The properties on differential-difference polynomials
The main aim of this paper is to improve some classical results on the distribution of zeros for differential polynomials and differential-difference polynomials. We present some results on the distribution of zeros of $[f(z)^nf(z + c)]^{(k)} \alpha (z)$ and $[f(z)^n(f(z + c) f(z))]^{(k)} \alpha (z)...
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Ukrains’kyi Matematychnyi Zhurnal| _version_ | 1860507504331456512 |
|---|---|
| author | Cao, T. B. Liu, K. Liu, X. L. Цао, Т. Б. Лю, К. Лю, X. Л. |
| author_facet | Cao, T. B. Liu, K. Liu, X. L. Цао, Т. Б. Лю, К. Лю, X. Л. |
| author_sort | Cao, T. B. |
| baseUrl_str | https://umj.imath.kiev.ua/index.php/umj/oai |
| collection | OJS |
| datestamp_date | 2019-12-05T09:23:35Z |
| description | The main aim of this paper is to improve some classical results on the distribution of zeros for differential polynomials and
differential-difference polynomials. We present some results on the distribution of zeros of $[f(z)^nf(z + c)]^{(k)} \alpha (z)$ and
$[f(z)^n(f(z + c) f(z))]^{(k)} \alpha (z)$ and give some examples to show that the results are best possible in a certain sense. |
| first_indexed | 2026-03-24T02:10:22Z |
| format | Article |
| fulltext |
UDC 517.5
K. Liu, T. B. Cao, X. L. Liu (Nanchang Univ., China)
THE PROPERTIES ON DIFFERENTIAL-DIFFERENCE POLYNOMIALS*
ВЛАСТИВОСТI ДИФЕРЕНЦIАЛЬНО-РIЗНИЦЕВИХ ПОЛIНОМIВ
The main aim of this paper is to improve some classical results on the distribution of zeros for differential polynomials and
differential-difference polynomials. We present some results on the distribution of zeros of [f(z)nf(z+ c)](k) - \alpha (z) and
[f(z)n(f(z + c) - f(z))](k) - \alpha (z) and give some examples to show that the results are best possible in a certain sense.
Основною метою роботи є полiпшення деяких класичних результатiв про розподiли нулiв для диференцiальних та
диференцiально-рiзницевих полiномiв. Нaведено деякi результати щодо розподiлiв нулiв для [f(z)nf(z + c)](k) -
- \alpha (z) та [f(z)n(f(z + c) - f(z))](k) - \alpha (z), а також деякi приклади, якi демонструють, що отриманi результати
є, в певному розумiннi, найкращими.
1. Introduction. We assume that the reader is familiar with standard symbols and fundamental
results of Nevanlinna theory [7, 11, 22]. A meromorphic function f means meromorphic in the
complex plane. If no poles occur, then f reduces to an entire function. Recall that a meromorphic
function \alpha (z) \not \equiv 0,\infty is a small function with respect to f(z), if T (r, \alpha ) = S(r, f), where S(r, f)
is used to denote any quantity satisfying S(r, f) = o(T (r, f)), and r \rightarrow \infty outside of a possible
exceptional set of finite logarithmic measure. The order of f(z) is defined by
\rho (f) := \mathrm{l}\mathrm{i}\mathrm{m} \mathrm{s}\mathrm{u}\mathrm{p}
r\rightarrow \infty
\mathrm{l}\mathrm{o}\mathrm{g}+ T (r, f)
\mathrm{l}\mathrm{o}\mathrm{g} r
.
A polynomial p(z) is called a Borel exceptional polynomial of f(z) when
\lambda (f(z) - p(z)) = \mathrm{l}\mathrm{i}\mathrm{m} \mathrm{s}\mathrm{u}\mathrm{p}
r\rightarrow \infty
\mathrm{l}\mathrm{o}\mathrm{g}+N
\biggl(
r,
1
f(z) - p(z)
\biggr)
\mathrm{l}\mathrm{o}\mathrm{g} r
< \rho (f),
where \lambda (f(z) - p(z)) is the exponent of convergence of zeros of f(z) - p(z). In this paper, we
assume that c is a nonzero complex constant, n is a positive integer, k is a nonnegative integer,
unless otherwise specified.
