$x_{n+1}=x_n+\cos\sqrt{|x_n|}$,How to prove $x_n$ is convergent? For iterative sequence, $x_{n+1}=f(x_n)=x_n+\cos\sqrt{|x_n|}$，$n=0,1,2\cdots$,prove:
$(1)$For any initial value $x_0\in \mathbb{R}$,the sequence $\{x_n\}$is convergent.
$(2)$Denote $\lim_{n\to \infty}x_n=x^*$,then there exists a constant $C>0$,such that $|x_n-x^*|\leqslant C|f'(x^*)|^n$.
My thought:It is easy to see that when $x\neq 0$,
$f'(x)=1-\frac{\sin \sqrt{|x|}}{2\sqrt{|x|}}\mathrm{sgn} x$ and $f(x)$ is an increasing function.I tried contraction mapping principle,but it seems that if $x_n$ is large enough,there is no way to prove the existence of the constant M,such that $|f'(x)|<M<1$.Besides,it is natural to try to prove that $x_n$ is a Cauchy sequence,but still I don't know how to proceed.Can anyone help me ?
 A: $(1)$When$x\neq 0$,obviously\begin{align*}f'(x)=1-\frac{\sin \sqrt{|x|}}{2\sqrt{|x|}}\mathrm{sgn} x>0,\end{align*}Thus $f(x)$is strictly increasing on $\mathbb{R}$.Consider the roots of $\cos \sqrt{|x|}$,we get $|x|=\left(k\pi+\frac{\pi}{2}\right)^2$,$k\in \mathbb{Z}$.For the convience of our discussion,we denote $E=\{x\in \mathbb{R}:|x|=\left(k\pi+\frac{\pi}{2}\right)^2,k\in \mathbb{Z}\}$.Then $\forall x\in E$,$f(x)=x$.We discuss the value of $x_0\in \mathbb{R}$as follows.
$1^\circ$When$x_0\in E$,$x_1=f(x_0)=x_0$,$x_2=f(x_1)=x_1=x_0$,$\cdots$,$x_n=x_0$,$\forall n\in \mathbb{N}$.In this case $\{x_n\}$is a constant sequence,thus it must be convergent.
$2^\circ$When$x_0\not\in E$,there must exists $s,t\in E$,$(s,t)\cap E=\varnothing$,and$x_0\in (s,t) $.By the strict increasing fact of $f$,we know that $f(s)<f(x_0)<f(t)$,or $s<x_1<t$.And then $s=f(s)<f(x_1)<f(t)=t$.Notice that $(s,t)\cap E=\varnothing$,we get$x_1\not\in E$,or$f(x_1)\neq x_1$.
$(\mathrm{i})$If $f(x_1)>x_1$,i.e.$x_2>x_1$.By the fact that $s<x_1<x_2<t$and the fact that $f$ is strict increasing,we can deduce that$f(s)<f(x_1)<f(x_2)<f(t)$,or$s<x_2<x_3<t$.Using the strict growth of $f$ repeatedly,we get\begin{align*}s<x_1<x_2<\cdots<x_n<\cdots <t.\end{align*}It indicates that$\{x_n\}$is strict increasing,and has upper bound $t$.By monotone convergence theorem,we can see that $\{x_n\}$must be convergent.
$(\mathrm{ii})$If$f(x_1)<x_1$,i.e.$x_2<x_1$.Similar to the discussion of $(\mathrm{i})$,we can prove that$\{x_n\}$is strict decreasing,and has lower bound $s$.By monotone convergence theorem again,we can see that $\{x_n\}$must be convergent.
Question $(1)$is solved,then how about $(2)$?How can I do it?
