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Fourier Series

  1. Nov 10, 2005 #1
    Find the FOurier Series in terms of [itex] \phi_{n} = \sin(nx) [/itex] of the step function

    f(x) = 0 for [itex] 0 \leq x \leq \frac{1}{2} \pi [/tex]
    f(x) =1 for [itex] \frac{1}{2} \pi < x \leq \pi [/itex]

    now i have no problem finding the series for each branch. But how would i combine them?

    for the 0 to 1/2 pi
    [tex] \frac{4}{\pi} \sum_{n=1}^{\infty} \sin nx [/tex]
    for the 1/2 pi to pi
    [tex] \frac{-4}{\pi} \sum_{n=1}^{\infty} \frac{1}{n} \left((-1)^n + \cos(\frac{n \pi}{2}) \right) [/tex]

    please help me on combining the two!

    Thank you for your help
  2. jcsd
  3. Nov 11, 2005 #2


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    Staff Emeritus
    Science Advisor

    Strictly speaking, you are not finding the Fourier series for two different functions, you are finding the cn in [itex]\Sum c_n sin(nx)[/itex] by integrating a single function.
    And, I might point out, you have NOT found the Fourier series on x between 0 and [itex]\frac{\pi}{2}[/itex]. f(x)= 0 there so the Fourier series is just 0.
    [tex]c_n= \frac{2}{\pi}\int_0^\pi f(x)sin(nx)dx= \frac{2}{/pi}\left(\int_0^{\frac{\pi}{2}}0 sin(nx)dx+ \int_{\frac{\pi}{2}}^\pi 1 sin(nx)dx\right)[/tex]

    That is just
    [tex]c_n= \frac{2}{\pi}\int_{\frac{\pi}{2}}^\pi sin(nx)dx[/tex]
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