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Fourier factor frustration

  1. Mar 30, 2010 #1
    1. The problem statement, all variables and given/known data

    I have conflicting notes and even textbooks about where i should and shouldn't have factors of 2 when doing fourier series. I just want to check once and for all i'm doing it right and not introducing erroneous factors of 2 or omitting any

    2. Relevant equations
    \tilde{f}(x)= \frac{a_0}{2} + \sum^{\inf}_{n=1} [a_n cos(\frac{n \pi x}{L}) + b_n sin(\frac{n\pi x}{L})]

    a_n = \frac{1}{L} \int^{L}_{-L} cos \frac{n\pi x}{L} f(x) dx

    b_n = \frac{1}{L} \int^{L}_{-L} sin(\frac{n\pi x}{L}) f(x) dx

    a_0 = \frac{1}{2L} \int^{L}_{-L} f(x) dx

    where the interval is [-L,L] and the period is 2L

    3. The attempt at a solution

    Is the above formulation correct (i'm particularly unsure about the 1/2L associated with a0)
    Last edited: Mar 30, 2010
  2. jcsd
  3. Mar 30, 2010 #2


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    Just plug in the series for f(x) into the integrals and see if the coefficients work out correctly.
  4. Mar 30, 2010 #3


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    You have an extra factor of 1/2 associated with [itex]a_0[/itex]. To see that this is the case, set [itex]f(x) = 1[/itex]. Then according to your formulas, [itex]a_0 = 1[/itex] and the Fourier series is

    [tex]\tilde{f}(x) = 1/2[/tex].

    There should be a 1/2 either on the defining integral for [itex]a_0[/itex] or on the [itex]a_0[/itex] term in the series expansion, but not both.

    You can check your other coefficients in a similar way by setting

    [tex]f(x) = \cos(n\pi x/L)[/tex]


    [tex]f(x) = \sin(n \pi x/L)[/tex]
  5. Mar 31, 2010 #4
    thank you guys!

    I had tried checking it but couldn't be certain it wasn't my dodgy arithmetic!
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