Paper 1, Section II, B

Numerical Analysis | Part IB, 2011

Consider a function f(x)f(x) defined on the domain x[0,1]x \in[0,1]. Find constants α,β\alpha, \beta, so that for any fixed ξ[0,1]\xi \in[0,1],

f(ξ)=αf(0)+βf(0)+γf(1)f^{\prime \prime}(\xi)=\alpha f(0)+\beta f^{\prime}(0)+\gamma f(1)

is exactly satisfied for polynomials of degree less than or equal to two.

By using the Peano kernel theorem, or otherwise, show that

f(ξ)f(0)ξ(αf(0)+βf(0)+γf(1))=0ξ(ξθ)H1(θ)f(θ)dθ+0ξθH2(θ)f(θ)dθ+ξ1ξH2(θ)f(θ)dθ\begin{aligned} f^{\prime}(\xi)-f^{\prime}(0)-\xi(\alpha f(0)&\left.+\beta f^{\prime}(0)+\gamma f(1)\right)=\int_{0}^{\xi}(\xi-\theta) H_{1}(\theta) f^{\prime \prime \prime}(\theta) d \theta \\ &+\int_{0}^{\xi} \theta H_{2}(\theta) f^{\prime \prime \prime}(\theta) d \theta+\int_{\xi}^{1} \xi H_{2}(\theta) f^{\prime \prime \prime}(\theta) d \theta \end{aligned}

where H1(θ)=1(1θ)20,H2(θ)=(1θ)20H_{1}(\theta)=1-(1-\theta)^{2} \geqslant 0, H_{2}(\theta)=-(1-\theta)^{2} \leqslant 0. Thus show that

f(ξ)f(0)ξ(αf(0)+βf(0)+γf(1))16(2ξ3ξ2+4ξ3ξ4)f.\left|f^{\prime}(\xi)-f^{\prime}(0)-\xi\left(\alpha f(0)+\beta f^{\prime}(0)+\gamma f(1)\right)\right| \leqslant\left.\frac{1}{6}\left(2 \xi-3 \xi^{2}+4 \xi^{3}-\xi^{4}\right)|| f^{\prime \prime \prime}\right|_{\infty} .

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