 Solution

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

a_{n + 1} = \frac{1}{2} (a_{n} + \frac{c}{a _{n}}) .

Since

a_{0} = 1 > 0

and

c > 0

, a positive

a_{n}

gives a positive

a_{n + 1}

a_{n} > 0 for all n \geq 0

, and also shows that every division in the recurrence relation is defined.

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