Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 67 / 6 / a

Solution

For this part relabel one consecutive block of

k + 1

distinct knots as

t_{0}, \dots, t_{k}

. This is a local reindexing; no extra global knot is being introduced. Let

g_{t} (u) = (u - t)_{+}^{k - 1}

ℓ (u) = \sum_{i = 0}^{k} g_{t} (t_{i}) \frac{ω ( u )}{( u - t _{i} ) ω ^{'} ( t _{i} )}, ω (u) = \prod_{j = 0}^{k} (u - t_{j}) .

ω (u) / (u - t_{i})

is monic of degree

k

[u^{k}] ℓ (u) = \sum_{i = 0}^{k} \frac{g _{t} ( t _{i} )}{ω ^{'} ( t _{i} )} .

The same leading coefficient is the order-

k

[t_{0}, \dots, t_{k}] g_{t}

, by the Newton interpolation polynomial representation. Multiplying by

k

therefore gives the explicit B-spline formula

M_{0} (t) = k i = 0 \sum k \frac{( t _{i} - t ) _{+}^{k - 1}}{ω ^{'} ( t _{i} )} .

Distinct knots make every denominator nonzero. For

k = 1

the degree-zero truncated power is interpreted as an interval-indicator convention; values at individual endpoints do not affect its integral.

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