Correct recursion for $|4-x_{n+1}|Concerning my question here, let $(x_n)_{n\in \mathbb N}$ be a sequence such that $$|4-x_{n+1}|<q |4-x_n|^2, \, q>0$$
I am looking for the correction recursion $$|4-x_{n+1}|<q^a|4-x_1|^b$$
(So I want to find $a$ and $b$). I tried the following but apparently it's not correct:
$|4-x_{n+1}|<q |4-x_n|^2<q\Big|q|4-x_{n-1}|^2\Big|^2=q^3|4-x_{n-1}|^4<...<q^{2^n-1}|4-x_1|^{2^n}$
 A: To simplify a chain of equations/inequalities/recurrence relations, it is usually a good idea to perform a change of variable to make the core part of the relations depend on as few parameters as possible.
In this case, let $u_n = q(4 - x_n)$, one will find $x_n$ converges to a limit iff $u_n$ converges to some other limit. Furthermore, the inequalities are simplified:
$$|4 - x_{n+1}| < q|4 - x_n|^2\quad\iff\quad |u_{n+1}| < |u_n|^2$$
From this, we can deduce
$$|u_{n+1}| 
< |u_{\color{red}{n}}|^{2^\color{red}{1}} 
< |u_{\color{green}{n-1}}|^{2^\color{green}{2}}
< |u_{\color{blue}{n-2}}|^{2^\color{blue}{3}}
< \cdots 
< |u_\color{magenta}{1}|^{2^\color{magenta}{N}}$$
It is clear the last term is $|u_1|$ raised to power $2^N$ for some integer $N$.
The problem is it may not be immediately obvious what $N$ should be. If that happens, a trick is look for invariant quantities in the intermediate terms.
Notice 


*

*$\color{red}{n + 1} = n+1$,

*$\color{green}{n-1 + 2} = n+1$,

*$\color{blue}{n-2 + 3} = n+1$
The sum of subscript of $u$ and the power of $2$ is an invariant and equals to $n+1$.  
This means $\color{magenta}{N + 1} = n + 1 \implies N = n$. As a result,
$$\begin{align} |u_{n+1}| < |u_1|^{2^n}
\iff & 
q|4-x_{n+1}| < |q(4-x_1)|^{2^n}\\
\iff &
|4-x_{n+1}| < q^{2^n-1} |4-x_1|^{2^n}\end{align}$$
What you have is correct.
