# Homework Help: Fourier Series

1. Jan 30, 2010

### libelec

1. The problem statement, all variables and given/known data

Using the Fourier cosine series for $$$f(x) = \left\{ \begin{array}{l} 1,x = 0 \\ 10,x = \pi \\ x,x \in (0,2\pi ) - \left\{ {0,\pi } \right\} \\ \end{array} \right.$$$, find a series that converges to $$$\int\limits_0^{2\pi } {{x^2}dx}$$$

3. The attempt at a solution

For the Fourier cosine series, I need the even extention of f(x), that is, $$$f(x) = \left\{ \begin{array}{l} - x,x \in ( - 2\pi ,0) - \left\{ {0, - \pi } \right\} \\ 10,x = - \pi \\ 1,x = 0 \\ 10,x = \pi \\ x,x \in (0,2\pi ) - \left\{ {0,\pi } \right\} \\ \end{array} \right.$$$. Now, $$$\int\limits_0^{2\pi } {{x^2}dx}$$$ = $$$2\left\| {f(x)} \right\|_2^2$$$, so I can use Parseval's equality, right?

But if that's correct, I'm unable to find a series that converges to that defined integral, since it has constant terms: $$$\frac{8}{3}{\pi ^3} = \frac{{{\pi ^3}}}{4} + \frac{8}{\pi }\sum\limits_{n = 1}^\infty {\frac{1}{{{{(2n - 1)}^4}}}}$$$

Evidently, I'm doing something wrong, but I don't know what. Is it the cosine series? Is it the convergence of the series? Or is it that I can't use Parseval's?

Thanks.

2. Jan 30, 2010

### libelec

OK, I just realized that the even extention is wrong. But that would solve anything if I still have the a0 term in the series.

3. Jan 31, 2010

### libelec

Anybody? I still can't solve it. And the even extension is OK.