- #1
libelec
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- 0
Homework Statement
Using the Fourier cosine series for [tex]\[f(x) = \left\{ \begin{array}{l}
1,x = 0 \\
10,x = \pi \\
x,x \in (0,2\pi ) - \left\{ {0,\pi } \right\} \\
\end{array} \right.\][/tex], find a series that converges to [tex]\[\int\limits_0^{2\pi } {{x^2}dx} \][/tex]
The Attempt at a Solution
For the Fourier cosine series, I need the even extention of f(x), that is, [tex]\[f(x) = \left\{ \begin{array}{l}
- x,x \in ( - 2\pi ,0) - \left\{ {0, - \pi } \right\} \\
10,x = - \pi \\
1,x = 0 \\
10,x = \pi \\
x,x \in (0,2\pi ) - \left\{ {0,\pi } \right\} \\
\end{array} \right.\][/tex]. Now, [tex]\[\int\limits_0^{2\pi } {{x^2}dx} \][/tex] = [tex]\[2\left\| {f(x)} \right\|_2^2\][/tex], 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: [tex]\[\frac{8}{3}{\pi ^3} = \frac{{{\pi ^3}}}{4} + \frac{8}{\pi }\sum\limits_{n = 1}^\infty {\frac{1}{{{{(2n - 1)}^4}}}} \][/tex]
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.