1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Numerical Analysis: Fixed Point Iteration

  1. Mar 23, 2006 #1
    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...
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted
Similar Discussions: Numerical Analysis: Fixed Point Iteration
  1. Fixed point iteration (Replies: 3)

  2. Fixed point iteration (Replies: 2)

Loading...