On generalizations of $p$-sets and their applications

TL;DR

This paper introduces generalized $p$-sets that are more flexible than traditional $p$-sets, providing new constructions with potential applications in numerical analysis and related fields, assuming Goldbach's conjecture.

Contribution

The paper develops generalized $p$-sets $\\mathcal{P}_{d,p}^{{\mathbf a},\epsilon}$ and constructs point sets ${\mathcal L}_{p,q}$ under Goldbach's conjecture, overcoming limitations of classical $p$-sets.

Findings

Upper bounds for exponential sums over the new $p$-sets.

Construction of point sets with arbitrary even cardinalities.

Potential applications in numerical integration, compressed sensing, and UQ.

Abstract

The $p$-set, which is in a simple analytic form, is well distributed in unit cubes. The well-known Weil's exponential sum theorem presents an upper bound of the exponential sum over the $p$-set. Based on the result, one shows that the $p$-set performs well in numerical integration, in compressed sensing as well as in UQ. However, $p$-set is somewhat rigid since the cardinality of the $p$-set is a prime $p$ and the set only depends on the prime number $p$. The purpose of this paper is to present generalizations of $p$-sets, say $\mathcal{P}_{d,p}^{{\mathbf a},\epsilon}$, which is more flexible. Particularly, when a prime number $p$ is given, we have many different choices of the new $p$-sets. Under the assumption that Goldbach conjecture holds, for any even number $m$, we present a point set, say ${\mathcal L}_{p,q}$, with cardinality $m-1$ by combining two different new $p$-sets, which overcomes a major bottleneck of the $p$-set. We also present the upper bounds of the exponential sums over $\mathcal{P}_{d,p}^{{\mathbf a},\epsilon}$ and ${\mathcal L}_{p,q}$, which imply these sets have many potential applications.

Equations

$${\mathcal{P}}_{d,p}=\left\{{\mathbf{x}}_0,\dots ,{\mathbf{x}}_{p-1}\right\}\subset [0,1{)}^d$$

$$\mathbf{x}_{j}=\Big{(}\left\{\frac{j}{p}\right\},\left\{\frac{j^{2}}{p}\right\},\ldots,\left\{\frac{j^{d}}{p}\right\}\Big{)}\in[0,1)^{d},\quad j\in\mathbb{Z}_{p},$$

$${\mathbb{Z}}_p^d:=\left\{\mathbf{a}=(a_1,\dots ,a_d)\in {\mathbb{Z}}^d:a_j\in {\mathbb{Z}}_p,j=1,\dots ,d\right\}.$$

$${\mathcal{P}}_{d,p}^{\mathbf{a},ϵ}:=\left\{{\mathbf{x}}_j^{\mathbf{a},ϵ}:j\in {\mathbb{Z}}_p\right\}$$

$$\mathbf{x}_{j}^{\mathbf{a},\epsilon}:=\Big{(}\left\{\frac{a_{1}j}{p}\right\},\left\{\frac{a_{1}^{\prime}j+a_{2}j^{2}}{p}\right\},\ldots,\left\{\frac{\sum_{h=1}^{d-1}a_{h}^{\prime}j^{h}+a_{d}j^{d}}{p}\right\}\Big{)}\in[0,1)^{d}$$

$$\mathcal{L}_{p,q}:=\left\{\begin{array}[]{c}\mathcal{P}_{d,p}\cup\mathcal{P}_{d,q},\quad p\neq q\\ \mathcal{P}^{\mathbf{a},\epsilon^{\prime}}_{d,p}\cup\mathcal{P}^{\mathbf{b},\epsilon^{\prime\prime}}_{d,p},\quad p=q,\end{array}\right.$$

$${\mathcal{Q}}_{p^2,d}$$

$${\mathcal{R}}_{p^2,d}$$

$$\Big{|}\sum_{x=1}^{p}e^{\frac{2\pi\mathbf{i}f(x)}{p}}\Big{|}\leq(d-1)\sqrt{p}.$$

$${\mathcal{P}}_{d,p}^{\mathbf{a},ϵ}:=\left\{{\mathbf{x}}_j^{\mathbf{a},ϵ}:j\in {\mathbb{Z}}_p\right\}$$

$$\mathbf{x}_{j}^{\mathbf{a},\epsilon}=\Big{(}\left\{\frac{a_{1}j}{p}\right\},\left\{\frac{a_{1}^{\prime}j+a_{2}j^{2}}{p}\right\},\ldots,\left\{\frac{\sum_{h=1}^{d-1}a_{h}^{\prime}j^{h}+a_{d}j^{d}}{p}\right\}\Big{)}\in[0,1)^{d}$$

$$\left|\sum_{\mathbf{x}\in\mathcal{P}^{\mathbf{a},\epsilon}_{d,p}}\exp(2\pi\mathbf{i}\mathbf{k}\cdot\mathbf{x})\right|=\Big{|}\sum_{j=0}^{p-1}\exp(2\pi\mathbf{i}\mathbf{k}\cdot\mathbf{x}_{j}^{\mathbf{a},\epsilon})\Big{|}\leq(d-1)\sqrt{p}.$$

$$g(j)=\sum _{\ell =1}^d{c}_{\ell }{j}^{\ell }$$

$$\Big{|}\sum_{j=0}^{p-1}\exp(2\pi\mathbf{i}\mathbf{k}\cdot\mathbf{x}_{j}^{\mathbf{a},\epsilon})\Big{|}=\Big{|}\sum_{j=0}^{p-1}\exp\Big{(}2\pi\mathbf{i}\frac{g(j)}{p}\Big{)}\Big{|}\leq(d-1)\sqrt{p}.$$

