A basic analog of a theorem of Polya

Math. Z. (2008) 258:363–379 DOI 10.1007/s00209-007-0176-6

Mathematische Zeitschrift

A basic analog of a theorem of Pólya M. H. Annaby · Z. S. Mansour

Received: 8 November 2006 / Accepted: 29 January 2007 / Published online: 25 April 2007 © Springer-Verlag 2007

Abstract We derive a q-analog of a theorem of George Pólya (1918), 0 < q < 1, concerning the zeros of basic cosine and sine transforms. The results are established without further restrictions on q. Keywords q-Functions · q-Fourier transforms · Zeros of entire functions Mathematics Subject Classification (2000) 30C15 · 33D15 · 33D60 · 42A38

1 Introduction and preliminaries In the following, q is a positive number which is less than 1. We are concerned with the derivation of a q-analog of the following theorem of Pólya [17]. Theorem A Let f (t) ∈ L 1 (0, 1) be a positive increasing function on [0, 1]. Then: 1.

the zeros of the entire functions of exponential type
1 U (z) =

f (t) cos(zt) dt, V (z) = 0

2.


1 f (t) sin(zt) dt

(1.1)

0

are real, infinite and simple. U (z) is an even function which has no zeros in [0, π2 ), and its positive zeros are situated in the intervals (πk − π/2, πk + π/2), 1
k < ∞, one zero in each interval. The odd function V (z) has only one zero in [0, π), and its positive zeros are situated in the intervals (πk, π(k + 1)), 1
k < ∞, one zero in each interval.

M. H. Annaby (B) · Z. S. Mansour Department of Mathematics, Faculty of Science, Cairo University, Giza, Egypt e-mail: [email protected] Z. S. Mansour e-mail: [email protected]

123

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M. H. Annaby, Z. S. Mansour

In [3], a basic analog of Theorem A has been introduced provided that further conditions are imposed on q. We aim in the present paper to give another q-analog without more restrictions on q. We first introduce some notations as well as the main results of [3]. The Jackson q-integration is defined by, cf. [13],
1 f (t) dq t := (1 − q)

∞ 

q n f (q n ),

(1.2)

n=0

0

provided that the series converges. By L q1 (0, 1) we mean the Banach space of all complex valued functions defined on [0, 1] such that
1  f  :=

| f (t)| dq t < ∞.

(1.3)

0

For f ∈ L q1 (0, 1), we denote by Ak ( f ) the q-moments of f , i.e.
1 Ak ( f ) :=

t k f (t) dq t, k ∈ N.

(1.4)

0

We denote by ck ( f ), c f , C f the constants ck ( f ) :=

A2k+2 ( f ) , k ∈ N, c f := inf ck ( f ), C f := sup ck ( f ) k∈N A2k ( f )(1 − q 2k+1 )(1 − q 2k+2 ) k∈N (1.5)

and by bk ( f ), b f , B f to bk ( f ) :=

A2k+3 ( f ) , k ∈ N, b f := inf bk ( f ), B f := sup bk ( f ). k∈N A2k+1 ( f )(1 − q 2k+3 )(1 − q 2k+2 ) k∈N (1.6)

The constants c f , C f , b f , B f exist and they are finite positive numbers, cf. [3]. The q-shifted factorial, see [7], is defined for a ∈ C to be ⎧ 1, n = 0, ⎪ ⎪ ⎪ ⎨ (a; q)n := n−1 (1.7)  ⎪ i ⎪ (1 − aq ), n = 1, 2, . . . . ⎪ ⎩ i=0

The limit lim (a; q)n always exists and will be denoted by (a; q)∞ . Moreover (a; q)∞ has n→∞ the following series representation, see e.g. [7, p. 11], (a; q)∞ =

∞  (−1)n q n(n−1)/2 n=0

an . (q; q)n

(1.8)

The third type of q-Bessel functions of Jackson is defined by, cf. e.g. [16], Jν(3) (z; q) = z ν

∞ n(n+1) (q ν+1 ; q)∞  z 2n (−1)n q 2 , ν > −1. (q; q)∞ (q; q)n (q ν+1 ; q)n n=0

123

(1.9)

A basic analog of a theorem of Pólya

365

This function is called in some literature the Hahn-Exton q-Bessel function, see [15,16,22]. It is also called the 1 φ1 q-Bessel function, cf. [14]. Since the other types of the q-Bessel (1) (2) functions, i.e. Jν (·; q), Jν (·; q), see e.g. [9,11,12,19], will not be considered throughout (3) this paper, we use the notation Jν (·; q) for Jν (·; q). The basic trigonometric functions cos(z; q) and sin(z; q) are defined on C by  2 2  1/2 q ; q ∞  −1/2 √ cos(z; q) :=  zq (1.10) (1 − q) J−1/2 z(1 − q)/ q ; q 2 , 2 q; q ∞  2 2  q ;q ∞ (z(1 − q))1/2 J1/2 z(1 − q); q 2 . sin(z; q) :=  2 q; q ∞

(1.11)

They are q-analogs of the cosine and sine functions, [2,7]. The θ -function is defined for z ∈ C\ {0} , |q| < 1 to be, cf. [10], θ (z; q) :=

∞ 

2

q n zn .

