# The Critical Strips of the Sums1+2z+⋯+nz

Authors: G. Mora; J. M. Sepulcre
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

We give a partition of the critical strip, associated with each partial sum1+2z+⋯+nz of the Riemann zeta function for Re z<−1, formed by infinitely many rectangles for which a formula allows us to count the number of its zeros inside each of them with an error, at most, of two zeros. A generalization of this formula is also given to a large class of almost-periodic functions with bounded spectrum.


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## Body

## 1. Introduction

Some industrial processes can be modeled [1] by functional equations of the form f(x)+f(2x)=0 or f(x)+f(2x)+f(3x)=0, x>0. The generalization of these functional equations to the complex plane is formally given by(1.1)f(z)+f(2z)+⋯+f(nz)=0,n≥2,
which admits analytic solutions of the form zα on the open set Ω=ℂ∖(-∞,0] if and only if α is a zero of(1.2)Gn(z)≡1+2z+⋯+nz.
For each integer n≥2, each function Gn(z) represents the nth partial sum of the Riemann zeta function ζ(z) on the half-plane Rez<-1, and it belongs to the class of the entire almost-periodic functions of exponential type. In [2], we can see a complete introduction devoted to the study of such class of functions. There, we can also find a theorem of Bohr [2, p. 270] that identifies the functions of the above class having their zeros in a strip (the critical strip) parallel to the real axis with those functions for which the upper and lower bounds of their spectra enter into the spectrum. That is, for instance, the case of the functions defined from (1.2) by means of a rotation of angle π/2(1.3)Hn(z)≡Gn(iz)=1+2iz+⋯+niz,
whose spectra, for each integer n≥2, are the finite sets(1.4){iln⁡k:k=1,…,n}.
Furthermore, the functions Hn(z) have the property consisting on the existence of some value of x=Rez, say x0, such that either ReHn(x0,y)≠0 or ImHn(x0,y)≠0 for all y∈ℝ. Indeed, x0=0 satisfies such property. Therefore, the Hn(z)’s belong to a very special class of almost-periodic functions whose study will be our main objective to determine a nonasymptotic formula that allows us to count the amount of zeros that they have inside the rectangles of a certain partition of their critical strips.One of the most important formulae [2, p. 277] on the number of roots of an almost-periodic function, with closed and bounded spectrum, say f(z), inside a rectangle in the strip where the zeros of f(z) are located, is given by(1.5)2πlim⁡x2-x1→∞N(x1,x2,y1,y2)x2-x1=φ′(y2)-φ′(y1),
where N(x1,x2,y1,y2) denotes the number of zeros of f(z) in the rectangle(1.6)x1<Rez<x2,y1<Im⁡z<y2,
and φ(y) is the mean function associated with ln|f(x+iy)| defined as(1.7)φ(y)=lim⁡T→∞12T∫-T+αT+αln⁡|f(x+iy)|dx.The formula (1.5) is of asymptotic type, and it is based on the assumption of the existence of derivative of φ(y). However, if the spectrum of f(z) is contained in the boundary of a bounded convex polygon of the complex plane and all the vertices of the polygon enter into the spectrum, there exists a formula [2, p. 298] much more explicit than (1.5). For instance, if the polygon is reduced to a segment of the imaginary axis, the formula is, for sufficiently large values of |y1| and |y2|,(1.8)N(x1,x2,y1,y2)=d2π(x2-x1)+O(1),
where d is the length of the segment.Formula (1.8) could be used, for instance, to estimate the number of zeros, of our functions Hn(z), inside the rectangle defined by the intersection of its critical strip with the strip x1≤Rez≤x2. Indeed, since the spectrum of Hn(z) is contained in the line segment [0,ilnn] of the imaginary axis, a simple application of (1.8) leads to the formula(1.9)N(x1,x2,y1,y2)=ln⁡n2π(x2-x1)+O(1).
