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Fourier (sin(2x)*x^2)

  1. Sep 9, 2010 #1
    Hello, I got a problem using Fourier series.

    f(x) = sin(2x)*(x^2) (-pi < x < pi)

    I got this answer, although it seems to be wrong.

    f(x) = (-16/9)*sin(x) + (-1/8)*sin(2x) + sum((2*(-1)^n/(2-n)^2 - 2*(-1)^n/(2+n)^2)*sin(nx) , n, 3, 100)
  2. jcsd
  3. Sep 10, 2010 #2
    Can you help me, please?
  4. Sep 10, 2010 #3
    [PLAIN]http://img826.imageshack.us/img826/6708/semttulocz.png [Broken]
    Last edited by a moderator: May 4, 2017
  5. Sep 10, 2010 #4
    What part are you having trouble with?
  6. Sep 11, 2010 #5
    Its hard to explain.
    I did this exercise a lot of times, every time I get the same answer.
    I can post the whole exercise to help me find what is wrong.
    That Bn is kind hard to get right.
  7. Sep 12, 2010 #6


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    I think your bn's are correct except for b2. I get

    [tex]b_2 = -\frac 1 8 +\frac {\pi^2} 3[/tex]

    And remember when plotting your graph, that your formula for your original function only works on (-pi,pi), so look at the convergence there.
  8. Sep 12, 2010 #7
    Thanks, the problem was at the B2.
  9. Sep 13, 2010 #8
    I got this:
    Bn = (2*(-1)^n)/(2-n)^2 - (2*(-2)^n)/(2+n)^2
    B2 = (2/(2-2)^2) - 1/8

    What did you make to get that: [tex]\frac {\pi^2} 3 [/tex]
    Last edited: Sep 14, 2010
  10. Sep 13, 2010 #9


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    That formula doesn't work for n = 2 so you have to do b2 separately. Just directly work out the integral

    [tex]\frac 2 \pi \int_0^\pi x^2 sin^2(2x)\, dx[/tex]
  11. Sep 14, 2010 #10
    Thanks, finally I got the right answer.
    You helped me a lot.
    Last edited: Sep 14, 2010
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