(1.12)

n=−∞

The following identity is introduced by Jacobi in 1829. It is called Jacobi’s triple product identity, see [7] ∞ 

q n z n = (q 2 ; q 2 )∞ (−qz; q 2 )∞ (−qz −1 ; q 2 )∞ , 2

(1.13)

n=−∞

where z ∈ C\ {0} and 0 < |q| < 1. The next two lemmas are of elementary nature and will be used throughout the paper. Lemma 1.1 For a positive sequence {z m }∞ m=1 , if z m −→ 1 as m −→ ∞, then inf m∈N z m > 0. Proof Since z m −→ 1 as m −→ ∞, then for 0 < ε < 1, there exists m 0 ∈ N such that |z m − 1| < ε for all m  m 0 . Hence z m  1 − ε > 0 for all m  m 0 . That is inf m
m 0 z m  1 − ε > 0. Therefore 

inf z m = min z 1 , z 2 , . . . , z m 0 , inf z m > 0. m∈N

m
m 0

  Lemma 1.2 If z ∈ C such that 0
|z| < 1, then log |1 − z|
log

1 . 1 − |z| Proof Let z ∈ C such that 0
|z| < 1. Since |1 − z|  1 − |z| > 0, then we have log |1 − z|  log(1 − |z|) = − log

1 . 1 − |z|

(1.14)

(1.15)

Also |1 − z|
1 + |z|. Then log {|1 − z|(1 − |z|)}
log(1 − |z 2 |)
0. Consequently log |1 − z|
− log(1 − |z|) = log Combining (1.15) and (1.16) we obtain (1.14).

1 . (1 − |z|)

(1.16)  

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M. H. Annaby, Z. S. Mansour

n Definition 1.3 Let f (z) := ∞ n=0 an z be an entire function. The maximum modulus is defined by, cf. [5], M(r ; f ) := sup {| f (z)| : |z| = r } . (1.17) The order of f , ρ( f ), is defined by ρ( f ) := lim sup r →∞

log log M(r, f ) n log n . = lim sup −1 log r n→∞ log |an |

(1.18)

Theorem B [5] If ρ( f ) is finite and is not equal to a positive integer, then f has infinitely many zeros. One can easily see that Jν (·; q 2 ) is entire function of order zero. Therefore from Theorem B, it has infinitely many zeros. Koelink and Swarttouw proved that Jν (·; q 2 ) has only (ν) real and simple zeros, cf. [15]. Let wm denote the positive zeros of Jν (·; q 2 ) in an increasing + order of m ∈ Z . In [1], Abreu et al. proved that if q 2ν+2 < (1 − q 2 )2 , then   2m+2ν log 1 − q1−q 2m (ν) , (q) = q −m+2m (ν) , 0 < m (ν) < wm 2 log q (1.19)  ∞  log q 2ν+2 ; q 2 ∞ . m (ν) = 4 log q m=0

(ν)

Moreover wm ∼ q −m when m → ∞ without the restriction q 2ν+2 < (1 − q 2 )2 . Applying these formulae to the cases ν = ∓1/2 respectively and using (1.10) and (1.11), we obtain the following estimates if q < (1 − q 2 )2 then xm =

q −m+1/2+2m (−1/2) , m  1, 1−q

(1.20)

if q 3 < (1 − q 2 )2 then ym =

q −m+2m (1/2) , m  1, 1−q

(1.21)

where hereafter xm and ym denote the positive zeros of cos(z; q) and sin(z; q) respectively. Moreover q −m+1/2 q −m xm ∼ , ym ∼ as m → ∞ (1.22) (1 − q) (1 − q) without further restrictions on q. The results for ym coincides with those of Bustoz and Cardoso [6]. A study of asymptotic formulae for the zeros of another class of basic trigonometric functions is given in [20,21]. Theorems 1.4 and 1.5 are the q-analogs of Theorem A which the authors introduced in [3]. Theorem 1.4 Let f ∈ L q1 (0, 1) be positive on {0, q n , n ∈ N}. If q −1 (1 − q)

cf > 1, Cf

(1.23)

then the zeros of the entire function of order zero
1 U f (z) :=

f (t) cos(t z; q) dq t, z ∈ C, 0

123

(1.24)