Nevertheless, it is well known that the term O(1) is an “obscure” function which we only know to represent a bounded quantity. In general, the term O(1) that appears in formula (1.8) depends on the function and the rectangle where we are counting the number of its zeros. Our aim is to give much more precise information about the expression O(1) when the function belongs to that special class of almost-periodic functions which contains, in particular, to our functions Hn(z). In fact, on this subject, we find in [3] the following result.“There exist infinitely many rectanglesx1<Rez<x2,y1<Imz<y2in the critical strip of the functionHn(z)for which the number of zeros ofHn(z)  is given by the formula(1.10)N(x1,x2,y1,y2)=ln⁡n2π(x2-x1)+Ωn,whereΩnis a real number with|Ωn|<1.”Now, by following the ideas exhibited in [3], our aim is to demonstrate that for the functions of that special class of almost-periodic functions, there exists a formula similar to that of (1.10) to determine the number of its zeros inside infinitely many rectangles in their critical strips with an error, at most, of two zeros.In particular, our main result will also prove that the boundn-1 which appears in the formula that determines the number of the zeros of an exponential polynomial of degree n inside certain rectangles of its critical strip can be substituted by a universal bound, namely, 2. In fact, to illustrate the scope of our result, we will start recalling an old theorem of Polya [4]:“ifz=x+iy(1.11)μ1<μ2<⋯<μl,andPν(z)is forν=1,2,…,la polynomial of degree≤mν-1with(1.12)m1+m2+⋯+ml=n,P1(z)Pl(z)≠0,then the numberN(gl,a,a+b)of the zeros (according to multiplicity) of the function(1.13)gl(z)=∑ν=1lPν(z)eiμνzin the infinite vertical strip(1.14)a≤x≤a+bsatisfies the inequality(1.15)|N(gl,a,a+b)-b2π(μl-μ1)|≤n-1.”Then, under the same hypotheses of the above theorem, our result could be stated as follows.Under the hypotheses of theorem of Polya with frequenciesμ1<μ2<⋯<μl linearly independent and m1=m2=⋯=ml=1,there exist infinitely many values fora,bsuch that, independently ofl, the inequality(1.16)|N(gl,a,a+b)-b2π(μl-μ1)|≤2holds.This result will be an immediate consequence of Lemma2.6 and Theorem 3.1 of the present paper.


## 2. Preliminaries

To prove our main theorem, we will use some elementary concepts and results such as the following.Definition 2.1.
A set{a1,a2,…,ak} of real numbers is said to be linearly independent if and only if any linear combination
(2.1)∑j=1knjaj=0,
with integers nj, implies that nj=0 for all j=1,…,k.For example, the set(2.2){ln⁡p1,ln⁡p2,…,ln⁡pk},
where p1,p2,…,pk are different prime numbers, is linearly independent. Nevertheless, for a given set of real numbers {x1,x2,…xl}, we can always suppose the existence of a basis {a1,a2,…,ak}. That is, on the one hand, {a1,a2,…,ak} is linearly independent and, on the other hand, for each m=1,…,l, we can write(2.3)xm=∑j=1knmjaj,
where the nmj are integers.An important result on linearly independent sets of real numbers is the famous theorem of Kronecker [5, p. 382] which will be used in the following form.Theorem 2.2 (Kronecker).
