the control input so that the controlled slave system achieves asymptotic synchronization with the reference system given that two systems start from different initial conditions, different parameters and/or different type of model. Using a gradient algorithm, the ideal controller gains which can stabilize the error equation are estimated. Simulation examples of two cases are given to demonstrate the validity of our proposed adaptive method. | s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882570977.50/warc/CC-MAIN-20220809124724-20220809154724-00714.warc.gz | CC-MAIN-2022-33 | 778 | 1 |
https://www.runningahead.com/groups/1500/forum/Post/398e2e3c01dd46f6a725ac4c7906081a | math | Roads were made for journeys...
Needs more cowbell!
• Do some dus...and some CX...and some tandem gravel...and some podiums...
• 130#s (or less)
• Stop letting my core go to hell
I had to do some surgery to move the 1000 mile bunny's tongue to the left...looking exhausted :-)
Noticed one more thing. How come 1500 mile bunny is behind? I know, I am picky. | s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698542648.35/warc/CC-MAIN-20161202170902-00273-ip-10-31-129-80.ec2.internal.warc.gz | CC-MAIN-2016-50 | 362 | 7 |
http://pages.rediff.com/baillie------psw-primality-test/1067326 | math | In number theory, the Baillie–PSW primality test is a probabilisticprimality testingheuristic algorithm: it determines if a number is composite or a probable prime.The authors of the test offered $30 for the discovery of a composite number that passed this test. , the value was raised to $620Guy, R. (1994). “Pseudoprimes. Euler Pseudoprimes. Strong Pseudoprimes.” §A12 in Unsolved Problems in Number Theory. 2nd ed., p. 28, New York: Springer-Verlag. ISBN 0-387-20860-7. consequently this can be considered a sound primality test on numbers below that upper bound.
A primality testing software PRIMO uses this algorithm to check for probable primes, and no certification of this test has yet failed. The author, Marcel Martin, estimates by those results that the test is accurate for numbers below 10000 digits. There is a heuristic argument suggesting that there may be infinitely many counterexamples. | s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084891530.91/warc/CC-MAIN-20180122173425-20180122193425-00032.warc.gz | CC-MAIN-2018-05 | 912 | 2 |
https://www.convertunits.com/from/decipoise/to/millipascal+second | math | How many decipoise in 1 millipascal second?
The answer is 0.1.
We assume you are converting between decipoise and millipascal-second.
You can view more details on each measurement unit:
decipoise or millipascal second
The SI derived unit for dynamic viscosity is the pascal second.
1 pascal second is equal to 100 decipoise, or 1000 millipascal second.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between decipoise and millipascal seconds.
Type in your own numbers in the form to convert the units!
1 decipoise to millipascal second = 10 millipascal second
5 decipoise to millipascal second = 50 millipascal second
10 decipoise to millipascal second = 100 millipascal second
15 decipoise to millipascal second = 150 millipascal second
20 decipoise to millipascal second = 200 millipascal second
25 decipoise to millipascal second = 250 millipascal second
30 decipoise to millipascal second = 300 millipascal second
40 decipoise to millipascal second = 400 millipascal second
50 decipoise to millipascal second = 500 millipascal second
You can do the reverse unit conversion from millipascal second to decipoise, or enter any two units below:
decipoise to pound/foot hour
decipoise to centipoise
decipoise to kilogram/meter hour
decipoise to pound/foot second
decipoise to poundal second/square foot
decipoise to pound-force second/square inch
decipoise to slug/foot second
decipoise to millipoise
decipoise to poundal hour/square foot
decipoise to poise
The SI prefix "deci" represents a factor of 10-1, or in exponential notation, 1E-1.
So 1 decipoise = 10-1 poise.
The definition of a poise is as follows:
The poise is a unit of dynamic viscosity in the CGS system of units. It is named after Jean Louis Marie Poiseuille. The analogous SI unit is the Pascal second. Water has a viscosity of 0.0089 poise at 25 °C, or 1 centipoise at 20 °C.
The SI prefix "milli" represents a factor of 10-3, or in exponential notation, 1E-3.
So 1 millipascal-second = 10-3 pascal-second.
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154878.27/warc/CC-MAIN-20210804142918-20210804172918-00612.warc.gz | CC-MAIN-2021-31 | 2,484 | 37 |
https://methods.sagepub.com/Reference/encyc-of-research-design/n336.xml | math | The laws of probability are a collection of equations that define probabilistic relationships among events. The validity of each equation, or probability law, often depends on the events having met certain conditions. Thus, each law is accompanied by a list of what, if anything, must be assumed about the events in question for the law to hold. There is no specific number or set of equations defined by the term laws of probability. A list that covers probabilistic relationships that are most often of interest, such as the probability that at least one of two events occurs, or the probability that two events occur together, is given here. Probability laws are employed frequently in quantitative research, as they are an essential piece of the foundation ...
Looks like you do not have access to this content. | s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320306346.64/warc/CC-MAIN-20220128212503-20220129002503-00178.warc.gz | CC-MAIN-2022-05 | 815 | 2 |
https://www.assignmentexpert.com/homework-answers/economics/microeconomics/question-99871 | math | When Mr Smith’s monthly income was €5,000, he usually went to the local coffee shop 12 times a month. Mr Smith had a pay rise and his current income is €5,500 a month. Now, he goes to the local coffee shop 10 times a month. Compute the income elasticity of demand using the midpoint method. Explain your answer. Is having a coffee at the local coffee shop normal or inferior good to Mr Smith? Illustrate your answer using an appropriate graph.
The income elasticity of demand using the midpoint method is:
Ei = (10 - 12)/(5,500 - 5,000)×(5,500 + 5,000)/(10 + 12) = -2/500×10,500/22 = -1.91, so a coffee at the local coffee shop is inferior good to Mr Smith. | s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662545548.56/warc/CC-MAIN-20220522125835-20220522155835-00086.warc.gz | CC-MAIN-2022-21 | 665 | 3 |
https://doaj.org/article/b0cfe43d5a25466a81856a3620e594da | math | Computer Science Journal of Moldova (Apr 2021)
A Sublinear Sudoku Solution in cP Systems and its Formal Verification
Sudoku is known as a NP-complete combinatorial number-placement puzzle. In this study, we propose the first cP~system solution to generalised Sudoku puzzles with $m \times m$ cells grouped in $m$ blocks. By using a fixed constant number of rules, our cP~system can solve all Sudoku puzzles in sublinear steps. We evaluate the cP~system and discuss its formal verification. | s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296943809.76/warc/CC-MAIN-20230322114226-20230322144226-00325.warc.gz | CC-MAIN-2023-14 | 489 | 3 |
http://experimental-theatre.blogspot.com/2011/06/ | math | We can express paradox in different ways. In literature, paradox is a mere figure of speech. In theology, a paradox is distinct from a contradiction; paradoxes in theology are truths that only seem contradictory, but further and careful scrutiny would show that they are not, whereas contradictions are impossibilities and untruths. In the mindblowing world of mathematics, paradoxes are contradictions that are possible and true (e.g. 1 = 0.99999…).
Indeed there are different treatments of paradox among different disciplines. But what is paradox in general? A paradox, according to wiki, is “a seemingly true statement or group of statements that lead to a contradiction or a situation which seems to defy logic or intuition.” So, as I want to interpret it, paradoxes can either be possible (which happens in theology and math) or impossible. Paradoxes can be seemingly contradictory but actually not or they can be outright contradictory.
I love paradoxes because they are awesome mindblowers.
|Warning: This may happen while reading this list|
Once you are ready for some massive brain-beating, meet me after the jump. | s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676593438.33/warc/CC-MAIN-20180722174538-20180722194538-00133.warc.gz | CC-MAIN-2018-30 | 1,129 | 5 |
https://www.statmodel.com/discussion/messages/11/8230.html?1590018965 | math | cd = (d0 * c0 - d1*c1)/(d0 - d1) = (951 x 5.750 - 950 x 5.722)/(951 - 950) TRd = (T0*c0 - T1*c1)/cd = (2728.083 x 5.750 - 2717.094 x 5.722)/cd
COMPARISON MODEL Loglikelihood H0 Value -154318.940 H0 Scaling Correction Factor 5.722 for MLR H1 Value -151645.790 H1 Scaling Correction Factor 2.887 for MLR ... Chi-Square Test of Model Fit Value 2717.094* Degrees of Freedom 950 P-Value 0.0000 Scaling Correction Factor 1.968 for MLR NESTED MODEL: Loglikelihood H0 Value -154323.099 H0 Scaling Correction Factor 5.750 for MLR H1 Value -151645.790 H1 Scaling Correction Factor 2.887 for MLR ... Chi-Square Test of Model Fit Value 2728.083* Degrees of Freedom 951 P-Value 0.0000 Scaling Correction Factor 1.963 for MLR
Thanks for your confirmation. Let me ask a follow-up question. TRd was found to be 4.304958... Since the material says, "For MLM and MLR the products T0*c0 and T1*c1 are the same as the corresponding ML chi-square values," am I supposed to use 3.841 as critical value to determine whether the calculated S-B scaled chi-square difference is significant at the level of .05 or not? That is, is the difference (4.304958...) significant since TRd > 3.841?
I forgot asking another question. The equality constraint was imposed on a single parameter (which measures the effect of child maltreatment on violent offenses) for two ethnic groups, whites and Asian Americans. In the comparison model, the coefficient was found to be .031 (SE = .027) for whites, whereas it was .654 (SE = 1.430). As you can see, both coefficients are not significant, although the S-B scaled chi-square difference is larger than 3.841. As I supposed to say the coefficient is significantly different between whites and Asian Americans even though the coefficient was found to be not significant in each ethnic group?
2nd post: Each coefficient being significantly different from zero or not is not the same as testing that they are the same. Typically, if you use the independent-sample z test of equality using your SEs, you get the same thing as the chi2.
Step 1 on the Mplus website (http://www.statmodel.com/chidiff.shtml) for Difference Testing Using the Loglikelihood is: 1. Estimate the nested and comparison models using MLR. The printout gives loglikelihood values L0 and L1 for the H0 and H1 models, respectively, as well as scaling correction factors c0 and c1 for the H0 and H1 models, respectively.
Does this refer to H0 and H1 values given for the SAME model (i.e., in the same output file); or for DIFFERENT models (estimated in separate runs, with separate output files)?
I ask because while I have seen BOTH H0 and H1 values in some output files, I only see H0 for in a model I estimated using a NBI dependent variable, as seen below. There is no H1 value offered. Can I still utilize the steps on the website to compare the fit of this model with that of another nested model, using the H0 values only (the ones provided for each distinct model--because I did not get H0 and H1 values together in one output file). Thanks.
MODEL FIT INFORMATION Number of Free Parameters 14 Loglikelihood H0 Value -1189.806 H0 Scaling Correction Factor for MLR 1.1902
To do difference testing you need to run two analyses. The first is the least restrictive model. It is referred to as H1 in the writeup. The nested model is referred to as H0 in the write up. In both cases, the H0 values are taken from the output to use in the computations.
EFried posted on Tuesday, April 02, 2013 - 1:55 pm
When comparing 2 models using the MLR estimator, each model provides 3 scaling correction factors and 2 loglikelihoods. I don't find it specified which one to use for model comparison (http://www.statmodel.com/chidiff.shtml).
Looks like the only difference is that model 2 has a direct effect from m to y2.
ri ri posted on Saturday, August 30, 2014 - 4:08 pm
Yes, as far as I know one Needs to do a chi square difference test to compare the two models. In regular way, one just uses the chi square values. But since I have categorical data, I shall do it differently I suppose? I used the difftest command, but could not find the scale correction to calculate the difference with the formula provided at the Website.
With only one parameter difference you can just look at the z-test for that parameter in the model that is less restrictive.
In the general case you use DIFFTEST, first running the less restrictive model and then the more restrictive model. You don't need the scaling correction factors or the computations on the website. DIFFTEST does it for you.
ri ri posted on Tuesday, September 02, 2014 - 12:16 am
I tried the DIFFTEST to compare the contrained and uncontrained model, it worked wonderfully!
Just I have another methodological question. In the user guide multiple Group Analysis, you wrote an example, Fixing the mean of the variables in Group 2 to Zero. If I compare constrained and unconstrained models, is it necessary to fix the mean to Zero? I also saw some People Center the means of the continous variables in order to minimize multicollinearity. If I compare two path models (such as the above mentioned model comparion), I wonder if mean centering is needed?
To compare means across groups, use the model with means zero in all groups versus the model with means zero in one group and free in the others. Centering is not needed.
Ari J Elliot posted on Wednesday, January 07, 2015 - 7:44 pm
Hello Drs. Muthen,
Regarding chi square difference testing with MLR, please confirm that the H0 scaling correction factor that should be used in the calculation is the one listed under Loglikelihood, NOT the one listed under the MLR chi-square test of model fit.
Thus in the following output for the nested model, I would use 1.5862 as the scaling correction factor.
H0 Value -15770.576 H0 Scaling Correction Factor 1.5862 for MLR H1 Value -15733.949 H1 Scaling Correction Factor 1.5549 for MLR ............................. .....Chi-Square Test of Model Fit
Value 50.133* Degrees of Freedom 5 P-Value 0.0000 Scaling Correction Factor 1.4612 for MLR
If you use chi-square for the difference testing, you should use the scaling correction factor under chi-square. If you use the loglikelihood for the difference testing, you should use the scaling correction factor under loglikelihood.
Ari J Elliot posted on Thursday, January 08, 2015 - 11:37 am
Ok thanks. To further clarify, the instructions on the webpage for difference testing using chi-square state: "Be sure to use the correction factor given in the output for the H0 model."
Under the chi square I only see one scaling correction factor, whereas for loglikelihood there are correction factors provided for both H0 and H1. Given that correction factors for both the nested and comparison models are used in the calculation I'm not sure to what the reference to the H0 model in the instructions refers.
You should use the H0 scaling correction factors. If only one is specified, it is the H0. Don't use the one for the baseline model. If you have further questions, send the output and your license number to firstname.lastname@example.org and we can tell you which number to use.
Dear Muthen, I have a SEM model with these relationships (1) one latent and its three observed variables (2) Five observed variables that involved in path relationships with the latent in (1) (3) A correlation in two of the observed variables in (2). I use MLR estimator and would like to know the chi-square p-value. I follow the instruction given in your website on difference testing on chi-square, to compute the scaled difference in chi-square. My question is for the H0 (restricted model), which relationships in (1)to(3) I need to constrain? Should I just constrain the path relationships in (2) which is my main interest? Our goal in this test is to get a non significant p-value right? like the ML estimator's chi-square result for model fit test?
is there any information how the Satorra-Bentler scaled chi-square difference test is influenced by high N´s (e.g. n > 30 000)? Is it influenced at all? I am investigating measurement invariance with large subpopulation-samples and complex survey data (students nested in teachers) and therefore I use the SB-chi-square to test nested models.
Dear Linda, In ESEM invariance test for 2 groups, when I compare the metric (loading) invariance model with the configural (baseline) model, by using MLR estimator, can I still calculate the chi-square difference in Mplus using MLR? I have a problem to determine the constraint/more restrictive model for my configural and metric invariance ESEM models. The same thing for other invariance test (scalar, error variance). Can you give some advises. Thanks.
Jiangang Xia posted on Thursday, November 02, 2017 - 9:41 am
Dear Linda, I am confused by the instructions from the website and your previous responses to some above questions regarding the "Difference Testing Using Chi-square".
For example, for the very first question in above, you said "Yes, you are doing this correctly." However, in the question, the "H0 Scaling Correction Factor for MLR" (5.722 and 5.750) under "Logliklihood" was used, not the "Scaling Correction Factor for MLR" (1.968 and 1.963)under "Chi-Square Test of Model Fit".
In another question in above that was posted by Ari J Elliot posted on January 07, 2015, you responded that "If you use chi-square for the difference testing, you should use the scaling correction factor under chi-square."
So I am not sure which "scaling correction factor" should we use for the "Difference Testing Using Chi-square".
In order to avoid the confusion, here I want to use the output from the first question. If I want to compare the two models using chi-square, should I use
H0 Scaling Correction Factor 5.722 for MLR ?
or should I use
Scaling Correction Factor 1.968 for MLR ?
A related question: when should we use chi-square and when should we use Loglikelihood?
The "Scaling Correction Factor for MLR" should be used with the "Difference Testing Using Chi-square".
There should be no confusion about this since that correction factor is printed just below the chi-square value.
In every case you can use chi-square or the log-likelihood. Both should give you exactly the same result (subject to round off error).
Derek Boy posted on Saturday, January 20, 2018 - 9:44 am
Dear Dr. Muthen In order to compare the two models as below, I have tried to do the deviance test using loglikelihood. However, I could not figure out the numbers of parameters (i.e., p0 and p1) which are required by the formula for computing the scaling correction (cd) = (p0*c0 - p1*c1)/(p0 - p1), where p0 is the number of parameters in the nested model and p1 is the number of parameters in the comparison model. Would you kindly help, please? Best regards. ------------------------------- MODEL A Number of Free Parameters 17 Loglikelihood H0 Value -215.987 H0 Scaling Correction Factor 0.4637 for MLR H1 Value -215.987 H1 Scaling Correction Factor 0.4637 for MLR … … … Chi-Square Test of Model Fit Value 0.000* Degrees of Freedom 0 P-Value 0.0000 Scaling Correction Factor 1.0000 for MLR ------------------------------- MODEL B Number of Free Parameters 19 Loglikelihood H0 Value -214.709 H0 Scaling Correction Factor 0.4195 for MLR H1 Value -214.715 H1 Scaling Correction Factor 0.4195 for MLR … … … Chi-Square Test of Model Fit Value 0.000* Degrees of Freedom 0 P-Value 1.0000 Scaling Correction Factor 1.0000 -------------------------------
You have zero degrees of freedom for both models so neither model is testable and no model comparison can be done.
Derek Boy posted on Saturday, January 20, 2018 - 6:37 pm
Dear Dr. Muthen So, my two models are saturated and having zero degrees of freedom. Many many thanks for your pointing it out to me. But, may I also ask whether fitting a saturated model is a bad thing to do? Should I do something to make it unsaturated? What is that something you would suggest me to do? Please kindly advice. With best regards.
I intended to compare two nested model with the Satorra-Bentler scaled chi-square difference test (both models used MLR as the estimator) and based on the formula, the computed SB-chi-square is a negative number. I wonder if this is possible and how should I resolve this issue?
I am comparing models in a TWOLEVEL MIXTURE (MLR) analysis to test if particular covariates should be included in the model. I am using the Loglikelihood to calculate the TRd value, as per the website (the Chi-square values do not appear in my output for some reason).
I can do the calculation, however, how do I calculate the degrees of freedom in order to obtain the critical value of the TRd value?
I comuted two CFAs, testing a three-fator model and a three-factor model with a general factor using Mplus editior 7.3. Now I'd like to do a DIFF test to see which model is better. However, my outputs do not give me a correction factor. How do I get the correction factor?
A quick look says that the only difference between the models is that you put a second-order factor behind the 3 first-order factors. If that's the case, the models are the same because 3 indicators of 1 factor is a just-identified model - you are not restricting the covariance matrix for the first-order factors. You need 4 or more first-order factors for that.
Q1: There is no way in Mplus or any software. The 2 models are the same when you have only 3 first-order factors. "The same" means that they produce the exact same covariances among the observed variables. The models have the same number of parameters and when you have estimated one, you can transform its parameter values to the parameter values of the other.
Q2: There is no change that you can make - just accept the fact that this can't be tested - you can present the second-order factor model but you can't say that it fits better or worse.
I am comparing two Path Analysis models using the chi-square difference testing (Satorra-Bentler scaled chi-square difference test) - I got the formula for it from this website: https://www.statmodel.com/chidiff.shtml - the first set of tests listed on the page.
