Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 303 / 2 / c

The coefficient $A$ comes from one sextic vertex with two external slow legs and two fast tadpole loops. The coefficient $B$ comes from one sextic vertex with four external slow legs and one fast tadpole. The coefficient $C$ comes from one quartic and one sextic vertex joined by two fast propagators, leaving six external slow legs.

Solved by gpt-5.6-sol high.

Choosing two slow legs in $h({\varphi }_{<}+{\varphi }_{>}{)}^6$ and contracting four fast legs gives $\left(\genfrac{}{}{0ex}{}{6}{2}\right)(3)h{I}_1^2{\varphi }_{<}^2$. Since the mass term is ${\mu }^2{\varphi }_{<}^2/2$,Choosing four slow legs and contracting the remaining pair gives $B=\left(\genfrac{}{}{0ex}{}{6}{4}\right)=15$. For the connected quartic-sextic cumulant, the factor is $\left(\genfrac{}{}{0ex}{}{4}{2}\right)\left(\genfrac{}{}{0ex}{}{6}{2}\right)2!=180$ and the cumulant has a minus sign, soAs a check, two quartic vertices give $-\frac{1}{2}{\left(\genfrac{}{}{0ex}{}{4}{2}\right)}^{2}2!=-36$, matching the supplied coefficient.

Solved by gpt-5.6-sol high.

The canonical eigenvalue of $h$ isThus ${\beta }_h=(-2+O(ϵ))h+O(g^3)+O(h^2)$. At a fixed point this forces $h_{\ast }=O(g_{\ast }^3)=O({ϵ}^3)$. Its feedback into the mass and quartic flows begins only at order ${ϵ}^3$, so the order-$ϵ$ fixed point from part (b)(iii) is unaffected.

Solved by gpt-5.6-sol high.