 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-1/33c/b/iii/solution

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

Under

a_{j} = \frac{1}{2} e^{u_{j} /2},

the chain rule gives

d o t a_{j} = (a_{j} /2) \overset{u}{˙}_{j}

. Part (ii) and

a_{j} > 0

therefore imply

\frac{1}{2} \overset{u}{˙}_{j} = a_{j + 1}^{2} - a_{j - 1}^{2} = \frac{1}{4} (e^{u_{j + 1}} - e^{u_{j - 1}}) .

Consequently

\overset{u}{˙}_{j} = \frac{1}{2} (e^{u_{j + 1}} - e^{u_{j - 1}}), 1 \leq j \leq n - 1.

The conventions

u_{0} = u_{n} = - \infty

give

e^{u_{0}} = e^{u_{n}} = 0

, exactly matching

a_{0} = a_{n} = 0

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