= Midpoint quadratic spline collocation
{title2=$A=\operatorname{tridiag}(1/8,3/4,1/8)$}
Sampling order-three <Cardinal B-splines> $N_j$ at $x_i=i+3/2$ gives $N_i(x_i)=3/4$, $N_{i-1}(x_i)=N_{i+1}(x_i)=1/8$, and all other entries zero. The <B-spline collocation matrix> therefore has row <strict diagonal dominance> margin at least $1/2$, giving inverse <operator norm> at most two. More precisely, with $q=3-2\sqrt2$, its absolute inverse row sums are $v_i=2[1-(q^i+q^{n+1-i})/(1+q^{n+1})]$. To see this, change signs by $D_{ii}=(-1)^i$: $DAD$ has negative off-diagonal entries and a nonnegative inverse given by a convergent <Neumann series>. Its inverse row sums solve $6v_i-v_{i-1}-v_{i+1}=8$, $v_0=v_{n+1}=0$. Thus $\|A^{-1}\|_\infty=v_{\lfloor(n+1)/2\rfloor}<2$, tending to two as the dimension grows.
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