                                            Zeros of Jensen polynomials and asymptotics for the Riemann xi
                                                                       function
                                                                                              Cormac O’Sullivan




arXiv:2007.13582v2 [math.NT] 9 Dec 2020
                                                                                                      Abstract
                                                    The classical criterion of Jensen for the Riemann hypothesis is that all of the associated Jensen poly-
                                                nomials have only real zeros. We find a new version of this criterion, using linear combinations of
                                                Hermite polynomials, and show that this condition holds in many cases. Detailed asymptotic expansions
                                                are given for the required Taylor coefficients of the xi function at 1/2 as well as related quantities. These
                                                results build on those in the recent paper of Griffin, Ono, Rolen and Zagier.


                                          1 Introduction
                                          The Riemann xi function, ξ(s) := 12 s(s−1)π −s/2 Γ(s/2)ζ(s), is entire of order 1. It satisfies ξ(1−s) = ξ(s)
                                          so that its development about the central point 1/2 is
                                                                                               ∞
                                                                                               X ξ (2n) (1/2)
                                                                                   ξ(s) =                            (s − 1/2)2n .                              (1.1)
                                                                                                       (2n)!
                                                                                               n=0

                                          The function Ξ(z) is defined as ξ(1/2 + iz). In our context it is useful to define another function
                                                                                                       √
                                                                                    Θ(z) := ξ(1/2 + z)

                                          which is entire of order 1/2. The Riemann hypothesis is equivalent to Ξ(z) having only real zeros and to
                                          Θ(z) having only negative real zeros. The central point is now at z = 0 and we write
                                                                             ∞
                                                                             X γ(m)                                             m! (2m)
                                                                   Θ(z) =                     zm,                   γ(m) :=         ξ   (1/2).                  (1.2)
                                                                                    m!                                        (2m)!
                                                                             m=0

                                          Note that ξ (2m) (1/2) is positive for all m since, as seen in (7.1), it may be expressed as the integral of a
                                          positive function. Hence any real zeros of Θ(z) are necessarily negative. We also see that Θ(z) is real which
                                          just means it maps R into R. It is convenient to label a function as hyperbolic if all of its zeros are real.
                                              Jensen describes the results of his research into the zeros of functions in [Jen13]. For any real entire
                                          function F (z) of genus at most 1, he associated a family of polynomials, now called Jensen polynomials,
                                          and showed that they are all hyperbolic if and only if F (z) is hyperbolic. He applied this idea to Ξ(z) in
                                          [Jen13, p. 189], giving a criterion for the Riemann hypothesis, and developed further equivalent conditions
                                          for hyperbolicity in [Jen13] and unpublished work; see the discussion in [Pól27].
                                              Following [DL11, GORZ19] we define the Jensen polynomial of degree d and shift n as
                                                                                                      d  
                                                                                       d,n
                                                                                                      X  d
                                                                                   J         (X) :=                  γ(n + j)X j .                              (1.3)
                                                                                                                j
                                                                                                      j=0

                                             Date: Nov 29, 2020.
                                             2020 Mathematics Subject Classification: 11M26, 11M06, 41A60.
                                             Support for this project was provided by a PSC-CUNY Award, jointly funded by The Professional Staff Congress and The City
                                             University of New York.


                                                                                                            1
This is associated to the nth derivative Θ(n) (z), and as we review in Corollary 3.2, Θ(n) (z) is hyperbolic
if and only if J d,n (X) is hyperbolic for all d ⩾ 1. Also, if Θ(z) is hyperbolic then all of its derivatives
must be hyperbolic as well; see Corollary 3.3. Hence we obtain the following extended criterion which has
presumably been known since the time of Jensen.

Theorem 1.1. The Riemann hypothesis is true if and only if J d,n (X) is hyperbolic for all d ⩾ 1, n ⩾ 0.

    Griffin, Ono, Rolen and Zagier revived interest in Theorem 1.1 when they showed in [GORZ19] that
a great many of the Jensen polynomials J d,n (X) are hyperbolic. For every fixed d, [GORZ19, Thm. 1]
states that J d,n (X) is hyperbolic for all sufficiently large n. This is made more explicit in [GOR+ ] where
their Theorem 1.1 shows that J d,n (X) is hyperbolic whenever n ≫ e8d/9 . In these papers, the results are
demonstrated by using precise asymptotics for γ(m) to show that renormalized versions of J d,n (X) may be
approximated by Hermite polynomials Hd (X) as n → ∞.
    We give a variant of Theorem 1.1 next by bringing in Hermite polynomials from the beginning. Set
                                                           d  
                                            d,n
                                                           X  d
                                        P         (X) :=                 γ(n + j)Hd−j (X).                   (1.4)
                                                                 j
                                                           j=0


Theorem 1.2. The Riemann hypothesis is true if and only if P d,n (X) is hyperbolic for all d ⩾ 1, n ⩾ 0.

    In fact, we see in Sect. 3 that Theorem 1.2 is a special case of a more general result where the Hermite
polynomials in (1.4) may be replaced by any Jensen polynomial associated to an element of the Laguerre-
Pólya class.
    Checking that the zeros of P d,n (X) are real seems easier than for J d,n (X). Combining the asymptotics
of γ(m) with a tailor-made theorem of Turán (that was also employed in [GOR+ ]) gives the next result
directly, showing that P d,n (X) has only real zeros for all but a relatively small number of shifts n.

Theorem 1.3. For all d sufficiently large, P d,n (X) is hyperbolic whenever n/ log2 n ⩾ d3/4 /2.

    Comparing J d,n (X) and P d,n (X) for specific d, n, we will see in Corollary 3.8 that J d,n (X) being hy-
perbolic implies that P d,n (X) is hyperbolic. Hence the results of [GORZ19, GOR+ ] also apply to P d,n (X).
Chasse in [Cha13, Sect. 3], proved that J d,n (X) is hyperbolic1 for all n ⩾ 0 and d ⩽ 2 × 1017 and so the
same is true for P d,n (X). As an example, it is easy to see that J 2,n (X) is hyperbolic if and only if

                                                   γ(n + 1)2 ⩾ γ(n)γ(n + 2).                                 (1.5)

This is the Turán inequality, necessary for the Riemann hypothesis, and first proved for all n ⩾ 0 in [CNV86].
However, P 2,n (X) is hyperbolic if and only if

                                          γ(n + 1)2 + 2γ(n)2 ⩾ γ(n)γ(n + 2),                                 (1.6)

and clearly (1.5) implies (1.6) but not the other way around.
    The key ingredient in the proof of Theorem 1.3 is the asymptotic expansion of the coefficients γ(n) in
(1.2) for large n. This is shown in [GORZ19, Thm. 9, Eq. (14)] with an application of Laplace’s method.
We give a more precise version of this result by including the usual error estimates and giving formulas
for all the coefficients. We also confirm a suggestion of Romik in [Rom, Sect. 6.1] that the answer can be
conveniently expressed in terms of W (2n/π), with W the Lambert function. Recall that this function is the
inverse to x 7→ xex and so satisfies

                                            W (xex ) = x,                W (x)eW (x) = x                     (1.7)

for at least x ⩾ 0. It is non-negative and increasing for x ⩾ 0 and W (x) ⩽ log x holds for x ⩾ e.
  1
      This range was extended in [GOR+ , Cor. 1.3] and may be further extended to d ⩽ 9 × 1024 using [PT].




                                                                     2
Theorem 1.4. Set w := W (2n/π). Then as n → ∞ we have
                               r           2
                                               n     K−1          K       !
                                   w     ew           X ck (w)     log  (n)
             γ(n) = 4π 2 e7w/4                    1+           +O
                                 w + 1 16ne2/w            nk          nK
                                                               k=1

for an implied constant depending only on K. Each ck (w) is a rational function of w with size O(logk (n))
and given explicitly in (7.13).
   The first coefficient is
                                               w4 + 58w3 + 29w2 − 24w − 16
                                  c1 (w) = −                               .                           (1.8)
                                                       192(w + 1)3
Table 1 compares the approximations of Theorem 1.4 with the actual value of γ(n) for n = 1000. All
decimals are correct to the accuracy shown. See also Table 2 in [GORZ19], (their γ(n) is 8 times larger).

                              K                Theorem 1.4
                              1      4.8460204243211378239 × 10−2568
                              3      4.8450426111532216799 × 10−2568
                              5      4.8450426112725884216 × 10−2568
                              7      4.8450426112725879772 × 10−2568
                                     4.8450426112725879772 × 10−2568       γ(1000)

                          Table 1: The approximations of Theorem 1.4 to γ(1000).

   Theorem 1.4 follows from the next theorem, giving the asymptotics of ξ (2n) (1/2) from (1.1).
Theorem 1.5. Set w := W (2n/π). Then as n → ∞ we have
                                   r                  K−1
                                                      X µk (w)     K      !
             (2n)           2 7w/4    2w  w 2n                   log (n)
           ξ      (1/2) = 4π e                     1+          +O                                      (1.9)
                                     w + 1 2e1/w          nk          nK
                                                      k=1

for an implied constant depending only on K. Each µk (w) is a rational function of w with size O(logk (n))
and given explicitly by (7.4) and (7.7).
     Theorem 1.5 is essentially a reformulated version of [GORZ19, Thm. 9] where they used the solution L
to the equation n = L(πeL + 3/4) instead of W (2n/π). The main term of the expansion (1.9) appears in
Thm. 6.1 of [Rom] and may also be compared with the weaker results of [Pus01] and [Cof09].
     The asymptotics of many related integrals are covered by our techniques in Sect. 4 and a general result
is formulated in Theorem 5.2. This gives the complete asymptotic expansion as n → ∞ of
                                         Z ∞
                           Iα (f ; n) :=     (log t)n e−αt f (t) dt      (α > 0)                     (1.10)
                                          1

for suitable functions f . For an application of this, also involving Hermite polynomials, recall Turán’s
expansion
                                               ∞
                                               X
                                       Ξ(z) =      (−1)n b2n H2n (z)                               (1.11)
                                                   n=0
for Ξ(z) = ξ(1/2 + iz). This series converges locally uniformly in C according to [Rom, Thm. 2.1]. We
may extend the asymptotics for b2n in [Rom, Thm. 2.7] with the next result.
Theorem 1.6. Set w := W (2n/π). Then as n → ∞ we have
                      7w/4−w 2 /16
                                   r                  K−1          3K      !
                     e                2w   w 2n     X τk (w)     log  (n)
          b2n = 4π 2                               1+          +O
                        (2n)!        w + 1 4e1/w          nk          nK
                                                      k=1

for an implied constant depending only on K. Each τk (w) is a rational function of w with size O(log3k (n))
and given by the formulas (5.4), (6.3) and (6.7).