Let{a1,a2,…,ak} be a linearly independent set of nonnull real numbers. For arbitrary numbers b1,b2,…,bk and T, ϵ>0, there exists a real number t>T and integers n1,n2,…,nk such that
(2.4)|taj-nj-bj|<ϵ,∀j=1,…,k.Given an entire almost-periodic functionf with closed and bounded spectrum, a rectangle in its critical strip will be defined as the intersection of the rectangle x<Rez<x+T, y1<Imz<y2, for some T>0 and sufficiently large values of |y1| and |y2|, with the strip where the zeros of f are situated, of course, by assuming that the critical strip is parallel to the real axis. Then, the number of zeros of f(z) in a rectangle in its critical strip will merely be denoted by N(f(z);x,x+T). Similarly, N(f(z);y,y+T) will denote the number of zeros of f(z) in a rectangle in its critical strip, provided that the critical strip of f(z) to be parallel to the imaginary axis. Nevertheless, noticing the change z by -iz transforms the zeros of a strip parallel to the real axis onto the zeros of a strip parallel to the imaginary axis and conversely, from now on, we will do our study on those functions by assuming that their critical strips are parallel to the imaginary axis.Because our aim is to study the number of zeros of almost-periodic functions, and noticing these functions, from Bochner’s theorem [2, p. 266], are characterized as uniform limits of exponential polynomials, we will start by demonstrating a formula of the type (1.10) assuming that they adopt the normalized form(2.5)P(z)=1+∑j=1nwjeμjz,
where the coefficients wj are nonnull complex numbers and the frequencies μj are positive real numbers so that(2.6)μ1<μ2<⋯<μn.
Then, a normalized exponential polynomial of the form (2.5), not affecting the zeros, will be considered as a prototype of an almost-periodic function whose definition [6, p. 101] we recall.Definition 2.3.
An entire functionf is said to be almost periodic if and only if for every ϵ>0 there exists a length l=l(ϵ) such that every interval b<y<b+l of length l on the imaginary axis contains at least one translation number τ associated with ϵ satisfying the inequality
(2.7)|f(z+iτ)-f(z)|≤ϵ,∀z∈C.From (2.7), we derive the notion of interval of almost periodicity.Definition 2.4.
Letf be an almost-periodic entire function on ℂ and ϵ>0. Then, any interval of length l, l=l(ϵ,f), will be called an ϵ-interval of almost periodicity of f.In each interval of almost periodicity of an exponential polynomialP(z), the solutions of the equations ReP(z)=0, ImP(z)=0 have a very special form, as we will prove in the following result.Lemma 2.5.
Let(2.8)P(z)=1+∑j=1nwjeμjz,wj∈C∖{0}
be an exponential polynomial with increasing positive frequencies μ1<⋯<μn. Then, there exist two real numbers x1, x2 such that all the zeros of P(z) are in the strip
(2.9)SP(z)={z:x1<Rez<x2}.
Furthermore, for n=1,2, there exists a value for y, say y0, such that either
(2.10){z:ReP(z)=0}∩{z:Im⁡z=y0}=∅
or
(2.11){z:Im⁡P(z)=0}∩{z:Im⁡z=y0}=∅.Proof.
Since(2.12)lim⁡x→-∞P(x+iy)=1,lim⁡x→+∞P(x+iy)wneμn(x+iy)=1,
for any value of y, there exist x1<0<x2 such that
(2.13)|P(z)-1|<1,∀zwithRez≤x1,|P(z)wneμnz-1|<1,∀zwithRez≥x2.
Hence, P(z) has no zero neither in the half-plane Rez≤x1 nor in the half-plane Rez≥x2. Consequently, all the zeros of P(z) are situated in the strip
(2.14)x1<Rez<x2.
To prove the second part of the lemma, we will only consider the case ImP(z)=0 (the case ReP(z)=0 is completely similar). In this case, for any positive integer n, the equation ImP(z)=0 can be explicitly written as
(2.15)∑j=1neμjx(αjsin(μjy)+βjcos⁡(μjy))=0,
where αj=Rewj and βj=Imwj. By defining
(2.16)Aj(y)≡αjsin(μjy)+βjcos⁡(μjy),foreachj=1,…,n,
equation (2.15) becomes
(2.17)∑j=1neμjxAj(y)=0.
On the other hand, it is plain that the set of the zeros of each function Aj(y), denoted by Bj, is given by
(2.18)Bj={1μj(πkj-arctanβjαj):kj∈Z},
where arctan(βj/αj) is taken as π/2 when αj=0. Since eμ1x>0 for all real x, the case n=1 easily follows by taking y=y0, for arbitrary y0∉B1.
