 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-4/39c/b/solution

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

h (t) = \sum_{n \in Z} h_{n} e^{iπn t}

into the rectangle rule. The finite geometric sum obeys

\frac{1}{N} \sum_{k = - N /2 + 1}^{N /2} e^{2 πink / N} = {1, 0, N ∣ n, N ∤ n .

Consequently the quadrature value is

\frac{2}{N} \sum_{k = - N /2 + 1}^{N /2} h (2 k / N) = 2 \sum_{j \in Z} h_{j N},

whereas

\int_{- 1}^{1} h (t) d t = 2 h_{0} .

e_{N} (h) = - 2 j \in Z j \neq = 0 \sum h_{j N} .

C > 0

and

0 < ρ < 1

such that

∣ h_{n} ∣ \leq C ρ^{∣ n ∣} .

∣ e_{N} (h) ∣ \leq 4 C \sum_{j = 1}^{\infty} ρ^{j N} = \frac{4 C ρ ^{N}}{1 - ρ ^{N}} .

Therefore

∣ e_{N} (h) ∣ = O (ρ^{N}),

which is exponential, and hence spectral, convergence.

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