Alphabeta Math
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Four two-node Gaussian rules give a sixteen-node tensor rule

Example

Take four coordinate moment data sets, each with a two-node Gaussian rule Qj(f)=wj,1f(xj,1)+wj,2f(xj,2). The tensor rule has node set {(x1,i1,x2,i2,x3,i3,x4,i4):ij∈{1,2}}, so it has 24=16 distinct nodes. It is exact for every coordinatewise polynomial P with degree at most three in each variable. In particular, Q⊗(x13x22x3x40)=∏j=14(∑i=12wj,ixj,iαj)=∏j=14∫xαj dμj(x),(α1,α2,α3,α4)=(3,2,1,0).

Facts & Assumptions

Given: Four order-three moment data sets and their two-node Gaussian rules.

[F1]

Tensor-product Gaussian quadrature: the tensor nodes are all coordinate tuples and the tensor weights are products of the coordinate weights.

[F2]

Coordinatewise product-moment functional for tensor polynomials: the product-moment functional on a coordinate monomial is the product of its coordinate moments.

[F3]

Tensor-product Gaussian quadrature preserves coordinatewise exactness: the tensor rule is exact for the stated coordinatewise degree ranges and has the product number of distinct nodes.

[F4]

Gaussian quadrature is exact through degree 2m−1 and has positive weights: each coordinate rule integrates every degree-at-most-three monomial exactly.

[F5]

The Gauss orthogonal polynomial has simple roots inside the interval: each two-node Gaussian rule has distinct nodes.

Proof

1.1F1construct

Since each coordinate has two distinct nodes by [F5], the sixteen coordinate tuples are distinct: two different tuples differ in a coordinate whose node entries are distinct.

2.1F1F2F3F4step 1.1algebra∎

The node-count statement of [F3] gives 2⋅2⋅2⋅2=16. For each coordinate degree αj≤3, [F4] evaluates the corresponding one-dimensional sum as its moment. Distributing the finite sums in [F1] therefore gives the product of those four moments, which is [F2]. Taking the specified exponents proves the displayed calculation and the general coordinatewise exactness claim.

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