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Fourier series

  1. Jun 28, 2004 #1
    I wondered if someone could help me to find the Fourier series for this function please. I believe it's an odd function.

    f(x) = 2x+e^x-e^-x (-1< x > 1)

    This is my first post, so I'm going to try this LayTex typing too! Here goes!

    [tex]f(x)=2x+e^x-e^-^x[/tex] (-1< x >1)

    Thanks
     
  2. jcsd
  3. Jun 28, 2004 #2
    sorry!

    (-1< x <1) Doh!
     
  4. Jun 28, 2004 #3

    mathman

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    You need to evaluate the integral of f(x)sin(pi)nx from -1 to 1. Look it up in an integral table.
     
  5. Jun 29, 2004 #4
    Here we are so far (can't get LaTex to work today!)

    (e^x - e^-x)/2 = sinh(x)
    so
    2x + e^x - e^-x = 2x + 2sinh(x)

    and for integration by parts
    INT u.dv = uv - INT v.du
    we have
    u=2x+2sinh(x)

    du=2+2cosh(x)

    dv=sin(j*pi*x)

    v=(1/j*pi)*cos(j*pi*x)

    how am I going?
     
  6. Jun 29, 2004 #5

    mathman

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    Try this: xsinax can be integrated by parts easily enough. For the other term, I would use sinax=(eiax-e-iax)/2i. Then all you have are the integrals of exponentials.
     
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