 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ii/paper-1/41b/c/solution

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

Write

q^{(0)} = \sum_{i} c_{i} q_{i}

with

c_{1} \neq = 0

. After

k

q_{1}

is at most

\frac{\sum _{i \geq 2} c _{i}^{2} ∣ λ _{i} ∣ ^{2 k}}{∣ c _{1} ∣∣ λ _{1} ∣ ^{k}} \leq \frac{λ _{2}}{λ _{1}}^{k} tan θ_{0} .

Since

∣ sin θ_{k} ∣ \leq tan θ_{k}

, the first bound follows. Also

λ^{(k)} - λ_{1} = \sum_{i \geq 2} (λ_{i} - λ_{1}) ∣ q_{i}^{T} q^{(k)} ∣^{2} .

Bound the coefficient by

max_{i \geq 2} ∣ λ_{i} - λ_{1} ∣

and the sum by

sin^{2} θ_{k}

, then use the squared first estimate to obtain the second bound.

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