The nested model has 8 degrees of freedom, whereas the comparison model is saturated and has 0 degrees of freedom and a chi square value of 0.
Is it okay to carry out the chi-square difference testing between the two models, given that one of the models is saturated?
If it is indeed okay, I wonder if there is a citation you are aware of that I could cite to justify doing this (I am getting pushback on this from others).
Hello, I am doing a Factor Analyisis and my goal is to test whether a hierachical model with a general factor or just a three-factor model fits my data best.
When performing the hierachical model the input reading terminates normally, however, my factor loadings and the std errors leave me with questions. Why are the loadings and errors so high? Is there a problem in my code?
Code: Data: FILE is Dr_E.inp; Variable: NAMES ARE EP1 EP2 EP3 EP4 EP5 EP6 EP7 EP8 EP9 EP10; USEVARIABLES ARE EP1 EP2 EP3 EP4 EP5 EP7 EP8 EP9 EP10; CATEGORICAL ARE EP1 EP2 EP3 EP4 EP5 EP7 EP8 EP9 EP10; MODEL: F1 BY EP1* EP2; F2 BY EP3* EP4 EP5; F3 BY EP7* EP8 EP9 EP10;[F1-F3@0];F1-F3@1; Y BY F1* F2 F3;[Y@0];Y@1; OUTPUT: SAMPSTAT tech1 MODINDICES; PLOT: TYPE=PLOT2; Output: Y BY F1 1.059 0.170 6.218 0.000 F2 5.140 7.062 0.728 0.467 F3 2.544 0.931 2.732 0.006 Means Y 0.000 0.000 999.000 999.000 Intercepts F1 0.000 0.000 999.000 999.000 F2 0.000 0.000 999.000 999.000 F3 0.000 0.000 999.000 999.000 Thank you for your insights!
I need to compare nested models. I am using ML and I got the following loglikelihood statistics for the two models I want to compare. As I understood correctly for ML I don't need the scalling correction factors:
Model 1: -6369.028 (67) Model 2: -6368.841 69
Loglikelihood ratio: .374 Difference in Degree of Freedom: 2
How do I now know if the difference is significant or not?
Thank you very much for the help. Best regards Dinah
With df=2, the 5% critical value for Chi-square is 5.991, so the difference is not significant at the 5% level. You know that 0.374 is small for df=2 because the expected value is 2 (the df).
mdehne posted on Wednesday, May 20, 2020 - 3:06 am
Hi Linda, Bengt, or Tihomir
I have got two questions:
1) Am I correct that the computation of the chi square difference test for MLR (http://www.statmodel.com/chidiff.shtml) does not ensure the positiveness as proposed by Satorra and Bentler (2010) and is thus based on the original paper (Satorra, 2000; Satorra & Bentler, 2001)?
If I am correct: 2) Is there any computational opportunity to use Satorra & Bentler's (2010) alternative? | s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487649688.44/warc/CC-MAIN-20210619172612-20210619202612-00414.warc.gz | CC-MAIN-2021-25 | 16,682 | 78 |
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https://edurev.in/course/quiz/attempt/-1_Test-Reflection-of-Light/3d175c97-8ea6-4e34-be85-6b97daef7381 | math | Raj is standing in front of a plane mirror. The distance between Raj and his image is 100cm. If the distance between the mirror and image is x. What could be the possible value of x ?
Distance between raj and image=100cm
Distance between image mirror =x
Distance between image and mirror= Distance between mirror and raj
Virtual image forms
A virtual image occurs when light rays do not actually meet at the image. Instead, you " See" the image because your eye projects light rays backward. In flat, or plane mirrors, the image isa virtual image, and is the same distance behind the mirror as the object is in front of the mirror.
Which among the following is not true for a plane mirror ?
The image formed by a plane mirror is always virtual,upright, and of the same shape and size as the object it is reflecting. A virtual image is a copy of an object formed at the location from which the light rays appear to come.
The focal length of a plane mirror is
The focal length of a plane mirror is infinity as the image can be formed at an infinite distance inside the mirror.
The image formed by plane mirror is
The image formed by a plane mirror is always virtual (meaning that the light rays do not actually come from the image), upright, and of the same shape and size as the object it is reflecting. A virtual image is a copy of an object formed at the location from which the light rays appear to come.
An observer moves towards a stationary plane mirror at a speed of 4m/s. The speed of image with respect to mirror is __________.
As the observer comes closer to the plane mirror, the image also comes closer. Also, in case of plane mirror, object distance = image distance. Hence, the speed of the image w.r.t. mirror is also 4 m/s, towards the mirror.
is the lateral inversion of
Lateral inversion is a phenomenon in which left appears to be right and vice versa. It is due to direction that light follows when it strikes a reflecting surface, generally a mirror.
Which of the following letter does not show lateral inversion ?
Lateral version of I is I
A medium through which light propagates easily is a
In the field of optics, transparency (also called pellucidity or diaphaneity) is the physical property of allowing light to pass through the material without being scattered.
A virtual image is defined as the opposite of a real image, therefore an image that cannot be obtained on a screen is reffered to as a virtual image. The rays of light that form a virtual image never converge therefore a virtual image can never be projected onto a screen.
A real image occurs where rays converge, whereas a virtual image occurs where rays only appear to converge. Real images can be produced by concave mirrors and converging lenses, only if the object is placed further away from the mirror/lens than the focal point and this real image is inverted.
What will be the angle of reflection if light falls normally on the reflecting surface?
For a plane mirror: If the incident ray falls along the normal, the angle of incidence is 0 degrees, NOT 90 degrees. The angle is measured with the normal to the mirror and not with the plane of the mirror. Hence the angle of reflection is equal to 0 degrees.
Angle of incidence is
Angle of incidence is the angle which an incident line or ray makes with a perpendicular to the surface at the point of incidence.
Angle of reflection is
Angle of reflection is the angle between ghe reflected ray and the normal angle of incidence is the angle between the incident ray and the normal always angle of incidence =angle of reflection
Which of the following letter shows lateral inversion ?
Letter L shows lateral inversion
Rahul is standing in front of a plane mirror. The distance between Rahul and his image is 150cm. If Rahul’s distance from the mirror is x. What could be the possible value of x ?
x = 75 cm, as for a plane mirror the distance between the object and the mirror is equal to the distance between the mirror and image.
Light always travels along a
Light follows rectilinear motion. It travels in straight line. A light wave can be considered to travel from one point to another along a straight line joining them.
Harry is standing in front of a plane mirror at a distance of 6 m. His image on the other side will form at a distance of
For reflection through a plane mirror, Image distance (v) = Object distance (u) = 6 m
In a lateral inversion
The left and right of a mirror image appear reversed ( lateral inversion) because of how we perceive the mirror image. The phenomenon where your left appears as the right and vice versa in a mirror is reffered to as lateral inversion.
Which among the following is true for a plane mirror ?
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https://www.analystforum.com/filed-under/121?page=6 | math | Study Session 13: Fixed Income: Topics in Fixed Income Analysis
In Credit Default Swaps, the recovery rate is the rate that the protection buyer gets in case of default let’s say it is 40%.
The payout ratio, is the amount that the protection seller pays. Based on the curriculum it is equivalent to 1-recovery rate, in that case it is equivalent to 60%.
My question is, should not the two be the same, as the amount paid by the seller is what the buyer gets??
I do not get this question!
[question removed by moderator]
Is not the total debt without defaulting $500,000, then why would I add the $200,000. As far as I understand, the $200,000 is the remaining debt that I will pay in case of default which is a part of the $500,000, so why would I add them together to get the value of the company’s total debt?
I’ve run into several questions that give you a binomial tree of interest rates and asks to calculate the interest rate call option value and provides a notion amount.
Let’s assume a 5% strike with a notion of $2M two year euro call option. The binomial tree is
4% 5.31% 8.3%
I am very confused with going Long CDS and going short CDS positions.
I have been doing some curriculum EOC questions and got incorrect answer. As I understand, protection buyers going Long on CDS position to transfer credit risk to protection seller, who going Short on CDS. However, the curriculum said something like this in its EOC solution: “shorting (buying protection) a long-term (20-year) CDX and going long (selling protection) a short-term (2-year) CDX”. Please correct my misunderstanding. I have spent hours on this.
The value of a straight bond does not change with interest rate volatility. Why???
When a company X make an unsolicited bid at a premium to acquire all the tradable shares of Y through issue of debt…say 5 n 10-Senior unsecured bonds. How can one profit from a equity-versus credit trade involving X and Y.
1. Short Y shares and short X 10 year CDS
2. Long Y shares and short X 5 yr CDS
3. Long X shares and Long X 5 year CDS
Is it not Long Y shares and Long X 5yr CDS?? The new debt will increase X’s credit spread as a result the CDS price must go up. However, the answer is option 2. How is this possible?
In reading 38, LOS D, we are given an intimidating Black Scholes model to price company debt based on an option analogy. However, the LOS says “Explain” not “Calculate”. So does this mean we can just understand the intuition of of what is happening and be fine or do we need to memorize this formula?
Can someone explain to me in which scenarios would you choose between a cash settlement and a physical settlement for a CDS using the cheapest to deliver bond? im just not getting the concept.
please help to clarify the below statement from p157 in Fixed Income topic (Interest rate risk of bonds with embedded options).
Compared side by side, putable bonds have more upside potential than otherwise identical callable bonds when interest rates decline. In contrast, when interest rates rise, callable bonds have more upside potential than otherwise identical putable bonds.
Study together. Pass together.
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http://www.biology.arizona.edu/biomath/tutorials/Quadratic/QuadraticFunctionApplications/LogPopMod.html | math | Scanning electron micrograph of the bacteria Heliobacterium chlorum. Photo credit: F. R. Turner, Indiana University, Bloomington/ Courtesy: National Science Foundation.
The logistic model of population growth assumes that the growth rate of the population decreases linearly with population size. In particular, the logistic equation gives the instantaneous rate of change of a population (ΔN) as,
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Number Sense and Operations, Algebra, symbols essay edit meaning Geometry, Measurement, and.
Tutorial videos on topics including arithmetic, algebra, trigonometry, and calculus Solutions in Saxon.
Algebra, i Unit #1. | s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232258849.89/warc/CC-MAIN-20190526045109-20190526071109-00405.warc.gz | CC-MAIN-2019-22 | 1,621 | 9 |
https://miraclelearningcentre.com/maths-lessons-topic-basic-concepts-of-time-speed-and-distance/ | math | 28 Apr Basic Concepts of Time, Speed and Distance
At Miracle Learning Centre, we teach you interesting stuff in math tuition. In one of our previous maths tuition, we had seen how to apply basic algebra in order to solve simple problems related to time, speed and distance. In this maths tuition, we are going further develop our understanding on the same topic and see how to apply algebra to solve some more difficult problems. Without any further delay, let us get started straight away.
This is a very common area of mistake. As a result, in this section of this math tuition, we are going to discuss how to compute the average speed.
So, consider the following problem. If you travel 100 kilometers at 25 kilometers/ hour and then travel another 100 kilometers at 20 kilometers/ hour, what is your average speed for the entire trip?
The formula for average speed goes as follows:
Average speed, vavg = (Total Distance Covered)/(Total TimeTaken)
So, in order to calculate the average speed, first we will have to figure out the total distance covered. In this case, the total distance covered = 100 + 100 = 200 kilometers.
Next, we are going to calculate the total time taken. The time taken for the first half of the trip:
t1 = 100/25 = 4 hours
The time taken for the second half of the trip:
t2 = 100/20 = 5 hours
So, total time taken = t = t1 + t2 = 4 + 5 = 9 hours.
As a result, the average trip for the entire trip = 200/9 = 22 2/9 kilometers/hours
Problems involving trains
Another popular set of problems related to time, speed and distance involves trains. In this section of this math tuition, we are going learn about the concepts which will help you to tackle these kinds of problems.
Let us discuss this with a sample problem. There is a tree just beside the railway tracks. How much time will a train which is 200 meters long and which is travelling at 10 meters/second, take to pass the tree?
The big idea for problems regarding trains is the fact that in order to pass something, the train must travel its own length plus the width of the object it is passing.
The width of the object the train is overtaking in this case, which is a tree, is negligible. As a result, we can simply calculate how long the train will take to cover a distance which is equal to its own length i.e. 200 meters in this case.
As a result, the time required = 200/10 = 20 seconds.
Now, suppose we are asked to calculate how long the same train above will take to pass platform which is 300 meters long.
In order to pass the platform completely, the train must cover a distance which is equal to the length of the train plus the length of the object which it has to pass i.e. the length of the platform.
As a result, the distance to be covered = 200 + 300 = 500 meters.
Therefore, required time to pass the platform = 500/10 = 50 seconds.
In this math tuition, we discussed two types of problems related to time, speed and distance and the related concepts using which we could tackle such problems. It is essential to have a solid understanding of the ideas discussed above in order to handle more complex problems related to time, speed and distance. As a result, it is extremely important that you have a thorough understanding of the above ideas. In case you face any difficulty in understanding any of the points discussed above, do not hesitate to talk to your teachers at math tuition or your peers. We hope you enjoyed this math tuition from Miracle Learning Centre. Look out for the next math tuition from Miracle Learning Centre. | s3://commoncrawl/crawl-data/CC-MAIN-2023-23/segments/1685224654606.93/warc/CC-MAIN-20230608071820-20230608101820-00261.warc.gz | CC-MAIN-2023-23 | 3,534 | 24 |
http://ctan.org/pkg/lshort-french | math | lshort-french – Short introduction to LaTeX, French translation
French version of A Short Introduction to LaTeX2ε.
|License||GNU General Public License|
Tobias Oetiker (inactive)
|Contained in||TeX Live as lshort-french|
|MiKTeX as lshort-french|
documentation in French
tutorial material for LaTeX
Download the contents of this package in one zip archive (2.1M).
Maybe you are interested in the following packages as well.
- lshort-japanese: Japanese version of A Short Introduction to LaTeX2ε
- lshort-vietnamese: Vietnamese version of the LaTeX introduction
- lshort-dutch: Introduction to LaTeX in Dutch
- lshort-zh-cn: Introduction to LaTeX, in Chinese | s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917122933.39/warc/CC-MAIN-20170423031202-00639-ip-10-145-167-34.ec2.internal.warc.gz | CC-MAIN-2017-17 | 661 | 14 |
https://www.coursehero.com/file/p6iuhhv/Click-Examine-and-move-the-mouse-across-the-force-vs-time-graph-When-the-force/ | math | 6.Click Examine, , and move the mouse across the force vs.time graph. When the force is maximum, is the acceleration maximum or minimum? To turn off Examine mode, click Examine, , again. 7.The graph of force vs.acceleration should appear to be a straight line. To fit a straight line to the data, click the graph, then click Linear Fit, . Record the equation for the regression line in the data table.8.Print copies of each graph.Trial II 9.Remove the Accelerometer, attach the 0.50 kgmass to the cart, and then re-attach theAccelerometer (see Figure 4). Record the totalmass of the cart, sensors, and additional mass inthe data table.10.Repeat Steps 4–8. DATA TABLETrial 1 Mass of cart with sensors (kg)Regression line for force vs. acceleration dataTrial IIMass of cart with sensors and additional mass (kg)Regression line for force vs. acceleration dataANALYSIS1.Are the net force on an object and the acceleration of the object directly proportional? Explain, using experimental data to support your answer.2.What are the units of the slope of the force vs.acceleration graph? Simplify the units of the slope to fundamental units (m, kg, s). 3.For each trial, compare the slope of the regression line to the mass being accelerated. What does the slope represent?4.Write a general equation that relates all three variables: force, mass, and acceleration.Figure 4
9 - 4 Physics with Vernier Computer 9 EXTENSION Use this apparatus as a way to measure mass. Place an unknown mass on the cart. Measure the acceleration for a known force and determine the mass of the unknown. Compare your answer with the actual mass of the cart, as measured using a balance.
You've reached the end of your free preview.
Want to read all 4 pages? | s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703496947.2/warc/CC-MAIN-20210115194851-20210115224851-00787.warc.gz | CC-MAIN-2021-04 | 1,731 | 4 |
https://edurank.org/engineering/petroleum/de/ | math | 10 Best universities for Petroleum Engineering in Germany
Below is a list of best universities in Germany ranked based on their research performance in Petroleum Engineering. A graph of 11.4K citations received by 759 academic papers made by 10 universities in Germany was used to calculate publications' ratings, which then were adjusted for release dates and added to final scores.
We don't distinguish between undergraduate and graduate programs nor do we adjust for current majors offered. You can find information about granted degrees on a university page but always double-check with the university website.
1. RWTH Aachen University
2. Clausthal University of Technology
3. University of Stuttgart
4. University of Potsdam
5. Karlsruhe Institute of Technology
6. Technical University of Munich
7. Kiel University
8. University of Tubingen
9. Dresden University of Technology
10. Leibniz University of Hanover
The best cities to study Petroleum Engineering in Germany based on the number of universities and their ranks are Aachen, Clausthal-Zellerfeld, Stuttgart, and Potsdam. | s3://commoncrawl/crawl-data/CC-MAIN-2023-23/segments/1685224648245.63/warc/CC-MAIN-20230602003804-20230602033804-00329.warc.gz | CC-MAIN-2023-23 | 1,084 | 14 |
https://nmfin.tech/forums/reply/re-how-to-compute-sharpe-ratio-for-a-high-frequency-multi-asset-longshort-strategy-4/ | math | Thank you Kevin for the ingenious insight.
If I understand you correct, to implement the SR computation, we need to keep track of a running capital [tex]V_t[/tex]. Then, we compute the return [tex]R_t[/tex] at some predetermined time points.
A few question –
Where does the [tex]$3.67[/tex] in your discussion come from? Sorry I couldn’t figure it out… :-[
How should we compute the SR for a basket? E.g., we trade a pair of [tex]Z = X – alpha Y[/tex]. Suppose X and Y are traded at different times, as in Ken’s example. So we can compute [tex]SR_X[/tex] and [tex]SR_Y[/tex], how do we “combine” them somehow to compute [tex]SR_Z[/tex]? Or, do we compute [tex]SR_Z[/tex] directly without computing for the individual assets separately? | s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400234232.50/warc/CC-MAIN-20200926040104-20200926070104-00595.warc.gz | CC-MAIN-2020-40 | 749 | 5 |
https://stepessays.com/psychology-construct-validity/ | math | Psychology Construct Validity
Topic: Construct Validity
Validity measures the usefulness of a test for specific purposes. For instance, tests of depression, such as the Beck Depression Inventory (BDI), correlate well with clinician assessments of depression, so they can be considered valid measures of depression. On the other hand, the BDI does not predict job performance well, so it would be invalid for that purpose.
Types of validity include criterion-related, content, construct, and face validity.
? •Criterion-related validity looks at the correlation between test scores and a criterion that the test scores could be expected to predict. For instance, SAT scores for high school seniors could be expected to correlate with their first-year college GPA.
? •Content validity generally depends on expert opinion and looks at whether the test adequately samples the content of interest. A math test for high school students should sample the types of math problems that are found in high school math textbooks and should cover the domains that high school math teachers say are important.
? •Face validity is important but is not validity in the technical sense. A test has face validity if it appears to measure what it purportedly measures. For instance, a math test for truck drivers should ask questions about gas mileage and load weight. If it contained questions about cake recipe proportions, it would lack face validity.
? •Construct validity is usually accumulated over time and represents the accumulation of many validity studies. It indicates whether the test appropriately measures a theoretical construct.
For this Discussion, you concentrate on construct validity. To prepare, choose a construct to use (one that was not used in the examples) and consider how you might assess the construct validity of a measure of that construct. Then think about how the reliability of your measure might influence its validity.
With these thoughts in mind:
? Post a detailed description of the construct you selected.
? Then explain how you might assess the construct validity of a measure of that construct.
? Finally, explain the influence of reliability of your measure on the magnitude of its validity.
