On the non-hyperbolicity of a class of exponential polynomials

In this paper we have constructed a class of non-hyperbolic exponential polynomials that contains all the partial sums of the Riemann zeta function. An exponential polynomial has been also defined to illustrate the complexity of the structure of the set defined by the closure of the real projections of its zeros. The sensitivity of this set, when the vector of delays is perturbed, has been analysed. These results have immediate implications in the theory of the neutral differential equations.


Introduction
We deal with exponential polynomials (EP for short) defined as h(z, a, r) The vectors a := (a 1 , a 2 , . . ., a N ), r := (r 1 , r 2 , . . ., r N ) are known as vector of coefficients, and vector of delays, respectively.The closure of the real projections of the zeros of h(z, a, r) is the set R h(z,a,r) = z : h(z, a, r) = 0 .
In the Example 2.1 of this document we have constructed an EP h(z, a, r) to illustrate, on one hand the complicate nature of R h(z,a,r) , and on the other hand how the stability of R h(z,a,r) is modified when the vector of delays is perturbed.The main result of the paper is the Theorem 3.3.There we have defined a class, say G, where any EP of G is non-hyperbolic, so any EP of G is not uniformly asymptotically stable (see, for instance, [1, Definitions 5.1, 5.2]), that contains to the family of EP having as components of the vectors of coefficients and delays the numbers a k := −1, r k := log(k + 1), k = 1, 2, . . ., N.

G. Mora
Therefore G contains all the partial sums of the Riemann zeta function, The main result is based on the fact that the point 0 belongs to the sets R g(z,a,r) when g(z, a, r) ∈ G. Indeed, we firstly prove that 0 ∈ R g(z,a,r) when g(z, a, r) is a partial sum ζ n (z), n ≥ 2.Then, by using a result of [1], the above property is also true for all the functions belonging to G.That is, 0 is a point of R g(z,a,r) for any g(z, a, r) ∈ G. Consequently, the nonhyperbolicity of any EP of G follows.Regarding the first question, it is important to stress that the vector of delays of ζ n (z), for each n ≥ 2, is (log 2, log 3, . . ., log n) so its components are not commensurable nor rationally independent (RI for short) for any n > 3. Thus, a priori, we have a new difficulty to add to the problem of determining the structure of the sets R g(z,a,r) when g(z, a, r) coincides with a partial sum ζ n (z) for n > 3. Indeed, besides the case that the components of the vector of delays are commensurable (see [1,Lemma 2.4]), mostly of the known results about the zeros of exponential polynomials, apply when the vector of delays has RI components (see, for instance, [1,3,8,9,[12][13][14] and [4,Chapter 3]).
The implications of the results of the present paper to the theory of functional difference equations and neutral functional differential equations are immediate.In effect, as we can see in [1], given the functional difference equation for any continuous function φ : [−ρ, 0] → R, where ρ ≥ max {r k : 1 ≤ k ≤ N}, there exists a unique solution x(φ) of (1.1), for t ≥ −ρ, which satisfies x(φ)(t) = φ(t) for all t ∈ [−ρ, 0].Therefore, by setting (S(t)φ)(u the set of operators is a strongly continuous semi-group of bounded linear operators on the space C([−ρ, 0], R) of continuous functions defined on [−ρ, 0] and valued in R.Moreover, if we define Therefore the location of the zeros of the EP h(z, a, r) gives information about the order β of the semi-group S(t).
The solution operator for the non-homogeneous neutral functional differential equation which usually appears in models of distributed networks [10,11] and in the control of structures through delayed forcing depending on the acceleration [2], can be written as a sum of a completely continuous operator and the operator S described above (see [5] and [1, p. 436].This gives information about the spectrum of the solution operator (see again [1, p. 436]).
Consequently, as it is noted in [12, Sect.1], h(z, a, r) = 0 determines the essential spectrum of the solution operator of (1.2) (a precise description of this property can be found in [6,Part 3]).The solutions of equation (1.2) satisfy a spectrum-determined growth condition (see, for instance, [7, Chapter 9, Corollary 3.1]), and the spectrum of the infinitesimal generator determines the stability of the zero solution, which can be sensitive to small changes in the delays (see again [12,Section 1]).Actually, could not be continuous with respect to the vector of delays r := (r 1 , . . ., r N ), like it is shown in [1, Example 2.1], and the same occurs in our example below.Therefore small changes in the delays can destabilise the equation, which is very important in control problems where usually there are slight delays in the application of the control action.
In the next section, we analyse the sensibility of certain exponential polynomial h(z, a, r) with respect to the vector of delays.

