 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2017/prelims/mathematics-ii/2/b/ii/solution

For every

n \geq 0

, direct algebra gives

a_{n + 1}^{2} - c = \frac{( a _{n}^{2} + c ) ^{2} - 4 c a _{n}^{2}}{4 a _{n}^{2}} = \frac{( a _{n}^{2} - c ) ^{2}}{4 a _{n}^{2}} \geq 0.

Thus

a_{n}^{2} \geq c for every n \geq 1

. This identity is the key estimate in Newton iteration for a square root.

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