A basic analog of a theorem of Pólya

367

simple are real,  and infinite. Moreover U f (z) is an even function with no zeros in the interval 0, q −1/2 / C f (1 − q) and its positive zeros lie in the intervals   q −r q −r −1  ,  , r = 0, 1, . . . , (1.25) (1 − q) C f (1 − q) C f one zero in each interval. Theorem 1.5 Let f ∈ L q1 (0, 1) be positive on {0, q n , n ∈ N}. If q −1 (1 − q)

bf > 1, Bf

(1.26)

then the zeros of the entire function of order zero
1 V f (z) :=

f (t) sin(t z; q) dq t, z ∈ C,

(1.27)

0

are  real, simpleand infinite. Moreover the odd function V f (z) has only one zero in the interval −1 and its positive zeros are located in the intervals 0, √ q B f (1−q)



q −r −1/2 q −r +1/2  ,  (1 − q) B f (1 − q) B f

 , r = 0, 1, . . . ,

(1.28)

one zero in each interval. The previous two theorems give a basic analog of Theorem A with the restrictions (1.23), (1.26) on q, respectively. These restriction makes q nearer to 0 than 1. For instance if ⎧ 1 ⎪ ⎪ , t = 1, ⎨ 1−q f (t) := ⎪ ⎪ ⎩ 0, otherwise, then V f (z) = sin(z; q) and we obtain the locations of the zeros of sin(z; q) for 0 < q < γ0 , where γ0 ≈ 0.429052 is the root of (1 − q)(1 − q 2 )(1 − q 3 ) − q, q ∈ (0, 1). This is more a restrictive condition than that of [6], which gives a similar result but for 0 < q < β0 , β0 ≈ 0.67104 is the root of (1 − q 2 )2 − q 3 , q ∈ (0, 1). The same for the zeros of cos(z; q), cf. [1,3]. However the results of [3] considered a more general situation. In the following we employ the technique of Bergweiler and Hayman [4] to derive a basic analog of Theorem A of Pólya without further restrictions on q. This is done in Sect. 3. Section 2 contains asymptotics of the zeros of q-cosine and q-sine functions.

2 Asymptotic relations In this section we study the asymptotic behavior of the zeros of the basic trigonometric functions cos(z; q) and sin(z; q). For convenience we define some notations concerning entire functions, cf. e.g. [5,8]. Let f, g be entire functions , we say that  f (z) = O g(z) , as z → ∞, (2.1)

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M. H. Annaby, Z. S. Mansour

 if f (z) g(z) is bounded in a neighborhood of ∞. We also write f (z) ∼ g(z), as z → ∞, if

lim

z→∞

| f (z)| = 1. |g(z)|

(2.2)

The following theorem is taken from [4] and will be needed in the sequel. It is concerned with the asymptotic behavior of the θ -function defined in (1.12). Theorem C Suppose that 0 < |q| < 1 and that z ∈ C, |z|  1. For m ∈ Z+ let Am be the annulus defined by   Am := z ∈ C, q −2m+2
|z| < |q|−2m . (2.3) Then we have, uniformly as m → ∞, log |θ (z; q)| =

−(log |z|)2 + log |1 + q 2m−1 z| + O(1), z ∈ Am . 4 log |q|

(2.4)

The theorem says that θ (z; q) is uniformly large except in small neighborhoods of its zeros −q 1−2m , m ∈ Z+ . To see this it suffices to take z = −q 1−2m in the right-hand side of (2.4) when m is sufficiently large and notice that it will be negative. Using the technique of Bergweiler and Hayman [4], we will derive explicit asymptotic formulae for the zeros of cos(z; q) and sin(z; q). Before this we state Rouché theorem which will be used in the sequel, see e.g. [18, p. 142]. Theorem D Suppose that φ(z) and ψ(z) are two functions analytic in a domain  of C and continuous on the closed domain . Assume that |φ(z)|
|ψ(z)| on the boundary of , ∂. Then the functions φ(z) and φ(z) + ψ(z) have the same number of zeros inside . Theorem 2.1 If {xm } and {ym } are the positive zeros of cos(z; q) and sin(z; q), respectively, then we have for sufficiently large m,  xm = q −m+1/2 (1 − q)−1 1 + O(q m ) , (2.5)  (2.6) ym = q −m (1 − q)−1 1 + O(q m ) . Proof We prove the theorem for the zeros of cos(z; q). The proof for the zeros of sin(z; q) is similar and is omitted. Let R(z) be the difference R(z) := cos(z; q) − Hence ∞ 

R(z) =

δ2k z 2k , δ2k :=

k=−∞

1 θ (−z 2 (1 − q)2 ; q), z ∈ C\ {0} . (q; q)∞

⎧ k2 2k  ⎪ k q (1 − q) ⎪ (−1) (q 2k+1 ; q)∞ − 1 , ⎪ ⎪ ⎨ (q; q)∞ ⎪ ⎪ ⎪ ⎪ ⎩

(2.7)

k  0, (2.8)

2

q k (1 − q)2k , (−1)k+1 (q; q)∞

k < 0.

