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Periodic function proof

  1. Sep 25, 2006 #1
    I am trying to prove that if f(x) and g(x) both have period P, then f(x)*g(x) also has period P.

    What I tried to do was let f(x) and g(x) be represented as Fourier series

    [tex]f(x)=a_{0,1}+\sum_{n=1}^{\infty}\left[a_{n,1}cos(\frac{2{\pi}n}{P}x)+b_{n,1}sin(\frac{2{\pi}n}{P}x)\right][/tex]
    [tex]g(x)=a_{0,2}+\sum_{n=1}^{\infty}\left[a_{n,2}cos(\frac{2{\pi}n}{P}x)+b_{n,2}sin(\frac{2{\pi}n}{P}x)\right][/tex]

    I then tried to multiply the right side of these 2 equations and then manipulate it to look like a Fourier series with constants [tex]a_{0,3},a_{n,3},b_{n_3}[/tex]. Multiplying it out became problematic when the last term became a product of two series:

    [tex]\left(\sum_{n=1}^{\infty}\left[a_{n,1}cos(\frac{2{\pi}n}{P}x)+b_{n,1}sin(\frac{2{\pi}n}{P}x)\right]\right)*\left(\sum_{n=1}^{\infty}\left[a_{n,2}cos(\frac{2{\pi}n}{P}x)+b_{n,2}sin(\frac{2{\pi}n}{P}x)\right]\right)[/tex]

    Is there any easy way to simplify this...If not, is there a better strategy for approaching this problem? Thanks.
     
    Last edited: Sep 25, 2006
  2. jcsd
  3. Sep 25, 2006 #2

    Galileo

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    It's much much easier than that. A function f is periodic if f(x+P)=f(x) for all x, where P is the period. Use that definition.
     
  4. Sep 25, 2006 #3

    shmoe

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    f and g don't necessarily equal their fourier series. A periodic function can be very nasty. Just take the worst function you can think of on the unit interval and extend it to the real line to make a nasty function of period 1.

    This is overly complicated in any case! This problem follows directly from the definition of periodic, what is f(x+nP)*g(x+nP) for an integer n?
     
  5. Sep 25, 2006 #4

    0rthodontist

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    Hmm... I think that depends on your definition of period. You can show that f(x + nP) * g(x + nP) = f(x) g(x) but to me that doesn't necessarily mean that the period is P, only that the period is at most P. A recent example on these forums was sin x cos x where each has period 2pi but their product, namely 1/2 sin 2x, has period pi. Would you say that the period is pi implies that the period is also 2pi, 3pi, etc.? It seems like an odd phrasing to me.
     
  6. Sep 25, 2006 #5
    so then would an appropriate 'proof' be:

    f(x)=f(x+nP), g(x)=g(x+nP)

    h(x)=f(x)g(x)
    h(x+nP)=f(x+nP)g(x+nP)
    h(x+nP)=f(x)g(x)=h(x)

    hence h(x) is also periodic with a period of P
     
  7. Sep 25, 2006 #6

    shmoe

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    Warr, that looks fine.

    Yes, I would. It's convenient to have your definitions set up so you don't have to worry that you have the minimal period and still be able to say your function has period P if that's all your require for whatever you are doing.
     
  8. Sep 25, 2006 #7

    0rthodontist

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    Oh, okay. .
     
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