$${\mathcal{P}}_{d,p}\cap {\mathcal{P}}_{d,q}=\left\{(0,\dots ,0)\right\}.$$

$$b_j{c}^j\equiv a_j(\mathrm{mod}p)forj=1,\dots ,d.$$

$${\mathcal{P}}_{d,p}^{\mathbf{a},ϵ}=\left\{{\mathbf{x}}_j^{\mathbf{a},ϵ}:j\in {\mathbb{Z}}_p\right\},$$

$$\mathbf{x}_{j}^{\mathbf{a},\epsilon}=\Big{(}\left\{\frac{a_{1}j}{p}\right\},\left\{\frac{a_{1}^{\prime}j+a_{2}j^{2}}{p}\right\},\ldots,\left\{\frac{\sum_{h=1}^{d-1}a_{h}^{\prime}j^{h}+a_{d}j^{d}}{p}\right\}\Big{)}.$$

$${\mathbf{x}}_{j_0}^{\mathbf{a},ϵ}$$

$${\mathcal{P}}_{d,p}^{\mathbf{a},ϵ}\subseteq {\mathcal{P}}_{d,p}^{\mathbf{b},ϵ}.$$

$${\mathbf{x}}_{j_0}^{\mathbf{b},ϵ}={\mathbf{x}}_{c^{-1}{j}_0}^{\mathbf{a},ϵ},$$

$${\mathcal{P}}_{d,p}^{\mathbf{b},ϵ}\subseteq {\mathcal{P}}_{d,p}^{\mathbf{a},ϵ}.$$

$${\mathcal{P}}_{d,p}^{\mathbf{a},ϵ}={\mathcal{P}}_{d,p}^{\mathbf{b},ϵ}.$$

$$a_1{j}_0\equiv b_1{k}_0(\mathrm{mod}p),$$

$$a_1^{\prime }{j}_0+a_2{j}_0^2\equiv b_1^{\prime }{k}_0+b_2{k}_0^2(\mathrm{mod}p),$$

$$⋮$$

$$\sum _{h=1}^{d-1}{a}_h^{\prime }{j}_0^h+a_d{j}_0^d\equiv \sum _{h=1}^{d-1}{b}_h^{\prime }{k}_0^h+b_d{k}_0^d(\mathrm{mod}p).$$

$$\mathcal{Z}:=\left\{j:{ϵ}_j^{\prime }\ne {ϵ}_j^{\prime \prime }\text{and}{a}_j^2+b_j^2\ne 0,1\le j\le d-1\right\},$$

$${\ell }_0:=\min \left\{j:j\in \mathcal{Z}\right\}.$$

$$a_j\equiv b_j{c}^j(\mathrm{mod}p)for{ϵ}_j^{\prime }={ϵ}_j^{\prime \prime }orj=d,$$

$$a_j=b_j=0for{ϵ}_j^{\prime }\ne {ϵ}_j^{\prime \prime },$$

Full text

On generalizations of $p$-sets and their applications

Heng Zhou

School of Sciences, Tianjin Polytechnic University, Tianjin 300160, China

E-mail address: zhouheng7598@sina.com.cn

 and 

Zhiqiang Xu

LSEC, Institute of Computational Mathematics, Academy of Mathematics and System Sciences, Chinese Academy of Sciences, Beijing 100190, China

E-mail address: xuzq@lsec.cc.ac.cn

Heng Zhou was supported by the National Natural Science Foundation of China (No. 61602341); Zhiqiang Xu was supported by NSFC grant (11422113, 91630203, 11331012) and by National Basic Research Program of China (973 Program 2015CB856000).

Abstract: The $p$-set, which is in a simple analytic form, is well distributed in unit cubes. The well-known Weil’s exponential sum theorem presents an upper bound of the exponential sum over the $p$-set. Based on the result, one shows that the $p$-set performs well in numerical integration, in compressed sensing as well as in UQ. However, $p$-set is somewhat rigid since the cardinality of the $p$-set is a prime $p$ and the set only depends on the prime number $p$. The purpose of this paper is to present generalizations of $p$-sets, say $\mathcal{P}_{d,p}^{\mathbf{a},\epsilon}$, which is more flexible. Particularly, when a prime number $p$ is given, we have many different choices of the new $p$-sets. Under the assumption that Goldbach conjecture holds, for any even number $m$, we present a point set, say ${\mathcal{L}}_{p,q}$, with cardinality $m-1$ by combining two different new $p$-sets, which overcomes a major bottleneck of the $p$-set. We also present the upper bounds of the exponential sums over $\mathcal{P}_{d,p}^{\mathbf{a},\epsilon}$ and ${\mathcal{L}}_{p,q}$, which imply these sets have many potential applications.

Key words and phrases $p$-set; Deterministic sampling; Numerical integral; Exponential sum

AMS Subject Classification 2000 11K38, 65C05, 11L05, 41A10, 65D32

1. Introduction

1.1. $p$-set

Let $p$ be a prime number. We consider the point set

[TABLE]

where

[TABLE]

$\mathbb{Z}_{p}:=\left\{0,1,\ldots,p-1\right\}$ and $\left\{x\right\}$ is the fractional part of $x$ for a nonnegative real number $x$. The point set $\mathcal{P}_{d,p}$ is called $p$-set and was introduced by Korobov [5] and Hua-Wang [4]. Recently, $p$-set attracts much attention since its advantage in numerical integration [1], in the recovery of sparse trigonometric polynomials [12] and in the UQ [13]. In [1], Dick presents a numerical integration formula based on $\mathcal{P}_{d,p}$ with showing the error bound of the formula depends only polynomially on the dimension $d$. In [12], Xu uses $\mathcal{P}_{d,p}$ to construct the deterministic sampling points of sparse trigonometric polynomials and show the sampling matrix corresponding to $\mathcal{P}_{d,p}$ has the almost optimal coherence. And hence, $\mathcal{P}_{d,p}$ has a good performance for the recovery of sparse trigonometric polynomials.

