function equation $f(x)+f(y)=f(\frac {x+y} {1+xy})$

Question

Is there exist $f:(-1,1)\to R$ such that $f(x)+f(y)=f(\frac {x+y} {1+xy})$? how to find $f$?

Answer 1

The functional equation for hyperbolic tangent is $$ g(x):=\tanh(x),\quad g(x+y) = \frac{g(x)+g(y)}{1+g(x)g(y)}. $$ The Wikipedia article Hyperbolic functions is one source for this.

The functional equation for its inverse function is $$ f(x):=\tanh^{-1}(x),\quad f(x)+f(y) = f\Big(\frac{x+y}{1+xy}\Big). $$

Note that, just like with Cauchy's functional equation, without any other assumptions such as continuity, and so on, there could be many strange solutions. Consult the Wikipedia article for details.

Answer 2

I will assume $f'(0)$ exist.

for $|x|<1$, we can find relatively small $H$ such that $|h|<H\,$ implies $1 - x^2 - xh\neq0$ .

I will define y = $\frac{h}{1-x^2 - xh}$.

Then $$ y - x^2 y = h + xyh = h(1+xy)\\ \Rightarrow \frac{y - x^2 y}{1+xy} = \boldsymbol{\frac{x+y}{1+xy} - x = h.} $$

Now, $f(x+h) - f(x) = f(y) = f(\frac{h}{1-x^2 - xh})$.

Moreover, $f(0) = 0$ is easily checked. ($x=y=0$ on original equation)

Thus $$ f'(x) =\lim_{h\to0} \frac{f(x+h) - f(x)}{h} = \lim_{h\to0} \frac{f(\frac{h}{1-x^2 - xh}) - f(0)}{h} = \lim_{h\to0} \frac{f(\frac{h}{1-x^2 - xh}) - f(0)}{\frac{h}{1-x^2 - xh}} \frac{1}{1-x^2 - xh}\\=\frac{f'(0)}{1-x^2} $$

Of course, $h$ varies in the range $|h|<H\,$.

now, for $|x|<1$ , $$f(x) = f(x)-f(0) = \int_{0}^{x} f'(t) dt = f'(0) \int_{0}^{x} \frac{1}{1-t^2} dt = \frac{C}{2} (\, \ln|1+x| - \,\ln|1-x|),$$

Where $C\in\mathbb{R}$.