Entire bivariate functions of unbounded index in each direction
We investigate a class of entire functions f(z₁, z₂) with property ∀b= (b₁, b₂) ∈ C² \ {0} ∀ z⁰₁, z⁰₂ ∈ C, the function f(z⁰₁ + tb₁, z⁰₂ + tb₂), as a function of one variable t ∈ C, has a bounded index but the function f(z₁, z₂) has an unbounded index in every direction b. In particular, we prove th...
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nasplib_isofts_kiev_ua-123456789-1773372025-02-09T17:51:53Z Entire bivariate functions of unbounded index in each direction Цілі функції від двох змінних необмеженого індексу за кожним напрямком Целые функции от двух переменных неограниченного индекса по каждому направлению Bandura, A.I. Skaskiv, O.B. We investigate a class of entire functions f(z₁, z₂) with property ∀b= (b₁, b₂) ∈ C² \ {0} ∀ z⁰₁, z⁰₂ ∈ C, the function f(z⁰₁ + tb₁, z⁰₂ + tb₂), as a function of one variable t ∈ C, has a bounded index but the function f(z₁, z₂) has an unbounded index in every direction b. In particular, we prove that, for an arbitrary even entire function f(t) that has an infinite sequence of complex zeros, the corresponding function f(√(z₁z₂)) has an unbounded index in every direction b. It improves our similar result [Bandura A. I. A class of entire functions of unbounded index in each direction // Mat. Stud. – 2015. – 44, № 1. – P. 107 – 112] proved for even entire functions f(t) with complex zeros ck such that c²k ∈ R. Дослiджується клас цiлих функцiй f(z₁, z₂) з такою властивiстю: ∀b= (b₁, b₂) ∈ C² \ {0} ∀ z⁰₁, z⁰₂ ∈ C, функцiя f(z⁰₁ + tb₁, z⁰₂ + tb₂), має обмежений iндекс як функцiя вiд однiєї змiнної t ∈ C, але функцiя f(z₁, z₂) є необмеженого iндексу за кожним напрямком b. Зокрема, доведено, що для довiльної парної цiлої функцiї f(t), яка має нескiнченну послiдовнiсть комплексних нулiв, вiдповiдна функцiя f(√(z₁z₂)) є необмеженого iндексу за кожним напрямком b. Це покращує подiбний результат з [Bandura A. I. A class of entire functions of unbounded index in each direction // Mat. Stud. – 2015. – 44,№ 1. – P. 107 – 112], доведений для парних цiлих функцiй f(t) з комплексними нулями ck такими, що c²k ∈ R. 2018 Article Entire bivariate functions of unbounded index in each direction / A.I. Bandura, O.B. Skaskiv // Нелінійні коливання. — 2018. — Т. 21, № 4. — С. 435-443 — Бібліогр.: 16 назв. — англ. 1562-3076 https://nasplib.isofts.kiev.ua/handle/123456789/177337 517.555 en Нелінійні коливання application/pdf Інститут математики НАН України |
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| description |
We investigate a class of entire functions f(z₁, z₂) with property ∀b= (b₁, b₂) ∈ C² \ {0} ∀ z⁰₁, z⁰₂ ∈ C, the function f(z⁰₁ + tb₁, z⁰₂ + tb₂), as a function of one variable t ∈ C, has a bounded index but the function f(z₁, z₂) has an unbounded index in every direction b. In particular, we prove that, for an arbitrary even entire function f(t) that has an infinite sequence of complex zeros, the corresponding function f(√(z₁z₂)) has an unbounded index in every direction b. It improves our similar result [Bandura A. I. A class of entire functions of unbounded index in each direction // Mat. Stud. – 2015. – 44, № 1. – P. 107 – 112] proved for even entire functions f(t) with complex zeros ck such that c²k ∈ R. |
| format |
Article |
| author |
Bandura, A.I. Skaskiv, O.B. |
| spellingShingle |
Bandura, A.I. Skaskiv, O.B. Entire bivariate functions of unbounded index in each direction Нелінійні коливання |
| author_facet |
Bandura, A.I. Skaskiv, O.B. |
| author_sort |
Bandura, A.I. |
| title |
