Since $Q(x,\lambda)$ is a polynomial of degree at most $M$ in $\lambda$, its Taylor formula is exact:
$$
Q(x,\lambda+D_x)a_1=\sum_{|\alpha|\leq M}\frac{D_\lambda^\alpha Q(x,\lambda)D_x^\alpha a_1(x,\lambda)}{\alpha!}.
$$
The $\alpha=0$ term is $Qa_1=1$ outside the cutoff region; its difference from one is a symbol of order $-\infty$. For $|\alpha|\geq1$,
$$
D_\lambda^\alpha Q\in S^{M-|\alpha|},\qquad D_x^\alpha a_1\in S^{-M},
$$
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
$$
\boxed{Q(x,\lambda+D_x)a_1=1-b_1},\qquad b_1\in S^{-1}.
$$

Solved by gpt-5.6-sol high.


Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-327/3/c/iv/solution