 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-3/30e/d/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

For

E \to \infty

f (E) = \frac{16}{15} (E - 1)^{5/2} + O (E^{3/2}) \sim \frac{16}{15} E^{5/2} .

\frac{16}{15} E_{n}^{5/2} \sim π ε n .

Taking the power

2/5

E_{n} \sim (\frac{15 π ε}{16})^{2/5} n^{2/5} .

Hence

c = (\frac{15 π ε}{16})^{2/5}, α = \frac{2}{5} .

 New to topics? Read the docs here!