OWR-9790358-011 · Partial answer: an explicit rule for n = 3, no symmetric rule for n = 4

Gaussian Cubature of Even Degree for the Product Chebyshev Weight in Three Variables: An Explicit Ten-Node Rule and No Twenty-Node Rule with a Sign Symmetry

Manuscript 15 October 2026 · Online 15 October 2026

math.NAmath.CAmath.ACUnrefereed preprint

Abstract

In Oberwolfach Report 7/2022 Y. Xu asked whether Gaussian cubature rules of degree \(2n-2\), with \(\dim\Pi^3_{n-1}=\binom{n+2}{3}\) real nodes, exist for the product Chebyshev weight of the second kind on the cube \([-1,1]^3\); existence is known only for \(n=2\) and \(n=3\). (The source prints the formula of the weight of the first kind under the name of the second kind; we follow the name.) We give partial results. From Xu's characterisation of Gaussian rules of even degree we derive that, for this weight in \(d\) variables, such a rule exists if and only if an explicit quadratic system has a real solution: for a function \(h\) on the multi-indices of length \(2n-1\), the matrix \(\bigl(\sum_{\beta}h(\alpha+\beta)h(\alpha'+\beta)-\delta_{\alpha\alpha'}\bigr)\) must depend only on \(\alpha+\alpha'\); the rules of Morrow and Patterson in two variables correspond to \(h=\delta_{(n,n-1)}\). For \(n=3\) we give an explicit ten-node rule with nodes and weights in \(\mathbb{Q}(\sqrt2,\sqrt3)\). For \(n=4\) we prove that no Gaussian rule of degree \(6\) has a node set invariant under a coordinate reflection, a half-turn about a coordinate axis, or \(x\mapsto-x\), hence under any non-trivial group of coordinate sign changes. The cases of one reflection and of a half-turn are computer-assisted, by exact Nullstellensatz certificates with rational coefficients, checked by independently written programs; two reflections and the central symmetry have short direct proofs. The case \(n=4\) remains open for node sets whose symmetry group contains no non-trivial sign change (for example a group generated by \((x,y,z)\mapsto(-y,-x,z)\) or by a rotation of order three, or the trivial group), and so do all \(n\ge5\); numerical searches for \(n=4\) and \(n=5\) found no rule, which is evidence only. The question and the theory of Gaussian cubature of even degree are due to Xu, building on Morrow, Patterson and Schmid; new, to our knowledge, are the exclusion of coordinate reflections and half-turns for \(n=4\) and the explicit rule for \(n=3\). This is an unrefereed note.

Record

Affiliation
Mercury Software GmbH
Result
Partial answer: an explicit rule for n = 3, no symmetric rule for n = 4
Categories
math.NA · math.CA · math.AC
Manuscript
15 October 2026
Online release
15 October 2026
Version
1.0
License
Creative Commons Attribution 4.0 International

Files and verification

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Citation

Alper Ferudun, “Gaussian Cubature of Even Degree for the Product Chebyshev Weight in Three Variables: An Explicit Ten-Node Rule and No Twenty-Node Rule with a Sign Symmetry,” EulerSolve Research Papers, OWR-9790358-011, 2026. https://doi.org/10.5281/zenodo.23399617.

BibTeX
@misc{Ferudun2026Owr9790358011,
  author = {Ferudun, Alper},
  title = {Gaussian Cubature of Even Degree for the Product Chebyshev Weight in Three Variables: An Explicit Ten-Node Rule and No Twenty-Node Rule with a Sign Symmetry},
  year = {2026},
  howpublished = {EulerSolve Research Papers},
  url = {https://eulersolve.org/papers/owr-9790358-011/},
  doi = {10.5281/zenodo.23399617},
  note = {OWR-9790358-011; unrefereed preprint}
}

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