Entire bivariate functions of unbounded index in each direction |
| title_short |
Entire bivariate functions of unbounded index in each direction |
| title_full |
Entire bivariate functions of unbounded index in each direction |
| title_fullStr |
Entire bivariate functions of unbounded index in each direction |
| title_full_unstemmed |
Entire bivariate functions of unbounded index in each direction |
| title_sort |
entire bivariate functions of unbounded index in each direction |
| publisher |
Інститут математики НАН України |
| publishDate |
2018 |
| url |
https://nasplib.isofts.kiev.ua/handle/123456789/177337 |
| citation_txt |
Entire bivariate functions of unbounded index in each direction / A.I. Bandura, O.B. Skaskiv // Нелінійні коливання. — 2018. — Т. 21, № 4. — С. 435-443 — Бібліогр.: 16 назв. — англ. |
| series |
Нелінійні коливання |
| work_keys_str_mv |
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2025-11-29T02:00:57Z |
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2025-11-29T02:00:57Z |
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| fulltext |
UDC 517.555
ENTIRE BIVARIATE FUNCTIONS OF UNBOUNDED INDEX IN EACH DIRECTION
ЦIЛI ФУНКЦIЇ ВIД ДВОХ ЗМIННИХ НЕОБМЕЖЕНОГО IНДЕКСУ
ЗА КОЖНИМ НАПРЯМКОМ
A. Bandura
Ivano-Frankivsk Nat. Techn. Univ. Oil and Gas,
Karpatska str., 15, Ivano-Frankivs’k, 76019, Ukraine
e-mail: andriykopanytsia@gmail.com
O. Skaskiv
Ivan Franko Nat. Univ. Lviv,
Universytetska str., 1, Lviv, 79000, Ukraine
e-mail: olskask@gmail.com
We investigate a class of entire functions f(z1, z2) with property ∀b=(b1, b2) ∈ C2 \{0} ∀ z01 , z02 ∈ C, the
function f(z01 + tb1, z
0
2 + tb2), as a function of one variable t ∈ C, has a bounded index but the function
f(z1, z2) has an unbounded index in every direction b. In particular, we prove that, for an arbitrary even
entire function f(t) that has an infinite sequence of complex zeros, the corresponding function f
(√
z1z2
)
has an unbounded index in every direction b. It improves our similar result [Bandura A. I. A class of
entire functions of unbounded index in each direction // Mat. Stud. – 2015. – 44, № 1. – P. 107 – 112]
proved for even entire functions f(t) with complex zeros ck such that c2k ∈ R.
Дослiджується клас цiлих функцiй f(z1, z2) з такою властивiстю: ∀b = (b1, b2) ∈ C2 \ {0} ∀ z01 ,
z02 ∈ C функцiя f(z01 + tb1, z
0
2 + tb2) має обмежений iндекс як функцiя вiд однiєї змiнної t ∈ C,
але функцiя f(z1, z2) є необмеженого iндексу за кожним напрямком b. Зокрема, доведено, що
для довiльної парної цiлої функцiї f(t) , яка має нескiнченну послiдовнiсть комплексних нулiв,
вiдповiдна функцiя f(
√
z1z2) є необмеженого iндексу за кожним напрямком b. Це покращує
подiбний результат з [Bandura A. I. A class of entire functions of unbounded index in each direction //
Mat. Stud. – 2015. – 44,№1. – P. 107 – 112], доведений для парних цiлихфункцiй f(t) з комплексними
нулями ck такими, що c2k ∈ R.
1. Introduction. To state a problem and a main result, we need some denotations. Let b ∈
∈ Cn \ {0} be a direction, L : Cn → R+ be a continuous function, F : Cn → C be an entire
function, gz0(t) := F
(
z0 + tb
)
, lz0(t) := L
(
z0 + tb
)
, t ∈ C.
Definition 1 [1, 2]. An entire function F (z), z ∈ Cn, is said to be of bounded L-index in
the direction b, if there exists m0 ∈ Z+ such that for all m ∈ Z+ and every z ∈ Cn the next
inequality is true:
1
m!Lm(z)
∣∣∣∣∂mF (z)∂bm
∣∣∣∣ ≤ max
{
1
k!Lk(z)
∣∣∣∣∂kF (z)∂bk
∣∣∣∣ : 0 ≤ k ≤ m0
}
, (1)
where
∂0F (z)
∂b0
= F (z),
∂F (z)
∂b
=
∑n
j=1
∂F (z)
∂zj
bj ,
∂kF (z)
∂bk
=
∂
∂b
(
∂k−1F (z)
∂bk−1
)
, k ≥ 2.
