 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/i/solution

For

f (x) = x^{2} - a

f^{'} (x) = 2 x

. Substitution into Newton's method gives

x_{n} = x_{n - 1} - \frac{x _{n - 1}^{2} - a}{2 x _{n - 1}} = \frac{1}{2} (x_{n - 1} + \frac{a}{x _{n - 1}}) .

Every positive input gives another positive input, so a starting value

x_{0} > 0

keeps the denominator nonzero at every step.

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