Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 327 / 3 / c / iv / Solution

Since $Q(x,\lambda )$ is a polynomial of degree at most $M$ in $\lambda$, its Taylor formula is exact:The $\alpha =0$ term is $Q{a}_1=1$ outside the cutoff region; its difference from one is a symbol of order $-\infty$. For $|\alpha |\ge 1$,so their product lies in $S^{-|\alpha |}\subset S^{-1}$. Defining $b_1$ as the negative of the cutoff remainder and these lower-order terms yields