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One-node Gauss data are midpoint data

Statement

Assume Countable Choice for the Lebesgue interval-measure formula. For Lebesgue measure on [0,1], the one-node Gaussian quadrature rule has node c1=1/2 and weight b1=1. Its one-stage collocation tableau therefore has Q11=1/2 and endpoint weight b1=1.

Facts & Assumptions

Given: Lebesgue measure on [0,1] and the one-node Gaussian quadrature rule from The m-node Gaussian quadrature rule for a positive measure.

[F1]

Lebesgue measure on the unit interval supplies Gaussian moment data: Lebesgue measure on [0,1] is valid one-node moment data through degree two, and its support has at least two points.

[F2]

Positive-measure Gaussian quadrature moment data: the order-two datum also meets the order-one branch used by the rule definition, because its measure, support, and moments through degree one satisfy that branch's conditions.

[F3]

Monic orthogonal polynomials exist uniquely through the quadrature order: for these data, the monic degree-one polynomial orthogonal to constants is unique.

[F4]

The m-node Gaussian quadrature rule for a positive measure and The Gaussian nodes, cardinal polynomials and weights are well-defined: the Gaussian node is the root of that polynomial and its weight is the integral of the one-node cardinal polynomial; the root and weight are defined.

[F5]

Cardinal-polynomial interpolation: the cardinal polynomial for one node is ℓ1(x)=1.

[F7]

Collocation tableau from distinct quadrature nodes: the one-stage collocation coefficient is Q11=∫0c1ℓ1(s) ds and the endpoint weight is b1=∫01ℓ1(s) ds.

[A1]

Countable Choice, through the Lebesgue interval-measure theorem used by [F1].

Proof

1.1A1F1F2F3F6algebraconstruct

By [F6], ∫01x dλ=∫01x dx=1/2 and ∫011 dλ=1. Thus p1(x)=x−1/2 is monic and satisfies ∫p1 dλ=0, the orthogonality condition against every degree-zero polynomial. The data are admissible by [A1, F1]–[F2], so uniqueness in [F3] identifies it with the one-node orthogonal polynomial.

2.1F4F5F7step 1.1algebra∎

The only root of p1 is 1/2, so [F4] gives c1=1/2. By [F5] its cardinal polynomial is 1, whose integral on [0,1] is 1 by step 1.1; hence b1=1. The tableau formula [F7] gives Q11=∫01/21 ds=1/2.

Depends on

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Sources