 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/iii/paper-45/4/b/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

Write

K = k_{m a x} r

ξ (r) = \frac{A}{2 π ^{2} r} \int_{0}^{k_{m a x}} k^{2} sin (k r) d k .

Two integrations by parts give the antiderivative

- k^{2} cos (k r) / r + 2 k sin (k r) / r^{2} + 2 cos (k r) / r^{3}

. Subtract its value at zero to obtain, for

r > 0

ξ (r) = \frac{A}{2 π ^{2} r ^{4}} [- K^{2} cos K + 2 K sin K + 2 (cos K - 1)] .

The origin is a removable limit. The bracket starts with

K^{4} /4

, so

ξ (0) = \frac{A k _{m a x}^{4}}{8 π ^{2}} .

This is also the density variance obtained directly by setting

sin (k r) / (k r) = 1

in the integral. The sharp spectral edge produces oscillatory correlations, with leading large-

r

term

- A k_{m a x}^{2} cos (k_{m a x} r) / (2 π^{2} r^{2})

. A negative correlation at some separation is compatible with a nonnegative power spectrum.

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