 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-2/41c/a/i/solution

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

e^{- im θ}

and summation over

m \in Z

give

\sum_{m} e^{- im θ} u_{m + k}^{n} = e^{ik θ} u^{n} (θ) .

Thus the recurrence becomes

A (θ) u^{n + 1} (θ) = B (θ) u^{n} (θ),

where

A (θ) = \sum_{k = r}^{s} a_{k} e^{ik θ}, B (θ) = \sum_{k = r}^{s} b_{k} e^{ik θ} .

Assuming

A (θ) \neq = 0

H (θ) = \frac{B ( θ )}{A ( θ )} = \frac{\sum _{k = r}^{s} b _{k} e ^{ik θ}}{\sum _{k = r}^{s} a _{k} e ^{ik θ}} .

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