TITLE: How to solve the functional equation $f(2x)=2f(x)/(1+(f(x))^2)$
QUESTION [5 upvotes]: solve the functional equation $f(2x)=\dfrac{2f(x)}{1+(f(x))^2}$

I've already known that $\tanh(2x)=2\tanh(x)/(1+(\tanh(x))^2)$.
hyperbolic tangent

REPLY [2 votes]: To use the identity you have observed i.e. $\tanh(2x)=2\tanh(x)/(1+(\tanh(x))^2)$
Let $ g(x)= \tanh^{-1} \circ f(x)$, so
\begin{align*}
\dfrac{2f(x)}{1+(f(x))^2}&= \dfrac{2\tanh\circ g (x)}{1+(\tanh\circ g (x))^2}                   \\
&=    \tanh(2g(x))   
\end{align*}
and
$$f(2x)=\tanh\circ g (2x) $$
Therefore,
$$2g(x)=g(2x).     \fbox{1} $$
Any $\mathbb{Q}-$linear function satisfies the previous. To show that $g$ is $\mathbb{R}-$ linear assume that $g$ is differentiable, and $g'$ is continuous at zero.
Differentiating $\fbox{1}$, we obtain $g'(x)=g'(2x)$. Inductively, we obtain 
$g'(x)= g'(x\cdot 2^{-n})\rightarrow g'(0)$. Therefore,
$$g(x)=g'(0)x.$$