Newton's method convergence

                It is interesting to calculate F¢(a) (see Aitkins.doc .htm for derivation of F') for Newtons method where

This does not mean that there is no error in the iterative procedure, but does mean that we need to take it to a higher approximation so that for Newton’s method

calculating the second derivative we find

the last term of which is zero for f(x)=0, but not the first two.

and hence we see the familiar quadratic convergence associated with Newton’s method.  For example with F¢¢(a)=10, and answer that is off by
10^-2 on step 5, will be off by 10^-4 ´ 10 = 10^-3  on step 6 and by 10^-5 on step 7.