How to solve these functional equations $f(x) = 1 + f(x) + f(x^2) + \ldots $ and $f(x) = 1 + f(x) + \big(f(x^2)\big)^2 + \ldots $? The identity $\frac{1}{1-f(x)}=1+f(x)+\big(f(x)\big)^2+\big(f(x)\big)^3 + \ldots$ is well known.
Is there somewhere in the literature a method allowing us to solve the following functional equations?
First equation:
$$f(x) = 1 + f(x) + f(x^2) + f(x^3) + \ldots $$
Second equation:
$$f(x) = 1 + f(x) + \big(f(x^2)\big)^2 + \big(f(x^3)\big)^3 + \ldots $$
 A: I think that there also isn't a solution for the second equation.
Plug in $x=0$. Then we get:
$
f(0) = 1 + f(0) + f(0)^2 + \cdots
$
So $f(0) = \frac{1}{1-f(0)}$ and $f(0)^2 - f(0) + 1 = 0$, and this has no roots over the reals. (Note that $f(0) \neq 1$ as then the equation for $f(0)$ won't converge).
Also note that for both equation you could do the exact same thing for $f(1)$.
A: There is no solution for your first functional equation. Indeed :
$f\left( x\right) =1+f\left( x\right) +f\left( x^{2}\right) +f\left(
x^{3}\right) +f\left( x^{4}\right) +f\left( x^{5}\right) +f\left(
x^{6}\right) +\cdots $
so
$-1=f\left( x^{2}\right) +f\left( x^{3}\right) +f\left( x^{4}\right)
+f\left( x^{5}\right) +f\left( x^{6}\right) +\cdots $
so for $x^{2}$ we have
$-1=f\left( x^{4}\right) +f\left( x^{6}\right) +f\left( x^{8}\right)
+f\left( x^{10}\right) +f\left( x^{12}\right) +\cdots $
then
$0=f\left( x^{3}\right) +f\left( x^{5}\right) +f\left( x^{7}\right) +f\left(
x^{9}\right) +\cdots $
so
$-1=f\left( x^{2}\right) +f\left( x^{4}\right) +f\left( x^{6}\right)
+f\left( x^{8}\right) +\cdots $
then $f\left( x^{2}\right) =0$ then $f\left( x^{2n}\right) =0$ so 
$-1=f\left( x^{3}\right) +f\left( x^{5}\right) +f\left( x^{7}\right)
+f\left( x^{9}\right) +\cdots $
so $-1=0$
