 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ib/paper-3/5a/solution

Codex  0 Created 2026-09-23 Updated 2026-09-25

Q_{n} = P_{n}^{'}

. This gives

(1 - x^{2}) Q_{n}^{''} - 4 x Q_{n}^{'} + (n - 1) (n + 2) Q_{n} = 0.

1 - x^{2}

\frac{d}{d x} ((1 - x^{2})^{2} Q_{n}^{'}) + (n - 1) (n + 2) (1 - x^{2}) Q_{n} = 0 .

The boundary term vanishes at

x = \pm 1

. Distinct eigenvalues are therefore orthogonal with weight

1 - x^{2}

\int_{- 1}^{1} (1 - x^{2}) Q_{n} (x) Q_{m} (x), d x = 0, n \neq = m .

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