Many mathematicians have made some works on the value distribution of differential polynomials
of different types. For example, some results on the zeros of f(z) and its derivatives can be found in
[10, 11, 13, 18, 23]. For the aim of the paper, we just recall partial results on the zeros distribution
of differential polynomial f(z)nf \prime (z) as follows. Hayman [10] (Theorem 10) proved that if f(z)
is a transcendental entire function, then fn(z)f \prime (z) - 1, n \geq 2, has infinitely many zeros. Mues
[20] proved the above result is also valid for f(z) is a transcendental meromorphic function and
n = 2. Bergweiler and Eremenko [1], Chen and Fang [3] considered the case n = 1 and f(z) is a
transcendental meromorphic function, they obtained the following result.
Theorem A. Let f be a transcendental meromorphic function. If n \geq 1, then fnf \prime - 1 has
infinitely many zeros.
* This work was partially supported by the NSFC (No. 11301260, 11661052, 11461042), the NSF of Jiangxi
(No. 20161BAB211005).
c\bigcirc K. LIU, T. B. CAO, X. L. LIU, 2017
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 1 71
72 K. LIU, T. B. CAO, X. L. LIU
Remark that [fn+1]\prime = (n+1)fnf \prime , Wang and Fang [21] (Corollary 1) improved Theorem A by
proving the following result.
Theorem B. Let f be a transcendental meromorphic function, n, k be two positive integers
with n \geq k + 1. Then (fn)(k) - 1 has infinitely many zeros.
Recently, there has been an interest to consider the zeros distribution of difference polynomials
(see, for example, [2, 4, 14 – 17, 24]). Laine and Yang [15] (Theorem 2) considered the zeros of
fn(z)f(z+c) - a, Liu and Yang [16] (Theorems 1.2 and 1.4) also considered the zeros of fn(z)f(z+
+ c) - p(z) and fn(z)\Delta cf - p(z), where \Delta cf := f(z+ c) - f(z) and p(z) is a nonzero polynomial.
These results can be viewed as the difference analogues of Theorem A, which can be summarized as
follows.
Theorem C. Let f be a transcendental entire function of finite order and p(z) be a nonzero
polynomial. If n \geq 2, then f(z)nf(z + c) - p(z) has infinitely many zeros. If f is not a periodic
function with period c, n \geq 2, then f(z)n\Delta cf - p(z) also has infinitely many zeros.
Difference analogues of Theorem B also deserve to be considered. Liu, Liu and Cao [17]
(Theorems 1.1 and 1.3) obtained the following result.
Theorem D. Let f be a transcendental entire function of finite order and \alpha (z) be a nonzero
small function with respect to f(z). If n \geq k+2, then [f(z)nf(z+ c)](k) - \alpha (z) has infinitely many
zeros. If f is not a periodic function with period c and n \geq k + 3, then [f(z)n\Delta cf ]
(k) - \alpha (z) has
infinitely many zeros.
Obviously, Theorem D is an extention of Theorem C of the case k = 0. Remark that n is related
to k in Theorem D. In the following, we will continue to consider what conditions on n and k
guarantee that [f(z)nf(z + c)](k) - \alpha (z) or [f(z)n\Delta cf ]
(k) - \alpha (z) can have infinitely many zeros?
Firstly, we consider the zeros distribution of [f(z)nf(z + c)](k) - \alpha (z).
Theorem 1.1. Let f be a transcendental entire function of finite order. If n \geq 1, k \geq 0 and
N
\biggl(
r,
1
f
\biggr)
= S(r, f), then [f(z)nf(z + c)](k) - \alpha (z) has infinitely many zeros, where \alpha (z) is a
nonzero small function with respect to f(z).
