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Homework Help: Fourier Series Expansions

  1. Apr 16, 2007 #1


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    1. The problem statement, all variables and given/known data
    Find the Fourier Series Expansion for:

    (a) f(x) = [pi-2x, 0 < x < pi | pi+2x, -pi < x < 0]

    (b) f(x) = [0, -pi < x < 0 | sin(x), 0 < x < pi]

    2. Relevant equations


    [tex]a_0=\frac{1}{\pi} \int_{- \pi}^{\pi}f(x)dx[/tex]

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

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

    3. The attempt at a solution

    For (a) my final answer was:


    and i think this is correct, but for (b) i got kind of a funny answer imo;


    if someone could work out b and see if they get the same answer i would appreciate it.

  2. jcsd
  3. Apr 16, 2007 #2
    Those are supposed to be piecewise functions, right?

    Hmm, in my denominator (for the second one) I got something different. Here's what I got
    [tex] f(x) = \sum_{n=0}^\infty \frac{cosnx}{\pi(1+1/n^2)}[/tex]

    Anybody else get something similar?
  4. Apr 18, 2007 #3


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    well heres my work for (b) so it can be checked:

    f(x)=[0, -pi < x < 0 | sin(x), 0 < x < pi]

    [tex]a_0= \frac{1}{\pi} \int_{- \pi}^{\pi}f(x)dx= \frac{1}{\pi} \int_{0}^{\pi}sin(x)dx= \frac{1}{\pi}(-cos(x))|_{0}^{\pi}=\frac{2}{\pi} \implies \frac{a_0}{2}=\frac{1}{\pi}[/tex]

    [tex]a_n=\frac{1}{\pi} \int_{- \pi}^{\pi}f(x)cos(nx)dx= \frac{1}{\pi} \int_0^{\pi}sin(x)cos(nx)dx=\frac{1}{\pi} \frac{cos(x)cos(nx)+nsin(x)sin(nx)}{n^2-1}|_0^{\pi}=\frac{1}{\pi}(\frac{1}{n^2-1}-\frac{cos(n \pi)}{n^2-1})=\frac{1}{\pi (n^2-1)}(1-(-1)^n)[/tex]
    [tex]\implies a_n=\frac{1}{\pi (n^2-1)}((-1)^{n+1}+1)[/tex]

    [tex] b_n=0[/tex]

    [tex]f(x)=\frac{1}{\pi}+\frac{2}{\pi}(\frac{cos(3x)}{8} +\frac{cos(5x)}{24}+...+\frac{cos(nx)}{n^2-1})[/tex]
    Last edited: Apr 18, 2007
  5. Apr 18, 2007 #4
    How did you go from the integral of sin(x) cos(nx) to what you have? A reduction formula?

    The b_n's are definitely zero. I forgot to put a_0 in my solution too.
  6. Apr 23, 2007 #5


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    I used the mathematica integrator at http://integrals.wolfram.com since from my experience it is pretty accurate.
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