Agostarawz

2022-06-29

Numerical Analysis - show something about the rate of convergence
We are given an iterative method for finding roots, ${x}_{n+1}=g\left({x}_{n}\right)$, we are given the rate of convergence of this method is $p$, and also that:
$\underset{n\to \mathrm{\infty }}{lim}\frac{|{e}_{n+1}|}{|{e}_{n}{|}^{p}}=c$
where ${e}_{n}=|x-{x}_{n}|$ (I'm assuming that $x$ is the value to which ${x}_{k}$ is converging)
Show that
$\underset{n\to \mathrm{\infty }}{lim}\frac{log|{e}_{n+1}|}{log|{e}_{n}|}=p$
Here's what I did:
$\underset{n\to \mathrm{\infty }}{lim}\frac{|{e}_{n+1}|}{|{e}_{n}{|}^{p}}=c$ implies
$log\left(\underset{n\to \mathrm{\infty }}{lim}\frac{|{e}_{n+1}|}{|{e}_{n}{|}^{p}}\right)=log\left(c\right)$
I assumed that $log\left(\underset{n\to \mathrm{\infty }}{lim}\frac{|{e}_{n+1}|}{|{e}_{n}{|}^{p}}\right)=\underset{n\to \mathrm{\infty }}{lim}log\left(\frac{|{e}_{n+1}|}{|{e}_{n}{|}^{p}}\right)$ is this true? if so why?, but suppose it is.
then
$\underset{n\to \mathrm{\infty }}{lim}log\left(\frac{|{e}_{n+1}|}{|{e}_{n}{|}^{p}}\right)=log\left(c\right)$ implies that
$p=\underset{n\to \mathrm{\infty }}{lim}\frac{log|{e}_{n+1}|-log\left(c\right)}{log|{e}_{n}|}$
And here I am stuck. Unless $log\left(c\right)=0$ then I did not solve the question.

### Answer & Explanation

Jenna Farmer

Use that for any $c>\epsilon >0$ and $n\ge {n}_{\epsilon }$ large enough
$c-\epsilon <\frac{|{e}_{n+1}|}{|{e}_{n}{|}^{p}}
to get useful inequalities.
And use $\underset{n\to \mathrm{\infty }}{lim}\mathrm{ln}\left(|{e}_{n}|\right)=-\mathrm{\infty }$

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Recalculate according to your conditions!