                                                         3
    We remark that Hermite polynomials have also appeared recently in connection with the asymptotics
of ζ(s) in [O’S], which gives a generalization of the Riemann-Siegel formula. Romik explores further
orthogonal polynomial expansions of Ξ[z] in [Rom]. Wagner, in [Wag20], extends the work in [GORZ19] to
large classes of L-functions.The techniques in this paper should also be useful for these generalizations.

    Acknowledgements. Thanks to Jacques Gélinas, Dan Romik, Tim Trudgian and both referees for their
helpful comments.


2 Preliminaries
The Hermite polynomials have generating function exp(2Xt − t2 ) and the explicit expression
                                                               ⌊d/2⌋
                                                               X          (−1)r
                                          Hd (X) = d!                              (2X)d−2r .
                                                                       r!(d − 2r)!
                                                               r=0

These polynomials are a special case of the Laguerre polynomials appearing in (3.8).
   The partial ordinary Bell polynomials B̂i,j are useful devices to keep track of power series coefficients.
With j ∈ Z⩾0 , we have the generating function definition
                                                                                ∞
                                                                       j       X
                               p 1 x + p 2 x2 + p 3 x3 + · · ·              =         B̂i,j (p1 , p2 , p3 , . . . )xi .        (2.1)
                                                                                i=j

Clearly B̂i,0 (p1 , p2 , p3 , . . . ) = δi,0 . The formulas
                                                                                          j!
                                                                                                      pℓ11 pℓ22 pℓ33 · · · ,
                                                                X
                      B̂i,j (p1 , p2 , p3 , . . . ) =                                                                          (2.2)
                                                                                   ℓ1 !ℓ2 !ℓ3 ! · · ·
                                                        1ℓ1 +2ℓ2 +3ℓ3 +···=i
                                                         ℓ1 +ℓ2 +ℓ3 +···=j
                                                               X
                                                  =                             pn1 pn2 · · · pnj                    (j ⩾ 1)   (2.3)
                                                        n1 +n2 +···+nj =i

hold, where the sum in (2.2) is over all possible ℓ1 , ℓ2 , ℓ3 , · · · ∈ Z⩾0 and the sum in (2.3) is over all possible
n1 , n2 , · · · ∈ Z⩾1 . For instance,

                                  B̂9,6 (p1 , p2 , p3 , . . . ) = 20p31 p32 + 30p41 p2 p3 + 6p51 p4 .

For j ⩾ 1 we see from (2.3) that B̂i,j (p1 , p2 , p3 , . . . ) is a polynomial in p1 , p2 , . . . , pi−j+1 of homogeneous
degree j with positive integer coefficients. See the discussion and references in [O’S19, Sect. 7], for example,
for more information.
    As an application we will need later, consider
                                                                       X  ∞
                                                         log(x + 1)
                                hu (x) := log 1 +                        =      ℓi (u)xi                             (2.4)
                                                                u
                                                                                               i=1

which is a holomorphic function of x ∈ C for |x| ⩽ 1/2 and u ⩾ 2, say. To find the coefficients ℓi (u) write
                                               ∞
                                                  (−1)j+1
                                         X
                              log(x + 1)
                    log 1 +                =               logj (x + 1)
                                   u                j · uj
                                                               j=1
                                                                ∞                      ∞
                                                               X       (−1)j+1 X
                                                           =                     B̂i,j (1, − 12 , 31 , . . . )xi
                                                                        j · uj
                                                               j=1                    i=j
                                                               ∞            i
                                                               X            X                                     (−1)j+1
                                                           =         xi           B̂i,j (1, − 12 , 13 , . . . )           .    (2.5)
                                                                                                                   j · uj
                                                               i=1          j=1



                                                                          4
Then
                                              i
                                              X                                     (−1)j+1
                                   ℓi (u) =         B̂i,j (1, − 12 , 31 , . . . )                          (2.6)
                                                                                     j · uj
                                              j=1

with
                      1                   1      1                 1    1      1
              ℓ1 (u) =  ,     ℓ2 (u) = −     −     ,     ℓ3 (u) =    +     +      ,    etc.       (2.7)
                      u                  2u 2u2                   3u 2u2 3u3
                                                                                    R∞
    We also record here a basic bound for the incomplete gamma function Γ(s, a) := a e−x xs−1 dx.
Lemma 2.1. For a, r ⩾ 0 we have

                                    Γ(r + 1, a) ⩽ 2r (ar + Γ(r + 1))e−a .

Proof. The result follows from
                   Z ∞                  Z a                   Z ∞                  
                        e−x xr dx = e−a      e−x (x + a)r dx +      e−x (x + a)r dx
                     a
                                        Z0 ∞                Z ∞a             
                                     −a        −x     r            −x     r
                                  ⩽e          e (2a) dx +         e (2x) dx .
                                               0                                 0

    Hence, for all a, r ⩾ 0 and c > 0
                                  Z ∞
                                        e−cx xr dx ≪ c−r−1 ((ac)r + 1)e−ac                                 (2.8)
                                    a

for an implied constant depending only on r.


3 Jensen polynomials and the Laguerre-Pólya class
3.1 Background
                                                                                                       2
In [Pól27] Pólya defines a function to be of genus 1∗ (“erhöhtem Genus 1”) if it is of the form e−αz f (z) for
α ⩾ 0 and f (z) an entire function of genus at most 1. So this class includes some entire functions of order
2 and, by Hadamard’s theorem, all entire functions of order < 2. It is easy to see that the product of two
functions of genus 1∗ is also of genus 1∗ . See for example [KK00, Sect. 1] for a discussion of Pólya’s long
study of genus 1∗ functions.
    The Laguerre-Pólya class consists of functions that are real, entire of genus 1∗ and that satisfy the added
condition of having only real zeros. In particular, the functions Ξ(z) and Θ(z) are both real entire functions
of genus 1∗ and they are in the Laguerre-Pólya class if and only if the Riemann hypothesis is true. Pólya and
Schur in [PS14] characterize the Laguerre-Pólya class in variousPways and we give their description in terms
of Jensen polynomials next. For a formal power series Φ(z) = ∞               j
                                                                     j=0 cj z /j! define

                                                              d  
                                                              X  d
                                          gd (Φ; x) :=                      cj xj .
                                                                       j
                                                              j=0

to be the Jensen polynomial of degree d associated to Φ. Let
                                                                           d  
                                                                           X  d
                                gd∗ (Φ; x) := xd gd (Φ; 1/x) =                           cj xd−j
                                                                                     j
                                                                           j=0

be the reciprocal polynomial. We have easily
                       d                                              d ∗              ∗
                         gd (Φ; x) = d · gd−1 (Φ′ ; x),                g (Φ; x) = d · gd−1 (Φ; x)          (3.1)
                      dx                                             dx d
and the right identity in (3.1) indicates that gd∗ (Φ; x) is an Appell polynomial. More properties of Jensen
polynomials are given in [CV90, Prop. 2.1].

                                                              5
Theorem 3.1 (Pólya-Schur [PS14]). Let Φ(z) = ∞                  j
                                                       P
                                                         j=0 cj z /j! be a formal power series with real coeffi-
cients. Then Φ(z) converges uniformly on compact sets in C to an entire function in the Laguerre-Pólya
class if and only if gd (Φ; z) is hyperbolic for all d ⩾ 1.

    Laguerre had seen cases of this theorem and Jensen proved it in [Jen13, pp. 183–187] with the assump-
tion that Φ(z) is a real entire function of genus at most 1. Pólya showed in [Pól27, Sect. 8] that Jensen’s
proof extends easily to Φ(z) being of genus 1∗ and we describe this briefly next.

Sketch of proof when Φ(z) is of genus 1∗ . In the easier direction, assume all gd (Φ; z) are hyperbolic. Then
since gd (Φ; z/d) → Φ(z) locally uniformly in C as d → ∞, it follows from Hurwitz’s theorem for example
that Φ(z) must be hyperbolic.
                d
    Let D = dz    . Then it can be shown to follows from Rolle’s theorem that if p(z) is a real hyperbolic
polynomial then (D − a)p(z) = p′ (z) − ap(z) is also hyperbolic for all a ∈ R. If f (z) and p(z) are real
hyperbolic polynomials then applying this argument deg(f ) times shows f (D)p(z) is hyperbolic. Assume
Φ(z) is hyperbolic. From its Weierstrass factorization it is possible to construct real hyperbolic polynomials
Φn (z) that converge uniformly to Φ(z) as n → ∞. Then each Φn (D)p(z) is hyperbolic and as n → ∞ they
converge to Φ(D)p(z). It follows that Φ(D)p(z) is hyperbolic whenever p(z) is. The final step is to note
that Φ(D)z d = gd∗ (Φ; z), making gd∗ (Φ; z) and hence gd (Φ; z) hyperbolic.