© A. Bandura, O. Skaskiv, 2018
ISSN 1562-3076. Нелiнiйнi коливання, 2018, т. 21, № 4 435
436 A. BANDURA, O. SKASKIV
The least such integer m0 is called the L-index in the direction b of F (z) and is denoted
by Nb(F,L). If such m0 does not exist, then we put Nb(F,L) = ∞ and F is said to be of
unbounded L-index in the direction.
If L(z) ≡ 1 then the function F is called of bounded index in the direction b and Nb(F ) ≡
≡ Nb(F, 1) is called the index in the direction b. If n = 1, b = 1 and L(z) = l(z), z ∈ C,
inequality (1) defines a bounded l -index with the l -index N(F, l) ≡ N1(F, l) [3]. And in the case
L(z) ≡ 1 we get a notion of bounded index with the index N(F ) ≡ N1(F, 1) [4].
These functions have been used in the theory value distribution and differential equations
([see bibliography in [5]). In particular, every entire function is a function of bounded value
distribution if and only if its derivative is a function of bounded index [6], and every entire
solution of the differential equation f (n)(t)+
∑n−1
j=0
ajf
(j)(t) = 0 is a function of bounded index
[7]. More general results for PDE’s are obtained by ours [1, 2, 8]. There are sufficient conditions
that every entire function satisfying some PDE is of bounded L-index in direction.
Another definition of bounded index in C2 is considered in the paper of F. Nuray and
R. F. Patterson [9]. Using this notion they presented a series of sufficient conditions that bivariate
entire function is of exponential type. The presented conditions are weaker than known necessary
and sufficient conditions of bounded index in joint variables. Besides, they [10] established the
relationship between the concept of bounded index and the radius of univalence, respectively
p-valence, of entire bivariate functions and their partial derivatives at arbitrary points in C2.
Recently we published the paper [11], which is devoted to interesting and important open
problems in the theory of entire functions of bounded index. In particular, there was formulated
the following
Problem 1 ([11], Problem 17). What are conditions on zero sets and growth of entire functi-
ons providing the bounded index of F (z01+b1t, z02+b2t) for every (z01 , z02) ∈ C2 and the unbounded
index of F (z1, z2) in the direction b = (b1, b2)?
For example, f(z1, z2) = cos
√
z1z2 has described properties [12]. It was proved by a
construction of some PDE and investigation of properties of its entire solutions.
We solved the mentioned problem [13], supposing c2k ∈ R, where ck, k ∈ N, are zeros of
entire function f(t). We got the next theorem as our decision.
Theorem 1 [13]. Let f(t), t ∈ C, be an even entire transcendental function of bounded
index. Then:
(i) for each direction b = (b1, b2) ∈ C2 \ {0} and for every fixed z01 , z
0
2 ∈ C the function
g(t) = f
(√(
z01 + b1t
) (
z02 + b2t
) )
is an entire function of bounded index t ∈ C;
(ii) if f(t) has no zeros or has a finite number of zeros, then f
(√
z1z2
)
is of unbounded index
in each direction b;
(iii) if {ck} is an infinite sequence of zeros of f(t), |c1| < |c2| < . . . < |ck| < . . . and for
every k ∈ N, c2k ∈ R, then the function f
(√
z1z2
)
is of unbounded index in each direction b.
The following remark will be useful in this paper.
Remark 1 [13]. The condition c2k ∈ R can be replaced by the condition that there exists an
infinite subsequence of zeros of the form c′k = |c′k| · eiθ, i.e., all c′k lay on a some ray. Then in the
case b1 6= 0, b2 6= 0 we choose ϕ = 2θ + arg(b1b2) and in the case b1 6= 0, b2 = 0 we choose
z01 = ei(2θ+ϕ). Other considerations are remained without changes in the proof of Theorem 1.
In 2015 at the Lviv seminar on the theory of analytic functions Prof. O. Skaskiv assumed that
condition “for every k ∈ N, c2k ∈ R” is excessive. It leads to a new next problem.
