An Extension of the Dirichlet Density for Sets of Gaussian Integers

# An Extension of the Dirichlet Density for Sets of Gaussian Integers

L. C. Rêgo L. C. Rêgo is with the Departamento de Estatística, Universidade Federal de Pernambuco. Email: leandro@de.ufpe.br    R. J. Cintra R. J. Cintra is with the Signal Processing Group, Departamento de Estatística, Universidade Federal de Pernambuco. Part of second author’s work was done during his sabbatical at the University of Calgary, Calgary, Canada. Email: rjdsc@de.ufpe.br
###### Abstract

Several measures for the density of sets of integers have been proposed, such as the asymptotic density, the Schnirelmann density, and the Dirichlet density. There has been some work in the literature on extending some of these concepts of density to higher dimensional sets of integers. In this work, we propose an extension of the Dirichlet density for sets of Gaussian integers and investigate some of its properties.

Keywords

Gaussian integers, Dirichlet density

## 1 Introduction

Several measures for the density of sets of integers have been discussed in the literature [1, 2, 3, 4, 5, 6, 7]. Presumably the most employed of such measures is the asymptotic density, also referred to as natural density [8, 1]. For a given set of integers , its asymptotic density is expressed by

 d(A)=limn→∞∥A∩{1,2,3,…,n}∥n,

provided that such a limit does exist. The symbol returns the cardinality of its argument.

In [2], Bell and Burris bring an ample exposition on the Dirichlet density, which is defined as follows.

###### Definition 1

The Dirichlet density of a subset of the positive integers is given by

 ∂(A) ≜lims↓1∑n∈A1nsζ(s),

if the limit does exist, for real . The quantity denotes the Riemann zeta function [9].

If the asymptotic density is well defined, then the Dirichlet density does also exist and assumes the same value [10, p. 10]. Since the converse is not always true, the Dirichlet density is a more encompassing tool when compared to the asymptotic density [10, p. 11]. Dirichlet density also admits lower and upper versions, which have been explored along with other densities to characterize primitive sets [11, 12, 13].

Gaussian integers are simply complex numbers of the form , where and are integers. Despite the considerable amount of development addressing densities for sets of positive integers [14], densities for sets of Gaussian integers appear to be an overlooked topic. However, a seminal paper by Cheo [15] investigated the question, suggesting an extension of the Schnirelmann density [3, 4]. Such extended definition applies to subsets of the nonzero Gaussian integers inclusively confined in the first quadrant of the complex plane.

Generalizations of Schnirelmann density for the -dimensional case were proposed in [16, 17]. Additionally, a modified Schnirelmann density was introduced in [18] and was generalized in [19] years later. In a comparable venue, Freedman [20, 21] advanced the concept of asymptotic density to higher dimensions.

In this context, the aim of the present work is to advance a method for evaluating the density of sets of Gaussian integers. To address this problem, a density based on Dirichlet generating functions is proposed.

For ease of notation, henceforth we identify a Gaussian integer with the pair of integers . All considered Gaussian integers and their sets are defined in , where is the set of strictly positive integers.

## 2 Definition and General Properties

The Gaussian integers can be realized as points over a square lattice in the complex plane. The square lattice is composed by an infinite array of Gaussian integers, set up in rows and columns. In addition, each lattice row or column can furnish sets of integers according to the following constructions: and .

Our goal is to investigate the properties of the following density for Gaussian integers, which we show to be a generalization of the Dirichlet density for sets of integers.

###### Definition 2

Let be a set of Gaussian integers. Admit and to be the indicator functions of the sets and , for , respectively. The proposed density for is given by

 dens(A)≜lims↓11ζ2(s)∞∑m=1∞∑n=1IA∗,n(m)IAm,∗(n)(mn)s,

provided that the limit exists.

From now on, we only consider sets whose referred densities are well-defined, i.e., the implied limits exist. Thus, we restrain ourselves of indicating in every instance that the results are valid only when the discussed limits exist. In account of the proposed definition, a series of consequences is listed below.

###### Proposition 1

Let and be two sets of Gaussian integers. The following assertions hold true:

1. .

2. .

3. if , then .

4. .

5. , where is the relative complement of in .

6. , where is the complement of .