    The simple idea behind our Theorem 1.2 is to see what happens when p(z) = z d , in the last sentence of
the above proof, is replaced by Hd (z). Also note that a similar limiting procedure allows the polynomial p(z)
to be replaced by more general functions; see Theorem 3.6. The full Theorem 3.1, requiring no conditions
on Φ(z), is stated in [PS14, p. 111] and proved there in Sections 2–4. See for example [CC89, Thm. 2.7] for
further characterizations of the Laguerre-Pólya class.
    In our previous notation, with (1.2) and (1.3),

                                            J d,n (x) = gd (Θ(n) ; x)                                               (3.2)

since the Taylor coefficients of the nth derivative Θ(n) (z) are shifted by n.

Corollary 3.2. The function Θ(n) (z) is hyperbolic if and only if J d,n (z) is hyperbolic for all d ⩾ 1.

Proof. We know that Θ(z) is real and entire of order 1/2. All of its derivatives must have the same properties
and this implies that Θ(n) (z) is real and entire of genus 1∗ for all n ⩾ 0. Therefore Θ(n) (z) is hyperbolic if
and only if it is the Laguerre-Pólya class. The corollary now follows from (3.2) and Theorem 3.1.

Corollary 3.3. If Θ(n) (z) is hyperbolic then Θ(m) (z) is also hyperbolic for all m > n.

Proof. By Corollary 3.2, if Θ(n) (z) is hyperbolic then J d,n (z) is hyperbolic for all d ⩾ 1. The identity on
the left of (3.1) implies that
                                        d d,n
                                           J (z) = d · J d−1,n+1 (z).
                                        dz
As we saw in the proof of Theorem 3.1, the derivative of a real hyperbolic polynomial must be hyperbolic.
Therefore J d,n+1 (z) is hyperbolic for all d ⩾ 1. Hence, by Corollary 3.2 again, Θ(n+1) (z) is hyperbolic.
Induction completes the proof.

3.2 Jensen polynomials of products
                                                                              P∞         j
Lemma 3.4. Let Φ and Ω be two formal power series with Φ(z) =                    j=0 cj z /j!. Then for all d ⩾ 0

                                                      d  
                                                      X  d
                                   gd∗ (Φ · Ω; x) =                       ∗
                                                                    cj · gd−j (Ω; x).                               (3.3)
                                                                j
                                                      j=0

Proof. This is an easy exercise.



                                                            6
    Replacing x by 1/x in (3.3) also yields
                                                          d  
                                                          X  d
                                      gd (Φ · Ω; x) =                   cj xj · gd−j (Ω; x).               (3.4)
                                                                    j
                                                          j=0
                                P∞     j                                        ∗
Theorem 3.5. Let Φ(z) =        j=0 cj z /j! be a real entire function of genus 1 and let Ω(z) be in the
Laguerre-Pólya class. Then Φ(z) is hyperbolic if and only if the polynomials
                                                  d  
                                                  X  d               ∗
                                                               cj · gd−j (Ω; x)                            (3.5)
                                                         j
                                                  j=0

are hyperbolic for all d ⩾ 1.

Proof. First note that Φ · Ω is of genus 1∗ because each of Φ and Ω are. The zeros of Ω are real since it is
in the Laguerre-Pólya class. Therefore Φ · Ω is in the Laguerre-Pólya class if and only if Φ(z) is hyperbolic.
By Theorem 3.1, Φ · Ω is in the Laguerre-Pólya class if and only if gd∗ (Φ · Ω; x) is hyperbolic for all d ⩾ 1.
Noting that gd∗ (Φ · Ω; x) equals (3.5) by Lemma 3.4 finishes the proof.

    Clearly we could replace (3.5) by (3.4) in the statement of Theorem 3.5 and get the same result. The
            2              2        z4
function e−z = 1 − 2!1 · z2! + 4!
                               2! · 4! − . . . is in the Laguerre-Pólya class and

                                      2  ⌊d/2⌋
                                             X d        (2j)! d−j
                                       −z
                               gd∗    e ;x =        (−1)j      x
                                                 2j         j!
                                                      j=0
                                                      ⌊d/2⌋
                                                      X        d!(−1)j d−j
                                                  =                       x = Hd (x/2).
                                                              (d − 2j)!j!
                                                      j=0
                                      P∞           j
By Lemma 3.4, for any Φ(z) =               j=0 cj z /j!, we then find

                                                        d  
                                           
                                                 −z 2
                                                       X  d
                                     gd∗    Φ · e ;x =        cj · Hd−j (x/2)
                                                           j
                                                               j=0

and in particular
                                       d  
                          
                            (n) −z 2
                                      X  d
                    gd∗    Θ · e ;x =        γ(j + n) · Hd−j (x/2) = P d,n (x/2)                           (3.6)
                                          j
                                                  j=0

with our notation (1.4).

Proof of Theorem 1.2. As seen in the proof of Corollary 3.2, Θ(n) (z) is real and entire of genus 1∗ for all
                                                                2
n ⩾ 0. Theorem 3.5 with Φ(z) = Θ(n) (z) and Ω(z) = e−z implies that Θ(n) (z) is hyperbolic if and only
if P d,n (x) as in (3.6) is hyperbolic for all d ⩾ 1. As in Corollary 3.3, the Riemann hypothesis is equivalent
to Θ(n) (z) being hyperbolic for all n ⩾ 0.

   We noted in the introduction that J d,n (x) being hyperbolic implies that P d,n (x) is hyperbolic. This is
shown next with the help of a general result of Pólya from [Pól15, p. 242].

Theorem 3.6. Let Φ(z) and Ω(z) be in the Laguerre-Pólya class with Φ(z) = ∞               j
                                                                               P
                                                                                   j=0 cj z /j!. Suppose

                                                        ∞
                                                        X cj
                                                                    Ω(j) (z)                               (3.7)
                                                               j!
                                                        j=0

converges when |z| < ρ for some ρ > 0. Then (3.7) represents a function in the Laguerre-Pólya class.

                                                                 7
                                                      2
    Following Pólya, the example of Ω(z) = e−z may be used in Theorem 3.6 along with the formula
                                                                  2
                                         Ω(j) (z) = (−1)j e−z Hj (z).

We obtain the following special case, which was also stated in Lemma II of [Tur59].

           3.7. Let m           j
                      P
Corollary
Pm                      j=0 aj z be a hyperbolic polynomial with real coefficients. Then the polynomial
  j=0 aj Hj (z) is also hyperbolic.

Corollary 3.8. If J d,n (x) is hyperbolic then P d,n (x) is hyperbolic.

Proof. If J d,n (x) is hyperbolic then so is the reciprocal polynomial xd J d,n (1/x). An application of Corol-
lary 3.7 to this reciprocal now shows that P d,n (x) is hyperbolic.

    Note that taking Φ(z) = J d,n (x) and Ω(z) = Hd (x/2) in Theorem 3.6 implies the similar polynomial
                                          d
                                          X        
                                                  d            Hd−j (x)
                                                      γ(n + j)
                                                  j            (d − j)!
                                          j=0

                                                            d j              d!
is also hyperbolic if J d,n (x) is. This uses the equality dx j Hd (x/2) = (d−j)! Hd−j (x/2).



3.3 Further examples of Jensen polynomials
Theorem 3.5 with Φ(z) = Θ(n) (z) may be used to give different criteria for the Riemann hypothesis, intro-
                                                                                                      2
ducing an interesting flexibility. For Ω(z) = 1 we obtain the polynomials J d,n (X) and for Ω(z) = e−z we
obtain P d,n (X). We could use cos(z) and sin(z) for Ω(z) for example, but we next focus on a family of
Bessel functions.
    For α ∈ R the Bessel function of the first kind has the series representation
                                            ∞
                                            X          (−1)m      z 2m+α
                                 Jα (z) =                                  .
                                                  m! Γ(m + α + 1) 2
                                            m=0

Then z −α Jα (2z) is entire and even and we may put
                                                        ∞
                                                  √     X            (−1)m z m
                            Jα (z) := z −α/2 Jα (2 z) =                           .
                                                              m=0
                                                                  m! Γ(m + α + 1)

A short calculation as in [DBC09, Sect. 3] finds
                                                           d!       (α)
                                      gd (Jα ; x) =                Ld (x)
                                                      Γ(d + α + 1)

with the generalized Laguerre polynomials given by
                                            d                     d 
                                                       (−1)k xk                    xk
                                                                            
                 (α)     Γ(d + α + 1)       X  d                    X   d+α
                Ld (x) =                                          =           (−1)k .                     (3.8)
                              d!                   k Γ(k + α + 1)       d−k        k!
                                            k=0                           k=0

These polynomials are orthogonal for α > −1 and therefore have only real roots when d ⩾ 1 for α in this
range. In fact the Laguerre polynomials are known to have real roots for α ⩾ −2. For d ⩾ 2 and α < −2,
  (α)
Ld (x) has non-real roots except possibly when α is an integer. See [Sze75, Eq. (5.2.1), Thm. 6.7.3] for
these results. It follows from Theorem 3.1 that Jα is in the Laguerre-Pólya class for all α ⩾ −2. Applying
Theorem 3.5 with Φ = Θ(n) and Ω = Jα , and using the alternate identity (3.4), proves an extended criterion
for the Riemann hypothesis involving the Laguerre polynomials:



                                                          8
Theorem 3.9. The Riemann hypothesis is true if and only if the polynomials
                                                     d     
                                    d,n,α
                                                     X   d+α                         (α)
                                Q           (x) :=                     γ(n + j)xj Ld−j (x)
                                                               j
                                                     j=0

are hyperbolic for all d ⩾ 1, all n ⩾ 0 and all real α ⩾ −2.
    For example, taking d = 2, the discriminant of Q2,n,α (x) is
                                                       h                      i
                     (α + 2)γ(n)2 + (α + 1)2 (α + 2)2 γ(n + 1)2 − γ(n)γ(n + 2) ,

and this is non-negative if α ⩾ −2 and the Turán inequality (1.5) holds.
     We note that Farmer, in the interesting preprint [Far], puts these types of Jensen polynomial approaches
to the Riemann hypothesis into context in the recent literature and comments on their likelihood of success.


