             ITERATED LAGUERRE AND TURÁN INEQUALITIES



                             Thomas Craven and George Csordas


        Abstract. New inequalities are investigated for real entire functions in the Laguerre-Pólya
        class. These are generalizations of the classical Turán and Laguerre inequalities. They provide
        necessary conditions for certain real entire functions to have only real zeros.




1. Introduction and notation.

                                                  P∞
Definition 1.1. A real entire function ϕ(x) := k=0 γk!k xk is said to be in the Laguerre-
Pólya class, written ϕ(x) ∈ L-P, if ϕ(x) can be expressed in the form

                                               ω 
                                               Y        
                                  n −αx2 +βx         x     − x
(1.1)                 ϕ(x) = cx e                 1+      e xk ,           0 ≤ ω ≤ ∞,
                                                     xk
                                               k=1

                                                                       P∞        2
where c, β, xk ∈ R, α ≥ 0, n is a nonnegative integer and                 k=1 1/xk < ∞. If ω = 0, then,
by convention, the product is defined to be 1.

   The significance of the Laguerre-Pólya class in the theory of entire functions stems from
the fact that functions in this class, and only these, are the uniform limits, on compact
subsets of C, of polynomials with only real zeros. For various properties and algebraic and
transcendental characterizations of functions in this class we refer the reader to Pólya and
Schur [11, p. 100], [12] or [9, Kapitel II].
                  P∞ γk k                                               2
   If ϕ(x) :=       k=0 k! x ∈ L-P, then the Turán inequalities γk − γk−1 γk+1 ≥ 0 and
the Laguerre inequalities ϕ(k) (x)2 − ϕ(x)(k−1) ϕ(k+1) (x) ≥ 0 are known to hold for all
k = 1, 2, . . . and for all real x (see [2] and the references contained therein). In this paper
we consider generalizations of both of these inequalities. For some of these generalizations
to hold, we must restrict our investigation to the following subclass of L-P.

  1991 Mathematics Subject Classification. Primary 26D05; Secondary 26C10, 30C15.
  Key words and phrases. Laguerre-Pólya class, multiplier sequence, Hankel matrix, real zeros.

                                                                                       Typeset by AMS-TEX
                                                       1
2                          THOMAS CRAVEN AND GEORGE CSORDAS

                                                  P∞
Definition 1.2. A real entire function ϕ(x) := k=0 γk!k xk in L-P is said to be in L-P+ if
γk ≥ 0 for all k. In particular, this means that all the zeros of ϕ lie in the interval (−∞, 0].

   Now if ϕ(x) ∈ L-P+ , then ϕ can be expressed in the form
                                     ω 
                                     Y           
                                n βx          x
(1.2)                ϕ(x) = cx e         1+        , 0 ≤ ω ≤ ∞,
                                              xk
                                        k=1
                                                        P∞
where c,P β ≥ 0, xk > 0, n is a nonnegative integer and  k=1 1/xk < ∞ [9, §9]. If
            ∞ γk k
ϕ(x) = k=0 k! x ∈ L-P, then, following the usual convention, we call the sequence of
coefficients, {γk }∞
                   k=0 , a multiplier sequence.

    In Section 2, the Laguerre inequality ϕ0 (x)2 − ϕ(x)ϕ00 (x) ≥ 0 is generalized to a system
of inequalities Ln (ϕ(x)) ≥ 0 for all n = 0, 1, 2, . . . and for all x ∈ R, where L1 (ϕ(x)) =
ϕ0 (x)2 − ϕ(x)ϕ00 (x) and ϕ(x) ∈ L-P (cf. [10, Theorem 1]). This system of inequalities
characterizes functions in L-P (Theorem 2.2). We show that the (nonlinear) operators
Ln satisfy a simple recursive relation (Theorem 2.1) and use this fact to give a different
proof of a result of Patrick [10, Theorem 1]. This, together with the converse of Patrick’s
theorem (cf. [5, Theorem 2.9]), yields a necessary and sufficient condition for a real entire
function (with appropriate restrictions on the order and type of the entire function) to
belong to the Laguerre-Pólya class (Theorem 2.2).

   In Section 3, we consider a different collection of inequalities based on the Laguerre
inequality, namely an iterated form of them. Our original proof of the second iterated
inequalities for functions in L-P+ [2, Theorem 2.13] was based on the study of certain
polynomial invariants. In Section 3, we give a shorter and a conceptually simpler proof of
these inequalities (Proposition 3.4 and Theorem 3.5). Moreover, the proof of Proposition
3.4 leads to new necessary conditions for entire functions to belong to L-P+ (Corollary
3.6). Evaluating these at x = 0 yields the classical Turán inequalities and iterated forms
of them, considered in Section 4. In Section 4, we show that for multiplier sequences
which decay sufficiently rapidly all the higher iterated Turán inequalities hold (Theorem
4.1). Our main result (Theorem 5.5) asserts that the third iterated Turán inequalities are
valid for all functions of the form ϕ(x) = x2 ψ(x), where ψ(x) ∈ L-P+ . An examination
of the proof of Theorem 5.5 (see also Lemma 5.4) shows that the restriction that ϕ(x)
has a double zero at the origin is merely a ploy to render the, otherwise very lengthy and
involved, computations tractable.

2. Characterizing L-P via extended Laguerre inequalities.

   Let ϕ(x) denote a real entire function; that is, an entire function with only real Tay-
lor coefficients. Following Patrick [10], we define implicitly the action of the (nonlinear)
operators {Ln }∞ n=0 , taking ϕ(x) to Ln (ϕ(x)), by the equation
                                                   ∞
                                                   X
                       2
(2.1)      |ϕ(x + iy)| = ϕ(x + iy)ϕ(x − iy) =            Ln (ϕ(x))y 2n,   (x, y ∈ R) .
                                                   n=0
                   ITERATED LAGUERRE AND TURÁN INEQUALITIES                                          3

In the sequel, it will become clear that Ln (ϕ(x)) is also a real entire function (cf. Remark
2.4). In [10], Patrick shows that if ϕ(x) ∈ L-P, then Ln (ϕ(x)) ≥ 0 for all n = 0, 1, 2, . . .
and for all x ∈ R. The novel aspect of our approach to these inequalities is based on
the remarkable fact that the operators Ln satisfy a simple recursive relation (Theorem
2.1). By virtue of this recursion relation, we obtain a short proof of Patrick’s theorem.
This, when combined with the known converse result [5, Theorem 2.9], yields a complete
characterization of functions in L-P (Theorem 2.2).

