Playing with my own question Got a factored version of the Taylor's series? I found that :
Define :
$$f(x)=\frac{1}{\sqrt{x-2/x}}$$
Then it seems we have :
$$\phi=f(f(\cdots f(\phi)\cdots))$$
Numerical checking is agree so some question :
edit :
I'm really interested by an generalization the next polynomial are :
$$x^4-2x^3+2x,x^5-3x^3+2x,...$$
How to (dis)prove it ? How fast is the convergence if true ? Is there a generalization if true ?
Let $f(x)=(x-2/x)^{-1/2}$, define $f_0=f$ and $f_n=f\circ f_{n-1}$.
Note that we do have $f_n(\phi)\to\phi$ simply because $f(\phi)=\phi$, i.e. $f_n(\phi)$ is a constant sequence. The more interesting question would be
$$\text{does $f_n(x)\to\phi$ for $x\neq\phi$ ?}$$
A simple find is $\psi=1-\sqrt5$ since $f(\psi)=\phi$. For all other $x\neq\phi$ or $\psi$, either $f_n$ diverges or $f_n$ is not well-defined. To see this, note that if $x\geq0$, then $f_2(x)$ will not be well-defined for $x\leq\sqrt2$ or $x\geq\sqrt3$, hence we shall only focus on $x\in(\sqrt2,\sqrt3)$. On this range, define the difference $$g(x)=|f(x)-\phi|-|x-\phi|=\big(2\phi-f(x)-x\big)\text{sgn}(x-\phi).$$ It is not hard to verify $g(x)>0$ for all $x\in(\sqrt2,\sqrt3)$ with $x\neq\phi$. This means every time we apply $f$, the distance to $\phi$ becomes larger and thus $f_n$ will diverge. To summarise, for positive $x\neq\phi$ either $f_n$ diverges or $f_n$ is not well-defined. As for all negative $x\neq1-\sqrt5$, we have $f(x)\geq0$ and $f(x)\neq\phi$, therefore the same argument for positive $x$ applies.
In addition, note that $\phi$ is the only possible limit point for $f_n$, since a limit point $x$ must satisfy $f(x)=x$ which yields only one solution $x=\phi$.