1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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]


    [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


    User Avatar
    Science Advisor
    Homework Helper

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook