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,

  • 1
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.