Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 2 / 31D (Asymptotic Methods)

For fixed $n$, define the remaindersThe hypotheses say $r_n=o({\varphi }_n)$ and $r_{n+1}=o({\varphi }_{n+1})$, while subtraction givesBecause $a_{n+1}\ne 0$,For $x$ sufficiently close to $x_0$ this denominator is therefore nonzero, andThus ${\varphi }_{n+1}=o({\varphi }_n)$ for every $n$, so $({\varphi }_n)$ is an asymptotic sequence.

Supposethroughout the punctured neighbourhood, and let $k$ be the least index with $c_k\ne 0$. Part (i) implies ${\varphi }_j=o({\varphi }_k)$ for every $j>k$. Dividing by ${\varphi }_k$ and taking $x\to x_0$ therefore givesa contradiction. Hence every $c_j$ vanishes and ${\varphi }_0,\dots ,{\varphi }_N$ are linearly independent.

Put $a=1+ϵ$ and, for $m\ge 2$, defineOne integration by parts givesStarting with $I=I_2$ and iterating $N+1$ times yieldswhereSince $0<I_{N+3}(ϵ)\le (1+ϵ{)}^{-1}$, the remainder is $O({ϵ}^{N+1})$, which is little-$o$ of the last retained term. ThereforeThe displayed terms form an asymptotic sequence because the ratio of successive magnitudes tends to zero.

Keeping the terms through $n=3$ in part (i) givesExpanding the four rational factors in powers of $ϵ$ and collecting terms gives