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Fourier series

  1. Nov 13, 2009 #1
    Does f(t)=1 have fourier series expansion or not?
     
    Last edited by a moderator: Nov 14, 2009
  2. jcsd
  3. Nov 13, 2009 #2
    Re: Fourier series expansion of Sin(x)

    Every (reasonably behaving) function has a fourier series. Some are just more boring than others.
     
  4. Nov 13, 2009 #3
    Re: Fourier series expansion of Sin(x)

    Fourier series is defined only for periodic signals.
    f(t)=1 is not a periodic signal-->no fourier series expansion??
     
  5. Nov 14, 2009 #4

    CompuChip

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    Re: Fourier series expansion of Sin(x)

    Depends on how you define "periodic" (if you want, it is periodic with any period P).

    Without arguing about definitions and semantics, you can note that indeed you can write f(t) as
    [tex]f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty \, [a_n \cos(nx) + b_n \sin(nx)][/tex]
    (copied from Wikipedia), where [tex]a_n = b_n = 0[/tex] except for [itex]a_0 = 2[/itex].
     
  6. Nov 14, 2009 #5

    HallsofIvy

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    The point is that sin(x)= (1)sin(x)+ 0 cos(x)+ (0)sin(2x)+ (0)cos(2x)+ ... is a perfectly good Fourier series!

    The function defined as f(t)= 1 for [itex]-\pi< x\le \pi[/itex] and continued periodically has a Fourier series expansion.
     
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