On zeros of some entire functions

On zeros of some entire functions Bing He

arXiv:1604.05179v1 [math.CA] 11 Apr 2016

College of Science, Northwest A&F University Yangling 712100, Shaanxi, People’s Republic of China [email protected]

Abstract. Let

A(α) q (a; z)

=

2 ∞ X (a; q)k q αk z k

k=0

(q; q)k

,

where α > 0, 0 < q < 1. In a paper of Ruiming Zhang, he asked under what con(α) ditions the zeros of the entire function Aq (a; z) are all real and established some (α) results on the zeros of Aq (a; z) which present a partial answer to that question. In the present paper, we will set up some results on certain entire functions which (α) includes that Aq (q l ; z), l ≥ 2 has only infinitely many negative zeros that gives a partial answer to Zhang’s question. In addition, we establish some results on zeros of certain entire functions involving the Rogers-Szeg˝o polynomials and the Stieltjes-Wigert polynomials. Keywords and phrases: Zeros of entire functions, P´olya frequence sequence, Vitali’s theorem, Hurwitz’s theorm, Rogers-Szeg˝o polynomials, Stieltjes-Wigert polynomials 2010 MSC: 30C15; 33D15; 30C10

1

Introduction

Recall that entire functions are functions that P are holomorphic in the whole comk plex plane. Given an entire function f (z) = ∞ k=0 ak z , then the order of f (z) can be computed by [5, (2.2.3)] ρ(f ) = lim sup k→∞

k log k . − log ak

(1.1)

(α)

Following [11], we define the entire function Aq (a; z) by A(α) q (a; z) =

2 ∞ X (a; q)k q αk z k

k=0

1

(q; q)k

,

where α > 0, 0 < q < 1 and (a; q)0 = 1, (a; q)k =

k−1 Y

(1 − aq j ) (k ≥ 1).

j=0

It is easily seen that 2

(1) Aq 2 (q −n ; z)

A(1) q (0; z)

k ∞ X 1 (q −n ; q)k q 2 z k = (q; q)n Sn (zq 2 −n ; q), = (q; q)k k=0 2 ∞ X qk zk = = Aq (−z), (q; q) k k=0

where Aq (z) and Sn (z; q) are the Ramanujan entire function and the Stieltjes(α) Wigert polynomial respectively [10]. So Aq (a; z) generalizes both Aq (z) and Sn (z; q). It is well-known that both of them have only real positive zeros. Therefore, Zhang in [18] asked under what conditions the zeros of the entire function (α) (α) Aq (a; z) are all real. In that paper, Zhang proved that Aq (−a; z) (a ≥ 0, α > (α) 0, 0 < q < 1) has only infinitely many negative zeros and Aq (q −n ; z) (n ∈ N, α ≥ 0, 0 < q < 1) has only finitely many positive zeros, which gave a partial answer to that question. In addition, Zhang obtained a result on the negativity of zeros of an entire function including many well-known entire functions. Our motivation for the present work emanates from Zhang’s question. In this paper, we will establish the following results which present a partial answer to Zhang’s question. Theorem 1.1. Let α > 0 and 0 < q < 1. Then (α) (i)if l ≥ 2 is an integer, then Aq (q l ; z) has only infinitely many real zeros and all of them are negative; (ii)if m and n are nonnegative integers such that at least one of them is positive, {lj }m j=1 are integers not less than 2, 0 < qj < 1 (1 ≤ j ≤ m) and νr > −1, 0 < qr < 1 (1 ≤ r ≤ n), then the function l ∞ Y m X (qjj ; qj )k k=0 j=1

2

q αk Qn zk (qj ; qj )k r=1 (qr , qrνr +1 ; qr )k

has only infinitely many real zeros and all of them are negative; (iii)if m ≥ 0 and n ≥ 1 are integers, {lj }m j=1 are integers not less than 2 and νr ≥ 0 (1 ≤ r ≤ m), then the function Qm ∞ X j=1 (lj )k Q zk n m+n (k!) (ν + 1) k r=1 r k=0 2

