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Proving Quadratic Convergence via Taylor Expansion

  1. Oct 10, 2011 #1
    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. jcsd
  3. Oct 11, 2011 #2
    Just a bump
     
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