Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 318 / 3

A B-spline $N_j$ is positive exactly in the interior of its support $(t_j,t_{j+k})$. If the B-spline collocation matrix $A_{\mathbf{x}}$ is invertible, its determinant contains a nonzero permutation term. Hence there is a permutation $\pi$ such that

If $x_i\le t_i$, then every one of the first $i$ points satisfies $x_r<t_i$. Any support containing such a point must have left endpoint $t_j<x_r$, hence $j<i$. Only $i-1$ B-splines are available to match these $i$ rows, contradicting that $\pi$ is a permutation. Thus $x_i>t_i$. Similarly, if $x_i\ge t_{i+k}$, each of the last $n-i+1$ points can only be matched to an index $j>i$, but only $n-i$ such indices exist. Therefore $x_i<t_{i+k}$. We have proved

The Schoenberg–Whitney theorem states, for strictly increasing knots and interpolation sites, thatIndeed its determinant is positive under these inequalities.

Write the interpolating spline asThe interpolation equations are $A_{\mathbf{x}}c=f_{\mathbf{x}}$, where $(f_{\mathbf{x}}{)}_i=f(x_i)$, soThe B-splines are nonnegative and form a partition of unity on the spline interval. ConsequentlyTaking the supremum over $t$ and then over $\Vert f{\Vert }_{\infty }\le 1$ proves the operator norm bound

For Cardinal cubic B-splines, the values at integer knots areand all other knot values vanish. Since $x_i=t_{i+2}$,This tridiagonal matrix is strictly diagonally dominant. The standard inverse bound for such a matrix givesThe minimum denominator is $4/6-2/6=1/3$, soPart (b) then yields