4 The asymptotics of Iα (n)
Before treating (1.10) we first look at the simpler case
                                                   Z ∞
                                         Iα (n) :=      (log t)n e−αt dt                                        (4.1)
                                                           1

which contains all the main ideas. Recalling (2.6) we define the rational functions
                               2r                   2 j+r
                               X   (2j + 2r − 1)!!    v
                     ar (v) :=                                  B̂2r,j (ℓ3 (v), ℓ4 (v), . . . )                 (4.2)
                                          j!        v+1
                                 j=0
                                                                   2v 4 + 9v 3 + 16v 2 + 6v + 2
                 so that      a0 (v) = 1,            a1 (v) =                                   ,       etc.    (4.3)
                                                                           24(v + 1)3
Theorem 4.1. Suppose α > 0 and set u := W (n/α). Then as n → ∞ we have
                                                             R−1
                                                                                               !
                                       √
                                                                           R
                                              un+1 eu−n/u    X ar (u)      u
                           Iα (n) =        2π p           1+       r
                                                                      + O    R
                                                                                                                (4.4)
                                                (1 + u)n     r=1
                                                                 n         n

where the implied constant depends only on R ⩾ 1 and α. Also ar (u) ≪ ur ≪ logr (n).
Proof. We give the proof in the rest of this section, following Laplace’s method as described in [FS09, Sect.
B6] for example. See also [GORZ19, Thm. 9] and [Rom, Sect. 2.4] for similar arguments. Let g(t) denote
the integrand in (4.1). Then                                   
                                           ′            n
                                          g (t) =            − α g(t)
                                                     t log t
and g′ (t0 ) = 0 for the unique t0 > 1 satisfying t0 log t0 = n/α. With (1.7) we have that
                                                                                                     n/α
               W (n/α) = W (t0 log t0 ) = log t0                   implies     t0 = eW (n/α) =              .   (4.5)
                                                                                                    W (n/α)
The largest contribution to the integral Iα (n) will be near the maximum of the integrand at t = t0 ≈
n/(α log(n/α)). We develop the integrand about this point:
                                   Z ∞                   
                                                    g(t)
                   Iα (n) = g(t0 )       exp log              dt
                                     1              g(t0 )
                                      Z ∞                             
                                                          g((x + 1)t0 )
                          = t0 g(t0 )          exp log                     dx
                                       1/t0 −1               g(t0 )
                                      Z ∞                                        
                                                                log(x + 1)
                          = t0 g(t0 )          exp n log 1 +                 − αt0 x dx.          (4.6)
                                       1/t0 −1                     log t0


                                                                9
For u = W (n/α) the integrand is
                                               
                                  log(x + 1)     x          h         x i
                    exp n log 1 +              −      = exp n hu (x) −                                  (4.7)
                                      u          u                     u

with hu (x) from (2.4) having a power series expansion for |x| ⩽ 1/2 and u ⩾ 2, say, with coefficients ℓi (u)
given in (2.6). After this point, the equality between u and W (n/α) may be ignored and we treat u as a free
parameter with u ⩾ 2, or later u ⩾ 1.
Lemma 4.2. Suppose |z| ⩽ 1/2 for z ∈ C and assume u ⩾ 2. Then for k ⩾ 0 we have
                                                 k
                                                 X
                                      hu (z) =         ℓi (u)z i + Rk (u, z)
                                                 i=1
                                           3 4 i                            12 4 k k+1
                                                                             
                     where      |ℓi (u)| ⩽       ,            |Rk (u, z)| ⩽       |z| .                 (4.8)
                                           u 3                              u 3

Proof. From the simple inequality | log(w + 1)| ⩽ 2|w| for |w| ⩽ 3/4, w ∈ C we obtain

                                |hu (w)| ⩽ 4|w|/u       for    |w| ⩽ 3/4, u ⩾ 2.
                      (i)
We may now bound hu (0) and the Taylor remainder using Cauchy’s estimates in the usual way, with the
bound |hu (w)| ⩽ 3/u for |w| = 3/4.

Lemma 4.3 (Neglecting the tails). Suppose 0 < δ ⩽ 1/100 and u ⩾ 2. Then
      Z ∞                                Z δ                                 2 
                    h         x i              h         x i                δ n
                exp n hu (x) −      dx =     exp n hu (x) −      dx + O u exp −                         (4.9)
        1/t0 −1                u          −δ                u                   4u

where the implied constant is absolute.

Proof. The elementary inequalities

                       log(1 + x) ⩽ x     (x ⩾ 0),            log(1 + y) < y/2 (y ⩾ 3)

imply that
                             hu (x) ⩽ log(1 + x/u) < x/(2u)           for      x ⩾ 3u.
Therefore           Z ∞                          Z ∞
                            h         x i              nx       2u −3n/2
                        exp n hu (x) −      dx ⩽     exp −     dx =   e      .                        (4.10)
                     3u                u          3u       2u       n
By design hu (x) − x/u is increasing for x < 0 and decreasing for x > 0. Then for x ⩾ δ we have

                    hu (x) − x/u ⩽ hu (δ) − δ/u
                                       u+1 2                    δ2               δ2
                                  ⩽−       δ + |R 2 (u, δ)| < −    (1 − 50δ) ⩽ −    .
                                       2u2                      2u               4u
We obtain the same bound for x ⩽ −δ and so
                   Z −δ       Z 3u !                                       2 
                                         h         x i                     δ n
                            +        exp n hu (x) −      dx ⩽ (1 + 3u) exp −     .
                    1/t0 −1    δ                    u                        4u

This is bigger than the bound in (4.10) and the lemma is proved.




                                                        10
      Put C := n(1 + u)/u2 so that n · ℓ2 (u) = −C/2 by (2.7). It follows from Lemma 4.2 that
                                                                               k
                                                                                          !
         Z δ     h         x i                      Z δ
                                                              −Cx2 /2
                                                                              X
                                                  k+1                                   i
             exp n hu (x) −      dx = exp O(nδ        )      e         exp n     ℓi (u)x dx.                                          (4.11)
          −δ                u                             −δ                                             i=3

Write the sum in (4.11) as y := ki=3 ℓi (u)xi . Then n|y| ≪ δ3 n/u by (4.8). We would like n|y| to be small
                                P

and so δ3 n should tend to 0 as n → ∞. Also, with the error in (4.9), δ2 n should tend to ∞. This means
choosing δ between n−1/2 and n−1/3 . We now fix

                            δ := n−2/5             so that              δ2 n = n1/5 ,
                                                                   δ3 n = n−1/5 .

With this choice of δ (and assuming k ⩾ 2) we have exp O(nδk+1 ) = 1 + O(nδk+1 ). Also, with this δ,
                                                                  

the integrand on the right of (4.11) is ≪k 1 and therefore
                                                           k
        Z δ                              Z δ                          !
                h             x i                2
                                                           X
            exp n hu (x) −          dx =     e−Cx /2 exp n   ℓi (u)xi dx + O(n3/5−2k/5 ).    (4.12)
         −δ                    u          −δ                                     i=3

      Without truncating the Taylor series (2.4) of hu (x), we may write
                              ∞                    ∞       i
                                            !
                             X                    X      X                                  nj x2j
                       exp      n · ℓi (u)xi =        xi     B̂i,j (ℓ3 (u), ℓ4 (u), . . . )
                                                                                              j!
                                 i=3                        i=0          j=0
                                                            X∞
                                                       =               er (n, u)xr                                                    (4.13)
                                                             r=0

for
                                                      ⌊r/3⌋
                                                      X                                           nj
                                       er (n, u) :=           B̂r−2j,j (ℓ3 (u), ℓ4 (u), . . . )      .                                (4.14)
                                                                                                  j!
                                                      j=0

It follows that er (n, u) is a polynomial in n of degree at most ⌊r/3⌋ with coefficients that are rational
functions in u. Using (2.3) and (4.8) shows

                                                        er (n, u) ≪r nr/3 .                                                           (4.15)

The next lemma gives a truncated version of (4.13).
Lemma 4.4. For all x with |x| ⩽ δ = n−2/5 and all u ⩾ 1 we have
                               k
                                               ! k−1                    
                              X
                                             i
                                                 X
                                                               r       1
                        exp       n · ℓi (u)x =      er (n, u)x + O k/15                                                              (4.16)
                              i=3                r=0
                                                                     n

for an implied constant depending only on k.

Proof. For a number c to be chosen later, the left side of (4.16) is
                      c
                      X
                        (ny)j /j! + O (n|y|)c+1
                                                
            exp(ny) =
                          j=0
                          c        k−2
                                                                  !j
                          X nj x2j X                                                      
                      =                          ℓi+2 (u)x    i
                                                                       + O (u · n1/5 )−c−1
                                 j!
                          j=0              i=1
                          c           (k−2)j
                                nj x2j X
                                                                                                                             
                          X
                                                                                                         i            1
                      =                          B̂i,j (ℓ3 (u), ℓ4 (u), . . . , ℓk (u), 0, 0, . . . )x + O                        .
                          j=0
                                 j!
                                          i=j
                                                                                                                   n(c+1)/5


                                                                        11
If we define er (n, u)k as in (4.14), but with ℓj (u) inside the Bell polynomial replaced by 0 when j ⩾ k + 1,
then we obtain
                                  k−1                    kc                              
                                  X
                                                   m
                                                        X
                                                                         m           1
                     exp(ny) =         em (n, u)x +          em (n, u)k x + O (c+1)/5
                                  m=0                   m=k
                                                                                 n

because em (n, u)k = em (n, u) for m ⩽ k − 1. As in (4.15) we have em (n, u)k ≪ nm/3 and hence
               kc
               X                                     kc
                                                     X                       kc
                                                                             X                        kc
                                                                                                      X
                     em (n, u)k xm ≪                       nm/3 xm ≪               (δ3 n)m/3 ≪               n−m/15 ≪ n−k/15 .
               m=k                               m=k                         m=k                      m=k

Choosing any c ⩾ k/3 − 1 completes the proof.

