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December 14

Slight variation on the angle sum formulae

I was looking over the angle sum formulae on List_of_trigonometric_identities#Angle_sum_and_difference_identities specifically at the one for arccos, Can the formulae be expressed more generally for say $A\arccos \alpha \pm B\arccos \beta$ ? I guess in theory its simply a case of repeated use of the formulae but my trig skills are severely lacking, Thanks--137.108.145.10 (talk) 12:28, 14 December 2010 (UTC)[reply]

Am i being an idiot or is it just

A*B\arccos(\alpha \beta \mp {\sqrt {(1-\alpha ^{2})(1-\beta ^{2})}})

--137.108.145.10 (talk) 17:53, 14 December 2010 (UTC)[reply]

No, it's not that. That would be

AB\arccos \alpha \pm AB\arccos \beta

. I'm not sure if there is a nice expression for what you want. --COVIZAPIBETEFOKY (talk) 18:07, 14 December 2010 (UTC)[reply]

Combining two low resolution images

Can mathematics give any clues about the improvement in resolution you could get from combining two different low resolution digital photos of a painting? Are there any general principals? If the (square?) pixels were not in perfect alignment but overlaped, what would be the best mathematical procedure to improve the resolution? Thanks 92.15.11.6 (talk) 18:28, 14 December 2010 (UTC)[reply]

This is called Super-resolution. Our article doesn't say much but you may find useful information in the references.

I think you need the photos taken from slightly different locations for it to work. The general approach is to express what each pixel of the low-res images would be for a given high-res object, and choose the values that give the best fit to the input low-res images. One implementation is to use sum of squared errors as a measure of fit, so the solution consists in solving a system of linear equations. -- Meni Rosenfeld (talk) 19:42, 14 December 2010 (UTC)[reply]

Registax can do this. 92.28.247.44 (talk) 22:39, 16 December 2010 (UTC)[reply]

Cluster points of uncountable sets

Does an uncountable subset of the reals have to have uncountably many cluster points? —Preceding unsigned comment added by 199.17.35.133 (talk) 20:15, 14 December 2010 (UTC)[reply]

Sure. Start with an uncountable set; it has to have a cluster point because there must be at least one interval [n,n+1] within which it has uncountably (and therefore infinitely) many points, and [n,n+1] is compact. Remove that cluster point; there are still uncountably many points left, so there must remain a cluster point. Iterate through all countable ordinals (what happens at limit ordinals should be obvious); at each step, you've removed only countably many points, so there are uncountably many left, so you can keep going. --Trovatore (talk) 20:49, 14 December 2010 (UTC)[reply]

My proof goes like this. We show by contradiction that for any such uncountable S, there exist two disjoint intervals in the reals that each contain uncountably many points from S. To do this, suppose otherwise, and follow along the proof of the Bolzano-Weierstrass Theorem, to show that S has only countably many points, which is untrue. Now that we know each uncountable S has uncountably many points in each of two disjoint intervals, we show it has uncountably many cluster points (in the same way one shows the Cantor set is uncountable). Eric. 82.139.80.124 (talk) 21:10, 14 December 2010 (UTC)[reply]

It's good to get in the mental habit of transfinite induction on problems like this, though. In this particular example it doesn't really buy you much, because the points removed don't really "interact" much (or their interaction is ultimately irrelevant). But it's a way of thinking that often makes the answer obvious after a small amount of thought. --Trovatore (talk) 22:43, 14 December 2010 (UTC)[reply]

Hmm, now that I think about it a little more carefully, my argument doesn't work as stated, because it's not clear what's meant by "removing a cluster point". --Trovatore (talk) 22:55, 14 December 2010 (UTC)[reply]

logical terminology

An implication like $A\Rightarrow B$ is equivalent to $\neg B\Rightarrow \neg A$ and this is called the contrapositive. The formula $(\forall x)\phi (x)$ is equivalent to $\neg ((\exists x)(\neg \phi (x)))$ . Is there a word like "contrapositive" that describes switching a quantifier like that? Thanks. 67.117.130.143 (talk) 20:45, 14 December 2010 (UTC)[reply]

Not exactly answering your question, and maybe you already thought of this this, but you're talking about something which already is basically a contrapositive.

