Agostarawz

2022-06-29

Numerical Analysis - show something about the rate of convergence

We are given an iterative method for finding roots, ${x}_{n+1}=g({x}_{n})$, 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

$log(\underset{n\to \mathrm{\infty}}{lim}\frac{|{e}_{n+1}|}{|{e}_{n}{|}^{p}})=log(c)$

I assumed that $log(\underset{n\to \mathrm{\infty}}{lim}\frac{|{e}_{n+1}|}{|{e}_{n}{|}^{p}})=\underset{n\to \mathrm{\infty}}{lim}log(\frac{|{e}_{n+1}|}{|{e}_{n}{|}^{p}})$ is this true? if so why?, but suppose it is.

then

$\underset{n\to \mathrm{\infty}}{lim}log(\frac{|{e}_{n+1}|}{|{e}_{n}{|}^{p}})=log(c)$ implies that

$p=\underset{n\to \mathrm{\infty}}{lim}\frac{log|{e}_{n+1}|-log(c)}{log|{e}_{n}|}$

And here I am stuck. Unless $log(c)=0$ then I did not solve the question.

We are given an iterative method for finding roots, ${x}_{n+1}=g({x}_{n})$, 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

$log(\underset{n\to \mathrm{\infty}}{lim}\frac{|{e}_{n+1}|}{|{e}_{n}{|}^{p}})=log(c)$

I assumed that $log(\underset{n\to \mathrm{\infty}}{lim}\frac{|{e}_{n+1}|}{|{e}_{n}{|}^{p}})=\underset{n\to \mathrm{\infty}}{lim}log(\frac{|{e}_{n+1}|}{|{e}_{n}{|}^{p}})$ is this true? if so why?, but suppose it is.

then

$\underset{n\to \mathrm{\infty}}{lim}log(\frac{|{e}_{n+1}|}{|{e}_{n}{|}^{p}})=log(c)$ implies that

$p=\underset{n\to \mathrm{\infty}}{lim}\frac{log|{e}_{n+1}|-log(c)}{log|{e}_{n}|}$

And here I am stuck. Unless $log(c)=0$ then I did not solve the question.

Jenna Farmer

Beginner2022-06-30Added 17 answers

Use that for any $c>\epsilon >0$ and $n\ge {n}_{\epsilon}$ large enough

$c-\epsilon <\frac{|{e}_{n+1}|}{|{e}_{n}{|}^{p}}<c+\epsilon $

to get useful inequalities.

And use $\underset{n\to \mathrm{\infty}}{lim}\mathrm{ln}(|{e}_{n}|)=-\mathrm{\infty}$

$c-\epsilon <\frac{|{e}_{n+1}|}{|{e}_{n}{|}^{p}}<c+\epsilon $

to get useful inequalities.

And use $\underset{n\to \mathrm{\infty}}{lim}\mathrm{ln}(|{e}_{n}|)=-\mathrm{\infty}$

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