Does the $x_{n+1}=2x_n^2-1$, $x_1=\frac45$ converge? The real sequence $x_n$  satisfies $x_{n+1}=2x_n^2-1$ and $x_1=\frac45$
I want to know if this sequence  convergences.
I think it converges to $-\frac12$; considering the graphs of $y=x$ and $y=2x^2-1$
Is it right? If so, how to prove that?
I tried to find a positive number $c$ $(0\lt c \lt1)$ such that $\forall n,\left | x_{n+2}-x_{n+1} \right | \leq c \left| x_{n+1}-x_{n} \right |$ which would prove this sequence is cauchy, but I failed.
 A: Let $f(x)=2x^2-1$.
If the sequence converges then the limit is a fixpoint of $f$, i.e. either $+1$ or $-\frac12$.
By induction, we have $|x_n|\le 1$ and $x_n\in\mathbb Q$ for all $n$.
One way of convergence would be that the sequence already lands on a fixed point after finitely many steps. But it could land on $+1$ only via $$ \tag1\ldots\to \pm\frac{\sqrt2}2\to 0\to -1\to1\to1\to\ldots$$
and it could land on $-\frac12$ only via
$$\tag2 \ldots\to\pm\frac{\sqrt{3}}2\to \frac12\to-\frac12\to-\frac12\to\ldots,$$
which is impossible as all $x_n$ are rational (and $x_1\notin\{-1,-\frac12,0,\frac12,1\}$).
Note that for $x\ne 1$ we have
$$\tag3 \frac{f(x)-1}{x-1}=2(x+1)>1 \quad\text{if }x>-\frac12$$
and for $x\ne-\frac12$
$$\tag4 \frac{f(x)+\frac12}{x+\frac12}=2x-1<-1 \quad\text{if }x<0.$$
We conclude that the sequence moves away from the fixpoints as soon as it is somewhat close to them. Hence no convergence.

Remarks: 


*

*We would have convergence with $x_1\in\{-1,-\frac12,0,\frac12,1\}$ and with any of the countably many irrational preimages that would extend $(1)$, $(2)$ to the left. As other hints suggest, these are the values $x_1=\cos(\alpha)$ such that $\cos(2^n\alpha)=1$ or
$\cos(2^n\alpha)=-\frac12$ for some $n\in\mathbb N_0$.

*The repelling property $(3)$ and $(4)$ is a consequence of $|f'(x)|=|4x|>1$ for the fixpoints $x=1$ and $x= -\frac12$.

A: Hint,Note
$$\cos{(2x)}=2\cos^2{x}-1$$
