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A four-dimensional product exactly integrated on a sparse node union

Example

Use the Clenshaw–Curtis levels with polynomial degrees 1 and 2 on [0,1] in four dimensions, and let Λ1={ℓ∈N04:∣ℓ∣1≤1}. For f(x)=x12x2x3x4, the Smolyak rule is exact with 48 distinct nodes; the full tensor product of level-one nodes has 81 nodes.

Facts & Assumptions

Given: f(x)=x12x2x3x4 on [0,1]4, with Q0:=Q1,[0,1] and Q1:=Q2,[0,1], the Clenshaw–Curtis rules of polynomial degrees one and two.

[F1]

Smolyak cubature on a finite downward-closed index set: QΛ1 is the sum of the tensor details at 0,e1,e2,e3,e4, and its node union is the union of the corresponding level-grid products.

[F2]

Deterministic node merging for Smolyak cubature: each distinct point in that union is counted once, including a point whose combined weight is zero.

[F3]

Clenshaw–Curtis quadrature with the midpoint zero-degree case: on [0,1], the nth rule samples at xj=(1+cos⁡(jπ/n))/2 for j=0,…,n; thus Q0 is level n=1 and Q1 is level n=2.

[F4]

Polynomial exactness and best-approximation error for Clenshaw–Curtis: the nth Clenshaw–Curtis rule integrates every polynomial of degree at most n exactly.

[F5]

Quarter-turn values and shifts by pi/2 and pi: cos⁡(π)=−1 and cos⁡(π/2)=0.

[F7]

The integral of a product function on a product rectangle is the product of the two integrals: the integral of a product over a product rectangle is the product of its coordinate integrals.

Verification

1.1givenF3F4F5F6F8algebra

The Given rules are the n=1 and n=2 rules. By [F3], their normalized nodes are (1+cos⁡(jπ/n))/2; [F5] and [F8] give X0={0,1} and X1={0,1/2,1}. By [F4], these rules are exact through degrees 1 and 2, respectively. Writing the level-zero weights as w0,w1, exactness on 1 and x gives w0+w1=1 and w1=1/2, hence Q0(x)=Q0(x2)=1/2. Also Q1(x)=1/2 and Q1(x2)=1/3 by exactness and [F6]. Thus Δ1(x)=0 and Δ1(x2)=−1/6.

2.1givenF1F6F7step 1.1algebra

Since f is a product, its tensor details factor into one-dimensional values. The zero-index term is Q0(x2)Q0(x)3=1/16. The e1 term is Δ1(x2)Q0(x)3=−1/48, and each ej term for j=2,3,4 vanishes because Δ1(x)=0. Therefore QΛ1(f)=1/16−1/48=1/24. By [F6]–[F7], I4(f)=(1/3)(1/2)3=1/24, so the cubature is exact.

3.1givenF1F2step 1.1algebra∎

The node union is X04 together with, for each coordinate j, the 8 points whose jth coordinate is 1/2 and whose other coordinates are endpoints. Those four midpoint sets are disjoint from one another and from the 16 corners, so W(Λ1)=16+4⋅8=48. The full tensor level-one grid has ∣X1∣4=34=81 nodes.

Depends on

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