(\forall x)\phi (x)

is the same as

x\in U\Rightarrow \phi (x)

, where U is the universe of discourse. Then the contrapositive of this is

\neg \phi (x)\Rightarrow \neg x\in U

, which means

\neg ((\exists x)(\neg \phi (x)))

. But I've never heard of a precise name for what you're talking about. Staecker (talk) 13:29, 15 December 2010 (UTC)[reply]

It's a variant of De Morgan's laws.—Emil J. 14:15, 15 December 2010 (UTC)[reply]

December 15

Something weird

In Lang's algebra page 813 he says "Let R be a principal ring. A module is flat iff its torsion free" and he says proof is "easy". Take the ring to be Z/4Z, then its flat over itself yet is not torsion free. Infact any principal non ideal domain will do. Am I missing something? Money is tight (talk) 02:29, 15 December 2010 (UTC)[reply]

I think Lang assumes that principal rings are domains. I can't say I like the terminology. Certainly your argument is valid. --Sam^Talk 18:45, 16 December 2010 (UTC)[reply]

Laws of Mathematics

Is it possible that some of the basic laws of logic and/or math are empirical in nature? That is, we take them to be true because they are intuitively obvious, but that they are only intuitively "obvious" because the universe is structured that way.

For example, if A = B, and B = C, then A = C. Can this be proved? If so, what assertions does the proof require, and can those assertions be proved?

I suspect these questions cannot be answered, but I'm curious to hear the thoughts of those more learned in math than myself. 74.15.138.27 (talk) 09:23, 15 December 2010 (UTC)[reply]

Any useful mathematical system must start with a set of axioms - propositions that are taken to be true, and do not need to be proved. These axioms do not have to be intuitively obvious - although sometimes they are - and neither do they have to correspond to the behaviour of anything in the real world - although sometimes they do this too. Sometimes we find, by accident or by design, that the mathematical system that is built on our axioms is a useful model of some aspect of the real world. But even in this case, I don't think the axioms are empirical, because an empirical proposition can be disproved or refined by the results of an experiment. The experimental results that validated relativity and disproved Newtonian mechanics, for example, did not disprove the mathematics behind Newtonian mechanics - they just showed that this mathematical structure is not always an accurate model of reality. Gandalf61 (talk) 12:13, 15 December 2010 (UTC)[reply]

There's a whole subject Philosophy of mathematics that deals with questions like that, but there aren't any real answers. It's certainly been proposed that the laws of mathematics are empirical. The article abductive reasoning might also interest you. 67.117.130.143 (talk) 19:45, 15 December 2010 (UTC)[reply]

You might like to read the article on equivalence relations. The standard equality that we know from arithmetic is an example of an equivalence relation. In the case of the standard equality, on the set of whole numbers {…, −2, −1, 0, 1, 2, …}, it is an equivalence relation because for any whole number n we have n = n. For any two whole numbers m and n: if m = n then n = m. For any three whole numbers k, m and n: if k = m and m = n then k = n. — Fly by Night (talk) 02:34, 17 December 2010 (UTC)[reply]

Series

If sum(x_n) is a convergent series, must it be that sum((xn)^3) is a convergent series? it is obviously so if all x_n > 0 but what if there are negative terms? thanks —Preceding unsigned comment added by 131.111.222.12 (talk) 10:38, 15 December 2010 (UTC)[reply]

No. Take

x_{n}={\begin{cases}{\frac {1}{\log n}}&3|n\\{\frac {-1}{2\log n}}&3\not |n\end{cases}}

-- Meni Rosenfeld (talk) 13:41, 15 December 2010 (UTC)[reply]

It works if the series is absolutely convergent, i.e. the sum of the absolute values of the terms is finite. Michael Hardy (talk) 21:13, 15 December 2010 (UTC)[reply]

Or

x_{n}={\frac {(-1)^{n}}{\sqrt {n}}}

. The point being to take advantage of the fact that squaring will turn an alternating series into an all-positive series.--71.175.63.136 (talk) 16:38, 15 December 2010 (UTC)[reply]

