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

A sequence $(a_n)$ has limit of a sequence $L$ ifFor the reciprocal rule, the required condition is $\boxed{L\ne 0}$. Indeed, eventually $|a_n-L|<|L|/2$, so $|a_n|\ge |L|/2>0$. The reciprocals are then defined on this tail, andA finite number of zero initial terms does not affect this tail limit of a sequence; if a reciprocal sequence is required at every index, those terms must be excluded or assigned arbitrary values separately.

Rewrite the recurrence relation asSince $a_0=1>0$ and $c>0$, a positive $a_n$ gives a positive $a_{n+1}$. Mathematical induction therefore proves $\boxed{a_n>0\text{ for all }n\ge 0}$, and also shows that every division in the recurrence relation is defined.

For every $n\ge 0$, direct algebra givesThus $\boxed{a_n^2\ge c\text{ for every }n\ge 1}$. This identity is the key estimate in Newton iteration for a square root.

For $n\ge 1$, the preceding parts give a positive denominator and a nonpositive numerator inHence $\boxed{a_{n+1}\le a_n\text{ for every }n\ge 1}$. The restriction $n\ge 1$ matters: if $c>1$, then $a_1=(1+c)/2>a_0$.

The tail $(a_n{)}_{n\ge 1}$ is a monotone sequence bounded below by zero. The bounded monotone sequence theorem therefore gives $a_n\to L\ge 0$. Taking the limit of a sequence in $a_n^2\ge c$ yields $L^2\ge c>0$, so $L>0$. The reciprocal rule from part (a) permits passage to the limit of a sequence in the recurrence relation:Multiplication by $2L$ givesThis proves existence of the positive square root without assuming that existence when constructing the sequence.

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.