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Fourier series of sin(x)

  1. Sep 16, 2007 #1
    I know it's trivial....but how do you find the Fourier series of sin(x) itself? I seem to get everything going to zero...
  2. jcsd
  3. Sep 16, 2007 #2
    You should. The Fourier series of sin(x) is simply sin(x). This is like asking what the Taylor series of y=x^2 is.
  4. Sep 16, 2007 #3
    Yes, but if the Fourier series of f(x) = ao/2 + sum over n from 1 to infinity (an*cos(n*x) + bn*sin(nx)) then if ao, an, and bn go to zero, you are left with simply zero as the Fourier series. Shouldn't this method still work?
  5. Sep 16, 2007 #4

    D H

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    What ever makes you think all of the terms go to zero? One term does not, the b1 term. You end up with sin(x).
  6. Sep 16, 2007 #5
    bn = (1/pi) int ( sin(x) * sin(n*x) dx) from -pi to pi becomes a multiple of sin (n*pi) which is always zero.
  7. Sep 16, 2007 #6


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    if n = 1, then it's the integral of sin^2x, which is certainly not zero over -pi to pi.
  8. Sep 16, 2007 #7
    That's true. Okay, but for bn where n isn't 1, bn is zero, correct?
  9. Sep 17, 2007 #8


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  10. Aug 20, 2010 #9
    Sorry to gravedig, but considering this is the first result for the fourier series of sin[x]. I had to correct it.

    A fourier series is a series that has a period equal to part of a qualified f(x) from -L to L or 0 to 2l.
    Yes, the fourier series of Sin[x] with L = 2n Pi where n is an element of Z is simply Sin[x].
    But the fourier series of Sin[x] with L =/= 2n Pi is NOT Sin[x].
    Look at the graph of Sin[x] from 0 to Pi/2. If you wanted a function that repeated that period, Sin[x] would no-where near work.
    The first few terms of that fourier series would be: (4 Cos[1/2] Sin[x])/(-1 + \[Pi]^2) + (
    8 Cos[1] Sin[2 x])/(-4 + \[Pi]^2) + (
    12 Cos[3/2] Sin[3 x])/(-9 + \[Pi]^2)
    According to mathematica. (It's slightly incorrect, since the outcome of the periods are off. Working on it)
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