where (a)k is defined by (a)0 = 1, (a)k = a(a + 1) · · · (a + k − 1) (k ≥ 1), has only infinitely many real zeros and all of them are negative. It should be mentioned that in [13, Theorem 4] Katkova et al. proved that (α) there exists a constant q∞ (≈ 0.556415) such that the function Aq (q; z) has only (α) real zeros if and only if q ≤ q∞ . So the similar result for Aq (q l ; z) does not hold for l = 1. The Gaussian binomial coefficients are q-analogs of the binomial coefficients, which are given by   (q; q)n n = . k q (q; q)k (q; q)n−k We now introduce the definition of the Rogers-Szeg˝o polynomials which were first investigated by Rogers [15] and then by Szeg˝o [16]. The Rogers-Szeg˝o polynomials are defined by n   X n k n−k x y . hn (x, y|q) = k q k=0

If q is replaced by q −1 in the Rogers-Szeg˝o polynomials, then we obtain the Stieltjes-Wigert polynomials (see [16]): gn (x, y|q) =

n   X n k=0

k

q k(k−n) xk y n−k .

q

From [18, Theorem 5], we know that hn (x|q) has only negative zeros for q ≥ 1 and gn (x|, q) has only negative zeros for 0 < q ≤ 1, where hn (x|q) and gn (x|, q) are defined by n   X n k x hn (x|q) := hn (x, 1|q) = k q k=0

and

gn (x|q) := gn (x, 1|q) =

n   X n k=0

k

q k(k−n) xk . q

Motivated by Zhang’s work, we will establish the following results on zeros of certain entire functions involving the Rogers-Szeg˝o polynomials and the StieltjesWigert polynomials. Theorem 1.2. Let 0 < q < 1. If α is positive number and 0 < x, y < 1, then ∞ X hn (x, y|q) n=0

(q; q)n 3

2

q αn z n

has infinitely many real zeros and all of them are negative; if −1 < x, y < 0 and α ≥ 12 , then ∞ X gn (x, y|q) αn2 n q z (q; q) n n=0 has infinitely many real zeros and all of them are positive.

Remark 1.1. (i) Applying the method which is used in the proof of Theorem 1.2, we can deduce the following results: let 0 < q < 1; if α is positive number and 0 < x < 1, then ∞ X hn (x|q) αn2 n q z (q; q)n n=0

has infinitely many real zeros and all of them are negative; if −1 < x < 0 and α ≥ 21 , then ∞ X gn− (x|q) αn2 n q z (q; q) n n=0 has infinitely many real zeros and all of them are positive, where gn− (x|q) = gn (x, −1|q). But we need the following results: − gn (x|q) hn (x|q) 1 1 (q; q)n ≤ (q, x; q)∞ , (q; q)n ≤ (q, |x|; q)∞

which can be derived easily. (ii) We can establish certain results on the Rogers-Szeg˝o polynomials and the Stieltjes-Wigert polynomials by using similar method. These are analogous to (ii) and (iii) of Theorem 1.1 We also set up the following result which is analogous to [18, Theorem 7]. Theorem 1.3. Suppose r and s are two positive integers, aj (1 ≤ j ≤ r) and 1 bk (1 ≤ k ≤ s) are r+s positive numbers and α > 0, 0 < q < 2− α . Then there exists K0 ∈ Z>0 such that for all integers K ≥ K0 , the function ∞ X (a1 )k (a2 )k · · · (ar )k αk2 k q z (b1 )k (b2 )k · · · (bs )k k=K

has only infinitely many real zeros and all of them are negative. In the next section, we will provide some lemmas which are crucial in the proof of Theorems 1.1 and 1.2. Section 3 is devoted to our proof of Theorems 1.1–1.3. 4

2

Preliminaries

In order to prove Theorems 1.1 and 1.2, we need some auxiliary results. We first recall from [7] that a real entire function f (z) is of Laguerre-P´olya class if m −αz 2 +βz

f (z) = cz e

∞  Y

k=1

z 1+ zk



e−z/zk ,

P −2 where c, β, zk ∈ R, α ≥ 0, m ∈ Z≥0 and ∞ k=1 zk < +∞. Let us recall that a real sequence {an }∞ olya frequence (or PF) n=0 is called a P´ ∞ sequence if the infinite matrix (aj−i )i,j=0 is totally positive, i.e. all its minors are nonnegative, where we use the notation ak = 0 if k < 0. This concept can be extended to finite sequences in the obvious way by completing the sequence with zero terms. Lemma 2.1. (SeeP[2])The sequence {ak }∞ k=0 is a PF sequence if and only if the k a z satisfies convergent series ∞ k=0 k ∞ X

k

m γz

ak z = cz e

k=0

∞ Y 1 + αk z

1 − βk z k=1

where c ≥ 0, γ ≥ 0, αk ≥ 0, βk ≥ 0, m ∈ Z+ and It was proved in [6] that

2 ∞ X qk

k=0

k!

