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Thread: Fourier-serie

  1. #1
    Newbie SADANE ANTAR's Avatar
    Joined
    Dec 2009
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    4

    Fourier-serie

    Calculate :
    $\displaystyle \begin{array}{l}
    P = \frac{\pi }{8}\left( {\frac{{\sin (x)}}{{1^3 }} + \frac{{\sin (3x)}}{{3^3 }} + \frac{{\sin (5x)}}{{5^3 }} + .......} \right) \\
    Q = \frac{\pi }{6} - \left( {\frac{{\cos (2x)}}{{1^2 }} + \frac{{\cos (4x)}}{{2^2 }} + \frac{{\cos (6x)}}{{3^2 }} + .....} \right) \\
    \end{array}$
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  2. #2
    Senior Member
    Joined
    Mar 2009
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    378
    I think you are looking for the following trick:
    $\displaystyle \int_{0}^{2\pi} \! P \sin m x \, dx = \frac{\pi}{8} \int _{0}^{2\pi} \! \left(\sum_{n=1}^{+\infty} \frac{\sin nx}{n^3}\right) \sin m x \, dx$
    $\displaystyle P_m \int_{0}^{2\pi} \! \sin m x \, dx = \frac{\pi}{8} \int _{0}^{2\pi} \! \frac{\sin^2 mx}{m^3} \, dx$
    $\displaystyle P_m = \frac{\frac{\pi}{8} \int _{0}^{2\pi} \! \frac{\sin^2 mx}{m^3} \, dx}{\int_{0}^{2\pi} \! \sin m x \, dx}$
    $\displaystyle P = \sum_{m=1}^{\infty} P_m$.
    Then use a similar trick for $\displaystyle Q$.
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