|Mar23-06, 08:59 AM||#1|
Numerical Analysis: Fixed Point Iteration
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...
|Similar Threads for: Numerical Analysis: Fixed Point Iteration|
|Fixed Point Iteration Requirements||General Math||4|
|Simple Fixed Point Iteration for Root-solving||Calculus||5|
|numerical analysis (composite numerical integration)||Engineering, Comp Sci, & Technology Homework||9|
|Fixed-Point Iteration for Nonlinear System of Equations||Calculus||2|
|Function has a fixed point?||Calculus||3|