Proving Quadratic Convergence via Taylor Expansion

[Scootertaj]

Homework Statement

The following is a modification of Newton's method:

x_n+1 = x_n - f(x_n) / g(x_n) where g(x_n) = (f(x_n + f(x_n)) - f(x_n)) / f(x_n)

Homework Equations

We are supposed to use the following method:

let En = x_n + p where p = root → x_n = p + En

Moreover, f(x_n) = f(p + En) = f(p) + f'(p)En + O(En²) = f'(p)En + O(En²) since f(p) = 0 at root

The Attempt at a Solution

Change the equation to x_n+1 - x_n = - f(x_n) / g(x_n) where g(x_n) = (f(x_n + f(x_n)) - f(x_n)) / f(x_n)
Then, LHS: E_n+1 - E_n
RHS broken up into parts:
f(x_n) = f(p + E_n) = f'(p)E_n + O(E_n²)
f(x_n + f(x_n)) = f(p + E_n + f(p + En)) = f(p + E_n + f'(p)E_n + O(E_n²)) = f(p) + f'(p)(E_n + E_nf'(p) + O(E_n²)) + O(E_n²) = f'(p)(E_n + E_nf'(p) + O(E_n²)) + O(E_n²)

So, f(x_n + f(x_n)) - f(x_n) = [ f'(p)(E_n + E_nf'(p) + O(E_n²)) + O(E_n²) ] - f'(p)E_n + O(E_n²)

= [ f'(p)(E_nf'(p) + O(E_n²)) + O(E_n²) ] - O(E_n²)

Also, f(x_n)² = (f'(p)En + O(En²))²

So, E_n+1 - E_n = (f'(p)En + O(En²))² / ( [ f'(p)(E_nf'(p) + O(E_n²)) + O(E_n²) ] - O(E_n²) )

So, from this somehow I'm supposed to get E_n+1 = O(E_n²)

 

[Scootertaj]

Just a bump