Lemma 4.5 (Central approximation and completing the tails). We have
               Z δ                           k
                                                                     !          √       ⌊(k−1)/2⌋ ∗            
                     −Cx2 /2
                                             X
                                                                 i                2πu      X a
                                                                                                  m         1
                     e            exp n                ℓi (u)x           dx = p                     + O                          (4.17)
                −δ                               i=3
                                                                                (u + 1)n m=0 nm           nk/15

when δ = n−2/5 , C = n(1 + u)/u2 , u ⩾ 1 and
                                                                                                 m
                                                                                            u2
                                                                                       
                                   a∗m = a∗m (n, u) := (2m − 1)!!                                     e2m (n, u).
                                                                                           u+1

Proof. Lemma 4.4 implies the left side of (4.17) equals
                                       k−1                   Z δ                                            
                                       X
                                                                         −Cx2 /2 r                    1
                                                 er (n, u)               e        x dx + O                       .               (4.18)
                                           r=0                −δ                                     nk/15

If we extend the integral in (4.18) to all of R then the difference is twice
                                   Z ∞                     Z ∞
                                          −Cx2 /2 r
                                        e         x dx ⩽        e−Cδx/2 xr dx.                                                   (4.19)
                                                 δ                                 δ

With (2.8) and the inequality C > n/u, (4.19) is bounded by a constant depending on r times
                                         r+1 !
                                2δr−1      2              2         1/5
                                      +              e−Cδ /2 ≪ e−n /(2u) .
                                  C       Cδ

For m ∈ Z⩾0 we use
                              Z ∞                                                            r
                                       −Cx2 /2 2m                Γ(m + 1/2)                       2π (2m − 1)!!
                                       e               x    dx =            =                                   .
                                  −∞                             (C/2)m+1/2                       C     Cm

Therefore
                             k
  Z δ                                            !
         −Cx2 /2
                             X
         e         exp n           ℓi (u)xi           dx
    −δ                       i=3
                         r        ⌊(k−1)/2⌋                                                              
                             2π     X                           (2m − 1)!!        1       k/3 −n1/5 /(2u)
                     =                               e2m (n, u)            + O k/15 + n e
                             C                                     Cm           n
                                    m=0
                                                                                    √        ⌊(k−1)/2⌋ ∗            
                                                                                      2πu       X a
                                                                                                        m          1
                                                                              =                             + O k/15
                                                                                    (u + 1)n m=0 nm
                                                                                 p
                                                                                                                 n

as required.

                                                                             12
   Each a∗m (n, u) is a polynomial in n of degree at most ⌊2m/3⌋ with coefficients in Q(u). Explicitly:
                ⌊2m/3⌋                                                2 m
                 X                                        (2m − 1)!!   u
        a∗m =            cm,j n   j
                                        for       cm,j :=                   B̂2m−2j,j (ℓ3 (u), ℓ4 (u), . . . ).                      (4.20)
                                                             j!       u+1
                 j=0

                                                                                                                      PLk      ∗   m
It is convenient to replace k by 2Lk + 1 with an L to be chosen later. Simplifying                                        m=0 am /n into a
polynomial in 1/n we find
                                      Lk                         Lk                                 2r
                                      X  a∗ (n, u)
                                              m
                                                                 X  âr (u)                         X
                                                          =                       for   âr (u) =         cr+j,j
                                              nm                        nr
                                  m=0                            r=0                                j=0

provided cm,j is set to zero for m > Lk. In other words

                                min{2r,Lk−r}                          2 j+r
                                        X            (2j + 2r − 1)!!   u
                    âr (u) =                                                 B̂2r,j (ℓ3 (u), ℓ4 (u), . . . )
                                                           j!         u+1
                                        j=0

and equation (4.17) becomes
          Z δ                           2Lk+1
                                                                 !            √           Lk                                
                 −Cx2 /2
                                         X
                                                             i                    2πu     X  âr (u)                  1
                e            exp n                 ℓi (u)x           dx = p                               +O                     .   (4.21)
           −δ                             i=3
                                                                           (u + 1)n r=0 nr                         n2Lk/15

   Now (4.12) and (4.21) can be combined to produce
  Z δ                             √
           h         x i          2πu
       exp n hu (x) −      dx = p
    −δ                u           (u + 1)n
                                    "k−1         Lk
                                                                                    !#
                                     X âr (u) X    âr (u)        n1/2       n1/2
                                 ×            +             +O             + 2Lk/15    . (4.22)
                                           nr         nr       n (4Lk−1)/5  n
                                     r=0        r=k

Comparing âr (u) with ar (u) defined in (4.2), we see they are equal when r < k and L ⩾ 3. Also, note that
ℓi (u) ≪ 1/u by (4.8) implies that B̂2r,j (ℓ3 (u), ℓ4 (u), . . . ) ≪ 1/uj by (2.3). Hence

                                                             ar (u), âr (u) ≪r ur .                                                 (4.23)

Choosing L = 15, for example, in (4.22) shows
                       Z δ                             √       "k−1         k #
                                h         x i          2πu    X ar (u)    u
                            exp n hu (x) −      dx = p                r
                                                                         +O k .
                         −δ                u           (u + 1)n r=0 n       n

Inserting this into Lemma 4.3 and recalling (4.6) gives
                                     √        "k−1                             !!#
                                       2πu     X ar (u)    uk    1/2      n1/5
                Iα (n) = t0 g(t0 ) p                    +O    + n u exp −         .
                                     (u + 1)n      nr      nk              4u
                                                                 r=0

Finally t0 = eu and g(t0 ) = un e−n/u by (4.5) gives (4.4) and completes the proof of Theorem 4.1.


5 Generalizing Theorem 4.1
Define                                              Z ∞
                                Iα (f ; n) :=             (log t)n e−αt f (t) dt                    (n, α > 0).
                                                      1



                                                                             13
As long as f (t) is reasonably well-behaved and can be developed in a power series about t = t0 , with
coefficients that are relatively small, then the proof of Theorem 4.1 should go through. The examples we
                                                                       2
have in mind for our applications are f (t) = tβ and f (t) = e−(log(t)) /16 . For the latter,
                                                                                   
                            f (t(x + 1))           v                1     2
                                         = exp − log(x + 1) −          log (x + 1)                  (5.1)
                                f (t)              8               16
with t = ev . Treating x as complex, we see (as in the proof of Lemma 4.2) that the right side of (5.1) is
bounded for |x| ⩽ 1/(2v) and v ⩾ 1, say. Then Taylor’s theorem and the usual estimates show
                              k−1
               f (t(x + 1))   X
                            =     fm (v)xm + O(v k |x|k )              for      |x| ⩽ 1/(2v), k ∈ Z⩾0
                   f (t)      m=0

and that fm (v) ≪ v m . The implied constants depend only on k and m, respectively.
    The coefficients fm (v) can be computed explicitly by combining the series for log(x + 1) and ex with
the Bell polynomials. Similarly to (2.5) we find
                                          ∞    i
                           v           X     X 1  v j
                       exp − log(x + 1) =   xi      −     B̂i,j (1, − 12 , 13 , . . . ),
                            8                    j!   8
                                                     i=0   j=0
                                       ∞     ⌊i/2⌋
                        
                        1
                                     X      X 1  1 j
                            2              i
                   exp − log (x + 1) =   x            −    B̂i,2j (1, − 21 , 31 , . . . ).
                        16                         j!   16
                                                     i=0   j=0

So with
               i                                                    ⌊i/2⌋             j
               X 1  v j                                           X 1          1
     pi (v) :=      −     B̂i,j (1, − 12 , 31 , . . . ),    qi :=              −           B̂i,2j (1, − 12 , 13 , . . . ),   (5.2)
                 j!   8                                                     j!   16
                j=0                                                 j=0

we find that fm (v) = m
                     P
                        i=0 pi (v) · qm−i is a polynomial in v of degree m.
    We may isolate the properties of this example into a definition.
Definition 5.1. A continuous function f : [1, ∞) → R is suitable if there exist real constants b, λ ⩾ 0 and
functions fm for m ∈ Z⩾0 so that the following conditions hold.
  (i) For t ⩾ 1 we have 0 < f (t) ≪ tb .

  (ii) With t = ev and v ⩾ 1 we have
                                     k−1
                      f (t(x + 1))   X
                                   =     fm (v)xm + O(v λk |x|k )             (|x| ⩽ 1/(2v λ ), k ∈ Z⩾0 ).                   (5.3)
                          f (t)      m=0


 (iii) Lastly, fm (v) ≪ v λm for all m ∈ Z⩾0 .
The implied constant in (ii) depends only on k and f ; the one in (iii) depends only on m and f .
                                            2
    As we have seen, f (t) = e−(log(t)) /16 is suitable with b = 0 and λ = 1. The example f (t) = tβ , for
                                              β
                                                
any real β, is suitable with b = β, fm (v) = m and λ = 0. It is also easy to show that products of suitable
functions are suitable. An example of a function that is not suitable is f (t) = e−t since f1 (v) = −ev is too
large.
    Define
                           X (2j + 2r − 1)!!  v 2 j+r 2r
                           2r                              X
             ar (f ; v) :=                                     B̂i,j (ℓ3 (v), ℓ4 (v), . . . ) · f2r−i (v), (5.4)
                                     j!        v+1
                            j=0                                  i=j

with ℓm (v) given in (2.6) as usual. Then the following theorem generalizes Theorem 4.1, and reduces to it
when f (t) = 1.