P∞

,

k=1 (αk

+ βk ) < +∞.

xk

is a real entire function and in Laguerre-P´olya class for |q| < 1 and x ∈ R. Then k2

by [7, Theorem C], we obtain that { qk! }∞ k=0 is a PF sequence for 0 < q < 1. n Lemma 2.2. (See [8, p. 1047] and [14]) Let {ak }m k=0 and {bk }k=0 be sequences of nonnegative numbers. Then m Pm(i) thek sequence {ak }k=0 is a a PF sequence if and only if the polynomial k=0 ak z has only nonpositive zeros; n (ii) if the sequences {ak }m k=0 and {bk }k=0 are PF sequences, then so is the sequence {ak · bk }∞ k=0 ; n (iii) if the sequences {ak }m k=0 and {bk }k=0 are PF sequences, then so is the ∞ sequence {k! · ak · bk }k=0 .

We also need the following results, namely, Vitali’s theorem [17] and Hurwitz’s theorm [1, §5, Theorem 2]. 5

Lemma 2.3. (Vitali’s theorem) Let {fn (z)} be a sequence of functions analytic in a domain D and assume that fn (z) → f (z) point-wise in D. Then fn (z) → f (z) uniformly in any subdomain bounded by a contour C, provided that C is contained in D. Lemma 2.4. (Hurwitz’s theorm) If the functions {fn (z)} are nonzero and analytic in a region Ω, and fn (z) → f (z) uniformly on every compact subset of Ω, then f (z) either identically zero or never equal to zero in Ω. We conclude this section with following result which is very important in the proof of Theorem 1.2. Lemma 2.5. Let x, y be two real numbers and α > 0, 0 < q < 1. Then the functions ∞ ∞ X X hn (x, y|q) αn2 n gn (x, y|q) αn2 n q z and q z (q; q)n (q; q)n n=0 n=0 are all entire functions.

Proof. We first consider the function by [9, (1.3.15)], we have |hn (x, y|q)| ≤ ≤

n   X n k=0 n X

k

k

P∞

n=0

|x| |y|

n−k

≤q

k=0

q −n . = (q; q)∞

n   X n k=0

n

(1 − q )(1 − q

n X

≤

q

k=0

−n

hn (x,y|q) αn2 n q z . (q;q)n

k

For |x| ≤ 1, |y| ≤ 1,

q

n−1

) · · · (1 − q n−k+1) k−n q (q; q)k

∞ X qk qk −n ≤q (q; q)k (q; q)k k=0

By the same arguments, we get for |x| > 1, |y| ≤ 1, |hn (x, y|q)| ≤

n   X n k=0

≤ |x|n

k

k

|y| |x|

= |x|

n

q

n   X n k=0

n  X k=0

n−k

 (|x|/q)n n ≤ ; k q (q; q)∞

6

k

q

|y|k |x|−k

for |x| ≤ 1, |y| > 1, |hn (x, y|q)| ≤ |y|

n

≤ |y|n

n   X n k=0 n  X k=0

k

|x|k |y|−k q

 (|y|/q)n n ≤ ; k q (q; q)∞

for |x| > 1, |y| > 1, |hn (x, y|q)| = |xy|

n

≤ |xy|n

n   X n k=0 n  X k=0

In any cases we obtain |hn (x, y|q)| ≤

k

|x|k−n |y|−k

q

 (|xy|/q)n n ≤ . k q (q; q)∞

an (q; q)∞

where a is positive number which depends on x, y and q. Then n αn2 hn (x, y|q) αn2 ≤ a q . q (q; q)n (q; q)2 ∞

so that

which proves that

1 hn (x, y|q) αn2 n q = 0, lim sup (q; q) n→∞ n ∞ X hn (x, y|q) n=0

(q; q)n

2

q αn z n

is an entire function. Similarly, we can deduce that ∞ X gn (x, y|q) n=0

(q; q)n

2

q αn z n

is also an entire function. This ends the proof of Lemma 2.5.

