 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-1/30k/c/solution

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

n

, the minimizing feedback from part (b) is

u_{n}^{*} = - \frac{X _{n - 1}}{N - n + 2} .

Starting with

u_{1}^{*} = - X_{0} / (N + 1)

and repeatedly substituting

X_{j} = X_{j - 1} + u_{j}^{*} + ξ_{j}

gives by induction

X_{n - 1} = \frac{N - n + 2}{N + 1} X_{0} + \sum_{j = 1}^{n - 1} \frac{N - n + 2}{N - j + 1} ξ_{j} .

Therefore

u_{n}^{*} = - \frac{X _{0}}{N + 1} - \frac{ξ _{1}}{N} - \frac{ξ _{2}}{N - 1} - \dots - \frac{ξ _{n - 1}}{N - n + 2},

as required.

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