2017
DOI: 10.1007/s00605-017-1078-2
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On the orthogonality of the Chebyshev–Frolov lattice and applications

Abstract: We deal with lattices that are generated by the Vandermonde matrices associated to the roots of Chebyshev-polynomials. If the dimension d of the lattice is a power of two, i.e. d = 2 m , m ∈ N, the resulting lattice is an admissible lattice in the sense of Skriganov [12]. These are related to the Frolov cubature formulas, which recently drew attention due to their optimal convergence rates [18] in a broad range of Besov-Lizorkin-Triebel spaces. We prove that the resulting lattices are orthogonal and possess a … Show more

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Cited by 9 publications
(22 citation statements)
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“…Note that the coefficient of n −λ−1/2+δ in the latter term is bounded under the assumptions we made, i.e., α > λ > 1/2 and 0 < δ < λ − 1/2, as well as (9). This completes the proof.…”
Section: The Randomized Lattice Algorithm Without Shiftsupporting
confidence: 60%
See 2 more Smart Citations
“…Note that the coefficient of n −λ−1/2+δ in the latter term is bounded under the assumptions we made, i.e., α > λ > 1/2 and 0 < δ < λ − 1/2, as well as (9). This completes the proof.…”
Section: The Randomized Lattice Algorithm Without Shiftsupporting
confidence: 60%
“…The advantage over previous related works lies in the potential for implementation. Other known algorithms which achieve the optimal error bounds, such as those based on Frolov's method (see, e.g., [6,9,11,28,29,30]), are very difficult to implement especially in high dimensions. For our algorithm, a simple probabilistic approach can be used to obtain suitable generating vectors, see Remark 10 below.…”
Section: Introductionmentioning
confidence: 99%
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“…To gain this, we use a reduction of the problem to theÅ s p,θ setting via tailored transformations in connection with the Frolov cubature formulae [12,36], which recently attracted significant interest [19,21,[44][45][46]. It has been already proved by Bykovskii [3] …”
Section: State Of the Art And Relevant Literaturementioning
confidence: 99%
“…We have 19) where the sum on the right-hand side is taken over all k ∈ N d 0 such that 0 ≤ k i ≤ m if i ∈ e 1 ( j, ) and k i = 0 otherwise. In addition, we put (k · h) = (k 1 h 1 , .…”
Section: Proof Of Theorems 41 and 42mentioning
confidence: 99%