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Howto understand this periodic fourier series

  1. Nov 19, 2009 #1
    1. The problem statement, all variables and given/known data

    I am given this function

    [tex]f(x) = \left\{\begin{array}{cccc} x^2 \ \mathrm{where} \ \frac{-\pi}{2} < x < \frac{\pi}{2} \\ \ \frac{1}{4}\pi^2 \ \mathrm{where} \ \frac{\pi}{2} < x < \frac{3\pi}{2} \end{array}[/tex]


    Doesn't this mean that the function is periodic fourier which is defined on

    [tex][-\frac{L}{2}, \frac{3L}{2}][/tex]???

    Anyway I have formula to find the corresponding fourier series had been defined on [-L,L] but do I still use this formula eventhough the interval is different?

    is then true that if p = pi/2

    then

    [tex]a_0: = \frac{1}{2\pi} \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} x^2 dx + \frac{1}{2\pi} \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \frac{1}{4} \cdot \pi^2 dx = \frac{\pi^2}{6}[/tex]

    /Susanne
     
    Last edited: Nov 19, 2009
  2. jcsd
  3. Nov 19, 2009 #2

    LCKurtz

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    The short answer to your question is yes. Your function is defined on [itex][-\pi/2,3\pi/2][/itex] which is an interval of length [itex]2\pi[/itex]. If you extend it periodically it will have period [itex]2\pi[/itex]. For a periodic function of period P, it is true that integrating it over any period will give the same result as integrating over any other period:

    [tex]\int_a^{a+P}f(x)\,dx = \int_b^{b+P}f(x)\,dx[/tex]

    So you don't have to integrate over [itex](-\pi,\pi)[/itex] and in fact you really want to integrate over [itex][-\pi/2,3\pi/2][/itex], because that is where you have the formulas given. If, for some reason, you decided to integrate from [itex](2\pi,4\pi)[/itex] you couldn't use [itex]x^2[/itex] because it would have to be translated.
     
  4. Nov 22, 2009 #3
    Hi again,

    Is my a_n then

    [tex]a_n: = \frac{1}{2\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cdot cos(\frac{nx\pi}{2\pi}) dx + \frac{1}{2\pi} \int_\frac{\pi}{2}^{\frac{3\pi}{2}} \frac{1}{4}\pi^2 cos(\frac{nx\pi}{2\pi}) dx[/tex]???

    and my b_n is

    [tex]b_n: = \frac{1}{2\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cdot sin(\frac{nx\pi}{2\pi}) dx + \frac{1}{2\pi} \int_\frac{\pi}{2}^{\frac{3\pi}{2}} \frac{1}{4}\pi^2 sin(\frac{nx\pi}{2\pi}) dx[/tex]???

    what makes this problem complicated is that most of the formulas I has deals with nice even interval :(
     
    Last edited: Nov 22, 2009
  5. Nov 22, 2009 #4

    LCKurtz

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    Why don't you cancel the [itex]\pi[/itex]'s?

    The period is [itex]2\pi[/itex] so [itex]2p=2\pi[/itex], so [itex] p=\pi[/itex] so you should just have nx inside the integrals.
     
  6. Nov 22, 2009 #5
    Okay, but besides from that I hopefully used the formel definition for a_n and b_n correctly ?

    [tex]a_n: = \frac{1}{\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cdot cos(\frac{nx\pi}{\pi}) dx + \frac{1}{\pi} \int_\frac{\pi}{2}^{\frac{3\pi}{2}} \frac{1}{4}\pi^2 cos(\frac{nx\pi}{\pi}) dx[/tex]???

    and my b_n is

    [tex]b_n: = \frac{1}{\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cdot sin(\frac{nx\pi}{\pi}) dx + \frac{1}{\pi} \int_\frac{\pi}{2}^{\frac{3\pi}{2}} \frac{1}{4}\pi^2 sin(\frac{nx\pi}{\pi}) dx[/tex]???
     
  7. Nov 22, 2009 #6

    LCKurtz

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    Yes. And of course a0 is different.
     
  8. Nov 22, 2009 #7
    thanks :cool:
     
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