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Math Help - Fourier sine and cosine series of x(Pi-x)

  1. #1
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    Fourier sine and cosine series of x(Pi-x)

    f(x) = x(π x) on [0, π]

    Find:
    a)Odd completion (sine series)
    f(x) ~ ∑bnsin(nx) where bn= 2/π 0∫π f(x)sin(nx)dx


    b)Even completion (cosine series)
    f(x) ~ a0/2 + ∑ancos(nx) where an= 2/π 0∫π f(x)cos(nx)dx and
    a0/2 = 1/π 0∫π f(x)dx

    c) Use the above results to show ∑1/n^2 = π^2 /6

    I get:
    Odd completion = ∑(2/n^3)(1 (-1)^n)sin(nx)

    Even completion = (π ^2 /6)+ ∑(-2/n^2)(1+(-1)^n)cos(nx)

    But I dont know how to manipulate these to show that part c) is true
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  2. #2
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    Durham student?

    I've been struggling to find the formulas for the odd/even completions and have no just realised i have them under a different name in my notes *d'oh*
    i'll have a look at that question again now.

    EDIT: where is your factor of 2 for An and Bn coming from?
    you seem to have An=2/π 0∫π f(x)cos(nx)dx I don't have the 2 in mine :s
    Last edited by doive; January 1st 2010 at 12:32 PM.
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  3. #3
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    let the fourier series = g(x)

    f(x)=g(x) since no discontinuities
    x(pi-x)=fourier
    RHS=0 when x=0 or pi
    putting those x values into both sides gives you 2 things in the sums both over n^2. One gives you the sum for even n the other for odd n. When you add the two together you get pi^2/6
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  4. #4
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    Thanks, I've solved this now, realised I forgot to apply my subsitution to both the top and bottom part of my sum
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