= Power-law crossover in a singular perturbation
{title2=$x^{-1/\epsilon}/x=x^{-(1+1/\epsilon)}$}
For $\epsilon>0$, the two terms in $y=x^{-1/\epsilon}+x$ are equal at $x=1$. Their ratio is larger than one for $0<x<1$ and smaller than one for $x>1$. As $\epsilon\to0^+$ it tends to infinity at fixed $x<1$ and to zero at fixed $x>1$. Comparable contributions require $|\log x|=O(\epsilon)$, equivalently $x=1+O(\epsilon)$.
To resolve this region exactly, introduce the <inner variable> $X=\log x/\epsilon$. Then $y=e^{-X}+e^{\epsilon X}\sim e^{-X}+1$ for bounded $X$. At fixed $x>1$, the first term $\exp[-(\log x)/\epsilon]$ is smaller than every positive integer power of $\epsilon$, so a <regular perturbation> in powers of $\epsilon$ misses it. At fixed $x<1$, that same term diverges. This explains why a <power series> approximation valid on one side need not be a uniform approximation across the crossover.
Back to article page