7

3

Proof of Theorems 1.1–1.3

Proof of Theorem 1.1. We first prove (i). According to the q-binomial theorem [3, 9], we obtain that for all complex numbers x and q with |x| < 1 and |q| < 1, there holds ∞ X (a; q)k k (ax; q)∞ x = . (3.1) (q; q)k (x; q)∞ k=0

Setting a = q l , x = zq −l in (3.1) gives ∞ X (q l ; q)k k=0

(q; q)k

(z; q)∞ 1 = . −l −l (zq ; q)∞ (zq ; q)l

q −lk z k =

Using Lemma 2.1, we get the sequence n  l (q ; q)k −lk q (q; q)k k=0 is a PF sequence. It follows from (i) of Lemma 2.2 that {ak }nk=0 is a a PF sequence if and only if {ck · ak }nk=0 is also a PF sequence for any c > 0. Hence,  l n (q ; q)k (q; q)k k=0 n

q αk k!

2

on

is a PF sequence. So from the fact is a PF sequence and (iii) of Lemma k=0 2.2, we arrive at the sequence )n ( 2 (q l ; q)k q αk (q; q)k k=0

is also a PF sequence, which, by (i) of Lemma 2.2, implies that the polynomial 2 n X (q l ; q)k q αk

k=0

(q; q)k

zk

has only nonpositive zeros. Here and below, set Ω = C −{x+ yi|x ∈ (−∞, 0], y = 0}. Then 2 n X (q l ; q)k q αk k z → Aαq (q l ; z) (q; q) k k=0 8

point-wise in Ω. It is easily seen that for 0 < q < 1, α ≥ 0 and each n ∈ N, z ∈ C, n ∞ l αk 2 X (q l ; q)k q αk2 k X (q ; q) q k z ≤ |z|k < +∞. (q; q) (q; q) k k k=0 k=0 We apply Lemma 2.3 to know that

2 n X (q l ; q)k q αk

k=0

(q; q)k

z k → Aαq (q l ; z)

uniformly on every compact subset of Ω and then apply Lemma 2.4 to see that Aαq (q l ; z) 6= 0 in Ω which means that Aαq (q l ; z) has no zeros outside the set {x + yi|x ∈ (−∞, 0], y = 0}. According to [10, Lemma 14.1.4], we have Aαq (q l ; z) has infinitely many zeros. Therefore, Aαq (q l ; z) has only infinitely many real zeros and all of them are negative, which proves (i). We next show (ii). According to [18], we know that the sequence 

1 (q, q ν+1 ; q)k

N

k=0

is a PF sequence for ν > −1, 0 < q < 1, which means that N  1 (qr , qrνr +1 ; qr )k k=0 are all PF sequences for 1 ≤ r ≤ n. Since ( lj )N  N (qj ; qj )k 1 (1 ≤ r ≤ n) (1 ≤ j ≤ m), (qj ; qj )k (qr , qrνr +1 ; qr )k k=0 k=0

n αk2 on and q k! are all PF sequences, we then apply (ii) and (iii) of Lemma 2.2 to k=0 find that )N ( m lj 2 Y (qj ; qj )k q αk Qn νr +1 ; q ) (q ; q ) r k j j k r=1 (qr , qr j=1 k=0

is also a PF sequence, which, by (i) of Lemma 2.2, implies that l N Y m X (qjj ; qj )k k=0 j=1

2

q αk z k Qn (qj ; qj )k r=1 (qr , qrνr +1 ; qr )k 9

has only negative zeros. For each positive integer N and z ∈ C, we have lj N Y m X αk 2 k (q ; q ) q z j k j Q n ν +1 r (qj ; qj )k r=1 (qr , qr ; qr )k k=0 j=1 ≤

l ∞ Y m X (qjj ; qj )k k=0 j=1

2

q αk |z|k Qn < +∞. (qj ; qj )k r=1 (qr , qrνr +1 ; qr )k

Similarly, we apply Lemmas 2.3 and 2.4 to establish that the function l ∞ Y m X (qjj ; qj )k k=0 j=1

