 Two-loop inverse-coupling logarithmic remainder

For the asymptotically free two-loop running coupling $da/d\log \mu =-{\beta }_0{a}^2-{\beta }_1{a}^3$, with ${\beta }_0>0$, let $y=1/a$, $c={\beta }_1/{\beta }_0$, and $L=\log (\mu /\Lambda )$. A choice of the scale $\Lambda$ gives the exact implicit equation $y-c\log ((y+c)/{\beta }_0)={\beta }_{0}L$ on the ultraviolet branch. Its expansion isThe remainder after the first two terms is therefore $O(\log L/L)$, and is not $O(1/L)$ when ${\beta }_1\ne 0$. A fixed rescaling of $\Lambda$ cannot change the coefficient of $\log L/L$. This distinction matters when quoting errors for asymptotic expansions.