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 $?

Question

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: $$f(x) = 1 + f(x) + \big(f(x^2)\big)^2 + \big(f(x^3)\big)^3 + \ldots $$

Answer 1

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 $f(1)$.

Answer 2

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 $-1=0$