From (1.8), we have for k ∈ Z+ ∞ ∞  (2k+1) j  q q j ( j−1)/2 2k+1
q 2k+1 ; q)∞ − 1 = (−1) j q j ( j−1)/2 (q (q; q) j (q; q) j j=1 j=1


∞ q 2k+1 q 2k+1  ( j−1)/2 q = . √ (q; q)∞ (1 − q)(q; q)∞ j=1

123

(2.9)

A basic analog of a theorem of Pólya

369

We conclude from (2.8) and (2.9) that q k +2k (1 − q)2k , k ∈ Z. |δ2k |
√ (1 − q)(q; q)2∞ 2

(2.10)

Thus for z ∈ C\ {0}, |R(z)|
=

∞  1 2 q k +2k (1 − q)2k |z 2k | √ (1 − q)(q; q)2∞ −∞

1 θ (q 2 (1 − q)2 |z|2 ; q). √ (1 − q)(q; q)2∞

(2.11)

Taking the logarithm of both sides of (2.11) yields log |R(z)|
log θ (q 2 (1 − q)2 |z|2 ; q) + log

1 , z ∈ C\ {0} . √ (1 − q)(q; q)2∞

(2.12)

We deduce from (2.4) that there exists m 0 ∈ N and a constant C > 0 such that if z ∈ Am , m  m 0 , then (log |z|)2 (log |z|)2 +log |1 + q 2m−1 z| − C
log |θ (z; q)|
− + log |1 + q 2m−1 z| + C. 4 log q 4 log q (2.13)  2 Thus if qz(1 − q) ∈ Am , m  m 0 , i.e. −

q −2m+2
q 2 |z|2 (1 − q)2 < q −2m , m  m 0 ,

(2.14)

then replacing z in (2.13) by q 2 (1 − q)2 |z|2 , we obtain log |θ (q 2 (1 − q)2 |z|2 ; q)|
−

(log |z|2 (1 − q)2 q 2 )2 + log 1 + q 2m+1 (1 − q)2 |z|2 + C. 4 log q (2.15)

Since log(1 + x) < x, x > 0, then (2.14) implies log 1 + q 2m+1 (1 − q)2 |z|2
q 2m+1 (1 − q)2 |z|2
q −1 .

(2.16)

Substituting in (2.15), we obtain for z ∈ C, which satisfies (2.14), log |θ (q 2 (1 − q)2 |z|2 ; q)|
−

(log |z|(1 − q)q)2 + q −1 + C. log q

Consequently, by (2.12), we obtain  2 log |z|(1 − q)q 1 log |R(z)|
− + q −1 + C + log √ log q (1 − q)(q; q)2∞  2 log |z|(1 − q) =− − log |z|2 + C1 , log q

(2.17)

(2.18)

where  √ C1 := C − log (1 − q)(q; q)2∞ − 2 log(1 − q) − log q + q −1 > 0.

(2.19)

123

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M. H. Annaby, Z. S. Mansour

m , m > m 0 , be the Notice that (2.18) holds uniformly for z satisfying (2.14), m  m 0 . Let A annulus defined by   m := z ∈ C : q −2m+2
|z|2 (1 − q)2
q −2m . (2.20) A m that, r := |z|, The left hand side of (2.13) implies for z ∈ A  2  − log r (1 − q)
log |θ (−(1 − q)2 z 2 ; q)| − log 1 − q 2m−1 z 2 (1 − q)2 + C. log q (2.21) m , Therefore by (2.18), z ∈ A

 2 2 log |R(z)|
log (q; q)−1 ∞ |θ (−(1 − q) z ; q)| − log 1 − q 2m−1 z 2 (1 − q)2 − log r 2 + C2 ,

(2.22)

m , then where C2 = C + C1 + log(q; q)∞ . If z ∈ ∂ A |1 − q 2m−1 z 2 (1 − q)2 |  q 2m−1 |z|2 (1 − q)2 − 1 = q −1 (1 − q), |z| = |1 − q 2m−1 z 2 (1 − q)2 |  1 − q 2m−1 |z|2 (1 − q)2 = (1 − q), |z| = That is

m . log |1 − q 2m−1 z 2 (1 − q)2 |  log(1 − q), z ∈ ∂ A

q −m , 1−q

q −m+1 . 1−q

(2.23) (2.24)

(2.25)

