# Proving Quadratic Convergence via Taylor Expansion

1. Oct 10, 2011

### Scootertaj

1. The problem statement, all variables and given/known data

The following is a modification of Newton's method:

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

2. Relevant equations

We are supposed to use the following method:

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

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

3. The attempt at a solution

Change the equation to xn+1 - xn = - f(xn) / g(xn) where g(xn) = (f(xn + f(xn)) - f(xn)) / f(xn)
Then, LHS: En+1 - En
RHS broken up into parts:
f(xn) = f(p + En) = f'(p)En + O(En2)
f(xn + f(xn)) = f(p + En + f(p + En)) = f(p + En + f'(p)En + O(En2)) = f(p) + f'(p)(En + Enf'(p) + O(En2)) + O(En2) = f'(p)(En + Enf'(p) + O(En2)) + O(En2)

So, f(xn + f(xn)) - f(xn) = [ f'(p)(En + Enf'(p) + O(En2)) + O(En2) ] - f'(p)En + O(En2)

= [ f'(p)(Enf'(p) + O(En2)) + O(En2) ] - O(En2)

Also, f(xn)2 = (f'(p)En + O(En2))2

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

So, from this somehow I'm supposed to get En+1 = O(En2)

2. Oct 11, 2011

Just a bump