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Finding Fourier coefficients and Fourier Series

  1. May 20, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the Fourier coefficients for the function

    *Should be a piecewise function, not sure how to write one in [itex /itex] tags*

    [itex]f(x) = [/itex]
    [itex]|x|, |x| < 1,[/itex]
    [itex]1, 1≤|x|< 2;[/itex]

    [itex]f(x+4) = f(x)[/itex]

    and

    Find the Fourier series for

    [itex]f(x) = cos1/2\pi x, -1≤x<1; f(x+2)=f(x)[/itex]

    Would be great if someone could explain how to solve these. It feels like our lecturer rushed over Fourier series / coefficients, without giving any similar examples. Thanks for any help
     
  2. jcsd
  3. May 20, 2013 #2

    HallsofIvy

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    The Fourier coefficients for a function, f, defined on interval [a, b]. are the coefficients [itex]A_i[/itex] and [itex]B_i[/tex] such that
    [tex]f(x)= A_0+ A_1 cos((2\pi/(b-a))x)+ B_1 sin((2\pi/(b-a))x)+ A_2 cos((4\pi/(b-a))x)+ B_2sin((4\pi/(b-a))x)+ \cdot\cdot\cdot= \sum_{i= 0}^\inft A_i cos((2i\pi/(b-a))x)+ B_i sin((2i\pi/(b-a))x)[/tex]

    The "theoretical" point is that the sines and cosines form an "orthonormal basis" for such functions and so we can find the coefficients by taking the "inner product" of f with each of those functions:

    [tex]A_i= \frac{2}{b- a}\int_a^b f(x)cos((2\pi/(b-a))x) dx[/tex]
    [tex]B_i= \frac{2}{b- a}\int_a^b f(x)sin((2\pi/(b-a))x) dx[/tex]

    Surely those formulas are in your text book?
     
  4. May 20, 2013 #3
    These are the equations we are given in our notes

    [itex]f(x) = \frac{a_{0}}{2} + \sum^{∞}_{m=1} [a_{m}cos\frac{mx\pi}{L} + b_{m}sin\frac{mx\pi}{L}][/itex]

    where [itex]a_{m}[/itex] and [itex]b_{m}[/itex] are the Fourier constants.

    To determine the Fourier coefficients, we are given

    [itex]a_{m} = <cos\frac{mx\pi}{L},f(x)>_{L} = \frac{1}{L} ∫^{L}_{-L}cos\frac{mx\pi}{L}f(x)dx[/itex]
    [itex]b_{m} = <sin\frac{mx\pi}{L},f(x)>_{L} = \frac{1}{L} ∫^{L}_{-L}sin\frac{mx\pi}{L}f(x)dx[/itex]

    L is calculated from [itex]T = 2L[/itex] where T is the period which is found from [itex]f(x+T) = f(x)[/itex] So L must be 2.

    Since the function involves a piecewise function, am I supposed to take the integral of the function over the intervals [0,1] and [1,2]?
     
  5. May 20, 2013 #4

    micromass

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    Yes. Except that your domain also has negative numbers. So you should also include [-1,0] and [-2,-1] somewhere.
     
  6. May 20, 2013 #5
    Ok so here is my attempt for [itex]a_{0}[/itex]

    [itex]a_{0} = \frac{1}{2}∫^{-1}_{-2}1dx + \frac{1}{2}∫^{0}_{-1}xdx + \frac{1}{2}∫^{1}_{0}xdx + \frac{1}{2}∫^{2}_{1}1dx[/itex]

    Evaluated, I get

    [itex]a_{0} = 1[/itex]

    Correct? Should be able to do the rest fine now if that's correct
     
  7. May 23, 2013 #6
    Can someone please confirm that this is incorrect / correct so that I know that I'm on the right track or not. Thanks
     
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