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Homework Help: Parseval's Theorem

  1. Apr 23, 2014 #1
    1. The problem statement, all variables and given/known data
    I'm given the following function
    [tex]f(x) = \begin{cases} x &-2<x<2\\ f(x+4) &\mbox{otherwise} \end{cases} [/tex]

    And I'm asked to find the Fourier sine series. Then I'm supposed to use Parseval's theorem to obtain a certain sum.

    2. Relevant equations
    Since I have a sine Fourier series, Parseval's Theorem for this says
    [tex] \frac{1}{b-a} \int_a^b |f(x)|^2 dx = \sum_{n=1}^\infty b_n^2 [/tex]

    3. The attempt at a solution
    So I worked through and got the Fourier sine series for this function which is
    [tex] f(x) = \frac{4}{\pi} \left[ \sin \left( \frac{\pi x}{2} \right) - \frac{1}{2} \sin \left( \frac{2 \pi x}{2} \right) + \frac{1}{3} \sin \left( \frac{3 \pi x}{2} \right) - \frac{1}{4} \sin \left( \frac{4 \pi x}{2} \right) + \ldots \right] [/tex]

    Now I apply Parseval's Theorem:
    [tex] \frac{1}{b-a} \int_a^b |f(x)|^2 dx = \sum_{n=1}^\infty b_n^2 \\
    \frac{1}{2-(-2)} \int_{-2}^2 x^2 dx = \sum_{n=1}^\infty b_n^2 \\

    On the left hand side:
    [tex]\frac{1}{4} \int _{-2}^{2} x^2 dx = \frac{16}{12} [/tex]

    On the right hand side:
    [tex] \left(\frac{4}{\pi} \right)^2 \sum_{n=1}^\infty \frac{1}{n^2} = \frac{16}{\pi ^2} \sum_{n=1}^\infty \frac{1}{n^2} [/tex]

    Now equating the left and right hand sides:
    [tex] \frac{16}{12} = \frac{16}{\pi ^2} \sum_{n=1}^\infty \frac{1}{n^2}\\ [/tex]
    [tex]\frac{\pi ^2}{12} \sum_{n=1}^\infty \frac{1}{n^2} [/tex]

    The answer I'm supposed to be getting is
    [tex]\frac{\pi ^2}{6} \sum_{n=1}^\infty \frac{1}{n^2} [/tex]

    So I'm off by a factor of a half somewhere but I can't figure it out. Some help would be appreciated. Thank you.
  2. jcsd
  3. Apr 24, 2014 #2


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    I don't think you have stated Parseval's theorem correctly for the sine series.

    Parseval's theorem for the complex Fourier series is as follows:
    $$\sum_{n=-\infty}^{\infty}c_n e^{2\pi i n x / P}$$
    in which case
    $$\sum_{n=-\infty}^{\infty}|c_n|^2 = \frac{1}{b-a}\int_{a}^{b} |f(x)|^2 dx$$
    If you have a real Fourier series:
    $$\frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos(2\pi n x / P) + \sum_{n=1}^{\infty} b_n \sin(2\pi n x / P)$$
    then how does ##c_n## relate to ##a_n## and ##b_n##?
    Last edited: Apr 24, 2014
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