# Show that, if $g(x)=\frac{1}{1+x^2}$ and $x_{n+1}:=g(x_n)$ then $x_n$ converges

Why does the sequence $\{x_n\}$ converge ?

If $x_{n+1}:=g(x_n)$, where $g(x)=\frac{1}{1+x^2}$

(We have a startpoint in $[0.5,1]$)

The sequence is bounded by $1$ independant of the startpoint (Is it necessary that $x_0\in[0.5,1]$ ?)

We have to show that the sequence is Cauchy

I compare $g(x_{n+1})$ with $g(x_n)$

$|g(x_{n+1})-g(x_n)|=|\frac{1}{1+(\frac{1}{1+x_n^2})^2}-\frac{1}{1+x_n^2}|=|\frac{x_n^4+2x_n^2+1}{x_n^4+2x_n^2+2}-\frac{1}{1+x_n^2}|$

Would this lead to an impasse, or how to continue ?

• Have you calculated $g([0.5,1])$? Is it the entire range or has it shrunk? Will it be shrunk further on every iteration? What is the only point that is mapped to itself? Is the interval shrinking towards that point? – Eric Towers Mar 19 '14 at 12:31
• @Eric Towers actually the task is to find roots of $x^3+x-1$. Since, $x^3+x-1=0\Leftrightarrow \frac{1}{1+x^2}=x$ i am searching for a fixed point now. – OBDA Mar 19 '14 at 12:34
• Have you plotted $\frac{1}{1+x^2}$ (on the same axes as $x$) to help your intuition? This plot makes it pretty clear that there is one fixed point, why there is only one, and why $g$ is a contraction mapping. – Eric Towers Mar 19 '14 at 12:36
• @EricTowers ah it is sufficient if $g$ is a contraction, and maps from a closed set, say $D$ to $D$. – OBDA Mar 19 '14 at 12:41

The function $g$ is monotone on the interval $J:=\bigl[{1\over2},1\bigr]$, and from $g\bigl({1\over2}\bigr)={4\over5}$, $\>g(1)={1\over2}$ it follows that $g(J)\subset J$.
When $x\in J$ then $$|g'(x)|={2x\over (1+x^2)^2}\leq{2x\over 1+2x^2}={1\over\sqrt{2}}\>{2\over{1\over \sqrt{2}\, x}+\sqrt{2}\,x}\leq{1\over\sqrt{2}}\ .$$ It follows that $g$ is contracting on $J$. Therefore by the general fixed point theorem $g$ has a unique fixed point $\xi\in J$, and any sequence $(x_n)_{n\geq0}$ of the considered kind will converge to $\xi$.
Solving the equation $g(x)=x$ numerically gives $\xi\doteq0.682328$.