Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 68 / 4

For fixed $t$, put $g_t(u)=(u-t{)}_{+}^{k-1}$, a truncated power function. Begin with distinct increasing knots. The two interpolation polynomials ${\ell }_i(\cdot ,t)$ and ${\ell }_{i+1}(\cdot ,t)$ agree at their $k-1$ shared knots. Their difference has degree at most $k-1$, soFor $k=1$ the empty product is one and the same assertion holds for constant interpolants. The leading coefficient of a degree-at-most-$k-1$ interpolant on $k$ nodes is its order-$k-1$ divided difference, by the Newton interpolation polynomial. ThereforeThis proves the Lee interpolation identityThe identity holds for every real $x,t$, with the usual convention for the order-zero truncated power. Admissible repeated knots are handled by the corresponding confluent divided differences and consistent one-sided limits at breakpoints; taking these limits preserves the Lee interpolation identity.

Sum over $i$ to telescope:If $t_k<t<t_{n+1}$, the first $k$ knot values of $g_t$ are zero, so ${\ell }_1=0$. At the last $k$ knots $t_{n+1},\dots ,t_{n+k}$, the function $g_t$ agrees with the polynomial $(u-t{)}^{k-1}$, so uniqueness of polynomial interpolation gives ${\ell }_{n+1}(x,t)=(x-t{)}^{k-1}$. Thus the Marsden identity is

Let $d=k-1$ and let $e_m$ denote the elementary symmetric polynomial of degree $m$ in the $d$ interior knots $t_{i+1},\dots ,t_{i+d}$, with $e_0=1$. Expand the two sides of the Marsden identity as polynomials in $x$:andComparing the coefficient of $x^{d-m}$ and cancelling the common sign gives the monomial B-spline coefficientsEquivalently,In particular $a_i^{(0)}=1$, giving partition of unity on the interior interval. For $k\ge 2$, $a_i^{(1)}=(t_{i+1}+\cdots +t_{i+k-1})/(k-1)$, and the highest-degree coefficient is the product of those interior knots. The formulas also cover $k=1,m=0$ through empty products.