 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-1/41a/a/i/solution

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

\frac{1}{2 π} \int_{- π}^{π} ∣ v (θ) ∣^{2} d θ = m, j \sum v_{m} \overline{v_{j}} \frac{1}{2 π} \int_{- π}^{π} e^{- i (m - j) θ} d θ = m \sum ∣ v_{m} ∣^{2} .

Finite truncations justify the calculation directly, and passage to the

ℓ^{2}

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