Математичнi Студiї. Т.13, №2
Matematychni Studii. V.13, No.2
УДК 517.53
I. N. Peresyolkova
ON DISTRIBUTION OF ZEROS OF GENERALIZED FUNCTIONS
OF MITTAG-LEFFLER’S TYPE
I. N. Peresyolkova. On distribution of zeros of generalized functions of Mittag-Leffler’s type,
Matematychni Studii, 13 (2000) 157–164.
The following two problems related to the entire function
Φρ1 ,ρ2 (z, µ1 , µ2 ) =
∞
X
zn
,
Γ(µ1 + n/ρ1 )Γ(µ2 + n/ρ2 )
n=0
ρ1 , ρ2 , µ1 , µ2 > 0,
are considered.
(i) For 0 < ρ1 ≤ 1/2, determine the values of µ1 , ρ2 , µ2 such that all zeros of Φρ1 ,ρ2 (z, µ1 , µ2 )
are negative.
(ii) For 1/2 < ρ1 < 1, determine the values of µ1 , ρ2 , µ2 such that all zeros of Φρ1 ,ρ2 (z, µ1 , µ2 )
are situated outside the angle {z : | arg z| ≤ π/(2ρ1 )}.
The results point out several values of µ1 and all values of ρ2 > 0, µ2 > 0 with the above
properties. Moreover, we obtain some results on a-points of the function Φρ1 ,ρ2 (z, µ1 , µ2 ).
И. Н. Пересёлкова. О распределении нулей обобщенных функций типа Миттаг-Леффлера // Математичнi Студiї. – 2000. – Т.13, №2. – C.157–164.
Рассматриваются следующие две задачи о целой функции
Φρ1 ,ρ2 (z, µ1 , µ2 ) =
∞
X
zn
,
Γ(µ1 + n/ρ1 )Γ(µ2 + n/ρ2 )
n=0
ρ1 , ρ2 , µ1 , µ2 > 0.
(а) Найти для 0 < ρ1 ≤ 1/2 такие значения µ1 , ρ2 , µ2 , что все нули Φρ1 ,ρ2 (z, µ1 , µ2 )
являются отрицательными.
(б) Найти для 1/2 < ρ1 < 1 такие значения µ1 , ρ2 , µ2 , что все нули Φρ1 ,ρ2 (z, µ1 , µ2 ) лежат
вне угла {z : | arg z| ≤ π/(2ρ1 )}.
Полученные результаты гарантируют выполнение этих свойств для многих значений µ1 и
любых ρ2 > 0, µ2 > 0. Кроме того, получены некоторые результаты об a-точках функции
Φρ1 ,ρ2 (z, µ1 , µ2 ).
1. Introduction. Following the definition of M. M. Dzhrbashyan [1], the function
Φρ1 ,ρ2 (z, µ1 , µ2 ) =
∞
X
n=0
zn
,
Γ(µ1 + n/ρ1 )Γ(µ2 + n/ρ2 )
(1)
2000 Mathematics Subject Classification: 30D35.
c I. N. Peresyolkova, 2000
158
I. N. PERESYOLKOVA
where ρ1 , µ1 and ρ2 , µ2 are arbitrary positive parameters, is said to be a generalized function
of Mittag-Leffler’s type.
Primary properties and different applications of the function Φρ1 ,ρ2 (z, µ1 , µ2 ) are summarized in [1]. The function Φρ1 ,ρ2 (z, µ1 , µ2 ) is a generalization of the well-known function
of Mittag-Leffler’s type Eρ (z, µ) ≡ Φρ,∞ (z, µ, 1) ≡ Φ∞,ρ (z, 1, µ) studied in [2] in detail. The
purpose of this paper is to extend some results [3] on distribution of zeros of the function of
Mittag-Leffler’s type Eρ (z, µ) to the zeros of generalized functions of Mittag-Leffler’s type
Φρ1 ,ρ2 (z, µ1 , µ2 ).
To state the first result, we define the following three transformations mapping the set
{(ρ, µ) : 0 < ρ < 1, µ > 0} into itself:
ρ
1
, µ ; B : (ρ, µ) →
,µ +
;
A : (ρ, µ) →
2
2
ρ
(
(ρ, µ − 1) for µ > 1;
C : (ρ, µ) →
(ρ, µ)
for 0 < µ ≤ 1.
ρ
Put
Wa = {(ρ, µ) : 1/2 < ρ < 1, µ ∈ [1/ρ − 1, 1] ∪ [1/ρ, 2] },
Wb = AWa ∪ BWa .
Denote by Wi the least set containing Wb and invariant with respect to A, B, C. Evidently,
the set Wi can be represented in the form
[
Wi = (Ak11 B k12 C k13 . . . Akn1 B kn2 C kn3 )Wb ,
where the union is taken over all n = 1, 2, . . . and over all 3n-tuples (k11 , k12 , k13 , . . . , kn1 ,
kn2 , kn3 ) of nonnegative integers.