  We remark that generalizations of the operators defined in (2.1) are given by Dilcher
                                                                        (m)
and Stolarsky in [6]. These authors study the distribution of zeros of Ln (ϕ(x)) for their
                        (m)
generalized operators Ln and certain functions ϕ(x).

Theorem 2.1. Let ϕ(x) be any real entire function. Then the operators Ln satisfy the
following:
   (1) Ln ((x + a)ϕ(x)) = (x + a)2 Ln (ϕ(x)) + Ln−1 (ϕ(x)), for a ∈ R and n = 1, 2, . . . ;
   (2) L0 (ϕ) = ϕ2 ;
   (3) Ln (c) = 0 for any constant c and n ≥ 1.


Proof. Parts (2) and (3) are clear from the definition. To check (1), we compute as follows:

                                                ∞
                                                X
 |(x + a + iy)ϕ(x + iy)|2 = ((x + a)2 + y 2 )         Ln (ϕ(x))y 2n
                                                n=0
                                       ∞
                                       X                        ∞
                                                                X
                          = (x + a)2         Ln (ϕ(x))y 2n +          Ln (ϕ(x))y 2n+2
                                       n=0                      n=0
                                       X∞                       X∞
                          = (x + a)2         Ln (ϕ(x))y 2n +          Ln−1 (ϕ(x))y 2n
                                       n=0                      n=1
                                                      ∞
                                                      X
                          = (x + a)2 L0 (ϕ(x)) +            [(x + a)2 Ln (ϕ(x)) + Ln−1 (ϕ(x))]y 2n,
                                                      n=1


from which (1) follows. 

   Using the recursion of Theorem 2.1, we obtain the following characterization of functions
in L-P.

                                                                                        2
Theorem 2.2. Let ϕ(x) 6≡ 0 be a real entire function of the form e−αx ϕ1 (x), where
α ≥ 0 and ϕ1 (x) has genus 0 or 1. Then ϕ(x) ∈ L-P if and only if Ln (ϕ) ≥ 0 for all
n = 0, 1, 2, . . . .

Proof. First assume that all Ln (ϕ) ≥ 0. If ϕ ∈
                                              / L-P, then ϕ has a nonreal zero z0 = x0 +iy0
4                        THOMAS CRAVEN AND GEORGE CSORDAS

with y0 6= 0. Hence
                                                  ∞
                                                  X
                             0 = |ϕ(z0 )|2 =            Ln (ϕ(x0 ))y02n .
                                                  n=0



Since all terms in the sum are nonnegative
                                   P∞      and y0 6= 0, we must have Ln (ϕ(x0 )) = 0 for all
n. But this gives |ϕ(x0 + iy)| = n=0 Ln (ϕ(x0 ))y 2n = 0 for any choice of y ∈ R, whence
                               2

ϕ itself must be identically zero.

   Conversely, assume that ϕ ∈ L-P. Since ϕ can be uniformly approximated on compact
sets by polynomials with only real zeros, it will suffice to prove that Ln (ϕ) ≥ 0 for poly-
nomials ϕ. For this we use induction on the degree of ϕ. From Theorem 2.1(2) and (3),
we see that Ln (ϕ) ≥ 0 for any n if ϕ has degree 0. If the degree of ϕ is greater than zero,
we can write ϕ(x) = (x + a)g(x), where a ∈ R and g(x) is a polynomial. By the induction
hypothesis, Ln (g(x)) ≥ 0 for all n ≥ 0 and all x ∈ R. Hence, Theorem 2.1(1) gives the
desired conclusion for ϕ. 


  Next we show that the explicit form of Ln (ϕ) given in [10] can also be obtained from
Theorem 2.1.


Theorem 2.3. For any ϕ ∈ L-P, operators Ln (ϕ) satisfying the recursion and initial
conditions of Theorem 2.1 are uniquely determined and are given by


                                    2n
                                    X          
                                       (−1)j+n 2n
(2.2)                 Ln (ϕ(x)) =                           ϕ(j) (x)ϕ(2n−j) (x) .
                                    j=0
                                          (2n)!         j




Proof. As before, it will suffice to prove the result for polynomials in L-P. It is clear that
the recursion formula of Theorem 2.1, together with the initial conditions given there,
uniquely determine the value of Ln on any polynomial with only real zeros. Thus it will
suffice to show that the formula given in (2.2) satisfies the conditions of Theorem 2.1. We
do a double induction, beginning with an induction on n.

  If n = 0, then L0 (ϕ) = ϕ2 by Theorem 2.1 and this agrees with (2.2). Assume that
n > 0 and that (2.2) holds for Ln−1 . Now we begin an induction on the degree of ϕ.

   If ϕ is a constant, then Ln (ϕ) = 0 by Theorem 2.1, which agrees with (2.2). Assume
the formula holds for polynomials of degree less than deg ϕ. Then we can write ϕ(x) =
(x + a)g(x), where a ∈ R and Ln−1 (g) and Ln (g) are given by (2.2). The conclusion now
                         ITERATED LAGUERRE AND TURÁN INEQUALITIES                                  5

follows from a computation using Theorem 2.1(1). Indeed,

                    Ln (ϕ(x)) = Ln ((x + a)g(x)) = (x + a)2 Ln (g(x)) + Ln−1 (g(x))
                                          X2n         
                                        2     (−1)j+n 2n (j)
                              = (x + a)                    g (x)g (2n−j) (x)
                                          j=0
                                               (2n)!   j
                                       2n−2
                                        X                    
                                              (−1)j+n−1 2n − 2 (j)
                                   +                            g (x)g (2n−2−j) (x).
                                               (2n − 2)!   j
                                       j=0


Also, using Leibniz’s formula for higher derivatives of a product,

            2n
            X          
               (−1)j+n 2n (j)
                          ϕ (x)ϕ(2n−j) (x)
                (2n)!   j
            j=0

                    X2n           
                        (−1)j+n 2n 
                  =                     j(2n − j)g (j−1) g (2n−1−j) + (x + a)jg (j−1)g (2n−j)
                    j=0
                          (2n)!    j
                                                                                  
