taylor expansion - Nested radical $sqrt{x+sqrt{x^2+cdotssqrt{x^n+cdots}}}$

I am studying the $f(x) = \sqrt{x+\sqrt{x^2+\cdots\sqrt{x^n+\cdots}}}$ for $x \in (0,\infty)$ and I am trying to get closed form formula for this, or at least some useful series/expansion. Any ideas how to get there?

So far I've got only trivial values, which are
$$f(1)=\sqrt{1+\sqrt{1+\cdots\sqrt{1}}}=\frac{\sqrt{5}+1}{2}$$
$$f(4) = 3$$

Second one follows from
$$2^n+1 = \sqrt{4^n+(2^{n+1}+1)} = \sqrt{4^n+\sqrt{4^{n+1}+(2^{n+2}+1)}} = \sqrt{4^n+\sqrt{4^{n+1}+\cdots}}$$

I have managed to compute several derivatives in $x_0=1$ by using chain rule recursively on $f_n(x) = \sqrt{x^n + f_{n+1}(x)}$, namely:

\begin{align*}
f^{(1)}(1) &= \frac{\sqrt{5}+1}{5}\\
f^{(2)}(1) &= -\frac{2\sqrt{5}}{25}\\
f^{(3)}(1) &= \frac{6\sqrt{5}-150}{625}\\
f^{(4)}(1) &= \frac{1464\sqrt{5}+5376}{3125}\\
\end{align*}

These gave me Taylor expansion around $x_0=1$

\begin{align*}
T_4(x) &= \frac{\sqrt{5}+1}{2} + \frac{\sqrt{5}+1}{5} (x-1) - \frac{\sqrt{5}}{25} (x-1)^2 + \frac{6\sqrt{5}-150}{3750} (x-1)^3 \\
&\ \ \ \ \ + \frac{61\sqrt{5}+224}{3125} (x-1)^4
\end{align*}

However this approach seems to be useful only very closely to the $x=1$. I am looking for something more general in terms of any $x$, but with my limited arsenal I could not get much further than this. Any ideas?

This was inspiring but kind of stopped where I did

http://integralsandseries.prophpbb.com/topic168.html

Edit:
Thanks for the answers, i will need to go through them, looks like the main idea is to divide by $\sqrt{2x}$, so then I am getting
$$\frac{\sqrt{x+\sqrt{x^2+\cdots\sqrt{x^n+\cdots}}}}{\sqrt{2x}} = \sqrt{\frac{1}{2}+\sqrt{\frac{1}{4}+\sqrt{\frac{1}{16x}+\sqrt{\frac{1}{256x^4}+.‌​..}}}}$$
Then to make expansion from this. This is where I am not yet following how to get from this to final expansion.