Contraction mapping of $f(x) = 2 + x^{-2}$ on $[2,\infty)$
Describe how this solution may be obtained by an iterative procedure, giving a bound for the error at the $n$’th stage in terms of the initial point of the iteration. If the initial guess is $x_0 = 2$, what is the first value of $n$ for which the iteration guarantees that $x_n$ approximates the fixed point to $3$ decimal places?
I have the solution but I'm having difficulty understanding it.
The contraction constant is $1/4$ by using derivatives and mean value theorem since $|f'(x)|<1/4$
Now we iterate. Let $x_0=2$ then $x_1=f(x_0)=2+1/4$ , $x_2=f(x_1)=2+16/81$.
If we let $a$ to be the fixed point, $$|a-x_n|\leq \frac{1}{4}^n\left(1-\frac{1}{4}\right)|x_1-x_0|$$
How do they achieve this? I managed to only show $$f(x_n)-f(x_{n-1})\leq \left(\frac{1}{4}\right)^n |x_1-x_0|$$
Start with $$ |x_n-a| = |f(x_{n-1})-f(a)|\leq \frac14 |x_{n-1}-a| \tag1 $$ Apply repeatedly to get
$$ |x_n-a| = \frac1{4^{n}} |x_0-a| \tag2 $$ Then estimate $|x_0-a|$ in terms of $|x_1-x_0|$, using the fact that $\lim_{n\to\infty} x_n = a$: $$ |x_0-a| \le \sum_{n=1}^\infty |x_n-x_{n-1}| \le \sum_{n=1}^\infty \frac1{4^{n-1}} |x_1-x_0| =\frac{1}{1-1/4}|x_1-x_0|\tag3 $$ Combine (2) and (3).