Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 32E (Asymptotic Methods)

For $n\ge 1$, put $r_n={\varphi }_n/{\varphi }_{n-1}$. Since $({\varphi }_n)$ is an asymptotic sequence, $r_n\to 0$. We can rewritesoTogether with ${\psi }_0={\varphi }_0$, this provesMoreover,Hence $({\psi }_n)$ is an asymptotic sequence. This is the harmonic-mean refinement of an asymptotic sequence.

First,For $n\ge 1$,Thus $({\chi }_n)$ is an asymptotic sequence.

Termwise asymptotic equivalence fails for every $n\ge 1$, becauseFor example, as $x\to 0$, the positive scale ${\varphi }_n(x)=|x{|}^n$ gives ${\chi }_n(x)=|x{|}^{n-1/2}$ for $n\ge 1$. This illustrates the geometric-mean refinement of an asymptotic sequence.

No. Let $x_0=0$ and, for $0<|x|<1$, definewhileBoth families are asymptotic sequences, and ${\psi }_n\sim {\varphi }_n$ for every $n$.

Take $f(x)=1$. It has the exact expansionSuppose it had an expansion $f\sim {\sum }_{n\ge 0}{b}_n{\psi }_n$. The order-zero condition forces $b_0=1$. At order one we would then require some constant $b_1$ such thatAfter division by $|x|$, the left side is $-|x{|}^{-1/2}-b_1$, which cannot tend to zero. This contradiction proves that no such coefficients exist. Thus termwise equivalent asymptotic scales need not preserve expansions.