                                                           14
Theorem 5.2. Let f be a suitable function associated with b, λ and coefficients fm . Suppose α > 0 and set
u := W (n/α). Then as n → ∞ we have
                                                        R−1
                                                                                       !!
                          √ un+1 f (eu )eu−n/u          X ar (f ; u)          uR(1+2λ)
              Iα (f ; n) = 2π p                     1+                +O                             (5.5)
                                     (1 + u)n                   nr               nR
                                                                       r=1
where the implied constant depends only on R ⩾ 1, α and b. Also ar (f ; u) is defined in (5.4) and satisfies
ar (f ; u) ≪ ur(1+2λ) .
Proof. We can reuse the proof of Theorem 4.1. As in (4.6),
                                          Z ∞          h         x i
                   Iα (f ; n) = t0 g(t0 )          exp n hu (x) −      f (t0 (x + 1)) dx.
                                           1/t0 −1                u
Using f (t) ≪ tb , t0 = n/(αu) and (2.8) we see that
  Z ∞          h           x i
           exp n hu (x) −       f (t0 (x + 1)) dx
   1/t0 −1                  u
                              Z δ                                                      2 
                                        h         x i                            b      δ n
                           =      exp n hu (x) −        f (t0 (x + 1)) dx + O u · n exp −     (5.6)
                               −δ                  u                                      4u
for 0 < δ ⩽ 1/100 and u ⩾ 2, where the implied constant depends only on α and b. This is as in Lemma
4.3 with an extra factor nb in the error. For er (n, u) defined in (4.14), set
                                                             r
                                                             X
                                             εr (n, u) :=          ej (n, u)fr−j (u).                       (5.7)
                                                             j=0

Lemma 5.3. For all x with |x| ⩽ δ = n−2/5 we have
                        k                                 k−1
                                        !                                        
                       X
                                       i f (t0 (x + 1))
                                                          X
                                                                        r       1
                exp        n · ℓi (u)x                  =     εr (n, u)x + O k/15
                                             f (t0 )      r=0
                                                                              n
                               i=3
as n → ∞ for an implied constant depending only on k and λ.
Proof. Recall from (4.5) that t0 = eu and u = W (n/α) = log n − log(αu). So u ⩾ 1/α implies u ⩽ log n.
Hence, for n large enough, n−2/5 ⩽ 1/(2uλ ) and we may use (5.3) with t = t0 and v = u.
   Next note the bounds
                                k
                                              !
                               X                           f (t0 (x + 1))
                        exp       n · ℓi (u)xi = O(1),                    = O(1)                 (5.8)
                                                               f (t0 )
                                      i=3

for |x| ⩽ δ = n−2/5 , where the left bound in (5.8) is shown after (4.11) and the right bound follows from
(5.3). Therefore, using Lemma 4.4 and (5.3),
          k
                          !                    k−1
                                                                ! k−1             !
                                                                                                    uλk
                                                                                                        
         X
                         i f (t0 (x + 1))
                                               X
                                                              r
                                                                   X
                                                                                m             1
   exp       n · ℓi (u)x                  =        er (n, u)x           fm (u)x      + O k/15 + 2k/5
                               f (t0 )                                                     n       n
         i=3                                   r=0                 m=0
                                          k−1                2k−2       k−1                              
                                          X
                                                        r
                                                             X
                                                                    r
                                                                        X                              1
                                       =      εr (n, u)x +        x           ej (n, u)fr−j (u) + O k/15 .
                                          r=0                 r=k     j=r−k+1
                                                                                                     n

Finally
  2k−2
  X             k−1
                X                              2k−2
                                               X            k−1
                                                            X
          xr             ej (n, u)fr−j (u) ≪          δr             nj/3 uλ(r−j)
   r=k         j=r−k+1                         r=k         j=r−k+1
                               2k−2
                               X
                           ≪          δr n(k−1)/3 uλ(k−1) ≪ δk n(k−1)/3 uλ(k−1) = n−k/15−1/3 uλ(k−1) ≪ n−k/15 ,
                               r=k

where we used the bound (4.15) for ej (n, u). This completes the proof.

                                                               15
   It is clear from (4.15), part (iii) of Definition 5.1 and (5.7) that

                                                   εr (n, u) ≪r nr/3 .

We may now continue the proof of Theorem 4.1 with er (n, u) replaced by εr (n, u). In (4.20), a∗m is replaced
             P⌊2m/3⌋
by a∗m (f ) = j=0 cm,j (f )nj for

                                        2 m 2m−2j
                            (2m − 1)!!   u     X
               cm,j (f ) :=                         B̂i,j (ℓ3 (u), ℓ4 (u), . . . )f2m−2j−i (u).
                               j!       u+1
                                                            i=j

                                     P2r
Then ar (u) becomes ar (f ; u) =        j=0 cr+j,j (f ) and so

                            2r                     2 j+r 2r
                            X     (2j + 2r − 1)!!   u     X
             ar (f ; u) =                                     B̂i,j (ℓ3 (u), ℓ4 (u), . . . )f2r−i (u)             (5.9)
                                        j!         u+1
                            j=0                                      i=j

agreeing with (5.4). The proof is finished as in Theorem 4.1 with the only difference that the bound (4.23) is
replaced by ar (f ; u) ≪ ur(1+2λ) which follows from (5.9).


6 Turán’s expansion of Ξ(z)
For t > 0 and x ∈ R set
                   X       2                         1
                       e−πk t ,                        3tθ ′ (t) + 2t2 θ ′′ (t) ,        Φ(x) := 2ex/2 ω(e2x ).
                                                                               
           θ(t) :=                       ω(t) :=                                                                  (6.1)
                                                     2
                    k∈Z

Recall Turán’s coefficients b2n from (1.11). As shown in [Rom, Eq. (2.1), Thm. 2.1], they have the formulas
                                              Z ∞
                                       1                  2
                           b2n = 2n−1              x2n e−x /4 Φ(x) dx
                                  2     (2n)! 0
                                              Z ∞
                                       1                                       2
                               = 4n−1             (log t)2n ω(t)t−3/4 e−(log t) /16 dt.
                                  2     (2n)! 1

Proof of Theorem 1.6. Write
                                       ∞
                                                                     2
                                       X
                                             2π 2 m4 t2 − 3πm2 t e−πm t
                                                                
                              ω(t) =                                                    (t > 0).
                                       m=1

For t ⩾ 1, ω(t) is dominated by its first two terms and it is easy to show that

                          ω(t) = 2π 2 t2 − 3πt e−πt + O t2 e−4πt
                                                                      
                                                                            (t ⩾ 1).

With                                             Z ∞
                                                                                    2
                                   Tα,β (n) :=         (log t)n e−αt tβ e−(log t) /16 dt,
                                                 1
we obtain
                                1          2
                                                                                         
                  b2n =                 2π   Tπ,5/4 (2n) − 3πTπ,1/4 (2n) + O T4π,5/4 (2n)  .                      (6.2)
                            24n−1 (2n)!
Theorem 5.2 may be used to find the asymptotics of Tα,β (n) since, by the discussion around Definition 5.1,
                    2
f (t) = tβ e−(log t) /16 is suitable with b = β, λ = 1 and
                                                            
                                                    X       β
                                       fm (v) =                pj2 (v)qj3 ,                           (6.3)
                                                            j1
                                                   j1 +j2 +j3 =m



                                                             16
using the formulas (5.2). We write ar (f ; v)β for the corresponding coefficients (5.4), adding a subscript to
keep track of the parameter β. Then for u := W (n/α) we have
                                                               R−1                   3R !
                          √ un+1 eu(β+1)−n/u−u2 /16             X ar (f ; u)β         u
                Tα,β (n) = 2π                               1+           r
                                                                               +O                        (6.4)
                                                                                      nR
                                      p
                                         (1 + u)n               r=1
                                                                       n

as n → ∞. For 2π 2 Tπ,5/4 (2n) − 3πTπ,1/4 (2n), u becomes w := W (2n/π) and in the main term of (6.4)
we have ew(β+1) for β = 5/4, 1/4. Using the identity ew = 2n/(πw) gives
                                                               πw
                                         ew(1/4+1) = ew(5/4+1)                                  (6.5)
                                                               2n
which lets us add the expressions. The result is

  2π 2 Tπ,5/4 (2n) − 3πTπ,1/4 (2n)
                                                                                   R−1              3R !
                                    5/2       w2n+1      9w/4−2n/w−w 2 /16
                                                                                   X     τr (w)     w
                             = 2π                        e                    1+             r
                                                                                                +O         (6.6)
                                                                                                    nR
                                          p
                                              (w + 1)n                                     n
                                                                                   r=1

where
                        τr (w) := 2−r−1 2ar (f ; w)5/4 − 3w · ar−1 (f, w)1/4 ≪ w3r .
                                                                            
                                                                                                            (6.7)
   Comparing (6.6) with the error
                                 w̃2n+1                   2
              T4π,5/4 (2n) ≪ p             e9w̃/4−2n/w̃−w̃ /16          for        w̃ := W (2n/(4π)),
                                 (w̃ + 1)n
in (6.2), we see the ratio is bounded by
                                           9(w̃ − w) w̃2 − w2    w̃ 1/w−1/w̃ 2n
                            r                                            
                          w̃ w + 1
                                     exp            −              e            .                           (6.8)
                          w w̃ + 1              4        16      w
The inequalities needed to estimate (6.8) are developed next; they will also be used in Sect. 8.
   Beginning with 1 + x ⩽ ex for x ⩾ 0, we obtain
                                              log(1 + x) ⩽ x      (x ⩾ 0)                                   (6.9)
and also (1 + x)1/x ⩽ e for x > 0. Hence
                                       a b
                                    1+       ⩽ ea     (a ⩾ 0, b > 0).                                      (6.10)
                                        b
Lemma 6.1. For x > 0 and c ⩾ 1 we have
                                        
                                     1
                               1−          log c ⩽ W (cx) − W (x) ⩽ log c.                                 (6.11)
                                   W (x)
Proof. By taking logs, the identities W (x)eW (x) = x and W (cx)eW (cx) = cx imply
                                          log W (x) + W (x) = log x,                                       (6.12)
                                     log W (cx) + W (cx) = log x + log c.
Subtracting the first equation from the second gives
                                                                        W (x)
                                     W (cx) − W (x) − log c = log
                                                                        W (cx)
and the right side is ⩽ 0 since the Lambert function is increasing. This proves the right inequality in (6.11).
Then                                                                         
                           W (cx)        W (x) + log c                  log c     log c
                       log         ⩽ log                 = log 1 +              ⩽
                            W (x)            W (x)                     W (x)      W (x)
where we used (6.9). The left inequality in (6.11) is then a result of:
                        log c         W (cx)       W (x)
                    −         ⩽ − log        = log        = W (cx) − W (x) − log c.
                        W (x)         W (x)        W (cx)

                                                             17
   Further inequalities for the Lambert function are contained in [HH08], for example. Also note that (6.12)
and W (e) = 1 imply that W (x) ⩽ log x for all x ⩾ e.
Lemma 6.2. For w = W (2n/π), w̃ = W (2n/(4π)) the expression (6.8) is ≪ e−2n/ log n for large n.

