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.

  • $\begingroup$ Why do you assume that it is possible to express $\mathcal F [\exp (f)]$ in terms of $\mathcal F [f]$? As you correctly note, the core difficulty stays in the fact that $\mathcal F$ behaves nicely with respect the underlying linear structure, but not with respect to multiplication. Your problem boils down to expressing $\mathcal F [f^2]$ in simple terms of $\mathcal F [f]$, which I do not think to be feasible. $\endgroup$ – Alex M. May 21 '15 at 17:26
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    $\begingroup$ I agree with your reduction of the problem to expressing $\mathcal{F}[f^2]$. However, $\mathcal{F}[f^2]$ should be $\mathcal{F}[f] \ast \mathcal{F}[f]$, so an expression in terms of $\mathcal{F}[f]$ is possible. I just thought there may be a nicer one. $\endgroup$ – bers May 21 '15 at 20:56
  • $\begingroup$ I think it would be difficult trying to express with the Taylor expansion as you do as iterated convolution will make stuff hairy. You may want to take a look at this question math.stackexchange.com/questions/171462/… $\endgroup$ – mathreadler Oct 2 '15 at 16:29
  • $\begingroup$ @mathreadler thanks, this other question looks very useful! $\endgroup$ – bers Oct 2 '15 at 17:24

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: $$3+2f+f^2+3f^3 = 3+f\cdot(2+f\cdot(1+f\cdot(3)))$$

  • $\begingroup$ Yes, I understand what you mean. I will think about it, but after a first look, it does not seem very helpful. I agree with the Horner analogy, but Horner's method is mainly used to efficiently evaluate a polynom numerically - I am not aware of any use this has to simplify other uses of the polynom. Thanks again, anyway! $\endgroup$ – bers Oct 2 '15 at 17:22
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    $\begingroup$ Yes, it is mainly a way to rewrite it to make it prettier or faster to compute. I can not say it helps me very much with the intuition either, but maybe it makes it easier for someone else who has seen something like that before. $\endgroup$ – mathreadler Oct 2 '15 at 17:28

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