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

• Are they part of the same problem or are they just two different functional equations? Commented Jul 2, 2017 at 14:50
• Do you assume $f(0)=1$ in your first equation? Commented Jul 2, 2017 at 14:53
• @kingW3 two different. I don't know if theirs solutions are related in some way... Commented Jul 2, 2017 at 15:04
• For the first in order for $f$ to converge $f(0)$ must be $0$ but then it must also be $1$ which is a contradiction. Commented Jul 2, 2017 at 15:06
• Note that both equations have a term $f(x)$ on both the left and right hand side. Is that intended or not? Commented Jul 2, 2017 at 15:09

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 $f(1)$.

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 $-1=0$

• 7th line is $0=f(x^2)+f(x^3)+f(x^5)+・・・$? Commented Jul 2, 2017 at 16:46
• Yes you're right Commented Jul 2, 2017 at 17:38