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Homework Help: How to check Fourier series solution (complex)

  1. Apr 6, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the complex Fourier series for:

    [tex]f(t)=t(1-t), 0<t<1[/tex]


    2. Relevant equations

    [tex]f(t)=\sum_{n=-\infty}^{\infty}c_n{e^{i\omega_n{t}}}[/tex]

    [tex]c_n=\frac{1}{\tau}\int_{t_0}^{t_0+\tau}e^{-i\omega_n{t}}f(t)dt[/tex]

    [tex]\omega_n=2\pi{n}\quad\tau=1[/tex]
    3. The attempt at a solution

    I solved for c_n. I want to check my answer. I can only think of checking it by graphing it out to a few (50 or so) terms. I tried to graph this in Maple with my value for c_n and it couldn't do it. After that, I tried to solve the entire problem in Maple and that also did not work.

    I have a few more of these to do, and I'd like to make sure I am doing this correctly before I move on. Does anyone know how to check my value for the coefficient?
     
  2. jcsd
  3. Apr 7, 2010 #2

    gabbagabbahey

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    Do your coefficients depend on your choice of [itex]t_0[/itex]? Because the relevant equations you posted are not correct.
     
  4. Apr 7, 2010 #3
    I found the coefficients, c_n, by integrating:

    [tex]c_n=\int_{0}^{1}t(1-t)e^{-i2\pi{n}t}dt[/tex]

    Are you saying that this is not the correct method to find c_n?
     
  5. Apr 7, 2010 #4

    gabbagabbahey

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    No, it isn't.

    The Fourier coefficients are given by

    [tex]c_n=\int_{-\frac{1}{2}}^{\frac{1}{2}}f(2\pi t)e^{-i\omega_n t}dt=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)e^{-int}dt[/tex]

    which is not equivalent to what you've posted. Where did you find that incorrect equation for the coefficients?
     
  6. Apr 7, 2010 #5
    Thanks for pointing that out. I'll recalculate my coefficient value later today.

    My tutorial lists the equation for c_n as:

    [tex]c_n=\frac{1}{\lambda}\int_{x_0}^{x_0+\lambda}e^{-i{k_n}x}f(x)dx[/tex]

    In a paragraph above this equation, it states that:

    "...But most applications involve either functions of position or of time. In the former case, the period of the function, [tex]\lambda[/tex], is more conventionally called the wavelength, and [tex]k_n=\frac{2\pi{n}}{\lambda}[/tex] is the wave number for the n'th mode. If time is the variable, however, the period is called, indeed, the period, and is usually represented by [tex]\tau[/tex]. The n'th mode has the angular frequency, or often simply the frequency, [tex]\omega_n=\frac{2\pi{n}}{\tau}[/tex]."

    It did not explicitly give the equation for c_n in this case. I must have made a mistake converting from a spatial to time variable.
     
  7. Apr 7, 2010 #6

    vela

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    Those expressions are for a function with period 2π. If you rescale for a function with period τ and allow for a shift (since it really doesn't matter which particular cycle you integrate over), you get PhysicsMark's integral.
     
  8. Apr 7, 2010 #7
    Thanks for the clarification, Vela. I spoke with my professor today, and he also said the original integral should be correct.

    He went on to say that I should be able to plot it in Maple. So, I guess that means I need some more practice in Maple (That should be no surprise to Vela...https://www.physicsforums.com/showthread.php?t=391887).

    Thanks to Vela and Gabbagabbahey for replying.
     
  9. Apr 7, 2010 #8

    vela

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    In case you want to check your answer, here's what I found for cn for [itex]n\ne 0[/itex]:

    [tex]c_n=-\frac{1}{2n^2\pi^2}[/tex]

    When you plot the series, you should only need to use a handful of terms -- 50 would be overkill -- to see if it's summing to what you expect. Just remember to match the negative n's with positive n's so that the sum is real.
     
  10. Apr 7, 2010 #9
    Thanks a lot! That is also the answer I got.

    I realized why maple wouldn't plot it, it was because I did not account for the "n" in the denominator. No wonder Maple was blabbering about a singularity.
     
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