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Using fourier series to sum 1/n^4

  1. Jun 12, 2010 #1
    [tex] \sum _{n=1}^{\infty } \frac{1}{n^4} [/tex]

    I can do 1/n^2 easily by using x^2 as a function but for this I try

    [tex] x^4=\frac{1}{2\pi }\int_{-\pi }^{\pi } x^4 \, dx+\sum _{n=1}^{\infty } \left(\frac{2}{\pi }\int _0^{\pi }x^4\text{Cos}[n x]dx\right)\text{Cos}[n x] [/tex]

    I arrive at:

    [tex] \pi ^4=\frac{\pi ^4}{5}+\sum _{n=1}^{\infty } \frac{8 \left(-6+n^2 \pi ^2\right)}{n^4 } [/tex]

    [tex] \frac{4\pi ^4}{5}=\sum _{n=1}^{\infty } \frac{8 \left(-6+n^2 \pi ^2\right)}{n^4 } [/tex]

    f(x) = x^4 does not seem helpful to sum 1/n^4, maybe I need a polynomial in x of degree 4?
     
  2. jcsd
  3. Jun 12, 2010 #2
    using [tex] x^4 - 2\pi^2x^2 [/tex]

    [tex] x^4 - 2\pi^2x^2 = \frac{1}{2\pi }\int_{-\pi }^{\pi } \left(x^4-2\pi ^2 x^2\right) \, dx+\sum _{n=1}^{\infty } \left(\int _0^{\pi }\left(x^4-2\pi ^2 x{}^{\wedge}2\right)\text{Cos}[n x]dx\right)\text{Cos}[n x][/tex]

    I end up with

    [tex] -x^4 = -\frac{7 \pi ^4}{15} -\sum _{n=1}^{\infty} \frac{48 }{n^4 } [/tex]

    [tex] \frac{8 \pi ^4}{720} = \sum _{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90} [/tex]
     
  4. Jun 13, 2010 #3
    You can use Fourier transform of x^2 to sum 1/n^4 too.

    Hint: compute the L^2 norm of the function x^2 in two ways, directly and using the Fourier coefficients.
     
  5. Feb 23, 2011 #4
    Sorry to drag up an old thread, but I'm working on the same problem and understand everything except where the x^4 - 2pi^2 * x^2 came from. Any help would be appreciated.
     
  6. Feb 24, 2011 #5

    dextercioby

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    The second post of the thread contains a trick which you don't need. The first post of the thread already contains the solution.
     
  7. Feb 24, 2011 #6
    Yes the second part isn't useful for fourier but is if you need to sum an infinite series. The function which you use needs to give you the series you want to sum and it is found with trial and error sometimes.
     
  8. Sep 26, 2011 #7
    one more solution is possible for the sum of 1/n^4

    . first calculate the fourier series of x^2...

    then integrate it ...3 times..

    on first and third time the coefficient of integration will be zero you can put
    x = 0;

    after that after third time integration you will be able to find
    the integration coefficient of 2nd time integration ...

    that will calculate to -7/180

    now put value x= pi in the second time integration ...


    you will get the desired result....
     
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