                      + (x + a)(2n − j)g (j) g (2n−1−j) + (x + a)2 g (j) g (2n−j)
                              2n
                              X            
                            2    (−1)j+n 2n (j)
                  = (x + a)                       g (x)g (2n−j) (x)
                                  (2n)!      j
                             j=0
                         2n−2
                          X                
                            (−1)j+n−1 2n − 2 (j)
                      +                       g (x)g (2n−2−j) (x),
                        j=0
                             (2n − 2)!   j

because the coefficient of (x + a) is shown to be zero by

  X2n          
      (−1)j+n 2n  (j−1) (2n−j)                                   
                     jg      g         + (2n − j)g (j)g (2n−1−j)
  j=0
       (2n)!    j
               " 2n                                 2n−1                                    #
        (−1)n X             2n                         X            2n
      =             (−1)j        jg (j−1) g (2n−j) +        (−1)j      (2n − j)g (j) g (2n−j−1)
         (2n)!               j                                       j
                j=1                                    j=0
               " 2n−1                 
        (−1)n       X            2n
      =         −       (−1)j            (j + 1)g (j) g (2n−j−1)
         (2n)!      j=0
                                j +1
                      2n−1                                      #
                       X           2n
                    +      (−1)j         (2n − j)g (j) g (2n−j−1)
                       j=0
                                     j
        =0
                      
         2n             2n
since        (j + 1) =     (2n − j). 
        j+1              j
6                            THOMAS CRAVEN AND GEORGE CSORDAS

   The main emphasis here has been that the result of Theorem 2.2 depends only on the
recursive condition of Theorem 2.1, and this seems to be the easiest way to prove Theorem
2.2. However, the operators Ln (ϕ) can be explicitly computed more easily than from the
recursive condition as was done in Theorem 2.3, as well as in greater generality.

Remark 2.4. For any real entire function ϕ, the operators Ln (ϕ) defined by equation
(2.1) are given by the formula
                                             2n
                                             X     
                                           (−1)j+n 2n (j)
                           Ln (ϕ(x)) =                ϕ (x)ϕ(2n−j) (x).
                                       j=0
                                            (2n)!   j



Proof. By Taylor’s theorem, for each fixed x ∈ R,
                                                                        ∞
                                                                        X h(2n) (0)
                 h(y) := |ϕ(x + iy)|2 = ϕ(x + iy)ϕ(x − iy) =                                  y 2n ,
                                                                        n=0
                                                                              (2n)!

where we have used the fact that h(y) is an even function (of y). Let Dy = d/dy denote
differentiation with respect to y. Then by Leibniz’s formula, for higher derivatives of a
product, we have
                                 2n 
                                 X      
                  (2n)                2n                                           
                 h       (0) =                    Dyk ϕ(x + iy) y=0 Dy2n−k ϕ(x − iy) y=0
                                         k
                                 k=0
                                 2n 
                                 X      
                                      2n
                             =                (−1)n+k ϕ(k) (x)ϕ(2n−k) (x)
                                         k
                                 k=0
                             =(2n)!Ln (ϕ(x)),

by the uniqueness of the Taylor coefficients.              

3. Iterated Laguerre inequalities.

Definition 3.1. For any real entire function ϕ(x), set
                     (1)
                  Tk (ϕ(x)) := (ϕ(k) (x))2 − ϕ(k−1) (x)ϕ(k+1) (x)             if    k ≥ 1,

and for n ≥ 2, set
        (n)                  (n−1)                     (n−1)        (n−1)
      Tk      (ϕ(x)) := (Tk            (ϕ(x)))2 − Tk−1 (ϕ(x)) Tk+1 (ϕ(x))                if     k ≥ n ≥ 2.


                                                                                   (n)                 (n)
  Remarks. (a) Note that with the notation above, we have Tk+j (ϕ) = Tk                                      (ϕ(j) ) for
k ≥ n and j = 0, 1, 2 . . . .
                    ITERATED LAGUERRE AND TURÁN INEQUALITIES                                7

   (b) The authors’ investigations of functions in the Laguerre-Pólya class ([2], [3]) have
led to the following problem.

  Open Problem If ϕ(x) ∈ L-P+ , are the iterated Laguerre inequalities valid for all
x ≥ 0? That is, is it true that
                       (n)
(3.1)                Tk       (ϕ(x)) ≥ 0   for all x ≥ 0 and k ≥ n?

                                                                       (n)
   (c) If we assume only that ϕ(x) ∈ L-P, then the inequality Tk (ϕ(x)) ≥ 0, x ≥ 0,
need not hold in general, as the following example shows. Consider, for example, ϕ(x) =
                                 (2)                                           (2)
(x−2)(x+1)2 ∈ L-P. Then T2 (ϕ(x)) = 216x(−2+3x+x3 ) and so we see that T2 (ϕ(x))
is negative for all sufficiently small positive values of x.

   (d) There are, of course, certain easy situations for which the iterated Laguerre in-
equalities can be shown to always hold. For example, if ϕ(x) = (x + a)ex , a ≥ 0 or
ϕ(x) = (x + a)(x + b)ex , a, b ≥ 0, this is true. Since the derivative of such a function again
                                                                                  (k)
has the same form, the remarks above indicate that it suffices to show that Tk (ϕ(x)) ≥ 0
for k = 1, 2, . . . and all x ≥ 0. For the quadratic case, we obtain
   (k)
 Tk (ϕ(x))
       ( k       k                                                     
         22 −2 e2 x (a + x)2 + (b + x)2 + 2(k − 1)(2x + a + b + k2 − k) , for k odd
     =     k     k
         22 −1 e2 x ((x + a)(x + b) + k(2x + a + b + k − 1)) ,            for k even

and each expression is clearly nonegative for all real x.

   (e) A particularly intriguing open problem is the case of ϕ(x) = xm in (3.1). Special
cases, such as the iterated Turán inequalities discussed in the next section, can be easily
                    (n)            n                              (n)
established (i.e. Tn (xn ) = (n!)2 ), but the general case of Tn (xn+k ), k = 0, 1, 2, . . . ,
seems surprisingly difficult.

  In [2, Theorem 2.13] it is shown that (3.1) is true when n = 2; that is the double
Laguerre inequalities are valid. Here we present a somewhat different and shorter proof
(which still depends on Theorems 2.2 and 2.3) in the hope that it will shed light on the
general case.

Proposition 3.4. If ϕ(x) is a polynomial with only real, nonpositive zeros and positive
leading coefficient (so that ϕ(x) ∈ L-P+ ∩ R[x]), then
                        (2)
(3.2)                 Tk (ϕ(x)) ≥ 0        for all   x≥0    and   k ≥ 2.