7. if , then .

8. .

Proof:  Follows directly from the definition.

The first three properties stated in the previous proposition are the same conditions that form an axiomatic definition of a probability measure, except for the -additivity axiom.

###### Proposition 2 (Cartesian Product)

Let and be two subsets of . Then the density of the Cartesian product satisfies

 dens(A×B)=∂(A)∂(B).

Proof:  We have that

 dens(A×B) =lims↓1∑(m,n)∈A×B1(mn)sζ2(s) =lims↓1∑m∈A1ms∑n∈B1nsζ2(s) =∂(A)∂(B).

###### Corollary 1 (Dirichlet Density)

Let be a set of positive integers. Then .

Proof:  This result is a direct consequence of the fact that  [14].

Given any set of Gaussian integers, let the horizontal and vertical axis sections be denoted by and , respectively.

###### Proposition 3

Let be a set of Gaussian integers. If or , then .

Proof:  For instance, assume that . Note that . Due to the monotonicity property, it follows that . Moreover, the property of the density of Cartesian products allows us to write . Applying the hypothesis, the result follows. The proof would be analogous in the case that .

###### Corollary 2 (Finite Axis Section)

If a set of Gaussian integers has a finite axis section, then .

Proof:  It is enough to observe that any finite set of integers has null Dirichlet density [14].

As a consequence, a finite set of Gaussian integers has null density, since both of its axis sections are finite. In particular, the density of a singleton is zero. On the other hand, nonzero density subsets must have infinite axis sections.

Let and . Next proposition states that, for density evaluation, the only relevant set elements are those located in the region defined by for any choice of and . This means that the “weight” of the set is located on its “tail”. Nevertheless, we need the result of the following lemma before.

###### Lemma 1

The set has unit density.

Proof:  Let be the complement of . Therefore, the set can be partitioned into . Then, it follows that . Notice also that . The union property allows us to state that . Since , , and have each at least one finite axis section, it follows that .

###### Proposition 4 (Heavy Tail)

Let be a set of Gaussian integers. Then, for any two given nonnegative integers and , we have

 dens(A)=dens(A∩Um0,n0).

Proof:  Observe that . Since we have a partition of , it follows that . But, , then .

###### Proposition 5 (Axis Independence)

If there is a pair such that, for every and , the functions and are independent of and , respectively, then

 dens(A)=∂(Am0,∗)∂(A∗,n0).

Proof:  Because of the assumed independence, we can write and , for and , respectively. Thus, for and , the set is indistinguishable of . But, the heavy tail property implies that

 dens(A) =dens(A∩Um0,n0) =dens((Am0,∗×A∗,n0)∩Um0,n0) =dens(Am0,∗×A∗,n0) =∂(Am0,∗)∂(A∗,n0).

Given a set of Gaussian integers and a Gaussian integer , let . This process is called a translation of by units [22, p. 49]. Now our goal is to show that the proposed density is translation invariant, i.e., , and . However, the proof that we will supply requires the following lemma.

###### Lemma 2 (Unitary Translation)

Let be a set of Gaussian integers, such as . Then

 dens(A⊕(1,0))=dens(A⊕(0,1))=dens(A).

Proof:  It suffices to show that , being the other case analogous. First, note that since

 ∑(m,n)∈A1(mn)s−∑(m,n)∈A1((m+1)n)s≥0,

it follows that . Also observe that

 sms+1≥∫m+1msxs+1dx=1ms−1(m+1)s≥0.

Thus, we have that

 ∑(m,n)∈A1(mn)s−∑(m,n)∈A1((m+1)n)s= ∑(m,n)∈A1ns(1ms−1(m+1)s) ≤ ∑(m,n)∈A1nssms+1 ≤ ∑n∈P1ns∑m∈suph(A)sms+1.

Dividing both sides by and letting , since the last series is convergent as , yields

 dens(A)−dens(A⊕(1,0))≤0.

###### Proposition 6 (Translation Invariance)

Let be a set of Gaussian integers. Then

 dens(A⊕(m,n))=dens(A),

where and are nonnegative integers.

Proof:  We have already proven that . Therefore, we have that

 dens(A⊕(m+1,n+1)) =dens(((A⊕(m,n))⊕(1,0))⊕(0,1)) =dens((A⊕(m,n))⊕(1,0)) =dens(A⊕(m,n)) =dens(A).

