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Fourier Analysis Question

  1. Apr 19, 2016 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    Consider a 2pi-periodic function f(x) = |x| for -pi ≤ x ≤ pi
    a) Compute the Fourier series of the function f.
    b) Prove that (from n=1 to n=infinity)∑ 1/(2k-1)^2 = pi^2/8.


    **note all "sums" from here on out will be defined from n = 1 to n=infinity
    2. Relevant equations


    3. The attempt at a solution

    For part a we start with the definition of the Fourier series
    f(x) = 1/2*a_0 + ∑(a_n*cos(n*x)+b_n*sin(n*x))

    Since f is an even function, we know that b_n = 0.

    a_n = (2/pi)*integral from 0 to pi of (x*cos(nx))dx
    a_n = 2((-1)^n-1)/(pi*n^2)

    a_0 = 1/pi * integral from -pi to pi of |x| dx = pi

    So we have the following fourier series for f(x) and the answer for part a:

    pi/2 + ∑2((-1)^n-1)/(pi*n^2) * cos(nx)



    For part b, we set x = 0, and find

    |x| = pi/2 + ∑2((-1)^n-1)/(pi*n^2) * cos(nx)
    => -pi/2 = ∑2((-1)^n-1)/(pi*n^2)
    => -pi^2/4 = ∑((-1)^n-1)/n^2
    Divide both sides by -2
    => pi^2/8 = ∑((-1)^n-1)/(-2*n^2)
    But I'm not quite sure how to get the right side (the sum) similar to ∑ 1/(2k-1)^2
     
    Last edited: Apr 19, 2016
  2. jcsd
  3. Apr 19, 2016 #2

    BvU

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    You wonder how ## {1\over 1} + 0 + {1\over 9} + 0 + {1\over 25} ... ## (n = 1, 2, 3, 4, 5) can be equal to
    ## {1\over 1} + {1\over 9} + {1\over 25} ... ## (k = 1, 2, 3) :smile: ?
     
  4. Apr 20, 2016 #3

    RJLiberator

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    Ah, I see.

    So, in the end, I was doing things right here.

    It was my inability to manipulate the problem to come to the conclusion.

    Tough, tough course.
     
  5. Apr 20, 2016 #4

    BvU

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    Good thing there is PF :wink:
     
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