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Pi appears a LOT in trigonometry, but only because of its 'circle-significance'. Does pi ever matter in things not concerned with circles? Is its only claim to fame the fact that its irrational and an important ratio?

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I'd rather this be a comment instead, so: $\pi$ turns up in the expression for the so-called "probability integral" (a.k.a. the "error function") among other things. How circles relate to this is a bit of a long-winded explanation though. –  J. M. Aug 26 '10 at 0:11
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Also, let's get one thing straight here: circles are eerily important. You will never stop running into circles in mathematics. –  Qiaochu Yuan Aug 26 '10 at 0:26
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(For example, although the Fourier transform is "concerned with circles" (functions on the circle being the same thing as periodic functions) it penetrates into the deepest parts of modern mathematics. Many appearances of pi are because of a Fourier transform lurking somewhere in the background. You might also want to read this MO thread where I asked a similar question: mathoverflow.net/questions/18180/…) –  Qiaochu Yuan Aug 26 '10 at 0:52
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Fundamental source of $\pi$ is circle nothing else. It may be difficult to find it but it is always there. –  Pratik Deoghare Aug 26 '10 at 12:43

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It is difficult to know if a circle is not lurking somewhere, whenever there is $\pi$, but the values of the Riemann zeta function at the positive even integers have a lot to do with powers of $\pi$: see here for the values.

For instance, you can prove that the probability that two "randomly chosen" positive integers are coprime is $\frac{1}{\zeta(2)} = \frac{6}{\pi^2}$.

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You had my upvote at "it is difficult to know if a circle is not lurking somewhere..." –  J. M. Aug 26 '10 at 0:22
    
Maybe that has something to do with angles and the 2D lattice. –  asmeurer Dec 20 '12 at 20:03
    
@asmeurer: Actually it has to do with the fact that $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$. –  ShreevatsaR Jun 9 '13 at 16:29
    
@ShreevatsaR you're just restating what $\zeta(2)$ is. That doesn't explain where the $\pi$ comes from, fundamentally. –  asmeurer Jun 9 '13 at 19:18
    
@asmeurer: True, but I'm saying it has nothing to do with "angles and the 2D lattice". Nothing I know of, anyway. The proof that $\zeta(2)$ has that value (see the linked article) doesn't use those, and the proof that the probability of two "randomly chosen" prime numbers is $\frac{1}{\zeta(2)}$ is straightforward. –  ShreevatsaR Jun 10 '13 at 2:01

$\pi$ appears in Stirling's approximation, which is not obviously related to circles. This means that $\pi$ appears in asymptotics related to binomial coefficients, such as

$$\displaystyle {2n \choose n} \approx \frac{4^n}{\sqrt{\pi n}}.$$

In other words, the probability of flipping exactly $n$ heads and $n$ tails after flipping a coin $2n$ times is about $\frac{1}{\sqrt{\pi n}}$. This asymptotic also suggests that on average you should flip between $n + \sqrt{\pi n}$ and $n - \sqrt{\pi n}$ heads.

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This is closely related to J. Mangaldan's comment about the probability integral. Somehow I think it all ties back to the fact that e^{-x^2} is its own Fourier transform. –  Qiaochu Yuan Aug 26 '10 at 0:31
    
Yes. Yes it does. :) –  J. M. Aug 26 '10 at 0:47
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I guess that comment is worth explaining: the relationship is that the constant in Stirling's approximation can be computed from the central limit theorem. This is explained at terrytao.wordpress.com/2010/01/02/… . –  Qiaochu Yuan Jan 14 '11 at 0:53
    
@QiaochuYuan You might be interested in Kunth's "Why Pi?" Lecture. He shows how this is related to cirlces! –  Pedro Tamaroff Feb 27 '12 at 5:19

Because of the formula $e^{i\pi}+1=0$ you will find $\pi$ appearing in lots of places where it's not clear there is a circle, e.g in the normal distribution formulae.

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Yes, the ratio $\pi$ of a circle's circumference to its diameter shows up in many, many places where one might not expect it!

One partial explanation (similar in spirit to "circles lurk everywhere") is that the equation for a circle is a $quadratic$ (eg. $x^2+y^2 = r^2$.) After nice linear functions, the next most commonly used functions are quadratic functions and everywhere one runs into a quadratic function, a trig substitution (e.g. $x = r \cos \theta; y=r\sin \theta$) may be useful, turning the quadratic function into something involving $\pi.$ This explains the antiderivative $\int \frac{1}{1+x^2} dx$ involving $\pi$, the sum of reciprocals of squares $\sum^\infty\frac{1}{k^2}$ involving $\pi$ and the area under the Gaussian distribution involving $\pi$. And so on....

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How does it explain the sums of reciprocals of squares involving pi? –  George Lowther Jan 13 '11 at 23:02
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@George: there are a few elementary proofs of sum 1/k^2 = pi^2/6 where pi creeps in for reasons at least analogous to a trig substitution: math.stackexchange.com/questions/8337/… –  Qiaochu Yuan Jan 14 '11 at 0:52
    
@Qiaochu: Most of the proofs I know apply equally well to evaluating $\sum 1/n^d$ (for d even) and even $\sum (-1)^d/n^d$ (for d odd), which also involve $\pi$. So, the fact that the terms are squares doesn't seem particularly significant to the appearance of $\pi$. –  George Lowther Jan 14 '11 at 1:41
    
I'll have a look through the alternative proofs in that link though. –  George Lowther Jan 14 '11 at 1:42
    
(I meant $\sum(-1)^d/(2n+1)^d$ above). I always thought of these sums involving $\pi$ for similar reasons, and not just the $d=2$ case in isolation. –  George Lowther Jan 14 '11 at 1:56

protected by t.b. Jul 20 '11 at 8:48

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