1.2. Extensions of $p$-set: $\mathcal{P}^{\mathbf{a},\epsilon}_{d,p}$ and $\mathcal{L}_{p,q}$

The $p$-set is in a simple analytic form and hence it is easy to be generated by computer. However, the $p$-set is somewhat rigid with the point set only depending on a prime number $p$. If the function values at some points in $p$-set are not easy to be obtained, one has to change the prime number $p$ to obtain a new point set which has the different cardinality with the previous one. Hence, in practical application, it will be better that one has many different choices. We next introduce a generalization of $p$-set.

Let

[TABLE]

Suppose that $\mathbf{a}=(a_{1},\ldots,a_{d})\in\mathbb{Z}^{d}_{p}$ and $\epsilon=(\epsilon_{1},\ldots,\epsilon_{d-1})\in\left\{0,1\right\}^{d-1}$. We set

[TABLE]

where

[TABLE]

and $a_{k}^{\prime}=\epsilon_{k}a_{k},k=1,\ldots,d-1$. We call $\mathcal{P}^{\mathbf{a},\epsilon}_{d,p}$ as the $p$-set associating with the parameter $\mathbf{a}$ and $\epsilon$. If we take $\mathbf{a}=(1,\ldots,1)$ and $\epsilon=(0,\ldots,0)$, then $\mathcal{P}^{\mathbf{a},\epsilon}_{d,p}$ is reduced to the classical $p$-set.

The $p$-set $\mathcal{P}^{\mathbf{a},\epsilon}_{d,p}$ associating with the parameters $\mathbf{a},\epsilon$ is more flexible. Given the prime number $p$, one can generate various point sets by changing the parameters $\mathbf{a}$ and $\epsilon$ with presenting an option set when the cardinality $p$ is given.

Note that the cardinality of both $\mathcal{P}^{\mathbf{a},\epsilon}_{d,p}$ and $\mathcal{P}_{d,p}$ is prime. Since the distance between adjacent prime can be very large, the cardinality of $p$-set does not change smoothly. Using the set $\mathcal{P}^{\mathbf{a},\epsilon}_{d,p}$, we next present a set with the cardinality being odd number. Suppose that $m\in 2\mathbb{Z}$ is given. The Goldbach conjecture, which is one of the best-known unsolved problem in number theory, says that $m$ can be written as the sum of two primes, i.e., $m=p+q$ where $p$ and $q$ are prime numbers. One has verified the conjecture up to $m\leq 4\cdot 10^{14}$ which is enough for practical application. We next suppose that $m=p+q$ with $p$ and $q$ being prime numbers. We set

[TABLE]

where $\mathcal{P}^{\mathbf{a},\epsilon^{\prime}}_{d,p}$ and $\mathcal{P}^{\mathbf{b},\epsilon^{\prime\prime}}_{d,p}$ are the $p$-sets that we have defined above and $\mathbf{a},\mathbf{b}\in\mathbb{Z}_{p}^{d},\epsilon^{\prime},\epsilon^{\prime\prime}\in\left\{0,1\right\}^{d-1}$. We call $\mathcal{L}_{p,q}$ the $(p,q)$-set. As shown later, $\mathcal{P}_{d,p}\cap\mathcal{P}_{d,q}=\left\{(0,\ldots,0)\right\}$ provided $p\neq q$. We can choose $\mathbf{a},\mathbf{b},\epsilon^{\prime}$ and $\epsilon^{\prime\prime}$ so that $\mathcal{P}^{\mathbf{a},\epsilon^{\prime}}_{d,p}\cap\mathcal{P}^{\mathbf{b},\epsilon^{\prime\prime}}_{d,p}=\left\{(0,\ldots,0)\right\}$. Hence, under the assumption of Goldbach conjecture, for any odd number, says $m-1$, there exist $p,q$ so that $|\mathcal{L}_{p,q}|=p+q-1=m-1$.

We would like to mention the following point sets with cardinality $p^{2}$ [5, 4] :

[TABLE]

The weighted star discrepancy of ${\mathcal{Q}}_{p^{2},d}$ and ${\mathcal{R}}_{p^{2},d}$ is given in [2]. Using a similar method with above, we can generalize ${\mathcal{Q}}_{p^{2},d}$ and ${\mathcal{R}}_{p^{2},d}$ to ${\mathcal{Q}}_{p^{2},d}^{\mathbf{a},\epsilon}$ and ${\mathcal{R}}_{p^{2},d}^{\mathbf{a},\epsilon}$, respectively. We will introduce it in Section 2.3 in detail.

