I am working on this problem to find all functions $f:\mathbb{N} \rightarrow \mathbb{R}$ such that $f(x) = f(2x)$. Specifically, I am trying to work out the generating function for this problem. I found the following relationship for the generating function $G(x^2) = \frac{1}{2} (G(x) + G(-x))$.
$$ G(x) = \sum_{k=1}^{\infty} f(k) x^k $$ $$ G(x)= \sum_{k=1}^{\infty} f(2k)x^{2k} + f(2k-1)x^{2k-1} $$ $$ G(x)= \sum_{k=1}^{\infty} f(k)x^{2k} + \sum_{k=1}^{\infty} f(2k-1)x^{2k-1} $$ $$ G(x)= G(x^2) + \frac{G(x)-G(-x)}{2} $$ $$ G(x^2) = \frac{1}{2} (G(x) + G(-x)) $$
I tried to find a way to solve this functional equation online but could not find anything. What is G(x), and what are the coefficients of the polynomial form of G(x)?
Edit
An example of a function that satisfys $f(x) = f(2x)$ is $$ f(x) = \sin(2 \pi log_2(x))) $$
Given $G(x)$ with $G(x^2)=\frac{G(x)+G(-x)}{2},$ define $H(x)=G(x)-G(x^2).$
Then $$\begin{align}H(-x)&=G(-x)-G(x^2)\\&=(G(-x)+G(x))-G(x)-G(x^2)\\&=2G(x^2)-G(x^2)-G(x)\\&=G(x^2)-G(x)\\&=-H(x).\end{align}$$
So $H$ is an odd function.
If $G$ is continuous at $0,$ then $H(x)\to 0$ as $x\to 0,$ so we get:
$$H(x)+H(x^2)+H(x^4)+H(x^{2^n})=G(x)-G(x^{2^{n+1}})\to G(x)-G(0)$$ as $n\to\infty.$
On the other hand, if $g_0$ is a real number and $H(x)$ is an odd function such that, on some interval around $0,$ $$G(x)=g_0+\sum_{j=0}^{\infty} H\left(x^{2^j}\right)$$ converges, then $G$ is such a function.
This reduces to your problem about $f.$ If $f(2k)=f(k),$ then $f$ is entirely determined by the value of $f$ on odd $k,$ and we can start with literally any function on the odd values.
Then $$H(x)=\sum_{k=0}^{\infty}f(2k+1)x^{2k+1}$$ is the odd function.
There are other $H$ which work which do not have power series near zero, however. For example, $H(x)=\sqrt[2n+1]x$ still has $G(x)$ converging on $(-1,1).$