Proof. First we prove (3.2) by induction, in the special case when k = 2. If deg ϕ = 0 or 1,
       (2)
then T2 (ϕ) = 0. Now suppose that (3.2) holds (with k = 2) for all polynomials g ∈ L-P+
8                       THOMAS CRAVEN AND GEORGE CSORDAS

of degree at most n. Let ϕ(x) := (x + a)g(x), where a ≥ 0. For notational convenience, set
          (1)
h(x) := T1 (g(x)) = (g 0 (x))2 − g(x)g 00 (x) and note that h(x) is just L1 (g(x)) in Theorem
2.3. Then some elementary, albeit involved, calculations (which can be readily verified
with the aid of a symbolic program) yield
                                 n                                                       o
              (2)           00              4 (1)
(3.3) ϕ(x)T2 (ϕ(x)) = ϕ (x) (x + a) T1 (h(x)) + ϕ(x) [12ϕ(x)L2 (g(x)) + A(x)] ,

where L2 (g(x)) is given by (2.2) and

                   A(x) = 8(g 0 (x))3 − 12g(x)g 0(x)g 00 (x) + 4g(x)2 g 000 (x).

Since ϕ(x), ϕ00 (x) ∈ L-P+ , ϕ(x) ≥ 0 and ϕ00 (x) ≥ 0 for all x ≥ 0. Also, by Theorem 2.2,
L2 (g(x)) ≥ 0 for all x ∈ R. Now, another calculation shows that
                                      (1)                      (2)
                             g 00 (x)T1 (h(x)) = g(x)T2 (g(x))
         (1)                                                                       (2)
and so T1 (h(x)) ≥ 0 for x ≥ 0, since by the induction assumption T2 (g(x))       Qn ≥ 0 for
x ≥ 0. Therefore, it remains to show that A(x) ≥ 0 for x ≥ 0. Let g(x) = c j=1 (x + xj ),
where c > 0 and xj ≥ 0 for 1 ≤ j ≤ n. Then using logarithmic differentiation and the
product rule we obtain
                           2
                                                      Xn
                       3 d      0       1             3         2
(3.4)   A(x) = 4g(x)          g   (x).        = 4g(x)                    ≥ 0 for all x > 0.
                        dx2            g(x)             j=1
                                                            (x + x j ) 3


                                                                                     (2)
Thus, the right-hand side of (3.3) is nonnegative for all x ≥ 0 and whence T2 (ϕ(x)) ≥ 0 if
                                                          (2)
x > 0. But then continuity considerations show that T2 (ϕ(x)) ≥ 0 for all x ≥ 0. Finally,
                                                          (n)        (n)
since L-P+ is closed under differentiation and since Tk+j (ϕ) = Tk (ϕ(j) ) for k ≥ n and
j = 0, 1, 2 . . . (see Remark (a)), we conclude that (3.2) holds. 

   Recall from the introduction, that if ϕ(x) ∈ L-P+ , then ϕ(x) can be expressed in the
form
                                         ω
                                         Y
                                    σx
(3.5)                    ϕ(x) = ce             (1 + x/xj ) ,    0 ≤ ω ≤ ∞,
                                         j=1
                                  P
where c ≥ 0, σ ≥ 0, xj > 0 and        1/xj < ∞. Now set
                                             min(N, ω)         
                                      σx N    Y           x
                        ϕN (x) = c 1 +                   1+       .
                                       N       j=1
                                                            x j


Then ϕN (x) → ϕ(x) as N → ∞, uniformly on compact subsets of C. Moreover, the
class L-P+ is closed under differentiation, and so the derivatives of ϕ(x) can also be
expressed in the form (3.5). Therefore, the following theorem is an immediate consequence
of Proposition 3.4.
                     ITERATED LAGUERRE AND TURÁN INEQUALITIES                                       9

Theorem 3.5. If ϕ(x) ∈ L-P+ , then for j = 0, 1, 2 . . . ,

                        (2)
                     Tk (ϕ(j) (x)) ≥ 0          for all    x≥0        and    k ≥ 2.


  In the course of the proof of Proposition 3.4, we have shown (see (3.4)) that for poly-
nomials g(x) ∈ L-P+ , the following inequality holds

(3.6)        2(g 0 (x))3 − 3g(x)g 0(x)g 00 (x) + g(x)2 g 000 (x) ≥ 0        for all x ≥ 0.

Next, we employ the foregoing limiting argument (see the paragraph preceding Theorem
3.5) and the fact that L-P+ is closed under differentiation, to deduce from (3.6) the fol-
lowing corollary.

Corollary 3.6. If
                                             ∞
                                             X γk
                                    ϕ(x) =              xk ∈ L-P+ ,
                                                   k!
                                             k=0

then for p = 0, 1, 2 . . . and for all x ≥ 0,
                       3                                  2
               (p+1)         (p)  (p+1)     (p+2)
(3.7)       2 ϕ      (x) − 3ϕ (x)ϕ      (x)ϕ      (x) + ϕ (x) ϕ(p+3) (x) ≥ 0.
                                                         (p)




   The interest in inequality (3.7) stems, in part, from the fact that for x = 0 it provides
a new necessary condition for a real entire function to belong to L-P+ . Indeed, for x = 0,
inequality (3.7) may be expressed in the form

                  2
                               
(3.8)      2γp+1 γp+1 − γp γp+2 ≥ γp (γp+1 γp+2 − γp γp+3 )                  (p = 0, 1, 2 . . . ).

                         2
The Turán inequalities γp+1 − γp γp+2 ≥ 0 imply that γp+1 γp+2 − γp γp+3 ≥ 0. Thus, if
γp > 0 for all p ≥ 0, then γp (γp+1 γp+2 − γp γp+3 ) /(2γp+1 ) is a nontrivial positive lower
                                 2
bound for the Turán expression γp+1  − γp γp+2 .