1.3. Organization

In Section 2, we present the upper bounds of the exponential sums over $\mathcal{P}^{\mathbf{a},\epsilon}_{d,p}$ and $\mathcal{L}_{p,q}$. Particularly, we present the condition under which $\lvert\mathcal{L}_{p,q}\rvert=p+q-1$ and also prove that $\mathcal{P}^{\mathbf{a},\epsilon}_{d,p}\cap\mathcal{P}^{\mathbf{b},\epsilon}_{d,p}=\left\{(0,\ldots,0)\right\}$ when $\mathcal{P}^{\mathbf{a},\epsilon}_{d,p}\neq\mathcal{P}^{\mathbf{b},\epsilon}_{d,p}$. We furthermore consider the generalization of the point sets ${\mathcal{Q}}_{p^{2},d}$ and ${\mathcal{R}}_{p^{2},d}$ and present the upper bounds of exponential sums over the new sets. The results in Section 2 show that the point sets presented in this paper have many potential applications in various areas. In Section 3, we choose $\mathcal{L}_{p,q}$ as a deterministic sampling set for the recovery of sparse trigonometric polynomials and then show their performance.

2. The exponential sums over $\mathcal{P}^{\mathbf{a},\epsilon}_{d,p}$ and $\mathcal{L}_{p,q}$

The aim of this section is to present the exponential sums over $\mathcal{P}^{\mathbf{a},\epsilon}_{d,p}$ and $\mathcal{L}_{p,q}$. To this end, we first introduce the well-known Weil’s formula, which plays a key role in our proof.

Theorem 2.1**.**

[11]** Suppose that $p$ is a prime number. Suppose $f(x)=\sum_{h=1}^{d}m_{h}x^{h}$ with $m_{h}\in\mathbb{Z}$ $(h=1,\ldots,d)$ and there is a $j\in\left\{1,2,\ldots d\right\}$, satisfying $p\nmid m_{j}$. Then

[TABLE]

2.1. The exponential sum over $\mathcal{P}^{\mathbf{a},\epsilon}_{d,p}$

Recall that

[TABLE]

and

[TABLE]

where $\mathbf{a}=(a_{1},\ldots,a_{d})\in[1,p-1]^{d}\cap\mathbb{Z}^{d}$, $a_{j}^{\prime}=\epsilon_{j}a_{j}$ and $\epsilon=(\epsilon_{1},\ldots,\epsilon_{d-1})\in\left\{0,1\right\}^{d-1}$. Note that $|\mathcal{P}^{\mathbf{a},\epsilon}_{d,p}|=p$. We next show the exponential sum formula over $\mathcal{P}^{\mathbf{a},\epsilon}_{d,p}$.

Theorem 2.2**.**

For any $\mathbf{k}\in[-p+1,p-1]^{d}\cap\mathbb{Z}^{d}$ and $\mathbf{k}\neq 0$, we have

[TABLE]

Proof.

Set

[TABLE]

where $c_{\ell}=k_{\ell}a_{\ell}+k_{\ell+1}a_{\ell}^{\prime}+\cdots+k_{d}a_{\ell}^{\prime}$. We set $j_{0}:=\max\left\{\ell:k_{\ell}\neq 0\right\}$. Then $c_{j_{0}}=k_{j_{0}}a_{j_{0}}$ and we have $p\nmid c_{j_{0}}$. According to Theorem 2.1, we obtain that

[TABLE]

∎

2.2. The exponential sum over $\mathcal{L}_{p,q}$

To this end, we consider the cardinality of $\mathcal{L}_{p,q}$. A simple observation is that $\lvert\mathcal{L}_{p,q}\rvert\leq p+q-1$. We would like to present the condition under which $\lvert\mathcal{L}_{p,q}\rvert=p+q-1$. We first consider the case where $p\neq q$.

Theorem 2.3**.**

Suppose that $p$ and $q$ are two distinct prime numbers. Then $|\mathcal{L}_{p,q}|=p+q-1$.

Proof.

According to (2), to this end, we just need show that

[TABLE]

We prove it by contradiction. Assume that $\mathcal{P}_{d,p}\cap\mathcal{P}_{d,q}\neq\left\{(0,\ldots,0)\right\}$, and then there exists $j\in\mathbb{Z}_{p}^{*}:=\mathbb{Z}_{p}\setminus\left\{0\right\}$ and $k\in\mathbb{Z}_{q}^{*}$ so that $\left\{\frac{j^{i}}{p}\right\}=\left\{\frac{k^{i}}{q}\right\}$, $i=1,\ldots,d$. Particularly, we have $\frac{j}{p}=\frac{k}{q}$, which is equivalent to $jq=kp$. Since $p$ and $q$ are different prime numbers, $j\in\mathbb{Z}_{p}^{*}$ and $k\in\mathbb{Z}_{q}^{*}$, we have $j|k$ and $k|j$, which means $j=k$ and hence $p=q$. A contradiction. ∎

We next consider the case where $p=q$, i.e., ${\mathcal{L}}_{p,q}=\mathcal{P}^{\mathbf{a},\epsilon^{\prime}}_{d,p}\cup\mathcal{P}^{\mathbf{b},\epsilon^{\prime\prime}}_{d,p}$. For the case where $\epsilon^{\prime}=\epsilon^{\prime\prime}$, we have

Theorem 2.4**.**

Suppose that $\epsilon\in\left\{0,1\right\}^{d-1}$ is a fixed vector and $\mathbf{a}=(a_{1},\ldots,a_{d})\in\mathbb{Z}_{p}^{d}$, $\mathbf{b}=(b_{1},\ldots,b_{d})\in\mathbb{Z}_{p}^{d}$.

1. (1)

$\mathcal{P}^{\mathbf{a},\epsilon}_{d,p}=\mathcal{P}^{\mathbf{b},\epsilon}_{d,p}$* if and only if there exists $c\in\mathbb{Z}_{p}^{*}$ such that*

[TABLE] 2. (2)

If $\mathcal{P}^{\mathbf{a},\epsilon}_{d,p}\neq\mathcal{P}^{\mathbf{b},\epsilon}_{d,p}$, then $\mathcal{P}^{\mathbf{a},\epsilon}_{d,p}\cap\mathcal{P}^{\mathbf{b},\epsilon}_{d,p}=\left\{(0,\ldots,0)\right\}$.

