Hi i was wondering something about the great Ramanujan :
I think moreover I'm not the only one who propose this kind of problem (so if you have a link related to this subject).
We have : $$3=\sqrt{1+2\sqrt{1+3\sqrt{1+\cdots}}}$$
Now the problem :
Let $f(x)$ be a positive,continuous and differentiable on function$(0,\infty)$ and non-constant then the functional equation : $$3=\sqrt{1+f(2)\sqrt{1+f(3)\sqrt{1+\cdots}}}$$ Have as unique solution $f(x)=x$
I have tried to build a counter-example without success .First it seems obvious for strictly increasing\decreasing functions such that $f(x)>x$ or $f(x)<x$ . I would like to create an counter-example of the form : $$f(x)=x+g(x)$$
Where $g(x)$ is a periodic function .
I have tried more general representation without success .
If it's true we can see how Ramanujan was great .
Any helps are highly appreciated .
Thanks a lot .
Define $f$ and $g$ as $$f\colon (0,\infty)\to\mathbb{R},\;x\mapsto x,$$ and $$g\colon (0,\infty)\to\mathbb{R},\;g(x)=0\;for\;x\in(0,2]\;and\;$$ $$g(x)=cos(2\pi x)-1\;for \;x\in(2,\infty).$$ Define $h$ as $$h\colon (0,\infty)\to\mathbb{R},\;x\mapsto g(x)+f(x).$$ $h$ is non-constant. For all $x\in(0,2]$, it holds, that $h(x)=f(x)\gt0$. For all $x\in(2,\infty)$, it holds, that $f(x)\gt2$ and $g(x)\ge-2$. Hence $h(x)\gt0$ for all $x\in(0,\infty)$. For all $x\in\mathbb N$ follows $g(x)=0$. $f$ and $g$ are differentiable over $(0,2)$ and $(2,\infty)$. Look at $$\lim_{k\to 0-0} \frac{g(2+k)-g(2)}{k}=\lim_{k\to 0-0} \frac{0-0}{k}=0$$ and $$\lim_{k\to 0+0} \frac{g(2+k)-g(2)}{k}=\lim_{k\to 0+0} \frac{cos(2\pi (2+k))-1-cos(2\pi\cdot2) +1}{k}=-2\pi sin(2\pi\cdot2)=0$$ and notice, that $g$ is differentiable at $x=2$. It follows, that $h=f+g$ is also differentiable as a sum of two differentiable functions. Hence there exists a positive, non-constant, differentiable function $h\neq f$, such that $$3=\sqrt{1+h(2)\sqrt{1+h(3)\sqrt{1+\cdots}}}.$$ Therefore this conjecture does not hold.