% \def\updated{14apr03} \magnification=1200\input carlformat\def\cite#1{[#1]}\parskip10pt % % %The perfect B-spline % %
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The perfect B-spline }%

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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 $$% M(\cdot|t_0,\ldots,t_k):x\mapsto k\dvd{t_0,\ldots,t_k}(\cdot-x)_+^{k-1} $$% whose knots $t_j:= \cos((k-j)\pi/k)$, $j=0{:}k$, are the extrema of the Chebyshev polynomial of degree $k$, i.e., of $T_k(x) = \cos(k\gth)$ with $x = :\cos\gth$. Hence, $\go(x) := \prod_j(x-t_j) = (x^2-1)DT_k(x)/c_k$ (with $c_k$ chosen to ensure that the right side has leading coefficient $1$), while, by [divided difference: basic formula], $$%
\dvd{t_0,\ldots,t_k} = \sum_j \dvd{t_j}/(D\go)(t_j). $$%
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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 $y:=\sin\gth$, $$%
2^{k-1}D\go(t_j) = -y^2(k\cos(k\gth)/y)*1/(-y) = (-)^j, $$%
while, for $j=0$ or $k$, $2^{k-1}D\go(t_j) = 2t_j U_{k-1}(t_j) = 2(-)^j$. %

Hence, $$%
2^{k-1}M(x|t_0,\ldots,t_k)/k = (-1-x)_+^{k-1}/2 + \sum_{j=1}^{k-1}(-)^j(t_j-x)_+^{k-1} + (-1)^k(1-x)_+^{k-1}/2, $$%
showing $|D^{k-1}M| = k/2^k$ on its support. %

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