= Solution
We derive the <Cox-de Boor recurrence> from the <Leibniz rule for divided differences>. Fix $t$, let $u_j=t_{i+j}$, and set
$$
q(s)=(s-t)_+^{k-2},\qquad h(s)=(s-t)q(s).
$$
Away from the <spline knots>, $h(s)=(s-t)_+^{k-1}$. Introduce
$$
D=[u_0,\ldots,u_k]q,\quad
L=[u_0,\ldots,u_{k-1}]q,\quad
R=[u_1,\ldots,u_k]q.
$$
The defining divided-difference recurrence says $(u_k-u_0)D=R-L$. Since $s-t=(s-u_0)+(u_0-t)$ and the first factor is linear, the <Leibniz rule for divided differences> gives
$$
[u_0,\ldots,u_k]h=R+(u_0-t)D.
$$
Multiplication by $u_k-u_0$ consequently yields
$$
(u_k-u_0)[u_0,\ldots,u_k]h
=(t-u_0)L+(u_k-t)R.
$$
Replace $L$ and $R$ by the corresponding lower-order <B-splines> divided by their support widths. This proves
$$
\boxed{
N_{i,k}(t)=\frac{t-t_i}{t_{i+k-1}-t_i}N_{i,k-1}(t)
+\frac{t_{i+k}-t}{t_{i+k}-t_{i+1}}N_{i+1,k-1}(t)}.
$$
The denominators are positive for the distinct <spline knots> of this question. The recursion starts with $N_{i,1}$, the interval indicator on $[t_i,t_{i+1})$. For $k\ge2$, values at the <spline knots> follow by the continuous extension of the left side; the usual half-open convention for the order-one factors gives the same result. This proves the recurrence rather than assuming it as a definition.
Back to article page
