Past exam of the perturbation methods and applied complex variables course of the University of Oxford / 2008 / A / 1

For each fixed $n$, the next term must be negligible relative to its predecessor:The ratios are understood where their denominators are nonzero near the limiting point, or equivalently through the little o notation definition. For example $1,ϵ,{ϵ}^2,\dots$ form an asymptotic sequence.

For every fixed truncation index $N$, the remainder must be smaller than the last retained term:This is an asymptotic expansion relative to the sequence in part (i). The number of retained terms is fixed before taking the limit. The associated series need not converge for any nonzero parameter; its role is to give successively smaller truncation errors. If $a_n$ is an asymptotic sequence, the same condition can be stated as a remainder $O(a_{N+1})$ when the next expansion condition is available, but the displayed little o notation formulation is the defining one.

Take $g(x)=x$ and $f(x)=x+1$. Then $f/g=1+1/x\to 1$, so $f\sim g$. However,for every $x$, and this ratio does not approach one. Thus asymptotic equivalence need not survive exponentiation. A small relative error between large exponents can leave a nonvanishing additive error. This is the distinction in asymptotic comparison of exponential functions.

The ratio can be evaluated directly:The assumption $|f-g|\to 0$ gives $f-g\to 0$, so continuity of the exponential function givesThe additional premise $f\sim g$ is not needed for this implication. The useful requirement is additive control of the exponents, as explained by asymptotic comparison of exponential functions.

First take the usual positive-parameter limit $ϵ\downarrow 0$. For real $x$, the equation implies $|x^2-1|\le ϵ$, so every root lies near $1$ or $-1$. Near the positive root,and hence $x^2=1-ϵ$ up to an exponentially small correction. Near the negative root, the corresponding limit is minus one, giving $x^2=1+ϵ$ to all algebraic orders. Expanding the square roots givesThere is one root in each neighborhood: the derivative $2x+{\mathrm{sech}}^2(x/ϵ)$ has the sign of $x$ there for sufficiently small positive $ϵ$. The bound above excludes further real roots.

If the source's unqualified $ϵ\to 0$ is read as a two-sided limit, use the exact identityThe two-sided versions therefore replace $ϵ$ by $|ϵ|$ in both displayed first corrections. The two one-sided algebraic expansions otherwise have the same reasoning.

The function $e^{-x^2}/x$ is strictly decreasing from infinity to zero on $x>0$, so there is a unique positive root, which tends to infinity. Let $L=\log (1/ϵ)$. Taking logarithms givesFirst $x^2\sim L$. Put $x=\sqrt{L}+d$ with $d=o(\sqrt{L})$, expand, and cancel $L$:The leading balance gives $d=-(\log L)/(4\sqrt{L})$. Substitution leaves a residual $O((\log L{)}^2/L)$; the derivative of $x^2+\log x$ is of order $\sqrt{L}$, so the root error is $O((\log L{)}^2/L^{3/2})$. ThusThe second term is logarithmic rather than a constant multiple of $L^{-1/2}$. This is logarithmic inversion of a Gaussian divided by its argument.