= Two-loop inverse-coupling logarithmic remainder
{title2=$a^{-1}=\beta_0L+(\beta_1/\beta_0)\log L+O(\log L/L)$}
For the <asymptotically free> two-loop <running coupling> $da/d\log\mu=-\beta_0a^2-\beta_1a^3$, 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 $y-c\log((y+c)/\beta_0)=\beta_0L$ on the ultraviolet branch. Its expansion is
$$
y=\beta_0L+c\log L+\frac{c^2}{\beta_0}\frac{\log L+1}{L}+O\!\left(\frac{(\log L)^2}{L^2}\right).
$$
The remainder after the first two terms is therefore $O(\log L/L)$, and is not $O(1/L)$ when $\beta_1\ne0$. A fixed rescaling of $\Lambda$ cannot change the coefficient of $\log L/L$. This distinction matters when quoting errors for <asymptotic expansions>.
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