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Complex Fourier series of sin (t)

  1. Nov 11, 2013 #1
    In finding Cn, I arrived at a different answer. I got an extra factor of (1/i) instead, which came when you do the integral of each exponential with respect to t; so you get a factor of 1/i(1-n) and 1/i(1+n) respectively..

    Did they intentionally leave that out?

    x1mz9j.png
     
    Last edited: Nov 11, 2013
  2. jcsd
  3. Nov 11, 2013 #2

    mathman

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    Last edited: Nov 11, 2013
  4. Nov 11, 2013 #3

    mathman

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    Addition thought. f(t) = sint is its own Fourier series. exp(it) = cost + isint. Therefore c1 needs a coefficient 1/i. I can't see what your disagreement is about. Perhaps you can give the details of your derivation.
     
  5. Nov 11, 2013 #4
    Sorry, I spotted my mistake!
     
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