 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ib/paper-3/7d/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Let

ω = e^{2 πi / N}

. Using the convention

x_{k} = \sum_{n = 0}^{N - 1} x_{n} ω^{- kn}, 0 \leq k < N,

x_{n} = \frac{1}{N} \sum_{m = 0}^{N - 1} (m N - 1) ω^{mn} .

\sum_{n = 0}^{N - 1} ω^{(m - k) n} = {N, 0, m = k, m \neq = k .

Since

m, k

both lie between zero and

N - 1

x_{k} = (k N - 1), k = 0, \dots, N - 1.

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