I am reading a proof and in the proof, it states, without details, that given a function $$\exp(\sum_{j=1}^{\infty}a_{j}\exp(ijλ))$$, $-π≤λ≤π$, the Fourier series expansion of the function is $\sum_{j=0}^{\infty}β_{j}\exp(ijλ)$ where $β_{0}=1$, $β_{1}=a_{1}$, $β_{2}=a_{2}+(1/2)a_{1}^2$, ...
Could someone please show me that how to compute the Fourier transform of the above function? Thank you very much.
Since the series $\psi(t):=\sum_{k\geq1} a_je^{ikt}$ only has terms with $k\geq1$ we can consider the analytic function $$f(z):=\sum_{k\geq1}a_kz^k\ .$$ The $\psi$ then encaptures the values of $f$ on the unit circle $\partial D$. The function $$g(z):=\exp\bigl(f(z)\bigr)=1+b_1 z+b_2z^2+\ldots$$ is then analytic as well, and is related to $f$ by $$g'(z)=f'(z)\,g(z)\ .\tag{1}$$ Now you can determine the Taylor coefficients $b_j$ of $g$ recursively from $(1)$. Finally write again $e^{it}$ for $z$, and you have the Fourier series of $e^\psi$.