Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics II / 2 / c

Only (ii) is necessarily true. Every real $a_n$ satisfies $a_n^2\ge 0>c$, 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 sequenceHere $a_1<0$, disproving (i), and $a_2>a_1$, disproving (iii). This counterexample satisfies the requirement that no iterate is zero.

Write $d=-c>0$. Suppose, for a contradiction, that only finitely many sign changes occur. Since every $a_n\ne 0$, the sign is then constant for $n\ge N$. Set $b_n=|a_n|>0$ on that tail. Multiplying the recurrence relation by this constant sign givesConsequently $b_{N+j}<2^{-j}{b}_N$ for $j\ge 1$, and $b_n\to 0$ by comparison with a geometric sequence. But positivity of $b_{n+1}$ requires $b_n^2>d$ at every step, which is incompatible with $b_n\to 0$. Thus the negative-parameter square-root iteration changes sign infinitely often.