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Fourier Series - proving a sum

  1. Jun 16, 2010 #1
    1. The problem statement, all variables and given/known data
    Let [itex]f(x)=x[/itex] on [itex] [-\pi,\pi) [/itex] and peridically extended. Compute the fourier series and hence show:

    [itex] \sum_{n \geq 1,nodd} \frac{1}{n^2} = \frac{\pi^2}{8} [/itex] and [itex] \sum_{n \geq 1} \frac{1}{n^2} = \frac{\pi^2}{6} [/itex]

    2. Relevant equations

    Parseval's equality

    3. The attempt at a solution
    I computed the fourier series to be [itex] -\sum_{n=1,nodd} \frac{4}{n^2 \pi} e^{inx}+\frac{\pi}{2} [/itex] (even terms [itex]\hat{f}(n)=0 [/itex]) and proved the first sum (letting x=0).
    How would I compute the second part? how do i get the whole sum from this? I tried to slipt the sum into even and odd parts, but i dont know how to compute the even sum when I dont have any terms for the even sum! thanks
  2. jcsd
  3. Jun 16, 2010 #2


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    You can't. If, in fact, all terms with even indices are 0, then "[itex]\sum_{n\ge 1, n odd}[/itex]" and [itex]\sum_{n\ge 1}[/itex] must be the same!
  4. Jun 16, 2010 #3
    Ok, so how would I use what I have to prove the second sum? That's assuming that fourier series is right - I think it is, because the first sum works out. On a related note, how would I go about proving that [itex] \sum_{n=0}^{\infty} \frac{1}{(2n+1)^4} = \frac{\pi^4}{96}[/itex]? I have started by saying that this is equivalent to the sum [itex] \sum_{n \geq 1, n odd} \frac{1}{n^4} [/itex]. I use a similar method as before and consider the function [itex] f(x) = |x^3| [/itex] on [itex] [-\pi, \pi) [/itex] but I dont seem to get the sum I require in the end...I get the sum of even numbers instead!
  5. Jun 16, 2010 #4


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    I don't get the same Fourier series for f(x)=x that you do.
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