(i) F is differentiable at least up to order 2,

(ii) F has a simple root z in (a,b) i.e. F(z)=0, and F'(z), F''(z) are different from zero.

Then there is an r>0 such that for every pair of start-values x

converges to z, the convergence being of order k = (1+sqrt(5))/2 = 1.61803 ...

Recall that the

In actual calculation of the sequence terms one should use the first formula, since the second may represent a division with a very small number.

The picture below illustrates the method by showing the first few terms, in the case of a quartic and arbitrary x

Notice that the method is not an

NewtonIterative.html

NewtonIterative2.html

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