= Solution
For integer knots, the recurrence simplifies to
$$
N_{i,k}(t)=\frac{t-i}{k-1}N_{i,k-1}(t)
+\frac{i+k-t}{k-1}N_{i+1,k-1}(t).
$$
Order-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 gives
$$
\begin{array}{c|ccc}
t&1&2&3\\ \hline
N_{0,3}(t)&1/2&1/2&0\\
N_{1,3}(t)&0&1/2&1/2.
\end{array}
$$
For example $N_{0,3}(1)=\frac12N_{0,2}(1)+N_{1,2}(1)=1/2$ and $N_{0,3}(2)=N_{0,2}(2)+\frac12N_{1,2}(2)=1/2$. Thus the <Cardinal cubic B-spline> values are
$$
\begin{aligned}
N_{0,4}(1)&=\frac13\frac12+\frac33\,0=\frac16,\\
N_{0,4}(2)&=\frac23\frac12+\frac23\frac12=\frac23,\\
N_{0,4}(3)&=\frac33\,0+\frac13\frac12=\frac16.
\end{aligned}
$$
Therefore
$$
\boxed{(N_{0,4}(1),N_{0,4}(2),N_{0,4}(3))=(1/6,\,2/3,\,1/6).}
$$
In particular its actual maximum is $2/3$ in this standard normalization; dividing by that maximum would change the requested recurrence normalization.
Back to article page
