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Fourier series

  1. Sep 17, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the Fourier series of the function [tex]\[f\in {{C}_{st}}\][/tex] that in the interval ]-pi, pi[ is given by:

    [tex]\[f\left( x \right)=\left\{ \begin{array}{*{35}{l}}
    0for\,-\pi <x\le 0 \\
    \cos \left( x \right)for\,0<x<\pi \\
    \end{array} \right.\][/tex]

    and give the sum of the series for x = p*pi for p [tex]$p\in Z$[/tex]

    2. Relevant equations

    [tex]\[{{a}_{k}}=\frac{1}{2\pi }\int_{-\pi }^{\pi }{f\left( x \right){{e}^{-ikx}}dxfor\,n\in Z}\][/tex]

    and

    [tex]\[{{f}_{N}}\left( x \right)=\sum\limits_{k=-N}^{N}{{{a}_{k}}{{e}^{ikx}}}\][/tex]


    3. The attempt at a solution

    Well, first I find the an simply by doing the integral, but only from 0 to pi, since it's 0 from -pi to 0.
    After that I insert that in the second equation, and get the partial sum.
    But it's the last bit I'm confused about. I know that p is element of Z, so therefor I should only check x for -1, 0 and 1 - I think. But am i done after I've done that, or...?

    Well, the last bit confuses me a bit.
    So anyone who can give me a hint ? :)


    Regards
     
    Last edited: Sep 17, 2009
  2. jcsd
  3. Sep 17, 2009 #2

    LCKurtz

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    Does your function satisfy the Dirichlet conditions? If so, to what does the FS converge. The multiples of pi are particularly interesting for the periodic extension of this function because....

    You don't even need to calculate the FS to answer what the sum is going to converge to.
     
  4. Sep 17, 2009 #3

    Mark44

    Staff: Mentor

    Fourier series
     
  5. Sep 17, 2009 #4
    Argh, too slow to delete it... :(
     
  6. Sep 17, 2009 #5
    Well, I guess it satisfies the Dirichlet conditions.
    And it converge to 1/4 ? :S
     
  7. Sep 17, 2009 #6

    LCKurtz

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    You didn't answer this (it isn't a rhetorical question):

    The multiples of pi are particularly interesting for the periodic extension of this function because....

    How did you get 1/4?

    What does the Dirichlet theorem say about the convergence? You need to apply it to your question.
     
  8. Sep 17, 2009 #7
    Because cos(k*pi) = 1 or -1 depending on what whole number k is ?

    And the theorem states that:
    If f satisfies Dirichlet conditions, then for all x, we have that the series obtained by plugging x into the Fourier series is convergent, and is given by

    [tex] \sum_{n = -\infty}^\infty a_n e^{inx} = \frac{1}{2}(f(x+) + f(x-)) [/tex],

    where


    [tex] f(x+) = \lim_{y \to x^+} f(y) [/tex]
    [tex] f(x-) = \lim_{y \to x^-} f(y) [/tex]


    And forget about the 1/4 :) Stupid integral I did.
     
  9. Sep 17, 2009 #8

    LCKurtz

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    It's not just that the cosine is +-1 at those points.

    Why does the theorem state the sum of the series as

    [tex]\frac {f(x^+)+ f(x^-)} 2[/tex]

    instead of just saying f(x)? Does it equal f(x)? Why or why not? Again, what is different about the function at the n pi values of x?
     
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