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Help with Fourier Sine Expansion.

  1. Mar 3, 2013 #1
    1. Find the Fourier sine expansion of [itex]\phi(x)=1[/itex]
    .


    2. Relevant equations.
    In the lecture the professor worked his stuff from scratch so I was trying to do it like his example.


    3. The attempt at a solution.
    I start with [itex]\phi(x)=A_1sin(\pi x)+A_2sin(2\pi x)+\cdots+A_nsin(n\pi x),[/itex] and then add multiply by [itex]A_msin(m\pi x)[/itex] term on each side and integrate from 0 to 1.
    So I have [itex]\int\phi(x)A_msin(m\pi x)=\int\left(A_1sin(\pi x)A_msin(m\pi x)+A_2sin(2\pi x)A_msin(m\pi x)+\cdots+A_nsin(n\pi x)A_msin(m\pi x)\right).[/itex]
    I know that due to orthogonality you can discard the terms where m is not equal to n (but I don't really understand why so if you can explain this I would appreciate it).
    Discarding those terms and using a trig relation I get, [itex]\int\phi(x)A_msin(m\pi x)=1/2\int\left(cos((m-n)\pi x)-cos((m+n)\pi x)\right).[/itex]
    I then solve the integral and try to get [itex]A_m[/itex] by alone, but I think I am doing something wrong cause what I get is terribly messy.
    The book answer is [itex]1=\frac{4}{\pi}\left(sin(\pi x)+\frac{1}{3}sin(3\pi x)+\frac{1}{5}sin(5\pi x) +\cdots\right).[/itex] Any and all help is greatly appreciated.
     
    Last edited: Mar 3, 2013
  2. jcsd
  3. Mar 3, 2013 #2

    cepheid

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    Welcome to PF,

    You can discard those terms because sin(nπx)sin(mπx)dx actually evaluates to zero for m ≠ n (with the appropriate limits of integration). Try it! That's how orthogonality of the basis functions is defined.

    I don't really understand what you did. If you discarded all but the m = n terms in the sum, then you should have had

    1 = Amsin(mπx)Amsin(mπx)dx


    = Am2sin2(mπx)dx

    That isn't the integral in your post, for some reason...
     
  4. Mar 3, 2013 #3
    I used the trig identity [itex]sin(x)sin(y)=\frac{cos(x-y)-cos(x+y)}{2}.[/itex] So my term becomes ##A_m \int \left(\frac{ cos(0)-cos(2m\pi x)}{2}\right) =A_m\left(\frac{1}{2}+\frac{sin(2m\pi)}{4m\pi} \right)## Is this not acceptable?

    I managed to obtain $$A_m=\frac{4}{\pi}\left(\frac{1}{m}\right)$$ but I can't understand why the answer only has terms for odd m's. Any idea?
     
    Last edited: Mar 3, 2013
  5. Mar 3, 2013 #4

    cepheid

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    Can you post your work? After integrating, you should have obtained something of the form

    some trig expression = 1

    which is only true for discrete values of m.

    Personally I would stick to the sine squared, since there is an easy trig identity for that.

    Since cos(2a) = cos2(a) - sin2(a)

    and cos2(a) = 1 - sin2(a),

    it follows that:

    cos(2a) = 1 - 2sin2(a)

    or sin2(a) = 1/2*[1 - cos(2a)]

    and this is REALLY easy to integrate.
     
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