And I misread that exponent as 2 instead of 3. Ignore me.--71.175.63.136 (talk) 16:39, 15 December 2010 (UTC)[reply]

centre of circle

Three points A(a,b) B(c,d) ,D(e,f) are three points on circle in a plane .is there any simple formula to find their centre (h,k) ,I have derived one ,but i think perhaps it will also be derived earlier .Is their short formula .how can i do research in a field of mathematics such that same can not be done in Mathematics. khan — Preceding unsigned comment added by True path finder (talk • contribs) 12:00, 15 December 2010 (UTC)[reply]

See if your formula agrees with the one given here: http://mathforum.org/library/drmath/view/55239.html. If you just want to check examples, Wolfram Alpha can do it: http://www.wolframalpha.com/input/?i=circle+through+%281%2C2%29+%280%2C0%29+%283%2C0%29.

Is your more general question about finding a research field where your work is unlikely to be already done by somebody else? If that's what you want to do, the traditional way is to work your way through a mathematics graduate school program, and build relationships with good mentors who can help you find open problems in their field which you could realistically work on. It is still possible for people outside of this traditional system to do new research in mathematics, but this is extremely rare. Staecker (talk) 13:40, 15 December 2010 (UTC)[reply]

There is a method from elementary geometry that you might be interested it. Connect the three points A, B, and C into a triangle. The circumcentre of the triangle will be the centre of the circle in which the triangle is inscribed. It is a trivial exercise of algebra to compute the perpendicular bisectors and their point of intersect. 24.92.70.160 (talk) 00:44, 16 December 2010 (UTC)[reply]

December 16

Probability

I did a puzzle recently that went like: 2010 points are distributed evenly on a circle. If three points are randomly selected, what is the probability that they form a right triangle? I reasoned that the converse of the Mean Value Theorem implies that on a circle there are chords with every possible slope. Therefore the first two points chosen do not matter, only the last point. However, there are two points on a circle that would form a chord at a right angle with the first; thus I answered 1/1004 (or 2/2008). However the answer the book had was 3/2009. Where was my reasoning faulty? 24.92.70.160 (talk) 00:41, 16 December 2010 (UTC)[reply]

First note that there are

{\tbinom {2010}{3}}\,\!

ways to choose 3 points on the circle. Recall that a right triangle inscribed in a circle has a diameter for its hypotenuse (Thales' theorem). So, when choosing any given point, we must also choose the point diametrically opposite if we wish to form a right triangle. After that we can select any of the 2008 other points. Then there are

2010(2008)/2

ways to form a right triangle from the 2010 points, where the division by two prevents double-counting. The probability is therefore

{\frac {2010(2008)/2}{\tbinom {2010}{3}}}={\frac {3}{2009}}.

Hope that helps. —Anonymous Dissident^Talk 01:15, 16 December 2010 (UTC)[reply]

Your error was forgetting that if the first two points are opposite (which has probability 1/2009) then any third point will work. The total probability becomes: (1/2009 × 1) + (2008/2009 × 2/2008) = 3/2009. PrimeHunter (talk) 02:16, 16 December 2010 (UTC)[reply]

Another (equivalent) way of looking at it is: Choose any point. the second must be its antipode, which has probability

1/2009

, or the third must be, which has probability

{\frac {2008}{2009}}{\frac {1}{2008}}

(where the first fraction is to ensure the second point is not opposite the first point), or the second must be opposite the third, which has probability

{\frac {2008}{2009}}{\frac {1}{2008}}

(see preceding parenthetical remark). Adding these you get

3/2009

.—msh210℠ 17:03, 16 December 2010 (UTC)[reply]

Calculus textbook question

After reading Spivak's Calculus, what is a good continuation into more advanced calculus (I'm looking for a book)? It should be no more rigorous than Spivak, and preferably slightly less rigorous. I am especially interested in partial derivatives and contour/path/line integrals. 24.92.70.160 (talk) 03:38, 16 December 2010 (UTC)[reply]

Options abound. There's Folland's Advanced Calculus, or Marsden and Tromba's Vector Calculus, both of which are pretty standard. I've heard very good things about Apostol's Calculus and Linear Algebra (the latter part of volume I and most of volume II should serve your need), although I've never read that one. I also recommend Div, Grad, Curl and all that as a very nice intuitive supplement. If you found Spivak's Calculus too rigorous for comfort, I would advise you to avoid Calculus on Manifolds (also by Spivak), or Rudin's Principles of Mathematical Analysis. Cheers, Ray^Talk 06:15, 16 December 2010 (UTC)[reply]

