From: spellucci@mathematik.tu-darmstadt.de (Peter Spellucci)
Subject: Re: rate of converge -- help
Date: 10 Feb 2000 10:17:48 GMT
Newsgroups: sci.math.num-analysis
Summary: [missing]

In article <38A2229E.F0714871@cae.ca>,
 dans <dans@cae.ca> writes:
|> Hi,
|> 
|>    For the following equation : x^n (x^2 - 9) = 0, where n=1,2,3 .....,
|> I want to calculate the rate of convergence using the Newton Method.
|> Now, I see that for n > 1 I have multiple root at x = 0.
|> The Newton method is : x(k+1) = x(k) - g(x(k))/g'(x(k)) where g'(x(k))
|> != 0.
|> Then I get the following equation:
|>                      (n+1) x(k)^3 - 9 (n -1) x(k)
|> x(k+1) =  -----------------------------------
|>                      (n+2) x(k)^2 - 9n
|> 
|> But from here, I am unable to see what rate of convergence .....
looks very much like students homework. hence i give you a hint only
(you should find this in your lecture notes)

theorem: let x* be a fixed point of the iteration map phi(x) and phi be
sufficiently often differentiable at x*. then if 
    0< |phi' (x*)| < 1 
then convergence is linear (Q-linear, i.e. |x_k-x*|<=L |x_{k-1}-x*| , 0<L<1)
but if 
    0=phi'(x*)=....=phi^{p}(x*) and phi^{p+1}(x*) <>0 , then the Q-order is p.
hint: if f is analytic and
    if x* is a k-th order zero of f then g(x)=f(x)/f'(x) has a single zero 
    at x* and is analytic in some neighborhood of x*
hope that helps
peter