 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-67/6/b/solution

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

The order-

k

M_{i}

is a finite linear combination of values at its knots. Hence integration can be interchanged with it directly. For a knot value

u \in [a, b]

\int_{a}^{b} (u - t)_{+}^{k - 1} d t = \frac{( u - a ) ^{k}}{k} .

\int_{a}^{b} M_{i} (t) d t = k [t_{i}, \dots, t_{i + k}] (u \mapsto \int_{a}^{b} (u - t)_{+}^{k - 1} d t) = [t_{i}, \dots, t_{i + k}] (u - a)^{k} = 1.

The final equality holds because the order-

k

k

\int_{a}^{b} M_{i} (t) d t = 1.

[t_{i}, t_{i + k}]

: below all its knots, the sampled truncated power agrees with a degree-

k - 1

polynomial, annihilated by the order-

k

divided difference; above all its knots, every sample is zero. No mass is omitted by integration over

[a, b]

Since

N_{i} = (t_{i + k} - t_{i}) M_{i} / k

, the corresponding partition-normalized B-spline has integral

(t_{i + k} - t_{i}) / k

. Its normalization is different from that of

M_{i}

 New to topics? Read the docs here!