 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-4/6b/a/solution

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

Q_{0}, \dots, Q_{n}

have distinct degrees and form an orthogonal basis of

P_{n}

p_{n}^{*} = \sum_{k = 0}^{n} \frac{⟨ f , Q _{k} ⟩}{⟨ Q _{k} , Q _{k} ⟩} Q_{k} .

For each

j \leq n

⟨ f - p_{n}^{*}, Q_{j} ⟩ = ⟨ f, Q_{j} ⟩ - \frac{⟨ f , Q _{j} ⟩}{⟨ Q _{j} , Q _{j} ⟩} ⟨ Q_{j}, Q_{j} ⟩ = 0.

Thus the residual

f - p_{n}^{*}

is orthogonal to all of

P_{n}

For any

p \in P_{n}

, write

f - p = (f - p_{n}^{*}) + (p_{n}^{*} - p) .

∥ f - p ∥^{2} = ∥ f - p_{n}^{*} ∥^{2} + ∥ p_{n}^{*} - p ∥^{2} \geq ∥ f - p_{n}^{*} ∥^{2} .

Equality holds only for

p = p_{n}^{*}

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