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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 , the one-node Gaussian quadrature rule has node and weight . Its one-stage collocation tableau therefore has and endpoint weight .
Facts & Assumptions
Given: Lebesgue measure on and the one-node Gaussian quadrature rule from The m-node Gaussian quadrature rule for a positive measure.
Lebesgue measure on the unit interval supplies Gaussian moment data: Lebesgue measure on is valid one-node moment data through degree two, and its support has at least two points.
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.
Monic orthogonal polynomials exist uniquely through the quadrature order: for these data, the monic degree-one polynomial orthogonal to constants is unique.
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.
Cardinal-polynomial interpolation: the cardinal polynomial for one node is .
Riemann and Lebesgue integrals agree for continuous data on [0,1] and The second fundamental theorem: if is differentiable on with and is integrable, then : for continuous and on , Lebesgue and Riemann integrals agree, and the Riemann integrals are computed by primitives.
Collocation tableau from distinct quadrature nodes: the one-stage collocation coefficient is and the endpoint weight is .
Countable Choice, through the Lebesgue interval-measure theorem used by [F1].
Proof
By [F6], and . Thus is monic and satisfies , 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.
The only root of is , so [F4] gives . By [F5] its cardinal polynomial is , whose integral on is by step 1.1; hence . The tableau formula [F7] gives .
Depends on
- Positive-measure Gaussian quadrature moment data
- The m-node Gaussian quadrature rule for a positive measure
- Monic orthogonal polynomials exist uniquely through the quadrature order
- The Gaussian nodes, cardinal polynomials and weights are well-defined
- Lebesgue measure on the unit interval supplies Gaussian moment data
- Riemann and Lebesgue integrals agree for continuous data on [0,1]
- The second fundamental theorem: if $G$ is differentiable on $[a,b]$ with $G' = f$ and $f$ is integrable, then $\int_a^b f = G(b)-G(a)$
- Cardinal-polynomial interpolation
- Collocation tableau from distinct quadrature nodes
- The Axiom of Countable Choice ($\mathrm{AC}_\omega$)
Used by
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Sources
- John A. Gubner, Gaussian Quadrature and the Eigenvalue Problem, §§2–5, 2020 (standard reference, not scraped)
- Ernst Hairer, Syvert P. Nørsett, and Gerhard Wanner, Solving Ordinary Differential Equations I, 2nd ed., Chapter II §II.7 (standard reference, not scraped)