2

q αk z k Qn (qj ; qj )k r=1 (qr , qrνr +1 ; qr )k

has no zeros outside the set {x + yi|x ∈ (−∞, 0], y = 0}. In view of [10, Lemma 14.1.4], this function has infinitely many zeros. Then (ii) is proved. Finally, we give a proof of (iii). Let qj = qr = q and α = n + m/2. Using the Hˆopital’s rule, we deduce that (q lj ; q)k (lj )k 1 (1 − q)2k = = . , lim ν +1 r q→1 (q; q)k k! q→1 (q, q ; q)k k!(νr + 1)k lim

Then lim

q→1

Qm

lj (n+m/2)k 2 j=1 (q ; q)k q Qn νr +1 ; q) (q; q)m+n k k r=1 (q

(1 − q)2kn

· zk

It is easy to see from

lq l−1 ≤

1 − ql = 1 + q + q 2 + · · · + q l−1 ≤ l 1−q

that l!q (2) ≤ l

Then

Qm k j=1 (lj )k · z Q . = (k!)m+n nr=1 (νr + 1)k

(q; q)l ≤ l!. (1 − q)l

l 2 q −(2 ) (1 − q)l q −l /2 1 ≤ ≤ ≤ . l! (q; q)l l! l! l

l

Combining this and the fact that (1−q) ≤ (1−q) for b ≥ 1 gives (q b ;q)l (q;q)l (1 − q)2kn Qm (q lj ; q)k q αk2 · z k |z|k j=1 Q ≤ n νr +1 ; q) k!m+2n (q; q)m+n k k r=1 (q 10

This, by Lemma 2.3, shows that Q Qm ∞ ∞ lj (n+m/2)k 2 k X (1 − q)2kn m · zk X j=1 (q ; q)k q j=1 (lj )k · z Q Q lim = . n n m+n m+n νr +1 ; q) q→1 (ν (k!) (q (q; q) r + 1)k k r=1 k r=1 k=0 k=0

converges uniformly in in any compact subset of C. It follows from Lemma 2.4 that the function Qm ∞ k X j=1 (lj )k · z Q (k!)m+n nr=1 (νr + 1)k k=0

has no zeros outside the set {x + yi|x ∈ (−∞, 0], y = 0}. Set Qm j=1 (lj )k Q ak = . n m+n (k!) r=1 (νr + 1)k It is easily seen from the Stirling’s formula [4] that − log ak = 2n k→∞ k log k lim

which, by (1.1), means that ρ

∞ X k=0

! Qm k (l ) · z 1 1 j k j=1 Qn = ≤ . m+n (k!) 2n 2 r=1 (νr + 1)k

Hence, by [10, Theorem 1.2.5], the function Qm ∞ k X j=1 (lj )k · z Q (k!)m+n nr=1 (νr + 1)k k=0

has infinitely many zeros. Then this function has only infinitely many real zeros and all of them are negative, which proves (iii). This completes the proof of Theorem 1.1. P hn (x,y|q) αn2 n z , where Proof of Theorem 1.2. We first consider the function ∞ n=0 (q;q)n q 0 < x, y < 1 and α > 0. From [12, Theorem 3.1, (3.1)], we know that ∞ X hn (x, y|q) n=0

(q; q)n

tn =

1 (xt, yt; q)∞

for max{|xt|, |yt|} < 1. Then, by Lemma 2.1, we have the sequence  ∞ hn (x, y|q) (q; q)n n=0 11

k2

is a PF sequence. It follows from the fact that { qk! }∞ k=0 is a PF sequence and (iii) in Lemma 2.2 that  N hn (x, y|q) αn2 q (q; q)n n=0 is also a PF sequence. So, by (i) of Lemma 2.2, we see that N X hn (x, y|q) n=0

(q; q)n

2

q αn z n

has only nonpositive zeros. We know that N X hn (x, y|q) n=0

(q; q)n

2

q αn z n →

∞ X hn (x, y|q) n=0

(q; q)n

2

q αn z n

point-wise in Ω. It is easy to see that for 0 < q < 1, α > 0 and each N ∈ N, z ∈ C, ∞ ∞ X hn (x, y|q) αn2 n X hn (x, y|q) αn2 n q z ≤ q |z| < +∞. (q; q) (q; q) n n n=0 n=0 Applying Lemma 2.3, we find that N X hn (x, y|q) n=0

