Functional equation $f(x+y)=\frac{f(x)+f(y)}{1-4f(x)f(y)}$ with $f'(1)=1/2$ Try to find the solution of  the functional equation $$f(x+y)=\frac{f(x)+f(y)}{1-4f(x)f(y)}$$ with $f'(1)=1/2$.
 A: HINT: Notice that the equation is equivalent to 
$$2 f(x+y) = \frac{2 f(x) + 2 f(y)} { 1 - (2 f(x)) \cdot (2 f(y))}$$
while we know that 
$$\tan(a(x+y)) = \frac{\tan(ax) + \tan(ay)}{1 - \tan(a x) \cdot \tan(a y)}$$
so we might just have $2 f(x) \equiv \tan a x$ for some convenient $a$.  
A: I am solving it with the assumption that $f'(0)=1/2$ and $f$ is continuously differentiable in an open neighbourhood containing $0$.
First, we observe that $f(0)=0$. 
Then,
$$
\frac{f(x+h)-f(x)}{h}=\frac{\frac{f(x)+f(h)}{1-4f(x)f(h)}-f(x)}{h}
=\frac{f(x)+f(h)-f(x)+4f(h)f^2(x)}{h\big(1-4f(h)f(x)\big)}
=\frac{f(h)\big((1+4f^2(x)\big)}{h\big(1-4f(h)f(x)\big)}\to f'(0)\big(1+4f^2(x)\big)=\frac{1}{2}\big(1+4f^2(x)\big).
$$
Hence, $f$ satisfies the IVP
$$
f'(x)=\frac{1}{2}\big(1+4f^2(x)\big), \quad f(0)=0.
$$
Hence
$$
\frac{d}{dx}\arctan\big(2f(x)\big)=\frac{2f'(x)}{1+4f^2(x)}=1.
$$
Thus $\arctan\big(2f(x)\big)=x+c$, for some $c\in\mathbb R$, and thus
$f(x)=\frac{1}{2}\tan (x+c)$. The initial data $f(0)=0$ forces $c=0$.
Thus $f(x)=\frac{1}{2}\tan x$.
Note. If $f'(1)=1/2$ is not a mistype in the OP, then we obtain instead that
$$
f(x)=\frac{1}{2}\tan\big(2f'(0)x\big),
$$
and 
$$
\frac{1}{2}=f'(1)=\frac{1}{2}\cdot 2f'(0)\cdot\frac{1}{\cos^2\big(2f'(0)\big)},
$$
and thus $2f'(0)= \cos^2\big(2f'(0)\big)$. Now the equation $x=\cos^2(x)$, has a unique solution $s_0\approx 0.6417$ and thus 
$$
f(x)=\frac{1}{2}\tan\big(s_0x\big).
$$
A: Define $g(x)=\arctan (2f(x))$ so $f(x)=\frac{1}{2}\tan g(x)$ and $$\tan g(x+y)=\frac{\tan g(x) + \tan g(y)}{1-\tan g(x) tan g(y)}=\tan (g(x)+g(y)).\quad(*)$$ Thus $$g(x+y)=g(x)+g(y)+\Pi(x,\,y),\quad(**)$$ where $\Pi(x,\,y)$ is a function with range $\subseteq \pi\mathbb{Z}$. Thus $g'(x+y)=g'(x)+g'(y)+\frac{\partial \Pi}{\partial x}$, where for the last term to be well-defined requires $\Pi$ to be constant for fixed $y$. A similar argument with $y$-derivatives shows $\Pi$ is constant; taking $x=y=0$ in $(**)$ gives $g(0)=-\Pi$. Since $\Pi$ is constant, define $G:=g-\Pi$ so $G$ satisfies the Cauchy relation $G(x+y)=G(x)+G(y)$. The only solution is $G(x)=Kx$ with $K$ constant. (The axiom of choice implies other solutions to the Cauchy relation exist, but they preclude f(0) being well-defined.) Thus $0=G(0)=g(0)-\Pi=-2\Pi$ and $\Pi=0$. Then $1=2f'(0)=g'(0) \sec^2 g(0)=K$ and $f(x)=\frac{1}{2}\tan x$.