About the title 'Another math question'. This is the maths reference desk so I would guess it was a maths question and since you started another topic I would guess it was 'another'. Please try and think of a better title when submitting questions. Dmcq (talk) 11:55, 16 December 2010 (UTC)[reply]

Probability and combinatorics

Hello. I'm looking for good formal introductions to probability and combinatorics – material that might be covered in an undergraduate course, for example. The topics are related but deserve separate treatment, of course, so I'm not looking for a single book for both. Thanks in advance for any recommendations. —Anonymous Dissident^Talk 11:04, 16 December 2010 (UTC)[reply]

Did you study our articles on probability and combinatorics? They have references to books. If you need something specific, ask again. Bo Jacoby (talk) 11:40, 18 December 2010 (UTC).[reply]

December 17

textbook equation doesn't work.

The equation: $2cos{\frac {\Pi }{6}}(5)+6\doteq 4.3$ is in my textbook. But I can't get any calculator to give me this results, radian mode or otherwise. Wolfram doesn't even return it. Any ideas? — Preceding unsigned comment added by 173.33.106.153 (talk • contribs) 02:18, 17 December 2010 (UTC)[reply]

Are you sure it's

2\cos({\frac {\pi }{6}})(5)+6=4.3

and not

2\cos({\frac {5\pi }{6}})+6=4.3

? 24.92.70.160 (talk) 01:24, 17 December 2010 (UTC)[reply]

I think the OP made a mistake too. In fact

\cos \left({\frac {5\pi }{6}}\right)=-{\frac {\sqrt {3}}{2}}\ \Rightarrow \ 2\cos \left({\frac {5\pi }{6}}\right)+6=6-{\sqrt {3}}=4.3\ (1\ {\mbox{d.p.}})

— Fly by Night (talk) 02:18, 17 December 2010 (UTC)[reply]

Are you sure you have your calculator set on radians rather than degrees? Michael Hardy (talk) 23:59, 18 December 2010 (UTC)[reply]

Figured it out- texbook's formula was $2cos{\frac {\pi }{6}}t+6$ , then they substituted in 5 to give $2cos{\frac {\pi }{6}}(5)+6\doteq 4.3$ . Some unclear notation on their part. Thanks! 173.33.106.153 (talk) 00:08, 19 December 2010 (UTC)[reply]

I don't see anything unclear about it. Michael Hardy (talk) 05:12, 20 December 2010 (UTC)[reply]

Derivative question

Is df(x)/dx in general equal to df(x²)/d(x²)? I would think so, but I'm having trouble proving it, and I used this for a math question and the answer I get seems wrong. 74.15.138.27 (talk) 04:33, 17 December 2010 (UTC)[reply]

No, it'll be the same derivative okay - but with x squared as the argument. Dmcq (talk) 08:19, 17 December 2010 (UTC)[reply]

Yeah, obviously, that was really stupid of me. Thanks. 74.15.138.27 (talk) 09:15, 17 December 2010 (UTC)[reply]

units in equations, cancelling

I realise this is a very basic question, but hopefully some of you will be willing to stoop to my level to explain it to me. I've always struggled with how units cancel in equations. Sometimes I get it right, other times I go horribly wrong and am never quite sure why. I'm studying solo and my textbook isn't explaining some of the basics (that I'm already supposed to have a clue about) and I have no teacher to ask. So, if I can get the formatting right, an example of my current confusion...

$v_{s}={\sqrt {\frac {\mu }{\rho }}}$

where μ is the rigidity modulus and

v s

is S wave speed and ρ is density.