(q; q)n

q

αn2 n

z →

∞ X hn (x, y|q) n=0

(q; q)n

2

q αn z n

uniformly on every compact subset of Ω and then applying Lemma 2.4, we deduce that the function ∞ X hn (x, y|q) αn2 n q z 6= 0 (q; q)n n=0 in Ω which means that

∞ X hn (x, y|q) n=0

(q; q)n

2

q αn z n

has no zeros outside the set {x + yi|x ∈ (−∞, 0], y = 0}. We use [9, (1.3.15)] to get n

0≤

hn (x, y|q) X xk y n−k = (q; q)n (q; q)k (q; q)n−k k=0 ≤

∞ X k=0

=

∞ xk X y k (q; q)k k=0 (q; q)k

1 . (x, y; q)∞ 12

Then, by Lemma 2.5 and [10, Lemma 14.1.4], we attain that the function ∞ X hn (x, y|q)

(q; q)n

n=0

2

q αn z n

has infinitely many zeros. Therefore, the function ∞ X hn (x, y|q)

(q; q)n

n=0

2

q αn z n

has infinitely many real zeros and all P of them are negative. gn (x,y|q) αn2 n z , where −1 < x, y < 0 We now investigate the function ∞ n=0 (q;q)n q 1 and α ≥ 2 . According to [12, Theorem 3.1, (3.2)], we have ∞ X

(−1)

n=0

n gn (x, y|q)q

(n2 )

(q; q)n

tn = (x, y; q)∞,

which, by Lemma 2.1, implies that (

gn (x, y|q)q ( 2 ) (−1)n (q; q)n n

)∞

n=0

is a PF sequence, namely, (

gn (x, y|q)q (−1)n (q; q)n

n2 2

)∞

n=0

k2

is a PF sequence. So, by the fact that { qk! }∞ k=0 is a PF sequence and (iii) in Lemma 2.2, ( )N αn2 g (x, y|q)q n (−1)n (q; q)n n=0

is a PF sequence, which, by (i) of Lemma 2.2, means that 2 N X gn (x, y|q)q αn n (−1)n z (q; q) n n=0

has only nonpositive zeros. It is obvious that N X n=0

∞

2

(−1)n

2

X gn (x, y|q)q αn n gn (x, y|q)q αn n z → (−1)n z (q; q)n (q; q) n n=0 13

point-wise in Ω. For 0 < q < 1, α > 0 and each N ∈ N, z ∈ C, ∞ ∞ X X αn2 αn2 g (x, y|q)q n n n gn (x, y|q)q n (−1) z ≤ |z|n < +∞. (−1) (q; q)n (q; q)n n=0

n=0

Then, by Lemma 2.3, N X

(−1)

n gn (x, y|q)q

n=0

αn2

(q; q)n

n

z →

∞ X

(−1)

n gn (x, y|q)q

n=0

(q; q)n

αn2

zn

uniformly on every compact subset of Ω. We apply Lemma 2.4 to derive that the function 2 ∞ X gn (x, y|q)q αn n z 6= 0 (−1)n (q; q) n n=0

in Ω. This shows that

∞ αn2 X n gn (x, y|q)q zn (−1) (q; q) n n=0

has no zeros outside the set {x + yi|x ∈ (−∞, 0], y = 0}. It is clear that n X g (x, y|q) |x|k |y|n−k n n (−1) ≤ (q; q)n k=0 (q; q)k (q; q)n−k

∞ ∞ X |x|k X |y|k ≤ (q; q)k k=0 (q; q)k k=0

=

1 . (|x|, |y|; q)∞

By Lemma 2.5 and [10, Lemma 14.1.4], the function ∞ αn2 X n gn (x, y|q)q zn (−1) (q; q) n n=0

has infinitely many zeros. Hence,

2 ∞ X gn (x, y|q)q αn n (−1)n z (q; q) n n=0

has infinitely many real zeros and all of them are negative, namely, 2 ∞ X gn (x, y|q)q αn

n=0

(q; q)n 14

zn

has infinitely many real zeros and all of them are positive. This finishes the proof of Theorem 1.2. Proof of Theorem 1.3. For 0 < q < 11 . Put 2α