4. Iterated Turán inequalities.

   Let Γ = {γk }∞ k=0 be a sequence of real numbers. We define the r-th iterated Turán
                    (0)                          (r)     (r−1) 2    (r−1) (r−1)
sequence of Γ via γk = γk , k = 0, . . . , and γk = (γk         ) −γk−1 γk+1 , k = r, r +1, . . . .
                          P                    (r)          (r)
Thus, if we write ϕ(x) =     γk xk /k!, then γk is just Tk (ϕ(x)) evaluated at x = 0. Under
certain circumstances, we can show that all of the higher iterated Turán expressions are
positive for a multiplier sequence. In Section 3 we mentioned some simple cases in which
we could, in fact, show that all of the iterated Laguerre inequalities hold. In this section we
establish the iterated Turán inequalities for a large class of interesting multiplier sequences.
10                     THOMAS CRAVEN AND GEORGE CSORDAS

Theorem 4.1. Fix c ≥ 1 and d ≥ 0. Consider the set Mc of all sequences of positive
numbers {γk }∞
             k=0 satisfying


(4.1)                                 γk2 − c γk−1 γk+1 ≥ 0,

for all k. Then

(4.2)        (γk2 − γk−1 γk+1 )2 − (c + d)(γk−1
                                            2                2
                                                − γk−2 γk )(γk+1 − γk γk+2 ) ≥ 0
                                                               √
                                                         3+     5 + 4d
for all k and all sequences in Mc if and only if c ≥                   .
                                                                2

Proof. To see necessity, consider the specific sequence γ0 = 1, γ1 = 1, γ2 = 1/b, γ3 =
1/(cb2 ), γk = 0 for k ≥ 4. This satisfies (4.1) for any b ≥ c. But (4.2) yields

                                     2                        
                             1      1                   1      1
                                − 2       − (c + d) 1 −
                             b2   cb                    b    c2 b4
                                                             
                                 1      2            c      d
                              = 2 4 c − 3c + 1 + − d +           .
                                c b                  b      b

Since b may
        √ be made as large as desired, this is only guaranteed to be nonnegative if
     3 + 5 + 4d
c ≥              , the larger root of c2 − 3c + 1 − d. The other alternative, of 1 ≤ c ≤
    √     2
3 − 5 + 4d
             does not occur for d > −1 (and, in particular, for d ≥ 0).
      2
                                                 √
                                             3 + 5 + 4d                          (1)
   Conversely, assume (4.1) holds with c ≥              . An upper bound for γk is γk2 .
                                                  2
From (4.1), we obtain the lower bound

                         (1)
                        γk = γk2 − γk−1 γk+1 ≥ (c − 1)γk−1 γk+1 .

Estimating the expression in (4.2), we obtain

              (1)               (1)   (1)
            (γk )2 − (c + d)γk−1 γk+1 ≥ [(c − 1)γk−1 γk+1 ]2 − (c + d)γk−1
                                                                       2    2
                                                                           γk+1
                                            = [c2 − 3c + 1 − d]γk−1
                                                                2    2
                                                                    γk+1 ≥0

by the condition on c. 

   The set M4 is of particular interest. Condition (4.1) forces the numbers γk to decrease
rather quickly, leading us to term such sequences rapidly decreasing sequences. They are
known to be multiplier sequences and were first investigated in some detail in [8]. These
interesting sequences are discussed at some length in [3, Section 4] and [4, Section 4].
                    ITERATED LAGUERRE AND TURÁN INEQUALITIES                                                  11
                                                                  √
                                                             3+ 5
Corollary 4.2. For a sequence as in (4.1) with c >                    ≈ 2.62, the corresponding
                                                                 2
                                                     (1)
constant for the sequence of Turán expressions γk = γk2 −γk−1 γk+1 is strictly greater than
                        2c2 − 3c + 2                                                    (2)
c by the amount d =                  . If we then iterate this, forming the sequence {γk }, the
                              2
corresponding constant again increases by more than d. After a finite number of steps, it
will reach 4 (in the normalized case γ0 = γ1 = 1) and the sequence of higher Turán expres-
         (r)
sions {γk }∞ k=r , for r fixed and sufficiently large, will be a rapidly decreasing sequence. In
particular, if the original sequence is a rapidly decreasing sequence, the sequence of Turán
inequalities is again a rapidly decreasing sequence and we obtain an infinite sequence of
multiplier sequences by iterating this process.

   Although the iterated Turán expressions seem to be positive for all multiplier sequences
(an open question in general), it follows from the Theorem 4.1 that inequality (4.1) with
c = 1 is not sufficient to achieve this since M1 contains sequences that fail to satisfy (4.2)
for d = 0. But then, the specific sequence used in the proof is not a multiplier sequence if
c = 1, as it violates condition (3.8) for p = 1.

5. The third iterated Turán inequality.
                                                                                  (3)
   In this section we establish the third iterated Turán inequality γk ≥ 0 (k = 3, 4, 5 . . . )
for multiplier sequences, {γk }∞k=0 , of the form γk = k(k − 1)αk , k = 1, 2, 3 . . . , where
{αk }∞
     k=0  is an arbitrary multiplier sequence. With the notation adopted in Section 4, we
have
                           (3)      (2)       (2)    (2)
(5.1)                    γk = (γk )2 − γk−1 γk+1 ,         k = 3, 4, 5, . . . ,
or equivalently
(5.2)                                                       
                                  2
      (3)                    (2)         (2)        (2)
    Tk (ϕ(x))          =   Tk (ϕ(x)) − Tk−1 (ϕ(x))Tk+1 (ϕ(x))                       ,   k = 3, 4, 5, . . . ,
                 x=0                                                         x=0
where
                                             ∞
                                             X      γk k
(5.3)                              ϕ(x) :=             x ∈ L-P+ .
                                                    k!
                                             k=0
Before embarking on the proof of the third iterated Turán inequality,
                                                                      we briefly
                                                                                  discuss
                                                          (3)    (3)
a representation of the third iterated Turán expression γk = Tk (ϕ(x))          in terms
                                                                                              x=0
of Wronskians and determinants of Hankel matrices (Proposition 5.1). We recall that
the (nth order) Wronskian (determinant) W (ϕ(x), ϕ0 (x), . . . , ϕ(n−1) (x)), where ϕ(x) is an
entire function, is defined as
                                                      ϕ(x)        ϕ0 (x)      ···       ϕ(n−1) (x)
                                                      ϕ0 (x)     ϕ(2) (x)     ···        ϕ(n) (x)
(5.4)     W (ϕ(x), ϕ0 (x), . . . , ϕ(n−1) (x)) :=        ..          ..                     ..     ,
                                                          .           .                      .
                                                    ϕ(n−1) (x)   ϕ(n) (x) · · ·         ϕ(2n−2) (x)
12                             THOMAS CRAVEN AND GEORGE CSORDAS

and that the (nth order) Hankel matrices, associated with the sequence {γk }∞
                                                                            k=0 , are ma-
                    (n)            n
trices of the form Hk = (γk+i+j−2 )i,j=1 , that is
                                                           
                γk                   γk+1   ...      γk+n−1
         (n)  γ                     γk+2   ...      γk+n+1 
        Hk =  k+1                                                (n = 1, 2, 3, . . . , k = 0, 1, 2, . . . ).
                                            ...
                     γk+n−1          γk+n   ...     γk+2n−2
                               (n)           (n)                                                                 (n)
We note that if we set Ak = det Hk , then W (ϕ(k) (0), ϕ(k+1) (0), . . . , ϕ(k+n−1) (0)) = Ak
and for n = 3 the following relation holds
                                                  (3)     (2)
(5.5)                                (−γk+2 ) Ak = γk+2         k = 0, 1, 2 . . . .

