 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2003/iii/paper-69/1/c/solution

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

Fix

m

and take

n \geq m

. Comparing the two products defining

f_{nm}

and the monomial

g_{m} (x) = x^{m}

, change one factor at a time. Every factor has absolute value at most one on

[0, 1]

∥ f_{nm} - g_{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} .

The contraction proved in (a) and the Bernstein falling-factorial identity from (b) now give

∥ B_{n} g_{m} - g_{m} ∥_{\infty} \leq ∥ B_{n} (g_{m} - f_{nm}) ∥_{\infty} + ∥ (f_{nm} (1) - 1) g_{m} ∥_{\infty} \leq \frac{m ( m - 1 )}{n} ⟶ 0 .

This proves uniform convergence for each fixed monomial; linearity then gives it for every fixed polynomial. The cases

m = 0, 1

have zero error for every

n

 New to topics? Read the docs here!