ISSN 1562-3076. Нелiнiйнi коливання, 2018, т. 21, № 4
ENTIRE BIVARIATE FUNCTIONS OF UNBOUNDED INDEX IN EACH DIRECTION 437
Problem 2. If f(t), t ∈ C, is an even entire transcendental function of bounded index, which
has an infinite number of zeros, then a function f
(√
z1z2
)
is of unbounded index in each direction
b. Is it true or false?
In this paper, we prove that mentioned proposition is true. Note that for the function
cos
(√
z1z2
)
there exists a partial differential equation and a positive continuous function L : C2 →
→ R+ with properties [14]:
(i) f
(√
z1z2
)
is its solution;
(ii) every entire solution of the PDE has bounded L-index in the direction b.
2. Auxiliary proposition. We need some notation. If for a given z0 ∈ Cn one has gz0(t) 6= 0
for all t ∈ C, then Gb
r
(
F, z0
)
:= ∅; if for a given z0 ∈ Cn we get gz0(t) ≡ 0, then
Gb
r
(
F, z0
)
:=
{
z0 + tb : t ∈ C
}
.
And if for a given z0 ∈ Cn we have gz0(t) 6≡ 0 and a0k are zeros of gz0(t), then
Gb
r
(
F, z0
)
:=
⋃
k
{
z0 + tb :
∣∣t− a0k∣∣ ≤ r} , r > 0.
Let
Gb
r (F ) =
⋃
z0∈Cn
Gb
r (F, z
0).
By n
(
r, z0, t0, 1/F
)
=
∑
|a0k−t0|≤r
1 we denote the counting function of the zero sequence
(
a0k
)
.
The following criterion is convenient for a proof of index boundedness in direction.
Theorem 2 [1, 2]. Let F (z) be an entire function in Cn. A function F (z) is of bounded
index in the direction b if and only if:
(i) for every r > 0 there exists P = P (r) > 0 that for each z ∈ Cn\Gb
r (F )∣∣∣∣ 1
F (z)
∂F (z)
∂b
∣∣∣∣ ≤ P ;
(ii) for every r > 0 there exists ñ(r) ∈ Z+ that for every z0 ∈ Cn satisfying F (z0+ tb) 6≡ 0,
and for all t0 ∈ C
n
(
r, z0, t0,
1
F
)
≤ ñ(r). (2)
Recently, we obtained weaker sufficient conditions [15, 16] than in Theorem 2 replacing the
universal quantifier by the existence quantifier. Besides, we need following property of bounded
index in direction.
Theorem 3 [1, 2]. Let m ∈ C \ {0}. An entire function F (z), z ∈ Cn, is of bounded index
in the direction b if and only if F (z) is of bounded index in the direction mb.
3. Main theorem.
Theorem 4. If f(t), t ∈ C, is an even entire transcendental function of bounded index,
which has an infinite number of zeros, then the function f
(√
z1z2
)
is of unbounded index in each
direction b.
ISSN 1562-3076. Нелiнiйнi коливання, 2018, т. 21, № 4
438 A. BANDURA, O. SKASKIV
Proof. Let (cl)∞l=1 be an infinite sequence of zeros of f(t), |c1| < |c2| < . . . < |ck| < . . . . In
view of Theorem 1 and Remark 1 we suppose that there are located a finite number of zeros ck
on every ray with origin. It remains to prove that function f
(√
z1z2
)
is of unbounded index in
the direction b. We show that condition (2) of Theorem 2 does not hold.
Obviously, for every ε > 0 there exists a sector with a vertex at the origin and a central angle
2ε where an infinite number of zeros ck is located inside. Let arg z = θ be a bisector of the
sector. In view of Remark 1 we suppose θ = 0 without loss of generality. Below in the proof
we consider only zeros cl containing in the specified sector. In fact, it shall be proven that the
condition (2) does not hold for zeros of function f
(√(
z01 + b1t
) (
z02 + b2t
) )
, generating by cl
from that sector.
Case 1. Let b1 6= 0, b2 = 0 and ak ∈ R+, ak → ∞. Later we will impose more conditions
on the sequence (ak)
∞
k=1. We put z0 =
(
z01 , z
0
2
)
, where
z01 = 1, z02 = a2k, t0 = 0. (3)
The zeros of the function f
(√(
z01 + b1t
) (
z02 + b2t
) )
are found from the equation(
z01 + b1t
)
z02 = z02b1t+ z01z
0
2 = c2l , l ∈ Z.
