 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-47/2/b/solution

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

The even time-series spectral density makes the sine part vanish. At lag zero its integral is the area of the triangle, so

γ_{0} = π^{2}

. For any nonzero integer

k

γ_{k} = 2 \int_{0}^{π} (π - λ) cos (kλ) d λ = \frac{2}{k} \int_{0}^{π} sin (kλ) d λ = \frac{2 ( 1 - cos kπ )}{k ^{2}} .

γ_{k} = ⎩ ⎨ ⎧ π^{2}, 4/ k^{2}, 0, k = 0, k odd, k \neq = 0 even .

No additional factor

1/ (2 π)

belongs in this inverse integral under the convention stated above.

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