question –
Find all continuous functions $f: \mathbb{R} \rightarrow \mathbb{R}$ which satisfy the equation
$$
f(x+y)=\frac{f(x)+f(y)+2 f(x) f(y)}{1-f(x) f(y)}
$$
for all $x, y$
my try –
i proved that f(0)=0 ..then hint says substitute $g(x)=f(x) /(1+f(x))$..
i am not getting how to use this hint…i substitute in original equation and simplified lots of thing but nothing seems useful..
any hints ??? …and how one can thought that we have to substitute this…
Best Answer
Using the given hint, we get, $$g(x)=\dfrac{f(x)}{1+f(x)}\\ \implies f(x)=\dfrac{g(x)}{1-g(x)}\\ \implies \dfrac{g(x+y)}{1-g(x+y)}=\dfrac{\dfrac{g(x)}{1-g(x)}+\dfrac{g(y)}{1-g(y)}+\dfrac{2g(x)g(y)}{(1-g(x))(1-g(y))}}{1-\dfrac{g(x)g(y)}{(1-g(x))(1-g(y))}}\\ \implies \dfrac{g(x+y)}{1-g(x+y)}=\dfrac{g(x)(1-g(y))+g(y)(1-g(x))+2g(x)g(y)}{1-g(x)-g(y)}\\ \implies\dfrac{g(x+y)}{1-g(x+y)}=\dfrac{g(x)+g(y)}{1-g(x)-g(y)}\\ \implies \dfrac{g(x+y)}{1-g(x+y)}+1=\dfrac{g(x)+g(y)}{1-g(x)-g(y)}+1\\ \implies \dfrac{1}{1-g(x+y)}=\dfrac{1}{1-g(x)-g(y)}\\ \implies \boxed{g(x+y)=g(x)+g(y)}$$
Now use this.