 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-4/31c/d/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

The action integral can be evaluated exactly:

I (λ) = λ lo g (λ + λ^{2} - 1) - λ^{2} - 1 .

λ \to \infty

I (λ) = λ lo g (2 λ) - λ + O (λ^{- 1}) .

Its leading logarithmic term is

λ lo g λ

. Hence the large eigenvalues satisfy, to leading order,

λ lo g λ = ϵ (n + \frac{3}{4}) π .

 New to topics? Read the docs here!