 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2014/discrete-mathematics-continuous-mathematics-and-probability/5/b/iii/solution

Subtract the previous iterate in Newton's method:

x_{n} - x_{n - 1} = \frac{a - x _{n - 1}^{2}}{2 x _{n - 1}} < 0 if x_{n - 1} > a .

The denominator is positive and the numerator is negative. If

x_{n - 1} = a

, the difference is zero and all subsequent iterates equal the root.

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