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.

Does it exist in the elementary functions? Does it exist elsewhere? Please help!

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    $\begingroup$ Have you tried Taylor series ? $\endgroup$ – Elad May 20 '17 at 20:59
  • $\begingroup$ Not sure how to use Taylor series to solve this... if you know, can you give me a hint as to how I might use them? $\endgroup$ – Frpzzd May 20 '17 at 21:05
  • $\begingroup$ I don't know it looks a bit like cushy furmula. But first try to take the derivative of both side and on the right hand side you will still have f(x). Then maybe with some algebra it will take you further. $\endgroup$ – Elad May 20 '17 at 21:10
  • $\begingroup$ 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 $(2^n-1)a_n=a_{n-1}$ $\endgroup$ – Hugh the Thistle May 20 '17 at 21:10
  • $\begingroup$ 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 $(-1)2^{-n}$. $\endgroup$ – 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 $$ f(x)=\exp\sum_{m\geq 1}\frac{(-1)^{m+1}x^m}{m(2^m-1)}. $$

  • $\begingroup$ Why assuming : $f(0) = 1$ ? Moreover isn't there a link between this FE and the $\Gamma$ function ? $\endgroup$ – J. OK May 29 '17 at 21:54
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    $\begingroup$ @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. $\endgroup$ – Jack D'Aurizio May 29 '17 at 22:08

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