# Find the function $f$ such that $f(x)=\frac{f(2x)}{x+1}$.

I've been trying to find a function with the property $$f(x)=\frac{f(2x)}{x+1}$$ using elementary functions only, and it has proven to be harder than I thought.

Does anyone know how to go about finding a function given one of its properties in a systematic manner?

I can already guess that this function will use a logarithm (base 2) somewhere in it, but I still can't find the function.

• using Elad's idea : if $f(x)$ can be written as $\sum \limits_{n=0}^{+\infty} a_n x^n$ then for all $n \geq 1$ one has $(2^n-1)a_n=a_{n-1}$ – Hugh the Thistle May 20 '17 at 21:10
• a function? Or all functions? $f(x) = 0;i \ne (-1)2^n$ is of course the simplest. $f(-1)$ is undefined so so must be $(-1)2^{-n}$. – fleablood May 26 '17 at 21:26
If we assume $f(0)=1$ we have that a solution of $$f(x)= \left(1+\frac{x}{2}\right)\cdot f\left(\frac{x}{2}\right)$$ over the interval $x\in(-2,2)$ is given by \begin{align} f(x)&=\prod_{n\geq 1}\left(1+\frac{x}{2^n}\right) \\ &=\exp\sum_{n\geq 1}\log\left(1+\frac{x}{2^n}\right) \\ &=\exp\sum_{n\geq 1}\sum_{m\geq 1}\frac{(-1)^{m+1}x^m}{m2^{mn}}, \end{align} i.e., by $$f(x)=\exp\sum_{m\geq 1}\frac{(-1)^{m+1}x^m}{m(2^m-1)}.$$
• Why assuming : $f(0) = 1$ ? Moreover isn't there a link between this FE and the $\Gamma$ function ? – J. OK May 29 '17 at 21:54
• @J.OK: if you assume something else you get the same solution up to a constant. I do not see any clear connection beween this function and the $\Gamma$ function. – Jack D'Aurizio May 29 '17 at 22:08