Consider its root
tl =
−(b2z01 + b1z
0
2) +
√
(b2z01 − b1z02)2 + 4c2l b1b2
2b1b2
.
The condition of a zero tl belongs to r -neighborhood of the point t0 has the form |tl−t0| < r.
Let ϕl = arg cl. It implies
∣∣∣∣ |cl|2e2iϕl − a2k
|b1| · a2k
∣∣∣∣ < r or
r|b1| a2k >
∣∣|cl|2e2iϕl − a2k
∣∣ = ∣∣|cl|2 cos 2ϕl − a2k + i|cl|2 sin 2ϕl
∣∣ =
=
√
(|cl|2 cos 2ϕl − a2k)2 + |cl|4 sin
2 2ϕl =
√
|cl|4 − 2|cl|2 cos 2ϕla2k + a4k.
Hence, we deduce a biquadratic inequality
|cl|4 − 2|cl|2 cos 2ϕl · a2k + a4k − r2|b1|2 · a4k < 0. (4)
Solving (4), we obtain an estimate for |cl| :
a2k
(
cos 2ϕl −
√
r2|b1|2 − sin2 2ϕl
)
< c2l < a2k
(
cos 2ϕl +
√
r2|b1|2 − sin2 2ϕl
)
. (5)
We choose r ≥ 1
|b1|
. It follows r2|b1|2 ≥ 1 or r2|b1|2 − sin2 2ϕl ≥ cos2 2ϕl. Therefore, the
left-hand side in (5) is nonpositive. Besides, we pick ε ∈
(
0,
π
4
)
, i.e. ϕl ∈ (−ε; ε) ⊂
(
−π
4
;
π
4
)
.
The right-hand side in (5) implies
a2k >
|cl|2
cos 2ϕl +
√
r2|b1|2 − sin2 2ϕl
.
ISSN 1562-3076. Нелiнiйнi коливання, 2018, т. 21, № 4
ENTIRE BIVARIATE FUNCTIONS OF UNBOUNDED INDEX IN EACH DIRECTION 439
For k + 1 zeros of f(t) we choose
a2k > max
1≤l≤k+1
|cl|2
cos 2ϕl +
√
r2|b1|2 − sin2 2ϕl
.
Thus, ∃r > 0
(
we have r ∈
[
1
|b1|
;∞
))
∀k ∈ N ∃z0 ∈ C2 ∃t0 ∈ C (see (3)) that
n
(
r, z0, t0,
1
f
)
≥ k + 1 > k.
Hence, the function f
(√
z1z2
)
is of unbounded index in the direction b.
Case 2. Let b1 6= 0, b2 6= 0 and ak ∈ R+, ak → ∞. Later we will impose more conditions
on the sequence (ak)
∞
k=1. By Theorem 3 an entire function F (z), z ∈ C2 is of bounded index
in direction (b1, b2) if and only if F (z) is of bounded index in direction
(
b1√
|b1b2|
,
b2√
|b2b2|
)
.
Hence, without loss of generality we suppose that |b1b2| = 1.
Put ϕ = arg (b1b2), z
0 =
(
z01 , z
0
2
)
, where z01 is arbitrary complex number,
z02 =
b2z
0
1 + (1− a2k)eiϕ/2
b1
, t0 =
a2ke
iϕ/2 − b2z01
b1b2
. (6)
The zeros of f
(√(
z01 + b1t
) (
z02 + b2t
) )
can be found from the equation(
z01 + b1t
)(
z02 + b2t
)
= b1b2t
2 +
(
z01b2 + z02b1
)
t+ z01z
0
2 = c2l , l ∈ N.
Consider its roots
tl =
−
(
b2z
0
1 + b1z
0
2
)
+
√(
b2z01 − b1z02
)2
+ 4c2l b1b2
2b1b2
.