Ak = Then

which means that

(a1 )k (a2 )k · · · (ar )k αk2 q . (b1 )k (b2 )k · · · (bs )k

r s Y A2k−1 ai + k − 2 Y bi + k − 1 −2α = q Ak Ak−2 a + k − 1 j=1 bi + k − 2 i=1 i

A2k−1 = q −2α > 4. k→∞ Ak Ak−2 So there exists a positive integer K0 such that lim

A2k−1 >4 Ak Ak−2 for k ≥ K0 . It follows from [7, Theorem B] that ∞ X (a1 )k (a2 )k · · · (ar )k αk2 k q z (b1 )k (b2 )k · · · (bs )k k=K

has only negative zeros for any K ≥ K0 . On the other hand, by the Stirling’s formula and (1.1), we have ! ∞ X (a1 )k (a2 )k · · · (ar )k αk2 k ρ = 0, q z (b 1 )k (b2 )k · · · (bs )k k=K which, by [10, Theorem 1.2.5], implies that the function ∞ X (a1 )k (a2 )k · · · (ar )k αk2 k q z (b 1 )k (b2 )k · · · (bs )k k=K

has infinitely many zeros. Hence, this function has only infinitely many real zeros and all of them are negative. This concludes the proof of Theorem 1.3. Acknowledgement. The author would like to thank Prof. Ruiming Zhang for his helpful discussion, comments and suggestions. This work was supported by the Initial Foundation for Scientific Research of Northwest A&F University (No. 2452015321). 15

References [1] L. Ahlfors, Complex analysis, 3rd edition, McGraw-Hill, 1979. [2] M. Aissen, A. Edrei, I.J. Schoenberg, and A.M. Whitney, On the generating functions of totally positive sequences, Proc. Nat. Acad. Sci. U.S.A. 37 (1951), 303–307. [3] G.E. Andrews, The Theory of Partitions, Cambridge University Press, Cambridge, 1998. [4] G.E. Andrews, R. Askey and R. Roy, Special functions, Cambridge University Press, Cambridge, 1999. [5] R.P. Boas, Entire Functions, Academic Press, 1954. [6] J. Carnicer, J.M. Pe˜ na and A. Pinkus, On some zero-increasing operators, Acta Math. Hungar. 94(2002) 173–190. [7] D.K. Dimitrov and J.M. Pe˜ na, Almost strict total positivity and a class of Hurwitz polynomials, Journal of Approximation Theory 132 (2005) 212–223. [8] K. Driver, K. Jordaan and A. Martin´ınez-Finkelshtein, P´ olya frequency sequences and real zeros of some 3 F2 polynomials, J. Math. Anal. Appl. 332 (2007), 1045– 1055. [9] G. Gasper and M. Rahman, Basic hypergeometric series, Cambridge University Press, Cambridge, 2004. [10] M.E.H. Ismail, Classical and quantum orthogonal polynomials in one variable, Cambridge University Press, Cambridge, 2005. [11] M.E.H. Ismail and R. Zhang, q-Bessel functions and Rogers-Ramanujan type identities, Proceedings of the American Mathematical Society, in production. [12] Z.-G. Liu, On the q-partial differential equations and q-series, The Legacy of Srinivasa Ramanujan, 213–250, Ramanujan Math. Soc. Lect. Notes Ser., 20, Ramanujan Math. Soc., Mysore, 2013. [13] O.M. Katkova, T. Lobova and A. Vishnyakova, On power series having sections with only real zeros, Comput. Methods Funct. Theory 3 (2003), 425–441. [14] G. P´ olya, G. Szeg˝ o, Problems and theorems in analysis II, Springer-Verlag, New York, 1976. [15] L.J. Rogers, On a three-fold symmetry in the elements of Heine’s series, Proc. London. Math. Soc. 24 (1893) 171–179. [16] G. Szeg˝ o, Ein Betrag zur Theorie der Thetafunktionen, Sitz. Preuss. Akad. Wiss. Phys. Math. 19 (1926) 242–252. [17] E.C. Titchmarsh, The theory of functions, corrected second edition, Oxford University Press, Oxford, 1964. [18] R. Zhang, Zeros of Ramanujan type entire functions, Proceedings of the American Mathematical Society, in production.

16