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Fourier coefficients

  1. Nov 30, 2009 #1
    1. The problem statement, all variables and given/known data
    Hi i would just like some fast hints, i'm doing the integrals wrong, im splitting up the integral below and get the wrong answer.

    well it's about finding the fourier series for f(t)={0 for -π<t<0 and sint for 0≤t≤π}

    2. Relevant equations
    [tex]a_{n} = \frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\cos(nt) dt , n\in Z_{+}[/tex]


    3. The attempt at a solution

    well i split the integral up in finding [tex]a_{n}[/tex] like

    [tex]\frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\cos(nt) dt = \frac{1}{\pi}\int_{-\pi}^{0} 0· dt+\frac{1}{\pi}\int_{0}^{\pi}\sin(t)\cos(nt) dt[/tex]
    Both of these elementary, but it fails to produce the right series.
    Hints anyone?
     
  2. jcsd
  3. Nov 30, 2009 #2

    LCKurtz

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    Did you forget the bn?
     
  4. Nov 30, 2009 #3
    no but that just get to zero
     
  5. Nov 30, 2009 #4

    LCKurtz

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    They can't be zero because the function you are expanding is not an even function.
     
  6. Nov 30, 2009 #5
    That's what i was thinking


    [tex]
    \frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\sin(nt) dt = \frac{1}{\pi}\int_{-\pi}^{0} 0· dt+\frac{1}{\pi}\int_{0}^{\pi}\sin(t)\sin(nt) dt
    [/tex]

    but the above is zero!

    i'm doing something wrong.
     
  7. Nov 30, 2009 #6

    LCKurtz

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    I have to run now. You didn't show your work but I'm guessing you need to look what happens when n = 1.
     
  8. Nov 30, 2009 #7
    halloo??

    any1 who knows this fourier series

    f(t)={0 for -π<t<0 and sint for 0≤t≤π}
     
  9. Nov 30, 2009 #8

    LCKurtz

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    Nobody is going to just give you the answer. Show us your work for the an and bn and we will help you find the mistake.
     
  10. Dec 1, 2009 #9
    Hey man chill.

    b_1=1/2 that was the problem.
     
  11. Dec 1, 2009 #10

    LCKurtz

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    Chill?? Surely you mean "Thanks for the suggestion, eh?"
     
  12. Dec 1, 2009 #11
    you were right LC, b_1 was the crucial step.

    thanx
     
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