I have been stuck in this problem for a while, given $f(z)$ as an exponential generating function (EGF), i.e. $f(z) = \{\frac{a_0z^0}{0!},\frac{a_1z^1}{1!},\frac{a_2z^2}{2!},\ldots\}$ for certain coefficients $\{a_0,a_1,a_2,\ldots\}$, I am supposed to find the solution to the following recurrence:
$$ f(z) = e^{-z}f(z/2)+e^{2z}-1 $$
The problem did not provide a value for $f(0)$. As an example of the type of solution am looking for, suppose that $f(0)=1$, if we set $f(z)$ as:
$$ f(z) = e^zf(z/2) $$
This equation can be iterated to $f(z)=e^ze^{z/2}e^{z/4}f(z/8) = \ldots = e^{2z}$, so that $2^n$ is the solution (i.e. EGF coefficient) to the recurrence:
$$ f_n = \sum_k \frac{n!}{k!(n-k)!}\frac{f_k}{2^k} = 2^n $$
But after several attempts, I could not find a solution of this sort for the first equation above... Any help ?
$f(z)=e^{2z}$ is a solution of $f(z)=e^{-z}f(z/2)+e^{2z}-1$; let $f(z)=e^{2z}+g(z)$ be an arbitrary solution. Then $g(z)=e^{-z}g(z/2)$, and iterating again, we get $g(z)=e^{-2z}g(0)$. Thus, the general solution is $$f(z)=e^{2z}+Ce^{-2z},\qquad C=\mathrm{const.}$$