Furthermore, if ϕ(x) ∈ L-P+ is given by (5.3) and if γk > 0, then by Theorem 3.5,

   (2)                                                                                  (3)
 Tk (ϕ(x))          ≥ 0 for x ≥ 0 (k = 2, 3, 4, . . . ) and whence, in light of (5.5), Ak ≤
             x=0
0 for k = 0, 1, 2 . . . . A straightforward, albeit lengthy, calculation yields the following
                                                           (3)
representation of the third iterated Turán expression γk .

                                            P∞   γk k
Proposition 5.1. Let ϕ(x) :=                 k=0 k! x be an entire function. Then for x ∈ R,

                                                                                            
          (3)                   (1)
        Tk (ϕ(x)) =           Tk (ϕ(x))           W ϕ(k−3) (x), ϕ(k−2) (x), ϕ(k−1) (x), ϕ(k) (x) ϕ(k) (x)2
(5.6)                         (2)           (2)          
                       T    (ϕ(x))Tk+1 (ϕ(x))
                      + k−1
                         ϕ(k−1) (x)ϕ(k+1) (x)

for k = 3, 4, 5 . . . . In particular, if x = 0 and k = 0, 1, 2 . . . , then
                                                                                       
              (3)           (3)                 2                        (4) 2    (3) (3)
(5.7)       γk+3 = Tk+3 (ϕ(x))              = (γk+3 − γk+2 γk+4 ) Ak γk+3      + Ak Ak+2 ,
                                            x=0

         (n)           (n)                                                                    (n)
where Ak       = det Hk        denotes the determinant of the Hankel matrix Hk .

Remarks 5.2. (a) Since the equalities (5.6) and (5.7) are formal identities, the assumption
that ϕ(x) is an entire function is not needed.

  (b) With the aid of some known identities (see, for example, [13, VII, Problem 19]),
equation (5.7) can be recast in the following suggestive form
                                             2
                               (3)        (3)    2    (3) (3)
(5.8)                         γk+3 =     Ak+1 γk+3 − Ak Ak+2 γk+2 γk+4 .

                           P∞ γk k         +                                  (3)
Now, suppose that ϕ(x) :=    k=0 k! x ∈ L-P . Then, by virtue of (5.8), γk+3 ≥ 0
              2
           (3)       (3) (3)
whenever Ak+1 − Ak Ak+2 ≥ 0, k = 0, 1, 2 . . . . However, this inequality is not valid,
                     ITERATED LAGUERRE AND TURÁN INEQUALITIES                                            13

                                                             γk k    2
                                                                              P∞
                                                                             11
in general, as the following example shows. Let ϕ(x) :=  k=0 k! x = x (x + 1) . Here,
                                                                                  2
                                                                               (3)
γ0 = γ1 = 0, γ2 = 2, γ3 = 66, γ4 = 1320, γ5 = 19800 and γ6 = 237600. Then A1          −
  (3) (3)
A0 A2 = −2718144.
                                                                      (3)     (3)
  (c) Let ϕ(x) ∈ L-P+ be given by (5.3). Since Ak Ak+2 ≥ 0 (cf. (5.5) and Theorem
                             (3)                              (4)                       (4)
3.5), (5.7) shows that γk+3 ≥ 0 whenever Ak ≥ 0. However, Ak may be negative,
as may be readily verified using the function ϕ(x) defined in part (b). Examples of this
sort are subtle as they depict a heretofore inexplicable phenomenon. The technique used
below sheds light on this and at the end of this paper we provide a sufficient condition
                         (4)
which guarantees that Ak < 0. In connection with the investigations of a conjecture of
S. Karlin, additional examples are considered in [7] and [1]. Furthermore, to highlight the
intricate nature of Karlin’s conjecture, it was pointed out in these papers, in particular,
       (4)
that Ak ≥ 0 if γk = αk /k!, k = 0, 1, 2 . . . , where {αk }∞
                                                           k=0 is any multiplier sequence.

  (d) In the sequel we will prove that if
                                             ∞
                                             X k(k − 1)αk
                                   ϕ(x) :=                          xk ∈ L-P+ ,
                                                         k!
                                             k=0

                                                                    (3)
where {αk }∞ k=0 is any multiplier sequence, then γ3      ≥ 0. It is not hard to see that this
is equivalent to proving the result only for multiplier sequences that begin with two zeros.
However, the assumption that {αk }∞  k=0 is a multiplier sequence is not necessarily required
to make the inequalities hold. To see this, we consider once again the example in part
                                                      γk
(b). Let α0 = α1 = 0 and for k ≥ 2, set αk = k(k−1)        . We claim that {αk }∞k=0 is not a
multiplier sequence. Indeed, consider the fourth Jensen polynomial (defined, for example,
in [2]) associated with the sequence {αk }∞k=0 , that is,

                                     4  
                                     X  4
                       g4 (x) =                    αk xk = 2x2 (3 + 22x + 55x2 ).
                                             k
                                     k=0

Since g4 (x) has two nonreal zeros, {αk }∞
                                         k=0 is not a multiplier sequence, though our main
theorem will establish the third iteration of the Turán inequalities for {γk }∞
                                                                               k=0 .
                                                                           
                                                                  (3)
   The proof of the main theorem requires that we express T3 (ϕ(x))               in terms of
                                                                                                   x=0
sums of powers of the logarithmic derivatives of ϕ(x). Accordingly, we proceed to establish
the following preparatory result.
                          Qn
Lemma 5.3. Let ϕ(x) = j=1 (x + xj ), xj > 0, j = 1, 2, . . . , n, be a polynomial in L-P+ .
                                                    1
For fixed x ≥ 0 and j = 1, 2, . . . , n, set aj := x+xj
                                                        and let
                     n
                     X                     n
                                           X                    n
                                                                X                           n
                                                                                            X
(5.9)         A :=         aj ,    B :=          a2j ,   C :=         a3j ,    and   D :=         a4j .
                     j=1                   j=1                  j=1                         j=1
14                                 THOMAS CRAVEN AND GEORGE CSORDAS

