For <Cardinal cubic B-splines>, the values at integer knots are
$$
N_i(t_{i+1})=\frac16,
\qquad
N_i(t_{i+2})=\frac46,
\qquad
N_i(t_{i+3})=\frac16,
$$
and all other knot values vanish. Since $x_i=t_{i+2}$,
$$
\boxed{A_{\mathbf x}=\frac16
\begin{pmatrix}
4&1&0&\cdots&0\\
1&4&1&\ddots&\vdots\\
0&\ddots&\ddots&\ddots&0\\
\vdots&\ddots&1&4&1\\
0&\cdots&0&1&4
\end{pmatrix}}.
$$
This <tridiagonal matrix> is strictly diagonally dominant. The standard inverse bound for such a matrix gives
$$
\|A_{\mathbf x}^{-1}\|_{\ell_\infty}
\leq\frac1{\min_i\left(|a_{ii}|-\sum_{j\ne i}|a_{ij}|\right)}.
$$
The minimum denominator is $4/6-2/6=1/3$, so
$$
\boxed{\|A_{\mathbf x}^{-1}\|_{\ell_\infty}\leq3}.
$$
Part (b) then yields
$$
\boxed{\|P_{\mathbf x}\|_{L_\infty}\leq3}.
$$

Solution

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

{wiki}

A tridiagonal matrix has zero entries outside its main diagonal and the two adjacent diagonals. Linear systems with a nonsingular tridiagonal coefficient matrix can be solved in linear time by the <Thomas algorithm>.

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