Exponential Generating Function Transformation

Question

Say we have an exponential generating function:

$$F(x) = \sum_{n\geq 0} f_n \frac{x^n}{n!}.$$

Is there a simple transformation from $F(x)$ to $G(x)$ where

$$G(x) = \sum_{n\geq 0} f_n \frac{x^{2n}}{(2n)!}?$$

Answer 1

If $F$ is exponentially bounded and entire, the Laplace transforms of $F(x)$ and $G(x)$ are $$ \mathcal L F(s) = \sum_{n=0}^\infty \int_0^\infty f_n \frac{x^n}{n!} e^{-sx}\; dx = \frac{1}{s} \sum_{n=0}^\infty f_n s^{-n} $$ and $$ \mathcal L G(s) = \sum_{n=0}^\infty \int_0^\infty f_n \frac{x^{2n}}{(2n)!} e^{-sx}\; dx = \frac{1}{s} \sum_{n=0}^\infty f_n s^{-2n} = s \mathcal L F(s^2) $$ for $\text{Re}(s)$ sufficiently large. That is, $G = \mathcal L^{-1} (s \mathcal L F(s^2))$.

Answer 2

Let's try an example ... $f_n = n!$ \begin{align} F(n) &= \sum_{n=0}^\infty \frac{n!\;x^n}{n!}= \sum_{n=0}^\infty x^n = \frac{1}{1-x} \\ G(x) &= \sum_{n=0}^\infty \frac{n! \;x^{2n}}{(2n)!} = 1+\frac{x\sqrt {\pi }\;{\rm erf} \left(x/2\right){{e}^{{x}^{2}/4}}}{2} \end{align} Is there a simple transformation from $F(x)$ to $G(x)$ in this case? I doubt it.