Let ϕl = arg cl. The condition of a zero tl belongs to r -neighborhood of the point t0 has the
form |tl − t0| < r or
r >
∣∣∣∣∣∣a2keiϕ/2 − b2z01 − −
(
2b2z
0
1 +
(
1− a2k
)
eiϕ/2
)
+
√(
a2k − 1
)2
eiϕ + 4eiϕ|cl|2
2
∣∣∣∣∣∣⇐⇒
⇐⇒ 2r >
∣∣∣∣a2k + 1 +
√(
a2k − 1
)2
+ 4|cl|2(cos 2ϕl + i sin 2ϕl)
∣∣∣∣ =
=
∣∣∣∣a2k + 1 +
√(
a4k − 2a2k + 1 + 4|cl|2 cos 2ϕl
)
+ 4i|c2l | sin 2ϕl
∣∣∣∣ . (7)
To calculate a square root of complex number we suppose x ∈ R, y ∈ R and use
√
x+ iy = ±
√√x2 + y2 + x
2
+ i
√√
x2 + y2 − x
2
. (8)
ISSN 1562-3076. Нелiнiйнi коливання, 2018, т. 21, № 4
440 A. BANDURA, O. SKASKIV
Applying (8) to (7), we obtain
2r >
∣∣∣∣∣a2k + 1± 1√
2
(√((
a2k − 1
)2
+ 4|cl|2 cos 2ϕl
)2
+ 16|cl|4 sin2 2ϕl +
+ 4|cl|2 cos 2ϕl +
(
a2k − 1
)2)1/2±
± i√
2
(√((
a2k − 1
)2
+ 4|cl|2 cos 2ϕl
)2
+ 16|cl|4 sin2 2ϕl −
− 4|cl|2 cos 2ϕl −
(
a2k − 1
)2)1/2∣∣∣∣∣. (9)
We choose aminus before square roots in (9). But for real x and y the inequality |x+iy| ≤ |x|+|y|
holds. We suppose that real and image parts of expression in modulus (9) don’t exceed r, i.e.,
√
2r >
∣∣∣∣∣√2 (a2k + 1
)
−
(
4|cl|2 cos 2ϕl +
(
a2k − 1
)2
+
+
√((
a2k − 1
)2
+ 4|cl|2 cos 2ϕl
)2
+ 16|cl|4 sin2 2ϕl
)1/2∣∣∣∣∣ (10)
and
1√
2
∣∣∣∣∣4|cl|2 cos 2ϕl + (a2k − 1
)2 −√((a2k − 1
)2
+ 4|cl|2 cos 2ϕl
)2
+ 16|cl|4 sin2 2ϕl
∣∣∣∣∣
1/2
< r.
(11)
At first we prove that for some r and some ε and for all ak the inequality (11) holds. From (11)
it follows that(
4|cl|2 cos 2ϕl + a4k − 2a2k + 1− 2r2
)2
<
(
a4k − 2a2k + 1 + 4|cl|2 cos 2ϕl
)2
+
+ 16|cl|2 sin2 2ϕl <
(
4|cl|2 cos 2ϕl + a4k − 2a2k + 1 + 2r2
)2
or (
4|cl|2 cos 2ϕl + a4k − 2a2k + 1
)2
+ 4r4 − 4r2
(
4|cl|2 cos 2ϕl + a4k − 2a2k + 1
)
<
<
(
a4k − 2a2k + 1 + 4|cl|2 cos 2ϕl
)2
+ 16|cl|2 sin2 2ϕl <
<
(
4|cl|2 cos 2ϕl + a4k − 2a2k + 1
)2
+ 4r4 + 4r2
(
4|cl|2 cos 2ϕl + a4k − 2a2k + 1
)
.
Simplifying and reducing by 4 we deduce
r4 − r2
(
4|cl|2 cos 2ϕl + a4k − 2a2k + 1
)
< 4|cl|2 sin2 2ϕl <
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ENTIRE BIVARIATE FUNCTIONS OF UNBOUNDED INDEX IN EACH DIRECTION 441
< r4 + r2
(
4|cl|2 cos 2ϕl + a4k − 2a2k + 1
)
. (12)
Now we choose ε ≤ π
4
, r < 3. Since ak → ∞ we suppose
(
a2k − 1
)2
> 9, i.e., |ak|2 > 4.
From here it follows that
r4 − r2
(
4|cl|2 cos 2ϕl + a4k − 2a2k + 1
)
= r2
(
r2 − 4|cl|2 cos 2ϕl −
(
a2k − 1
)2)
< 0
for |ϕl| <
π
4
. Therefore, the left-hand side in (12) is negative.
The right-hand side in (12) is equivalent to
4|cl|2
(
sin2 2ϕl − r2 cos 2ϕl
)
< r4 + r2
(
a2k − 1
)2
. (13)
For validity of (13) we choose r ∈ (2; 3).
