6

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.

bers
  • 552
  • 1
  • 4
  • 17
  • 1
    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. – Alex M. May 21 '15 at 17:26
  • 1
    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. – bers May 21 '15 at 20:56
  • 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 http://math.stackexchange.com/questions/171462/fourier-transform-of-function-composition – mathreadler Oct 02 '15 at 16:29
  • @mathreadler thanks, this other question looks very useful! – bers Oct 02 '15 at 17:24

1 Answers1

2

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

mathreadler
  • 25,824
  • 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! – bers Oct 02 '15 at 17:22
  • 1
    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. – mathreadler Oct 02 '15 at 17:28