In [3], the following theorem is proved.
Theorem A ([3, Theorem 3]). If (ρ, µ) ∈ Wi , then all zeros of Eρ (z, µ) are negative and
simple.
The following theorem is a generalization of Theorem A.
Theorem 1. Assume (ρ1 , µ1 ) ∈ Wi ∪ Wj , where Wi is the set described above and
Wj =
∞ [
n=1
where
−n A
1
,
∪ 2
, 0 < µ1 < 1 + 1/ρ1 ,
(ρ1 , µ1 ) : ρ1 ∈
5 · 2n−2 2n−1
1 1
A = − arctan
2 π
10 3
ln
9π 2
(≈ 0.45).
Then, for any ρ2 > 0, µ2 > 0, all zeros of the entire function Φρ1 ,ρ2 (z, µ1 , µ2 ) are real and
negative.
In [3], the following theorem was also proved.
Theorem B ([3, Theorem 1]). Assume that one of the following conditions is satisfied:
(i) ρ > 1, µ ∈ [1, 1 + 1/ρ];
159
DISTRIBUTION OF ZEROS OF MITTAG-LEFFLER’S TYPE FUNCTIONS
(ii) ρ = 1, µ ∈ (0, 2);
(iii) 1/2 < ρ < 1, µ ∈ [1/ρ − 1, 1] ∪ [1/ρ, 2].
n
Then all zeros of Eρ (z, µ) are situated outside the closed angle z : | arg z| ≤
π
2ρ
o
.
The following theorem is a generalization of Theorem B (iii).
Theorem 2. Assume 1/2 < ρ1 < 1, µ1 ∈ [1/ρ1 − 1, 1] ∪ [1/ρ1 , 2]. Then, for any ρ2 > 0,
µ2 > 0, all zeros of the entire function Φρ1 ,ρ2 (z, µ1 , µ2 ) are situated outside the closed angle
π
z : | arg z| ≤
2ρ1
.
(2)
2. Proof of Theorem 1. We use the following well-known Laguerre theorem.
Theorem C ([4, p. 269]). Let ϕ(ω) be an entire function of genus 0 or 1, real–valued
for real values of ω, whose all zeros are real and negative. Let f (z) be an entire function
representable in the form
az+b
f (z) = e
∞ Y
z
,
1+
zn
n=1
where a and all zn are positive. If
f (z) =
∞
X
an z n ,
n=0
then
g(z) =
∞
X
an ϕ(n)z n
n=0
is an entire function whose all zeros are real and negative.
Set
f (z) = Eρ1 (z, µ1 ) =
∞
X
n=0
ϕ(ω) =
zn
,
Γ(µ1 + n/ρ1 )
1
.
Γ(µ2 + ω/ρ2 )
(3)
(4)
Show that if the values of parameters ρ1 , µ1 , ρ2 , µ2 are taken from the statement of Theorem 1, then the conditions of Theorem C are satisfied, so, by virtue of formulas (1) and (4),
Φρ1 ,ρ2 (z, µ1 , µ2 ) =
∞
X
n=0
ϕ(n)z n
Γ(µ1 + n/ρ1 )
(5)
is an entire function whose all zeros are real and negative.
Consider the function f (z). By Theorem A, if (ρ1 , µ1 ) ∈ Wi , then all zeros of this function
are real and negative. By [3, Theorem 2] and [5, Theorem 1], if (ρ1 , µ1 ) ∈ Wj , then all zeros
160
I. N. PERESYOLKOVA
of this function are also real and negative. Since Eρ1 (z, µ1 ) is an entire function of order ρ1
(but Wi and Wj have only ρ1 ≤ 1/2), by the Hadamard theorem, we have
∞ 1 Y
z
1+
,
Eρ1 (z, µ1 ) =
Γ(µ1 ) n=1
zn
where all zn are positive.
Thus, the function f (z) satisfies the conditions of Theorem C.
The function ϕ(ω) is an entire function of genus 1 vanishing at the points ωk = −ρ2 (k +
µ2 ), k ∈ Z+ . All of them are real and negative. Moreover, since ϕ(R) ⊆ R, the function
ϕ(ω) satisfies the conditions of Theorem C. This completes the proof of Theorem 1.
3. Proof of Theorem 2. We use the following two lemmas. The first one was also used
in [3].
