let say I have a periodic function I can write $f(t)=a_0+ a_1 \cos(\omega t)+a_2 \cos(2\omega t)...+b_1\sin(\omega t)+b_2\sin(2\omega t)$ that is a fourier series. Is there a link between the coefficients $a_i,b_i$ of $f(t)$ and the $a_i,b_i$ of $\sin(f)(t)$ and $\cos(f)(t)$?
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Do you mean $\sin(f(t))$ ? If so then try first with $f(t)^2$, then with $f(t)^k$, and use $\sin z$'s Taylor series to conclude. Of course replacing your series in $\sin(n \omega t)$,$\cos(n \omega t)$ by series in $e^{i n \omega t}$ will make things much easier. – reuns Nov 25 '17 at 00:46
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Well 3 months ago I would have written it like that, but I read some mathematics books since and it seems $\sin(f)(t)$ is a more appropriate form and I think I can see why. Though I tried this with the square, but that's quite a long calculation. – kalish Nov 25 '17 at 22:09
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$f(t) = \sum_n c_n e^{i n\omega t}$. $f(t)^2 = \sum_n d_n e^{in\omega t}$ where $d_n = \sum_m c_m c_{n-m}$. What do you get for $f(t)^3 $ ? – reuns Nov 26 '17 at 00:52
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Well we should have $\sum_n g_n e^{i n\omega t}$ with $g_n=\sum_m^n d_m c_{n-m}$ but that's really long, like...infinitely long. Problem being I don't know which coefficient will be of importance. Sorry for the delay. – kalish Dec 03 '17 at 20:14