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Calculating fourier Series

  1. Dec 5, 2011 #1
    1. The problem statement, all variables and given/known data

    15qyqe9.jpg

    2. Relevant equations
    Equations to compute fourier series given in question


    3. The attempt at a solution

    Ok so I've just done this problem and I'm having some trouble computing the fourier series of [itex]x^{2}[/itex] between -∏ < ∏

    So first I calculate [itex]a_{0}[/itex] and that turns out to be [itex]\frac{∏^{2}}{3}[/itex], which according to wolfram is right.

    Next [itex]a_{n}[/itex], which is the one I'm having trouble with. Essentially you have to do integration by parts twice (right?) and I end up getting something quite messy, as does wolfram

    http://www.wolframalpha.com/input/?i=integrate+x^2+cos+%28n+x%29+dx+from+x%3D-pi+to+pi

    Is there any way to clean this up for when I plug [itex]a_{n}[/itex] into the final formula?

    [itex]b_{n}[/itex] is 0 obviously so there's no issue there (since it's an even function times an odd function).

    Thanks in advance everyone. :)
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 5, 2011 #2

    vela

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    You should find it simplifies down to [itex]a_n = (-1)^n\frac{4}{n^2}[/itex]. Since you know n is an integer, you should find some terms vanish, etc.

    You can get the answer from Wolfram Alpha by entering "FourierTrigSeries[x^2, x, 10]".
     
  4. Dec 5, 2011 #3
    Thanks for the response.

    So let me get this right... [itex]cosn∏ = (-1)^n[/itex] and [itex]sinn∏ = (-1)^n[/itex]?

    Sorry still a bit confused!
     
  5. Dec 5, 2011 #4

    vela

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    Not quite. You might want to take a look at a plot of the trig functions.
     
  6. Dec 5, 2011 #5
    Whoops, meant to say [itex]sinn∏=0[/itex], is this correct?
     
  7. Dec 5, 2011 #6

    vela

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    Yes, that's right.
     
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