Remark 1.1. (1) Theorem 1.1 is not true for entire function with infinite order. It can be seen by
f(z) = zee
z
, ec = - n, \alpha (z) = p(z) is a nonconstant polynomial. Thus [f(z)nf(z+ c)](k) - p(z) =
= [zn(z + c)](k) - p(z) has finitely many zeros.
(2) The condition \alpha (z) \not \equiv 0 can not be deleted, which can be seen by f(z) = ez, ec = 2, thus
[f(z)nf(z + c)](k) = 2(n+ 1)ke(n+1)z has no zeros.
(3) The condition N
\biggl(
r,
1
F
\biggr)
= S(r, f) can not be removed, which can be seen by f(z) = z(ez -
- 1) and ec = - 1. It is easy to get N
\biggl(
r,
1
F
\biggr)
\not = S(r, f). Thus [f(z)f(z+c)]
\prime - 2z - c = - [2z2+(2c+
+ 2)z + c]e2z has only two zeros. [f(z)f(z + c)]
\prime \prime - 2 = - [4z2 + (4c+ 6)z + 3c+ 2]e2z has only
two zeros.
Theorem 1.2. Let f be a transcendental entire function of finite order. If n \geq k
2
+ 1, k \geq 0
and f has infinitely many multiorder zeros, then [f(z)nf(z+ c)](k) - p(z) has infinitely many zeros,
where p(z) is a nonzero polynomial.
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 1
THE PROPERTIES ON DIFFERENTIAL-DIFFERENCE POLYNOMIALS 73
Chen, Huang and Zheng [4] considered the value distribution of f(z)f(z + c) - a if f has a
Borel exceptional value and obtained the following result.
Theorem E ([4], Corollary 1.3). Let f(z) be a transcendental entire function of finite order. If
f(z) has a Borel exceptional value 0, then f(z)f(z+ c) takes every nonzero value a infinitely often.
If f has a Borel exceptional polynomial, we can obtain the following results.
Theorem 1.3. Let f be a transcendental entire function of finite order and p(z) be a nonzero
polynomial and n \geq 1, k \geq 0. If f has a Borel exceptional polynomial q(z), then [f(z)nf(z +
+ c)](k) - p(z) has infinitely many zeros, except that f(z) = q(z) + Aq(z)e\alpha z, n = 1 and p(z) =
= [q(z)q(z + c)](k), where e\alpha c = - 1 and A is nonzero constant.
Remark 1.2. From the example of Remark 1.1 (3), we know that the exceptional case can occur.
Corollary 1.1. Under the assumptions of Theorem 1.3. If f(z) - q(z) has no zeros, then
[f(z)nf(z + c)](k) - p(z) has infinitely many zeros, except that f(z) = q + he\alpha z, n = 1, q(z) = q,
p(z) = q2, where e\alpha c = - 1 and h is nonzero constant.
Remark 1.3. From Theorems 1.1 – 1.3, if n \geq 2, k \geq 1 and [f(z)nf(z + c)](k) - p(z) has
finitely many zeros, then f(z) should satisfy: (i) f has infinitely many zeros, (ii) all zeros of f are
simple, (iii) f has no Borel exceptional polynomial.
The following theorems are related to the zeros of [f(z)n\Delta cf ]
(k) - \alpha (z).
Theorem 1.4. Let f be a transcendental entire function of finite order, not a periodic function
with period c. If n \geq 1, k \geq 0 and N
\biggl(
r,
1
F
\biggr)
= S(r, f), then [f(z)n\Delta cf ]
(k) - \alpha (z) has infinitely
many zeros.
Remark 1.4. The condition N
\biggl(
r,
1
f
\biggr)
= S(r, f) can not be deleted in Theorem 1.4, which
can be seen by function f(z) = ez + z, ec = 1, it is easy to see N
\biggl(
r,
1
f
\biggr)
\not = S(r, f), thus
[f(z)(f(z+ c) - f(z))]\prime - c = cez has no zeros. The condition \alpha (z) \not \equiv 0 can not be removed, which
can be seen by function f(z) = ez, ec = 2, thus [f(z)nf(z+ c)](k) = (n+1)ke(n+1)z has no zeros.