Substituting in (2.22) we obtain  2 2 2  log |R(z)|
log (q; q)−1 ∞ |θ (−(1 − q) z ; q)| − log(1 − q) − log r + C 2 , z ∈ ∂ Am . (2.26) We can also choose m 0 sufficiently large such that − log(1 − q) − log r 2 + C2 < 0. Hence for sufficiently large m  2 2  |R(z)|
(q; q)−1 (2.27) ∞ |θ (−(1 − q) z ; q)| , z ∈ ∂ Am . Now applying Rouché theorem with  2 2 φ(z) = (q; q)−1 ∞ θ −(1 − q) z ; q , ψ(z) = R(z),

(2.28)

m , m > m 0 , we conclude that the functions φ(z) and φ(z) + ψ(z) = cos(z; q)  = A m . Since θ (−(1 − q)2 z 2 ; q) has two simple symmetric have the same number of zeros in A m , we deduce that cos(z; q) has exactly two zeros in A m . Since cos(z; q) and zeros in A cos(z; q) = cos(z; q) are even functions, then these two zeros are real, symmetric and simple. m . Then from (2.7) Assume now that xm , m > m 0 , is the positive zero of cos(z; q) in A 2 2 |R(xm )| = (q; q)−1 ∞ θ (−(1 − q) x m ; q) and we obtain from (2.22) that log 1 − q 2m−1 xm2 (1 − q)2
−2 log xm + C2 .

(2.29)

In other words log 1 − q m−(1/2) xm (1 − q) + log 1 + q m−(1/2) xm (1 − q)
−2 log xm + C2 . (2.30)

123

A basic analog of a theorem of Pólya

Hence

371

2 log 1 − q m−(1/2) xm (1 − q)
−2 log xm + C2 , i.e. C2 log 1 − q m−(1/2) xm (1 − q)
− log xm + . 2

Then

(2.31) (2.32)

1 − q m−(1/2) xm (1 − q) = O(x −1 ). m

(2.33)

This implies that xm ∼ q −m+1/2 (1 − q)−1 for sufficiently large m. More precisely  xm = q −m+1/2 (1 − q)−1 1 + O(q m ) , as m → ∞.

(2.34)  

3 The main results In this section we derive the main results of this paper. We give a q-analog of the result of Pólya mentioned in Sect. 1. For this reason, we derive some asymptotic relations involving basic cosine and sine functions. In Theorem C, the asymptotic behavior of log |θ (z; q)| as z → ∞ is studied in the set of annuli {Am , m  1} defined in (2.3) when m is large enough. We shall study the asymptotic behavior of | cos(z; q)| and | sin(z; q)| in suitable sets of annuli. We begin with the q-cosine function. Let  αm := log |xm /xm+1 | log q, m ∈ Z+ . (3.1) Then αm > 0 for all m ∈ Z+ . Moreover, from (2.5) xm  = q 1 + O(q m ) x m+1 for sufficiently large m. That is lim αm = 1. Therefore from Lemma 1.1 m→∞

0 < α := inf αm
1. m∈Z+

∞ We define two sequences, {am }∞ m=1 , {bm }m=1 , to be α +α m , if αm  = α, am := 2 α if αm = α, 2,

and α b1 := , bm+1 := 2 where m  1. Then 

∞

α −α m , if αm  = α, 2 α , if αm = α, 2

xm q −am = xm+1 q bm+1 , m  1.

(3.2)

(3.3)

(3.4)

(3.5)

The set of annuli Acm m=1 where we derive an asymptotic relation for log | cos(z; q)| is defined to be   Acm := z ∈ C : xm q bm
|z|
xm q −am , m  1, (3.6)   dividing the region z ∈ C : |z|  q α/2 x1 into annuli.

123

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M. H. Annaby, Z. S. Mansour

Theorem 3.1 Assume that |z|  q x1 and Acm , m  1, be the annulus defined by (3.6). Then we have the asymptotic relation log | cos(z; q)| = −

z2 (log |z(1 − q)|)2 + log |1 − 2 | + O(1), z ∈ Acm , log q xm

(3.7)

uniformly as m → ∞. Proof First of all, it should be noted that z ∈ Acm , m  1, if and only if |z| = xm q gm (t) , gm (t) := −(am + bm )t + bm , t ∈ [0, 1].

(3.8)

Let m  1 be fixed and z ∈ Acm . Then there exists t ∈ [0, 1] such that |z| = xm q gm (t) = xm q −(am +bm )t+bm .

(3.9)

Since cos(z; q) is an entire function of order zero, then, see e.g. [5],  ∞   z2 1− 2 . cos(z; q) = xn

(3.10)

n=1

Thus

∞ cos(z; q) m−1 2 2   1 − z + 1 − z = log log log 1 − ( xz )2 xn2 xn2 m n=1

=

m−1  n=1

Set

n=m+1

z log | |2 + xn

m−1  n=1

∞  x2 z2 log 1 − n2 + log 1 − 2 . z xn n=m+1

1

γn := q n− 2 (1 − q)xn − 1, n = 1, 2, . . . . From (2.5), γn =

O(q n )

(3.11)

(3.12)

as n → ∞. Define the constants l, L to be l := inf 1 + γn , n∈Z+

L := sup 1 + γn .

