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PDE: How to use Fourier Series to express a real function?

  1. Mar 11, 2015 #1
    actually have two questions:

    here we have a fourier series..
    $$f(t) = \sum c_k e^{2\pi ikt}$$ (c is complex)
    if we're trying to express a real function via fourier series, and we do it the following way..

    Impose condition: $$\overline{c_k} = c_{-k}$$

    $$f(t) = \sum\limits_{k= -n}^n c_k e^{2\pi ikt}$$
    then, although the constants would become real, wouldn't the exponents go to zero? why not?



    could someone give me more of an intuitive understanding for the "orthogonality" of sine functions? i know the integral definition, but how exactly does this integral dictate what can & can't be expressed in terms of what (ie could you relate the integral definition to the classical sense of the word 'orthogonal')
     
    Last edited: Mar 11, 2015
  2. jcsd
  3. Mar 11, 2015 #2
    Maybe you should try and explain why the exponent would go to zero. Or why the constants become real for that matter. I dont understand you last paragraph.
     
  4. Mar 11, 2015 #3
    (i accidentally left out the k)

    actually.. you'd essentially be having this

    $$c(u+iv)e^{2\pi i k t} +c(u-iv)e^{-2\pi i k t}$$ (since the condition above would imply k is part of v)

    so i don't even know how you'd add the two to get a real anymore..
     
  5. Mar 11, 2015 #4
    Try and write out the exponentials as cos and sin. Also complex number plus its conjugate is real.
     
  6. Mar 11, 2015 #5
    $$c(u+iv)(cosg+ising)+ c(u-iv)(-cosg-ising) = (cosg+ising)(c(u+iv)-c(u-iv))$$

    yes, but instead of a plus i get a minus, what'd i do wrong?
     
  7. Mar 11, 2015 #6
    exp(-ix) = cos(x) - isin(x)
     
  8. Mar 11, 2015 #7
    oops, okay and after that, since all the sin terms go to zero they're irrelevant. thanks
     
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