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Fouiers series Calculate sum

  1. Nov 3, 2011 #1
    1. The problem statement, all variables and given/known data
    Consider the 2[itex]\pi[/itex]-periodic function f(t) = t t in [-Pi;Pi]
    a) show that the real fouier series for f(t) is:
    [itex]f(t) ~ \sum\limits_{n=1}^{\infty}\frac{2}{n}(-1)^{n+1}\sin nt[/itex]
    Use the answer to evaluate the following : [itex]\sum\limits_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{2n-1}[/itex]
    Hint: Use Fouier's law with t = [itex]\pi[/itex]/2

    2. Relevant equations
    Fouiers Law? I'm danish, and therefore i'm not really sure what it's called.

    3. The attempt at a solution
    Part a i have done by finding the coefficients.
    Part b) I can't see where the problem in part b and the answer to a relates. I've tried with Maple 15 to calculate the value and i'm getting Pi/4, but i keep getting something different for the series from a)
    Please Help me, i would really like to understand this as i'm studying physics.
    Last edited: Nov 3, 2011
  2. jcsd
  3. Nov 3, 2011 #2
    The answer should be pi/4 ...

    from fourier series we got ...

    t = [itex]\sum\frac{2}{n}(-1)n+1sinnt[/itex]

    for t = [itex]\frac{\pi}{2}[/itex] ,

    [itex]\frac{\pi}{2}[/itex] = 2 - 2/3 + 2/5 - 2/9 + ......

    which followed ,

    [itex]\frac{\pi}{2}[/itex] = 2 * [itex]\sum\frac{(-1)n+1}{2n - 1}[/itex]

    hence, [itex]\sum\frac{(-1)n+1}{2n - 1}[/itex] = [itex]\frac{\pi}{4}[/itex]
  4. Nov 3, 2011 #3
    Hmm, isn't that a backwards proof? Can't i use something like the even terms of sin(n*Pi/2) = 0 and the odd ones alternates between -1 and 1

    And how is t = ... isn't it f(t)?
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