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Estimating the sum of reciprocal powers using a given fourier series

  1. Oct 15, 2008 #1
    1. The problem statement, all variables and given/known data

    Let f(x) be defined by the following Fourier series for [tex] \left|x\right|[/tex]:

    [tex] f(x) = \frac{\pi}{2} - \frac{4}{\pi}\sum_{1,3,...}\frac{cos\left(nx\right)}{n^{2}}[/tex]

    Show that

    [tex] \sum_{1,3,...}\frac{1}{n^{2}} = \frac{\pi^{2}}{8}[/tex]

    and

    [tex] \sum_{1,2,3,...}\frac{1}{n^{2}} = \frac{\pi^{2}}{6}[/tex]

    3. The attempt at a solution

    I was able to find the first sum by letting x = 0. I don't now how to approach the second part since the sum consists of the odd and even n integers, but the Fourier series is only comprised of the odd integers.
     
  2. jcsd
  3. Oct 15, 2008 #2
    I think I figured it out, or at least obtained a really close approximation.

    Multiplying 1/4 to both sides of

    [tex] \sum_{1,3,...}\frac{1}{n^{2}} = \frac{\pi^{2}}{8} [/tex]

    will yield the sum for the terms 2, 6, 10, ...

    [tex] \sum_{2,6,10,...}\frac{1}{n^{2}} = \frac{\pi^{2}}{32} [/tex]

    Now multiplying this sum by 1/4 gives the terms 4, 8, 12,...

    [tex] \sum_{4,8,12,...}\frac{1}{n^{2}} = \frac{\pi^{2}}{128} [/tex]

    Combining all of the sums together gives a close approximation:

    [tex] \sum_{n}\frac{1}{n^{2}} = \frac{21\pi^{2}}{128} \approx \frac{\pi^{2}}{6} [/tex]
     
  4. Oct 15, 2008 #3

    Dick

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    If S is the sum over all n (even or odd) then (1/4)S is the sum over all evens. S-(1/4)S is then the sum over all odds which is pi^2/8. Solve for S.
     
  5. Oct 17, 2008 #4
    EDIT: I see now. Duh! Thanks.
     
    Last edited: Oct 17, 2008
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