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Today I learned a beautiful proof of the identity $\zeta(2) = \frac{\pi^2}{6}$.

The idea is strikingly simple: we show that $\pi^2(x^2 - x + \frac{1}{6}) = \sum_{k=1}^\infty \frac{\cos(2\pi k x)}{k^2}$ everywhere by calculating the Fourier coefficients of the left-hand-side, and then using a lemma from Fourier analysis that says that if the Fourier coefficients are absolutely summable, then the Fourier series converges uniformly and absolutely.

We then simply evaluate at $x = 0$ to get the desired result!

I was left with a nagging question though: If one wanted to calculate $\zeta(2)$, how would one know to look at the polynomial $\pi^2(x^2 - x + \frac{1}{6})$?

It seems somewhat natural to look at series like $\sum_{k=1}^\infty \frac{\cos(2\pi k x)}{k^2}$, since this is a way of turning a discrete series into something smooth enough to use analytic tools (although you might try several series before finding this particular one). But once you have guessed a series that might be helpful, is there any easy way to find candidate functions for which your series might be the Fourier series?

whpowell96
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stillconfused
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    I'd assume it comes from the well-known approach to the Basel problem of finding the Fourier Series of $x^2$, which itself likely arose from clever manipulation of that series. By adding lower order terms in a precise manner, one could arrive at the LHS you described. – Leonidas Lanier Aug 30 '24 at 17:32
  • Computing $\zeta(4)$ using Fourier series was an exam question in one of my differential equations classes and that is exactly how it is done. You start with $x^4$ then adjust with lower order terms until the numerator of the coefficients is 1 – whpowell96 Aug 30 '24 at 17:41
  • I edited your title to better reflect the nature of your question for other users – whpowell96 Aug 30 '24 at 17:46
  • This answer may help (see the examples if you don't enjoy distributions) – Raymond Manzoni Aug 31 '24 at 08:48
  • @LeonidasLanier - your comment seems to be the closest thing to an answer to the question that I asked so far. If you care to expand into an answer, with explanations of the manipulations you mention and how to find the right lower order terms, I would accept! – stillconfused Sep 02 '24 at 19:36
  • @stillconfused I'll work on it and get back to you. – Leonidas Lanier Sep 05 '24 at 12:01

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the polynomial is not unique. I remember seeing a similar proof by considering the fourier coefficents of $f(t) = t$ (maybe it is less clear why the fourier series is pointwise converging in $(-\pi, \pi)$, but it is more natural to look at that function (we can even say something stringer; the fourier series is uniformly converging to $f$ in $[-r, r]$ for all $r<\pi$)). If one wanted to compute $\zeta(2)$, he would probably not look at $\pi^2\cdot (x^2-x-\frac{1}{6})$, but will probably look at the fourier series of 1, $x$, $x^2$, since those are the simplest functions you can think of. The key observation here is that inner product is bilinear; once you know that the fourier of those polynomials, seeing that the fourier coefficients of $x^2$ are similar to $\frac{1}{k^2}$, you can easily 0bserve (using some theorems of fourier analysis) that you can get the wnated equality

Koren Parkhov
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