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$f(x) = \pi^2 - x ^2$ on $|x|<\pi$ and $f(x+2\pi)=f(x)$.

I did find its Fourier expansion

$f(x)=\frac{2}{3}\pi^2$+$\sum_{n \geq 1}\frac{4}{n^2}(-1)^{n+1}\cos x$

And by putting $x=\pi$, I got the zeta of $2$ , $\zeta(2) = \sum_{n \geq 1}\frac{1}{n^2} = \frac{\pi^2}{6}$.

Similarly, if i try do this process about $f(x) = \pi^4-x^4$ then I could get the value $\zeta(4)=\frac{\pi^4}{90}$ if so, why can't I get the value of $\zeta(3)$ with $f(x) = \pi^3 - x ^3$ ? I wonder the principle that I can't do in general, why can't we get the value the zeta of odd $n$?

user26857
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J.U.math
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1 Answers1

8

Of course I don't have a proof that $\zeta(3)$ is a completely new constant, but I think this might be interesting :

  • Let $\displaystyle f(z) = \frac{\pi^2}{\sin^2(\pi z)}$ and $\displaystyle g(z) = \frac{d^2}{dz^2} \log(\Gamma(z))$ $= \psi'(z)$.

    Using Liouville's theorem (and showing $\Gamma(z)$ has no zeros) you get that $$f(z) = \sum_{n=-\infty}^\infty \frac{1}{(z+ n)^2}, \qquad\quad g(z) = \sum_{n=0}^\infty \frac{1}{(z+n)^2}$$

  • Differentiating : $\displaystyle f^{(k)}(z) = (-1)^k \frac{(2+k)!}{2}\sum_{n=-\infty}^\infty \frac{1}{(z+ n)^{2+k}}$ so that $$f^{(2k+1)}(1/2) = 0, \qquad f^{(2k)}(1/2) = (2+2k)!(2^{2k+2}-1)\zeta(2k+2)$$

$$g^{(k)}(1/2) = \frac{(2+k)!}{2}(2^{k+2}-1)\zeta(k+2)$$

  • Finally, evaluating $\zeta(2k)$ is as easy as evaluating the derivatives of $\sin(\pi z)$ at $z=1/2$,

    while evaluating $\zeta(2k+1)$ is as hard as evaluating the derivatives of $\Gamma(z)$ at $z=1/2$.

reuns
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