Support your response using current sources. Your post should be three solid paragraphs.
0ne page 2, or more sources | s3://commoncrawl/crawl-data/CC-MAIN-2023-50/segments/1700679100164.87/warc/CC-MAIN-20231130031610-20231130061610-00141.warc.gz | CC-MAIN-2023-50 | 2,339 | 15 |
https://www.cuemath.com/questions/what-is-11-percent-as-a-fraction-in-simplest-form/ | math | from a handpicked tutor in LIVE 1-to-1 classes
What is 11% as a fraction in simplest form?
To write a fraction in its simplest form means that the numerator and denominator can no longer be divided by the same whole number exactly or evenly.
Answer: The value of 11% as a fraction in the simplest form is 11/100.
Let us convert 11% into a fraction in the simplest form.
Step 1: Convert 11% to a fraction form by removing the percent sign and dividing it by 100.
11% = 11/100
Therefore, the GCF is the 1.
Hence we cannot further reduce the fraction 11/100.
Thus, 11% as a fraction in the simplest form is 11/100. | s3://commoncrawl/crawl-data/CC-MAIN-2023-40/segments/1695233511351.18/warc/CC-MAIN-20231004020329-20231004050329-00228.warc.gz | CC-MAIN-2023-40 | 611 | 10 |
https://www.gmtlight.com/sell/show/2026/ | math | [product model] 3D exaggerated iron blood warrior latex mask
[commodity color] multi color mask
[commodity material] latex
[item size] adult size, 12 * 16cm
[single weight] 120g
[packaging] 100 pieces in a box, 53x40x40cm, 13kgs
Tip: the product is latex products. If the taste of latex is normal, the taste will gradually decrease after placing it under wind. | s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499919.70/warc/CC-MAIN-20230201081311-20230201111311-00219.warc.gz | CC-MAIN-2023-06 | 360 | 7 |
http://www.studymode.com/essays/Effect-Of-Mobile-Phones-On-Life-1273675.html | math | Little's Theorem (sometimes called Little's Law) is a statement of what was a "folk theorem" in operations research for many years:
N = λT
where N is the random variable for the number of jobs or customers in a system, λ is the arrival rate at which jobs arrive, and T is the random variable for the time a job spends in the system (all of this assuming steady-state). What is remarkable about Little's Theorem is that it applies to any system, regardless of the arrival time process or what the "system" looks like inside. Proof: Define the following:
α ( t ) ≡ number of arrivals in the interval (0,t ) δ ( t ) ≡ number of departures in the interval (0,t )
N ( t ) ≡ number of jobs in the system at time t = α (t ) − δ( t )
γ ( t ) ≡ accumulated customer - seconds in (0,t )
These functions are graphically shown in the following figure:
The shaded area between the arrival and departure curves is γ (t ) .
λ t = arrival rate over the interval (0,t )
α (t ) t
N t = average # of jobs during the interval (0,t ) =
γ (t) t
Tt = average time a job spends in the system in (0,t )
γ (t) α (t)
⇒ γ ( t ) = Ttα ( t ) T α (t ) ⇒ Nt = t = λt Tt t
Assume that the following limits exist:
€ lim λt = λ
lim Tt = T
lim N t = N
also exists and is given by N = λT .
Keywords: Little's Law Little's Theorem Steady state
Page 2 of 2
Cite This Essay
(2012, 12). Effect of Mobile Phones on Life. StudyMode.com. Retrieved 12, 2012, from http://www.studymode.com/essays/Effect-Of-Mobile-Phones-On-Life-1273675.html
"Effect of Mobile Phones on Life" StudyMode.com. 12 2012. 12 2012 <http://www.studymode.com/essays/Effect-Of-Mobile-Phones-On-Life-1273675.html>.
"Effect of Mobile Phones on Life." StudyMode.com. 12, 2012. Accessed 12, 2012. http://www.studymode.com/essays/Effect-Of-Mobile-Phones-On-Life-1273675.html. | s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368705618968/warc/CC-MAIN-20130516120018-00097-ip-10-60-113-184.ec2.internal.warc.gz | CC-MAIN-2013-20 | 1,834 | 26 |
https://plainmath.net/45287/a-quarterback-takes-the-ball-from-the-line-of-scrimmage-runs-backward | math | A quarterback takes the ball from the line of scrimmage, runs backwards for 10.0 yards, then runs sideways parallel to the line of scrimmage for 15.0 yards. At this point, he throws a 50.0-yard forward pass straight downfield, perpendicular to the line of scrimmage. What is the magnitude of the football’s resultant displacement?
Answer & Explanation
Let us denote the displacements of the ball made by the quarterback as:
We can calculate the resulting displacement,
Most Popular Questions | s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711637.64/warc/CC-MAIN-20221210005738-20221210035738-00261.warc.gz | CC-MAIN-2022-49 | 493 | 5 |
https://tethys-engineering.pnnl.gov/publications/oscillating-wave-energy-converter-nonlinear-snap-through-power-take-systems-regular | math | Floating oscillating bodies constitute a large class of wave energy converters, especially for offshore deployment. Usually the Power-Take-Off (PTO) system is a directly linear electric generator or a hydraulic motor that drives an electric generator. The PTO system is simplified as a linear spring and a linear damper. However the conversion is less powerful with wave periods off resonance. Thus, a nonlinear snap-through mechanism with two symmetrically oblique springs and a linear damper is applied in the PTO system. The nonlinear snap-through mechanism is characteristics of negative stiffness and double-well potential. An important nonlinear parameter γ is defined as the ratio of half of the horizontal distance between the two springs to the original length of both springs. Time domain method is applied to the dynamics of wave energy converter in regular waves. And the state space model is used to replace the convolution terms in the time domain equation. The results show that the energy harvested by the nonlinear PTO system is larger than that by linear system for low frequency input. While the power captured by nonlinear converters is slightly smaller than that by linear converters for high frequency input. The wave amplitude, damping coefficient of PTO systems and the nonlinear parameter γ affect power capture performance of nonlinear converters. The oscillation of nonlinear wave energy converters may be local or periodically inter well for certain values of the incident wave frequency and the nonlinear parameter γ, which is different from linear converters characteristics of sinusoidal response in regular waves. | s3://commoncrawl/crawl-data/CC-MAIN-2024-10/segments/1707947473360.9/warc/CC-MAIN-20240221002544-20240221032544-00157.warc.gz | CC-MAIN-2024-10 | 1,648 | 1 |
https://www.arxiv-vanity.com/papers/hep-th/0307287/ | math | IFUP-TH/2003-23, ITEP-TH-42/03, TIT-HEP/506
NONABELIAN SUPERCONDUCTORS: VORTICES AND CONFINEMENT
Roberto AUZZI , Stefano BOLOGNESI ,
Jarah EVSLIN , Kenichi KONISHI , Alexei YUNG
Scuola Normale Superiore - Pisa , Piazza dei Cavalieri 7, Pisa, Italy
Dipartimento di Fisica “E. Fermi” – Università di Pisa ,
Istituto Nazionale di Fisica Nucleare – Sezione di Pisa ,
Via Buonarroti, 2, Ed. C, 56127 Pisa, Italy
Dept. of Physics, Tokyo Inst. of Technology , 2-12-1 Oh-okayama, Meguro-ku, Tokyo 152-8551 Japan
Petersburg Nuclear Physics Institute , Gatchina, 188300 St. Petersburg, Russia
Institute of Theoretical and Experimental Physics , B. Cheryomushkinskaia 25, 117259 Moscow, Russia
We study nonabelian vortices (flux tubes) in gauge theories, which are responsible for the confinement of (nonabelian) magnetic monopoles. In particular a detailed analysis is given of SQCD with gauge group deformed by a small adjoint chiral multiplet mass. Tuning the bare quark masses (which we take to be large) to a common value , we consider a particular vacuum of this theory in which an subgroup of the gauge group remains unbroken. We consider flavors so that the sub-sector remains non asymptotically free: the vortices carrying nonabelian fluxes may be reliably studied in a semi-classical regime. We show that the vortices indeed acquire exact zero modes which generate global rotations of the flux in an group. We study an effective world sheet theory of these orientational zero modes which reduces to an sigma model in (1+1) dimensions. Mirror symmetry then teaches us that the dual group is not dynamically broken.
1. Introduction and Discussion
Some sort of nonabelian vortices are believed to be responsible for confinement in QCD. Although in string theory these objects appear naturally, they turn out to be somewhat elusive in four-dimensional field theories. The existing literature on the subject certainly provides an incomplete picture.
There are several reasons for this unsatisfactory situation. One of the reasons is that boundstates of vortices are not generally stable. An example is the case of an gauge theory (e.g., gauge theory with all fields in the adjoint representation) broken completely by a Higgs mechanism, where possible vortices represent nontrivial elements of the fundamental group
-charged objects cannot be BPS saturated [1, 2], and this fact, together with the unknown dependence of their properties on the form of the potential, number of the fields, etc., has obstructed investigations of such vortices.
Secondly, often these theories become strongly coupled at low energies and therefore an analytical study of the vortex configurations is very difficult. For instance, confinement in QCD may be due to the vortices of electric fields appearing in a dual (magnetic) ( or ?) theory. Unfortunately, neither the true nature of the effective magnetic degrees of freedom nor their form of interactions is known at the moment. ’t Hooft’s suggestion that they be abelian monopoles of a gauge-fixed theory , must still be verified. On the other hand, there is no experimental indication that the gauge group is dynamically broken to .
Finally, in the examples of classical solutions for “nonabelian vortices” discussed so far in the literature the vortex flux is actually always oriented in a fixed direction in the Cartan subalgebra, showing that they are basically abelian.
Useful hints come from the detailed study of a wide class of softly broken supersymmetric gauge theories where the dynamics appears particularly transparent. It was shown that, in fact, different types of confining vacua are realized in these models [5, 6, 7, 8, 9]. It is possible that in some cases confinement is due to the condensation of monopoles associated with the maximally abelian subgroup (a dual Meissner effect), as in the vacua surviving the adjoint mass perturbation in the pure SYM [5, 7, 8]. These cases provided the first examples of four-dimensional gauge theory models in which the ’t Hooft-Mandelstam mechanism of confinement is realized and can be analysed quantitatively. A detailed study of these cases has shown however that dynamical abelianization takes place there, with a characteristically richer meson spectrum [8, 1, 10, 11]. Indeed the low-energy effective gauge group of the theory is and the meson spectrum is classified according to the number of possible abelian strings via
(cfr. (1.1)). Thus vorties and therefore mesons come in infinite towers, a feature not expected in the real world QCD.
However, such is not the typical situation in softly broken theories with fundamental matter fields (quarks) [12, 9]. Confining vacua in , and gauge theories with quark flavor, are typically described by effective nonabelian dual gauge theories. For instance, in the so-called -vacua of gauge theory with flavors and vanishing bare quark masses, the low-energy effective theory is a dual theory. Addition of the adjoint chiral multiplet mass term breaks supersymmetry to , and the dual quarks in the of condense. These “dual quarks” have been recently identified as the quantum Goddard-Nuyts-Olive-Weinberg monopoles [14, 15]. Their condensation is believed to give rise to nonabelian confinement via formation of nonabelian flux tubes.
In fact, the problem of nonabelian vortices is very closely related to (in a sense, it is one and the same problem as) that of the nonabelian monopoles on which they end. A key feature found in is that the quantum behavior of the nonabelian monopoles, and in fact the vacuum properties themselves depend critically on the presence of massless flavors of matter. We shall find below that the existence of nonabelian vortices similarly requires the presence of massless flavors in the underlying theory.
Inspired by these developments, and based on a work by Marshakov and one of the authors (A.Y.) , we present in this paper a study of nonabelian superconductors, concentrating our attention on the properties of the vortices appearing in these systems. In a companion paper , we shall explore more extensively the properties of nonabelian monopoles themselves.
Our analyses are done in a context where the dynamics of the model is well understood and the transition from a theory with abelian vortices to one with nonabelian vortices can be studied in a weakly coupled semi-classical regime throughout. The model we consider is probably the simplest of such models, QCD with gauge group and hypermultiplets of fundamental matter (quarks). Upon deformation of this theory via a small mass term for the adjoint chiral multiplet, , the Coulomb branch of the theory shrinks to a number of isolated vacua.
Generically the vacuum expectation value (VEV) of the adjoint field breaks the gauge symmetry down to . However, it was shown in [12, 9] (see also ) that some of the vacua of QCD preserve a nonabelian subgroup. These vacua are classified by an integer . In a semiclassical regime, which is valid at large bare quark masses
the adjoint scalar VEVs in those vacua take the form,
where quark masses out of possible masses are chosen to satisfy the vacuum equations.
When the quark masses are tuned to a common value , the pattern of the spontaneous breaking changes to
The sector is a pure Yang Mills theory 111Recall that the quark masses come from the superpotentials and becomes strongly interacting at low energies and gets dynamically broken to . The sector, on the other hand, having massless flavors, remains weakly coupled as long as .
Furthermore, in the presence of the aforementioned adjoint mass perturbation, the light squark fields acquire VEVs of color-flavor diagonal form (“Color-Flavor Locking”),
which breaks the gauge group completely at scales far below the bare quark masses: . The theory is now in the Higgs phase, and develops vortex configurations, representing nontrivial elements of
The key fact is that the system has an exact global symmetry, respected both by the interactions and by the scalar VEVS (1.4) and (1.6). A given vortex configuration however breaks this symmetry: it turns out that the symmetry is broken as (see below.) As a result, exact orientation zero modes of are generated.
To work things out concretely, we analyse the case of the vacua of the gauge theory ( above) in detail in the main body of this paper. The subgroup, classically restored in the limit of equal quark masses, stays unbroken in the full quantum theory, as the relevant sector of the theory is infrared free if , or is conformal invariant if . On the other hand, of course, the underlying gauge theory is asymptotically free for , so we shall take to be either or .
This is one of the important points of our analysis: by working in the regime in which the interactions remain weak at all scales, the continuous transition from the theory with abelian vortices (unequal quark masses) to the theory with nonabelian vortices which are qualitatively different, can be studied explicitly and reliably.
The unbroken gauge group is further broken at a much lower mass scale, yielding vortices representing the nontrivial elements of
Indeed, as the bare quark masses are tuned to a common value, , starting from unequal and generic values, the low-energy gauge group gets enhanced from to . The set of abelian vortices appearing in the unequal mass cases acquires a certain degeneracy and at the same time some orientation (in the color space) zero modes appear which relate the vortices of the same tension by global rotations. These zero modes are associated with the diagonal global subgroup of color crossed with the flavor which is an exact symmetry of the system. More precisely, the vortex zero modes parametrize as each vortex solution breaks the exact symmetry to a subgroup.
We then work out the effective world-sheet theory of the vortex zero modes, and show that it reduces to the sigma model in (1+1) dimensions. Classically the sigma model has spontaneous symmetry breaking and appears to yield massless Goldstone fields. In terms of strings in four dimensions this would mean that is spontaneously broken and the string flux is oriented in some particular direction inside the gauge subgroup.
However the quantum physics of the sigma model in (1+1) is quite different. It is well understood using the mirror map , which relates it to a sine-Gordon theory. In particular it is known that the model has a mass gap and no spontaneous symmetry breaking. In terms of strings in 4D this means that the string is not oriented in any particular direction inside group. This ensures that our vortices are truly nonabelian. The sine-Gordon superpotential is generated dynamically in the effective (1+1)-dimensional worldsheet theory which produces exactly two vacua.
Our considerations can be straightforwardly generalized to the vacua of the theory with , with unbroken group, although our analysis in these more general cases is less complete. In particular, in the case of an theory broken to the zero modes of the vortex are described by a 2-dimensional CP sigma model whose mirror is an affine Toda theory with the desired vacua.
The vortices studied in this paper, though stable in the low-energy theory, are strictly speaking metastable as the underlying gauge group (e.g., ) is simply connected. Their decay rates are however small, being exponentially suppressed by ratios of heavy monopole masses squared to the string tensions [19, 20] .
Our result provides, albeit indirectly, a couterexample to the no-go theorem on the existence of monopoles with nonabelian charges discussed earlier . These nonabelian monopoles do exist in our theory as stable solitons and act as the sources of the nonabelian vortices considered here, and are actually confined by them. We exhibit here explicitly the transformations among the vortices, which imply certain non-local transformations for their sources. We will see that the zero modes of the vortices are normalizable. To calculate the zero mode of a single monopole, which necessarily sources an infinite vortex, we must integrate that of the vortex along its infinite length. Thus we find, as was seen in the flavorless cases of Refs. , that the zero mode of a single monopole is nonnormalizable. In a color-neutral configuration of monopoles the total length of the vortices may be taken to be finite and so the integral is finite, yielding normalizable zero modes which again generalize those known to exist in the flavorless case.
Throughtout this paper we limit ourselves to cases with large bare quark masses where the original electric subgroup remains weakly coupled. When the bare quark masses are tuned to small values or even to zero, the low-energy system is weakly coupled when described in terms of the magnetic variables instead of the electric ones. The excitations which are quarks in the electric description at large quark masses become monopoles in the magnetic description at small quark masses [6, 22]. The properties of the corresponding - vacua have been studied in detail in , and in the case of a SCFT vacua of theory, in . The properties of these corresponding vacua are closely related by holomorphy.
The organization of the paper is as follows. In Sect. 2 we review QCD with equal quark masses, work out its low-energy description, vacuum structure and the low-energy spectrum. In Sect. 3 we derive nonabelian Bogomolny equations, construct vortices and study their zero modes. We discuss the generalization to the more general case of breaking in Sect. 4. In Sect. 5 we work out the effective world sheet theory for orientational zero modes and discuss its physics.
While this work was in preparation Ref. appeared which considers vortices in the very similar three-dimensional theory with an FI term. While these vortices are not strings but particles, the worldvolume theories appear to be related by dimensional reduction, and the vacuum structures and spectra appear to be the same. Thus many of our results as well as an extensive analysis of the relevent moduli spaces may be found there.
2.1. The Model
The field content of QCD with the gauge group SU(3) and flavors of chiral multiplets is as follows. The vector multiplet consists of the gauge field , two Weyl fermions , and the scalar field , all in the adjoint representation of the gauge group. Here is a spinor index while all adjoint fields are matrices in the Lie algebra .
The chiral multiplets of the theory consist of complex scalar squarks and and Weyl fermion quarks and , all in the fundamental representation of the gauge group. Here is a color index while is a flavor index, .
This theory has a Coulomb branch on which the adjoint scalar acquires the vacuum expectation value (VEV)
generically breaking the gauge group down to . Here and are the Gell-Mann matrices of the Cartan subalgebra.
In this paper we consider the special vacua for which
For these vacua the low-energy gauge group is , at least classically.
We perturb the above theory by adding a small mass term for the adjoint matter via the superpotential
Generally speaking, the superpotential breaks down to . The Coulomb branch shrinks to a number of isolated vacua [12, 9]. In the limit these vacua correspond to special singular points on the Coulomb branch in which pairs of monopoles/dyons or quarks become massless. Three of these points are always at strong coupling. They correspond to vacua of the pure gauge theory. The massless quark points are at weak coupling if the quark masses are large, . The vacua in which quarks become massless will be referred to as the quark vacua. We shall be mainly interested in these quark vacua.
It is important to note that supersymmetry is not broken to the leading order in the parameter in the effective theory [1, 11]. In the effective low-energy theory the superpotential (2.3) gives rise to a superpotential linear in plus higher order corrections. If only the linear term in ’s in the superpotential is kept and if we restrict our attention to the special vacua (2.2), then it reduces to a Fayet-Iliopoulos term which does not break the supersymmetry.
2.2. symmetric low-energy theory
The gauge group is broken down to by the VEV of the adjoint scalar (2.1) at generic values of quark masses. However, in the equal quark mass limit () which we shall consider from now on, the VEV of the field vanishes (see Sec. (2.3.)), and the low-energy gauge group is . W-bosons which are charged with respect to both factors of the low-energy group acquire a large mass of order . The third color components of quarks also become heavy in this vacua with masses of order of .
