I've been studying the function \(\displaystyle \rho (x) = \sqrt{x+\sqrt{x^2+\sqrt{x^3+...}}}\), defined on \(\displaystyle x=[0, \infty]\), and it has proved itself very uncooperative. So far I've gotten some pretty basic bounds; \(\displaystyle \sqrt{2x}\leq \rho(x)\leq \sqrt{3x}\) which works for all values of x bigger than one I think.

Additionally, I've managed to get a couple explicit values for \(\displaystyle \rho(x)\). \(\displaystyle \rho(0)\) is trivially equal to 0, and \(\displaystyle \rho(1)=\phi\) is a relatively famous result as well (in the context of the golden ratio itself). I've also managed to discover a value of \(\displaystyle \rho'(x)\), the derivative of the function in question. If we use the chain rule recursively, we get the expansion

\(\displaystyle \rho'(x)=\frac{1}{2\sqrt{x+\sqrt{x^2+...}}}\left ( 1+\frac{1}{2\sqrt{x^2+\sqrt{x^3+...}}}\left ( 2x+\frac{1}{2\sqrt{x^3+\sqrt{x^4+...}}}\left ( 3x^2+... \right ) \right ) \right )\).

Plugging in x=1 and multiplying out the terms, we can see that \(\displaystyle \rho'(1)=\sum_{k=0}^{\infty}k(2\phi)^{-k} = \frac{1+\sqrt5}{5}\).

That's all I've got though. Anybody else here think they can make any headway? :0

EDIT: sorry, it looks like the brackets get a bit messy when this transfers over to the board. I hope that part with the derivative isn't too hard to read!