 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/iii/paper-62/1/c/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

Fix

m

and take

n \geq m

. On

[0, 1]

, every factor

x - r / n

and every factor

x

has absolute value at most one. Replacing the factors successively and applying the triangle inequality gives

∥ f_{nm} - x^{m} ∥_{\infty} \leq \sum_{r = 0}^{m - 1} \frac{r}{n} = \frac{m ( m - 1 )}{2 n}, ∣ f_{nm} (1) - 1∣ \leq \frac{m ( m - 1 )}{2 n} .

∥ B_{n} (x^{m}) - x^{m} ∥_{\infty} \leq ∥ B_{n} (x^{m} - f_{nm}) ∥_{\infty} + ∥ B_{n} (f_{nm}) - x^{m} ∥_{\infty} \leq ∥ x^{m} - f_{nm} ∥_{\infty} + ∣ f_{nm} (1) - 1∣ \leq \frac{m ( m - 1 )}{n} .

For

m = 0, 1

, the monomials are reproduced exactly. For every fixed

m

, $B_{n} (x^{m}) \to x^{m}$ uniformly on $[0, 1]$ .

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