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Peano kernel identity for one Simpson panel

Statement

Let a<b, put L=b−a and m=(a+b)/2, and let f∈C4([a,b]). Define the one-block Simpson value and its kernel by

S(f;a,b):=L6(f(a)+4f(m)+f(b)),Ka,b(x):={(x−a)3(3(x−a)−2L)72,a≤x≤m,(b−x)3(3(b−x)−2L)72,m≤x≤b.

Then

∫abf(x) dx−S(f;a,b)=∫abKa,b(x)f(4)(x) dx,

and Ka,b≤0 with ∫ab∣Ka,b(x)∣ dx=L5/2880. The value S(f;a,b) is the one-block specialization of Composite Simpson rule on an even uniform grid.

Facts & Assumptions

Given: Reals a<b, L=b−a, m=(a+b)/2, and f∈C4([a,b]).

[F1]

Composite Simpson rule on an even uniform grid: on two subintervals of equal width L/2, the Simpson output is L6(f(a)+4f(m)+f(b)).

[F2]

Higher derivatives and the classes Ck and C∞: C4([a,b]) means all derivatives through order four exist and are continuous, with endpoint derivatives taken from within the interval.

[F3]

If u,v are differentiable on [a,b] with u′,v′ integrable, then ∫abuv′=u(b)v(b)−u(a)v(a)−∫abu′v: for differentiable u,v with integrable derivatives, ∫uv′=[uv]−∫u′v on a closed interval.

[F4]

Every continuous function on a closed nondegenerate rectangle in Rm is Riemann integrable: every continuous function on a closed nondegenerate interval is Riemann integrable.

[F6]

Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function: constants and the identity are continuous; sums, scalar multiples and products preserve continuity, so the polynomial pieces of Ka,b and their products with f(4) are continuous on each half-panel.

Proof

1.1givenF2F4F6algebra

On the left half, Ka,b(x)=(3y4−2Ly3)/72 with y=x−a; hence K(4)=1, K(a)=K′(a)=K′′(a)=0, and at m its value and first two derivatives are −L4/1152, 0, and L2/24, while K′′′(m−)=L/3. The right-half formula gives K(4)=1, K(b)=K′(b)=K′′(b)=0, and K′′′(m+)=−L/3. Thus K,K′,K′′ match at m. Each polynomial formula for K and its product with the continuous f(4) is continuous on its closed half by [F2] and [F6]. Both formulas agree at m, so K and Kf(4) are continuous on [a,b]; also f is continuous by [F2]. These functions and their restrictions are integrable by [F4].

2.1F1F2F3F4F6F10step 1.1algebra

On each closed half, every derivative K(r) for 0≤r≤4 is polynomial and continuous by [F6], and every f(s) for 0≤s≤4 is continuous by [F2]. They are all integrable by [F4], so the four applications of integration by parts [F3] satisfy their derivative-integrability hypotheses. They give ∫Ka,bf(4)=[B]+∫f, where B=Kf′′′−K′f′′+K′′f′−K′′′f and the bracket is the endpoint difference for that half. Summing the brackets, the terms at m involving K,K′,K′′ cancel, while the jump in K′′′ contributes −2Lf(m)/3; the outer endpoints contribute −L(f(a)+f(b))/6. Additivity [F10] combines both half-integrals of Kf(4) and of f; the remaining integral is ∫abf because K(4)=1. By [F1], this is precisely ∫abf−S(f;a,b), proving the identity.

3.1F5F7F8F9F10step 1.1algebra∎

Since 0≤x−a≤L/2 on the left half, 3(x−a)−2L<0, so Ka,b≤0 there; the right formula gives the same sign. Define H(x)=35(x−a)5−L2(x−a)4 on [a,m]. The power rule [F7], chain rule [F9], and sum/scalar rules [F8] give H′(x)=3(x−a)4−2L(x−a)3. By the FTC [F5], additivity [F10], and symmetry of the two kernel pieces, and H(m)−H(a)=−L5/80, ∫ab∣Ka,b∣=−2(H(m)−H(a))/72=L5/2880.

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