## Computing zeta(k), for k odd, using Fourier coefficients

I'm not really sure what topics exactly this falls under, so I apologize if I've misclassified this question.

There is a neat way of computing $\sum_{n=1}^{\infty}\frac{1}{n^{2}}$ using Fourier analysis: Compute the Fourier series of t^2 extended 2π-periodically, which turns out to be

By Fejer's theorem (I think), the Fourier series around π converges, so we get an equation that can be solved for the zeta(2).

I think a similar approach can be taken for zeta(2k) by taking t^{2k} extended 2π-periodically, but all my attempts to do something like this for odd integers fail.

On the other hand, since {1/n^k} for k odd is in l^2, there should be an L^2 function that has the sequence as its Fourier coefficients. Can one be explicitly constructed? What if we allow the entries in the sequence to alternate, or let finitely many of them deviate from 1/n^k?

Basically, I want to find an L^2, 2π-periodic function whose Fourier coefficients would give a relatively straightforward computation of zeta(k) when k is odd.

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Are you aware that there is no relatively straightforward computation of zeta(k) when k is odd, at least in terms of other constants? – Qiaochu Yuan Nov 6 2009 at 4:19
You should also clarify what you mean by "explicitly constructed." The function with Fourier coefficients 1/n^k is perfectly well-defined as is. – Qiaochu Yuan Nov 6 2009 at 4:33

Well, I wish you the best. But I don't think you are going to succeed for two reasons:

(1) Proving almost anything about $\zeta(2k+1)$ is hard.

(2) There are nice formulas for $\sum_{n=1}^{\infty} \cos (n \theta)/n^{2k}$ and for $\sum_{n=1}^{\infty} \sin (n \theta)/n^{2k+1}$. There are not particularly nice formulas for $\sin$ with even powers or $\cos$ with odd. One way to think about this is that the imaginary part of $\log (1-e^{i \theta})$ has a simple formula but the real part does not.

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 I think you can compute zeta(2) using the series 1 - 1/3^2 + 1/5^2 + ..., which you can sum using (2), but doing the same for zeta(3) doesn't work. – Akhil Mathew Nov 6 2009 at 12:25

Qiaochu, Parseval does apply, it just doesn't yield anything elementary. :) Still, it is easy to deduce (for example) the identity

but its not clear what good such an identity is, relative to the original problem.

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The difficulty becomes apparent if you play around a little bit: the function whose nth Fourier coefficient is n-3 is a twice-iterated indefinite integral of

which will be some highly mysterious transcendental function (a "trilogarithm") whose values are just as obscure as those of .

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 Actually, I regret using the word "obscure": the study of polylogarithms and their special values is a venerable field which is enjoying a current vogue and turns out to be related to some quite high-brow things. – David Hansen Nov 6 2009 at 17:31

A point that hasn't been made yet is that one reason the sums for k even are easy to evaluate is that they occur, not only as values of periodic functions, but as $L^2$ norms of those functions. One can then apply Parseval's identity and evaluate the sum of the squares of 1/n^k as a definite integral, as is done in the Wikipedia article. Parseval's identity does not, on the other hand, apply to the odd values.

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This is very open, from what I know. I think there was a recent (last 15 years) result where it was proved that zeta(3) is transcendental (or maybe just irrational?) but my understanding is that virtually nothing is known about odd zeta values, and someone needs to do something new in order to get any useful information about them.

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Apery proved zeta(3) was irrational in 1979: 30 years ago. – Kevin Buzzard Nov 6 2009 at 12:39
Ahh, then I am misremembering what someone told me about how long ago anything was proved, as some googling reveals that it's open if zeta(3) is transcendental. The point of the answer: that we know virtually nothing about zeta(2n+1) stands, though. – Charles Siegel Nov 6 2009 at 13:20
There was a development in the last 15 years: in 2000, Tanguy Rivoal showed that there are that infinitely many odd irrational zeta values. More precisely, he showed that the dimension of the $\mathbb Q$ vector space spanned by the first $n$ odd zeta values has dimension at least $log(n)$. Applied to the even zeta values, this shows that $\pi$ is transcendental, but the odd zeta values are supposed to be algebraically independent. – Ben Wieland Jul 5 2010 at 23:29

What you easily get for any integer $k \ge 2$ is $\sum 1/n^k$ summed over all nonzero integers $n$, both positive and negative. When $k$ is even, we can use this to evaluate $\zeta(k)$. When $k$ is odd, we get the (correct, but uninteresting) result $0$.

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