Proof.

(1) We first suppose that there exists $c\in\mathbb{Z}_{p}^{*}$ such that $b_{j}c^{j}\equiv a_{j}\,\,(\mathrm{mod}\quad p)$ for $j\in\left\{1,\ldots,d\right\}$. Recall that

[TABLE]

where

[TABLE]

For any $j_{0}\in\mathbb{Z}_{p}$, we take $k_{0}\equiv cj_{0}\quad(\mathrm{mod}\quad p)$. Then

[TABLE]

which implies that

[TABLE]

Here we use $b_{j}^{\prime}c^{j}\equiv a_{j}^{\prime}\,\,(\mathrm{mod}\,\,p)$ which follows from $b_{j}c^{j}\equiv a_{j}\,\,(\mathrm{mod}\,\,p)$.

Since $p$ is a prime number, there exists $c^{-1}\in\mathbb{Z}_{p}^{*}$ so that $c^{-1}c\equiv 1\quad(\mathrm{mod}\,\,p)$. Then we have $b_{j}\equiv a_{j}c^{-j}\quad(\mathrm{mod}\quad p),j=1,\ldots,d$. Then, similarly, for any $j_{0}\in\mathbb{Z}_{p}$,

[TABLE]

which implies that

[TABLE]

Then we arrive at

[TABLE]

We next suppose that $\mathcal{P}^{\mathbf{a},\epsilon}_{d,p}=\mathcal{P}^{\mathbf{b},\epsilon}_{d,p}$. Then there exist $j_{0},k_{0}\in\mathbb{Z}_{p}^{*}$ so that $\mathbf{x}_{j_{0}}^{\mathbf{a}}=\mathbf{x}_{k_{0}}^{\mathbf{b}}$, i.e.,

[TABLE]

[TABLE]

[TABLE]

[TABLE]

We set $c:\equiv k_{0}j_{0}^{-1}\quad(\mathrm{mod}\quad p)$, where $j_{0}^{-1}\in\mathbb{Z}_{p}$ so that $j_{0}^{-1}j_{0}\equiv 1\quad(\mathrm{mod}\quad p)$. Then (5) implies that $a_{1}\equiv b_{1}c\quad(\mathrm{mod}\quad p)$. Combining (5) and (6), we have $a_{2}j_{0}^{2}\equiv b_{2}k_{0}^{2}\quad(\mathrm{mod}\quad p)$ which implies that $a_{2}\equiv b_{2}c^{2}\quad(\mathrm{mod}\quad p)$. Similarly, we can obtain that $a_{j}\equiv b_{j}c^{j}\quad(\mathrm{mod}\quad p)$ for $3\leq j\leq d$.

(2) We prove it by contradiction. Assume that $\mathcal{P}_{d,p}\cap\mathcal{P}_{d,q}\neq\left\{(0,\ldots,0)\right\}$, and then there exist $j_{0},k_{0}\in\mathbb{Z}_{p}^{*}$ so that $\mathbf{x}_{j_{0}}^{\mathbf{a},\epsilon}=\mathbf{x}_{k_{0}}^{\mathbf{b},\epsilon}$. Similarly with the above proof, we can find a $c:\equiv k_{0}j_{0}^{-1}\quad(\mathrm{mod}\quad p)$ so that $a_{j}\equiv b_{j}c^{j}\quad(\mathrm{mod}\quad p)$ for $j=1,\ldots,d$. It leads to $\mathcal{P}^{\mathbf{a},\epsilon}_{d,p}=\mathcal{P}^{\mathbf{b},\epsilon}_{d,p}$ by (1) of Theorem 2.4, which is impossible by the assumption in (2). ∎

We next consider the case where $\epsilon^{\prime}\neq\epsilon^{\prime\prime}$.

Theorem 2.5**.**

Suppose that $\epsilon^{\prime},\epsilon^{\prime\prime}\in\left\{0,1\right\}^{d-1}$ with $\epsilon^{\prime}\neq\epsilon^{\prime\prime}$. Set

[TABLE]

and

[TABLE]

Then the followings hold.

1. (1)

$\mathcal{P}^{\mathbf{a},\epsilon^{\prime}}_{d,p}=\mathcal{P}^{\mathbf{b},\epsilon^{\prime\prime}}_{d,p}$* if and only if there exists a $c\in\mathbb{Z}_{p}^{*}$ so that*

[TABLE]

where $\mathbf{a},\mathbf{b}\in\mathbb{Z}_{p}^{d}\setminus\left\{(0,\ldots,0)\right\}$ with $a_{1}\neq 0$ being given. 2. (2)

Assume that $\mathcal{P}^{\mathbf{a},\epsilon^{\prime}}_{d,p}\neq\mathcal{P}^{\mathbf{b},\epsilon^{\prime\prime}}_{d,p}$ where $\mathbf{a},\mathbf{b}\in\mathbb{Z}_{p}^{d}$. If ${\mathcal{Z}}=\emptyset$ then we have

[TABLE]

If ${\mathcal{Z}}\neq\emptyset$, then $|{\mathcal{P}^{\mathbf{a},\epsilon^{\prime}}_{d,p}\cap\mathcal{P}^{\mathbf{b},\epsilon^{\prime\prime}}_{d,p}}|\leq r+1$, where

[TABLE] 3. (3)

Assume that ${\mathcal{Z}}\neq\emptyset$, and $a_{1}\neq 0$. If $a_{\ell_{0}+1}b_{1}^{\ell_{0}+1}\equiv a_{1}^{\ell_{0}+1}b_{\ell_{0}+1}\quad(\mathrm{mod}\quad p)$, then $\mathcal{P}^{\mathbf{a},\epsilon^{\prime}}_{d,p}\cap\mathcal{P}^{\mathbf{b},\epsilon^{\prime\prime}}_{d,p}=\left\{(0,\ldots,0)\right\}$.

