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Fourier Series and the Riemann-Zeta Function

  1. Apr 9, 2007 #1
    1. The problem statement, all variables and given/known data
    Use the Fourier series technique to show that the following series sum to the quantities shown:
    [tex]\sum_{n=1}^\infty \frac{1}{n^4}=\frac{\pi^4}{90}[/tex]
    [tex]1+\frac{1}{3^2}+\frac{1}{5^2}+...=\frac{\pi^2}{8}[/tex]

    2. Relevant equations
    I know the Riemann-Zeta function is
    [tex]\zeta (m)=\sum_{n=1}^\infty \frac{1}{n^m}[/tex]

    3. The attempt at a solution
    For the first series, I know that it's the evaluation of [itex]\zeta (4)[/itex]. The problem is getting the derivation down. I've found an example of it being solved and proven at [itex]\zeta (2)[/itex] here: http://planetmath.org/?op=getobj&from=objects&name=ValueOfTheRiemannZetaFunctionAtS2 . The problem is, I don't really understand what to define as the initial function [itex]f(x)[/itex].

    For the second series, I think I've found it to be:
    [tex]\frac{1}{(2n-1)^2}[/tex]
    After that, I don't know how to go about solving it using the Fourier method. Any help would be greatly appreciated, thanks!
     
  2. jcsd
  3. Apr 9, 2007 #2
    By "Fourier series technique", do you mean expand the 1/n4 or the 1/(2n - 1)2 into its Fourier series?
     
  4. Sep 25, 2008 #3
    A general method appears in the paper 'Recursive Formulas for [itex]\zeta(2k)[/tex] and [itex]L(2k-1)[/tex]' by Xuming Chen.

    Generally, try [itex]f(x) = x^{2k}[/tex] for your function (an even one on x here) in a Fourier expansion to get [itex]\zeta(2k)[/tex], and consider the derivative of the expansion to get [itex]L(2k-1)[/tex] (the L function here is the analogue of the Riemann zeta on odd numbers only).
     
  5. Dec 20, 2008 #4
    but still you ll not express

    [itex]\zeta(2k+1)[/itex]
     
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