For $h(x)=e_p(q(x))$, the product in the $U^3$ cube average is
$$
e_p\!\left(
\sum_{\epsilon\in\{0,1\}^3}(-1)^{|\epsilon|}
q(x+\epsilon\mathbin\cdot h)
\right).
$$
The expression in parentheses is the third additive derivative of the <quadratic form> $q$, so it vanishes. Every cube contributes one and therefore the <quadratic phase> satisfies
$$
\|h\|_{U^3}=1.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 129 / 4 / b / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-129/4/b/ii/solution