Proof.

(1) We assume that (7) holds. Take

[TABLE]

Noting that $a_{j}=b_{j}=0$ provided $\epsilon_{j}^{\prime}\neq\epsilon_{j}^{\prime\prime}$, we have $\mathcal{P}^{\mathbf{a},\epsilon}_{d,p}=\mathcal{P}^{\mathbf{a},\epsilon^{\prime}}_{d,p}$ and $\mathcal{P}^{\mathbf{b},\epsilon}_{d,p}=\mathcal{P}^{\mathbf{b},\epsilon^{\prime\prime}}_{d,p}$. Theorem 2.4 implies that $\mathcal{P}^{\mathbf{a},\epsilon}_{d,p}=\mathcal{P}^{\mathbf{b},\epsilon}_{d,p}$ and hence $\mathcal{P}^{\mathbf{a},\epsilon^{\prime}}_{d,p}=\mathcal{P}^{\mathbf{b},\epsilon^{\prime\prime}}_{d,p}$.

We next assume that $\mathcal{P}^{\mathbf{a},\epsilon^{\prime}}_{d,p}=\mathcal{P}^{\mathbf{b},\epsilon^{\prime\prime}}_{d,p}$ which is equivalent to that there exists a permutation $\left\{k_{0},k_{1},\ldots,k_{p-1}\right\}$ of $\left\{0,1,\ldots,p-1\right\}$ so that

[TABLE]

This is equivalent to

[TABLE]

and

[TABLE]

for $i=2,\ldots,d$. Since $a_{1}\neq 0$, by (9) we have

[TABLE]

Set $j_{0}:=\min\left\{i:\epsilon_{i}^{\prime}\neq\epsilon_{i}^{\prime\prime}\right\}$. Using the same argument with the one in Theorem 2.4 we have $a_{i}\equiv b_{i}k_{1}^{i}\quad(\mathrm{mod}\quad p)$, $i=1,\ldots,j_{0}$. Combining (10) for $i=j_{0}+1$ and (11) we have

[TABLE]

Without loss of generality, we can assume $\epsilon_{j_{0}}^{\prime}=1$ and $\epsilon_{j_{0}}^{\prime\prime}=0$ and then

[TABLE]

which implies $a_{j_{0}}=b_{j_{0}}=0$ since $k_{2}\equiv 2k_{1}\quad(\mathrm{mod}\quad p)$ and $a_{j_{0}}\equiv b_{j_{0}}k_{1}^{j_{0}}\quad(\mathrm{mod}\quad p)$.

(2) We first assume that ${\mathcal{Z}}=\emptyset$ which implies $a_{j}=b_{j}=0$ provided $\epsilon_{j}^{\prime}\neq\epsilon_{j}^{\prime\prime}$. Take

[TABLE]

Noting that $a_{j}=b_{j}=0$ provided $\epsilon_{j}^{\prime}\neq\epsilon_{j}^{\prime\prime}$, we have $\mathcal{P}^{\mathbf{a},\epsilon}_{d,p}=\mathcal{P}^{\mathbf{a},\epsilon^{\prime}}_{d,p}$ and $\mathcal{P}^{\mathbf{b},\epsilon}_{d,p}=\mathcal{P}^{\mathbf{b},\epsilon^{\prime\prime}}_{d,p}$. The (2) of Theorem 2.4 implies that $\mathcal{P}^{\mathbf{a},\epsilon^{\prime}}_{d,p}\cap\mathcal{P}^{\mathbf{b},\epsilon^{\prime\prime}}_{d,p}=\left\{(0,\ldots,0)\right\}$. We next consider the case where ${\mathcal{Z}}\neq\emptyset$. Suppose that $\mathcal{P}^{\mathbf{a},\epsilon^{\prime}}_{d,p}$ and $\mathcal{P}^{\mathbf{b},\epsilon^{\prime\prime}}_{d,p}$ have a common nonzero point. Then, there exist $j,k\in\mathbb{Z}_{p}^{*}$ so that

[TABLE]

Note that $a_{h}=b_{h}=0$ when $h\leq r-1$ and $a_{h}^{\prime}=b_{h}^{\prime}=0,h\leq r-1$ . The (12) implies that

[TABLE]

[TABLE]

Without loss of generality, we can assume $\epsilon_{\ell_{0}}^{\prime}=1$ and $\epsilon_{\ell_{0}}^{\prime\prime}=0$. By (14), we have

[TABLE]

Taking $h=r$ in (13), we have $a_{r}\equiv b_{r}(kj^{-1})^{r}\quad(\mathrm{mod}\quad p)$ where $j^{-1}\in\mathbb{Z}_{p}$ satisfies $j^{-1}j\equiv 1\quad(\mathrm{mod}\quad p)$. Since $a_{r}^{2}+b_{r}^{2}\neq 0$, we have $a_{r}\neq 0$. Set $x_{0}=kj^{-1}$. Then $x_{0}$ satisfies

