Numerical Analysis: Fixed Point Iteration

by Goomba
Tags: analysis, fixed, iteration, numerical, point
Goomba is offline
Mar23-06, 08:59 AM
P: 11
Consider the fixed point iteration formula:
*x_(n+1) = (2/3)[(x_n)^3 - 1] - 3(x_n)^2 + 4x_n = g(x)

*Note: "_" precedes a subscript and "^" precedes a superscript

(a) Find an interval in which every starting point x_0 will definitely converge to alpha = 1.

(b) Show that the order of the above fixed point iteration formula is 2 (quadratic convergence).


For (a), I took the derivative of g(x) and set it equal to zero. I found that when g'(x) = 2x^2 - 6x +4 = (2x - 2)(x - 2)= 0, x = 1, 2.

But g'(alpha) = g'(1) = 2 - 6 + 4 = 0...?

I want to say that the interval is (1,2]...

For (b), I tried |alpha - x_(n + 1)| <= c|1 - x_n|^p, where p is the order and c is some constant >= 0. And Newton's method usually converges quadratically... I ended up with:

|-(2/3)(x_n)^3 + 3(x_n)^2 - 4x_n + (5/3)| <= c|1 - x_n|^p

I don't know how to conclude that p must be 2... or if this is even right...
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