TITLE: Showing that the sequence $x_n=\frac{1}{3}x_{n-1}(4+x_{n-1}^3)$ where $x_0=-0.5$ quadratically converges
QUESTION [5 upvotes]: I am stuck at a point in solving this problem:

Show that the sequence defined by:
For all $$n\in\mathbb{N}, 
x_n =
\begin{cases}
-\frac{1}{2},  & \text{if $n=0$} \\
x_n=\frac{1}{3}x_{n-1}(4+x_{n-1}^3), & \text{if $n>0$}
\end{cases}\,\,\,$$ quadratically converges
In other words:
Show that there exist some positive constant $\lambda$ such that $\lim_{n\to \infty}\frac{|x_{n+1}-p|}{|x_n-p|^2}=\lambda$ where $$p=\lim_{n\to \infty}x_{n}.$$

(Note: I've got this sequence by applying Newton's method to the function $f(x)=\frac{1}{x^3}+1$ on the interval $[-2,0)$ with initial approximation $x_0=-\frac{1}{2}$)
I was able to show that $p=\lim_{n\to \infty}x_n=-1$.
But when I try to calculate $\lim_{n\to \infty}\frac{|x_{n+1}-p|}{|x_n-p|^2}$ I got stuck.
Thanks for any hint/help.

REPLY [4 votes]: You may notice that you are just applying Newton's method to $f(x)=1+\frac{1}{x^3}$, since:
$$ x-\frac{f(x)}{f'(x)}  = \frac{x(4+x^3)}{3}.$$
We have that $f(x)$ is a decreasing and concave function on $\mathbb{R}^-$, with a simple zero at $x=-1$, hence quadratic (and monotonic) convergence is ensured by the properties of Netwon's method:
$$ \forall x\in(-1,0),\qquad x-\frac{f(x)}{f'(x)}\leq -1+2(x+1)^2 $$
implies:
$$ 0\leq (x_{n+1}+1) \leq 2(x_n+1)^2. $$