Then
                 ϕ0 (x)              ϕ00 (x)                       ϕ000 (x)
(5.10)                  = A,                 = A2 − B,                      = A3 − 3AB + 2C         and
                 ϕ(x)                ϕ(x)                           ϕ(x)

                                   ϕ(4) (x)
                                            = A4 − 6A2 B + 3B 2 + 8AC − 6D.
                                    ϕ(x)


Proof. Logarithmic differentiation yields
                                                                                  
            0         n
                      X                                  Xn                               X   n
           ϕ (x)              1            00      0                         1                1
                 =                    and ϕ (x) = ϕ (x)                            − ϕ(x)              .
           ϕ(x)             x + xj                                          x + xj          (x + xj )2
                      j=1                                             j=1                     j=1

Hence,
                                                                     
                                                         n
                                                           X                  n
                              00
                            ϕ (x)           0
                                           ϕ (x)                  1  X      1
                                  =                                     −
                            ϕ(x)           ϕ(x)                  x + xj   (x + xj )2
                                                           j=1               j=1
                                                          2
                                           n
                                           X            Xn
                                                 1             1
                                     =               −                2
                                                                         = A2 − B.
                                           j=1
                                               x + xj   j=1
                                                            (x + x j )

Continuing in this manner, similar calculations yield
                        3                                                          
    000       Xn                  Xn              Xn                    X n
  ϕ (x)            1                   1              1         +2         1        
         =                  − 3                                 2                      3
   ϕ(x)       j=1
                  x + xj          j=1
                                      x + x j     j=1
                                                      (x + x j )         j=1
                                                                             (x + x j )

                =A3 − 3AB + 2C

and
                      4                   2                                          2
     (4)  X n                 X n                X n                      Xn
 ϕ (x)           1                 1                  1                        1
       =                  −6                                    2
                                                                   + 3
                                                                                          2
                                                                                            
  ϕ(x)     j=1
                x + xj         j=1
                                   x +  x  j      j=1
                                                      (x + x j )          j=1
                                                                              (x +  x j )
                                                                             
            X n            Xn                       Xn               Xn
                    1            1       − 6           1               1      
       +8                               3
                 x + xj        (x + xj )                x + xj           (x + xj )4
                      j=1                  j=1                          j=1             j=1
                  4          2         2
                =A − 6A B + 3B + 8AC − 6D.

     
                                                                   
                                                      (3)   (3)
     The next lemma gives an explicit expression for γ3 = T3 (ϕ(x))                                  , where ϕ(x) is
                                                                                               x=0
of the form ϕ(x) = x2 ψ(x). While the verification involves only simple algebraic manipu-
lations, the expression obtained is sufficiently involved to warrant the use of a computer.
                        ITERATED LAGUERRE AND TURÁN INEQUALITIES                                                                   15

                                     P∞      αk k
Lemma 5.4. Let ψ(x) :=                   k=0 k! x be an entire function. Let
                                                                         ∞
                                                                         X
                                                          2                γk
                                             ϕ(x) = x ψ(x) =                          xk ,
                                                                                 k!
                                                                         k=0

so that γ0 = γ1 = 0 and γk = k(k − 1)αk−2 , for k = 2, 3, . . . . Then
                                                                    
               (3)     (3)                          2
(5.11)        γ3 = T3 (ϕ(x))         = 768 3 ψ 0 (0) − 2 ψ(0) ψ 00 (0) E(0),
                                                 x=0

where
                    6                                 4                              2            2             2              3    3
E(x) :=729 ψ 0 (x) − 1458 ψ(x) ψ 0(x) ψ 00 (x) + 324 ψ(x) ψ 0 (x) ψ 00 (x) + 216 ψ(x) ψ 00 (x)
                   2         3                                 3
        +54x ψ(x) ψ 0 (x) ψ (3) (x) − 360 ψ(x) ψ 0 (x) ψ 00 (x) ψ (3)(x)
                   4             2                    4
        +100 ψ(x) ψ (3) (x) − 90 ψ(x) ψ 00 (x) ψ (4)(x).

  Preliminaries aside, we are now in a position to prove the principal result of this section.
                                      P∞         αk k      +
Theorem 5.5. Let ψ(x) :=                     k=0 k! x ∈ L-P . Let
                                                                         ∞
                                                                         X
                                                          2                γk
                                             ϕ(x) = x ψ(x) =                          xk ,
                                                                                 k!
                                                                         k=0

so that γ0 = γ1 = 0 and γk = k(k − 1)αk−2 , for k = 2, 3, . . . . Then
                                                
                               (3)     (3)
(5.12)                        γ3 = T3 (ϕ(x))           ≥ 0.
                                                                             x=0


                                                                                                             
                                                                                                      (1)
Proof. In view of (5.11) of Lemma 5.4, since ψ(x) ∈ L-P+ ,                                          Tk (ψ(x))                 ≥ 0 (k =
                                                                                                                        x=0
                                                                                                                          +
1, 2, 3 . . . ), we only need to establish that E(0) ≥ 0. Also, since ψ(x) ∈ L-P , ψ(x) can
be uniformly approximated, on compact subsets of C, by polynomials having Q       only real,
                                                                                    n
nonpositive zeros. Therefore, it suffices to prove inequality (5.12) when ψ(x) = j=1 (x +
                                                                                  6
xj ), (xj ≥ 0), is a polynomial in L-P+ . Now if ψ(0) = 0, then E(0) = 729 ψ 0 (0) ≥ 0 and
so in this case inequality (5.12) is clear. Thus, henceforth we will assume that ψ(0) 6= 0
and, for fixed x ≥ 0, consider E(x) as given in Lemma 5.4. We will prove a stronger result,
namely, that for all x ≥ 0,
                                6               4
           E(x)     729 ψ 0 (x)     1458 ψ 0 (x) ψ 00 (x)
                  =          6    −             5
          (ψ(x))6     ψ(x)                ψ(x)
                                     2            2                      3                            3
                        324 ψ 0 (x) ψ 00 (x)              216 ψ 00 (x)           540 ψ 0 (x) ψ (3) (x)
                   +                     4            +              3       +                            4
                                 ψ(x)                         ψ(x)                               ψ(x)
                                                                                         2
                        360 ψ 0 (x) ψ 00 (x) ψ (3)(x)              100 ψ (3) (x)                 90 ψ 00 (x) ψ (4) (x)
                   −                         3                +                  2           −                      2    ≥ 0.
                                     ψ(x)                                ψ(x)                                 ψ(x)
16                                THOMAS CRAVEN AND GEORGE CSORDAS

For fixed x ≥ 0, by Lemma 5.3 with ψ in place of ϕ, we obtain

      E(x)
             =729A6 − 1458A4 (A2 − B) + 324A2 (A2 − B)2
     (ψ(x))6
            +216(A2 − B)3 + 540A3 (A3 − 3AB + 2C) − 360A(A2 − B)(A3 − 3AB + 2C)
            +100(A3 − 3AB + 2C)2 − 90(A2 − B)(A4 − 6A2 B + 3B 2 + 8AC − 6D)

or
                   E(x)
                         6
                           =A6 + 12 A4 B − 18 A2 B 2 + 54 B 3 + 40 A3 C
                  (ψ(x))
                                  +240 A B C + 400 C 2 + 540 A2 D − 540 BD
                                                          2          !
                                                                     2
                                                       3 B      63 B
                                  =A6 + 12 B    A2 −          +
                                                        4        16
                                                                            
                                  +40 A3 C + 240 A BC + 400 C 2 + 540 A2 − B D.

