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Fourier coefficients

  1. Dec 4, 2012 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    I've reached a relation but then I need to obtain the coefficients ##A_m## and ##B_m##'s, those are the only unknowns.
    Here's the expression: ##\sum _{m=0}^\infty a^m [A_m \cos (m \theta ) + B_m \sin (m \theta )]=T_0\sin ^3 \theta##.

    2. Relevant equations
    Fourier coefficients.
    Result is probably ##B_1=\frac{3T_0}{4a}## and ##B_3=-\frac{T_0}{4a^3}##.
    All the other coefficients being 0.

    3. The attempt at a solution
    So I am not really sure. I have something of the form ##\sum _{m=0}^\infty a^m [A_m \cos (m \theta ) + B_m \sin (m \theta )]=f(\theta) ##.
    So in order to get let's say the ##A_m## first, I guess I must multiplicate both sides by ##cos (m \theta )## and integrate from ... 0 to ##2 \pi##?
    I'd get ##\sum _{m=0}^\infty a^m A_m \int _0^{2\pi } \cos ^2 (m \theta ) d \theta + \sum _{m=0}^\infty a^m B_m \int _0^{2\pi } \sin (m \theta ) \cos (m\theta ) d \theta=\int _0 ^{2\pi } f(\theta ) d \theta##. Where an integral vanishes.
    I'd be left with ##\sum _{m=0}^\infty a^m A_m \int _0^{2\pi } \cos ^2 (m \theta ) d \theta =\int _0 ^{2\pi } f(\theta ) d \theta##.
    I don't really know how to evaluate those integrals nor how to isolate ##A_m##.
     
  2. jcsd
  3. Dec 4, 2012 #2

    vela

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    You can't multiply by ##\cos m\theta## because ##m## is your dummy index. Also, you forgot to multiply the righthand side. You can multiply by ##\cos n\theta## and then do the integrals. Only the term with n=m will survive.

    But for this problem, it's probably better to use trig identities to express ##\sin^3 \theta## in terms of sines and cosines without exponents.
     
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