Bonjour,
Find all continuous functions, $f$, such that $f(x)-1999f\big(\frac{2x}{1-x^2}\big)=18$ for $|x|\neq 1$.
My try: taking $x=\tan{h}$ leads to: $f(\tan{4h})-f(\tan{2h})=\frac{1}{1999}(f(\tan{2h})-f(\tan{h}))$. So by induction: $f(\tan{h})-f(\tan{\frac{h}{2}})=\frac{1}{1999^n}(f(\tan{\frac{h}{2^n}})-f(\tan{\frac{h}{2^{n+1}}}))$ for $n$ positive. Using continuity and limit we get: $f(\tan{h})=f(\tan{h/2})$. With same argument as before we end showing that $f(x)=f(0)$. Since $f(0)=-\frac{1}{111}$ therefore $f(x)=-\frac{1}{111}$
$f(x)-1999f\left(\dfrac{2x}{1-x^2}\right)=18$
$f(\tan x)-1999f\left(\dfrac{2\tan x}{1-\tan^2x}\right)=18$
$f(\tan x)-1999f(\tan2x)=18$
$1999f(\tan2^{x+1})-f(\tan2^x)=-18$
$f(\tan2^x)=1999^{-x}\Theta(x)-\dfrac{18x-9}{1999}$ , where $\Theta(x)$ is an arbitrary periodic function with unit period (according to http://eqworld.ipmnet.ru/en/solutions/fe/fe1102.pdf)
$f(x)=1999^{-\log_2\tan^{-1}x}\Theta(\log_2\tan^{-1}x)-\dfrac{18\log_2\tan^{-1}x-9}{1999}$ , where $\Theta(x)$ is an arbitrary periodic function with unit period