Lemma A (Gauss’s lemma). Let f (z) 6≡ 0 be a meromorphic function representable in the
form
∞
γ0 X γk
+
,
f (z) =
z
z
−
a
k
k=1
where
∞
X
γk
<∞.
|a
k|
k=1
γk ≥ 0 (k = 0, 1, 2, ...),
If the sequence {ak }∞
k=1 is contained in the open angle
Aη,δ = {z : η < arg z < η + δ} ,
η ∈ [0, 2π) ,
0<δ≤π,
(6)
then f (z) does not vanish outside Aη,δ .
The following lemma is a slight generalization of the Laguerre theorem mentioned above.
Lemma 1. Let f (z) be an entire function of order ρ < 1, nonvanishing outside the open
angle Aη,δ defined by formula (6). Let ϕ(ω) be an entire function of genus 0 or 1, real-valued
for real values of ω, whose all zeros are real and negative. If
f (z) =
∞
X
an z n ,
n=0
then
g(z) =
∞
X
an ϕ(n)z n
n=0
is an entire function whose all zeros are contained in the open angle Aη,δ .
Proof. First of all, g(z) is an entire function. Indeed, by the Hadamard theorem
ωb
ϕ(ω) = ae
∞ Y
n=1
where αn > 0 for all n.
1+
ω − αω
e n,
αn
DISTRIBUTION OF ZEROS OF MITTAG-LEFFLER’S TYPE FUNCTIONS
161
Since (1 + x)e−x ≤ 1 for x ≥ 0, we have |ϕ(n)| ≤ |a|enb , and hence the series for g(z)
converges everywhere.
Consider the function
0
f (z)z α1 f (z)
z 0
+
.
g1 (z) = f (z) + f (z) =
α1
α1
z
f (z)
Show that all zeros of g1 (z) are contained in Aη,δ .
By the Hadamard theorem
∞ Y
z
f (z) = c
1−
,
a
k
k=1
where
ak ∈ Aη,δ (k = 1, 2, ...) ,
Therefore
∞
X
1
<∞.
|ak |
k=1
∞
0
f (z) X 1
=
.
f (z)
z
−
a
k
k=1
0
(z)
does not vanish outside Aη,δ . Since, moreover, g1 (0) 6= 0,
By Lemma A, the function αz1 + ff (z)
we conclude that g1 (z) does not vanish outside Aη,δ , as well.
Iterating this argument we establish that for any n = 2, 3, ...
gn (z) = gn−1 (z) +
z 0
g (z)
αn n−1
does not vanish outside Aη,δ .
On the other hand,
1
p
g1 (z) = a0 + a1 1 +
z + ... + ap 1 +
z p + ... ,
α1
α1
1
p
1
p
g2 (z) = a0 + a1 1 +
1+
z + ... + ap 1 +
1+
z p + ... ,
α1
α2
α1
α2
............
............
............
1
1
p
p
... 1 +
z + ... + ap 1 +
... 1 +
z p + ... .
gn (z) = a0 + a1 1 +
α1
αn
α1
αn
Therefore
gn
n
X
exp b −
αi −1 z̃ =
1
= a0 + a1 e
b
n Y
i=1
1
1+
αi
− α1
e
i
z̃ + ... + ap e
pb
n Y
i=1
p
1+
αi
− αp
e
and also all zeros of this function are situated inside the angle Aη,δ .
Let
n Y
z
− z
bz
Gn (z) = e
1+
e αi .
αi
i=1
i
z̃ p + ... ,
162
I. N. PERESYOLKOVA
Then
n
X
gn exp b −
αi−1 z̃ = a0 + a1 Gn (1)z̃ + ... + ap Gn (p)z̃ p + ... .
1
Since
Gn (j) →
ϕ(j)
a
(n → ∞) ,
j∈N,
we have
gn
n
X
g(z̃)
1 z̃ →
exp b −
αi
a
1
uniformly in any finite domain.
In virtue of the Hurwitz theorem,P
all zeros of
the function g(z̃)/a are exactly limits of
n
−1
the zeros of the function gn exp b − 1 αi z̃ . Hence all zeros of g(z) are contained in
the open angle Aη,δ .
Let us immediately prove Theorem 2. Let f (z) and ϕ(ω) be the functions defined by
formulas (3) and (4), respectively. Show that if the values of parameters ρ1 , µ1 , ρ2 , µ2 are
taken from the statement of Theorem 2, then the conditions of Lemma 1 are satisfied, hence,
in virtue of formula (5), Φρ1 ,ρ2 (z, µ1 , µ2 ) is an entire function whose all zeros are situated in
the same domain where the zeros of f (z) are.
Since 1/2 < ρ1 < 1, µ1 ∈ [1/ρ1 −1, 1]∪[1/ρ1 , 2], by Theorem B(iii), we have that all zeros
of Eρ1 (z, µ1 ) are situated outside the closed angle (2). Therefore f (z) is an entire function
of order ρ1 < 1 nonvanishing outside the angle Aη,δ , where
η = π/(2ρ1 ) ∈ [0, 2π) ,
0 < δ = 2π − π/ρ1 ≤ π .
