On the zeros of certain polynomials and entire functions

Entire functions whose coe ﬃ cients are polynomials having real and negative roots are built from Touchard polynomials. The particular set of polynomials n


Introduction and results
The aim of this note is to prove some results about the zeros of certain polynomials and Fourier transforms.We add an appendix showing the connection of some of our results to certain approximations in number theory.Each section is independent to read.
The location of roots of polynomials and entire functions is an old subject.Classical texts are [9], [12].The reader may consult [6] for an abundant bibliography up to 2011 on zeros of Fourier transforms.
We recall that an entire function f (z) is said to be in the Laguerre-Pólya class if it is the uniform limit on compact sets of polynomials with real roots.This is known [6] to be equivalent to the fact that there exists β, α, α n real, 0 ≤ α with We will use Hurwtiz theorem in its simplest form: if f n (z), f(z) are entire functions and f n (z) → f (z) uniformly on compact sets of the complex plane where the roots of f n (z) are real then either f (z) is identically zero or it has only real roots.Then p n−1 (x), n = 1, 2, . . ., is a polynomial with positive coefficients of n − 1 degree with only real negative roots.Also the polynomials q f n (x), n = 0, 1, 2, . . ., have only real roots.ii) Set

Use of Touchard polynomials.
. ., is a polynomial with positive coefficients of n − 1 degree with only real negative roots.
Proof.i) The Touchard polynomials T n (x) have only real roots.These roots are all negative except a simple root at x = 0.They can be defined using the generating function (see [8]) Next observe that by Laguerre's theorem ( [12] (II), problem 67, part V) and Hurwitz theorem if a polynomial p(x (2.1) has real roots.Here the equality follows from Cauchy's formula where γ is a curve enclosing the origin.
If fact from careful examination of the proof given in [12] if a polynomial p(z) has real and negative roots then (2.1) has only real and non positive roots (or more simply observe that a i > 0, T m (i) > 0 if i = 1, 2, . . ., and Finally observe that e e t −1 − 1 = c 1 t + c 2 t 2 + . . .with c i > 0 for all i.This implies that exp where p n−1 (x) is a polynomial of n − 1 degree with real and negative roots (observe that e x xp n−1 (x) = q n (x)).
If f (z) is in the Laguerre-Pólya class then it can be approximated on compact sets by polynomials p(x) = a 0 + a 1 x + a 2 x 2 + • • • + a k x k and the proof is exactly the same.ii) Again by Laguerre's theorem if a polynomial p(x) = a 0 + a 1 x + a 2 x 2 + • • • + a k x k has real and negative roots then similarly as in (i) where p * n−1 (x) is a polynomial with positive coefficients (and therefore with negative roots).The proof is complete.
2 Iterated exponentials have been studied by Bell [1] and also by Ramanujan ([2] see Chapter 4 and bibliography therein).

On the zeros of a class of polynomials
The main result of this section is the following theorem.
have all its roots on the line z = it, t ∈ R.

Proof.
Given a function f (z), we define the operator T f(z) := f (z + 1) + f (z − 1) and let g m (z) be a polynomial of m + 1 degree defined by The following lemma is the key to the proof.
with m a non negative integer, ∆ j > 0 and 0 ≤ r (in case r = 0 the last product is set equal to 1).
If m = 0 or m = 1 then T f(z) and T 2 f (z) have all their roots on the line In order to prove Lemma 1 we shall need the following lemma (compare with Lemma 1 in [4]).

Proof.
The lemma readily follows by observing that, for z

2
We shall prove Lemma 1 for m ≥ 2 (the proof for cases m = 0, 1 is implicitly contained in what follows).
We have that where We will show that all the roots of E(z) are on the line it, t ∈ R. In fact, if E(z) = 0, then | and, by Lemma 2, the same happens to each term in the product.A similar reasoning for the case <(z) < 0 yields that all the roots of E(z) are on the line it, t ∈ R.
As E(z) has degree 2r + 1, E(0) = 0, E 0 (0) 6 = 0 and Applying T one more time, one gets where then the same argument using Lemma 2 yields that all the roots of E * (z) are on the line it, t ∈ R. As E * (z) has degree 2r + 2, E * (0) 6 = 0 and This ends the proof of the Lemma 1.

Remarks on the zeros of certain Fourier transforms
Lemma 3. Assume that P N k=0 a k,N z k is a family of polynomials with its roots on |z| = 1.Furthermore assume that there is a continuous function b(x) : [0, 1] → R such that |b(k/N ) − a k,N | can be made arbitrarily small for sufficiently large N , uniformly in k.
If p(z) is a real polynomial with only real roots then has only real zeros or it is identically equal to zero. Proof.
In fact one knows that q N,λ (z) := P N k=0 a k,N p(z + (2k − N )iλ) has real roots, Theorem 4 pg.201 [4].Note that as N → ∞ The result follows changing variables 2τ − 1 = u 2 Lemma 4. Assume that b(x) is as in the last lemma.Then has only real roots.

