I am having trouble showing the following and i was wondering if anyone can point out my mistake.

let f be a 2pi periodic function and

$$f_m(t)=f(mt)$$

show the n-th fourier coefficient

$$\hat{f_m}(n)=\hat{f}(n/m)$$ if m divides n $$\hat{f_m}(n)=0$$ if m does not divides n

my attempt:

$$\hat{f_m}(n)=\frac{1}{2\pi}\int_0^{2\pi}f(mt)e^{-int}dt$$

substitute x=mt we get

$$=\frac{1}{2\pi}\int_0^{2\pi}f(x)e^{-i\frac{n}{m}x}\frac{dx}{m}$$

as you can see, im having a extra m on the denominator. Furthermore, i do not see why it should equal 0 if m does not divide n

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You forgot to substitute the integral limits. Also I think your description of the problem is missing some premise -- is $f$ periodic with period $2\pi$?
@jack: You can use the fact that $f$ is periodic with period $2\pi$. That means that in a sense you have $m$ times the same integral with different factors, and the result is determined by how the factors add up. – joriki Mar 16 '11 at 1:59
@jack: To see that the factors cancel if $m\nmid n$, you can either use what you know about orthogonality of Fourier modes, or you can apply the formula for the partial sum of a geometric series. – joriki Mar 16 '11 at 2:02
sorry i still dont see it. i have $\frac{1}{2\pi}\int_0^{2\pi}f(x)e^{-i\frac{n}{m}x}dx$ – jack Mar 16 '11 at 2:21
@jack: The integral goes from $0$ to $m\cdot2\pi$. That covers $m$ periods of $f$. In each of those periods, the integrand is the same except that the exponential changed by a certain factor. So you can express the integral over each period as some factor times the integral that you wrote above, then factor out that integral and figure out the sum of the factors, which will depend on whether $m$ divides $n$. – joriki Mar 16 '11 at 2:29
so its equal to $\sum_{j=0}^{k-1} \frac{1}{2\pi}\int_0^{2\pi}f(x)e^{-i\frac{n}{m}(x+2\pi j)}dx$. is that right? – jack Mar 16 '11 at 2:59