Delay perturbations
As in [1], we introduce the notation and where x ∈ R, and A, B are bounded subsets of R.
From [1, Lemma 2.5], R h(z,a,r) is always lower semicontinuous in r at any vector r 0 , that is, lim However, in general, (2.1) is not true if we substitute δ by the Hausdorff distance δ H . Indeed, it occurs, for instance in an example of EP, given by Silkowski [15] and analysed in [1, Example 2.1], with a vector of delays r having two commensurable components.In our example, it is defined an EP h(z, a, r) with a vector of delays having three components being RI two of them.As we prove below, the set R h(z,a,r) is the union of an isolated point and a closed interval.

Example 2.1.
A study on the sensitivity of the exponential polynomial The EP (2.2) is of the form 1 − ∑ N k=1 a k e −zr k , with N = 3 and a = (3, 1, 3), r = (log 27, log 64, log 216), as vectors of coefficients and delays, respectively.Since log 216 = log 27 + 1 2 log 64, the components of r are linearly dependent over the rationals, but log 27, log 64 are RI.Consider the sequence of vectors of delays r n := (log 27, log 64, 1  n + log 216), n = 1, 2, . . .First we claim the components of r n are RI for any fixed n ≥ 1.Otherwise, for some It means that the number e would be an algebraic number, which is a contradiction because e is transcendental.Consequently the claim follows.Now, let us define the sequence of exponential polynomials Our aim is now to find the sets R h(z,a,r n ) .To do it we introduce a new sequence of EP and then, by using the relation R h(z,a,r n ) = −R H(z,a,r n ) , we have enough to find the sets R H(z,a,r n ) .By (2.3), so, according to [13, Theorem 9], we firstly need to prove that the intermediate equations x n , (2.4) do not have any real solution.Indeed, since 64 x ≤ 1 for any x ≤ 0, we get x n , for all n ≥ 1.
If x > 0, since 64 x < 216 x 3e x n , it follows x n , for all n ≥ 1.
If x > 0, since 27 x 3 < 216 x 3e x n , we get x n , for all n ≥ 1.
Consequently (2.5) has no real solution whether x ∈ R \ [−1/3, 0] for all n ≥ 1.It only remains to prove that the equation (2.5) has no real solution in the interval [−1/3, 0], for all n ≥ 1.Indeed, we write (2.5) as Then, for any n ≥ 1, since e −1 3n ≤ e x n for all x ∈ [−1/3, 0], we have Now we claim that Indeed, by means of the change of variable 3x + 1 = u, and defining A n := e −1 3n , the inequation (2.8) becomes (2.9) For each n ≥ 1, it is not hard to check that the function attains its maximum value at a point, say u n , satisfying A n 2 2 u n = log 3 log 6 .Therefore we have and taking into account that c ≈ 0.7752037030, it follows which means that f n (u) < 1, for any u ∈ [0, 1], for all n ≥ 1.
On the other hand, the function is strictly increasing on [0, 1] and then its minimum value is g(0) = 5 4 .This proves (2.9), so (2.8) follows too.Then the claim follows.Therefore, taking into account (2.7), the equation (2.6) has no real solution in the interval [−1/3, 0] for any n ≥ 1.Consequently, (2.5) does not have any real solution.This implies, by virtue of [13,Theorem 9], that R H(z,a,r n ) has no gap for all n ≥ 1 and then On the other hand, (2.2) can be written as a product, that is, . Since all the zeros of h(z, a 1 , r 1 ) are imaginary, R h(z,a 1 ,r 1 ) = {0}.Regarding the set R h(z,a 2 ,r 2 ) , we claim that Indeed, as done earlier, we define H(z, a 2 , r 2 ) := h(−z, a 2 , r 2 ).Then H(z, a 2 , r 2 ) = 1 − 8 z − 27 z 3 = 1 − e z log 8 − 3e z log 27 , so the vector r 2 = (log 8, log 27) has RI components.It is immediate that the equation 8 x = 1 + 27 x 3 has no real solution.Then, because of [13,Theorem 9], it follows that R H(z,a 2 ,r 2 ) = [α , β ] for some real numbers α , β .From [13, (4.1)], α , β are the unique real solutions of the equations respectively.An easy computation gives us the approximate values α ≈ −0.47 and β ≈ −0.17, so R H(z,a 2 ,r 2 ) ≈ [−0.47, −0.17].Noticing H(z, a 2 , r 2 ) := h(−z, a 2 , r 2 ), we have as claimed.Consequently However, from (2.10), R h(z,a,r n ) ≈ [−0.22,0.47] for n sufficiently large and then, taking into account (2.12), it is evident that This means that the continuity of R h(z,a,r) with respect to the Hausdorff metric at the vector r = (log 27, log 64, log 216) fails.In other words, the perturbation of the vector of delays has destabilised the closure of the set of the real part of the zeros of h(z, a, r).
) can be also written of the form g(z, a, s) where and γ k • s is the inner product in R 2 of γ k by s, for k = 1, 2, 3. Since the components of s are RI, by designating by t = (t 1 , t 2 ) a generic vector of delays, we could apply [1, Theorem 2.2] ("If s is a fixed vector of (R + * ) M , where M > 1 is an integer, R + * := (0, +∞) and the components of s are RI, then R g(z,a,t) → R g(z,a,s) in the Hausdorff metric as t → s") obtaining lim t→s δ H (R g(z,a,t) , R g(z,a,s) ) = 0, which means that the perturbation of the vector of delays s does not destabilise the set R g(z,a,s) .It is important to stress that, in spite of R h(z,a,r) = R g(z,a,r) , the vectors r, s, used in each representation of (2.2), are distinct: r ∈ R 3 and it has components rationally dependent whereas s ∈ R 2 with RI components.

