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Fourier series (I need help here)

  1. Apr 11, 2006 #1
    Here is the question:
    At [itex] x= \frac{\pi}{2} [/itex] the square wave equals 1. From the Fourier series at this point find the alternating sum that equals [itex] \pi [/itex].
    [tex] \pi = 4(1 - \frac{1}{3}+\frac{1}{5}-\frac{1}{7} + \ldots [/tex]

    I do not understand what the question is asking. I'm not knowledgeable enough with Fourier series to understand it I believe. This is my best guess:

    I need to find the fourier series of the square wave starting at [itex] \frac{\pi}{2} [/itex] fand this will "magically" yield the alternating series for [itex] \pi [/itex].

    But I don't understand what "the square wave equals 1" part means. Do I define the square wave to jump to 1 at [itex] \frac{pi}{2} [/itex] instead of the typical wave (typical for me) where the wave is -1 from -pi to 0, and jumps to 1 at 0 to pi ?

    Any help clarifying would be swell. Thanks :)

    What's up with the LaTeX by the way?
  2. jcsd
  3. Apr 11, 2006 #2


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    It says consider a square wave that is 1 at pi/2. The most casual one would be the one you mention later: the wave is -1 from -pi to 0, and jumps to 1 at 0 to pi. The fourier series of a function at a point, if it converges towards the function, equals the function evaluated at that point. See what you get when you evaluate the fourier series of the square wave at x=pi/2.
  4. Apr 11, 2006 #3
    For a problem before this one, I just got done evaluating the Fourier series for a square wave with properties:

    -1 , -pi to 0
    1 , 0 to pi

    [tex] \hat f(x) = \frac{4}{\pi}\left(\frac{\sin x}{1}+\frac{\sin 3x}{3}+\frac{\sin 5x}{5}+\ldots\right) [/tex]

    Evaluating the Fourier series at [itex] \frac{\pi}{2} [/itex] yields:
    (It took me so long to realize what this means. :)

    [tex] \hat f(\frac{\pi}{2}) = \frac{4}{\pi}\left(\frac{\sin \frac{\pi}{2}}{1}+\frac{\sin \frac{\pi}{2}}{3}+\frac{\sin \frac{\pi}{2}}{5}+\ldots \right) [/tex]

    [tex] \hat f(\frac{\pi}{2}) = \frac{4}{\pi}\left(1 + \frac{1}{3} + \frac{1}{5} + \ldots \right)[/tex]

    We know that the square wave at pi/2 is equal to 1 so:
    [tex] 1 = \frac{4}{\pi}\left(1 + \frac{1}{3} + \frac{1}{5} + \ldots \right) [/tex]

    This can then be solved for [itex] \pi [/itex]. But! This is not correct. So is my Fourier series for the square wave wrong?
  5. Apr 12, 2006 #4


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    sin(3*pi/2) is not sin(pi/2)!!
  6. Apr 12, 2006 #5
    yeah, you are right !

    man, I took this stupid cold medicine. I'm just going to go to bed and work on this in the morning, because obviously I'm making some really dumb mistakes!

    thanks you guys :)
    that was actually an interesting problem.
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