= Solution
Let $d=k-1$ and let $e_m$ denote the <elementary symmetric polynomial> of degree $m$ in the $d$ interior knots $t_{i+1},\ldots,t_{i+d}$, with $e_0=1$. Expand the two sides of the <Marsden identity> as <polynomials> in $x$:
$$
(x-t)^d=\sum_{m=0}^d(-1)^m\binom dm t^m x^{d-m},
$$
and
$$
\omega_i(x)=\prod_{j=1}^d(x-t_{i+j})
=\sum_{m=0}^d(-1)^m e_m(t_{i+1},\ldots,t_{i+d})x^{d-m}.
$$
Comparing the coefficient of $x^{d-m}$ and cancelling the common sign gives the <monomial B-spline coefficients>
$$
\boxed{a_i^{(m)}=
\frac{e_m(t_{i+1},\ldots,t_{i+k-1})}{\binom{k-1}{m}},
\qquad 0\leq m\leq k-1.}
$$
Equivalently,
$$
a_i^{(m)}=\binom{k-1}{m}^{-1}
\sum_{1\leq j_1<\cdots<j_m\leq k-1}
t_{i+j_1}\cdots t_{i+j_m}.
$$
In particular $a_i^{(0)}=1$, giving <partition of unity> on the interior interval. For $k\geq2$, $a_i^{(1)}=(t_{i+1}+\cdots+t_{i+k-1})/(k-1)$, and the highest-degree coefficient is the product of those interior knots. The formulas also cover $k=1,m=0$ through empty products.
Back to article page