The non-hyperbolicity of the class G
Let us first recall that an EP h(z, a, r) is said to be hyperbolic at a vector r 0 if 0 / ∈ R h(z,a,r 0 ) (see, for instance, [1,Definition 5.1]).In this section we prove the non-hyperbolicity of a class of EP, denoted by G, that contains all the partial sums of the Riemann zeta function.The functions of G will be of the form (2.13) and all them will be denoted as g(z, a, s).Therefore we begin by expressing the partial sums ζ n (z) under the form (2.13) with the peculiarity that the vectors of delays will have RI components.To do it, for each n ≥ 2, it is enough to introduce vectors, say Γ j , for j = 1, . . ., n − 1, as follows.
Given the integer n ≥ 2, let k n be the number of primes not exceeding n.For each j = 1, . . ., n − 1, the vector Γ j := (Γ jl ) l=1,2,...,k n of R k n has components Γ jl defined as the unique non-negative integers such that each j + 1 is expressed of a unique form as by virtue of the fundamental theorem of arithmetic.Then, and so on.The vectors Γ j allow us to write ζ n (z) under the form where Γ j • p denotes the usual inner product in R k n of Γ j by the vector p defined as p := (log 2, log 3, log 5, . . ., log p k n ).
That is, the components of p are the logarithms of all the prime numbers not exceeding n, so p k n denotes the last prime such that p k n ≤ n, and, consequently, p has RI components.We define the class where R + * := (0, +∞).Therefore G contains all the partial sums of the Riemann zeta function In order to facilitate the reading of the manuscript, we state two results that will be used below.
k n , where p k n is the last prime not exceeding n > 2. If a real number c is such that the vertical line where N, M are positive integers, a j ∈ R for all 0 ≤ j ≤ N, the vectors γ j ∈ R M , 1 ≤ j ≤ N, have components which are non-negative integers and r is a vector of R M with positive components.Then, the following statements are equivalent: (i) 0 ∈ R g(z,a,r 0 ) for some r 0 with RI components; (ii) 0 ∈ R g(z,a,r) for all r with RI components.2) and a j = 1 for all 0 ≤ j ≤ N, by applying Corollary 3.2, we deduce that 0 ∈ R g(z,a,r) for any g(z, a, r) ∈ G. Consequently, from [1, Definition 5.1], it follows that any EP of G is non-hyperbolic.The proof is now complete.