Now, assume thatn=2. If the sets B1 and B2 are distinct, suppose that there exists some y0∈B1 such that y0∉B2. Then, the right-line of equation y=y0 does not meet ImP(z)=0. Indeed, if for some real x the point (x,y0) satisfies the equation ImP(z)=0, then, from (2.17) and taking into account that y0∈B1, it necessarily would have A2(y0)=0 and, therefore, y0∈B2, which is a contradiction. Consequently, the lemma follows for the value y0. Finally, we analyse the case B1=B2. This case means that for each integer k1 there exists another integer k2 such that
(2.19)1μ1(πk1-arctanβ1α1)=1μ2(πk2-arctanβ2α2),
and reciprocally. By defining the numbers
(2.20)μ≡μ1μ2,b≡arctan⁡(β1/α1)-μarctan(β2/α2)π,
equality (2.19) can be written as
(2.21)k1-μk2=b,
which represents an equation with infinitely many solutions for integers k1 and k2. Let k1, k2 and k1′, k2′ be integers verifying (2.21). Then, by subtracting in (2.21), one has
(2.22)(k1-k1′)-μ(k2-k2′)=0,
which implies that μ must be necessarily a rational number (observe that it means, in particular, that the frequencies μ1, μ2 are linearly dependent) and, because of 0<μ1<μ2, the number μ is a positive rational less than 1. On the other hand, since
(2.23)|arctanβ1α1|,|arctanβ2α2|≤π2,b is a rational number verifying
(2.24)|b|<1.
Now, suppose the value k2=0 is given. Then, there exists an integer k1 satisfying (2.19) and, according to (2.21), it follows that k1=b. Hence, b is an integer and then, noticing (2.24), b=0. Consequently, k1=0. Since B1=B2, let y1 be the point of B1=B2 corresponding to the values k1=k2=0. Then, from (2.18), one has
(2.25)y1=-1μ1arctanβ1α1=-1μ2arctanβ2α2.
Now, assume that, for any real number y, there exists a value of x such that
(2.26){z=x+iy:Im⁡P(x,y)=0}∩{z:Im⁡z=y}≠∅.
Thus, in particular, given y1, there exists a1 such that ImP(a1,y1)=0. On the other hand, as the set B1=B2 is discrete, there exists an open interval (v,y1) such that one has y∉B1=B2 for any y∈(v,y1) and, therefore, Aj(y)≠0 for j=1,2. Then, by assuming (2.26), if we divide (2.17) by eμ1xA1(y) one has the following property.
For eachy∈(v,y1),there existsxsuch that the relation(2.27)1+A2(y)A1(y)e(μ2-μ1)x=0,holds.
Now, by taking the limit in (2.27), when y→y1, it follows that the point (a1,y1) satisfies
(2.28)1+lim⁡y→y1A2(y)A1(y)e(μ2-μ1)a1=0.
However, since
(2.29)lim⁡y→y1A2(y)A1(y)=μ2|w2|μ1|w1|
is positive, by substituting in (2.28), we are led to a contradiction. Consequently, the case B1=B2 follows, and the proof of the lemma is now completed.When the frequencies are linearly independent, the preceding lemma is valid for arbitraryn.Lemma 2.6.
Letn be an arbitrary positive integer and
(2.30)P(z)=1+∑j=1nwjeμjz,wj∈C∖{0}
an exponential polynomial with increasing positive frequencies μ1<⋯<μn forming a linearly independent set. Then, there exists a value for y, say y0, such that either
(2.31){z:ReP(z)=0}∩{z:Im⁡z=y0}=∅
or
(2.32){z:Im⁡P(z)=0}∩{z:Im⁡z=y0}=∅.Proof.