Let us consider now the scales of order , which are well below W-boson masses ( is taken small, ). There the low-energy theory contains the following light fields of the vector multiplet: four complex scalar light fields and where is the color index, one gauge field and one gauge field together with their fermionic superpartners. For example the gauge fields are defined as follows:
where our notation corresponds to expanding gauge and adjoint fields in the orthogonal basis of the Gell-Mann matrices, being the first three Gell-Mann matrices normalized as
Light quark multiplets contain complex scalar -doublets , together with their fermionic superpartners, .
The bosonic part of the low-energy effective theory then acquires the form
Here is the covariant derivative in the adjoint representation of gauge subgroup, while
where we suppress the color indices and are Pauli matrices. The coupling constants and correspond to and sectors respectively. The potential in the Lagrangian (2.5) is given by the D and F terms
where other D-terms involving the adjoint scalar fields and () (which vanish at and ) are left implicit. The term in the second line arises when we expand fields and in the superpotential (2.3) around their VEV’s and keep only terms linear in fluctuations of these fields. As we have already noted, this means that the theory in (2.5) is a bosonic part of a supersymmetric theory. In particular this ensures that our theory has BPS vortices [1, 11, 16] (see also the seventh ref. in ).
Below the scale the gauge group is broken and we have two coupling constants and which run according to the and renormalization group flows respectively. Note that with a logarithmic accuracy we can neglect mixing of these two coupling constants. In the case with four flavors the coupling does not run ( theory with is conformal) and is given by its value at the scale
Since at large the sector is weakly coupled, it remains so at low energies.
The coupling undergoes an additional renormalization from the scale to the scale determined by the masses of light states in the low-energy theory (which are of the order of , see next subsection). Thus we have
where we use the fact that the one loop coefficient of the -function for theory is and substitute and the electric charge , see (2.6). Clearly, this coupling is even smaller than the one in the sector.
If the number of the quark flavors is taken to be five, the gauge coupling constant also runs to smaller values towards the infrared. In general therefore one has , both small, and we shall not need more details in the analyses below.
2.3. Vacuum structure and low-energy spectrum
In this subsection we review the vacuum structure and low-energy mass spectrum of QCD [12, 9] generalizing the analysis made in to the case of the low-energy group. To find the vacua of the effective theory (2.5) we have to look for the zeros of the potential (2.7). At generic large values of quark masses solutions have the following structure [9, 16]. Besides the three strong coupling vacua which exist already in the pure gauge theory there are vacua and vacua, were is the number of quark flavors which develop non-zero VEV’s.
Here we are mostly interested in vacua, which have an unbroken gauge group which becomes exact in the case of equal quark masses. Clearly the minimal number of flavors for which we can have a vacuum is . Let us consider this case first.
The adjoint scalar matrix is given by
where is the common mass of both flavors. In the above notation (2.10) reads
For real values of and we can use gauge rotations to make squark VEV’s real. We write the squark field as a matrix where is a color index and is a flavor one. Then the squark VEV’s are given by
where we have used color-flavor mixed matrix notation, and we have introduced
which acts as the Fayet-Iliopoulos parameter of the . sets the scale of the low-energy theory (2.5). Only the two upper color components and the first two flavors are shown in Eq.(2.12): all other components have vanishing VEVS.
Now consider the spectrum of light fields in this vacuum. The low-energy gauge group is broken completely by squark VEV’s and all gauge bosons acquire masses. The mass matrix for the gauge fields , can be read off of the kinetic terms for the quarks in (2.5). It turns out that it is diagonal in the basis , . In particular, the mass of is given by
while the mass of the W-boson is
The mass matrix for squarks is now of size including four real components of complex fields and for each color and flavor. It has four zero eigenvalues associated with the four states “eaten” by the Higgs mechanism for and gauge factors and two non-zero eigenvalues coinciding with gauge boson masses (2.14) and (2.15). The eigenvalue (2.14) corresponds to three squark eigenvectors while the one in (2.15) corresponds to nine squark eigenvalues.
Altogether we have one long multiplet with mass (2.14), containing eight bosonic states (3 states of the massive field plus 2 states of plus 3 squark states) and eight fermionic states. In addition we have three long multiplets with mass (2.15) labeled by the color index also containing eight bosonic and eight fermionic states each 222 See for a discussion of the emergence of long multiplets in Seiberg-Witten theory upon adjoint mass term deformation.. Note that no Nambu-Goldstone multiplets appear in this vacuum: all phases associated with broken symmetries are ”eaten” by Higgs mechanism.
Actually, in the theory with discussed above, the gauge interactions become strong below the scale , and the properties of the theory at low energies (at mass scales of order of ) cannot be determined from the Lagrangian (2.5) only.
For this reason, we introduce more flavors into our theory and consider the theory with or . The low-energy then remains in the weak coupling regime.
This theory has vacua of the type described above, for unequal quark masses. Each of these vacua corresponds to choosing two flavors out of which develop VEV’s. This gives choices for . In the limit of equal masses all six vacua coalesce and a Higgs branch develops from the common root. The dimension of this Higgs branch is [12, 16]. To see this note that we have real variables subject to four -term and eight -term conditions in the potential (2.7). Also 3+1 gauge phases are eaten by the Higgs mechanism. Thus we have remaining degrees of freedom.
We consider below a special submanifold of the Higgs branch which admits BPS flux tubes (cf. [1, 25, 16, 26]). This base submanifold is compact and has the minimal value of the quark condensate . One point on this submanifold which corresponds to non-zero VEV of the first flavor and non-zero VEV of the second flavor while all other components are zero is given in (2.12).
Other points on the base of the Higgs branch are given by a flavor rotation of (2.12). The dimension of the base submanifold of the Higgs branch is . To see this note that VEV’s of two flavors break symmetry down to . Thus the number of ”broken” generators is and also we have to subtract four phases “eaten” by the Higgs mechanism.
Other points on the dimensional Higgs branch correspond to non-zero VEV’s of massless moduli fields, and these points do not admit BPS strings. In particular, the ANO strings on the Higgs branch were studied in [28, 26], they correspond to a limiting case of type I strings with the logarithmically thick tails associated with massless scalar fields. We shall not discuss here strings at generic points on the Higgs branch.
Before ending this subsection, we need to comment on the soliton sector. In the monopole sector, all solitonic states associated with the symmetry breaking (2.10) are massive. In particular, one finds an exactly degenerate doublet of BPS monopoles of minimum mass .
Apparently, a set of “monopole” states become massless as the bare quark masses are tuned to a common value, , at which point the low-energy gauge group gets enhanced from to . For instance, the BPS monopole carrying magnetic charge (1,-1) with respect to two factors above has mass proportional to , and appears to become massless in the limit of restoration. Classically this “state” becomes infinitely extended in space in such a limit, and at the same time the fields degenerate into trivial vacuum configuration 333This is analogous to the fate of the ’t Hooft - Polyakov monopole of the spontaneously broken theory, in the limit . . More importantly, as the topological structure of the theory changes in the restoration limit (from Eq.(3.6) to Eq.(3.13), see below) such a “massless monopole” is no longer topologically stable.
3. Non-abelian Vortices
We will now construct (BPS) vortex solutions in the theory described above and show that they possess exact zero modes.
3.1. Non-abelian Bogomolny Equations
As we have already anticipated, by restricting ourselves to a particular base submanifold of the Higgs branch of the theory with four flavors, we are able to deal with BPS strings throughout. By gauge and flavor rotations the squark VEVS can be taken to be of the form (2.12). Then classically only the two flavors which develop VEV’s will play a role in the vortex solution. Other flavors remain zero on the solution, and one can consider the squark fields to be matrices. Note however that the additional two flavors are crucial in the quantum theory, in keeping the interactions weakly coupled 444In fact the additional flavors are important even classically. In the presence of additional flavors strings can turn into semilocal strings, see for a review on semilocal strings. We shall not study this issue here..
Let us make an ansatz,
and a convenient redefinition of the squark fields The low-energy action (2.5) then reduces ( and stand for the and coupling constants at the scale , respectively) to
where , leading to the following first order equations for strings
Here is the sign of the total flux specified below.
The string solutions found in the case of unequal quark masses can be readily recognized as particular solutions of these equations. To construct them we further restrict the gauge field to the single color component (by setting ), and consider only squark fields of the color-flavor diagonal form:
by setting all other components to zero. For unequal masses the relevant topological classification was
and the allowed strings formed a lattice labeled by two integer winding numbers. In particular, assume that the first flavor winds times while the second flavor winds times and look for solutions of (3.4) using the following ansatz 555We use a notation slightly different from the one used in : instead of ; instead of . The cylindrical coordinates are here denoted as ), the vortex center extending along the axis.
where are polar coordinates in the (1,2) plane while the profile functions , for scalar fields and , for gauge fields depend only on .
The profile functions in these equations are determined by the following boundary conditions
for the gauge fields, and the requirement that the squark fields be everywhere regular. The behavior of the latter at ,
and that at (e.g., if ), follow from these requirements. Here the sign of the string flux is
The tension of a -string for the case of equal quark masses is determined by the flux of the gauge field alone and is given by
Note that and -strings are exactly degenerate.
Note also that does not enter the central charge of the algebra and so does not affect the string tension. The stability of the string in this case is due to the factor of the low-energy group only.
The equations (3.4) represent a nonabelian generalization of the Bogomolny equations for the ANO string . For a generic -string equations (3.8) do not reduce to the standard Bogomolny equations. For instance, for the -string these equations reduce to two Bogomolny equations while for the and strings they do not.
The charges of -strings can be plotted on the Cartan plane of the algebra. We shall use the convention of labeling the flux of a given string by the magnetic charge of the monopole which produces this flux and must be attached to each end. This is possible since both string fluxes and monopole charges are elements of the group . This convention is convenient because specifying the flux of a given string automatically fixes the charge of the monopole that it confines.
Our strings are formed by the condensation of squarks which have electric charges equal to the weights of algebra. The Dirac quantization condition tells us that the lattice of -strings is formed by roots of the algebra . The lattice of -strings is shown in Fig. 3. Two strings and are the “elementary” or “minimal” BPS strings. If we plot two lines along charges of these “elementary” strings (see Fig. 3) they divide the lattice into four sectors. It turns out that the strings in the upper and lower sectors, which are labeled by black circles in Fig. 3, are BPS but they are marginally unstable at real quark mass ratios. Instead, strings in the right and left sectors, which are labeled in Fig. 3 by white circles, are bound states of the “elementary” ones but they are not BPS.
3.2. Minimal vortex of generic orientation: zero modes
Actually, the relevant homotopy group here is
instead of (3.6), as we are working with the case of equal quark masses where the low-energy gauge group is . The generator of the fundamental group is a loop which encircles the once , and thus to calculate the tension of a string, or to determine whether it is stable, it suffices to simply count the winding number around this circle. This means that the lattice of -strings reduces to a tower labeled by one integer . For instance, the -string becomes completely unstable as it winds forward once and then backward once, and so there is no net winding and so no topological charge. On the restored group manifold it is also trivial, as it goes half way around the equator and then goes back. The string goes all of the way around the equator, making a contractible loop, but is stable because it wraps the twice (it wraps the original once).
On the other hand, the and strings cannot be shrunk because they correspond to a half circle along the equator. They have the same tension (see (3.12)) for equal quark masses and thus apparently belong to doublet of an .
In general non-BPS strings on the -lattice (see Fig. 3) become unstable as they have tensions above their BPS bounds and we are left with BPS strings at each winding number . The reduction of the string lattice is illustrated Fig. 4.
Most importantly, this suggests that there be a continuously infinite number of vortices of minimum winding and with the same tension,
of which the and vortices discussed above are just two particular cases (Fig. 5).
Below we show that this is indeed correct, by a continuous deformation of the -string solution transforming it into a -string. This deformation leaves the string tension unchanged and therefore corresponds to an orientational zero mode.
First let us separate physical variables from the gauge phases eaten by the Higgs mechanism in the quark fields. To do so we use the following parametrization of the quark matrix
Here and are matrices from the and gauge factors respectively while and are real. The parametrization (3.15) represents eight real variables in terms of 3+1=4 gauge phases eaten by the Higgs mechanism and four physical variables and . In particular, (2.12) corresponds to
Now let us fix the unitary gauge (at least globally, which is enough for our purposes) by imposing the condition that squark VEV’s are given precisely by (3.16) and so all gauge phases are zero. Now transform the -string solution (3.7) into unitary gauge, which corresponds to the singular gauge in which the string flux comes from the singularity of the gauge potential at zero. In this gauge the solution (3.7) for the -string takes the form
Note that a global diagonal subgroup in the product of gauge and flavor symmetries is not broken by the squark VEV. Namely,
where is a global rotation in while the squark VEV matrix is given by (2.12). We call this unbroken group .
Now let us apply this global rotation to the string solution (3.17). We find
where we use a matrix notation for the gauge field, . Using the representation 666Explicitly, if , the rotation matrix is given by .
where is a unit vector on , , we can rewrite the gauge potential of (3.19) in the form
Since the symmetry is not broken by squark VEV’s it is physical and does not correspond to any of the gauge rotations eaten by the Higgs mechanism. To see this explicitly let us rewrite the quark field of our solution (3.19) using the parametrization (3.15). We get
We see that all gauge phases are zero while physical variables acquire an -dependence. Clearly the solution (3.19) interpolates between and strings. In particular it gives a -string for and a -string for .
The symmetry is exact and the tension of the string solution (3.19) is independent of :
see (3.12). However, an explicit vortex solution breaks the exact as
the two angles associated with vector - two orientational bosonic zero modes of the string - parametrize the quotient space .
In the regular gauge, the minimal nonabelian vortex of generic orientation (3.19) takes the form
where is given by Eq. (3.20) and the profile functions are solutions of Eq.(3.8) for . In this gauge it is particularly clear that this solution smoothly interpolates between the and solutions: if the first flavor squark winds at infinity while for the second flavor does.
To further convince ourselves that the rotation considered above corresponds to physical zero modes we can construct a gauge invariant operator which has -dependence on our solution. One example is
which is a matrix in flavor indices. Inserting the solution (3.25) this operator reads
We see that is a gauge invariant operator which has -dependence localized near the string axis where is non-zero.
As we have already mentioned the central charge of the algebra reduces to the component of the flux
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Additional resources for Algebraic K-theory: Proceedings of a conference held at Oberwolfach, June 1980, Part I
C pC1 p Á2 ; p; Pp1;01 . / to each element. t u Moreover, we construct a posteriori error estimates for 1. We replace u D ek C Uk in (1a), multiply by a test function V and, integrate over Œxk ; xkC1 to obtain Superconvergence of Discontinuous Galerkin Method to Nonlinear Differential. . x/ is obtained by mapping Ä PpC1 . / D . 1/ Pp1;0 . / C . p C 1 2 1;0 / Pp 1 . / p to Œxk ; xkC1 . Testing against V D PQpC1 and solving (10) for ˛pC1;k . we measure the accuracy of a posteriori error estimates using the effectivity index defined by s ÂDs N P kD0 N P kD0 jjEk jj2k jju : Uk jj2k A posteriori error estimates are considered asymptotically exact, if the effectivity index converges to 1 under x or p refinements.
Xi /j . x1 ; x2 //: Fig. K. Balyan We now describe the terms which shall be needed in sequel. Oum /Á Áx2 : Let s be a side common to the elements m and n : We may assume that s is the image of Á D 1 under the mapping Mm which maps S to m and also the image of Á D 1 under the mapping Mn which maps S to n : We now define the jumps at the inter-element Boundaries: kŒuk20; s D kOum . ; 1/ uO n . um /x1 . un /x1 . um /x2 . un /x2 . ; 1/k21=2;I : Here I D (-1,1). Now we define Z Z j Lui j2 dx1 dx2 D j Lui j2 Ji d dÁ: (5) S i Here Ji .
T / C0 e t : The proof is detailed in . Proposition 4. For f0 0 we suppose: i) f0 2 B 1 . t /: Proof. i i /. B. Rejeb Definition 5. x; t / 2 0; Œg; where A1 and A2 are positive constants. x; y/. g/ We have to prove that ˆ is a continuous map from S to itself, which will prove the existence of a solution in S . g/ T 1=2 by virtue of (16) and the condition imposed to f0 in proposition (4), it’s easy to see that f 2 S if g 2 S . Q / ! 0 when n ! C1: Lemma 6. Q / The proof is detailed in . On Classical Solution in Finite Time of BGK-Poisson’s Equations 19 Theorem 7. | s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084886758.34/warc/CC-MAIN-20180116224019-20180117004019-00744.warc.gz | CC-MAIN-2018-05 | 3,810 | 16 |
https://brainmass.com/business/income-statement/pg9 | math | Austin Grocers recently reported the following income statement (in millions of dollars): Sales $700 Operating costs 500 EBIT $200 Interest 40 EBT $160 Taxes (40%) 64 Net income $ 96 ------ Dividends $ 32 Addition to retained $ 64 earnings This ye
I do not understand Pro-forma income statements & I have to create one. Here is the problem: Style Right Company makes hair dryers. During the past few days, its accountants have been preparing the master budget for the coming year, 2006. To date, they have gathered the following projected data: Sales revenue (at $20 per u
Uncle Ralph's sells cookies in a large shopping mall. The following multi-step income statement was prepared for the year ending December 31, 2004. Sales $36,000 Cost of Good Sold 4,000 Gross Profit
The net income reported on the income statement for the current year was $275,000. Depreciation recorded on fixed assets and amortization of patents for the year were $40,000 and $9,000, respectively. Balances of current asset and current liability accounts at the end and at the beginning of the year are as follows:
Preparation of Income Statement (What are the Net Income and Earning per Share?) and Retained Earnings Statement (What is the Retained earnings on June 30, 2006?). Prepare an income statement and a statement of retained earnings for Big Sky Corporation for the year ended June 30, 2006, based on the following information---see at
Please assist with the attached problem. The following data pertain to Brown Brother's Company operations for July. Total Product X Product Y Number of units sold... 10,000 12,000 Selling price per unit... $20,000 $25.00 Variable cost per unit: Production... $9 $10.00 Selling & administration... $3
The following selected information is taken from the records of Beckstrom Corporation. Accounts payable . . . . . . . . . . . . . . . . . . . . . . . . $ 35,000 Accounts receivable . . . . . . . . . . . . . . . . . . . . . . 65,000 Advertising expense . . . . . . . . . . . . . . . . . . . . . . 15,000 Cash . . . . . . . . .
Find the most recent annual report of a company of your interest and answer the following questions regarding the company's Income Statement
Please see the attached file. ? add a copy of the Income Statement you used In the library or on the Internet, find the most recent annual report of a company of your interest and answer the following questions regarding the company's Income Statement: What accounting period(s) does the Income Statement (IS) cove
Please help me with the following 3 problems. Fundamentals of Financial Management 3-1 Income Statement little Books Inc. recently reported $3 million of net income. Its EBIT was $6 million, and its tax rate was 40 percent. What was its interest expense? [Hint: Write out the headings for an income statement and then fill
Following are three separate events affecting the managerial accounting systems for different companies. Match the management concept(s) that the company is likely to adopt for the event identified.There is some overlap in the meaning of customer orientation and total quality management and, therefore, some responses can include
I need help with Exercise 18-5 (Management concepts) and questions 1 and 2 of Problem 18-8A (Manufacturing and income statements; inventory analysis.) Please see attached document for full description. Thank you!!