Theorem 1.5. Let f be a transcendental entire function of finite order, not a periodic function
with period c. If n \geq k
2
+ 1, k \geq 0 and f has infinitely many multiorder zeros, then [f(z)n(f(z +
+ c) - f(z))](k) - p(z) has infinitely many zeros.
Theorem 1.6. Let f be a transcendental entire function of finite order, not a periodic function
with period c, let n \geq 1, k \geq 0. If f has a Borel exceptional polynomial q(z), then [f(z)n(f(z +
+ c) - f(z))](k) - p(z) has infinitely many zeros, except that f(z) = q(z) + he\alpha z, n = 1 and
p(z) = [q(z)(q(z + c) - q(z))](k), where e\alpha c = 1.
Remark 1.5. The exceptional case also can occur in Theorem 1.6, which can be seen by f(z) =
= ez + z2 and e\alpha c = 1. Then\bigl[
f(z)(f(z + c) - f(z))
\bigr] \prime - [z2(2zc+ c2)]\prime = (2zc+ c2 + 2c)ez
has only one zero.
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 1
74 K. LIU, T. B. CAO, X. L. LIU
2. Some lemmas. The difference logarithmic derivative lemma, given by Halburd and Korhonen
[8] (Theorem 2.1), [9] (Theorem 5.6), Chiang and Feng [6] (Corollary 2.5) plays an important part
in considering the difference analogues of Nevanlinna theory. Here, we state the version of [9]
(Theorem 5.6).
Lemma 2.1. Let f be a transcendental meromorphic function of finite order. Then
m
\biggl(
r,
f(z + c)
f(z)
\biggr)
= S(r, f).
Lemma 2.2 ([6], Theorem 2.1). Let f(z) be a transcendental meromorphic function of finite
order. Then
T (r, f(z + c)) = T (r, f) + S(r, f).
Lemma 2.3 ([22], Theorem 1.22). Let f(z) be a transcendental meromorphic function, n be a
positive integer. Then
T (r, f (n)) \leq T (r, f) + nN(r, f) + S(r, f).
Using Lemmas 2.1 and 2.2 and the Valiron – Mohon’ko theorem [19], we obtain the following
lemma.
Lemma 2.4. Let f(z) be a transcendental meromorphic function of finite order with N(r, f) =
= S(r, f), and let F = f(z)nf(z + c). Then
T (r, F ) = (n+ 1)T (r, f) + S(r, f).
Remark 2.1. If f(z) and F (z) satisfy the conditions of Lemma 2.4, then \rho (F ) = \rho (f) =
= \rho (F (k)). However, the above equation is not valid for arbitrary meromorphic functions, see an
example in Remark 1.1 (1).
Lemma 2.5. Let f(z) be a transcendental meromorphic function of finite order with N(r, f) +
+N
\biggl(
r,
1
f
\biggr)
= S(r, f). Then
T (r, f(z)n[f(z + c) - f(z)]) = (n+ 1)T (r, f) + S(r, f). (2.1)
Proof. Let G(z) = f(z)n[f(z + c) - f(z)]. Then
1
f(z)n+1
=
1
G
\biggl[
f(z + c) - f(z)
f(z)
\biggr]
.
Using the first main theorem of Nevanlinna theory, Valiron – Mohon’ko theorem [19] and Lemma 2.1,
we get
(n+ 1)T (r, f) \leq T (r,G(z)) + T
\biggl(
r,
f(z + c) - f(z)
f(z)
\biggr)
+O(1) \leq
\leq T (r,G(z)) +m
\biggl(
r,
f(z + c) - f(z)
f(z)
\biggr)
+N
\biggl(
r,
f(z + c) - f(z)
f(z)
\biggr)
+O(1) \leq
\leq T (r,G(z)) +N
\biggl(
r,
1
f(z)
\biggr)
+N(r, f(z)) +N(r, f(z + c)) + S(r, f) \leq
\leq T (r,G(z)) + S(r, f),
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 1
THE PROPERTIES ON DIFFERENTIAL-DIFFERENCE POLYNOMIALS 75
thus,
T (r,G) \geq (n+ 1)T (r, f) + S(r, f). (2.2)
On the other hand, from Lemma 2.1, we obtain
T (r, f(z)n[f(z + c) - f(z)]) = m(r, f(z)n[f(z + c) - f(z)]) + S(r, f) \leq
\leq m
\biggl(
r, f(z)n+1
\biggl[
f(z + c)
f(z)
- 1
\biggr] \biggr)
+ S(r, f) \leq
\leq (n+ 1)T (r, f) + S(r, f). (2.3)
Combining (2.2) with (2.3), (2.1) follows.
