= Solution
Put $s_j=t_{i+j}$ for $0\le j\le k$, with the knots in increasing order. The explicit formula for a <divided difference> at distinct nodes gives
$$
M_i(t)=k\sum_{j=0}^k\frac{(s_j-t)_+^{k-1}}{\prod_{\ell\ne j}(s_j-s_\ell)}.
$$
Each <truncated power function> is a <piecewise polynomial function> of degree $k-1$ and is $C^{k-2}$ for $k\ge2$. Therefore their sum has the same global smoothness. The jump in the $(k-1)$st <derivative> at $s_j$ equals
$$
\frac{k(-1)^k(k-1)!}{\prod_{\ell\ne j}(s_j-s_\ell)},
$$
which is nonzero. Thus each $s_j$ really is a knot, and generally there is no additional smoothness.
If $t<s_0$, the values being differenced are the values of the degree-$k-1$ <polynomial> $(s-t)^{k-1}$, so their $k$th <divided difference> vanishes. If $t>s_k$, every sampled <truncated power function> is zero. To check that both support endpoints occur, for $s_0<t<s_1$ subtract the vanished full <polynomial> divided difference to obtain
$$
M_i(t)=\frac{k(t-s_0)^{k-1}}{\prod_{\ell=1}^k(s_\ell-s_0)}>0.
$$
For $s_{k-1}<t<s_k$, only the last term survives and is positive. The nonnegativity from the recurrence in (b) actually gives positivity throughout the interior: for each $s_0<t<s_k$, at least one lower-order <B-spline> covering $t$ has a positive recurrence coefficient, and induction starts with positive order-one indicators. Hence \b[$\operatorname{supp}M_i=[t_i,t_{i+k}]$ and $M_i\in C^{k-2}$ for $k\ge2$], with degree $k-1$ on its knot intervals. For $k=1$, the <B-spline> is the step function $1_{[s_0,s_1)}(t)/(s_1-s_0)$, so no continuity is asserted.
The <unit-integral normalization of a B-spline> also follows directly: integrating $k(s_j-t)_+^{k-1}$ over $s_0\le t\le s_k$ gives $(s_j-s_0)^k$. The $k$th <divided difference> of this monic degree-$k$ <polynomial> is one. Thus $\int M_i(t)\,dt=1$, explaining its normalization.
Back to article page
