Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 68 / 6

For order $k\ge 2$, set $g(u)=(u-t{)}_{+}^{k-2}$. Then $(u-t)g(u)=(u-t{)}_{+}^{k-1}$. Apply the given divided difference product identity with nodes $t_i,\dots ,t_{i+k}$ and multiply by the length $t_{i+k}-t_i$ of their knot interval:Using the definition of each lower-order normalized B-spline, the two divided differences are $N_{i,k-1}(t)/(t_{i+k-1}-t_i)$ and $N_{i+1,k-1}(t)/(t_{i+k}-t_{i+1})$. Hence the Cox-de Boor recurrence isFor completeness, the product identity itself follows fromand the ordinary divided difference recurrence. Substitution yields the stated convex combination of the two order-$k-1$ divided differences.

The initial order-one splines are the indicators $N_{i,1}(t)=1_{[t_i,t_{i+1})}(t)$. For repeated knots, a term with zero knot-span denominator is defined as zero; this agrees with the standard limiting basis convention. The recurrence uses the partition of unity normalization with $0\le N_{i,k}\le 1$, not a separate rescaling of each spline to make its actual supremum exactly one. The original PDF recurrence has the denominators above; several of these indices and signs were corrupted in the converted TeX.

For integer knots, the recurrence simplifies toOrder-two splines are triangular: $N_{i,2}(i)=N_{i,2}(i+2)=0$ and $N_{i,2}(i+1)=1$. Applying the recurrence once more givesFor example $N_{0,3}(1)=\frac{1}{2}{N}_{0,2}(1)+N_{1,2}(1)=1/2$ and $N_{0,3}(2)=N_{0,2}(2)+\frac{1}{2}{N}_{1,2}(2)=1/2$. Thus the Cardinal cubic B-spline values areThereforeIn particular its actual maximum is $2/3$ in this standard normalization; dividing by that maximum would change the requested recurrence normalization.