Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 67 / 5 / b / Solution

The B-spline collocation matrix in the cubic case is the finite tridiagonal matrixFor $n=1$, this is simply the one-entry matrix $(2/3)$.

To evaluate the inverse norm exactly, let $D=\mathrm{diag}((-1{)}^1,\dots ,(-1{)}^n)$. Then $B_n=D{A}_{n}D=(4I-C)/6$, where $C$ has ones immediately above and below the diagonal and zeros elsewhere. Since $\Vert C/4{\Vert }_{{\ell }^{\infty }}\le 1/2$, the convergent Neumann serieshas nonnegative entries and proves invertibility. Since $A_n^{-1}=D{B}_n^{-1}D$, its absolute entries equal those of $B_n^{-1}$. The absolute row sums are therefore the entries of $u=B_n^{-1}1$.

The equation $B_{n}u=1$ becomesIts constant particular solution is $3$; the two characteristic roots of the homogeneous recurrence are $\rho =2-\sqrt{3}$ and ${\rho }^{-1}=2+\sqrt{3}$. Imposing the two boundary values givesThis can also be checked by direct substitution, since $\rho +{\rho }^{-1}=4$. The numerator is smallest at the central index or central pair of indices: its successive difference is $(1-\rho )({\rho }^{n-i}-{\rho }^i)$. Thus, with $j=\lfloor (n+1)/2\rfloor$, the exact inverse norm of midpoint cubic spline collocation isEquivalently, if $n=2m-1$ the fraction is $2{\rho }^m/(1+{\rho }^{2m})$, and if $n=2m$ it is $({\rho }^m+{\rho }^{m+1})/(1+{\rho }^{2m+1})$. The finite norm is not exactly three: it equals $3/2$ for $n=1$, two for $n=2$, and tends to three as $n$ grows. Hence three is the sharp dimension-independent bound for these inverse matrix norms.