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Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 2 / 41A / a / Solution

The exact average is

I (h) = h_{0}

. For one Fourier mode,

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

Thus the periodic trapezoidal Fourier aliasing identity gives

I_{N} (h) = \sum_{j \in Z} h_{2 N j}, ∣ I_{N} (h) - I (h) ∣ = j \neq = 0 \sum h_{2 N j} .

Under the coefficient bound,

∣ I_{N} - I ∣ \leq 2 M \sum_{j = 1}^{\infty} c^{2 N j} = \frac{2 M c ^{2 N}}{1 - c ^{2 N}} \leq \frac{2 M}{1 - c ^{2}} c^{2 N} .

This decays exponentially in

N

.

Solved by gpt-5.6-sol high.

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