Proof. Lemma 6.1 implies that (1 − 1/w̃) log 4 ⩽ w − w̃ ⩽ log 4. If we write w̃ = w − δ then, by choosing
n large enough, we can certainly ensure 1 < δ < 2. We have

                         9(w̃ − w) w̃2 − w2
           r                                                               
         w̃ w + 1                                           9δ δ(2w − δ)
                   exp              −             ≪ exp − +                      ≪ eδw/8 < ew/4 .
         w w̃ + 1             4           16                 4         16

Also
                                      w = W (2n/π) ⩽ log(2n/π) < log n,
so that ew/4 < n1/4 . For the remaining factor in (6.8)
                                              2n                 w·2n/w            
                                w̃ 1/w−1/w̃                     δ                   2δn
                                  e                  =       1−               exp −       .             (6.13)
                                w                               w                   ww̃

The bound                                  w
                                        δ
                                     1−          ⩽ e−δ         (w > 0, 0 ⩽ δ/w < 1)
                                        w
follows by taking logs and using the inequality log(1 − x) ⩽ −x for 0 ⩽ x < 1. Therefore (6.13) is at most
                                                                       
                                     2δn 2δn                  2n      2n
                            exp −        − 2 < exp −              −         .
                                      w      w               log n log2 n
                                                     2
Simplifying the final bound with n1/4 e−2n/ log n ≪ 1 completes the proof of the lemma.

   We have shown that the error term in (6.2) is exponentially small
                                                                  p compared to the main term, and so it
may be included inside the error in (6.6). Simplifying with ew/2 = 2n/(πw) gives the final result

                            7w/4−w 2 /16
                                           r                  K−1          3K !
                          2e                    2w  w 2n                 w
                                                              X τk (w)
               b2n = 4π                                    1+       k
                                                                       +O                               (6.14)
                                (2n)!          w + 1 4e1/w        n        nK
                                                                              k=1

and this completes the proof of Theorem 1.6.

   The main term of (6.14) agrees with [Rom, Thm. 2.7]. We could also replace 1/(2n)! in (6.14) with its
asymptotic expansion (7.12). The numbers τr (w) are given explicitly by (6.7), using (5.2), (6.3) to define its
components fm (v) and then ar (f ; v)β with (5.4). For example

                                      −3w6 + 78w5 + 217w4 + 468w3 + 284w2 − 32
                       τ1 (w) = −                                              .
                                                    768(w + 1)3

An example of the accuracy of Theorem 1.6 for 2n = 2000 and different values of K is displayed in Table
2.

                                K                 Theorem 1.6
                                1       2.3786738117568138992 × 10−5738
                                3       2.3732111799604212549 × 10−5738
                                5       2.3732111791829324664 × 10−5738
                                7       2.3732111791829329059 × 10−5738
                                        2.3732111791829329059 × 10−5738                 b2000

                            Table 2: The approximations of Theorem 1.6 to b2000 .



                                                              18
7 The Riemann ξ function
Recall the definitions of ω and Φ in (6.1). The identity
                                                    Z ∞
                                 (2n)          1−2n
                               ξ      (1/2) = 2          (log t)2n ω(t)t−3/4 dt                             (7.1)
                                                         1

follows easily from Riemann’s formulas of 1859. It also follows from the elegant expression
                                       Z ∞
                           (n)
                          ξ (1/2) =         Φ(y) · y n dy           (n ∈ Z⩾0 )                              (7.2)
                                             −∞

by changing variables and recalling that Φ(−y) = Φ(y); see [Rom, Eqs. (1.9), (6.1)]. The identity (7.2) is
implicit in Jensen’s formula from [Jen13, p. 189]:
                                 d                        Z ∞
                      ∗
                                 X    d j (j)
                     gd (Ξ; x) =         i ξ (1/2) · xd−j =     Φ(y) · (x + iy)d dy.
                                      j                      −∞
                                 j=0

Proof of Theorem 1.5. We may follow the proof of Theorem 1.6 in Sect. 6. Setting
                                              Z ∞
                                  Iα,β (n) :=     (log t)n e−αt tβ dt,
                                                     1

we obtain from (7.1), as in (6.2),
                   ξ (2n) (1/2) = 21−2n 2π 2 Iπ,5/4 (2n) − 3πIπ,1/4 (2n) + O I4π,5/4 (2n) .
                                                                                                
                                                                                                             (7.3)
                                                                                     β
We noted after Definition 5.1 that f (t) = tβ is suitable with b = β, fm (v) = m         and λ = 0. From (5.4) we
find
                             2r                  2 j+r X    2r                                      
                            X   (2j + 2r − 1)!!      v                                             β
            ar (f ; v)β =                                        B̂i,j (ℓ3 (v), ℓ4 (v), . . . )              (7.4)
                                      j!          v+1                                            2r − i
                          j=0                                   i=j

and by Theorem 5.2
                                                           R−1
                                                                                           !
                                 √     un+1 eu(β+1)−n/u    X ar (f ; u)β    R
                                                                            u
                    Iα,β (n) =       2π p               1+         r
                                                                         +O R                               (7.5)
                                            (1 + u)n       r=1
                                                                 n          n

as n → ∞ for u := W (n/α). The asymptotics for the terms Iπ,5/4 (2n) and Iπ,1/4 (2n) may be combined,
as in (6.6) using the identity (6.5), to produce
                                                   2n+1 e9w/4−2n/w    R−1
                                                                      X µr (w)      R !
                                                 w                                   w
       2π 2 Iπ,5/4 (2n) − 3πIπ,1/4 (2n) = 2π 5/2 p                 1+       r
                                                                               +O               (7.6)
                                                      (w + 1)n            n          nR
                                                                            r=1

where w := W (2n/π) and
                        µr (w) := 2−r−1 2ar (f ; w)5/4 − 3w · ar−1 (f, w)1/4 ≪ wr .
                                                                            
                                                                                                            (7.7)
    Comparing (7.6) with the error
                                       w̃2n+1
                  I4π,5/4 (2n) ≪ p               e9w̃/4−2n/w̃         for   w̃ := W (2n/(4π)),
                                       (w̃ + 1)n
in (7.3), we see the ratio is bounded by

                                                            w̃ 1/w−1/w̃ 2n
                                  r                                 
                                w̃ w + 1       9(w̃ − w)
                                         exp                  e            .                                (7.8)
                                w w̃ + 1            4       w

Then (7.8) is smaller than (6.8) and so ≪ e−2n/ log n for large n by Lemma 6.2. This   p exponentially small
error may therefore be included in the error in (7.6). Rearranging slightly with ew/2 = 2n/(πw) completes
the proof of Theorem 1.5.

                                                         19
    The coefficient µ1 (w) may be computed as

                                                         w4 + 66w3 + 53w2 − 8
                                        µ1 (w) = −                            .
                                                             192(w + 1)3

Proof of Theorem 1.4. Recall that
                                                               n! (2n)
                                                γ(n) :=            ξ   (1/2).                                 (7.9)
                                                             (2n)!
The asymptotic expansion of the gamma function goes back to Laplace, with
                              √     n n                                     
                                                κ1   κ2          κk           1
                  Γ(n + 1) = 2πn            1+     + 2 + · · · + k + O k+1                                   (7.10)
                                     e          n    n           n          n

often called Stirling’s formula. There are many ways to express the coefficients κm and a simple example is
                                            2m                                                  
                                            X    −m − 1/2                         2 2 2 2
                      κm = (2m − 1)!!                                     B̂2m,j − , , − , , · · ·           (7.11)
                                                             j                    3 4 5 6
                                            j=0

from [O’S19, Eq. (8.1)]. The same coefficients appear in the expansion of the reciprocal of gamma but
with alternating signs. One way to see this is to note that the well-known asymptotic series for log Γ(n + 1)
involves an odd function; see also [FS09, Sect. VIII.3]. Then we have
                                                                                       
                    1           1  e n          κ1     κ2              k κk         1
                          =√                 1−      + 2 − · · · + (−1) k + O k+1             .        (7.12)
                Γ(n + 1)       2πn n               n     n                 n       n

    Let
                                 k
                                 X                                                  k
                                                                                    X
                        κ∗k :=         (−2)−j κj · κk−j ,             ck (w) :=           µj (w) · κ∗k−j .   (7.13)
                                 j=0                                                j=0

Formulas (7.10), (7.12) imply

                                            κ∗1  κ∗2         κ∗k
                                                                       
                        n!    1  e n                                 1
                            =√           1+     + 2 + · · · + k + O k+1     .                                (7.14)
                      (2n)!    2 4n         n    n           n       n

Using (7.14) and Theorem 1.5 in (7.9) then proves Theorem 1.4.

