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Fourier series of a lineer function

  1. Oct 11, 2005 #1
    Hello,

    My QP homework involves (not is) Fourier expansion. i think i'm done with the physics part and for the answer, i need to expand a function to fourier series and solve it. So far well, but I couldn't solve that simple function:

    f(x) = x (in -1,1 interval)

    I've found various series, but when I graph them, it doesn't match the original function.
    I've tried:

    [tex]a_0 = \frac{1}{T} \int_0^T {f(x) dx}[/tex]
    with T = 4 (well, is it 2 or 4!?), and got 8. similarly
    [tex]a_n = \frac{1}{T} \int_0^T{f(x)cos(\frac{2 \pi n}{T}) dx}[/tex]
    and evertime I tried to solve, I've just messed it up.
    Can someone help?...
     
  2. jcsd
  3. Oct 11, 2005 #2

    mathman

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    You've got the integration domain screwed up.
    Your interval should be (-1,1) not (0,T). Furthermore since your function (x) is odd, a0 and all the cos coefficients will be 0. Your series will have only sin terms.
     
  4. Oct 11, 2005 #3

    Tom Mattson

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    And on top of that, you shouldn't expect the graph of a Fourier series to match that of the original function. To do that a necessary (but not sufficient!) condition is that you have to include all of the infinitely many terms of the series.
     
  5. Oct 12, 2005 #4
    thanks!
    I know that I'll need infinite elements to get the original graph. I was just looking for similarity. But what about 1/T? should it be 1/2 or 1/1?

    And by the way, there's also a function in the form of [tex]e^{-bx^2}[/tex] that should also be expanded to Fourier series. As far I know, there's no analytic solution for the intergral for that function. So how am I supposed to write a Fourier series???
     
  6. Oct 12, 2005 #5

    Tide

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    Regarding [itex]e^{-bx^2}[/itex], do you mean Fourier series or Fourier transform? If the latter then you can certainly do the integration.
     
  7. Oct 12, 2005 #6
    I mean Fourier transfrom... well, since there's no analytic integral, it cannot be solved analyitcally?
     
  8. Oct 13, 2005 #7

    Tide

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    You certainly can evaluate the integral analytically:
    [tex]\int_{-\infty}^{\infty}e^{-a x^2 + i k x}dx[/tex]
    Just complete the square in the exponential and you essentially have the integral of the Gaussian function for which you can obtain an analytic expression.
     
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