 Solution

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

Only (ii) is necessarily true. Every real

a_{n}

satisfies

a_{n}^{2} \geq 0 > c

, so (ii) holds automatically. For a single counterexample to (i) and (iii), choose

c = - 3

and

a_{0} = 1

. The recurrence relation then gives the nonzero periodic sequence

a_{2 j} = 1, a_{2 j + 1} = - 1 (j \geq 0) .

Here

a_{1} < 0

, disproving (i), and

a_{2} > a_{1}

, disproving (iii). This counterexample satisfies the requirement that no iterate is zero.

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