Numerical Analysis Grinshpan Complete Cubic Spline Let x1,...,xn ...

Proof. <strong>Let</strong> f(x) satisfy the conditions of the theorem and let∆(x) = s(x) − f(x). Then ∆(x) has a root at each x k and∆ ′ (a) = ∆ ′ (b) = 0. Observe that∫ bas ′′ (x)∆ ′′ (x)dx = s ′′ (x)∆ ′ (x) ∣ ∫ bb − s ′′′ (x)∆ ′ (x)dxaTherefore∫ ba= −∫ ban−1as ′′′ (x)∆ ′ (x)dx (because ∆ ′ (a) = ∆ ′ (b) = 0)∑∫ xk+1= − c k ∆ ′ (x)dx (because s ′′′ (x) ∣ (xkis constant),x k+1 )x kk=1∑n−1[= − c k ∆(xk+1 ) − ∆(x k ) ]k=1= 0 (because ∆(x k ) = 0).f ′′ (x) 2 dx ==≥∫ ba∫ ba∫ ba(s ′′ (x) − ∆ ′′ (x)) 2 dxs ′′ (x) 2 dx +s ′′ (x) 2 dx.∫ ba∆ ′′ (x) 2 dxEquality holds if and only if ∆(x) is linear. But then∆(a) = ∆ ′ (a) = 0 implies that ∆(x) is zero.Error estimateIt has been shown (Hall 1968, Hall & Meyer 1976) that, if y(x) is fourtimes continuously differentiable on [a, b] and δ max = max k δ k , then|y(x) − s(x)| ≤ 5max384[a,b]|y (4) (x)| δ 4 max, a ≤ x ≤ b.5This estimate is optimal in the sense that the constantsmallest possible.384is the3□