###### Corollary 3

The proposed density is not -additive.

Proof:  This result follows directly from Propositions 1 and 6.

Now consider the set operation defined as , where is a Gaussian integer. This construction can be interpreted as a dilation on the elements of . The following proposition relates the density of a set of Gaussian integers with the density of its dilated form.

###### Proposition 7 (Dilation)

Let be a set of Gaussian integers and let be any Gaussian integer. Then

 dens((a,b)⊗A)=1abdens(A).

Proof:  This result follows directly from the definition of the proposed density:

 dens((a,b)⊗A) =lims↓1∑(m,n)∈A1(ambn)sζ2(s) =lims↓11(ab)s∑(m,n)∈A1(mn)sζ2(s) =1abdens(A).

## 3 Density of Particular Sets

In this section, we evaluate the density of some particular sets of Gaussian integers.

### 3.1 Cartesian Product of Arithmetic Progressions

Let be an integer. The set constitutes an arithmetic progression with Dirichlet density . Furthermore, the Cartesian product of two arithmetic progressions generates a rectangular lattice denoted by , where and are positive integers. Then it follows from Proposition 2 that . Let us investigate the density of sets that are intersections of particular Cartesian products of arithmetic progressions.

###### Proposition 8 (Intersection)

For any positive integers , , and , we have that

 dens(M(p,q)∩M(s,t))=dens(M(lcm(p,s),lcm(q,t)))=1lcm(p,s)lcm(q,t),

where denotes the least common multiple of its arguments.

Proof:  First, note that . Therefore,

 M(p,q)∩M(s,t) =((p,q)⊗P2)∩((s,t)⊗P2)

Applying on both sides of above equation and invoking the dilation property, we obtain the desired result.

###### Corollary 4

Let be a Gaussian integer. Admit also that and , where returns the greatest common divisor of its arguments. Then

 dens(M(mp,nq)∩M(ms,nt))=1mndens(M(p,q)∩M(s,t)).

Proof:  Follows directly from Proposition 8.

### 3.2 Sets Delimited by Functions

Let us consider a set of Gaussian integers defined as where and are functions such that for every integer . Functions and delimit the set , confining the set elements in between. Figure 1 illustrates a possible configuration for the set . By definition, the proposed density of is given by

 dens(C)=lims↓11ζ2(s)∞∑m=11ms⌊g(m)⌋∑n=⌈f(m)⌉1ns,

where and represent the usual ceiling and floor functions, respectively.

Let us establish upper and lower bounds for the double summation. Initially, notice that the inner summation satisfies the following bounds:

 ⌊g(m)⌋∑n=⌈f(m)⌉1ns =1⌈f(m)⌉s+⌊g(m)⌋∑n=⌈f(m)⌉+11ns ≤1+∫⌊g(m)⌋⌈f(m)⌉1xsdx ≤1+∫g(m)f(m)1xsdx =1+1−s+1(g(m)−s+1−f(m)−s+1).

Thus, an upper bound for the double summation is expressed by

 ∞∑m=11ms⌊g(m)⌋∑n=⌈f(m)⌉1ns =ζ(s)+1−s+1∞∑m=11ms(g(m)−s+1−f(m)−s+1).

Performing analogous manipulations, we obtain the following lower bound for the inner summation:

 ⌊g(m)⌋∑n=⌈f(m)⌉1ns ≥−1(⌈f(m)⌉−1)s+∫⌊g(m)⌋+1⌈f(m)⌉−11xsdx ≥−1+∫g(m)f(m)1xsdx =−1+1−s+1(g(m)−s+1−f(m)−s+1).

This implies that the double summation is lower bounded by:

 ∞∑m=11ms⌊g(m)⌋∑n=⌈f(m)⌉1ns ≥∞∑m=11ms(−1+1−s+1(g(m)−s+1−f(m)−s+1)) =−ζ(s)+1−s+1∞∑m=11ms(g(m)−s+1−f(m)−s+1).

The upper and lower bounds present similar formulations, inviting an application of the squeeze theorem. Thus, after dividing both expressions by and taking the limit as , minor manipulations furnish

 dens(C)=lims↓11ζ(s)∞∑m=11ms(f(m)−s+1−g(m)−s+1).