Please assist with this problem to include reference sources. Mohican Inc. has the following items on its GAAP financial statements: Net income per the financial statements is $5,000,000. A $7,500,000 contingent liability and related expense for legal damages is included in the financial statements. The footnote ind
Pro-Forma Income Statement Silver Company has asked you to prepare a pro-forma income statement for the coming year.The following information is available: Expected sales revenue . . . . . . . . . . . . . . . .. . . $1,240,000 Manufacturing costs: Variable cost of goods sold . . . . . . . . . . . . . . 625,000 Fixed ove
The income statement and supplemental information for Xtra Company are provided below: (see attachment) a. Acquired equipment for $87,000 by paying cash of $75,750 and is issuing a note payable for the difference. b. Beginning cash balance, $15,020. Increase in cash, $86,490 c. Collections from customers were $5,250 more than
The sustainable Chemical Corporation produced a specialty chemical called SC. At the beginning of each year, the company estimated what the cost of SC would be for the coming year as one factor in the development of its pricing and promotion strategies and as a benchmark against which to compare the actual costs of production. T
Please help me prepare an income statement and answer the following questions for this situation: Diane Maynard was grateful for the balance sheets that her friend prepared. ( I attached the balance sheet I did) In going over the numbers, she remarked, "it's sort of surprising that cash increased by $31677, but net income was
2. An S corporation's profit and loss statement shows net profits of $90,000(book income). The corporation has three equal shareholders. From supplemental data, you obtain the following information about some items that are included in the $90,000. Selling expenses (21,200) Municipal bond interest income 2,000 Dividends re
Listed below are the results of Rulon Candies' operations for 2005 and 2006. (Assume 4,000 shares of outstanding stock for both years.) 2006 2005 Sales $300,000 $350,000 Utilities expenses 15,000 8,500 Employee salaries 115,000 110,000 Advert
The trial balance as at 31 March 2007 (1) Office equipment, at cost 2,750,000 (DR) Motor vehicles, at cost 937,500 (DR) Land, at valuation 2,487,500 (DR) Accumulated depreciation at 1 April 2006 Office equipment 2,016,800 (CR) Motor vehicles 126,572 (CR) Additional information relates to the above: Depreciation is t
A contribution income statement for the Nantucket Inn is shown below. (Ignore income taxes). Revenue.............................................$500,000 Less: Variable expenses............................300,000 Contribution margin................................$200,000 Less: Fixed expenses..............................
1. Build a Statement of Cash Flows using the Indirect Method 2. Determine Income Tax owed and amount paid to Vendors Please see attachment.
My next question is on Sales. I need to list the Total Sales (as revenue) at the beginning of my Income Statement. Will that include the following? Sales Cash (from start up in November 2005 Balance sheet) $10,172 accounts receivable $870 minus ending cash (what shows up as cash on hand in March 20
I do not understand the role of the income statement on an accrual basis. If you are given a balance sheet from November 2005 (which is the company start up), and one from March 2006 (when the company is no longer open), how do you calculate the income statement. Do you use the information given from the November 2005 as the beg
Problem 16-6 Contribution Margin and Functional Income Statements The following information is available for Dabney Company for 2006: Sales revenue (at $20 per unit) $151,200 Fixed manufacturing costs 24,000 Variable manufacturing costs (at $8 per unit) 60,480 Fixed selling expenses 70,000 Variable selling expenses (at $2
See attached file for full problem description.
Burrel Manufacturing Company established the following standard price and cost data Sales price 7.50 per unit Variable Manufacturing cost 3.00 per unit fixed manufacturing cost 3,000 total fixed selling and administrative cost 1,2
Using the trial balance given at the top of the next page, prepare an income statement. Debit
Prepare an income statement and a statement of retained earnings for Big Sky Corporation forthe year ended June 30, 2006, based on the following information: Capital stock (1,500 shares @ $100) . . . . . . . . . . . . . . . . . . . . . . . . . . $150,000 Retained earnings, July 1, 2005 . . . . . . . . . . . . . . .
Income statement for Shirley Company for 2005 and 2006 follow: 2006 2005 Sales 240,000 200,000 Cost of Goods Sold 147,900 108,000 Selling Expenses 40,100
I am having trouble preparing this income statement can you please sow me how to do it and explain why it is done this way? Problem 2-5 Income Statement Preparation The following information is taken from the records of Hill, Dunn, & Associates for the year ended December 31, 2006. P.A.S.S. Power Accounting System Softwar | s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818686983.8/warc/CC-MAIN-20170920085844-20170920105844-00411.warc.gz | CC-MAIN-2017-39 | 8,474 | 31 |
https://www.abebooks.com/9780883850275/Knot-Theory-Mathematical-Association-America-0883850273/plp | math | This specific ISBN edition is currently not available.View all copies of this ISBN edition:
Knot Theory, a lively exposition of the mathematics of knotting, will appeal to a diverse audience from the undergraduate seeking experience outside the traditional range of studies to mathematicians wanting a leisurely introduction to the subject. Graduate students beginning a program of advanced study will find a worthwhile overview, and the reader will need no training beyond linear algebra to understand the mathematics presented. The interplay between topology and algebra, known as algebraic topology, arises early in the book, when tools from linear algebra and from basic group theory are introduced to study the properties of knots, including one of mathematics' most beautiful topics, symmetry. The book closes with a discussion of high-dimensional knot theory and a presentation of some of the recent advances in the subject - the Conway, Jones and Kauffman polynomials. A supplementary section presents the fundamental group, which is a centerpiece of algebraic topology.
"synopsis" may belong to another edition of this title.
Knot Theory, a lively exposition of the mathematics of knotting, will appeal to a diverse audience of mathematical readers, from undergraduates to professionals. The author introduces tools from linear algebra and basic group theory and uses these to study the properties of knots, high-dimensional knot theory and the Conway, Jones and Kauffman polynomials.Review:
'The author's book would be a good text for an undergraduate course in knot theory ... The topics in the book are nicely tied together ... The topics and the exercises together can provide an opportunity for many undergraduates to get a real taste of what present day mathematics is like.' Mathematical Reviews
'Get knotted ... ' Scouting for Boys
"About this title" may belong to another edition of this title.
Book Description The Mathematical Association o, 1993. Hardcover. Condition: New. Never used!. Seller Inventory # P110883850273
Book Description The Mathematical Association of America, 1993. Condition: New. book. Seller Inventory # M0883850273 | s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794867277.64/warc/CC-MAIN-20180526014543-20180526034543-00140.warc.gz | CC-MAIN-2018-22 | 2,157 | 9 |
https://nrich.maths.org/9333 | math | Every third page number in this book has been omitted. Can you work out what number will be on the last page?
MatildaMatildaMatil... What is the 1000th letter?
Consider two arithmetic sequences: 1998, 2005, 2012,... and 1996, 2005, 2014,... Which numbers will appear in both?
How many squares are needed to continue this pattern?
Walking up a steep hill, I pass 10 equally spaced street lamps. How long do I take to walk from the first lamp to the last?
Leonard writes down a sequence of numbers. Can you find a formula to predict the seventh number in his sequence?
A pattern repeats every six symbols. What are the 100th and 101st symbols?
Trinni rearanges numbers on a clock face so each adjacent pair add up to a triangle number... What number did she put where 6 would usually be?
This grocer wants to arrange his fruit in a particular order, can you help him?
How many hexagons are required for the perimeter of the whole shape to have length 1002cm?
Grannie's watch gains 30 minutes every hour, whilst Grandpa's watch loses 30 minutes every hour. What is the correct time when their watches next agree?
This pattern repeats every 12 dots. Can you work out what a later piece will be?
A robot moves along the number line. Where will it be after 2011 slides?
Can you work out what fraction of this grid is shaded?
By drawing 5 horizontal and four vertical lines, one can form 12 rectangles. What is the greatest number of rectangles that can be formed by drawing 15 lines?
This tiled floor has 109 purple tiles. How many tiles are there altogether?
Weekly Problem 51 - 2016
Pegs numbered 1 to 50 are placed in a row. Alternate pegs are knocked down, and this process is repeated. What is the number of the last peg to be knocked down?
Consider all of the five digit numbers which we can form using only the digits 2, 4, 6 and 8. If these numbers are arranged in ascending order, what is the 512th number?
Knights always tell the truth. Knaves always lie. Can you catch these knights and knaves out?
How many numbers less than 2017 are both the sum of two consecutive integers and the sum of five consecutive integers?
Given that the number 2008 is the correct answer to a sum, can you find n?
How many diagonals does a regular icosagon (20 sides) have?
In a supermarket, there are two lines of tightly packed trolleys. What is the length of one trolley?
When will 2000 appear in this sequence?
Repeat a pattern of numbers to form a larger number. Can you find the sum of all the digits?
From only the page numbers on one sheet of newspaper, can you work out how many sheets there are altogether?
Can you work out which number will appear directly below 400 in this pattern?
If a number is even, halve it; if odd, treble it and add 1. If a sequence starts at 13, what will be the value of the 2008th term?
A sequence is generated using these rules. For which values of n is the nth term equal to n? | s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499710.49/warc/CC-MAIN-20230129080341-20230129110341-00284.warc.gz | CC-MAIN-2023-06 | 2,902 | 30 |
http://cosupperclub.com/read/abelian-group-theory-proceedings-of-the-2-nd-new-mexico-state-university | math | By Robert B. Warfield Jr. (auth.), David M. Arnold, Roger H. Hunter, Elbert A. Walker (eds.)
Read or Download Abelian Group Theory: Proceedings of the 2nd New Mexico State University Conference, Held at Las Cruces, New Mexico, December 9–12, 1976 PDF
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Extra info for Abelian Group Theory: Proceedings of the 2nd New Mexico State University Conference, Held at Las Cruces, New Mexico, December 9–12, 1976
C. Megibben, On mixed groups of torsion-free rank one, Ill, J. Math. ii (1967), 134-144. 28. C. Megibben, Modules over an incomplete discrete valuation ring, Proc. Amer. Math. Soc. 19 (1968), 450-452. 29. R. J. Nunke, Modules of extensions over Dedekind rings, Ill. J. Math. 3 (1959), 222-241. 30. R. J. Nunke, Homology and direct sums of countable Abelian groups, Math. Zeits° 191 (1967), 182-212. 31. R. J. Nunke, Uniquely elongating modules, Proc. Conf. on Abelian Groups, (November, 1972), Symposia Mathematica 13, Academic Press, London, 1974, ppl 315-330.
In attempting to generalize Ulm's theorem, care must be taken to preserve relative Ulm invariants. Warfield . This necessitates the following definition, due to 53 Definition. A decomposition basis X of a group A is a lower decomposition basis if, whenever the relative Ulm invariant f(a,p,A,[X]) is infinite (for a an ordinal or ~), then f(a,p,A) = f(a,p,A,[X]). Lemma 9. If A has a decomposition basis X, then there is a decomposition basis Y subordinate to X which is a lower decomposition basis.
41. E. A. Walker, Ulm's theorem for totally projective Amer. Math. Soc. 37 (1973), 387-392. 42. K. Alg. 17 (1971), 482-488. 43. R. B. , Homomorphisms and duality for torsion-free groups, Math. Zeits. 107 (1968), 189-200. 44. R. B. , An isomorphic refinement groups, Pacific J. , 34 (1970), 237-255. 45. R. B. , Invariants and a classification theorem for modules over a discrete valuation ring, University of Washington notes, 1971. 46. R. B. , Classification theorems for p-groups modules over a discrete valuation ring, Bull. | s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583512679.76/warc/CC-MAIN-20181020080138-20181020101638-00305.warc.gz | CC-MAIN-2018-43 | 3,135 | 13 |
http://www.solutioninn.com/suppose-that-n-independent-trials-each-of-which-results-in | math | Question: Suppose that n independent trials each of which results in
Suppose that n independent trials, each of which results in any of the outcomes 0, 1, or 2, with respective probabilities p0, p1, and p2,
are performed. Find the probability that outcomes 1 and 2 both occur at least once.
Relevant QuestionsConsider a collection of n individuals. Assume that each person’s birthday is equally likely to be any of the 365 days of the year and also that the birthdays are independent. Let Ai,j, i ≠ j, denote the event that ...Let A ⊂ B. Express the following probabilities as simply as possible: P(A|B), P(A|Bc), P(B|A), P(B|Ac) Consider a school community of m families, with ni of them having i children, Consider the following two methods for choosing a child: 1. Choose one of the m families at random and then randomly choose a child from that ...Five distinct numbers are randomly distributed to players numbered 1 through 5. Whenever two players compare their numbers, the one with the higher one is declared the winner. Initially, players 1 and 2 compare their ...Two coins are to be flipped. The first coin will land on heads with probability .6, the second with probability .7. Assume that the results of the flips are independent, and let X equal the total number of heads that ...
Post your question | s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818688926.38/warc/CC-MAIN-20170922074554-20170922094554-00272.warc.gz | CC-MAIN-2017-39 | 1,318 | 5 |
https://justaaa.com/physics/154473-you-push-a-32-kg-block-against-a-horizontal | math | You push a 3.2 kg block against a horizontal spring, compressing the spring by 16 cm. Then you release the block, and the spring sends it sliding across a tabletop. It stops 62 cm from where you released it. The spring constant is 170 N/m. What is the coefficient of kinetic friction between the block and the table?
Gravitational acceleration = g = 9.81 m/s2
Mass of the block = m = 3.2 kg
Force constant of the spring = k = 170 N/m
Compression of the spring = X = 16 cm = 0.16 m
Coefficient of kinetic friction between the block and table =
Friction force on the block = f = mg
Distance traveled by the block from release point before coming to rest = L = 62 cm = 0.62 m
By conservation of energy the initial potential energy of the spring is converted into work done against friction as the block comes to rest again.
kX2/2 = fL
kX2/2 = mgL
(170)(0.16)2/2 = (3.2)(9.81)(0.62)
Coefficient of kinetic friction between the block and table = 0.112
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https://yoyouheard.com/show-that-the-complex-number-000-i-has-the-following-properties-a-0zz-and-z0z-for-all-c/ | math | Show that the complex number 0(=0+0 i) has the following properties. (a) 0+z=z and z+0=z, for all complex numbers z Hint: Let …
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https://surface.syr.edu/che/9/ | math | Assuming a one-parameter model for the two-particle distribution function of a surface, one can choose a value of the "width parameter" to yield a correct surface tension. In order to then identify the width of the surface layer with the value of the width parameter, as is often done, one can check the validity of the model by verifying that other properties (e.g., surface energy) are correctly calculated, or, as proposed herein, by demonstrating that different formulas for the surface tension give identical results. A new formula is derived, and the relation between the different formulas discussed. Calculations are performed for the Ar system. For the model used, no choice of the parameter can yield identical results for both ways of calculating surface tension. This points to problems in the interpretation of the width parameter. Misleading results may be obtained if one- and two-particle distributions not related by the Born-Green-Yvon equation are used together.
Goodisman, Jerry, "Determination of Surface Thickness Assuming a Linear-Density Profile" (1979). Chemistry - Faculty Scholarship. 9.
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http://www.reference.com/browse/wiki/East_India_(disambiguation) | math | East Indian may refer to
Note: It would be atypical to refer to Native Americans (sometimes called "Indians") from the East as "East Indians". It would be more common to call them "Eastern Indians".
The "East Indies" were thus called to distinguish them from the West Indies, i.e. the Caribbean. | s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1386163051684/warc/CC-MAIN-20131204131731-00018-ip-10-33-133-15.ec2.internal.warc.gz | CC-MAIN-2013-48 | 295 | 3 |
https://cs.stackexchange.com/questions/151720/is-the-set-of-languages-satisfying-the-pumping-lemma-closed-under-concatenation | math | Let $L$ be the set of all languages that satisfy the pumping lemma, including non-regular languages that satisfy it. Is the set $L$ closed under concatenation?
I couldn’t prove it or find a counterexample.
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Suppose that $L_1$ satisfies the pumping lemma: there exists $p_1$ such that every word $w \in L_1$ of length at least $p$ can be decomposed as $w = xyz$, where $|xy| \leq p_1$, $y \neq \epsilon$, and $xy^iz \in L_1$ for all $i \geq 0$. Similarly, suppose that $L_2$ satisfies the pumping lemma, with constant $p_2$.
I claim that $L = L_1L_2$ satisfies the pumping lemma with constant $p = p_1 + p_2$. Suppose we are given a word $w = w_1w_2 \in L$ of length at least $p$, where $w_1 \in L_1$ and $w_2 \in L_2$. Then either $|w_1| \geq p_1$ or $|w_2| \geq p_2$. You take it from here. | s3://commoncrawl/crawl-data/CC-MAIN-2023-50/segments/1700679100769.54/warc/CC-MAIN-20231208180539-20231208210539-00756.warc.gz | CC-MAIN-2023-50 | 984 | 5 |
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Guess it, press it and check it with a cool calculator the maths calculator book is packed with cool maths, perplexing puzzles, and brilliant brain- teasers, complete with a colourful calculator to check your maths. On a calculator it is the " log" button. 1 maths log book easons for primary & secondary school books log tables approved for use in the state examinations.
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https://www.arxiv-vanity.com/papers/1110.3661/ | math | Affine Mirković-Vilonen polytopes
Each integrable lowest weight representation of a symmetrizable Kac-Moody Lie algebra has a crystal in the sense of Kashiwara, which describes its combinatorial properties. For a given , there is a limit crystal, usually denoted by , which contains all the other crystals. When is finite dimensional, a convex polytope, called the Mirković-Vilonen polytope, can be associated to each element in . This polytope sits in the dual space of a Cartan subalgebra of , and its edges are parallel to the roots of . In this paper, we generalize this construction to the case where is a symmetric affine Kac-Moody algebra. The datum of the polytope must however be complemented by partitions attached to the edges parallel to the imaginary root . We prove that these decorated polytopes are characterized by conditions on their normal fans and on their -faces. In addition, we discuss how our polytopes provide an analog of the notion of Lusztig datum for affine Kac-Moody algebras. Our main tool is an algebro-geometric model for constructed by Lusztig and by Kashiwara and Saito, based on representations of the completed preprojective algebra of the same type as . The underlying polytopes in our construction are described with the help of Buan, Iyama, Reiten and Scott’s tilting theory for the category . The partitions we need come from studying the category of semistable -modules of dimension-vector a multiple of .
Let be a symmetrizable generalized Cartan matrix, with rows and columns indexed by a set . We denote by the Kac-Moody algebra defined by . It comes with a triangular decomposition , with a root system , and with a Weyl group . The simple roots are indexed by and the group is a Coxeter system, generated by the simple reflections . We denote the length function of by and the set of positive (respectively, negative) roots by (respectively, ). The root lattice is denoted by and we set . The canonical pairing between and its dual will be denoted by angle brackets. Lastly, we denote by the set of linear combinations of the simple roots with nonnegative coefficients and we set .
The combinatorics of the representation theory of is captured by Kashiwara’s theory of crystals. Let us summarize quickly this theory; we refer the reader to the nice survey for detailed explanations.
A -crystal is a set endowed with maps , , , and , for each , that satisfy certain axioms. This definition is of combinatorial nature and the axioms stipulate the local behavior of the structure maps around an element . This definition is however quite permissive, so one wants to restrict to crystals that actually come from representations.
In this respect, an important object is the crystal , which contains the crystals of all the irreducible lowest weight integrable representations of (see Theorem 8.1 in ). This crystal contains a lowest weight element annihilated by all the lowering operators , and any element of can be obtained by applying a sequence of raising operators to .
The crystal itself is defined as a basis of the quantum group in the limit . Working with this algebraic construction is cumbersome, and there exist other, more handy, algebro-geometric or combinatorial models for .
One of these combinatorial models is Mirković-Vilonen (MV) polytopes. In this model, proposed by Anderson , one associates a convex polytope to each element . The construction of is based on the geometric Satake correspondence. More precisely, the affine Grassmannian of the Langlands dual of contains remarkable subvarieties, called MV cycles after Mirković and Vilonen . There is a natural bijection from onto the set of all MV cycles [11, 12, 23], and is simply the image of by the moment map.