(3.13)

n∈Z+

Since 1 + γn > 0, for all n ∈ Z+ and 1 + γn → 1 as n → ∞, then l > 0. We define constants K 1 , K 2 , K 3 as follows K 1 :=

L , l

K 2 := q −a , a = sup an , n∈Z+

K 3 := sup αn − α. n∈Z+

(3.14)

Then q gn (t)
q −an
K 2 for all n  1 and t ∈ [0, 1]. Let n ∈ Z+ , n  m + 1. From (3.12), we have for z ∈ Acm z xm q gm (t) 1 + γm gm (t) = =q n−m q
K 1 q −am q n−m
K 1 K 2 q n−m . (3.15) x x 1 + γn n n Consequently ∞  n=m+1

 2  ∞ ∞   z 2 z z 2 log 1 − 2
log 1 +
x xn xn n n=m+1


K 12 K 22

n=m+1

∞  n=m+1

123

q 2(n−m)
K 12 K 22

q2 . 1 − q2

(3.16)

A basic analog of a theorem of Pólya

373

As for n = 1, 2, . . . , m − 1, xn xn q −gm (t) 1 + γn −gm (t) = = q m−n q
K 1 q −bm q m−n . z xm 1 + γm

(3.17)

From (3.3) and (3.4) we obtain q −bm = q −αm−1 q am−1
q −α0 q α = q −K 3 , α0 := sup αn . n
n 0

xn
K 1 q −K 3 q m−n , n = 1, 2, . . . , m − 1. z

Therefore

(3.19)

xn xn q −gm (t) xm−1 q −gm (t) =
, n = 1, 2, . . . , m − 1. z xm xm

Moreover

(3.18)

(3.20)

But from (3.5) xm−1 = q am−1 +bm and q −gm (t) = q am t−(1−t)bm
q −bm . xm xn
q am−1 < q α < 1, z ∈ Ac , n = 1, 2, . . . , m − 1. m z

Then

(3.21)

So from (1.14) we obtain ⎞ ⎛ ∞ 2  x 1 x n log 1 − n
log   = log ⎝1 + | |2 j ⎠ xn2 z 2 z 1 − | z2 | j=1  = log 1 +

| xzn |2 1 − | xzn |2




| xzn |2

| x n |2

z
1 − | xzn |2 1 − q 2α

(3.22)

K 12 q −2K 3 2(m−n) q 2(m−n) 1 + γn 2 −2gm (t)
q
q . 1 − q 2α 1 + γm 1 − q 2α Hence m−1  2 −2K 3 m−1 2  K 12 q −2K 3 q 2 x K q log 1 − n2
1 2α q 2(m−n)
. z 1−q (1 − q 2α )(1 − q 2 ) n=1

(3.23)

n=1

Combining (3.11), (3.16), and (3.23) we end for z ∈ Acm with 2 cos(z; q) m−1 z  + O(1), as m → ∞. = log log z x 2 1 − (x ) n m

(3.24)

n=1

123

374

M. H. Annaby, Z. S. Mansour

Now m−1  n=1

2 m−1 m−1   z log = log |z|2 − log |xn |2 xn n=1

n=1

m−1 

= 2(m − 1) log |z(1 − q)| − 2

(−n + 1/2) log q − 2

n=1

m−1 

log |1 + γn |

n=1

= 2(m − 1) log |z(1 − q)| + (m − 1)2 log q − 2

m−1 

log |1 + γn |.

(3.25)

n=1

Since γn = O(q n ) as n → ∞, then there exists a constant K 4 > 0 and n 0 ∈ N such that |γn |
K 4 q n , for all n  n 0 . Let   K 5 := max q −1 |γ1 |, . . . , q −m 0 +1 |γm 0 −1 |, K 4 . Thus |γn |
K 5 q n for all n ∈ Z+ , and m−1 

log |1 + γn |


n=1

m−1 

 log 1 + |γn |

n=1




m−1 

|γn |


n=1

∞ 

|γn |
K 5

n=1

∞  n=1

(3.26) q . q
K5 1−q n

Since |z(1 − q)| = q −m+1/2 |1 + γm |q gm (t) , then m−1=−

log |z(1 − q)| + dm , dm = −1/2 + log(1 + γm ) + gm (t) log q. log q

(3.27)

A simple computation yields 2(m − 1) log |z(1 − q)| + (m − 1)2 log q = −

(log |z(1 − q)|)2 + dm2 log q log q

(log |z(1 − q)|)2 =− + O(1), log q

(3.28)

as z → ∞. The theorem follows by combining (3.24), (3.25), and (3.28). Corollary 3.2 For r := |z| −→ ∞ we have 



(log r (1 − q))2 M(r ; cos(z; q)) = O exp − log q

 

 .

