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Square of Fourier series

  1. Nov 26, 2014 #1
    1. The problem statement, all variables and given/known data
    Evaluate following series:
    [tex]\sum_{n=1}^\infty \frac{1}{(4n^2-9)^2} [/tex]
    by finding the Fourier series for the [itex]2\pi[/itex]-periodic function
    [tex]f(x) =
    \begin{cases}
    sin(3x/2) & 0<x<\pi \\
    0 & otherwise
    \end{cases}
    [/tex]
    2. Relevant equations
    [tex]a_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x)cos(nx)dx = -\frac{6}{\pi(4n^2-9)} [/tex]
    [tex]b_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x)sin(nx)dx = \frac{4ncos(\pi n)}{\pi(4n^2-9)} [/tex]
    [tex]f(x) = \frac{1}{2}a_0 + \sum_{n=1}^\infty \left(a_ncos(nx)+b_nsin(nx)\right) \\ = \frac{1}{2}\frac{2}{3\pi} + \frac{1}{\pi}\sum_{n=1}^\infty \left( \frac{4ncos(\pi n)sin(nx)-6cos(nx)}{4n^2-9}\right)[/tex]

    3. The attempt at a solution
    I have found the Fourier series and evaluated 1/(4n^2-9) as it was the first part of this exercise. However, I am not sure how to evaluate it for the square. I figured I just needed to square the answer but it turns out it's that simple .

    I found Parseval's formula in my book
    [tex] \frac{1}{\pi}\int_T |f(x)|^2dx = \frac{1}{2}|a_0|^2 + \sum_{n=1}^\infty (|a_n|^2+|b_n|^2) [/tex].
    which I tried using, but the [itex]b_n[/itex] term gives me an ugly expression in the numerator.
     
    Last edited: Nov 26, 2014
  2. jcsd
  3. Nov 26, 2014 #2

    pasmith

    User Avatar
    Homework Helper

    [itex]b_n[/itex] should not be a function of [itex]x[/itex]. Take another look at it, and remember that [itex]\cos(\pm n\pi) = (-1)^n[/itex].
     
  4. Nov 26, 2014 #3
    Oh yes, my mistake. I meant cos(pi n) not cos(xn).
     
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