Consider a function f(x) defined on the domain x∈[0,1]. Find constants α,β, so that for any fixed ξ∈[0,1],
f′′(ξ)=αf(0)+βf′(0)+γf(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(ξ−θ)H1(θ)f′′′(θ)dθ+∫0ξθH2(θ)f′′′(θ)dθ+∫ξ1ξH2(θ)f′′′(θ)dθ
where H1(θ)=1−(1−θ)2⩾0,H2(θ)=−(1−θ)2⩽0. Thus show that
∣f′(ξ)−f′(0)−ξ(αf(0)+βf′(0)+γf(1))∣⩽61(2ξ−3ξ2+4ξ3−ξ4)∣∣f′′′∣∣∣∣∣∞.