Substituting in my figures, I end up with:

$v_{s}={\sqrt {\frac {3.0\times 10^{10}{\text{ kg m s}}^{-2}{\text{m}}^{-2}}{2.70\times 10^{3}{\text{ kg m}}^{-3}}}}$

I know that, for the final answer, I should get m s^–1. And my textbook tells me that, when I come to do the square root, I'll be working with the units m² s^–2, which is exactly what I want to get m s^–1. But how did they get to m² s^–2? They've cancelled out the kg top and bottom, which I'd have done, but I first tried to also use the m^–3 to cancel out the m and the m^–2. Why don't they cancel (apart from the fact that they can't because then I'd have no metres!)? But the bigger question is what has happened to make the answer at square root stage m²?! I have no idea how to get to m². I'm lost. What am I missing? I feel dreadfully stupid! I realise I should know this and one must wonder why I don't, but I guess I've always managed to blunder through and get by on knowing what the final units should be. Phantasten (talk) 13:28, 17 December 2010 (UTC)[reply]

In this case it's just about paying attention to the signs of the exponents.

m^{-2}={\frac {1}{m^{2}}}

and

{\frac {1}{m^{-3}}}=m^{3}

, so

{\frac {m\cdot m^{-2}}{m^{-3}}}=m\cdot m^{-2}\cdot {\frac {1}{m^{-3}}}=m\cdot {\frac {1}{m^{2}}}\cdot m^{3}={\frac {m\cdot m^{3}}{m^{2}}}={\frac {m^{4}}{m^{2}}}=m^{2}

.

Of course, this can be done with many less steps once you get the hang of it. -- Meni Rosenfeld (talk) 14:02, 17 December 2010 (UTC)[reply]

Another way to do it is to add all the (meter exponent) numerators together and subtract all the (meter exponent) denominators, being careful to maintain the signs:

(+1) + (-2) - (-3) = 1 - 2 + 3 = 2

So, this gives us m² (in the numerator). StuRat (talk) 17:59, 17 December 2010 (UTC)[reply]

You've both been really really helpful, thank you so much! Just what I needed. Phantasten (talk) 17:42, 18 December 2010 (UTC)[reply]

Rational points on a circle

I am solving a question which goes like this: Maximum number of points with rational co-ordinates on a circle whose center is $({\sqrt {3}},0)$ is (a) 1, (b) 2, (c) 4, (d) ∞. This question does not seem complete to me. Do we also need some information on the radius whether it is rational or irrational, or can we say about such points without knowing anything about radius. Thanks. - DSachan (talk) 23:02, 17 December 2010 (UTC)[reply]

That's why the word "maximum" is in the question. You're being asked for the maximum over all possible radii. Algebraist 23:07, 17 December 2010 (UTC)[reply]

Oh wow, very nice question. Thanks, I got the answer 2. - DSachan (talk) 00:45, 18 December 2010 (UTC)[reply]

December 18

Calculus practice

Hello. Where can I find some extra practice with the concepts introduced in Spivak's Calculus, the book used in my class, so I can solidify my understanding? (I'm looking for a book or a website) These should have answers and solutions easily accessible. I'm looking for a coverage that is about the same scope as the textbook, but more problems; Spivak doesn't really give practice in the problems at the end of the chapters, he uses them to make you think or introduce new concepts, but I need to gain a grasp on the old concepts first! 24.92.70.160 (talk) 01:30, 18 December 2010 (UTC)[reply]

That is a pretty theoretical book, sort of halfway between a calculus book and one on real analysis. What concepts are you having trouble with, and what kind of trouble are you having? If you're having calculational difficulties, try something like a Schaum's outline (they are often better than they sound). Wikibooks also has a calculus book in progress, b:Calculus, some parts of which are pretty good. 67.117.130.143 (talk) 09:09, 18 December 2010 (UTC)[reply]

Nonogram constant

Is there a “nonogram constant”, which appears in the order of growth of the number of all n×n-nonograms with a unique solution? --84.61.182.248 (talk) 12:18, 18 December 2010 (UTC)[reply]

December 19

Unclear about an equation in an article

Hi there, in Central_limit_theorem#Relation_to_the_law_of_large_numbers, what are $\varphi$ , $O$ , $\Xi$ and so on? Thanks. It Is Me Here ^{t / c} 11:36, 19 December 2010 (UTC)[reply]

I'm not quite clear - do you want to know about the symbols or about their meaning?

\varphi

(phi) and

\Xi

(xi) are greek letters, used by mathematicians in moderate distress because they ran out of Latin letters (mathematicians in high distress turned to Hebrew, and eventually just made up new symbols ;-). --Stephan Schulz (talk) 11:43, 19 December 2010 (UTC)[reply]

No, no, hehe, I mean: what do these letters represent in the given article section? It Is Me Here ^{t / c} 11:53, 19 December 2010 (UTC)[reply]

The part before "Informally, ..." is just an introduction about asymptotics, where f is an arbitrary function and

\phi _{i}

are also somewhat arbitrary functions that give an expansion of f.

