Is it possible, solely with the function $f(x) = \sum_{n>0} a_nx^n$, to obtain the function $\sum_{n>0} \frac{a_n}{n!} x^n$?

Question

It popped in my head these functions seems fairly independent even tho the first one kinda should determine the other.

for $a_n = 1$ we have $\sum x^n = \frac1{1-x}$ and $\sum \frac{x^n}{n!} = e^x$

doesn't look so trivial neither easy to generalize the arbitrary case.

Let's assume nice convergence on an interval of reals and maybe continously differentitation.

EDIT: ok so Greg gave a great idea on the comments:

$f(x) = \sum_{n>0} a_n x^n$ and $g(x) = \sum_{n>0} \frac{a_n}{n!} x^n$

So:

$f(x) = \sum_{n>0} \frac{a_n}{n!} \int_0^{\infty}e^{-t}t^ndt x^n = \int_0^{\infty} e^{-t} g(tx)dt $

which reminds us of Laplace's Transform!

This solves the problem of given $g(x)$ to find $f(x)$ but as I want the inverse it seemed to me it was better to change the variables of integration:

$xf(x) = \int_{s=0}^{\infty} e^{-\frac sx} g(s)ds$ thus $xf(x) = G(\frac 1x)$ where $G$ is the Laplace transform of $g$

So maybe if $H = L^{-1}$ and doing $u = \frac1x$

$g(u) = H(\frac{f(\frac1u)}u)$

Answer 1

as I proved in the own question:

$g(t) = \mathcal L^{-1} (\frac{f(\frac1s)}{s})$

where $\mathcal L^{-1}$ is the inverse of the Laplace transform.

If $f(s) = \frac1{1-s}$ then $g(t) = \mathcal L^{-1} (\frac1{s-1}) = e^t$ correct.

If $f(s) = 1 \implies g(t) = 1$ correct.

And so on