Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 69 / 5 / a / Solution

Fix $t$ and let $F_t(u)=(u-t{)}_{+}^{k-1}$. Initially assume distinct spline knots, so the ordinary polynomial interpolation definitions are unambiguous. The two interpolants ${\ell }_i(\cdot ,t)$ and ${\ell }_{i+1}(\cdot ,t)$ agree with $F_t$ at their $k-1$ shared nodes $t_{i+1},\dots ,t_{i+k-1}$. Their difference has degree at most $k-1$ and all these roots, henceFor $k=1$ the product is empty and the same statement concerns constants. The leading coefficient of an interpolating polynomial is its highest-order divided difference, so the divided difference recurrence givesThis proves Lee's formula:Summing over $i$ telescopes to ${\ell }_{n+1}(x,t)-{\ell }_1(x,t)$. If $t_k<t<t_{n+1}$, the first interpolation nodes $t_1,\dots ,t_k$ are all smaller than $t$, so their truncated-power data vanish and ${\ell }_1=0$. The last nodes $t_{n+1},\dots ,t_{n+k}$ are larger than $t$, so their data are values of the polynomial $(u-t{)}^{k-1}$ and uniqueness gives ${\ell }_{n+1}(x,t)=(x-t{)}^{k-1}$. Thus the telescoped identity isthe Marsden identity, valid for every real $x$.

If repeated spline knots are allowed, ordinary value interpolation at repeated nodes does not specify ${\ell }_i$ uniquely. The standard extension uses confluent divided differences and Hermite interpolation, or limits of distinct knots: the same polynomial identity then persists wherever these quantities are defined. At a maximally repeated knot where the truncated power lacks the required derivative, the appropriate one-sided spline convention must be fixed; an unspecified classical derivative is not part of the proof. For distinct knots, the printed identity requires no such qualification.