Remark 2.2. Remove the condition N
\biggl(
r,
1
f
\biggr)
= S(r, f), for entire function f(z), we only get
nT (r, f) + S(r, f) \leq T (r, f(z)n[f(z + c) - f(z)] \leq (n+ 1)T (r, f) + S(r, f).
Following Hayman [12, p. 75, 76], we define an \varepsilon -set E to be a countable union of discs not
containing the origin, and subtending angles at the origin whose sum is finite. If E is an \varepsilon -set, then
the set of r \geq 1 for which the circle S(0, r) meets E has finite logarithmic measure.
Lemma 2.6 [2]. Let g(z) be a transcendental meromorphic function of order \sigma (f) < 1, h > 0.
Then there exists an \varepsilon -set E such that
g\prime (z + c)
g(z + c)
\rightarrow 0 and
g(z + c)
g(z)
\rightarrow 1 as z \rightarrow \infty in \BbbC \setminus E,
uniformly in c for | c| \leq h. Further, E may be chosen so that for large z \not \in E, the function g has no
zeros or poles in | \zeta - z| \leq h.
The following result is needed to prove Theorems 1.1 and 1.4.
Lemma 2.7 ([23], Lemma 1). Let f be a nonconstant meromorphic function and \alpha (z) be a
small function of f such that \alpha (z) \not = 0,\infty . Then
T (r, f) \leq N(r, f) +N
\biggl(
r,
1
f
\biggr)
+N
\biggl(
r,
1
f (k) - \alpha
\biggr)
- N
\Biggl(
r,
1\bigl(
f (k)/\alpha
\bigr) \prime
\Biggr)
+ S(r, f).
Lemma 2.8 ([22], Theorem 1.62). Let fj(z) be a meromorphic functions, fk(z), k = 1, 2, . . . , n -
- 1, are not constants, satisfying
\sum n
j=1
fj = 1 and n \geq 3. If fn(z) \not \equiv 0, and
n\sum
j=1
N
\biggl(
r,
1
fj
\biggr)
+ (n - 1)
n\sum
j=1
N(r, fj) < (\lambda + o(1))T (r, fk),
where \lambda < 1, k = 1, 2, . . . , n - 1, then fn(z) \equiv 1.
Lemma 2.9 ([22], Theorem 1.51). Let fj(z), j = 1, 2, . . . , n, n \geq 2, be a meromorphic func-
tions, gj(z), j = 1, 2, . . . , n, be entire functions satisfying:
(i)
\sum n
j=1
fj(z)e
gj(z) \equiv 0,
(ii) when 1 \leq j < k \leq n, gj(z) - gk(z) is not a constant,
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 1
76 K. LIU, T. B. CAO, X. L. LIU
(iii) when 1 \leq j \leq n, 1 \leq h < k \leq n, T (r, fj) = o(T (r, egh - gk)) (r \rightarrow \infty , r \not \in E), where
E \subset (1,\infty ) is of finite linear measure or finite logarithmic measure.
Then fj(z) \equiv 0, j = 1, 2, . . . , n.
The following results are related to the growth of solutions of linear difference equations which
are needed for the proofs of Theorems 1.3 and 1.6. Here, we give the version with small changes of
the type of equations, the proofs are similar.