For the sake of brevity, we will prove the lemma in the caseImP(z)=0 (the case ReP(z)=0 is completely similar). Consider the coefficients wj of the exponential polynomial P(z), since all them are nonnull, the set J={1,2,…,n} can be partitioned in the following four disjoint sets (some of them could be eventually empty)
(2.33)J1={j∈J:αj≥0,βj>0},J2={j∈J:αj>0,βj≤0},J3={j∈J:αj≤0,βj<0},J4={j∈J:αj<0,βj≥0}.
Now, define the numbers
(2.34)aj≡μj2π,∀j∈J,bj≡{18,ifj∈J1,38,ifj∈J2,58,ifj∈J3,78,ifj∈J4.
Let us pick an arbitrary real number T and a positive ϵ such that ϵ<1/4π. Then, by applying Theorem 2.2, there exists t>T and integers nj such that
(2.35)|taj-nj-bj|<ϵ,∀j∈J.
Hence, by substituting the values of aj and bj in the preceding inequality and multiplying by 2π, one has
(2.36)tμj={π4+ηj+2πnj,ifj∈J1,3π4+ηj+2πnj,ifj∈J2,5π4+ηj+2πnj,ifj∈J3,7π4+ηj+2πnj,ifj∈J4.
where the ηj’s are real numbers such that |ηj|<1/2. Then, according to the definition of the Jk’s, it is clear that
(2.37)αjsin(tμj)+βjcos⁡(tμj)>0,∀j∈J.
Consequently,
(2.38)Im⁡P(x,t)=∑j=1neμjx(αjsin(μjt)+βjcos⁡(μjt))>0,∀x∈R,
and then the lemma follows by taking y0=t.Corollary 2.7.
Let(2.39)P(z)=1+∑j=1nwjeμjz,wj∈C∖{0}
be an exponential polynomial with increasing positive frequencies μ1<⋯<μn forming a linearly independent set. Then, there exist infinitely many rectangles {Rk} in the critical strip of P(z) such that either ReP(z) or ImP(z) is always positive at every point of the sides of each Rk that are parallel to the real axis.Proof.
By applying Lemma2.5, determine the right lines of equations x=x1, x=x2 that define the strip x1<Rez<x2 where all the zeros of P(z) are comprised. Let m be an arbitrary integer, by taking T1=m and by applying Theorem 2.2, just as we have done in Lemma 2.6, we determine a value tm>T1 such that ImP(x,tm)>0 for all x∈ℝ. Now, again from Theorem 2.2, for T2=tm there exists a value tm+1>T2 such that ImP(x,tm+1)>0. Then, the four right lines of equations x=x1, y=tm, x=x2, y=tm+1 define a rectangle, say Rm, such that ImP(z) is positive when z lies on any of the sides of Rm that are parallel to the real axis. By reiterating this process, we will obtain the infinitely many rectangles {Rk:k≥m} desired. A completely analogous result we would have obtained if we had considered ReP(z) instead of ImP(z). Our corollary is then proved.


## 3. A Class of Almost-Periodic Functions with Bounded Spectrum  Containing the Partial Sums of the Riemann Zeta Function

In this section, we are going to generalize the preceding results to the class of almost-periodic functionsf(z) with bounded spectrum having the property of the existence of some value of y=Imz, say y0, such that either Ref(x,y0)≠0 or Imf(x,y0)≠0 for all x∈ℝ. By denoting this class by 𝒜𝒮, it follows that 𝒜𝒮 is nonvoid. Indeed, from lemmas 2.5 and 2.6, 𝒜𝒮 contains all exponential polynomials of degree n=1,2 with increasing positive frequencies and all exponential polynomials of arbitrary degree with linearly independent positive frequencies. Then, in particular, G2(z)≡1+2z belongs to 𝒜𝒮, and, since the frequencies log2<log3 are linearly independent, the function G3(z)≡1+2z+3z is also in the class 𝒜𝒮. However, although for any n≥4, the frequencies log2<log3<⋯<logn are always linearly dependent, all the approximants(3.1)Gn(z)≡1+2z+⋯+nz
of the Riemann zeta function ζ(z), in the half-plane Rez<-1, belong to the class 𝒜𝒮. Likewise, all the derivatives of Gn(z) are in the class 𝒜𝒮. To see that it is enough to check that for the value y0=0, ReGn(k)(x,0)>0, for all x∈ℝ and every k=0,1,2,…. Then, we are going to obtain our formula to count the zeros of the functions Gn(z) as a consequence of a general result on the functions of the class 𝒜𝒮.Theorem 3.1.
