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

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 $C\in\mathbb{R}$.

Answer 2

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.