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I have some troubles with the following problem:

Suppose for any $x_0$ in reals, the sequence$\{x_n\}_{n=0}^\infty$ satisfies the inequality $|x_{n+1}−z|≤4|x_n−z|^2$ for n= 0,1,2,...Thus if $x_0$ is sufficiently close to $z$, the sequence ${x_n}$ converges to $z$. Find the largest value of $\alpha$ such that if $x_0\in(z−\alpha,z+\alpha)$, then the sequence converges. Prove that if $x_0\in(z−\alpha,z+\alpha)$ then the sequence converges.

The idea: WLOG $\alpha > 0$ $z > 0$,

then $x_0<z+\alpha$ and $\frac{|x_{n+1}−z|}{4|x_n−z|^2}≤1$

if $g(x_n)=x_{n+1}$, then by Taylor's expansion: $g(x_n)=g(z)+g'(z)(z-x_n)+\frac{g''(z)}{2}(z-x_n)^2+\frac{g'''(\xi)}{2}(z-x_n)^3$ for $\xi$ in between $x_n$ and $z$. Since $z$ is a fixed point of the convergence,

$x_{n+1}=z+g'(z)(z-x_n)+\frac{g''(z)}{2}(z-x_n)^2+\frac{g'''(\xi)}{2}(z-x_n)^3$. Here I am not sure but $g'(z)=0$ , $g''(z)\neq 0$ and $g'''(z)=0$ since it seems we have a quadratic convergence (am I right?).

Then $x_{n+1}-z=\frac{g''(z)}{2}(z-x_n)^2$ and

$\frac{|x_{n+1}−z|}{|x_n−z|^2}≤|\frac{g''(z)}{2}|$ So $\frac{g''(z)}{2}=4$

But this leads me nowhere. Isn't this suppose to be less than 1 to converge? Also I am not sure how to get $\alpha$ from here. It seems like it should be easier:

To converge $|x_{1}−z|≤4|z+\alpha−z|^2<1$ so $\alpha<1/2$ but this may not converge in the next iterations.

Thanks and Regards,

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    $\begingroup$ There is no guarantee that you can Taylor expand $g(x)$ around $x = z$. eg. $x_n$ generated by the map $g(x) = z -4(x-z)|x-z|$ satisfies the inequality while $g'(z)$ doesn't exist. In fact, same function shows that maximum $\alpha \le \frac14$ as $x_0 = z + \frac14 \implies x_n = z + \frac{(-1)^n}{4}$ is a period 2 cycle! $\endgroup$ Oct 18 at 16:26
  • $\begingroup$ Thank you, for clarifying it. I realized my calculation was wrong in the previous comment and I deleted it. Thus to converge $\alpha<1/4$. I cannot calculate the maximum $\alpha≤1/4$. I tried $𝑔(𝑧+\alpha)=𝑧−4\alpha^2$ so the maximum .... $\endgroup$
    – Mihai.Mehe
    Oct 18 at 17:20
  • $\begingroup$ I finally seem to have figure it out (definitely maybe): $x_1<x_0$ so $x1=g(x_0)=z+4\alpha^2<x_0=z+\alpha$ so $4\alpha<1$. $\endgroup$
    – Mihai.Mehe
    Oct 18 at 17:44
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If you have some recursive inequality $|e_{n+1}|\le |e_n|^2$, you get $$ |e_n|\le |e_0|^{2^n}. $$ This converges to zero if $|e_0|<1$. This is quadratic convergence.

You can easily rescale the given inequality into this form.

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