 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-2/11f/b/v/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

True. By the Weierstrass approximation theorem, choose polynomials

q_{n}

with

q_{n} \to f^{(m)}

uniformly. Define

P_{n}

by integrating

q_{n}

exactly

m

times and choose the

m

integration constants so that

P_{n}^{(r)} (0) = f^{(r)} (0), 0 \leq r < m .

Then

P_{n}^{(m)} = q_{n}

, and repeated use of the fundamental theorem of calculus gives

∥ P_{n}^{(r)} - f^{(r)} ∥_{\infty} \leq \frac{∥ q _{n} - f ^{(m)} ∥ _{\infty}}{( m - r )!} (0 \leq r < m) .

Every derivative therefore converges uniformly as required.

 New to topics? Read the docs here!