Paper 3, Section II, D

Numerical Analysis
Part IB, 2018

Taylor's theorem for functions fCk+1[a,b]f \in C^{k+1}[a, b] is given in the form

f(x)=f(a)+(xa)f(a)++(xa)kk!f(k)(a)+R(x).f(x)=f(a)+(x-a) f^{\prime}(a)+\cdots+\frac{(x-a)^{k}}{k !} f^{(k)}(a)+R(x) .

Use integration by parts to show that

R(x)=1k!ax(xθ)kf(k+1)(θ)dθR(x)=\frac{1}{k !} \int_{a}^{x}(x-\theta)^{k} f^{(k+1)}(\theta) d \theta

Let λk\lambda_{k} be a linear functional on Ck+1[a,b]C^{k+1}[a, b] such that λk[p]=0\lambda_{k}[p]=0 for pPkp \in \mathbb{P}_{k}. Show that

λk[f]=1k!abK(θ)f(k+1)(θ)dθ\lambda_{k}[f]=\frac{1}{k !} \int_{a}^{b} K(\theta) f^{(k+1)}(\theta) d \theta

where the Peano kernel function K(θ)=λk[(xθ)+k].[K(\theta)=\lambda_{k}\left[(x-\theta)_{+}^{k}\right] . \quad[ You may assume that the functional commutes with integration over a fixed interval.]

The error in the mid-point rule for numerical quadrature on [0,1][0,1] is given by

e[f]=01f(x)dxf(12)e[f]=\int_{0}^{1} f(x) d x-f\left(\frac{1}{2}\right)

Show that e[p]=0e[p]=0 if pp is a linear polynomial. Find the Peano kernel function corresponding to ee explicitly and verify the formula ( \dagger ) in the case f(x)=x2f(x)=x^{2}.