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Math Help - Fourier series help (odd/even)

  1. #1
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    Question Fourier series help (odd/even)

    Q1, 2: Options are Odd or Even extensions.

    My attempt.
    1. Odd extension
    2. Even extension (guessing, just looking at graph)

    Please help me out, cheers.
    Attached Thumbnails Attached Thumbnails Fourier series help (odd/even)-mathq19.png  
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  2. #2
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    Re: Fourier series help (odd/even)

    I think you have it wrong. With (A), f(-t)=f(t), so it is an even extension, and with (B), f(-t)=-f(t), so it is an odd extension.

    - Hollywood
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  3. #3
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    Re: Fourier series help (odd/even)

    Quote Originally Posted by hollywood View Post
    I think you have it wrong. With (A), f(-t)=f(t), so it is an even extension, and with (B), f(-t)=-f(t), so it is an odd extension.

    - Hollywood
    1. Even
    2. odd
    3. 2/n
    4. pi
    5. 4/((pi)n^2) or 0

    I tried this but got it wrong, please help me out
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  4. #4
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    Re: Fourier series help (odd/even)

    I'm pretty sure #4 should be the average value of the function, which is \frac{pi}{2}. How did you calculate #3 and #5?

    - Hollywood
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  5. #5
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    Re: Fourier series help (odd/even)

    Quote Originally Posted by hollywood View Post
    I'm pretty sure #4 should be the average value of the function, which is \frac{pi}{2}. How did you calculate #3 and #5?

    - Hollywood
    3) b(n) = (2/π) ∫(t = 0 to π) (π - t) sin(nπt/π) dt
    ............= (2/π) [(π - t) * -cos(nt)/n - (-1)(-sin(nt)/n^2)] {for t = 0 to π}
    ............= 2/n.


    4) a(0) = (2/π) ∫(t = 0 to π) (π - t) dt
    ............= (2/π) [(πt - (1/2)t^2) {for t = 0 to π}
    ............= π.

    5 )For n > 0:
    a(n) = (2/π) ∫(t = 0 to π) (π - t) cos(nπt/π) dt
    .......= (2/π) [(π - t) * sin(nt)/n - (-1)(-cos(nt)/n^2)] {for t = 0 to π}
    .......= (2/π) * -((-1)^n - 1)/n^2
    .......= 0 if n is even and 4/(πn^2) if n is odd.
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  6. #6
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    Re: Fourier series help (odd/even)

    It looks like you got past this part of the problem, based on your other post about this problem.

    - Hollywood
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