Lemma 2.10 ([6], Theorem 9.2). Let A0(z), . . . , An(z) be a entire functions such that there
exists an integer l, 0 \leq l \leq n, such that
\rho (Al(z)) > \mathrm{m}\mathrm{a}\mathrm{x}
0\leq l\leq n,j \not =l
\rho (Aj(z)).
If f(z) is a meromorphic solution of
An(z)y(z + cn) + . . .+A1(z)y(z + c1) +A0(z)y(z) = 0,
then \rho (f) \geq \rho (Al(z)) + 1.
Lemma 2.11 ([5], Theorem 1.2). Let P0(z), . . . , Pn(z) be a polynomials such that Pn(z)P0(z) \not \equiv
\not \equiv 0 and satisfy
\mathrm{d}\mathrm{e}\mathrm{g}(Pn(z) + . . .+ P0(z)) = \mathrm{m}\mathrm{a}\mathrm{x}\{ \mathrm{d}\mathrm{e}\mathrm{g}Pj(z) : j = 0, . . . , n\} \geq 1.
Then every finite order meromorphic solution f(z)(\not \equiv 0) of
Pn(z)f(z + cn) + . . .+ P1(z)f(z + c1) + P0(z)f(z) = 0
satisfies \rho (f) \geq 1.
3. Proofs of Theorems 1.1 and 1.4. Let F (z) = f(z)nf(z+ c). From the conditions, it is easy
to get N(r, f) +N
\biggl(
r,
1
f
\biggr)
= S(r, f), thus
N(r, F ) +N
\biggl(
r,
1
F
\biggr)
= S(r, f). (3.1)
From Lemmas 2.4, 2.7 and (3.1), we get
(n+ 1)T (r, f) + S(r, f) = T (r, F ) \leq
\leq N(r, F ) +N
\biggl(
r,
1
F
\biggr)
+N
\biggl(
r,
1
F (k) - \alpha
\biggr)
+ S(r, F ) \leq
\leq N
\biggl(
r,
1
F (k) - \alpha
\biggr)
+ S(r, f).
Hence, [f(z)nf(z + c)](k) - \alpha (z) has infinitely many zeros.
Let G(z) = f(z)n[f(z + c) - f(z)]. Since N(r, f) +N
\biggl(
r,
1
F
\biggr)
= S(r, f), then
N(r,G) +N
\biggl(
r,
1
G
\biggr)
\leq N
\biggl(
r,
1
f(z + c) - f(z)
\biggr)
\leq T (r, f) + S(r, f).
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 1
THE PROPERTIES ON DIFFERENTIAL-DIFFERENCE POLYNOMIALS 77
From Lemmas 2.5, 2.7 and above inequality, we get
(n+ 1)T (r, f) + S(r, f) = T (r,G) \leq N(r,G) +N
\biggl(
r,
1
G
\biggr)
+N
\biggl(
r,
1
G(k) - \alpha
\biggr)
+ S(r,G) \leq
\leq T (r, f) +N
\biggl(
r,
1
G(k) - \alpha
\biggr)
+ S(r, f).
Then nT (r, f) \leq N
\biggl(
r,
1
G(k) - \alpha
\biggr)
+S(r, f). Since n \geq 1, then [f(z)n(f(z+ c) - f(z))](k) - \alpha (z)
has infinitely many zeros.
