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).
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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) <= c1  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) <= c1  x_n^p
I don't know how to conclude that p must be 2... or if this is even right...