(3.29)

Proof From (3.7) we conclude that there exists m 0 ∈ N such that for all m  m 0 we have −

(log |z(1 − q)|)2 z2 + log |1 − 2 | − δ
log | cos(z; q)| log q xm


−

(log |z(1 − q)|)2 z2 + log |1 − 2 | + δ, log q xm

z ∈ Acm . From (3.6) we obtain q 2bm


123

|z|2
q −2am . xm2

(3.30)

(3.31)

A basic analog of a theorem of Pólya

Hence

375

  z 2 |z|2 log 1 − 2
log 1 + 2
q −2am , m  1. xm xm

(3.32)

Since the sequence {am }∞ m=1 is bounded, then there exists a constant R > 0 such that log | cos(z; q)|
− Thus

(log |z(1 − q)|)2 + R, z ∈ Acm , m  m 0 . log q

(log |z(1 − q)|)2 log q | cos(z; q)|
e R e , z ∈ Acm , m  m 0 , −

(3.33)

(3.34)  

proving the corollary.

The same study may be carried out for sin(z; q). For this aim we define another set of annuli. Let (3.35) βm := log |ym /ym+1 |/ log q, m ∈ Z+ . ∞ Then βm → 1 and β := inf βm > 0. Let {cm }∞ m=1 and {dm }m=1 be the sequences defined m∈Z+

by

⎧β +β m ⎪ , ⎨ 2 cm := ⎪ ⎩β 2,

if βm  = β, (3.36) if βm = β,

⎧ ⎨ βm − β β , d1 := , dm+1 := 2 ⎩ β, 2 2

and

if βm  = β,

(3.37)

if βm = β,

where m  1. Then we have ym q −cm = ym+1 q dm+1 , m  1.  ∞ We define the set of annuli Asm m=1 to be   Asm := z ∈ C : ym q dm
|z|
ym q −cm , m  1,

(3.38)

(3.39)



 dividing the region z ∈ C : |z|  q β/2 y1 . Theorem 3.3 Assume that |z|  q x1 . Let Asm , m  1, be the annulus defined in (3.39). Then we have, uniformly as m → ∞ log | sin(z; q)| = −

(log |z(1 − q)|)2 z2 + log |1 − 2 | + O(1), z ∈ Asm . log q ym

(3.40)

The proof is similar to that of Theorem 3.1, and it is omitted. Also we have the following corollary Corollary 3.4 For r := |z| −→ ∞ we have 



(log r (1 − q))2 M(r ; sin(z; q)) = O exp − log q

 .

(3.41)

123

376

M. H. Annaby, Z. S. Mansour

Now we state and prove a q-analog of Pólya’s theorem in the case of the basic cosine transform. Theorem 3.5 Let f ∈ L q1 (0, 1) and U f (z) be defined for z ∈ C by
1 U f (z) :=

f (t) cos(t z; q) dq t, 0 < q < 1.

(3.42)

0

Then U f (z) has at most a finite number of non-real zeros and it has an infinite number of real zeros {±ζm }∞ m=1 , ζm > 0, such that ζm ∼ x m as m → ∞. More precisely ζm = xm (1 + O(q m )) as m → ∞.

(3.43)

Proof From the definition of the q-integral (1.2), U f (z) can be written as U f (z) = H (z) + R(z),

(3.44)

where H (z) := (1 − q) f (1) cos(z; q), and R(z) :=

∞ 

q k (1 − q) f (q k ) cos(q k z; q).

(3.45)

k=1

From (3.29) we can find constants r0 , C > 0 such that max | cos(z; q)|
Ce−(log r (1−q))

2 /log q

r =|z|

, for all r > r0 .

(3.46)

Let z ∈ C such that r := |z| > q −1 r0 , then using the maximum modulus principle we have (log qr (1 − q))2 ∞  log q q k (1 − q)| f (q k )| |R(z)|
Ce −

k=1

= q −1 C

e

−(log r (1−q))2 / log q

r 2 (1 − q)2

Thus

(3.47)


q | f (t)| dq t. 0



2 log r (1 − q) log |R(z)|
C1 − 2 log r − , (3.48) log q ⎛ ⎞
q where C1 := log ⎝q −1 (1 − q)−2 C | f (t)| dq t ⎠. Let Acm be the annuls defined in (3.6) 0

and m is sufficiently large. Then 

2 log r (1 − q) z2 log | cos(z; q)| = − + log 1 − 2 + O(1), z ∈ Acm , log q xm

(3.49)

uniformly as m → ∞. Hence there exists a constant C2 > 0 and m 0 ∈ N such that for all z ∈ Acm , m  m 0 we have