Since xj > 0 for j = 1, 2 . . . , n, we have A, B, C, D > 0 and all the derivatives of ψ(x)
are positive for x ≥ 0. Therefore we also have A2 − B = ψ 00 (x)/ψ(x) > 0, and thus
E(x) > 0 for x ≥ 0. 

Remarks 5.6. (a) We wish to point out that in Theorem 5.5 we introduced the factor x2
in order to simplify the ensuing algebra. In the absence of this factor we would have to
            (5)                 (6)
calculate ϕϕ(x) (x)
                    as well as ϕϕ(x)(x)
                                        (see the proof of Lemma 5.3). Then, as in the proof
of Theorem 5.5, we would obtain an expression, analogous to E(x), which has 112 terms
rather than nine. Nevertheless, it seems that the technique developed above, should yield
the desired result (5.12) for an arbitrary multiplier sequence rather than one with the first
two terms equal to zero.

   (b) We briefly indicate here how the foregoing technique can be used to derive a sufficient
                                       (4)      (4)
condition which guarantees that A0 = det H0 < 0. Let ϕ(x) := x2 ψ(x), where ψ(x) =
Q n                                              +                                   th
  j=1 (x + xj ), xj > 0, is a polynomial in L-P . Then, the determinant of the 4        order
                   (i+j−2)     4
Hankel matrix (ϕ           (0))i,j=1 reduces to
                    (4)
                  A0 =W (ϕ(0), ϕ0 (0), ϕ00 (0), ϕ000 (0))
                                                4                      2                            2         2
(5.13)                    =48 (27 ψ 0(0) − 54 ψ(0) ψ 0(0) ψ 00 (0) + 12 ψ(0) ψ 00 (0)
                                          2                                3
                          +20 ψ(0) ψ 0 (0) ψ (3) (0) − 5 ψ(0) ψ (4) (0)).

Guided by (5.13) and the argument used in the proof of Theorem 5.5, we form the expres-
sion
                              4                 2                              2
                27 ψ 0 (x)            54 ψ 0 (x) ψ 00 (x)       12 ψ 00 (x)            20 ψ 0 (x) ψ (3) (x)       5 ψ (4) (x)
       K(x) =             4       −                 3       +              2       +               2          −
                  ψ(x)                        ψ(x)                ψ(x)                       ψ(x)                   ψ(x)
                      ITERATED LAGUERRE AND TURÁN INEQUALITIES                                       17

and with the aid of Lemma 5.4, for fixed x ≥ 0, we obtain that

                                       K(x) = 3(10 D − B 2 ),
                           Pn                   Pn
where the quantities B = j=1 a2j and D = j=1 a4j have the same meaning as in (5.9).
Thus, we readily infer that if the the zeros of the polynomial ψ(x) ∈ L-P+ are distributed
                                                  (4)
such that 10 D < B 2 holds at x = 0, then A0 < 0. By way of illustration, consider
ϕ(x) = x2 ψ(x) = x2 (x + a)12 , where a > 0. Then for x = 0, we find that 10D = 120/a4 <
                                                (4)
144/a4 = B 2 , and whence by our criterion, A0 < 0. Indeed, a direct computation yields
      (4)
that A0 = −3456a44 .

   The authors wish to thank the referee for extensive comments and suggestions.


                                            References

 1. T. Craven, G. Csordas, Karlin’s conjecture and a question of Pólya (to appear).

 2. T. Craven and G. Csordas, Jensen polynomials and the Turán and Laguerre inequalities, Pacific J.
    Math. 136 (1989), 241–259.

 3. T. Craven and G. Csordas, Problems and theorems in the theory of multiplier sequences, Serdica Math.
    J. 22 (1996), 515–524.

 4. T. Craven and G. Csordas, Complex zero decreasing sequences, Methods Appl. Anal. 2 (1995), 420–
    441.

 5. G. Csordas and R. S. Varga, Necessary and sufficient conditions and the Riemann Hypothesis, Adv.
    in Appl. Math. 11 (1990), 328–357.

 6. K. Dilcher and K. B. Stolarsky, On a class of nonlinear differential operators acting on polynomials,
    J. Math. Anal. Appl. 170 (1992), 382–400.

 7. D. K. Dimitrov, Counterexamples to a problem of Pólya and to a problem of Karlin, East J. Approx.
    4 (1998), 479–489.

 8. J. I. Hutchinson, On a remarkable class of entire functions, Trans. Amer. Math. Soc. 25 (1923),
    325–332.

 9. N. Obreschkoff, Verteilung und Berechnung der Nullstellen reeller Polynome, VEB Deutscher Verlag
    der Wissenschaften, Berlin, 1963.

10. M. Patrick, Extensions of Inequalities of the Laguerre and Turán type, Pacific J. Math. 44 (1973),
    675–682.

11. G. Pólya, Collected Papers, Vol. II Location of Zeros, (R. P. Boas, ed.), MIT Press, Cambridge, MA,
    1974.

12. G. Pólya and J. Schur, Über zwei Arten von Faktorenfolgen in der Theorie der algebraischen Gle-
    ichungen, J. Reine Angew. Math. 144 (1914), 89–113.
18                        THOMAS CRAVEN AND GEORGE CSORDAS

13. G. Pólya and G. Szegö, Problems and Theorems in Analysis, Vols. I and II, Springer-Verlag, New
    York, 1976.


     Department of Mathematics, University of Hawaii, Honolulu, HI 96822



     Department of Mathematics, University of Hawaii, Honolulu, HI 96822

     E-mail address: tom@math.hawaii.edu, george@math.hawaii.edu