Letf(z) be a function of the class 𝒜𝒮. Then, there exist infinitely many rectangles {Rk} in the critical strip of f(z) such that the number of zeros inside each rectangle Rk, N(f(z);Rk), satisfies(3.2)|N(f(z);Rk)-μhk2π|<2,
where μ denotes the difference between the upper and the lower bounds of the spectrum of f(z) and hk is the height of Rk.Proof.
Asf(z) is an entire function of exponential type with bounded spectrum, from the Bohr theorem [2, p. 270], its zeros are all in a critical strip which we can suppose parallel to the imaginary axis (otherwise, we would consider the function g(z)≡f(iz)). Hence, without loss of generality, we can assume the existence of two real numbers x1, x2 such that all the zeros of f(z) are located in the strip
(3.3){z:x1<Rez<x2}.
On the other hand, sincef(z) is an almost-periodic function, let ∑j=1∞Ajeμjz be the Dirichlet series [7, p. 77] associated with f(z), denoted by
(3.4)f(z)~∑j=1∞Ajeμjz.
Then, the set of the Fourier exponents of the above series, also called Dirichlet exponents,
(3.5){μj:j∈N},
forms the spectrum of f(z), say Sf(z). Now, because the lower and the upper bounds of the spectrum of f(z) enter in the spectrum, let us define
(3.6)μ1≡min⁡{μj:j∈N},μ∞≡max⁡{μj:j∈N},μ≡μ∞-μ1,
with μ1,μ∞∈Sf. Then, by considering the two almost-periodic functions
(3.7)f(z)A1eμ1z-1,f(z)A∞eμ∞z-1,
from (3.4), it follows that these functions have associated Dirichlet series
(3.8)∑j=1∞Aj+1A1e(μj+1-μ1)z,∑j=1∞Aj+1A∞e(μj+1-μ∞)z,
respectively. Now, as the Dirichlet exponents of the above series, μj+1-μ1 and μj+1-μ∞ are strictly positive and negative, respectively, because of [7, Theorem 3.21], the functions
(3.9)f(z)A1eμ1z-1,f(z)A∞eμ∞z-1
tend to zero as Rez→-∞ and Rez→+∞ uniformly with respect to y, respectively. Therefore, there exist two reals a, b with a≤x1<x2≤b such that
(3.10)|f(z)A1eμ1z-1|<1,forRez≤a,|f(z)A∞eμ∞z-1|<1,forRez≥b.
On the other hand, as f is of class 𝒜𝒮, let y0 be a value of y=Imz such that, for instance,
(3.11)Im⁡f(x+iy0)>0,∀x∈R.
Then, from continuity, given the interval K≡[a,b] there exists δ>0 such that
(3.12)Im⁡f(x+iy0)≥δ,∀x∈K.
Now, from Definition 2.4, by taking ϵ1=δ/2, let (0,l1) be the ϵ1-interval of almost periodicity of f(z), with l1>0. Then, Definition 2.3 involves the existence of a translation number τ1∈(0,l1) such that
(3.13)|f(z+iτ1)-f(z)|≤ϵ1,∀z=x+iy0,withx∈K.
According to (3.12), inequality (3.13) implies
(3.14)Im⁡f(x+i(y0+τ1))≥ϵ1,∀x∈K.
Then, the four right lines of equations x=a, y=y0, x=b, y=y0+τ1 define a rectangle S1 such that, from (3.12) and (3.14), Imf(z) is positive when z lies on any of the sides of S1 parallel to the real axis. Furthermore, from (3.10), one has f(z)≠0 on the sides of S1 parallel to the imaginary axis.