   The coefficients ck (w) are completely described by the formulas (2.6), (7.4), (7.7), (7.11) and (7.13).
The first was given in (1.8) and the next one is
                                 8          7            6            5         4            3        2
              c2 (w) = − 1295w +7804w +21682w +40124w  +29911w +13712w +2080w −768w−256
                                                   73728(w+1)6                          .


8 Zeros of P d,n (X)
In this section we develop a condition for P d,n (X) to have only real roots. It is based on the next result, due
to Turán from [Tur59, Thm. III].

Theorem 8.1. Let G(z) = dj=0 cj Hj (z) for real numbers cj and Hj (z) the jth Hermite polynomial. Then
                            P
the zeros of G(z) are real and simple when
                                                d−2
                                                X
                                                      2j j!c2j < 2d (d − 1)!c2d .                             (8.1)
                                                j=0




                                                                 20
                                                                                                 d
Proof of Theorem 1.3. The condition (8.1) for P d,n (X) to be hyperbolic, with cj =              j γ(n + d − j), is

                                           d
                                              d d γ(n + j)2
                                           X       
                                                            < 1.                                                  (8.2)
                                             2j j! j  γ(n)2
                                           j=2

It follows from Theorem 1.4 with K = 1 and w := W (2n/π) that
                                                                                     n
                                                                               ew2
                                                                 r      
                               log(n)                         2 7w/4     w
         γ(n) = Ψ(n) 1 + O                   for   Ψ(n) := 4π e                         .
                                  n                                    w + 1 16ne2/w

As a consequence, for any ǫ > 0 there exists an N so that
                                     s                   !n
  γ(n + j)   Ψ(n + j)                   1 + 1/w wj2 n
           ⩽          (1 + ǫ) =
    γ(n)      Ψ(n)                     1 + 1/wj w2 n + j
                                                                                  !j
                                                                        ewj2
                                                                
                                            7(wj − w) 2n(wj − w)
                                    × exp            +                               (1 + ǫ) (8.3)
                                                4         wwj      16(n + j)e2/wj

for all n ⩾ N and all j ⩾ 0, where we set wj := W (2(n + j)/π).
Lemma 8.2. Suppose c ⩾ 1. For n large enough and 0 ⩽ j ⩽ nc − n, we have
                                                                  j
                                                      3c log2 (n)
                                                      2
                                      γ(n + j)
                                               < 1.01                .
                                        γ(n)             16n

Proof. By Lemma 6.1 we have wj − w ⩽ log(1 + λ) for λ := j/n. Examining the components of (8.3) we
first find, using (6.9), (6.10),
         w 2n                           2n                            w·2n/w
            j             w + log(1 + λ)                    log(1 + λ)
                  ⩽                              =       1+                          ⩽ elog(1+λ)·2n/w ⩽ e2j/w .
          w                     w                               w

Next,
                                                                                                   
                    7(wj − w) 2n(wj − w)                             7 log(1 + λ) 2n log(1 + λ)
                exp          +                               ⩽ exp               +
                        4        wwj                                       4           w2
                                                                              
                                                                     7j     2j
                                                             ⩽ exp       + 2 .
                                                                     4n w

Bounding trivially gives            s                           n
                                        1 + 1/w           n               p
                                                                      ⩽    1 + 1/w.
                                        1 + 1/wj         n+j
                                                           p
Assembling our bounds, and with n large enough that (1 + ǫ) 1 + 1/w ⩽ 1.01, we have shown
                                                                      !j
                                         wj2
                                                    
                      γ(n + j)                         7  2  2   2
                               ⩽ 1.01           exp      + +   −   +1     .                                       (8.4)
                        γ(n)          16(n + j)       4n w w2 wj

For n sufficiently large the exponential term can be made close to e < 3 and (8.4) is at most
                                                     !j                              j
                                        3wj2                         3 log2 (2nc /π)
                                                                   
                                1.01                         < 1.01                     .
                                     16(n + j)                            16n




                                                               21
    Applying Lemma 8.2 with c = 2, (it will become apparent that any c ⩾ 4/3 will do), the left side of
(8.2) is
                        d                  d                             j
                           d d γ(n + j)2 X d dj               9 log4 (n)
                        X                                 
                                         <            · 1.03
                          2j j! j  γ(n)2     2j j! j!            16n2
                        j=2                             j=2
                                                            d                      j
                                                                      9d log 4 (n)
                                                                    
                                                            X   d
                                                     = 1.03                           .                      (8.5)
                                                              (j!)2      32n2
                                                                j=2

If we choose n large enough so that
                                                                2
                                                  9d log4 (n)
                                              
                                                                          3
                                                                      ⩽
                                                     32n2                 d
then (8.5) is at most
                                 d            √ !j         ∞
                                 X   d          3         X   3j/2
                            1.03              √    < 1.03           < 0.94 < 1.
                                   (j!)2        d             (j!)2
                                 j=2                      j=2

This verifies condition (8.2) and Theorem 1.3 follows.


References
[CC89]     Thomas Craven and George Csordas. Jensen polynomials and the Turán and Laguerre inequalities. Pacific
           J. Math., 136(2):241–260, 1989.
[Cha13]    Matthew Chasse. Laguerre multiplier sequences and sector properties of entire functions. Complex Var.
           Elliptic Equ., 58(7):875–885, 2013.
[CNV86]    George Csordas, Timothy S. Norfolk, and Richard S. Varga. The Riemann hypothesis and the Turán
           inequalities. Trans. Amer. Math. Soc., 296(2):521–541, 1986.
[Cof09]    Mark W. Coffey. Asymptotic estimation of ξ (2n) (1/2): on a conjecture of Farmer and Rhoades. Math.
           Comp., 78(266):1147–1154, 2009.
[CV90]     George Csordas and Richard S. Varga. Necessary and sufficient conditions and the Riemann hypothesis.
           Adv. in Appl. Math., 11(3):328–357, 1990.
[DBC09]    Dimitar K. Dimitrov and Youssèf Ben Cheikh. Laguerre polynomials as Jensen polynomials of Laguerre-
           Pólya entire functions. J. Comput. Appl. Math., 233(3):703–707, 2009.
[DL11]     Dimitar K. Dimitrov and Fábio R. Lucas. Higher order Turán inequalities for the Riemann ξ-function.
           Proc. Amer. Math. Soc., 139(3):1013–1022, 2011.
[Far]      David W. Farmer.     Jensen polynomials are not a viable route to proving the Riemann Hypothesis.
           arXiv:2008.07206.
[FS09]     Philippe Flajolet and Robert Sedgewick. Analytic combinatorics. Cambridge University Press, Cambridge,
           2009.
[GOR+ ]    Michael Griffin, Ken Ono, Larry Rolen, Jesse Thorner, Zachary Tripp, and Ian Wagner. Jensen polynomi-
           als for the Riemann xi function. arXiv:1910.01227.
[GORZ19] Michael Griffin, Ken Ono, Larry Rolen, and Don Zagier. Jensen polynomials for the Riemann zeta function
         and other sequences. Proc. Natl. Acad. Sci. USA, 116(23):11103–11110, 2019.
[HH08]     Abdolhossein Hoorfar and Mehdi Hassani. Inequalities on the Lambert W function and hyperpower func-
           tion. JIPAM. J. Inequal. Pure Appl. Math., 9(2):Article 51, 5, 2008.
[Jen13]    J. L. W. V. Jensen. Recherches sur la théorie des équations. Acta Math., 36(1):181–195, 1913.
[KK00]     Haseo Ki and Young-One Kim. On the number of nonreal zeros of real entire functions and the Fourier-
           Pólya conjecture. Duke Math. J., 104(1):45–73, 2000.
[O’S]      Cormac O’Sullivan. A generalization of the Riemann-Siegel formula. Submitted, arXiv:1811.01130.


                                                         22
[O’S19]    Cormac O’Sullivan. Revisiting the saddle-point method of Perron. Pacific J. Math., 298(1):157–199, 2019.
[Pól15]   G. Pólya. Algebraische Untersuchungen über ganze Funktionen vom Geschlechte Null und Eins. J. Reine
           Angew. Math., 145:224–249, 1915.
[Pól27]   G. Pólya. Über die algebraisch-funktionentheoretischen Untersuchungen von J.L.W.V. Jensen. Kgl
           Danske Vid Sel Math-Fys Medd, (7):3–33, 1927.
[PS14]     G. Pólya and J. Schur. Über zwei Arten von Faktorenfolgen in der Theorie der algebraischen Gleichungen.
           J. Reine Angew. Math., 144:89–113, 1914.
[PT]       Dave Platt and Tim Trudgian. The Riemann hypothesis is true up to 3 · 1012 . arXiv:2004.09765.
[Pus01]    L. D. Pustyl’nikov. Asymptotics of the coefficients of the Taylor series of the ξ(s) function. Izv. Ross.
           Akad. Nauk Ser. Mat., 65(1):93–106, 2001.
[Rom]      Dan Romik. Orthogonal polynomial expansions for the Riemann xi function. arXiv:1902.06330.
[Sze75]    Gábor Szegő. Orthogonal polynomials. American Mathematical Society, Providence, R.I., fourth edition,
           1975. American Mathematical Society, Colloquium Publications, Vol. XXIII.
[Tur59]    P. Turán. To the analytical theory of algebraic equations. Bulgar. Akad. Nauk. Otd. Mat. Fiz. Nauk. Izv.
           Mat. Inst., 3:123–137, 1959. Reprinted in the Collected Papers of Paul Turán, Ed. P. Erdős, Vol. 2, pp.
           1080–1090. Akadémiai Kiadó, 1990.
[Wag20]    Ian Wagner. The Jensen–Pólya program for various L-functions. Forum Math., 32(2):525–539, 2020.


D EPT. OF M ATH , T HE CUNY G RADUATE C ENTER , 365 F IFTH AVENUE , N EW YORK , NY 10016-4309, U.S.A.
E-mail address: cosullivan@gc.cuny.edu




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