Then sin2 2ϕl − r2 cos 2ϕl < sin2 2ϕl − 4 cos 2ϕl. Now we require sin2 2ϕl − 4 cos 2ϕl < 0. It
means that
cos2 2ϕl + 4 cos 2ϕl − 1 > 0.
Its solution is
cos 2ϕl < −2−
√
5 or cos 2ϕl > −2 +
√
5.
Hence, for ε < 1
2
arccos
(
−2 +
√
5
)
<
π
4
and r ∈ (2; 3) the inequality (11) is valid for all
|ak|2 > 4.
Now we shall choose r and ε and construct sequence (ak) such that inequality (10) is true.
That inequality is equivalent to the following estimate:
√
2(a2k + 1− r) <
<
(
4|cl|2 cos 2ϕl +
(
a2k − 1
)2
+
√((
a2k − 1
)2
+ 4|cl|2 cos 2ϕl
)2
+ 16|cl|4 sin2 2ϕl
)1/2
<
<
√
2(a2k + 1 + r)⇐⇒ 2
(
a2k + 1− r
)2
<
< 4|cl|2 cos 2ϕl +
(
a2k − 1
)2
+
√((
a2k − 1
)2
+ 4|cl|2 cos 2ϕl
)2
+ 16|cl|4 sin2 2ϕl <
< 2
(
a2k + 1 + r
)2
. (14)
Putting r = 2 in (14), we strengthen that inequality:
2
(
a2k − 1
)2
<
< 4|cl|2 cos 2ϕl +
(
a2k − 1
)2
+
√((
a2k − 1
)2
+ 4|cl|2 cos 2ϕl
)2
+ 16|cl|4 sin2 2ϕl <
< 2
(
a2k + 3
)2 ⇐⇒
⇐⇒
(
a2k − 1
)2
< 4|cl|2 cos 2ϕl +
√((
a2k − 1
)2
+ 4|cl|2 cos 2ϕl
)2
+ 16|cl|4 sin2 2ϕl <
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442 A. BANDURA, O. SKASKIV
< a4k + 14a2k + 17. (15)
But
4|cl|2 cos 2ϕl +
√((
a2k − 1
)2
+ 4|cl|2 cos 2ϕl
)2
+ 16|cl|4 sin2 2ϕl >
>
(
a2k − 1
)2
+ 4|cl|2 cos 2ϕl >
(
a2k − 1
)2
.
Thus, the left inequality in (15) is obvious. The middle expression in (15) does not exceed(
a2k − 1
)2
+ 8|cl|2 cos 2ϕl + 4|cl|2| sin 2ϕl|. We obtain inequality
a4k − 2a2k + 1 + 4|cl|2(2 cos 2ϕl + | sin 2ϕl|) < a4k + 14a2k + 17⇐⇒
⇐⇒ 4|cl|2(2 cos 2ϕl + | sin 2ϕl|)− 16 < 16a2k.
Hence, we must choose ak such that
a2k >
1
4
|cl|2(2 cos 2ϕl + | sin 2ϕl|)− 1.
For k + 1 zeros of f(t) we pick ak satisfying this condition
a2k > max
1≤l≤k+1
|cl|2
4
(2 cos 2ϕl + | sin 2ϕl|)− 1.
Thus, ∃r > 0 (we have r ∈ (2; 3)) ∀k ∈ N ∃z0 ∈ C2 ∃t0 ∈ C (see (6)) that
n
(
r, z0, t0,
1
f
)
≥ k + 1 > k.
We conclude that function f
(√
z1z2
)
is of unbounded index in direction b.
Combining Theorem 1 and Theorem 4 in one statement, we get the following
Theorem 5. Let f(t), t ∈ C, be an even entire transcendental function of bounded index.
Then:
(i) for each direction b = (b1, b2) ∈ C2 \ {0} and for every fixed z01 , z
0
2 ∈ C the function
g(t) = f
(√(
z01 + b1t
) (
z02 + b2t
))
is an entire function of bounded index (t ∈ C);
(ii) the function f
(√
z1z2
)
is of unbounded index in each direction b.
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Received 10.03.2017,
after revision — 18.10.2018
ISSN 1562-3076. Нелiнiйнi коливання, 2018, т. 21, № 4
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