Using Berenstein and Zelevinsky’s work , the second author showed in that these MV polytopes can be described in a completely combinatorial fashion: these are the convex lattice polytopes whose normal fan is a coarsening of the Weyl fan in the dual of , and whose -faces have a shape constrained by the tropical Plücker relations. In addition, the length of the edges of is given by the Lusztig data of , which indicate how , viewed as a basis element of at the limit , compares with the PBW bases.
1.2 Generalization to the affine case
This paper aims at generalizing this model of MV polytopes to the case where is an affine Kac-Moody algebra.
Obstacles pop up when one tries to generalize the above constructions of to the affine case. Despite difficulties in defining the double-affine Grassmannian, the algebro-geometric model of using MV cycles still exists in the affine case, thanks to Braverman, Finkelberg and Gaitsgory’s work ; however, there is no obvious way to go from MV cycles to MV polytopes.
On the algebraic side, several PBW bases for have been defined in the affine case by Beck , Beck and Nakajima , and Ito , but the relationship between the different Lusztig data they provide has not been studied111Recently, Muthiah and Tingley have considered this problem in the case . They have shown that the resulting combinatorics matches that produced in the present paper, in the sense that the MV polytopes coming from the Lusztig data provided by the PBW bases match those defined here. It should be easy to extend this result to the case of an arbitrary symmetric affine Kac-Moody algebra..
As recalled above, in finite type, the normal fan of an MV polytope is a coarsening of the Weyl fan, so the facets of an MV polytope are orthogonal to the rays in the Weyl fan. Therefore an MV polytope is determined just by the position of theses facets, which form a set of numerical values dubbed “Berenstein-Zelevinsky (BZ) data”. In the case , a combinatorial model for an analog of these BZ data was introduced by Naito, Sagaki, and Saito [51, 52]. Later, Muthiah related this combinatorial model to the geometry of the MV cycles. However, the complete relationship between this combinatorial model and our affine MV polytopes is not yet clear.
1.3 The preprojective model
Due to the difficulties in the MV cycle and PBW bases models, we are led to use a third construction of , recently obtained by the first two authors for the case of a finite dimensional . This construction uses a geometric model for based on quiver varieties, which we now recall.
This model exists for any Kac-Moody algebra (not necessarily of finite or affine type) but only when the generalized Cartan matrix is symmetric. Then is the incidence matrix of the Dynkin graph ; here our index set serves as the set of vertices and is the set of edges. Choosing an orientation of this graph yields a quiver , and one can then define the completed preprojective algebra of .
A -module is an -graded vector space equipped with linear maps. If the dimension-vector is given, we can work with a fixed vector space; the datum of a -module then amounts to the family of linear maps, which can be regarded as a point of an algebraic variety. This variety is called Lusztig’s nilpotent variety; we denote it by , where is the dimension-vector. Abusing slightly the language, we often view a point as a -module.
For , let be the set of irreducible components of . We set . In , Lusztig endows with a crystal structure, and in , Kashiwara and Saito show the existence of an isomorphism of crystals from onto . This isomorphism is unique since has no non-trivial automorphisms.
Given a finite-dimensional -module , we can consider the dimension-vectors of the -submodules of ; they are finitely many, since they belong to a bounded subset of the lattice . The convex hull in of these dimension-vectors will be called the Harder-Narasimhan (HN) polytope of and will be denoted by .
The main result of is equivalent to the following statement: if is finite dimensional, then for each , the set contains a dense open subset of . In other words, is the general value of the map on .
This result obviously suggests a general definition for MV polytopes. We will however see that for of affine type, another piece of information is needed to have an complete model for ; namely, we need to equip each polytope with a family of partitions. Our task now is to explain what our polytopes look like, and where these partitions come from.
1.4 Faces of HN polytopes
Choose a linear form and let denote the maximum value of on . Then is a face of . Moreover, the set of submodules whose dimension-vectors belong to has a smallest element and a largest element .
The existence of and follows from general considerations: if we define the slope of a finite dimensional -module as , then is the semistable subquotient of slope zero in the Harder-Narasimhan filtration of . Introducing the abelian subcategory of semistable -modules of slope zero, it follows that, for each submodule ,
In other words, the face coincides with the HN polytope of , computed relative to the category , and shifted by .
Our aim now is to describe the normal fan to , that is, to understand how , and depend on . For that, we need tools that are specific to preprojective algebras.
1.5 Tits cone and tilting theory
One of these tools is Buan, Iyama, Reiten and Scott’s tilting ideals for . Let be the simple -module of dimension-vector and let be its annihilator, a one-codimensional two-sided ideal of . The products of these ideals are known to satisfy the braid relations, so to each in the Weyl group of , we can attach a two-sided ideal of by the rule , where is any reduced decomposition of . Given a finite-dimensional -module , we denote the image of the evaluation map by .
Recall that the dominant Weyl chamber and the Tits cone are the convex cones in the dual of defined as
We will show the equality for any finite dimensional -module , any and any linear form . This implies that is a vertex of and that the normal cone to at this vertex contains . This also implies that is contained in
When runs over the Tits cone, it generically belongs to a chamber, and we have just seen that in this case, the face is a vertex. When lies on a facet, is an edge (possibly degenerate). More precisely, if lies on the facet that separates the chambers and , with say , then . Results in and moreover assert that is the direct sum of a finite number of copies of the -module .
There is a similar description when is in ; here the submodules of that come into play are the kernels of the coevaluation maps , where again .
1.6 Imaginary edges and partitions (in affine type)
From now on in this introduction, we focus on the case where is of symmetric affine type, which in particular implies is of untwisted affine type.
The root system for decomposes into real and imaginary roots ; the real roots are the conjugate of the simple roots under the Weyl group action, whereas the imaginary roots are fixed under this action. The Tits cone is .
We set . The projection maps onto the “spherical” root system , whose Dynkin diagram is obtained from that of by removing an extending vertex. The rank of is , which is also the multiplicity of the imaginary roots.
The vector space identifies with the hyperplane of the dual of . The root system defines an hyperplane arrangement in , called the spherical Weyl fan. The open cones in this fan will be called the spherical Weyl chambers. Together, this fan and the hyperplane arrangement that the real roots define in make up a (non locally finite) fan in the dual of , which we call the affine Weyl fan and which we denote by .
Each set of simple roots in is a basis of ; we can then look at the dual basis in , whose elements are the corresponding fundamental coweights. We denote by the set of all fundamental coweights, for all possible choices of simple roots. Elements in are called spherical chamber coweights; the rays they span are the rays of the spherical Weyl fan.
Now take a -module . As we saw in the previous section, the normal cone to at the vertex (respectively, ) contains (respectively, ). Altogether, these cones form a dense subset of the dual of : this leaves no room for other vertices. This analysis also shows that the normal fan to is a coarsening of .
Thus, the edges of point in directions orthogonal to one-codimensional faces of , that is, parallel to roots. In the previous section, we have described the edges that point in real root directions. We now need to understand the edges that are parallel to . We call these the imaginary edges.
More generally, we are interested in describing the faces parallel to . Let us pick and let us look at the face . As we saw in section 1.4, this face is the HN polytope of , computed relative to the category . It turns out that and only depend on the face of the spherical Weyl fan to which belongs. We record this fact in the notation by writing for .
We need one more definition: for , we say that a -module is a -core if it belongs to for all sufficiently close to . In other words, the category of -cores is the intersection of the categories , taken over all spherical Weyl chambers such that .
For each , the set of indecomposable modules is a constructible subset of . It thus makes sense to ask if the general point of an irreducible subset of is indecomposable. Similarly, the set of modules that belong to is an open subset of , so we may ask if the general point of an irreducible subset of is in . In section 7.4, we will show the following theorems.
For each integer and each , there is a unique irreducible component of whose general point is an indecomposable -core.
We denote by this component.
Let be a positive integer and let be a spherical Weyl chamber. There are exactly irreducible components of whose general point is an indecomposable module in . These components are the , for .
In Theorem 1.2, the multiplicity of the root materializes as a number of irreducible components.
Now let and pick in a spherical Weyl chamber . Let be a general point of and let , an object in . Write the Krull-Schmidt decomposition of as , with , …, indecomposable; then each is in , so for a certain integer . Moreover, it follows from Crawley-Boevey and Schröer’s theory of canonical decomposition that each is the general point of an irreducible component . Using Theorem 1.2, we then see that each is a component for a certain . Gathering the integers according to the coweights , we get a tuple of partitions . In this context, we will show that the partition depends only on and , and not on the Weyl chamber .
We are now ready to give the definition of the MV polytope of : it is the datum of the HN polytope , for general in , together with the family of partitions defined above.
1.7 -faces of MV polytopes
Let us now consider the -faces of our polytopes . Such a face is certainly of the form
where belongs to a -codimensional face of . There are three possibilities, whether belongs to , or .
Suppose first that . Then the root system is finite of rank , of type or type . More precisely, let be of minimal length such that and let ; then the element maps onto the root system . The full subgraph of defined by gives rise to a preprojective algebra . The obvious surjective morphism induces an inclusion , whose image is the category . Further, the tilting ideals provide an equivalence of categories
whose action on the dimension-vectors is given by . Putting all this together, we see that is the image under of the HN polytope of the -module . In addition, genericity is preserved in this construction: if is a general point in an irreducible component of a nilpotent variety for , then is a general point in an irreducible component of a nilpotent variety for . When is of type , this implies that the -face obeys the tropical Plücker relation from .
A similar analysis can be done in the case where is in . It then remains to handle the case where , that is, where belongs to a face of codimension one in the spherical Weyl fan. Here is an affine root system of type . The face separates two spherical Weyl chambers of , say and , and there are spherical chamber coweights and such that and .
Choose and . Assume that is the general point of an irreducible component. As we saw in section 1.6, the modules and are then described by tuples of partitions and , respectively. Both these modules are subquotients of , so this latter contains the information about the partitions for all .
Let be the polytope obtained by shortening each imaginary edge of the -face by . Then , equipped with the two partitions and , is an MV polytope of type .
A partition can be thought of as an MV polytope of type , since the generating function for the number of partitions equals the graded dimension of the upper half of the Heisenberg algebra. Thus, the family of partitions can be thought of as an MV polytope of type . We can therefore regard the datum of the face and of the partitions as an MV polytope of type .
Theorem 1.3 will be proved in section 7.5. Our method is to construct an embedding of into , where is the completed preprojective algebra of type ; this embedding depends on and its essential image is large enough to capture a dense open subset in the relevant irreducible component of Lusztig’s nilpotent variety. In this construction, we were inspired by the work of I. Frenkel et al. who produced analogous embeddings in the quiver setting.
So the final picture is the following. Let be the set of all lattice convex polytopes in , equipped with a family of partitions , such that:
The normal fan to is a coarsening of the Weyl fan .
To each spherical Weyl chamber corresponds a imaginary edge of ; the difference between the two endpoints of this edge is equal to .
A -face of is an MV polytope of type , or ; in the type case, this means that its shape obeys the tropical Plücker relation.
At the end of section 1.6, we associated an element of to each .
The map is bijective.
1.8 Lusztig data
As explained at the end of section 1.1, for a finite dimensional , the MV polytope of an element geometrically encodes all the Lusztig data of .
In more detail, let be the number of positive roots. Each reduced decomposition of the longest element of provides a PBW basis of the quantum group , which goes to the basis at the limit . To an element , one can therefore associate many PBW monomials, one for each PBW basis. In other words, one can associate to many elements of , one for each reduced decomposition of . These elements in are called the Lusztig data of . A reduced decomposition of specifies a path in the 1-skeleton of that connects the top vertex to the bottom one, and the corresponding Lusztig datum materializes as the lengths of the edges of this path.
With this in mind, we now explain that when is of affine type, our MV polytopes provide a fair notion of Lusztig data.
To this aim, we first note that a reasonable analog of the reduced decompositions of is certainly the notion of “total reflection order” (Dyer) or “convex order” (Ito), see [15, 28]. By definition, this is a total order on such that
(Unfortunately, the convexity relation implies that for any positive integers and . We therefore have to accept that is only a preorder; this blemish is however limited to the imaginary roots.)
A convex order splits the positive real roots in two parts: those that are greater than and those that are smaller. One easily shows that the projection maps onto a positive system of . Thus, there exists such that
Given such a convex order , we will construct a functorial filtration on each finite dimensional -module , such that each is a vertex of . The family of dimension-vectors are the vertices along a path in the -skeleton of connecting the top vertex and bottom vertices. The lengths of the edges in this path form a family of natural numbers , defined by the relation . Further, if we choose satisfying (1.1), then and .
Fix and take a general point in . Besides the family of natural numbers mentioned just above, we can construct a tuple of partitions by applying the analysis carried after Theorem 1.2 to the module , where is the spherical Weyl chamber containing . To , we can thus associate the pair consisting in the two families and . All this information can be read from . We call the Lusztig datum of in direction .
Let us denote by the set of all partitions and by the set of finitely supported families of non-negative integers.
The map is bijective.
The MV polytope contains the information of all Lusztig data of , for all convex orders. This is in complete analogy with the situation in the case where is finite dimensional. The conditions on the -faces given in the definition of say how the Lusztig datum of varies when the convex order changes; they can be regarded as the analog in the affine type case of Lusztig’s piecewise linear bijections.
The knowledge of a single Lusztig datum of , for just one convex order, allows one to reconstruct the irreducible component . This fact is indeed an ingredient of the proof of injectivity in Theorem 1.4.
Through the bijective map , the set acquires the structure of a crystal, isomorphic to . This structure can be read from the Lusztig data. Specifically, if is the smallest element of the order , then is the -coordinate of , and the operators and act by incrementing or decrementing this coordinate.
As mentioned at the beginning of section 1.3, Beck in , Beck and Nakajima in , and Ito in construct PBW bases of for of affine type. An element in one of these bases is a monomial in root vectors, the product being computed according to a convex order . To describe a monomial, one needs an integer for each real root and a -tuple of integers for each imaginary root , so in total, monomials in a PBW basis are indexed by . Moreover, such a PBW basis goes to at the limit . (This fact has been established in for Beck’s bases, and the result can probably be extended to Ito’s more general bases by using or .) In the end, we get a bijection between and . We expect that this bijection is our map .
1.9 Plan of the paper
Section 2 recalls combinatorial notions and facts related to root systems. We emphasize the notion of biconvex subsets, which is crucial to the study of convex orders and to the definition of the functorial filtration mentioned in section 1.8.
Section 3 is devoted to generalities about HN polytopes in abelian categories.
In section 4, we recall known facts about preprojective algebras and Lusztig’s nilpotent varieties. We also prove that cutting a -module according to a torsion pair is an operation that preserves genericity.
In section 5, we exploit the tilting theory on to define and study the submodules and mentioned in section 1.5. An important difference with the works of Iyama, Reiten et al. and of Geiß, Leclerc and Schröer is the fact that we are interested not only in the small slices that form the categories (notation of Iyama, Reiten et al.) or (notation of Geiß, Leclerc and Schröer), but also at controlling the remainder. Moreover, we track the tilting theory at the level of the irreducible components of Lusztig’s nilpotent varieties and interpret the result in term of crystal operations.
In section 6, we construct embeddings of into , where is the completed preprojective algebra of type . The data needed to define such an embedding is a pair of rigid orthogonal bricks in satisfying . The key ingredient in the construction is the -Calabi-Yau property of .
The final section 7 deals with the specifics of the affine type case. All the results concerning the imaginary edges, the cores, or the partitions are stated and proved there.
We thank Claire Amiot for suggesting to us that the reflection functors in are related to those in , which allowed us to take the current literature [14, 27, 59] into account in section 5. We also thank Thomas Dunlap for sharing his ideas about affine MV polytopes and for providing us with his PhD thesis . We thank Alexander Braverman, Bernhard Keller, Bernard Leclerc, and Dinakar Muthiah for very helpful discussions. Finally, we thank two anonymous referees for thorough and insightful reports which led to significant improvements in the presentation.
P. B. acknowledges support from the ANR, project ANR-09-JCJC-0102-01. J. K. acknowledges support from NSERC. P. T. acknowledges support from the NSF, grants DMS-0902649, DMS-1162385 and DMS-1265555.
1.11 Summary of the main notations
the set of partitions.
the Grothendieck group of , an essentially small abelian category.
the set of isomorphism classes of simple objects in .
the HN polytope of an object .
the face of a HN polytope , where and is the maximal value of on .
the base field for representation of quivers and preprojective algebras.
a finite graph, wihout loops (encoding a symmetric generalized Cartan matrix).
the corresponding symmetric Kac-Moody algebra.
the upper nilpotent subalgebra of .
an orientation of (thus is a quiver).
the set of edges of the double quiver .
the source and target maps.
the completed preprojective algebra of .
the category of finite dimensional left -modules.
the root system of .
the standard basis of .
the positive and negative roots with respect to this basis.
the Weyl group.
the length function.
the root lattice.
, the -vector space with basis .
the dual vector space.
the -th coordinate on ; thus is the dual basis of .
the -invariant symmetric bilinear form (real roots have square length ).
the dominant Weyl chamber.
the Tits cone.
, for .
and , the root subsystem and the parabolic subgroup defined by .
the longest element in , when the latter is finite.
the linear form such that for each .
, for ; thus for any reduced decomposition .
the preprojective algebra of type .
In the case of an affine root system:
the positive primitive imaginary root.
the projection modulo .
the spherical (finite) root system.
the “minimal” lift, a right inverse of .
the rank of .
the dual of .
the set of all spherical chamber coweights.
the Weyl fan on , completed on by the spherical Weyl fan.
the coroot lattice, spanned over by the elements .
the translation, for ; thus for each .
the image of in .
And after having chosen an extending vertex in the extended
the vertices of the (finite type) Dynkin diagram.
a preferred system of simple roots for .
the spherical fundamental coweights, a basis of .
the dominant spherical Weyl chamber.
The set of irreducible components of a topological space is denoted by . If is an irreducible topological space, then we say that a propriety depending on a point holds for general in if the set of points of at which holds true contains a dense open subset of . We sometimes extend this vocabulary by simply saying “let be a general point in ”; in this case, it is understood that we plan to impose finitely many such conditions .
2 Combinatorics of root systems and of MV polytopes
In this section, we introduce our notations and recall general results about root systems and biconvex subsets. Starting from section 2.3 onwards, we focus on the case of an affine root system.
2.1 General setup
Let be a finite graph, without loops: here is the set of vertices and is the set of edges. We denote by the free abelian group on and we denote its canonical basis by . We endow it with the symmetric bilinear form , given by for any , and for , is the negative of the number of edges between the vertices and in the graph . The Weyl group is the subgroup of generated by the simple reflections ; this is in fact a Coxeter system, whose length function is denoted by . Lastly, we denote by the set of all linear combinations of the with coefficients in and we denote by the linear form that maps each to .
The matrix with entries is a symmetric generalized Cartan matrix, hence it gives rise to a Kac-Moody algebra and a root system . The latter is a -stable subset of , which can be split into positive and negative roots and into real and imaginary roots .
Given a subset , we can look at the root system . Its Weyl group is the parabolic subgroup of . If is finite, then it has a longest element, which we denote by . An element is called -reduced on the right if for each . If is -reduced on the right, then for all . Each right coset of in contains a unique element that is -reduced on the right.
The Weyl group acts on and on its dual . The dominant chamber and the Tits cone are the convex cones in defined as
The closure is the disjoint union of faces
for . The stabilizer of any point in is precisely the parabolic subgroup . Thus
The disjoint union on the right endows with the structure of a (non locally finite) fan, which we call the Tits fan.
To an element , we associate the subset . If is a reduced decomposition, then
The following result is well-known (see for instance Remark in ).
For , the following three properties are equivalent:
Let and let . If is -reduced on the right, then .
Let and let be such that . Then there exists such that . Since is a nonnegative linear combination of the roots for , it follows that there exists such that . For this , we have , whence by Lemma 2.1. Therefore is not -reduced on the right.
