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First nonzero terms of Fourier sine series

  1. Dec 17, 2012 #1
    1. The problem statement, all variables and given/known data

    Consider the function φ(x) ≡ x on (0, l). Find the sum of the first three (nonzero) terms of its Fourier sine series.

    Reference: Strauss PDE exercise 5.1.3

    2. Relevant equations



    3. The attempt at a solution

    I have found the coefficient without difficulty, it is A[itex]_{n}[/itex]= -2l/n[itex]\pi[/itex] cos (n[itex]\pi[/itex]) + 2/n[itex]^{2}[/itex][itex]\pi[/itex][itex]^{2}[/itex] sin (n[itex]\pi[/itex]).

    When I plug in n=1,2,3; I obtain 2l/[itex]\pi[/itex], -2l/[itex]\pi[/itex] and 2l/3[itex]\pi[/itex], respectively.

    However the answer is given by: 2l/[itex]\pi[/itex] sin ([itex]\pi[/itex]x/l) - 2l/[itex]\pi[/itex] sin (2[itex]\pi[/itex]x/l) + 2l/3[itex]\pi[/itex] sin (3[itex]\pi[/itex]x/l)

    What I found comes from A[itex]_{n}[/itex] for n=1,2,3 but I do not understand which those sine terms come from.

    Thanks in advance
     
  2. jcsd
  3. Dec 17, 2012 #2

    Dick

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    So far you've calculated the coefficients of the sine series. Even saying 'coefficients' means they should multiply something. The sum of the sine series is a function of x. The thing the coefficients multiply is exactly those sine functions your are integrating against. Look at equation 6 here http://mathworld.wolfram.com/FourierSeries.html
     
  4. Dec 17, 2012 #3
    Thank you for your reply.
    So that would be sin (n[itex]\pi[/itex]x/l) for each corresponding n? Do I need multiply the coefficients by sine because I am asked for a sine series?
     
  5. Dec 17, 2012 #4

    Dick

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    Yes, it's asking you for the sine series. Not just the coefficients.
     
  6. Dec 17, 2012 #5
    Alright, so if I get this right, to find the first 3 nonzero terms, I plug in my results for n=1,2,3 in equation 6 and disregard the A[itex]_{n}[/itex] cos (nx) terms since I am asked for the sine series, and A[itex]_{0}[/itex] vanishes so I'm left with B[itex]_{n}[/itex] sin (nx) (A[itex]_{n}[/itex] in my answer). Thanks again
     
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