Replacing $\phi_k$ by $-\phi_k$ also replaces $a_k$ by $-a_k$, so both $\mathbb E[a_kb_l]$ and $\phi_k(s)$ change sign and their product is unchanged. Replacing $u_l$ by $-u_l$ similarly replaces $b_l$ by $-b_l$. Every summand, and hence $\beta$, is independent of all eigenfunction sign choices.


Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 225 / 4 / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-225/4/ii/solution