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Complex Fourier Series

  1. Aug 10, 2011 #1
    What is the complex fourier series of f(x)=x from -pi to pi?


    I'm in a complex variables class and we have an extra credit assignment to figure out the complex fourier series of f(x)=x from -pi to pi. We only vaguely covered the topic in class and our book is not very good so I'm not entirely sure what to do. Please help!
     
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  3. Aug 10, 2011 #2

    jbunniii

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    What's the formula for computing the Fourier coefficients?
     
  4. Aug 10, 2011 #3
    C_n = 1/2pi int(f(x)*e^-inx)dx
     
  5. Aug 10, 2011 #4

    jbunniii

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    OK, so you have to calculate

    [tex]\frac{1}{2\pi} \int_{-\pi}^{\pi} x e^{-inx} dx[/tex]

    What have you tried so far?
     
  6. Aug 10, 2011 #5
    I think I have the correct integration..I have 1/2pi[((x*e^-inx)/-in)+(1/in*1/-in)*e^-inx].
    Sorry I'm new and don't know how to properly insert formulas yet!
     
  7. Aug 10, 2011 #6

    jbunniii

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    Yes, that looks right for the indefinite integral. What's the definite integral from [itex]-\pi[/itex] to [itex]\pi[/itex]?
     
  8. Aug 10, 2011 #7
    I have 1/2pi[((pi*e^-in*pi)/-in)+(1/n^2)(e^-inx)] - [((-pi*e^in*pi)/-in)+(1/n^2)*e^in*pi]
    which would then simplify down to (I think)

    1/2pi[(pi/-in)+(1/n^2)(-1^n)]-[(pi/in)+(1/n^2)(1^n)]

    And then I'm stuck, I'm not sure what it simplifies down to from here
     
  9. Aug 10, 2011 #8

    jbunniii

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    You can simplify quite a bit. Do you know Euler's formula:

    [tex]e^{ix} = \cos(x) + i \sin(x)[/tex]

    and therefore

    [tex]\frac{1}{2}(e^{ix} + e^{-ix}) = \cos(x)[/tex]
    [tex]\frac{1}{2i}(e^{ix} - e^{-ix}) = \sin(x)[/tex]

    You can make heavy use of these identities here.

    By the way, there's a short cut which involves using Euler's formula before you integrate. You will get the same answer, but with less effort.
     
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