For sake of simplicity $f$ is assumed to be $2π-$periodic function on $\Bbb R$. Given the fourier series of $f$, which coefficients are
$$\widehat{f}(k)= \frac{1}{2π}\int_0^{2π}f(t)e^{-ikt}dt,~~~~~~k\in \Bbb Z. $$
how to calculate the fourier series coefficients of $(f(x))^m?$
If I know the fourier series expansion of a function $f(x)$ is it possible to express the fourier series coefficients of $(f(x))^m$ as functions of fourier series coefficients of $f(x)?$
Is there any general algorithm to calculate this?
I think by a recursive formula it suffices to find the result for $m=2$. Does anyone has an idea?
If $$f(x)=\sum_{m=-\infty}^\infty \hat f(m)e^{imx}$$ then $$f(x)^2=\sum_{m,n=-\infty}^\infty\hat f(m)\hat f(n)e^{i(m+n)x}$$ Letting $m=M-n$ gives $$\sum_{M,n=-\infty}^\infty\hat f(M-n)\hat f(n)e^{iMx}$$ so $$\widehat{f^2}(M)=\sum_{n=-\infty}^\infty \hat f(M-n)\hat f(n)=\hat f\ast \hat f,$$ where $\ast$ denotes the convolution of two sequences. In general, $\widehat {f^m}=\hat f\ast \hat f\ast\cdots\ast \hat f$ $m$ times.