4. Proofs of Theorems 1.2 and 1.5. We first to give the proof of Theorem 1.2. Let F (z) =
= f(z)nf(z+ c). Assume that F (k)(z) - p(z) has finitely many zeros, from Hadamard factorization
theorem, then
F (k)(z) - p(z) = h(z)eq(z), (4.1)
where h(z) is a nonzero polynomial, q(z) is a nonconstant polynomial, otherwise, if q(z) = A,
where A is a constant, then F (k)(z) - p(z) = h(z)eA. It implies that F (z) = f(z)nf(z + c) is also
a polynomial, which is a contradiction with Lemma 2.4. Differentiating (4.1), we get
F (k+1)(z) - p\prime (z) = [h\prime (z) + h(z)q\prime (z)]eq(z). (4.2)
Combining (4.1) with (4.2), eliminating eq(z), we have
F (k+1)(z)
F (k)(z)
=
h\prime (z) + h(z)q(z)
h(z)
+
\biggl[
p\prime (z) - h\prime (z) + h(z)q(z)
h(z)
p(z)
\biggr]
1
F (k)(z)
. (4.3)
From the left-hand side of (4.3), we remark that the poles of
F (k+1)(z)
F (k)(z)
must be simple. If f has
infinitely many multiorder zeros and n \geq k
2
+ 1, we can find z0 which is a zero of f and not the
zero of h(z) and p\prime (z) - h\prime (z) + h(z)q(z)
h(z)
p(z), then the poles of right-hand side of (4.3) must be
multiorder, a contradiction. Thus, we have the proof of Theorem 1.2.
Using the similar method as above, we can get the proof of Theorem 1.5.
5. Proof of Theorem 1.3. Since q(z) is a Borel exceptional polynomial of f(z), thus the
value 1 is a Borel exceptional value of
f(z)
q(z)
, then f(z) must have positive integer order [22, p. 106]
(Corollary). Assume that \rho (f) = s, s is a positive integer, then transcendental entire function f(z)
can be written as f(z) = q(z) + h(z)e\alpha z
s
, where \alpha is a nonzero constant and h(z) is a nonzero
entire function with \lambda (h) \leq \rho (h) < \rho (f) = s. Hence,
f(z + c) = q(z + c) + h(z + c)e\alpha (z+c)s = q(z + c) + h1(z)e
\alpha zs , (5.1)
where
h1(z) = h(z + c)e\alpha (C
1
s z
s - 1c+C2
s z
s - 2c2+...+Cs - 1
s zcs - 1+cs). (5.2)
Suppose that
\bigl[
f(z)nf(z+c)
\bigr] (k) - p(z) has finitely many zeros, from Hadamard factorization theorem
and \rho
\bigl(
f(z)nf(z + c)
\bigr)
= \rho (f) = \rho
\bigl(
(f(z)nf(z + c))(k)
\bigr)
, then we assume that
[f(z)nf(z + c)](k) - p(z) = A(z)e\beta z
s
, (5.3)
where A(z) is an entire function with \rho (A) < s and has finitely many zeros, \beta is a nonzero constant.
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78 K. LIU, T. B. CAO, X. L. LIU
Case 1. If k = 0, then n = 1 from Theorem C, hence,
f(z)f(z + c) - p(z) = A(z)e\beta z
s
. (5.4)
Substituting f(z) = q(z) + h(z)e\alpha z
s
into (5.4), we get
h(z)h1(z)e
2\alpha zs + (h(z)q(z + c) + q(z)h1(z))e
\alpha zs =
= p(z) - q(z)q(z + c) - A(z)e\beta z
s
. (5.5)
Combining (5.5) with the definition of type of entire function f(z), that is
\tau f = \mathrm{l}\mathrm{i}\mathrm{m} \mathrm{s}\mathrm{u}\mathrm{p}
r\rightarrow \infty
\mathrm{l}\mathrm{o}\mathrm{g}M(r, f)
r\rho (f)
,
we get \beta = 2\alpha , where M(r, f) = \mathrm{m}\mathrm{a}\mathrm{x}| z| =r | f(z)| .
Case 1.1. If p(z) - q(z)q(z + c) \equiv 0, from Lemma 2.9 and (5.5), then
(h(z)q(z + c) + q(z)h1(z))e
\alpha zs \equiv 0.
It implies that
q(z + c)h(z) + q(z)h(z + c)e\alpha (C
1
s z
s - 1c+C2
s z
s - 2c2+...+Cs - 1
s zcs - 1+cs) \equiv 0.
