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Homework Help: Foorier represtaion of this function

  1. Dec 23, 2009 #1
    [tex]f(x)=\sin(\frac{px}{2})\\[/tex]

    [tex]a_0=0[/tex]

    [tex]a_n=0[/tex]

    [tex]b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}\(\sin(\frac{px}{2}))\sin(nx)dx=\frac{1}{2\pi}[\sin(\frac{p\pi}{2}-n\pi)-\sin(\frac{-p\pi}{2}+n\pi)]+\frac{1}{2\pi}[\sin(\frac{p\pi}{2}+n\pi)-\sin(\frac{-p\pi}{2}-n\pi)][/tex]




    i used trig identetied to splt into two cosines

    andi solved

    but i got sines

    i need an expression of cosines to do cos nx=(-1)^n



    i need to have a simple linear fracture without cosines or sines



    i cant transform it here in the needed form

    ?

    and if thinking thurely then i see that i have a trig function on a simetric period
    so its zero

    so the foorier representation of the given function is zero
    ?

    where is the mistake
    it cant be zero
     
  2. jcsd
  3. Dec 23, 2009 #2

    tiny-tim

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    Hi nhrock3! :smile:

    These trigonometric identities need a factor of 1/2 inside the RHS brackets. :wink:
     
  4. Dec 23, 2009 #3
    i took the 0.5 out side of the integral
     
  5. Dec 23, 2009 #4

    tiny-tim

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    i'm getting confused :redface:

    let's start again … shouldn't there be factors of 1/(p/2 ± n) after the integration?
     
  6. Dec 23, 2009 #5

    whs

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    foorier... Really? Did you really just do that? I'd attempt a response but I can't read whatever the hell you typed.
     
  7. Dec 24, 2009 #6

    LCKurtz

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    You need to state your problem more clearly. Are you trying to find a Fourier series that represents your function for all x? If not that, on what interval? Why are you choosing [itex](-\pi,\pi)[/itex]? Depending on the value of p, the periodic extension of your function from that interval may have discontinuities. Do your care about that? Do you want a half range expansion? A more careful statement of the problem please.

    And what is "a simple linear fracture without sines or cosines?"
     
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