%
\def\updated{14apr03} \magnification=1200\input carlformat\def\cite#1{[#1]}\parskip10pt % % %The perfect B-splines, i.e., B-splines whose piecewise constant derivative is absolutely constant, make their first appearance in Schoenberg71d, as the solution to an optimal control problem. %
These are, up to a scalar multiple,
the B-splines
$$%
Now observe that $c_kD\go(x) = 2x DT_k(x) + (x^2-1)D^2T_k(x)$, while
$DT_k(x) = k\sin(k\gth)/\sin\gth$, hence $DT_k(1) = k^2$ and $DT_k(-1) =
(-)^{k+1}k^2. So, for $j=0$ or $k$,
$2^{k-1}D\go(t_j) = 2t_j DT_k(t_j) = 2k^2(-)^j$, while
for $0\lessthan j\lessthan k$,
$c_kD\go(t_j) = (t_j^2-1) D^2T_k(t_j)$, hence, with $x := t_j =:
\cos\gth$ and $y:=\sin\gth$,
$$%
Hence,
$$%
\vfill\rightline{\updated}
%
%
\bye
\dvd{t_0,\ldots,t_k} = \sum_j \dvd{t_j}/(D\go)(t_j).
$$%
%
2^{k-1}D\go(t_j) = -y^2(k\cos(k\gth)/y)*1/(-y) = (-)^j,
$$%
while, for $j=0$ or $k$, $2^{k-1}D\go(t_j) = 2t_j U_{k-1}(t_j) = 2(-)^j$.
%
2^{k-1}M(x|t_0,\ldots,t_k)/k = (-1-x)_+^{k-1}/2 + \sum_{j=1}^{k-1}(-)^j(t_j-x)_+^{k-1}
+ (-1)^k(1-x)_+^{k-1}/2,
$$%
showing $|D^{k-1}M| = k/2^k$ on its support.
%