As above, by now takingϵ2=δ/22, in the ϵ2-interval of almost periodicity of f, (0,l2), with l2>0, there exists a translation number τ2∈(0,l2) such that
(3.15)|f(z+iτ2)-f(z)|≤ϵ2,∀z=x+i(y0+τ1),withx∈K.
Then, inequalities (3.14) and (3.15) imply
(3.16)Im⁡f(x+i(y0+τ1+τ2))≥ϵ2,∀x∈K.
Therefore, the four right lines of equations x=x1, y=y0+τ1, x=x2, y=y0+τ1+τ2 define a rectangle S2 such that Imf(z) is positive on the sides of S2 that are parallel to the real axis, and f(z)≠0 on the sides of S2 that are parallel to the imaginary one. Then, continuing in this way, we obtain a family of rectangles
(3.17){Sk:k∈N},
with the property consisting of Imf(z)>0 on the sides that are parallel to the real axis and f(z)≠0 on the sides that are parallel to the imaginary one, for each rectangle Sk. Now, the intersection of each Sk with the critical strip of f, defined by (3.3), is a new rectangle, say Rk. Then, we claim that the set
(3.18){Rk:k∈N}
is the desired family of rectangles. Indeed, firstly, we observe that the number of zeros of f(z) inside each rectangle Rk is equal to the number of zeros inside each rectangle Sk. Secondly, Rk and Sk have the same height hk. Then, the variation of the argument of f(z) on each rectangle Sk is held to the following considerations:(1)
sinceImf(z)>0 on the sides of Sk that are parallel to the real axis, the variation of the argument of f(z) on these sides is less than π;(2)
from (3.10), the variation of the argument of f(z) on each side of Sk defined by the lines x=a and x=b is
(3.19)μ1hk+Θa,with|Θa|<π,μ∞hk+Θb,with|Θb|<π,
respectively. Therefore, noticing the previous considerations, the total variation of the argument of f(z) on each rectangle Sk is
(3.20)(μ∞-μ1)hk+Φf(z),k,with|Φf(z),k|<4π.
Consequently, the number of zeros of f(z) inside each 𝒜𝒮 satisfies the inequality
(3.21)|N(f(z);Rk)-μhk2π|<2.
Therefore, the Rk’s are the desired rectangles, as we claimed. Now, the proof of the theorem is completed.Corollary 3.2.
The critical strip associated with each partial sum of the Riemann zeta function on the half-planeRez<-1, Gn(z)≡1+2z+⋯+nz, n≥2, can be partitioned in infinitely many rectangles {Rk:k∈ℤ} such that the number of its zeros inside each rectangle Rk, N(Gn(z);Rk), satisfies
(3.22)|N(Gn(z);Rk)-hkln⁡⁡n2π|<2,
where hk is the height of Rk.Proof.
Firstly, from Lemma2.5, for each integer n≥2, there exist two real numbers an, bn such that the critical strip of the zeros of Gn(z) is defined by
(3.23)SGn(z)={z:an<Rez<bn}.
Now, starting from y0=0 and taking into account that Gn(z)∈𝒜𝒮, determine the family of rectangles {Rk:k=0,1,2,…} whose existence is guaranteed from the preceding theorem. It is plain that this family of rectangles forms a partition of the upper critical strip
(3.24){z∈SGn(z):0≤Im⁡z}.
Now, defining the rectangle R-k as the conjugate of Rk-1 for k=1,2,…, the desired partition of SGn(z) is formed by the rectangles of the family {Rk:k∈ℤ}. Finally, noticing Gn(z)=0 if and only if Gn(z¯)=0, one has
(3.25)N(Gn(z);R-k)=N(Gn(z);Rk-1),k=1,2,…,
and, since the set {lnk:1≤k≤n} is the spectrum of each Gn(z), inequality (3.22) follows. The corollary is then proved.



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*Source: 909674-2011-05-05.xml*