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Fourier Series

  1. Jan 16, 2016 #1
    1. The problem statement, all variables and given/known data
    The odd 2π-periodic function f(x) is defined by
    f(x) = x2 π > x > 0
    -x2 −π<x<0
    Find the coefficient bn in the Fourier series
    f(x) = a0/2 + ∑(an cos(nx) + bn sin(nx)).
    What are the values of the coefficients a0 and an and why?

    2. Relevant equations
    bn = 1/π ∫ f(x)sin(nx)
    an = 1/π ∫f(x)cos(nx)
    3. The attempt at a solution

    I'm unsure what to do as the function changes so you cannot integrate between pi and -pi like you would normally. Also as the function is even would bn = 0?
     
  2. jcsd
  3. Jan 16, 2016 #2
    The function doesn't "change" - it is defined in a piecewise manner. So, you just have to break up the integration interval accordingly.

    Are you sure the function is even?
     
  4. Jan 16, 2016 #3

    Samy_A

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    Redudant with @Fightfish 's post.
     
  5. Jan 16, 2016 #4

    Ray Vickson

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    What are the integration limits? [You should always indicate this; it is an important part of your analysis.] Have you never before performed integrations of piece-wise funcctions (defined by different formulas in different regions? Why do you say in your final sentence that the function is even?
     
  6. Jan 16, 2016 #5
    so would i just break it up to get bn = 1/π[∫π0 x2sin(nx) + ∫0 -x2sin(nx)]
     
  7. Jan 16, 2016 #6

    Samy_A

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    Yes, and if you give it a little thought, computing one of the integrals (for each n) is all the work you need to do.
     
    Last edited: Jan 16, 2016
  8. Jan 16, 2016 #7
    so when integrating the 2 integrals are the same except for the sign changes. This means when n = odd I get bn = 2nπ2 - 8n3
    and when n = even bn = -2nπ2. Does that sound feasible?
     
  9. Jan 16, 2016 #8

    Samy_A

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    As your function is odd, ##\displaystyle b_n=\frac{2}{\pi}\int_0^{\pi} x²\sin(nx) \, dx##.
    I get something different for the ##b_n##, but I can make a computation error just as you can.
     
  10. Jan 16, 2016 #9
    I've re-done it to get (-1)n(-π2n-2n3)-(2n3) Thank you for all your help. Also would an = 0?
     
  11. Jan 16, 2016 #10

    Samy_A

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    Still different from what I got, but again, you may have the correct result. Difficult to check without seeing the calculation.
    ##a_n## is the definite integral of an odd function over ##(-\pi,+\pi)##. The integrals between ##(-\pi,0]## and ##[0,+\pi)## cancel out. So yes, ##a_n=0##.
     
  12. Jan 17, 2016 #11

    Samy_A

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    Come to think of it, the Fourier coefficients of a continuous function (among others, but that is of no importance here) on ##[-\pi, +\pi]## satisfy a number of convergence properties.
    For example, ##\displaystyle \lim_{n\rightarrow +\infty}{a_n}=0,\ \lim_{n\rightarrow +\infty}{b_n}=0,\ \sum_{n=0}^\infty { |a_n|}^{2}<\infty, \sum_{n=1}^\infty { |b_n|}^{2}<\infty##.

    Clearly the ##b_n## you got do not satisfy these properties.
     
    Last edited: Jan 17, 2016
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