# Fourier transform of exponential of a function

I am wondering what $\mathcal{F}[\exp(f)]$ is in terms of $\mathcal{F}[f]$. The farthest I have got is using the series expansion of $\exp$, such that I end up with

$\mathcal{F}[\exp(f)] = \sum_{k=0}^\infty \frac1{k!} \mathcal{F}[f^k]$,

but $\mathcal{F}[f^k]$ is a $k-1$-fold convolution of $\mathcal{F}[f]$ with itself, which turns out not to be too useful in my context.

#### Solutions Collecting From Web of "Fourier transform of exponential of a function"

We can at least use linearity one more step, using the notation $a^{\otimes k} = (\underset{\overbrace{\text{k times}}}{a*\cdots *a})$: $$\sum \mathcal{F}(f^k) = \sum \mathcal{F}(f)^{\otimes k} = |\text{linearity of \mathcal{F} and distributivity of *}| = S_k(\mathcal{F}(f))$$
Where $S_k$ is a recursive summation such that: $S_k(a) = a*(S_{k-1}(a)+\delta)$

Does this make sense? What I’m trying to do is basically the convolutional equivalent of a Horner’s method for evaluating polynomials / power series.

Horners method on an ordinary polynomial is this factoring: $$3+2f+f^2+3f^3 = 3+f\cdot(2+f\cdot(1+f\cdot(3)))$$