\xi

is just the name given to the limit distribution, which the text says must be normal.

\Xi

is the name given to the limit in the case that the initial distribution does not have a finite mean\variance, which the text says can take various forms but must be stable. -- Meni Rosenfeld (talk) 12:01, 19 December 2010 (UTC)[reply]

And the O is as in Big O notation especially the "Infinitesimal asymptotics" section. This is basically saying there are other terms in f which we can ignore when taking the limit.--Salix (talk): 13:25, 19 December 2010 (UTC)[reply]

Fourier transform of a multivariate gaussian

I'm trying to compute the Fourier transform of a multivariate Gaussian like

e^{-{\frac {1}{2}}x^{T}C^{-1}x}

where C is a matrix.

Does anybody know a reference where I can see this computation?--Pokipsy76 (talk) 13:57, 19 December 2010 (UTC)[reply]

I've seen several books where it's done, but none comes immediately to mind. I'll see if I can find one. The Fourier transform is itself a Gaussian function. Michael Hardy (talk) 17:31, 19 December 2010 (UTC)[reply]

I'd guess it must be of the form a exp(–½b ξ^TCξ), where a and b are constants involving 2 and π that will depend on your chosen convention for the Fourier transform --Qwfp (talk) 11:26, 20 December 2010 (UTC)[reply]

Over the weekend I worked it out for symmetric non-positive-definite matrices C. And the guess above is right. Except that I relied on something that I needed to stop and remember requires proof. That is that the variance of the probability distribution whose density is proportional to

\exp \left({\frac {-x^{T}C^{-1}x}{2}}\right),

actually is the matrix C. Qwfp is right except that I used the probabilists' convention where the exponent in the definition of the Fourier transform has a "+". Details later maybe........ Michael Hardy (talk) 19:38, 20 December 2010 (UTC)[reply]

Can someone give a simpler explanation of ultraradicals to me?

I'm trying to understand quintics and I'm not understanding the ultraradical function (it's a function, right?). What sort of function is it? Is it differentiable? What's its y=f(x) look like when graphed? Thanks.--70.122.122.203 (talk) 21:39, 19 December 2010 (UTC)[reply]

It's the inverse of g(x)=-x⁵-x, so wouldn't the graph of y=f(x) be the graph of x=g(y)? The implicit function theorem implies that f will be differentiable except where g′(y)=0. So it's differentiable on the real line but there are branch points in the complex plane.--RDBury (talk) 03:58, 20 December 2010 (UTC)[reply]

December 20

Binomial distribution for small p, high right tails, mental calculation?

Suppose I'm interested in a quantity distributed as a binomial distribution with small p and large N, and I want to get an estimate of the probability that the quantity is, say, four standard deviations above the mean. According to binomial distribution, the skewness is (1-2p)/sqrt(np(1-p)), which will be positive, meaning fat right tails, if I have that right. So I probably can't just treat it as a normal variate with mean Np and variance Np. Is there any fast, mental-arithmetic-type way of estimating the probability of such an extreme value? --Trovatore (talk) 10:13, 20 December 2010 (UTC)[reply]

(P.S. for why I'm interested, see Talk:Congenital insensitivity to pain with anhidrosis#contradiction template.) --Trovatore (talk) 10:32, 20 December 2010 (UTC)[reply]

For small p and large n the appropriate approximation to the Binomial distribution Bin(n, p) is the Poisson distribution Poi(λ = np). See Binomial distribution#Poisson approximation. The cumulative distribution function of the Poisson appears a bit hard to evaluate using mental arithmetic, though you could investigate the asymptotics of the Incomplete gamma function. For your first case of interest (US), n = 3×10⁸ and p = 1/1.25×10⁸ giving np = 300/125 = 2.4. Stata and R then both give P(X ≥ 80) = 3.4×10^–90 for a Poisson with λ = 2.4. Your second case of interest (Japan) leads to arithmetic underflow. For the whole world n = 6×10⁹, np = 48 gives P(X ≥ 80) = 1.5×10^–5. --Qwfp (talk) 11:08, 20 December 2010 (UTC)[reply]

