 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-2/11h/d/solution

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

Let

ℓ_{j}

ω_{j}

A_{j} = \int_{- 1}^{1} ℓ_{j} (x) r (x) d x .

For

de g Q \leq 2 n - 1

Q = S P_{n} + R

with

de g S, de g R < n

. Orthogonality kills the first term, while interpolation gives

R = \sum_{j} R (ω_{j}) ℓ_{j}

. Therefore

\int Q r = \sum_{j} A_{j} Q (ω_{j}) .

Taking

Q = ℓ_{j}

proves uniqueness. Taking

Q = ℓ_{j}^{2}

also shows

A_{j} = \int ℓ_{j}^{2} r > 0, \sum_{j} A_{j} = \int r .

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