The function ϕ(ω) also satisfies the conditions of Lemma 1. These conditions can be
verified by reasoning as in the proof of Theorem 1. Thus Theorem 2 is proved.
4. In conclusion we obtain some results on a-points of the function Φρ1 ,ρ2 (z, µ1 , µ2 ),
which are more general than the results on a-points of the function Eρ (z, µ) obtained in [5].
Namely, we obtain analogues of the following theorems.
Theorem D ([5, Theorem 3]). Assume ρ > 1/2, a ∈ {z ∈ C : |z| ≤ Rρ }, where Rρ =
min{2ρ − 1, 1}. Then all a-points of the function Eρ (z, 1) are situated outside the closed
angle
{z : | arg z| ≤ π/(2ρ)}
(if a = 1, then the point 0 is excluded from the angle).
Corollary A ([5, Corollary 1]). Assume a ∈ (−1, 1). If (1 + |a|) · 2−n−1 ≤ ρ < 2−n (n ∈ N),
then all a-points of the function Eρ (z, 1) are strictly negative.
Theorem E ([5, Theorem 4]). Assume 1/2 < ρ < 1, a ∈ {z ∈ C : |z| ≤ ρ}. Then all
a-points of the function Eρ (z, 2) are situated outside the closed angle
n
|a| o
1
.
z : | arg z| ≤ arccos
ρ
ρ
We need the following lemma, which is a generalization of Lemma 1.
DISTRIBUTION OF ZEROS OF MITTAG-LEFFLER’S TYPE FUNCTIONS
163
Lemma 1. Let f (z) be an entire function of order ρ < 1 whose all a-points are contained
in the open angle Aη,δ defined by formula (6). Let ϕ(ω) be an entire function of genus 0 or
1, real-valued for real values of ω, whose all zeros are real and negative. If
f (z) =
∞
X
an z n ,
n=0
then
g(z) =
∞
X
an ϕ(n)z n
n=0
is an entire function whose all aϕ(0)-points are contained in the open angle Aη,δ .
Proof. Consider the function f (z) − a. This function satisfies the conditions of Lemma 1.
Since
∞
X
g(z) − aϕ(0) =
an ϕ(n)z n + (a0 − a)ϕ(0),
n=1
in virtue of Lemma 1 we have that all zeros of the function g(z) − aϕ(0) are contained in the
same domain where the zeros of the function f (z) − a are, i. e. in the open angle Aη,δ .
Theorem 3 (Analogue of Theorem D). Assume 1/2 < ρ1 < 1, ρ2 > 0, µ2 > 0, a ∈ {z ∈ C :
|z| ≤ Rρ1 }, where Rρ1 = 2ρ1 − 1. Then all a/Γ(µ2 )-points of the function Φρ1 ,ρ2 (z, 1, µ2 ) are
situated outside the closed angle
z : | arg z| ≤ π/(2ρ1 )
(if a = 1, then the point 0 is excluded from the angle).
Corollary 1 (Analogue of Corollary A). Assume a ∈ (−1, 1). If (1 + |a|) · 2−n−1 ≤ ρ1 <
2−n (n ∈ N), ρ2 > 0, µ2 > 0, then all a/Γ(µ2 )-points of the function Φρ1 ,ρ2 (z, 1, µ2 ) are
strictly negative.
Theorem 4 (Analogue of Theorem E). Assume 1/2 < ρ1 < 1, ρ2 > 0, µ2 > 0, a ∈ {z ∈
C : |z| ≤ ρ1 }. Then all a/Γ(µ2 )-points of the function Φρ1 ,ρ2 (z, 2, µ2 ) are situated outside
the closed angle
n
|a| o
1
arccos
.
z : | arg z| ≤
ρ1
ρ1
To prove these Theorems, it suffices to apply Lemma 2 to the functions
f (z) = Eρ1 (z, µ1 ),
ϕ(ω) =
1
.
Γ(µ2 + ω/ρ2 )
Reasoning as in the proof of Theorem 1, we can prove that the function ϕ(ω) satisfies the
conditions of Lemma 2 for any ρ2 > 0, µ2 > 0. If the values of parameters ρ1 , µ1 are taken
from the statements of Theorems 3 and 4, then the function f (z) also satisfies the conditions
of Lemma 2, as, by Theorems D and E, all its a-points are situated in the corresponding open
angles. The author is deeply grateful to Prof. I. V. Ostrovskii for formulating the problem.
164
I. N. PERESYOLKOVA
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Department of Mechanics and Mathematics, Kharkov National University,
Svobody Sq., 4, 310077, Kharkov, Ukraine
[email protected] ; [email protected]
Received 13.10.1998