Numerical Analysis: Fixed Point Iteration

[Goomba]

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...

 

[The Bill]

For anyone reading this thread in 2022 or after, the * on the second line isn't part of the equation, it appears to just be a "footnote," not mathematical notation. If anyone wants to use a similar footnote, in this example it would have been better to have it on the first line like this:

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

As for the OP's question, here are some sources to read and work through if you find yourself with a similar problem:

https://math.stackexchange.com/ques...the-interval-of-convergence-in-Newtons-method

https://en.wikipedia.org/wiki/Newton's_method

This next source is a video which goes into the weeds a bit farther than the OP's question, but the first eleven minutes or so are an excellent lecture on the "Calculus 1" type of Newton's method problems. Also, the weeds the video gets into after that are very pretty, so one might want to stick around for them anyway:

 

[pasmith]

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...

Setting $x_n = 1 + \epsilon_n$ should be the first step, and leads to $$\epsilon_{n+1} = 3\epsilon_n^2 + \tfrac23 \epsilon_n^3.$$ For $|\epsilon_n| < \frac32$ this leads to the bound $$|\epsilon_{n+1}| \leq 3|\epsilon_n|^2 + \tfrac23|\epsilon_n|^3 < 4|\epsilon_n|^2.$$

We also have $|3\epsilon^2 + \frac23\epsilon^3| < |\epsilon|$ when $$\left|\left(\frac94 + \epsilon\right)^2 - \frac{81}{16}\right| < \frac32$$ which yields $$-0.363 \approx \frac{\sqrt{57} - 9}4 < \epsilon < \frac{\sqrt{105} - 9}{4} \approx 0.312.$$ Provided we start with $\epsilon = x - 1$ in this interval, the iteration will converge to $x = 1$.