Parseval's Theorem and Fourier series

[Niles]

Homework Statement

Hi all.

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

http://img143.imageshack.us/img143/744/picture2ao8.png

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?

 

[cellotim]

What is the series?

 

[Niles]

It is given by:

$$f(x) = 1 + \sum\limits_{n = 1}^\infty {\left( {\frac{{\cos nx}}{{3^n }} + \frac{{\sin nx}}{n}} \right)}$$

 

[cellotim]

Do you mean $$f(x) = 1 + \sum_{n=1}^\infty \frac{\cos(\pi nx/L)}{3^n} + \frac{\sin(\pi nx/L)}{n}$$?

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 $$a_0,a_n,b_n,n=1,\dots,\infty$$
$$||f(x)||^2 = L|a_0|^2 + L/2\sum_{n=1}^\infty |a_n|^2 + |b_n|^2$$.

 

[Niles]

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

http://planetmath.org/encyclopedia/LyapunovEquation.html

- under "Parseval's Theorem".

 

[cellotim]

The Riesz-Fischer theorem should give you the proof.

 

[Niles]

Hmm, ok.. I will try and look into it. Thanks

 

[cellotim]

Sorry, I didn't have much time to post last time. The Riesz-Fischer theorem essentially says (among other things) that, given $$a_0,a_n,b_n,n=1,\dots,\infty$$, if $$|a_0|^2 + \sum_{i=1}^\infty |a_n|^2 + |b_n|^2 < \infty$$, then there must exist a function $$f$$ such that $$a_0,a_n,b_n$$ are its Fourier coefficients, and this is its L2 norm. Since you have a function with those coefficients already, square integrability should follow.