 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-2/41e/a/solution

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

Reverse the second half of the DFT sum by putting

m = 2 N - n - 1

. Since

ω_{2 N}^{2 N k} = 1

Y_{k} = n = 0 \sum N - 1 x_{n} (ω_{2 N}^{nk} + ω_{2 N}^{- (n + 1) k}) = 2 ω_{2 N}^{- k /2} n = 0 \sum N - 1 x_{n} cos [\frac{π}{N} (n + \frac{1}{2}) k] .

This is the required identity.

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