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Fourier Series

  1. Jan 30, 2010 #1
    1. The problem statement, all variables and given/known data

    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]


    3. 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.
     
  2. jcsd
  3. Jan 30, 2010 #2
    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.
     
  4. Jan 31, 2010 #3
    Anybody? I still can't solve it. And the even extension is OK.
     
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