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Fourier series of a periodic function not starting at x=-L

  1. Dec 3, 2013 #1
    1. The problem statement, all variables and given/known data
    In "oppgave 4" http://www.math.ntnu.no/emner/TMA4120/2011h/xoppgaver/tma4120-2010h.pdf
    you have a periodic function which is NOT periodic from ##x=-L=-\pi## to ##x=L=\pi##, but at ##x=0## and ends at ##x=2 \pi=2L##.

    The formulas I have (like these http://tutorial.math.lamar.edu/Classes/DE/FourierSeries.aspx) for building up a fourier transform assumes the function is periodic and begins its period at ##x=-L## and ends at ##x=L##.

    What am I to do?

    3. The attempt at a solution

    My gut feeling tells me I should just modify them so that the integration starts at ##x=0## and ends at ##x=2 \pi##.

    Is this correct? why?

    EDIT: Obviously I've missed some important theory on fourier analysis, so I hope you guys could enlighten me a bit? As a guess: The formulas I already got are built up using that cos and sin are perpendicular functions when their product is integrated from -pi to pi. Since the same holds if they're integrated from 0 to 2pi, all the existing formulas can painlessly be modified to also be valid for a function periodic from x=0 to x=2L? Or, x=x_0 to x=x_0 + 2L, for that matter?
     
    Last edited: Dec 3, 2013
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  3. Dec 3, 2013 #2

    HallsofIvy

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    If f(x) has period from 0 to [itex]2\pi[/itex] then f(x- \pi) has period from [itex]-\pi[/itex] to [itex]\pi[/itex].
     
  4. Dec 3, 2013 #3
    Ahh, so if I just do that transformation (##x \rightarrow x' - \pi##) on all the formulas, I'll end up with the limits going from ##x'=0## to ##x'=2 \pi##? Alright, thanks :)

    So, for example: ##a_0 = \frac{1}{2 \pi} \int_{x=-\pi}^{x=\pi} f(x) dx = \frac{1}{2 \pi} \int_{x'=0}^{x'=2 \pi} f(x'-\pi) dx'##, and ditto for ##a_n## and ##b_n##. But what about the ##(x'-\pi)## input of ##f## in the second integral? the function is shifted to the right?

    How would I use the above result to find the fourier series of the function from the OP, using the original formulas?
     
    Last edited: Dec 3, 2013
  5. Dec 3, 2013 #4

    LCKurtz

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    Your function is periodic of period ##2\pi##. You can use the usual formulas on ##[-\pi,\pi]## as long as you have the right function ##f(x)## on that interval. If there is some reason given in the problem that you can't do this, well, I don't read German (or whatever that is).
     
  6. Dec 3, 2013 #5
    I'm sorry, I formulated myself horribly in the OP. Should've read it before posting.


    What I meant was:

    f(x) is periodic, but its period starts on x=0 and ends on x=2pi, instead of starting at x=-pi and ending at x=pi like I am used to.
    The question: How am I supposed to find the fourier series of f(x) in oppgave 4, when all the formulas I know presume that the period of a periodic function f(x)=f(x+2pi) starts on -pi and ends at pi?

    I tried transforming the coordinates of my usual formulas to fit the situation for f(x):

    But I am doing it wrong.

    So could somebody help to clear this up? I'd really appreciate it.
     
    Last edited: Dec 3, 2013
  7. Dec 3, 2013 #6

    LCKurtz

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    You don't have to shift anything. Write the equation of that line segment in your picture on ##[-\pi,0]## and the equation of the line segment on ##[0,\pi]##. Use those two equations in a two piece formula ##f(x)## and use your standard formulas.
     
  8. Dec 3, 2013 #7
    Could you write down the expressions for ##a_0##,##a_n## and ##b_n## as an example?

    The function from oppgave 2 is this:

    ##f(x) = 2 \pi -x## for ##0 <x \leq \pi## and ##f(x) = 3 \pi - 2x## for ## \pi <x<2\pi##, while ##f(x) = f(x+ 2 \pi)##.
     
  9. Dec 3, 2013 #8

    LCKurtz

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    OK. You already have ##f(x)## on ##(0,\pi)##. What is the equation of that line in your picture going from ##(-\pi,\pi)## to ##(0,-\pi)##? It's just a straight line segment. Write its equation and it is the other piece of ##f(x)##. Then use your standard formulas for the coefficients. You are going to have to break up the integrals like$$
    \int_{-\pi}^\pi = \int_{-\pi}^0+\int_0^\pi$$because of the two piece formula. Is that what is bothering you?
     
  10. Dec 3, 2013 #9
    oh crap, nevermind. I thought you only could approximate a periodic function with a fourier expansion if the function is smooth over the interval you use to create the expansion. I realize that's incorrect now.

    OK thanks for the help! :)

    EDIT: But btw, it would be possible to also use the interval going from ##x=0## to ##x=2 \pi## to create a fourier series of the function, right?
     
    Last edited: Dec 3, 2013
  11. Dec 3, 2013 #10

    LCKurtz

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    Yes. You can use any ##2\pi## interval for the integration. If you use ##(0,2\pi)## you can use the two piece formulas you are given. For a function of period ##P##$$
    \int_0^P f(x)~dx = \int_a^{a+P}f(x)~dx$$for any ##a##.
     
  12. Dec 3, 2013 #11
    ahh yes, of course. How silly of me. OK, thanks :)
     
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