How can I prove that the product of two periodic functions also has a period P?

[Warr]

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.

 

[Galileo]

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.

 

[shmoe]

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?

 

[0rthodontist]

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.

 

[Warr]

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

 

[shmoe]

Warr, that looks fine.

Would you say that the period is pi implies that the period is also 2pi, 3pi, etc.?

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.

 

[0rthodontist]

Oh, okay. .