123

A basic analog of a theorem of Pólya

377



2 log r (1 − q) z 2 + log 1 − 2 − C2
log | cos(z; q)| − log q xm  2 log r (1 − q) z2 + log 1 − 2 + C2 .
− log q xm

(3.50)

Consequently by (3.48) for z ∈ Acm , m  m 0 , we obtain

z2 log |R(z)|
C1 + C2 − 2 log r + log | cos(z; q)| − log 1 − 2 xm 2 z = log |H (z)| − 2 log r − log 1 − 2 + C3 , x

(3.51)

m

where C3 := C1 + C2 − log(1 − q)| f (1)|. Let Dm 0 be the disk defined by   Dm 0 := z ∈ C : |z| < q bm 0 xm 0 ,

(3.52)

where bm 0 is defined in (3.4) above. Clearly C is split into two disjoint parts via C := Dm 0 ∪ ∪m
m 0 Acm .

If z ∈ ∂ Dm 0 , i.e. |z| = q bm 0 xm 0 , then by (3.51)

log |R(z)|
log |H (z)| − 2 log r − log 1 − q 2bm 0 | + C3 . (3.53) We choose m 0 sufficiently large such that −2 log r − log 1 − q 2bm 0 + C3 < 0. That is log |R(z)| < log |H (z)|, |z| = q bm 0 xm 0 .

(3.54)

Applying Rouché theorem on  = Dm 0 , we conclude that H (z) and U f (z) have the same number of zeros inside Dm 0 . Since cos(z; q) has exactly 2m 0 − 2 zeros inside Dm 0 , then so has U f (z). Consequently it has at most 2m 0 − 2 zeros inside Dm 0 . It remains to search for the zeros of U f (z) in ∪m
m 0 Acm . Indeed, if z ∈ ∂ Acm , m  m 0 , then 2 2 1 − z  1 − q 2bm , |z| = q bm xm , 1 − z  q −2am − 1, |z| = q −am xm . (3.55) xm2 xm2 Thus

  z2 log 1 − 2  min log(1 − q 2bm ), log(q −2am − 1) , z ∈ ∂ Acm . xm

(3.56)

Since the sequence {am }∞ m=1 is a bounded positive sequence, there exists a constant C 4 > 0 such that z2 log 1 − 2  C4 , z ∈ ∂ Acm , m  m 0 . (3.57) xm That is log |R(z)|
log |H (z)| − 2 log r − C4 + C3 . Again, we choose the m 0 large enough such that if r = |z|, z ∈

Acm ,

(3.58) m  m 0 , then

−2 log r + C3 − C4
0. That is |R(z)|
|H (z)|, z ∈ ∂ Acm , m  m 0 .

123

378

M. H. Annaby, Z. S. Mansour

Applying Rouché theorem again on Acm , m  m 0 , we conclude that H (z) and U f (z) have the same number of zeros inside Acm . But cos(z; q) has two simple symmetric zeros there. So we deduce that the even function U f (z) has only two real, symmetric and simple zeros inside Acm , m  m 0 . Now we give the asymptotic behavior of such zeros. Let ζm , be a positive zero of U f (z) in Acm , m is sufficiently large. Then log |H (ζm )| = log |R(ζm )|. Consequently by (3.51) ζm2 (3.59) log 1 − 2
C3 − 2 log |ζm |, m  m 0 . xm Thus

Hence

ζm C3
− log |ζm |. log 1 − xm 2

(3.60)

 1 − ζm = O |ζm |−1 , as m → ∞, xm

(3.61)

i.e. ζm ∼ xm for sufficiently large m. Hence ζm = xm (1 + O(q m )), as m → ∞. Noting that the non-real zeros, if any, lie in Dm 0 , completes the proof.

(3.62)  

Similarly we have the following theorem for the q-sine transforms. Theorem 3.6 If f ∈ L q1 (0, a) and V f (z) be defined for z ∈ C by
1 V f (z) :=

f (t) sin(t z; q) dq t, 0 < q < 1,

(3.63)

0

then V f (z) has at most a finite number of non-real zeros and it has an infinite number of real + simple zeros {±ηm }∞ m=1 , ηm > 0, for all m ∈ Z such that ηm ∼ ym as m → ∞. In other words ηm = ym (1 + O(q m )) as m → ∞. (3.64) Remark The above theorems are q-counterparts of Theorem A above together with that of Pólya, see [18, p. 143], where positivity and monotonicity conditions are omitted. Instead, the condition 0 < | f (1)| < | f (0)| is added. We do not need such a condition in our study since cos(z; q) and sin(z; q) are not bounded. It should be also noted that in Pólya’s result [18, p. 143] no asymptotic behavior of zeros is given. This makes our analogs mixtures of Theorem A of Sect. 1 and problem 199 of [18, p. 143]. Acknowledgements

The authors thank the referee for the constructive comments.

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