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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 $2\pi$?

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    @jack: Also remember that you can only apply the geometric sum formula if the factor isn't $1$; else you just get $0/0$.2011-03-16