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%B-splines for the Bernstein knots
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Click here 
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B-splines for the Bernstein knots
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Since
$$%
1 = (t + (1-t))^{k-1} = \sum_{j=0}^{k-1} {k-1\choose j} t^j(1-t)^{k-1-j},
$$%
the (nontrivial) normalized B-splines for the Bernstein knots
$$%
\openB := (\ldots, 0, 0, 0, 1, 1, 1, \ldots)
$$%
are
$$%
N(t|\underbrace{0,\ldots,0}_{k+1-\mu\,\rm times},
\underbrace{1,\ldots,1}_{\mu\,\rm times})
= {k-1\choose \mu-1} t^{\mu-1} (1-t)^{k-\mu}, \quad \mu=1,\ldots,k.
$$%
Those normalized to integrate to 1 therefore are
$$%
M(t|\underbrace{0,\ldots,0}_{k+1-\mu\,\rm times},
\underbrace{1,\ldots,1}_{\mu\,\rm times})
= k! \braket{t}^{\mu-1} \braket{1-t}^{k-\mu}, \quad \mu=1,\ldots,k
$$%
(with $\braket{t}^j:= t^j/j!$ the normalized power function).
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It follows that, for any smooth $f$,
$$%
[\underbrace{0,\ldots,0}_{k+1-\mu\,\rm times},
\underbrace{1,\ldots,1}_{\mu\,\rm times}] f
= \int_0^1\braket{t}^{\mu-1}\braket{1-t}^{k-\mu} D^kf(t)\dd t.
$$%
Since, by the Genocchi formula,
$$%
[t_0,\ldots,t_k]f = \int_{[t_0,\ldots,t_k]} D^kf,
$$%
a linear change of variables finally gives
$$%
\int_{[\underbrace{x,\ldots,x}_{k+1-\mu\,\rm times},
\underbrace{y,\ldots,y}_{\mu\,\rm times}]} g
=
\int_0^1\braket{t}^{\mu-1}\braket{1-t}^{k-\mu} g(x + t(y-x))\dd t.
$$%
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\bye