The probability that the quantity is four standard deviations above the mean is 1−e^−pNΣ[0≤i<pN+4√(pN)](pN)ⁱ/i!, where [] is an Iverson bracket. The finite sum can be computed for small values of pN, and otherwise the normal distribution approximation is excellent. Bo Jacoby (talk) 11:44, 20 December 2010 (UTC).[reply]

The Poisson distribution does seem appropriate to the context. Regardless of the skewness of the binomial distribution, for any non-negative valued random variable, Markov's inequality applies, and in this case that's enough to tell you that being that far above average is quite improbable. (Markov's inequality is what says no more that 1/n of the population can have more than n times the average income (if incomes are non-negative), etc.) Michael Hardy (talk) 19:33, 20 December 2010 (UTC)[reply]

Here's a handwaving version that gives approximately the right order of magnitude using only mental arithmetic: for a Poisson distribution, using of the third result under Incomplete gamma function#Asymptotic behavior and then Stirling's approximation:

\operatorname {P} (X\geq x)\sim \operatorname {P} (X=x)={\frac {e^{-\lambda }\lambda ^{x}}{x!}}\sim {\frac {e^{-\lambda }\lambda ^{x}}{e^{-x}x^{x}}}=e^{-\lambda }(e\lambda /x)^{x}

where

~

means something like "is asymptotically of approximately the order of magnitude of". With

\lambda =2.4

and

x=80

,

e^{-\lambda }(e\lambda /x)^{x}\approx 0.1\times 0.1^{80}=10^{-81}.

--Qwfp (talk) 21:23, 20 December 2010 (UTC)[reply]

Table of marks

Hi all,

I'm trying to figure out how the table of marks is calculated. For those of you unfarmiliar with it it is defined on the wikipedia page

http://en.wikipedia.org/wiki/Burnside_ring here(under the section marks), the important bit being:

For each pair of subgroups $H,K\leq G$ define

$m(K,H)=\left|[G/K]^{H}\right|=\#\left\{gK\in G/K\mid HgK=gK\right\}.$ Then finding the groups subgroups, and then putting them into conjugacy classes of subgroups, we get a string of representative subgroups $G_{1},\dots ,G_{N}$ where $G_{1}$ is the trivial group and $G_{N}$ is the whole group. Then we can make a N x N matrix where the i,j th entry is $m(G_{i},G_{j})$

So for $S_{3}$ ${\begin{array}{ccccc}S_{3}&1&Z_{2}&Z_{3}&S_{3}\\S_{3}/1&6&.&.&.\\S_{3}/Z_{2}&3&1&.&.\\S_{3}/Z_{3}&2&0&2&.\\S_{3}/S_{3}&1&1&1&1\end{array}}$

I can calculate the top left entry obviously, and the bottom row, but I literally have no idea how to calculate the others. If someone could talk me through how to calculate another more difficult entry for S3, that is a difficult $m(G_{i},G_{j})$ , it would be of great help.

Many thanks in advance for any help, and have a merry christmas everyone! —Preceding unsigned comment added by 86.139.252.252 (talk) 12:46, 20 December 2010 (UTC)[reply]

Euclid's Algorithm

Can someone give me some advice on how you would use Euclid's algorithm to find the solutions in x for $ax\equiv b$ (mod m)? I have thought of two possible approaches, both of which seem useless as I end up using Bezout's Theorem rather than Euclid. For the first, I simply say we must have ax-b=km for some integer k, then rearrange to have ax-km=b, define k'=-k to give ax+k'm=b and apply Bezout (after checking that hcf(a,m) divides b). The second approach I thought of is simply find some t such that $ta\equiv b$ (mod m), which will allow me to multiply both sides of the original congruence to give $x\equiv bt$ (mod m). To find t, I would need to apply Bezout to ta-b=pm for some integer p, so these two methods are essentially the same. I don't see Euclid in either of them. Any ideas? Thanks 92.11.32.186 (talk) 21:49, 20 December 2010 (UTC)[reply]

Bezout's identity is a corollary of the extended Euclidean algorithm. Aenar (talk) 00:07, 21 December 2010 (UTC)[reply]