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Finding the value of pi/4 using Fourier series

  1. Jun 17, 2013 #1
    1. The problem statement, all variables and given/known data

    The function f(x) is defined by:

    [itex]f(x) = -1[/itex] when [itex]\pi < x <0[/itex] and [itex]0[/itex] when [itex]0<x<\pi[/itex]

    Show that [itex]\sum^{∞}_{0}\frac{(-1)^n}{2n+1}=\frac{\pi}{4}[/itex]

    2. Relevant equations

    Fourier series for a function of period [itex]2\pi[/itex] [itex]= a_{0} + \sum^{∞}_{1}a_{n}cos(nx) + b_{n}sin(nx)[/itex]

    [itex]a_{0}=\frac{1}{2\pi}\int_{-\pi}^{\pi}cos(nx)dx[/itex]

    [itex]a_{n}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)cos(nx)dx[/itex]

    [itex]b_{n}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)sin(nx)dx[/itex]

    3. The attempt at a solution

    f(x)=0 between 0 and pi, so I can ignore that interval in all of the integrals and integrate from -pi to pi.

    Doing this, I get [itex]a_{0}=-1/2[/itex] and [itex]a_{n}=0[/itex].

    Then I do [itex]b_{n}=\frac{1}{\pi}\int_{-\pi}^{0}-sin(nx)dx=\frac{1}{\pi}[\frac{1-cos(n\pi)}{n}][/itex] since f(x)=-1 in the relevant interval.

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

    I plug this value of [itex]b_{n}[/itex] into the Fourier series, I get [itex] -1/2 + \sum_{0}^{∞}\frac{2}{\pi(2n+1)}sin(nx)[/itex].

    I can turn this into a series similar to the alternative harmonic series by letting [itex]x=3\pi/2[/itex], giving me [itex]-1/2 + \sum_{0}^{∞}\frac{2(-1)^{n}}{\pi(2n+1)}[/itex]. Now the missing piece is showing that this is equal to 0, which I don't know how to do.
     
  2. jcsd
  3. Jun 17, 2013 #2
    But your original function is not defined when [itex]x=3\pi/2[/itex].
     
  4. Jun 17, 2013 #3
    Oh, sorry, I forgot to add that [itex]f(x+2\pi)=f(x)[/itex]
     
  5. Jun 17, 2013 #4
    Well then, what is f(3π/2)?
     
  6. Jun 17, 2013 #5
    Ah, right. 0. I got caught up thinking about what sin(nx) was doing instead of f(x). Embarrassing! Thanks though.
     
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