Showing a sequence $x_{n+1} = Tx_n$ forms a contraction mapping

I want to show that the sequence given by $$x_{n+1} = Tx_n = x_n-\frac{(x_n^2-2)}{x_n+x_{n-1}}$$ forms a contraction mapping. That is $$|Tx_1-Tx_2|\leq c|x_1-x_2|.$$ Where $c$ is to be determined. I am also unsure of what the conditions for convergence are. So far I have

\begin{align*}|Tx_1-Tx_2| &= \bigg|(x_1-\frac{(x_1^2-2)}{x_1+x_0})-(x_2-\frac{(x_2^2-2)}{x_2+x_1})\bigg| \\ &= \bigg|\frac{(x_1^2-2)(x_0-x_2)}{(x_1+x_0)(x_1+x_2)}\bigg| \end{align*} and I am not really sure where to go from here.

I have also thought about $$|Tx_1-Tx_2| = \bigg|(x_1-x_2)-\frac{(x_1^2-2)}{x_1+x_0}+\frac{(x_2^2-2)}{x_2+x_1}\bigg|$$ and factoring out the $(x_1-x_2)$ term but I am also unsure of where to go from there.

I should note that this sequence comes from the secant method with $f(x)=x^2-2$. Would it be better to show that the secant method forms a contraction mapping? If so how would one do that?

Any help and comments would be appreciated. Thank you.

• One problem: your 'T' as you've written it isn't strictly an operator (function) on individual $x$, but rather on the sequence $\bar{x}\equiv\{x_i\}$, because of the dependence on $x_{n-1}$ listed there. It's not even immediately clear to me what's meant by 'being a contraction mapping' in this context, because one can't speak of $|Ta-Tb|$ for arbitrary $a,b\in \mathbb{R}$. – Steven Stadnicki Oct 23 '13 at 0:09
• I would try eliminating $x_2$ from both sides by writing $x_2 = Tx_1$ and then simplifying. – user7530 Oct 23 '13 at 0:09
• I guess what I mean by contraction mapping is that $$|x_{n+1}-x_n|\leq c|x_n-x_{n-1}|$$ where $0\leq c <1$. – RDizzl3 Oct 23 '13 at 0:17
• It is not true that $|x_3- x_2|\le c|x_1-x_2|$, no matter what (fixed) $c$ you take. As an example, take $x_1=-2$ and $x_2=2-\epsilon$ where $\epsilon$ is small. The secant will be nearly horizontal, and therefore $x_3$ will be huge. It is true that things get better after that, but the failure of the contractive property at the beginning shows that your algebraic manipulation could not possibly work. – user103402 Nov 18 '13 at 6:30