2.2 Biconvex sets (general type)
A subset is said to be clos if the conditions , , imply (see , chapitre 6, §1, nº 7, Définition 4). A subset is said to be biconvex if both and are clos. We denote by the set of all biconvex subsets of and endow it with the inclusion order.
An increasing union or a decreasing intersection of biconvex subsets is itself biconvex.
Each finite biconvex subset of consists of real roots and is a , with (see , Proposition 3.2). For convenience, we will say that a biconvex set is cofinite if its complement is finite. Given , we set and | s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487648373.45/warc/CC-MAIN-20210619142022-20210619172022-00025.warc.gz | CC-MAIN-2021-25 | 29,618 | 160 |
https://www.omnicalculator.com/statistics/relative-standard-deviation | math | Relative Standard Deviation Calculator
Once you know the standard deviation and mean of your dataset, the RSD calculator (relative standard deviation calculator) helps you make decisions about that data. Is the variance 'great' or 'small'? How does your data compare to other datasets? Read on to learn more.
What is the relative standard deviation?
The relative standard deviation, or RSD, is a form of standard deviation where it is presented as a percentage of the mean. The RSD is always positive.
Using the relative standard deviation formula, you can easily calculate RSD by dividing the standard deviation by the absolute value of the mean, then multiplying the result by 100 to represent it as a percentage:
relative standard deviation = (standard deviation / |mean|) * 100%
It is common to write the relative standard deviation after the mean and with a plus-minus sign, e.g., 25 ± 2%, where ± 2% is the relative standard deviation.
Common uses for the RSD calculator
Relative standard deviation puts the standard deviation into perspective by comparing it to the mean. Viewing standard deviation as RSD helps people make decisions in a variety of situations, including:
- Quality assurance, e.g., a grocery store may require the RSD of all fruit sizes to be less than 10%.
- Evaluating the volatility of stock prices.
- In analytical chemistry, to express the precision of an assay.
- Comparing the variation of two different datasets.
When not to use relative standard deviation
It is not appropriate to use the relative standard deviation calculator for situations in which 0 does not represent an absence of quantity, such as temperature in degrees Celsius or Fahrenheit.
For example, if the average temperature was 12 ± 3°C on one day and 1 ± 3°C the next day, the RSD would be 25% on the first day and 300% (an enormous increase!) on the second day.
Since 0°C is arbitrary, having a mean temperature closer to 0 shouldn't necessarily make deviations from the mean seem greater. Instead, relative standard deviation could be calculated for temperatures expressed in Kelvin.
Relative standard deviation and coefficient of variation
If you check our coefficient of variation calculator, you'll figure out its formula is very similar to the relative standard deviation formula. The only difference is that in the relative standard deviation formula, we divide the standard deviation by the absolute value of the mean, while the coefficient of variation divides the standard deviation by the mean. As a result, the coefficient of variation can be either positive or negative, while the relative standard deviation is always positive.
Comparing apples and oranges (and pineapples!) - An example
David wants to stock his fruit stand with a new and exciting item. He can choose from a box of apples, a box of oranges, or a box of pineapples. However, he wants to make sure that the fruits are consistent in weight for easy pricing. The mean and standard deviation of weight for each fruit is:
- Apples: 100 ± 5 g.
- Oranges: 120 ± 30 g.
- Pineapples: 2 ± 0.5 lbs.
Which fruit has the most consistent weight? For apples, the calculation would be:
(5 g / 100 g) × 100% = 5%
Try using the RSD calculator to convert the standard deviations into relative standard deviations.
Now we can see the mean and RSD of weight for each fruit:
- Apples: 100 g ± 5%
- Oranges: 120 g ± 25%
- Pineapples: 2 g ± 25%
With the standard deviation expressed as relative standard deviation, David can now see that apples have the most consistent weight. | s3://commoncrawl/crawl-data/CC-MAIN-2023-40/segments/1695233510319.87/warc/CC-MAIN-20230927171156-20230927201156-00552.warc.gz | CC-MAIN-2023-40 | 3,550 | 32 |
http://www.easyunitconversion.com/en/area/square-poles-to-square-millimeters | math | Square poles to Square millimeters conversion
Enter square poles value you wish to convert:
Square poles conversion
Square pole is a unit of area defined as 30.25 square yards.
Square millimeters conversion
Square millimeter is a prefixed form of square meter used to denote sub multiples of the square meter.
1 square millimeter = 10-6 square meters | s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376830305.92/warc/CC-MAIN-20181219005231-20181219031231-00523.warc.gz | CC-MAIN-2018-51 | 350 | 7 |
http://www.fluentstrategygroup.com/index.php/pdf/algebraic-geometry-2-sheaves-and-cohomology-translations-of-mathematical | math | By Kenji Ueno
Sleek algebraic geometry is equipped upon basic notions: schemes and sheaves. the idea of schemes was once defined in Algebraic Geometry 1: From Algebraic types to Schemes, (see quantity 185 within the comparable sequence, Translations of Mathematical Monographs). within the current booklet, Ueno turns to the speculation of sheaves and their cohomology. Loosely conversing, a sheaf is a fashion of keeping an eye on neighborhood info outlined on a topological area, resembling the neighborhood holomorphic services on a posh manifold or the neighborhood sections of a vector package deal. to check schemes, it truly is beneficial to review the sheaves outlined on them, specially the coherent and quasicoherent sheaves. the first instrument in knowing sheaves is cohomology. for instance, in learning ampleness, it's often priceless to translate a estate of sheaves right into a assertion approximately its cohomology.
The textual content covers the real issues of sheaf idea, together with different types of sheaves and the basic operations on them, similar to ...
coherent and quasicoherent sheaves. right and projective morphisms. direct and inverse photographs. Cech cohomology.
For the mathematician unexpected with the language of schemes and sheaves, algebraic geometry can look far away. besides the fact that, Ueno makes the subject look common via his concise variety and his insightful causes. He explains why issues are performed this fashion and supplementations his factors with illuminating examples. therefore, he's capable of make algebraic geometry very available to a large viewers of non-specialists.
Read Online or Download Algebraic geometry 2. Sheaves and cohomology PDF
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Written for graduate scholars and researchers alike, this set of lectures presents a established creation to the idea that of equidistribution in quantity conception. this idea is of becoming significance in lots of parts, together with cryptography, zeros of L-functions, Heegner issues, best quantity concept, the speculation of quadratic types, and the mathematics facets of quantum chaos.
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The concept that of a interval of an elliptic crucial is going again to the 18th century. Later Abel, Gauss, Jacobi, Legendre, Weierstrass and others made a scientific learn of those integrals. Rephrased in sleek terminology, those provide how to encode how the advanced constitution of a two-torus varies, thereby exhibiting that definite households include all elliptic curves.
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Additional info for Algebraic geometry 2. Sheaves and cohomology
A general series of equivalence is then defined as an aggregate of virtual sets obtained by addition and subtraction of a finite number of elementary series. Series (of points) and systems (of curves) of equivalence on a threefold are defined in an analogous manner. The theory of equivalence on a threefold V which, in all that follows, is assumed to be non-singular, requires for its development a knowledge of the corresponding results for curves and surfaces; in particular, the establishment of the invariant series and systems of V rests on the theory of invariants of curves and surfaces.
Chapter III. Systems of Surfaces. 1. The RIEMANN-RoCH theorem. We consider m this section the problem of determining the freedom of the complete linear system characterised by a given non-singular surface on a non-singular threefold V. If C is such a surface, with virtual characters n, n, p, we define the virtual freedom d of the system ICI by the formula d = n - n + p - Pa + 2, where Pa denotes the arithmetic genus of V. In the case where C is non-special, with effective freedom r = d, we say that ICI is regular.
For example, B. SEGRE [IJ uses the present methods to obtain equivalences for the invariant series of any surface of the form c1 Sl + C2 S2 ' where Cv C2 are integers, in terms of the invariant series of Sl and S2' In the same work SEGRE finds the covariant systems of one or two nets of surfaces on V, of two or more pencils, and also of linear systems of freedom three or four, thereby establishing many interesting relations between the entities in question. One of the most striking of these is the following: given two pencils IAI, IBI of general character, generically situated, the number of pairs A, B which have stationary contact with one another is 48(P,. | s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267159359.58/warc/CC-MAIN-20180923114712-20180923135112-00383.warc.gz | CC-MAIN-2018-39 | 5,096 | 18 |
https://www.studysmarter.us/textbooks/physics/fundamentals-of-physics-10th-edition/conduction-of-electricity-in-solids/q53p-at-what-pressure-in-atmospheres-would-the-number-of-mol/ | math | At what pressure, in atmospheres, would the number of molecules per unit volume in an ideal gas be equal to the number density of the conduction electrons in copper, with both gas and copper at temperature T =300K?
The pressure value that would give the number of molecules per unit volume in an ideal gas equal to the number density of conduction electrons in copper is .
a) The number of conduction electrons in copper,
b) The temperature value, T =300 K
The equation that relates the characteristic properties of an ideal gas is known as Ideal Gas Equation. The ideal gas equation can be given as-
Here P is the pressure,V is the volume, n id the number of moles and k is the Boltzmann constant.
Using the given condition and data in equation, we can get the value of the pressure in atmospheres as follows:
Hence, the value of the pressure is .
Pure silicon at room temperature has an electron number density in the conduction band of about and an equal density of holes in the valence band. Suppose that one of every silicon atoms is replaced by a phosphorus atom. (a) Which type will the doped semiconductor be, n or p? (b) What charge carrier number density will the phosphorus add? (c) What is the ratio of the charge carrier number density (electrons in the conduction band and holes in the valence band) in the doped silicon to that in pure silicon?
The compound gallium arsenide is a commonly used semiconductor, has an energy gap . Its crystal structure is like that of silicon, except that half the silicon atoms are replaced by gallium atoms and half by arsenic atoms. Draw a flattened-out sketch of the gallium arsenide lattice, following the pattern of Fig. 41-10a.What is the net charge of the (a) gallium and (b) arsenic ion core? (c) How many electrons per bond are there? (Hint: Consult the periodic table in Appendix G.)
94% of StudySmarter users get better grades.Sign up for free | s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296948951.4/warc/CC-MAIN-20230329054547-20230329084547-00349.warc.gz | CC-MAIN-2023-14 | 1,902 | 11 |
https://projecteuclid.org/journals/notre-dame-journal-of-formal-logic/volume-37/issue-3/Connection-Structures-Grzegorczyks-and-Whiteheads-Definitions-of-Point/10.1305/ndjfl/1039886519.full | math | Whitehead, in his famous book Process and Reality, proposed a definition of point assuming the concepts of "region" and "connection relation" as primitive. Several years after and independently Grzegorczyk, in a brief but very interesting paper, proposed another definition of point in a system in which the inclusion relation and the relation of being separated were assumed as primitive. In this paper we compare their definitions and we show that, under rather natural assumptions, they coincide.
"Connection Structures: Grzegorczyk's and Whitehead's Definitions of Point." Notre Dame J. Formal Logic 37 (3) 431 - 439, Summer 1996. https://doi.org/10.1305/ndjfl/1039886519 | s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046153803.69/warc/CC-MAIN-20210728220634-20210729010634-00096.warc.gz | CC-MAIN-2021-31 | 675 | 2 |
https://www.hindawi.com/journals/jfs/2021/5137845/ | math | Recent Advances in Function Spaces and its Applications in Fractional Differential Equations 2021View this Special Issue
A Conservative Crank-Nicolson Fourier Spectral Method for the Space Fractional Schrödinger Equation with Wave Operators
In this paper, the Crank-Nicolson Fourier spectral method is proposed for solving the space fractional Schrödinger equation with wave operators. The equation is treated with the conserved Crank-Nicolson Fourier Galerkin method and the conserved Crank-Nicolson Fourier collocation method, respectively. In addition, the ability of the constructed numerical method to maintain the conservation of mass and energy is studied in detail. Meanwhile, the convergence with spectral accuracy in space and second-order accuracy in time is verified for both Galerkin and collocation approximations. Finally, the numerical experiments verify the properties of the conservative difference scheme and demonstrate the correctness of theoretical results.
The Schrödinger equation is one of the most basic equations in quantum mechanics, which was proposed by Austrian physicist Schrödinger in 1926. The equation can correctly describe the quantum behaviors of wave function, which has made great contributions to the study of quantum mechanics. Since then, the Schrödinger system has attracted a large number of mathematicians and physicists to explore the characteristics of its solution and physical applications. The study of conservative methods for the Schrödinger equation is one of the most popular research fields.
Over the past decades, most of the researches on the conservative method of the Schrödinger equation focus on the integer-order Schrödinger equation (e.g., see Refs. [1–7]). As models of science and engineering are needed to be more realistic, the fractional-order Schrödinger equation becomes one of the most important models in the fields of Bose-Einstein condensation, plasma, nonlinear optics, fluid dynamics [8, 9], etc. However, few studies have been investigated on conservative methods for the fractional Schrödinger equation. Besides that, most of the existing fractional-order conservative methods are finite element and finite difference methods [10, 11].
From the viewpoint of mathematics, the solution of the Schrödinger system has important geometric structures such as energy conservation and multisymplectic structure. Therefore, these properties should be maintained as much as possible in the construction of numerical methods. In this paper, we consider the following nonlinear fractional Schrödinger equation: subject to the boundary condition and the initial conditions where and are positive real constants, , and . and are given real functions. The fractional Laplacian operator can be defined as a pseudo-differential operator with the symbol : where is the Fourier transform and is the Fourier transform of .
The spectral method is a generalization of a standard separation variable method, for which Chebyshev polynomials and Legendre polynomials are generally used as the basic functions of approximate expansions. And the Fourier series is convenient to deal with the periodic boundary conditions. Bridges and Reich first put forward the Hamiltonian system using the Fourier spectrum discrete method in 2001. Based on their theoretical ideas, Chen and Qin in the same year proposed the Fourier pseudo-spectral method for the Hamiltonian partial differential equation and used it to integrate the nonlinear Schrödinger equation with periodic boundary conditions. For more comprehensive work on the different conservative Fourier pseudo-spectral methods, refer to [2, 14–16] and their references.
Since the equation is calculated on a finite interval , it is converted into periodic boundary conditions in this paper and studied on and below. Let
The outline of the remainder of this paper is as follows. In Section 2, a conserved Crank-Nicolson Fourier Galerkin method and a conserved Crank-Nicolson Fourier collocation method are constructed to discrete time variables and spatial variables. Energy-preserving and mass-preserving properties of the new method are investigated, and the error estimate is derived in Section 3. In Section 4, numerical experiments are presented to illustrate the theoretical results. Finally, the conclusions are given in Section 5.
2. Crank-Nicolson Fourier Spectral Method and Conservation Laws
Let be the set of all complex-valued and -periodic -functions on . Denote as the inner product on the space with the norm (abbreviated as ):
For as a nonnegative real number, let be the closure of . Note that . For any function , the following equations can be developed easily: where the Fourier coefficients are arranged as
For the Fourier transform of fractional Laplacian , we have
In order to discretize the equation in the temporal direction, the time step is defined by . Denote difference operator where is a positive integer (). Therefore, the Crank-Nicolson method was used to discretize equation (6) in the time axis.
2.1. Crank-Nicolson Fourier Galerkin Method
For positive even number , the basis function space can be constructed as where the norm and seminorm of are characterized by
The orthogonal operators are defined as follows:
Lemma 1 [18, 19]. Suppose that for all ; it holds that Denote The time variables of equation (6) are discretized by the Crank-Nicolson method. And the discrete Fourier Galerkin approximation for equation (6) has a modified scheme as follows: where , .
2.2. Crank-Nicolson Fourier Collocation Method
For positive even number , consider the points , , as collocation nodes. The discrete Fourier coefficients of a function on with respect to the collocation points are the following form:
Using the inversion formula, we have
Define the interpolation operator at the collocation points:
According to (27),
Corollary 3. For any , , there exists a constant independent of and , such that Using the Fourier collocation method to discrete the spatial variables of the equation, we get the fully discrete scheme for equations (6)–(8) as the following forms: Applying the Fourier transformation to (24), we get the following form: where .
2.3. Theory Analysis of Conservation
Proof. We derive the full discrete Fourier Galerkin method:
Let in equation (37); it holds that
Taking the imaginary part of equation (38), due to
The above equality indicates that the method (24) maintains the conservation of the discrete mass. The following items consider the conservation of the discrete energy.
Let ; according to equation (37), we also get Taking the real part of (42), due to therefore, using (43)–(46), we obtain thus, Based on the above analysis, the method (24) also maintains the conservation of the discrete energy.☐
3. Theory Analysis of Convergence
In order to simplify the notation, we always assume that is a positive constant in this article, which might be different in every formula.
Lemma 5 . For any discrete function , it holds that
Lemma 6. For , there exists a positive constant , such that
Proof. Using Theorem 4, it yields thus, Because of , it satisfies Sum the inequalities of Lemma 5 from 0 to yields Adding (53) and (54), we can obtain the following items: For is sufficiently small (), this implies According to the discrete Gronwall’s inequality, there is Therefore, ☐
Proof. Let , , , and ; then, . From triangle inequality and Lemma 1, it yields According to the orthogonality of the projection operator , we get The authors derive the full discrete Fourier Galerkin method: Subtracting equation (62) from equation (61), due to thus, According to the orthogonality of operator , i.e., . Therefore, Let in (64), and taking the real part, due to therefore, using (66)-(68), this implies where Thus, according to Lemma 6, we can get Note Lemma 1; it gives that Then, Thus, (69) becomes Because of and from Lemma 5, it gives that Then, combining (74) and (76) leads to Summing above inequalities (77) from 1 to yields Hence, using the discrete Gronwall’s inequality gives thus, Substituting (80) into (60) can yield which immediately gives conclusion.☐
Similar to the proof of Theorem 7, we can obtain the following theorem.
4. Numerical Example
Numerical examples will be proposed in this section to verify the correctness of the theoretical analysis, that is, the convergence of the numerical method and its ability to maintain discrete mass and discrete energy.
Example 1. Consider the nonlinear fractional Schrödinger equation with the wave operator: Let , , and . Figures 1 and 2 present the numerical solutions for and . We can find that the order of will affect the shape of the solution.
There is no exact solution of (83) known for . Therefore, numerical solution calculated by the method (24) with and is taken as the reference solution. Let be the numerical solution, and calculate the error at in the sense of the discrete norm: | s3://commoncrawl/crawl-data/CC-MAIN-2023-50/segments/1700679518883.99/warc/CC-MAIN-20231211210408-20231212000408-00856.warc.gz | CC-MAIN-2023-50 | 8,941 | 41 |
https://scholar.archive.org/search?q=On+the+Orphans+and+Covering+Radius+of+the+Reed-Muller+Codes. | math | A copy of this work was available on the public web and has been preserved in the Wayback Machine. The capture dates from 2017; you can also visit the original URL.
The file type is
Cai and VS. Pless, Orphan structure of the first-order Reed-Muller codes, Discrete Mathematics 102 (1991) 239-247. We investigate a method of combining two codes which we call the outer product. ... First-order Reed-Muller codes are outer products of a number of copies of the full binary space of length 2, and we apply our results to obtain cosets of the Reed-Muller codes which have no ancestors, ... The first-order Reed-Muller code R(1, m) is the outer product of m first order Reed-Muller codes R(1, 1). ...doi:10.1016/0012-365x(92)90118-y fatcat:ldwuv7uotbfltgmqq2f73mmmg4
A deep result about the Reed Muller codes, proved by Mykkeltveit in 1980, is that the covering radius of the Reed Muller code R(1, 7) equals 56. ... Among the results about \(1, m), that \(1, 7)=56, proved by Mykkeltveit , is one of the most difficult. ... Introduction Let R(r, m) be the rth order Reed Muller code of length 2 m , and let \(r, m) be its covering radius. ...doi:10.1006/jcta.1996.0055 fatcat:aphetd6usbay7dkzi52mfsrk3i
We characterize cosets that are orphans, and then prove the existence of a family of orphans of first order Reed-Muller codes R(1,m). ... Richard A. (1-WI); Pless, Vera S. (1-ILCC) Orphans of the first order Reed-Muller codes. ...
