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Fourier Series and the first term

  1. Mar 21, 2012 #1
    I wasn't really sure where to post this because I am covering this in 2 classes (Math and Physics). Figured this would be my best bet.

    The fourier series of some Function is [itex]a_{0}/2+etc...[/itex]. I've looked in several textbooks but none explain why the [itex]1/2[/itex] is there, and not in any of the other terms of the summation.

    I do have a homework problem concerning this, but my professor said it's ok to not explain this part of the Fourier series. I'm intrigued by this now, so I'd like to know.
  2. jcsd
  3. Mar 21, 2012 #2
    Remember that the general term of a Fourier series is


    where the [itex]e_i[/itex] is normalized (has norm 1)

    We want to define

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

    If n>0, then this holds. The [itex]1/\pi[/itex] comes from normalizing the cosine. That is, the above is actually equal to

    [tex]<f,\cos nx>[/tex]

    but cos(nx) does not have norm 1, but rather pi. So we must divide by pi to normalize.

    We want the formula for an to hold for n=0 as well. But in this case, we have


    and the 1 is not normalized and has norm 2pi. So in order to normalize the thing, we must divide by 2pi. Division by pi is already taken care of in the definition of an, so we must also divide by 2.
  4. Mar 21, 2012 #3
    So, why doesn't the 2 tag along with the rest of the terms? In this case, it seems like there is a piecewise function under conditions n=0 and n>0.

    Sorry if I'm asking a dumb question.
  5. Mar 21, 2012 #4


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    Integrating from -pi to pi:
    The function f=1 yields 2pi
    The function f=cos2(nx) yields pi.
    Last edited: Mar 21, 2012
  6. Mar 21, 2012 #5


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    You should have cos2(nx), not cos(nx).
  7. Mar 21, 2012 #6


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    Thanks. I edited my original post to avoid confusing anybody
  8. Mar 25, 2012 #7
    I watched a video by MIT's OCW for 18.03 when I was doing this a while back, and the prof derived the Fourier series and explained the 1/2 quite well
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