= Solution
At fixed regulator abbreviate
$$
A(\epsilon)=\frac{a_1}{\epsilon}+a_2+a_3\epsilon+\cdots,\qquad
C(\epsilon)=\frac{z_1}{\epsilon}+z_2+z_3\epsilon+\cdots.
$$
The <coupling> expansion is $g=g_0+Ag_0^2+O(g_0^3)$, whose inverse is $g_0=g-Ag^2+O(g^3)$. Using part c before removing the regulator,
$$
\begin{aligned}
\beta(g)&=-\epsilon g_0(1+2Ag_0)+O(g_0^3)\\
&=-\epsilon g-\epsilon A g^2+O(g^3),\\
\gamma(g)&=-\epsilon g_0\frac{C}{1+Cg_0}+O(g_0^2)
=-\epsilon Cg+O(g^2).
\end{aligned}
$$
Therefore the regulator-free coefficients at the orders determined by the supplied expansions are
$$
\boxed{\beta(g)=-a_1g^2+O(g^3),\qquad\gamma(g)=-z_1g+O(g^2).}
$$
The canonical term $-\epsilon g$ vanishes, and the factors of $\epsilon$ cancel the simple poles. The finite coefficients $a_2,z_2$ and higher positive powers of $\epsilon$ do not enter these leading regulator-free terms. Inverting the <coupling> is essential: substituting $g_0=g$ too early would incorrectly give $-2a_1g^2$.
The calculation establishes finiteness only at the displayed perturbative orders. Higher-order pole cancellations require the higher-order coefficients, as the next part makes explicit.
Back to article page