Cosets of the Reed-Muller code R(m-3,tn) are classified under the actions of GL(m,2) and GA(m, 2), the latter being the automorphism group of R(m -3, m) for m > 4. ... The number of cosets in each class is calculated. Orphans of R(m-3, m) are identified, and the normality of R(m-3, m) is established. ... G-equivalent cosets have the same coding theoretic properties. In this paper, we concentrate on the cosets of the (m -3)rd order Reed-Muller code R( m -3, m) of length 2". ...doi:10.1016/0012-365x(94)90113-9 fatcat:5rkol7flgfgibg7yfabbzxao7a
We study a family of particular cosets of the first-order Reed-Muller code R(1; m): those generated by special codewords, the idempotents. ... Thus we obtain new maximal weight distributions of cosets of R(1; 7) and 84 distinct almost maximal weight distributions of cosets of R(1; 9), that is, with minimum weight 240. ... Charpin and C. Carlet for their motivating discussions and N. Sendrier for valuable improvements concerning the implementation. We wish to thank H. F. Mattson and E. F. ...doi:10.1109/18.761276 fatcat:rfor5ftkqjhb5i5iteq3ukmhl4
cosets of the Reed-Muller code R(m — 3, m). ... The covering radius is defined as the weight of the coset of greatest weight; it is the maximum distance of any vector in the space from the code. In a paper on RM, G. Seroussi and A. ...
II (224-233); Philippe Langevin, On the orphans and covering radius of the Reed-Muller codes (234-240); Yuan Xing Li and Xin Mei Wang, A joint authentication and encryption scheme based on algebraic coding ... Janwa, On the parameters of algebraic geometric codes (19-28); Erich Kaltofen and B. David Saunders, On Wiedemann’s method of solving sparse linear systems (29-38); Simon N. Litsyn [S. N. ...
(English summary) 94b:94033 Langevin, Philippe On the orphans and covering radius of the Reed-Muller codes. (English summary) (see 94b:68002) Litsyn, S. ... ., 94m:94030 Hou, Xiang Dong Some results on the covering radii of Reed-Muller codes. (English summary) 94f:94020 Katsman, G. L. Bounds on covering radius of dual product codes. ...
Orphans of the first order Reed-Muller codes. 91b:94044 Bruck, Jehoshua (with Naor, Moni) The hardness of decoding linear codes with preprocess- ing. (Not in MR) Bussemaker, F.C. ... A new table of constant weight codes. 91h:94028 Brualdi, Richard A. (with Pless, Vera S.) On the length of codes with a given covering radius. 91¢:94025 — (with Pless, Vera S.) ...
On the covering radius of Reed-Muller codes. (English summary) 93¢e:94016 Conway, J. H. (with Pless, Vera S.; Sloane, N. J. A.) ... Orphan structure of the first-order Reed-Muller codes. 93b:94021 Cai, Ning see Brualdi, Richard A.; et al., 93b:94021 Calderbank, A. R. Covering machines. ...
Proceedings of 1995 IEEE International Symposium on Information Theory
The proofs of Propositions 2.5 and 2.6, by induction on t, are based on the fact that codes with length n, covering radius t, and containing K(n, t) words (i.e., optimal covering codes) cannot be too unbalanced ... Table A gives bounds for K(The so-called sphere-covering bound states that if C is a code of length n and covering radius t, then the volume of a sphere of radius t, multiplied by the cardinality of C, ... Specific Classes of Codes Covering Radius of Reed-Muller Codes Reed-Muller (RM) codes are among the most interesting families in the study of covering radius. ...doi:10.1109/isit.1995.535749 fatcat:vqartrp3offzdj6ols6lnqeht4
(From the text) 91a:94027 94B05 — (with Pless, Vera S.) Orphans of the first order Reed-Muller codes. JEEE Trans. Inform. Theory % (1990), no. 2, 399-401. ... On the length of codes with a given covering radius. Coding theory and design theory, Part I, 9-15, IMA Vol. Math. Appl., 20, Springer, New York, 1990. ...
On the length of codes with a given covering radius. Coding theory and design theory, Part I, 9-15, IMA Vol. Math. Appl., 20, Springer, New York, 1990. ... (From the text) 91a:94027 94B05 — (with Brualdi, Richard A.) Orphans of the first order Reed-Muller codes. JEEE Trans. Inform. Theory % (1990), no. 2, 399-401. ...
Boolean Models and Methods in Mathematics, Computer Science, and Engineering
Acknowledgement We thank Caroline Fontaine for her careful reading of a previous draft of this chapter. ... of the covering radius of the Reed-Muller code of order 1 if n is even; indeed, in the case of the Reed-Muller code of order 1, the covering radius coincides with the maximum nonlinearity of Boolean functions ... (the nonlinearity of f ) equals 2 n−1 − 2 n/2−1 (the covering radius of the Reed-Muller code of order 1). ...doi:10.1017/cbo9780511780448.011 fatcat:dtgopxbkmjgahepxw46xgo6rd4
Langevin, Philippe On the orphans and covering radius of the Reed-Muller codes. ... (see 94i:00015) 65M06 (76M25) Langsetmo, Lisa The K-theory localization of loops on an odd sphere and applications. Topology 32 (1993), no. 3, 577-585. ...
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https://treulatevabestia.cat/how-to-find-value-of-x-and-y-in-triangle | math | B) 6x + 6y = 36. How do you find the value of x in a triangle using pythagorean?
Now, as we know the sum of internal angles of a triangle is 180.
How to find value of x and y in triangle. The hypotenuse is 2 times the length of either leg, so y =72. In ∆abc, ∠a = x ∠b = 50° ∠acb = y ∠acd = 120° we know that, exterior angle is sum of interior opposite angles ∠acd = ∠a + ∠b 120° = x + 50° 120° − 50° = x 70° = x x = 70° we know that, sum of angles of a triangle is 180° ∠a + ∠b + ∠acd = 180° x + 50° + y = 180°. Since the length of the hypotenuse is 1 and it is twice the length of the shorter leg, x, we can say that 1 x = 2.
266 chapter 4 congruent triangles guided p ractice for examples 3 and 4 5. To find x, take the square root of both sides. Sum the x values and divide by 3.
To find the value of x, use #gfj. An exterior angle of a triangle is equal to the sum of the two opposite interior angles. Find the value of x and y using substitution method:
X x y e since only a negative divided by a negative will result in a positive. Find the values of x and y in the following triangle. Find the values of the unknowns x and y in the following diagrams:
Length of midsegment and base of triangle are 3 x + 5 and 12 x − 8 respectively. Find the values of x and y in the following triangle. Thus, the angle e is in the iii quadrant.
Find the values of the unknowns x and y in the following diagrams: Find the values of x and y in the diagram. Divide both sides by 5, you get x is equal 143/5, which we can just leave as an improper fraction.
#sin30° =5/y# using the table of trig values, we find that #sin30° = 1/2# so, this gives us: 0 y y x x x x x x thus, the common quadrant is the iii quadrant. Right triangle trigonometry special right triangles examples find x and y by using the theorem above.
Take equation b) from equation a) to eliminate the x component. These are the three possible values of x, given the information that they gave us right up there. The square root of x² is x, so the answer is that x = √(81/5).
X = 76 subtract 104 from each side. 0 or tan e ! Find the values of x and y.
X y 1 1 30o 3 2 1 2 1 60o (3 1) 22, x 1 1 x y 60o 1 Y can also be solved using the law of sines. X notice in the right triangle, x is the opposite side of the given angle and the given value of 15 meters is the adjacent side of the given angle.
Let's find the length of #y# first. How do you find the pythagorean triple? Identify the centroid coordinate, (4.
The exterior angle of a triangle is 120°. If you square each number, subtract one square from the square greater than it, then square root this number, you can find pythagorean triples. #1/2 = sin30° = 5/y#
Special values of trigonometric functions. A balloon is 150 feet above the ground. A) 6x + 10y = 52.
Since the longer leg, y, is 3 times the length of the shorter leg, we can say that 1 y = 2 3, or equivalently, 3 y = 2. Multiply equation a) by 2 and equation b) by 3 to form the following equations. Write answers in simplest radical form.
X − y = 2 and 2 x − y = 9 Right triangle trigonometry special right triangles examples find x and y by using the theorem above. Next, i found z using the law of sines.
The pythagorean theorem, the trigonometric ratio: To find the values of x and y, you will need to use. Alternatively, you could observe that the right triangle with hypotenous of 10 and leg of 6 is a 3:4:5 triangle and solve for x using that ratio.
Reasoning use parts (b) and (c) in example 4 and the sss congruence postulate to give a different proof that nqps >npqr. You could write it as a mixed number or however else you might want to write it. So that gets us to 143.
B) 6x + 6y = 36. We need to find a way to equate either the x terms of the y terms in each equation. Ex 6.3, 2 (i) find the values of the unknowns x and y in the following diagrams:
Tan 22q | 0.4040262258 answer: The pythagorean theorem, the trigonometric ratio: Sin e 0 cot e !
Learning to find the value of x and y from congruent triangles. The angle of elevation from an observer on the ground to the balloon is We again want to find the values of x and y.
⇒ y + x = 180° ⇒ 140° + y = 180° subtract 140° from both sides. The legs of the triangle are congruent, so x =7. This is because it is half of the larger triangle's side length.
So it's not going to divide nicely. Thus, we have that x < 0 and y < 0. I) as we know, in an isosceles triangle, two sides and the angles they make with the third side are equal.
Find the value of x in the following triangle. Then, = 50 o + x = 120 o. Thus, tan 22q 15 x x 15 tan 22q | 6.06 note:
Sum the y values and divide by 3. The hypotenuse is 2 times the length of either leg, so To find the value of y, look at &fjh.it is a straight angle.
First find x using the pythagorean theorem with the other two legs of the right triangle. | s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662532032.9/warc/CC-MAIN-20220520124557-20220520154557-00658.warc.gz | CC-MAIN-2022-21 | 4,891 | 27 |
https://www.textkit.com/greek-latin-forum/viewtopic.php?f=13&t=1596&start=80 | math | Why don't you know whether or not you have this disease? What is creating the uncertainty?
You must have been missing the entire point of my stand then. The uncertainty is created by other possibilities. It is possible that this could have happened or this could have happened. There is no reason to doubt that. For example, we could say that it is possible that aliens took people from the Bermuda Triangle, or it is possible that a giant sea creature swallowed them, or that it is possible that the ships went through a worm hole. There is no reason to deny it. They are just unlikely things to happen. There Is no reason to deny that the possibilities are boundless. "It could ahve been anything." "Anything is possible." Even thogh what the person meant by these things may have been for effort purposes, the literal take on these sentences is true. Why would anything not be possible?
Now since there are many many possibilities in our world, it creates this uncertainty I'm talking about. If there was another course of action that COULD HAVE taken place, then we cannot be absolutly certain about our beliefs. For example, it is possible that I have that disease that impairs my senses to sense things that are not there. It may not be probable, but it is definitly possible. Why is it possible? This is not a good answer, but my thinking is because it can be thought of. If I can think of a logical flow of events, each event not having to make sense in the first place, then I can create uncertainty in anything. The reason that each event does not have to make sense is because some of the things involved in it may not be understandable or we do not understand yet. In our disease example, I could say that it only affects me when there is a half moon and an eclipse. Why not? It is possible. Now I am haven't really seen an eclipse so I wouldn't have shown signs yet. I do not need to explain why this may be because it could be about something that humans are completely unfamiliar with. Now let us say that at 12:15 tomorrow, there will be an eclipse and the moon is a half moon. Well, at the time, which happens to be our break, I would start sensing things that are not there and tasting things that are not there. It would be strange. In this example, I have this disease and I was affected by it. But now, you would say, it does not affect me now. But the other thing is that it is possible that this 'trigger' could be anything. So the trigger could be oxygen! It is also possible that every time I was affected, which would be forever, all the things I sensed and stuff would make sense to people watching me and would not suspect a problem.
I hope you understand this because you are right. I have said this many many times and I'm not sure if you have acknowleded it or not. Here is some proof if you will...
How do you know that your brain is misinterpreting your senses and making you believe that you are a male? How do you know that everyone knows this but just won't tell you so you can have a good life? How do you know that you understand the meaning of english? You cannot!
Every time I say how do you know, it could be substited for it is possible that...
Yes it is possible for this to happen, but the fact of the matter is that it can nver be proven that that is reality because our senses may cause us to believe the wrong things.
I say our senses may be. That means it is possible that...
However, this infinite amount of knoledge could also be equal to an infinite amount of possibilities. If there are an ifinite amount of possibilities, then that must create some uncertainty in our thoughts. The fact of the matter is that it IS possible to know things, but we can NEVER know them WITH absolute certainty. This is because there is uncertainty in our thoughts. Even though we believe what these things are to be true, we still have uncertainty.
I am not saying that you misunderstand english, but merely saying that there is a chance.
I do not think I do because no symptoms have shown. Therefore I probably do not, but possibily have one that has not shown symptoms yet.
I also said there that it is possible that I DO have a disease, but am unaware of it.
smell what I call dirt and grass, but I do not know that it is real because there is a possibility that my senses are detecting things that are not there. However, I believe that it is grass and dirt.
If I did taste the ground, I think I would taste dirt. However, I cannot be sure that the dirt is real because there is a possibility that I am tasting something that is not there. This possibility makes me so I am uncertain.
I har things, but I cannot be absolutely certain that they are real. This is because there is a possibility that my senses are sensing something that is not real. However, I do believe that I hear real sounds.
I don't think you have realized how great possibilities are. | s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583658662.31/warc/CC-MAIN-20190117000104-20190117022104-00619.warc.gz | CC-MAIN-2019-04 | 4,893 | 16 |
https://www.coursehero.com/file/5820079/Lec15/ | math | This preview shows pages 1–3. Sign up to view the full content.
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Unformatted text preview: I E/Stat 265, Fall 2009 I E/Stat 265, Fall 2009 O /Stat 65, a 009 O /Stat 65, a 009 Lecture #15: Lecture #15: Estimation Concepts and Methods Estimation Concepts and Methods- Concepts of Point Estimation s a o C o p s a d o d s s a o C o p s a d o d s 6 1 Concepts of Point Estimation A. Biased Vs. Unbiased Estimators B. Minimum Variance Estimators C. Standard Error of a Point Estimate 6-2 Point Estimation Methods A. Method of Moments (MOM) 1 B. Method of Maximum Likelihood (MLE) C. Method of Bayes (Not testable this term) 1 Point Estimates 6.1 Point Estimates Objective obtain an estimate of a population parameter from a sample (e.g. sample mean, , is point estimate of population mean x ) x Applications: Parameter Estimation Hypothesis Testing (and other inferential statistics methods) 2 arameter Estimation Example Parameter Estimation Example uppose you wish to estimate the population Suppose you wish to estimate the population mean, ,. Some possible estimators include: Sample Mean, Sample Median, Sample Trimmed Mean, Sample Geometric Mean, Sample Harmonic Mean, (Max X + Min X)/2 Recall example from descriptive statistics, which of the following is the best estimator? ample Mean = 965 0 Sample Median = 1009 5 Sample Mean = 965.0 Sample Median = 1009.5 Sample Trim Mean = 971.4 3 arameter Estimates it Parameter Estimates - Fit Estimates, , always have measurement error. hy? estimation error = + Why? Because is a function of observed x i s (r.v.) that change with n We identify the best estimator as the one with: inimum Bias (unbiased) vg stimation error Minimum Bias (unbiased) avg estimation error = 0 Minimum Variance of estimation error 0 as n More consistent predictions 4 p Increases the likelihood the parameter estimate represents the true parameter iased Vs Unbiased Estimator Biased Vs. Unbiased Estimator Bias - difference between the expected value of the statistic and the true parameter nbiased Estimator: Bias = E( ) Unbiased Estimator: Example: is an unbiased estimated of (Bias = 0) Suppose X 1 , X 2 , .. X n are iid rvs, with E(x i ) = x n i i 1 E(x ) n E(x) E( ) = = = = = 9 n n nbiased Estimator of Variance Unbiased Estimator of Variance hich of the following formulas is an unbiased Which of the following formulas is an unbiased estimator of variance, 2 ? n n 2 2 i i 2 2 i 1 i 1 1 2 (X X) (X X) s n 1 n = = = = = = ( ) ( ) = = 2 2 2 2 n 1 E s E n What happens to the bias effect as n becomes large?...
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This note was uploaded on 03/17/2010 for the course IOE 265 taught by Professor Garyherrin during the Fall '09 term at University of Michigan-Dearborn.
- Fall '09 | s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501170884.63/warc/CC-MAIN-20170219104610-00574-ip-10-171-10-108.ec2.internal.warc.gz | CC-MAIN-2017-09 | 2,823 | 6 |
https://elitewriters.net/big-als-pizza-inc-needs-a-cash-budget-for-year-3-and-has-provided-you-with-the-following-information-2/ | math | Big Als Pizza, Inc. needs a cash budget for year 3 and has provided you with the following information. Sales are all on account and are estimated to be collected over a three-month period, with 70 percent collected in the month of sale, 25 percent collected in the next month and 4 percent collected in the third month. The remaining1 percent is estimated to be uncollectible. December and November sales from the previous year were $201,638 and $185,000 respectively. Because of the lag in collecting cash from sales on account, Big Als delays payment on some of its purchases of materials. Big Als estimates that 60 percent of each months material purchases are paid in the month of purchase and 40 percent in the following month. The accounts payable balance for materials at the end of the previous year was $20,000. Big Als also requires a minimum balance of $40,000 in cash at the end of each month. The company will use its line of credit when needed to bring the balance up to that minimum level. For any money borrowed, the interest rate is 6 percent compounded annually. For simplicity, you can assume that cash is borrowed on the fi rst day of the month and that loan repayments are made at the end of the month. Big Als plans to exercise the option on the leased production equipment in March (as described in Part Seven). The purchase price on the equipment will be $153,450, with payments of $3,260.36 per month. Big Als also plans on expanding the existing production space in May at a cost of $200,000. Big Als would like to fi nance the expansion out of current earnings and so will use the line of credit, if necessary, in May. The expansion will cause fi xed manufacturing overhead to increase by $10,000 per month, starting in May. Required: A. Prepare a cash receipts budget for year 3, assuming estimated sales of 385,000 meat pizzas and 30,000 veggie pizzas and the following monthly distribution of sales.January 8.3% July 8.5% February 9.2 August 9.8 March 10.3 September 7.5 April 7.6 October 9.1 May 8.0 November 7.2 June 6.9 December 7.6Direct material costs per unit are $.74 per meat pizza and $.62 per veggie pizza. Direct labor costs are $2.51 per meat pizza and $2.78 per veggie pizza. Monthly fi xed selling and administrative costs are $15,300, while monthly fi xed manufacturing overhead is $2,851. The variable overhead cost is $.55 per pizza. The sales price for veggie pizzas is $5.25 per pizza and the sales price for meat pizzas is $5.00. B. Prepare a cash disbursements budget for the year. C. Prepare a summary cash budget for the year, showing any borrowing and repayment of debt with interest. Discuss Big Als ability to repay the expansion loan. Include a discussion of the feasibility of the project. Include qualitative factors to be considered. D. What if the sales forecast was increased by 50 percent? What impact does that have on the budget, and what is the potential impact on the company? | s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400245109.69/warc/CC-MAIN-20200926200523-20200926230523-00673.warc.gz | CC-MAIN-2020-40 | 2,944 | 1 |
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