Proof.
By the last lemma the integral ´(z + iu) n du has real roots as does z n I(1/z) =: J(z).Now J(z/n) tends to the integral (4.1) as n → ∞. 2 Theorem 3. Assume that b(x) satisfies one of the following two conditions: If the function (4.1) is not identically zero then it has only real roots.
Proof.i) Assume for the moment that f (x) is continuous.As 0 ≤ f (x) is non-decreasing by the Eneström-Kakeya theorem the polynomial q(z) = P N k=0 f (k/N )z k has its roots in |z| ≤ 1.Then by a theorem of Schur [13] we know that q(z) + z N q(1/z) = P N k=0 {f (k/N ) + f (1 − k/N )} z k has its roots on |z| = 1.We are in the conditions of the last lemma with b(x) = f (x) + f (1 − x) and then (4.1) has real roots.
The general theorem follows using a limiting argument and Hurwitz's theorem.ii) Again assume that b(x) is continuous and that 0 < α = k 0 /N 0 < 1/2 for some positive integers k 0 , N 0 .By Theorem 3 of [5] the polynomial q(z) =

. (diadic subdivision
).Thus using the above lemmas then the function (4.1) has real roots.
Again a limit argument yields the general result.2

Appendix
Here we show a connection between the polynomials given in Theorem 2 with some results proved in [11] about certain approximations of the Riemann's zeta function (see [14] or [7]).For the sake of completeness, we provide ad-hoc proofs, correct a minor error that appear in [11] and we add some new results.
n s is the Riemann's zeta function, then in the strip 0 < <(s) < 1, where As it is shown below, the function g(s) is approximated in the strip 0 < <(s) < 1 by where, for λ ≥ 0 and n ≥ 0, A n (λ) and B n (λ) are defined in terms of the (Jacobi) polynomials Here (α) n stands for α(α + 1) The key observation is that, if we set z = 2λ, then which is the polynomial appearing in Theorem 2.
It turns out that A n and B n satisfy a three-term recurrence relation (Theorem 5).Since polynomials B n (λ) appear in these formulas as denominators, we state some of their properties in Theorem 6.In Theorem 7 we give a closed form formula for g m (s) using B n (λ).
The two main ingredients of the proof below are the Euler's hypergeometric formula and the orthogonality of the F n (x) with respect to a power weight (see Lemma 5).Although orthogonality of Jacobi polynomials F n (x) is well known, we provide a simple proof that only uses Cauchy's residue theorem, following the lines of [3,Ch. 3].

Statement of the results
With the definitions and notation given we have the following results.
Note: a constant 1 2π was stated in [11] instead of the correct 2 .

Previous lemmas
The next two lemmas will be used in the proof of Theorem 4 and 5.The first one is about the polynomials F n (x) defined in (5.4).
Proof.Using residues one easily gets where γ is a curve enclosing 0, 1, ..., n.By taking γ to be contained in <(t) > −1/2 we can applied Fubini's Theorem to obtain The last integral is zero since, after simplifying the fraction, γ can be changed by a circumference of arbitrary large radius.Also, taking γ as above, we have, for j ≥ 0, , which turns out to be (j+1)n (λ+n+j+1) n+1 since the integrand has residue 0 at infinity.
2 For λ ≥ 0 we define Then we have the following lemma.
Lemma 6.Let s be in the strip 0 < <(s) < 1, then, with the above definition,

Proof.
One may prove this formula by making the change of variable which is valid for 0 < <(s) < 1, and adding in n.

Proof of Theorem 4
Let 0 < < 1/2 and s = σ + it with ≤ σ ≤ 1 − .First we observe that g n (s) are well defined since the rational functions A n (λ) Bn(λ) are continuous on [0, +∞) and By Lemma 6 we have to estimate where we have used that |B n (λ)| is increasing on [0, +∞), which follows from (5.6).Now recalling formulas (5.4), (5.5), (5.6) the key observation is that which will allow us to prove The proof of this inequality is as follows.From Lemma 5, and recalling the Euler's formula for the hypergeometric function 2 F 1 (a, b, c, z), with

Proof of Theorem 5
Observe that the coefficient x n of F n (x) is non-zero.Therefore one can find constants α n , β n so that F n (x) + (α n x + β n )F n−1 (x) is a polynomial of degree at most n−2, which in turn can be written as a linear combination of Coefficients α n , β n , γ n can be calculated equating the coefficients of x n , x n−1 , x n−2 and using (5.4).This yields (6.2).
To prove that A n also verify (6.1) we observe that The last integral above is zero for n ≥ 2 by Lemma 5.This completes the proof.

Theorem 1 .
i) Assume that f (z) is in the Laguerre-Pólya class and that exp

1 t
m+1 dt has only real and non positive roots.Letting k → ∞ in this last formula and writing exp(xe e t −1 ) = e t −1 ) 1 t m+1 dt, has only real and non positive roots.
non positive roots.Now taking p(x) = (1 + x/k) k , writing exp µ xe e e t −1 −1 ¶ = ∞ X n=0 r n (x)t n , and making k → ∞ yields that the function r m (x) = 1 2πi Z γ exp(xe e t −1 ) 1 t m+1 dt, has only real and non positive roots.But then it can be proved in a similar way as before that exp µ x ½ e e e t −1 −1 − 1