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Fourier series proof question

  1. Oct 22, 2012 #1
    The problem statement, all variables and given/known data
    Suppose, in turn, that the periodic function is symmetric or antisymmetric about the point x=a. Show that the Fourier series contains, respectively, only cos(k_{n}(x-a)) (including the a_0) or sin(k_{n}(x-a)) terms.


    2. Relevant equations

    The Fourier expansion for the periodic function, f(x):
    f(x)=a_0 + \sum_{n=1}^{\infty}(a_{n}cos(k_{n}x) + b_{n}sin(k_{n}x))

    a_0=\frac{1}{\lambda}\int_{x_0}^{x_0+\lambda}f(x) dx


    a_n=\frac{2}{\lambda}\int_{x_0}^{x_0+\lambda}f(x)c os(k_{n}x) dx, n\neq0

    b_n=\frac{2}{\lambda}\int_{x_0}^{x_0+\lambda}f(x)s in(k_{n}x) dx, (b_0=0)

    k= \frac{2n\pi}{\lambda}\qquad \lambda=period

    Hint: An integral centered on a point where the integrand is antisymmetric will vanish.

    3. The attempt at a solution

    I believe I understand what they are saying, but I do not know how to show/prove it. I know that cosine is symmetric about a point, while sine is not. Initially I was thinking that since sine is antisymmetric, for f(x) to be symmetric, the sine terms cannot exist in the expansion (which is exactly what the problem states). I was thinking the same for the opposite case regarding cosines.

    The hint leads me to believe I am supposed to set up an integral. However, I do not understand how.

    If I am to set up an integral, the best I can figure is that the lower limit should be a-\lambda and the upper limit should be a+\lambda

    I think this because for the integral to be centered at point "a", we would need to go 1 full period in both directions.

    I do not know what to do next
     
  2. jcsd
  3. Oct 22, 2012 #2

    LCKurtz

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    Just fixed your tex to make it readable. You should preview your posts before submitting them.
     
  4. Oct 22, 2012 #3
    Put "itex" "/itex" [in brackets] around your codes to make them readable.

    Edit: KCKurtz beat me to it, and then some.
     
  5. Oct 23, 2012 #4
    As a more useful hint, first try the [itex] a_{0} [/itex] term, and use a function about the point x = 0. What is the integral of a function that is asymmetric about the origin, from [-L, L] ?

    Then, if you multiply a function that is asymmetric with one that is symmetric? What kind of symmetry does the resulting function have? What would be the integral of this function?

    Do the same thing for asymmetric times asymmetric and symmetric times symmetric. What does this tell you about the integrals used to find your Fourier coefficients?

    Example: f(x) = x, which is asymmetric about the origin (i.e. it's odd):
    I'm just going to use a period of 2 so the integrals are easier. This would also be called the "odd extension of x." Here are the integrals you would set up:

    [itex] a_{0} = \frac{1}{2} \int_{-1}^{1} x dx [/itex]

    [itex] a_{n} = \frac{1}{2} \int_{-1}^{1} x cos(\frac{n\pi x}{2}) dx [/itex]

    [itex] b_{n} = \frac{1}{2} \int_{-1}^{1} x sin(\frac{n\pi x}{2}) dx [/itex]

    Do the necessary integration by parts, and you'll see what happens, then you could try [itex] f(x) = |x| [/itex] for an even function. (you can make it piecewise to evaluate the integrals)

    These relationships can be generalized to any point x = a about which a periodic function is symmetric/asymmetric, as the question suggests, but I'll leave it up to you to show that.

    Also, nice username.
     
    Last edited: Oct 23, 2012
  6. Oct 23, 2012 #5
    thank u sir, with an awesome username too !

    I understand everything that you're saying, im just confused how the final answer is supposed to have [itex]cosk_n(x-a)[/itex] in it, wheres that come from?
     
  7. Oct 23, 2012 #6
    That comes from the fact that we also need cosine and sine to be symmetric/asymmetric about x=a. For example, note that sin and cosine have no particular symmetry about say, [itex] x = 5 [/itex]. The (x-a) term essentially shifts the functions from the origin, x=0, to x=a, so that they will exhibit the necessary symmetries at the same point as whatever f(x) we have.
     
    Last edited: Oct 23, 2012
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