# If $x_0$ is sufficiently close to $z$, the sequence ${x_n}$ converges to $z$

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 $$x_0 and $$\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,

• 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! Oct 18 at 16:26
• 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 .... Oct 18 at 17:20
• 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 $4\alpha<1$. Oct 18 at 17:44

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.