TITLE: Approximating an iteratively defined function
QUESTION [6 upvotes]: Let $f_0,f_1,\ldots$ be a sequence of functions $f_n : [0,1] \rightarrow R$ defined as follows:
$$f_0(x) =1+2x$$
$$f_{n}(x) := \left\{\frac{5+t}{2} :  \text{ where t solves } f_{n-1}\left(\frac{x}{t}\right) = \frac{5+t}{2}\right  \}$$
In other words, $f_n(x)$ is equal to the intercept (with respect to $y$) of the functions $f_{n-1}(x/t)$ and $\frac{5+t}{2}$. 
It isn't hard to prove by induction that each $f_n$ (for $n>0$) will be a continuous increasing function such that $f_n(0)=5/2$ and $f_n(1)=3$. If we fix $0<x<1$ and consider the function $f_{n-1}(x/t)$ as a function of $t$, we see that it is decreasing as $t$ increases and $f_{n-1}(x/t)=3$ for $t=x$, and  $f_{n-1}(x/t) <3$ for $y=1$. From this it follows that there is a unique intercept and that $f_n$ is well defined.
One can show that:
$$f_1(x)= \frac{1}{4}(\sqrt{16x+9}+7),$$
$$f_2(x) = \{\frac{5+t}{2} : \text{ where t solves } x = (t^3+3t^2)/4 \}.$$


Is there a method for finding a good approximation (upper and lower bounds) by elementary functions of the $n$-th iterate $f_n(x)$ in a neighborhood of $x=1$? 


I'm interested in both a solution to this particular problem, as well as understanding methods that work in similar situations.

REPLY [3 votes]: From your implicit iteration, $f_n(x)=f_{n-1}\big(\frac{x}{2f_n(x)-5}\big)$, it follows that $f_n$ is the inverse function to the function $g_n:[5/2,3]\to[0,1]$ defined by $g_n(y):=\frac{1}{2}(y-1)(2y-5)^n$.  
If we put  $z=z_n(x):=(2f_n(x)-5)/3$   the above relation writes $z(1+z)^{1/n}=4^{\frac{1}{n}} 3^{-\frac{n+1}{n}} x^{\frac{1}{n} }$.
The local inverse at $0$ of $z\mapsto z(1+z)^{1/n}$ is an analytic function whose expansion at $z=0$ is
$$F_n(z):=\sum_{k=1}^\infty\frac{1}{k}{-\frac{k}{n}\choose k-1}z^k\; ,$$
so we have $$f_n(x)=\frac{5}{2}+\frac{3}{2}F_n\big( 4^{\frac{1}{n}} 3^{-\frac{n+1}{n}} x^{\frac{1}{n} }\big)  \; .$$ 
It should be easy (I did not try) to compute the radius of convergence of the above power series expansion, to check whether it also cover $x=1$. For $n=1$ the radius of convergence is $1/4$, so it does not cover $F_n(4/9)$, needed for $f_n(1)$, but for larger $n$ it could be better.