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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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    ah i see. so that upper limit of the integral should be $m*2\pi$. is that right?2011-03-16
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    @jack: Yes, that's right. Does that allow you to solve it?2011-03-16
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    i still dont see why it equals 0 when m does not divide n2011-03-16
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    @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.2011-03-16
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    @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.2011-03-16
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    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$2011-03-16
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    @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 $n$.2011-03-16
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    now im really lost. in each period, the integrand stays as $f(x)e^{-i\frac{n}{m}x}/m$ right?2011-03-16
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    @jack: No -- $f$ stays the same, but the exponential doesn't. $x$ increases by $2\pi$ per period, but the argument of the exponential increases by $\frac{n}{m}2\pi$, which may or may not be a multiple of $2\pi$.2011-03-16
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    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?2011-03-16
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    @jack: That's almost right, except that the factor $1/m$ is missing and a new variable $k$ has appeared out of nowhere :-)2011-03-16
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    sorry should be m. then i can factor it out and use the geometric sum formula. i get $\frac{1-(e^{-i\frac{n}{m}2\pi})^m}{1-e^{-i\frac{n}{m}2\pi}}$. i still dont see how this is equal to 0 ;(2011-03-16
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    nm, i can use euler's id2011-03-16
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    @jack: Now you can use $(a^b)^c=a^{(b\cdot c)}$.2011-03-16
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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