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Fourier expansion (i cant understand a thing! darn)

  1. Sep 13, 2007 #1
    1. The problem statement, all variables and given/known data

    what is the fourier expansion of the periodic function whose definition in one period is

    f(t)= {[tex]^{0}_{1} ^{-\pi \leq t < 0}_{sint 0\leq t \leq \pi}[/tex]

    uh sorry about the small font
    i dont know how to make it bigger

    about the question,

    as much as i would like to even attempt a solution but the fact is, i can't understand the concept behind the fourier expansion!!!

    i really need some help.. its our quiz tomorrow.. and

    the only thing i figured out is what equation to use if a function is odd or even

    beyond that, nada! zip!...

    i would really appreciate your help
    a gazillion thanks everybody..

    just a little bit of enlightenment about this expansion would really help......its not really about the question...

    a gazillion thanks again
  2. jcsd
  3. Sep 13, 2007 #2
    uhmmm pls help me!!!!
  4. Sep 13, 2007 #3


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    Staff Emeritus
    Science Advisor

    It's not at all clear what you are asking!

    A Fourier expansion is a sum of sin(nx) and cos(nx) for all positive integer n:
    [tex]f(x)= \sum_{n=0}^\infty A_n cos(nx)+ B_n sin(nx)[/tex]
    (Of course, for n=0, sin(nx)= sin(0)= 0 for all x so you can ignore B0.)

    Here your interval, from [itex]-\pi[/itex] to [itex]\pi[/itex] has length [itex]2\pi[/itex] which is the standard example. I could look up the formulas for the An and Bn but they should be in your text book.
  5. Sep 13, 2007 #4
    "The concept behind Fourier expansion"... Hmm, that's a tough one. The abstractions could just go on and on... One way I used to visualise them was as an alternative way to approximate things, apart from a Taylor expansion. So a Taylor expansion about a point gives successively better approximations of a function near a particular point, by using terms such as [tex](x-x_0)^n[/tex], and you work out the coefficients by doing differentiation. A Fourier series is another way to approximate a function, but more globally -- so each term reduces the error, not only near some expansion point, but all over the entire function; the terms are cos and sin, and the coefficients are calculated using integration.

    As far as even/odd goes, you know that sin is odd and cos is even? Therefore, a sum of sin's is still odd, and sum of cos's is even. So even functions would only have cos terms, and odd functions sin's.
  6. Sep 13, 2007 #5
    i just want to know, is finding the fourier expansion just the same as fourier transformation?
  7. Sep 13, 2007 #6
    A Fourier transformation is what you get when you relax the condition that the function is periodic. Alternatively, you take the limit where the period goes to infinity. There are also other ways to characterise it, but this is probably the most relevant for you.
  8. Sep 13, 2007 #7
    yes it is.....thank you..
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