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Summation formula for trig functions

  1. May 3, 2009 #1
    Does anyone know if there is a summation formula to find the sum of an expression with n as an argument in a trig function? I'm asking this because I'm learning about Fourier series/analysis but it seems that once we have the Fourier series we only sum for n=1,n=2,n=3... We never sum there series all the way to infinity. Is there a way to do that, simular to the way an integral is defiened as the Riemann sum of an infinate number of points?

    If anyone can point me in a good dirrection I would sincerely appreciate it.

    Thanks a lot
     
  2. jcsd
  3. May 3, 2009 #2

    Ben Niehoff

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    Gold Member

    There are some tricks that can be used, but not every series can be summed explicitly. As an example, take

    [tex]\sum \frac1n \sin nx[/tex]

    Using complex variables, we can write this as

    [tex]\Im \sum \frac1n e^{inx}[/tex]

    Now let [itex]z = e^{ix}[/itex], so we can write

    [tex]\Im \sum \frac1n z^n[/tex]

    And now use the series for [itex]\ln (1+z)[/itex]:

    [tex]\ln (1+z) = z - \frac{z^2}2 + \frac{z^3}3 - \frac{z^4}4 + \ldots[/tex]

    to write the sum as

    [tex]\Im (- \ln (1-z)) = - \arg (1-z) = - \arctan \frac{\Im (1-z)}{\Re (1-z)}[/tex]

    Finally, put in the definition of z:

    [tex]\sum \frac1n \sin nx = \arctan \frac{\sin x}{1 - \cos x}[/tex]

    If you plot this, it reproduces the periodic ramp function represented by the original series.

    To learn how it works, why don't you try working it out for a square wave? Basically all you do is use the ln(1+z) series in creative ways.
     
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