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Parseval's Theorem and Fourier series

  1. Oct 3, 2008 #1
    1. The problem statement, all variables and given/known data
    Hi all.

    Please take a look at the lowest equation in this picture:


    This is Parselvals Identity.

    Let us say that I am given a Fourier series of f(x), and I want to calculate the integral of f(x)^2 from -L to L. In order to do this, I use Parsevals Identity. But the requirement for me to use Parsevals Identity is that the series is well-defined and square integrable. How do I show this?
  2. jcsd
  3. Oct 3, 2008 #2
    What is the series?
  4. Oct 3, 2008 #3
    It is given by:

    f(x) = 1 + \sum\limits_{n = 1}^\infty {\left( {\frac{{\cos nx}}{{3^n }} + \frac{{\sin nx}}{n}} \right)}
  5. Oct 3, 2008 #4
    Do you mean [tex]f(x) = 1 + \sum_{n=1}^\infty \frac{\cos(\pi nx/L)}{3^n} + \frac{\sin(\pi nx/L)}{n}[/tex]?

    In any case, it doesn't matter that much. The way to solve this is to recall that the set of Fourier basis functions are orthogonal; doing out the integral multiplying all terms, it's not hard to show that, for a Fourier series with coefficients [tex]a_0,a_n,b_n,n=1,\dots,\infty[/tex]
    [tex] ||f(x)||^2 = L|a_0|^2 + L/2\sum_{n=1}^\infty |a_n|^2 + |b_n|^2 [/tex].
  6. Oct 3, 2008 #5
    In my first post, the L's are supposed to be switched with pi's.

    How can I show that f(x) is a well-defnied, square integrable function on [-pi; pi] so that I am allowed to use Parsevals Identity?

    EDIT: See here


    - under "Parseval's Theorem".
    Last edited: Oct 3, 2008
  7. Oct 3, 2008 #6
    The Riesz-Fischer theorem should give you the proof.
  8. Oct 3, 2008 #7
    Hmm, ok.. I will try and look into it. Thanks
  9. Oct 3, 2008 #8
    Sorry, I didn't have much time to post last time. The Riesz-Fischer theorem essentially says (among other things) that, given [tex]a_0,a_n,b_n,n=1,\dots,\infty[/tex], if [tex]|a_0|^2 + \sum_{i=1}^\infty |a_n|^2 + |b_n|^2 < \infty[/tex], then there must exist a function [tex]f[/tex] such that [tex]a_0,a_n,b_n[/tex] are its Fourier coefficients, and this is its L2 norm. Since you have a function with those coefficients already, square integrability should follow.
    Last edited: Oct 3, 2008
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