Now let us analyze the density of a set in the light of the asymptotic behavior of the delimiting functions.

###### Proposition 9 (Asymptotics)

Let and be delimiting functions that are always greater or equal to one. If and , then

 dens(C)=lims↓11ζ(s)∞∑m=11ms(u(m)−s+1−v(m)−s+1). (1)

Proof:  By the definition of the -notation [23, p. 434], there exist a quantity , such that, for every , both functions and satisfy:

 c1u(m)≤f(m)≤c2u(m), c3v(m)≤g(m)≤c4v(m),

where , , , and are positive constants. Moreover, notice that

 dens(C)= lims↓11ζ(s)∞∑m=11ms(f(m)−s+1−g(m)−s+1) = lims↓11ζ(s)[m0−1∑m=11ms(f(m)−s+1−g(m)−s+1)] +lims↓11ζ(s)[∞∑m=m01ms(f(m)−s+1−g(m)−s+1)] = lims↓11ζ(s)∞∑m=m01ms(f(m)−s+1−g(m)−s+1).

Thus, for , we have that

 ∞∑m=m0 1ms((c2u(m))−s+1−(c3v(m))−s+1) ≤∞∑m=m01ms(f(m)−s+1−g(m)−s+1) ≤∞∑m=m01ms((c1u(m))−s+1−(c4v(m))−s+1).

Now we show that after dividing by and letting , both upper and lower bounds above have the same limit. Since and , it follows that for arbitrary positive constants and :

 ∞∑m=m01ms ((k1u(m))−s+1−(k2v(m))−s+1)= k−s+11∞∑m=m01msu(m)−s+1−k−s+12∞∑m=m01msv(m)−s+1.

Thus, since both and tend to one as , we have that

 lims↓11ζ(s)∞∑m=m01ms((k1u(m))−s+1−(k2v(m))−s+1) =lims↓11ζ(s)∞∑m=m01ms(u(m)−s+1−v(m)−s+1).

Therefore, we maintain that

 lims↓11ζ(s)∞∑m=m01ms(f(m)−s+1−g(m)−s+1) =lims↓11ζ(s)∞∑m=m01ms(u(m)−s+1−v(m)−s+1).

Finally, since

 lims↓11ζ(s)m0−1∑m=11ms(u(m)−s+1−v(m)−s+1)=0,

the proposition is proven.

We now supply two examples. But, the following lemma is needed before.

###### Lemma 3

For ,

Proof:  Taking into account the substitution , it follows that:

 lims↓1ζ((α+1)s−α)(s−1) =limt↓1ζ(t)(t+α1+α−1) =limt↓1ζ(t)t−11+α =11+α.

###### Example 1

Let us examine the density of the set , where and , for real quantities . In order to compute such density we need the previous lemma. Thus, by invoking Equation 1, it follows that the sought density is given by

 dens(Cpow) =lims↓11ζ(s)∞∑m=11ms((mα)(−s+1)−(mβ)(−s+1)) =lims↓11ζ(s)(ζ((α+1)s−α)−ζ((β+1)s−β)) =11+α−11+β.

In particular, if , we have .

###### Example 2

Consider the set , where , for . Thus, by Equation 1,

 dens(Cexp)=lims↓11ζ(s)(ζ(s)−∞∑m=1(am)−s+1ms).

Then, note that for each , there is a quantity such that implies that . By Example 1, we know that

 lims↓11ζ(s)∞∑m=1(mβ)−s+1ms=11+β.

Moreover, since , it follows that

 lims↓11ζ(s)∞∑m=M(mβ)−s+1ms=11+β.

Thus,

 dens(Cexp) ≥lims↓11ζ(s)(ζ(s)−(M−1∑m=1(am)−s+1ms+∞∑m=M(mβ)−s+1ms)) =1−(0+11+β)=ββ+1.

Finally, letting yields .

## Acknowledgements

The second author acknowledges partial financial support from the Department of Foreign Affairs and International Trade (DFAIT), Canada, and the Conselho Nacional de Desenvolvimento Científico e Tecnológico (CNPq), Brazil.

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