 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-1/19/a/solution

The Newton root-finding iteration follows the current tangent line to its horizontal-axis intersection. Since

f^{'} (x) = 2 x

, it gives

x_{n + 1} = x_{n} - \frac{x _{n}^{2} - a}{2 x _{n}} = \frac{1}{2} (x_{n} + \frac{a}{x _{n}}), x_{n} \neq = 0.

A positive initial guess is appropriate for the positive square root; negative starting values instead target the negative root.

 New to topics? Read the docs here!