Real-valued function solving $f(x+y)=\frac{f(x)+f(y)}{1-f(x)f(y)}$

Question

Let $f$ be a real valued differentiable function of a single variable s.t $$f(x+y)=\frac{f(x)+f(y)}{1-f(x)f(y)},$$ $\forall x,y$ with $f(x)f(y)$ is not equal to $-1$ & $f(0)=0.$ Then prove that $f(t)=\tan(gt)$, $g$ being a constant.

Answer 1

Let

$$f(x+y)=\frac{f(x)+f(y)}{1-f(x)f(y)} \tag{1}$$

We note that $f$ is odd; since

\begin{align*} f(x-x) = \frac{f(x) + f(-x)}{1 - f(x) f(-x)} &\Rightarrow f(0) = \frac{f(x) + f(-x)}{1 - f(x) f(-x)}\\ &\Rightarrow f(x) + f(-x) =0 \\ &\Rightarrow f(-x) = - f(x) \end{align*}

Since $f$ is differentiable the limit $\displaystyle \lim_{x \rightarrow x_0} \frac{f(x) - f(x_0)}{x-x_0} \in \mathbb{R}$. Hence,

\begin{align*} \lim_{x \rightarrow x_0} \frac{f(x) - f(x_0)}{x-x_0} &= \lim_{h \rightarrow 0} \frac{f \left ( x_0 + h \right ) - f(x_0)}{h} \\ &=\lim_{h \rightarrow 0} \frac{\frac{f(x_0) + f(h)}{1-f(x_0)f(h)} - f(x_0)}{h} \\ &=\lim_{h \rightarrow 0} \frac{\frac{f(x_0) + f(h) - f(x_0) \left ( 1 - f(x_0) f(h) \right )}{1-f(x_0) f(h)}}{h} \\ &= \lim_{h \rightarrow 0} \frac{f(x_0) + f(h) - f(x_0) + f^2(x_0) f(h)}{h\left ( 1 - f(x_0) f(h) \right )} \\ &= \lim_{h \rightarrow 0} \frac{f(h) + f^2(x_0) f(h)}{h \left ( 1 - f(x_0) f(h) \right )} \\ &= \lim_{h \rightarrow 0} \frac{f(h) \left ( 1 + f^2(x_0) \right )}{h \left ( 1 - f(x_0) f(h) \right )} \\ &= \lim_{h \rightarrow 0} \frac{f(h)}{h} \cdot \frac{1 + f^2(x_0)}{1 - f(x_0) f(h)} \\ &= f'(0) \left ( 1 + f^2(x_0) \right ) \end{align*}

Therefore ,

$$f'(x) = f'(0) + f’(0) f^2(x) \quad \forall x $$

The last differential equation is classical. Can you take it from here?