Order of convergence of fixed-point to Newton's method

Hi,

In a problem I first show that the order of convergence of simple iteration is 1 and that in order for it to converge I need $\displaystyle |g(x)|<1$ for all $\displaystyle x$.

From this I must show that Newton's method has an order of convergence of 2. I usually show this by starting with a Taylor expansion etc.

I'm thinking that I could perhaps start by saying that fixed-point iteration will converge if $\displaystyle |g'(x)|<1$ for all $\displaystyle x$ and since Newton's method has that ,

$\displaystyle |g(x)|=\bigg|x-\frac{f(x)}{f'(x)}\bigg|,$

I get,

$\displaystyle |g'(x)| = \bigg|1 - \frac{[f'(x)]^2-f(x)f''(x)}{[f'(x)]^2}\bigg|=\bigg|\frac{f(x)f''(x)}{[f'(x)]^2}\bigg|<1$

Since Newton's method is defined as

$\displaystyle x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)} \Rightarrow f'(x_n)=-\frac{f(x_n)}{(x_{n+1}-x_n)}$,

we get that,

$\displaystyle |g'(x_n)| = \frac{f''(x_n)(x_{n+1}-x_n)^2}{f(x_n)}<1$

Then,

$\displaystyle (x_{n+1}-x_n)^2 \rightarrow 0$ as $\displaystyle n \rightarrow \infty$ but that doesn't show that Newton's is a second order method..

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I've also started by saying that since simple iteration is of order 1, we have that,

$\displaystyle \frac{|x_{n+1}-\xi|}{|x_n-\xi|} = |g'(c)| $

where $\displaystyle c \in [x_n,\xi]$ and $\displaystyle |g'(c)|\in(0,1)$,

but it does not lead me to anything good.

A nice hint would be great! Thanks.