[TABLE]

Each nonzero point in $\mathcal{P}^{\mathbf{a},\epsilon^{\prime}}_{d,p}\cap\mathcal{P}^{\mathbf{b},\epsilon^{\prime\prime}}_{d,p}$ corresponds to a solution to

[TABLE]

Note that $a_{r}\equiv b_{r}x^{r}\quad(\mathrm{mod}\quad p)$ has at most $r$ solutions. Hence,

[TABLE]

(3) We prove it by contradiction. Assume that $\mathcal{P}_{d,p}^{\mathbf{a},\epsilon^{\prime}}\cap\mathcal{P}_{d,q}^{\mathbf{a},\epsilon^{\prime\prime}}\neq\left\{(0,\ldots,0)\right\}$, and then there exist $j_{0},k_{0}\in\mathbb{Z}_{p}^{*}$ so that $\mathbf{x}_{j_{0}}^{\mathbf{a},\epsilon^{\prime}}=\mathbf{x}_{k_{0}}^{\mathbf{b},\epsilon^{\prime\prime}}$. Particularly, we have

[TABLE]

[TABLE]

Without loss of generality, we can assume $\epsilon_{\ell_{0}}^{\prime}=1$ and $\epsilon_{\ell_{0}}^{\prime\prime}=0$. By (17), we have

[TABLE]

By (16) with $h=1$, we have

[TABLE]

according to $a_{\ell_{0}+1}b_{1}^{\ell_{0}+1}\equiv a_{1}^{\ell_{0}+1}b_{\ell_{0}+1}\quad(\mathrm{mod}\quad p)$. By (18), we have $a_{\ell_{0}}j_{0}^{\ell_{0}}\equiv 0\quad(\mathrm{mod}\quad p)$, which implies that $a_{\ell_{0}}\equiv 0\quad(\mathrm{mod}\quad p)$ or $j_{0}\equiv 0\quad(\mathrm{mod}\quad p)$. This is impossible by the assumption.

∎

In the following, we choose the appropriate vectors $\mathbf{a},\mathbf{b}$ so that $|\mathcal{L}_{p,q}|=q+p-1$. We now state the inequalities for exponential sums over ${\mathcal{L}}_{p,q}$, which is the main result of this subsection.

Theorem 2.6**.**

Suppose $p$ and $q$ are odd prime numbers and set $m=p+q$. Recall that

[TABLE]

We assume that $\lvert\mathcal{L}_{p,q}\rvert=p+q-1$. Then, for any $\mathbf{k}\in[-p+1,p-1]^{d}\cap[-q+1,q-1]^{d}\cap\mathbb{Z}^{d}$ and $\mathbf{k}\neq 0$, we have

[TABLE]

Proof.

We first consider the case where $p=q$. We have

[TABLE]

Recall that

[TABLE]

Then

[TABLE]

Here, in the last inequality, we use Theorem 2.2. We next consider the case where $p\neq q$. When $p\neq q$, $\mathcal{L}_{p,q}=\mathcal{P}_{d,p}\cup\mathcal{P}_{d,q}$. Then we have

[TABLE]

∎

2.3. The exponential sums over ${\mathcal{Q}}_{p^{2},d}^{\mathbf{a},\epsilon}$ and ${\mathcal{R}}_{p^{2},d}^{\mathbf{a},\epsilon}$

Suppose that $\mathbf{a}\in{\mathbb{Z}}_{p}^{d}$ and $\epsilon\in\left\{0,1\right\}^{d-1}$. We set

[TABLE]

[TABLE]

and

[TABLE]

[TABLE]

The ${\mathcal{Q}}_{p^{2},d}^{\mathbf{a},\epsilon}$ and ${\mathcal{R}}_{p^{2},d}^{\mathbf{a},\epsilon}$ can be considered as the generalization of the $p$-sets given in $(\ref{eq:p2set})$. Based on the Lemma 5 and Lemma 6 in [2], we can obtain the following inequalities for exponential sums over ${\mathcal{Q}}_{p^{2},d}^{\mathbf{a},\epsilon}$ and ${\mathcal{R}}_{p^{2},d}^{\mathbf{a},\epsilon}$.

Theorem 2.7**.**

Suppose that $\mathbf{a}\in{\mathbb{Z}}_{p}^{d}$ and $\epsilon\in\left\{0,1\right\}^{d-1}$. Then, for any $\mathbf{k}=(k_{1},\ldots,k_{d})\in[-p+1,p-1]^{d}\cap\mathbb{Z}^{d}$ and $\mathbf{k}\neq 0$, we have

[TABLE]

Proof.

Set

[TABLE]

where $c_{\ell}=k_{\ell}a_{\ell}+k_{\ell+1}a_{\ell}^{\prime}+\cdots+k_{d}a_{\ell}^{\prime}$. We set $j_{0}:=\max\left\{\ell:k_{\ell}\neq 0\right\}$. Then $c_{j_{0}}=k_{j_{0}}a_{j_{0}}$ and we have $p\nmid c_{j_{0}}$. According to Lemma 5 in [2], we have

[TABLE]

∎

Theorem 2.8**.**

Suppose that $\mathbf{a}\in[1,p-1]^{d}\cap\mathbb{Z}^{d}$. Then, for any $\mathbf{k}\in[-p+1,p-1]^{d}\cap\mathbb{Z}^{d}$ and $\mathbf{k}\neq 0$, we have

[TABLE]

Proof.

Set

[TABLE]

and $c_{\ell}:=k_{\ell+1}a_{\ell+1}+k_{\ell+2}a_{\ell+1}^{\prime}+\cdots+k_{d}a_{\ell+1}^{\prime}$. We set $j_{0}:=\max\left\{\ell:k_{\ell}\neq 0\right\}$. Then $c_{j_{0}-1}=k_{j_{0}}a_{j_{0}}$ and we have $p\nmid c_{j_{0}-1}$. Using Lemma 6 in [2], we have

[TABLE]

∎

3. The applications of $\mathcal{P}_{d,p}^{\mathbf{a},\epsilon}$ and

${\mathcal{L}}_{p,q}$

Based on the exponential sum formula in Section 2, the new point sets are useful in numerical integration [8, 1], in UQ [13] and in the recovery of sparse trigonometric polynomials [12]. We just state the results for the recovery of sparse trigonometric polynomials in detail.

We start with some notations which go back to [12]. Set

[TABLE]

Note that $\Pi_{s}^{d}$ is a linear space with the dimension $D:=(2s+1)^{d}$. For

[TABLE]

we set $\mathbf{T}:=\left\{\mathbf{k}:c_{\mathbf{k}}\neq 0\right\}$ which is the support of the sequence of coefficients $c_{\mathbf{k}}$, and set

[TABLE]

where $\Pi_{\mathbf{T}}$ denotes the space of all trigonometric polynomials whose coefficients are supported on $\mathbf{T}$. When $M\ll D$, we call the trigonometric polynomials in $\Pi_{s}^{d}(M)$ as $M$-sparse trigonometric polynomials.

The recovery of sparse trigonometric polynomials is an active topic recently. The main aim of this research topic is to design a sampling set $X=\left\{\mathbf{z}_{j}\right\}_{j=1}^{N}$ so that one can recover $f\in\Pi_{s}^{d}(M)$ from $f(\mathbf{z}_{j}),\mathbf{z}_{j}\in X$ [12, 9, 6]. We state the problem as follows. Assume the sampling set is $X=\left\{\mathbf{x}_{j}\in[0,1)^{d},j=1,\ldots,N\right\}$. Then our aim is to solve the following programming:

[TABLE]

Denote by $\mathbf{F}_{X}$ the $N\times D$ sampling matrix with entries

[TABLE]

Let $\mathbf{a}_{\mathbf{k}}=(\exp(2\pi\mathbf{i}\mathbf{k}\cdot\mathbf{x}_{j}))_{j=1}^{N}$ denote a column of $\mathbf{F}_{X}$ with $\mathbf{k}\in[-s,s]^{d}\bigcap\mathbb{Z}^{d}$. A simple observation is that $\|\mathbf{a}_{\mathbf{k}}\|_{2}=\sqrt{N}$. Set

[TABLE]

which is called the mutual incoherence of the matrix $\mathbf{F}_{X}/\sqrt{N}$. Theorem 2.5 in [6] shows that if $\mu<1/(2M-1)$ then the Orthogonal Matching Pursuit Algorithm (OMP) and the Basis Pursuit Algorithm (BP) can recover any $M$-sparse trigonometric polynomials in $\Pi_{s}^{d}(M)$. Therefore, our aim is to choose the sampling set $X$ so that $\mu$ is small and hence OMP and BP can recover $M$-sparse trigonometric polynomials. Based on Theorem 2.2 and Theorem 2.6 respectively, the following results give upper bounds of $\mu$ with taking $X=\mathcal{P}_{d,p}^{\mathbf{a},\epsilon}$, and $X={\mathcal{L}}_{p,q}$, respectively.

Lemma 3.1**.**

1. (1)

Suppose that $X=\mathcal{P}_{d,p}^{\mathbf{a},\epsilon}$ where $\mathbf{a}\in[1,p-1]^{d}\cap\mathbb{Z}^{d}$ and $p\geq 2s+1$ is a prime number. Then

[TABLE] 2. (2)

Suppose that $p,q\geq 2s+1$ are prime numbers and $\mathbf{a},\mathbf{b}\in[1,p-1]^{d}\cap\mathbb{Z}^{d}$. Recall that

[TABLE]

Set $X=\mathcal{L}_{p,q}$ and $m=p+q$. Then

[TABLE]

As said before, if $\mu<1/(2M-1)$ then OMP (and also BP) can recover every $M$-sparse trigonometric polynomials. Then we have the following corollary:

Theorem 3.2**.**

1. (1)

Suppose that $p>\max\left\{2s+1,(d-1)^{2}(2M-1)^{2}+1\right\}$ is a prime number and $\mathbf{a}\in[1,p-1]^{d}\cap\mathbb{Z}^{d}$ . Then OMP (and also BP) recovers every $M$-sparse trigonometric polynomial $f\in\Pi_{s}^{d}(M)$ exactly from the deterministic sampling $\mathcal{P}_{d,p}^{\mathbf{a},\epsilon}$. 2. (2)

Under the condition in (2) of Lemma 3.1. Suppose that

[TABLE]

Then OMP (and also BP) recovers every $M$-sparse trigonometric polynomial $f\in\Pi_{s}^{d}(M)$ exactly from the deterministic sampling set $\mathcal{L}_{p,q}$ .

Proof.

We first consider (1). Note that $p\geq(d-1)^{2}(2M-1)^{2}+1$ implies that $(d-1)/\sqrt{p}<1/(2M-1)$. According to (1) in Lemma 3.1, if $(d-1)/\sqrt{p}<1/(2M-1)$ then $\mu<1/(2M-1)$ and hence the conclusion follows. Similarly, we can prove (2). ∎