We affirm that s = 1, otherwise, if s \geq 2, from Lemma 2.10, then \rho (h) \geq s follows, which is a
contradiction with \rho (h) < s. Thus,
q(z + c)h(z) + q(z)h(z + c)e\alpha c \equiv 0. (5.6)
Combining (5.6), \rho (h) < 1 with Lemma 2.11, we get the degree of q(z + c) + q(z)e\alpha c must be less
than the degree of q(z). If q(z) is not a constant, then e\alpha c = - 1. Hence,
q(z + c)h(z) - q(z)h(z + c) \equiv 0,
which implies that
h(z + c)
q(z + c)
=
h(z)
q(z)
, thus H(z) =
h(z)
q(z)
is a periodic function with period c. Since
\rho (H) = \rho (h) < 1, thus H(z) must be a constant A, which implies that h(z) must be a polynomial
with the form h(z) = Aq(z).
If q(z) is a constant, we have h(z)+h(z+c)e\alpha c \equiv 0. From Lemma 2.6, we get e\alpha c = - 1. Since
that \rho (h) < s = 1, then h(z) must be a constant. Thus, we get h(z) and q(z) are two constants,
which can be written as h(z) = Aq(z).
Thus, we get f(z) = q(z) +Aq(z)e\alpha z, where e\alpha c = - 1.
Case 1.2. If p(z) - q(z)q(z + c) \not \equiv 0, then
[h(z)h1(z) - A(z)]e2\alpha z
s
+ (h(z)q(z + c) + q(z)h1(z))e
\alpha zs =
= p(z) - q(z)q(z + c).
Let f1(z) =
[h(z)h1(z) - A(z)]
p(z) - q(z)q(z + c)
e2\alpha z
s
and f2(z) =
(h(z)q(z + c) + q(z)h1(z))
p(z) - q(z)q(z + c)
e\alpha z
s
. Thus, f1(z)+
+ f2(z) = 1. From the second main theorem, we get
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THE PROPERTIES ON DIFFERENTIAL-DIFFERENCE POLYNOMIALS 79
T (r, f1) \leq N(r, f1) +N
\biggl(
r,
1
f1
\biggr)
+N
\biggl(
r,
1
f1 - 1
\biggr)
+ S(r, f1) \leq
\leq N
\biggl(
r,
1
f1
\biggr)
+N
\biggl(
r,
1
f2
\biggr)
+ S(r, f1) \leq
\leq O(rs - 1+\varepsilon ) + S(r, f1),
which is a contradiction with \rho (f1) = s. Thus, [f(z)nf(z + c)](k) - p(z) has infinitely many zeros.
Case 2. If k \geq 1, from (5.1) and (5.3), we have\bigl[
(q(z) + h(z)e\alpha z
s
)n(q(z + c) + h1(z)e
\alpha zs)
\bigr] (k) - A(z)e\beta z
s
= p(z),
which implies that \bigl[
q(z)nq(z + c) +B1(z)e
\alpha zs + . . .+Bj(z)e
j\alpha zs + . . .
. . .+Bn(z)e
n\alpha zs + h(z)h1(z)e
(n+1)\alpha zs
\bigr] (k) - A(z)e\beta z
s
= p(z), (5.7)
where
Bj(z) = Cj
nq(z)
n - jq(z + c)h(z)j + Cj - 1
n q(z)n - j+1h1(z)h(z)
j - 1,
\rho (Bj(z)) < s, j = 1, . . . , n. For any integer k, from (5.7), we obtain
D1(z)e
\alpha zs +D2(z)e
2\alpha zs + . . .+Dj(z)e
j\alpha zs + . . .
. . .+Dn(z)e
n\alpha zs +Dn+1(z)e
(n+1)\alpha zs - A(z)e\beta z
s
= p(z) - [q(z)nq(z + c)](k), (5.8)
where Dj(z) are differential polynomials of h(z), h1(z), q(z), q(z+ c) and their powers derivatives
and \rho (Dj(z)) < s, j = 1, . . . , n+ 1. In the following, we will consider two cases.
Case 2.1. If p(z) - [q(z)nq(z+c)](k) \equiv 0, from Lemma 2.9, then all Dj(z) \equiv 0, j = 1, 2, . . . , n,
and Dn+1(z) - A(z) \equiv 0. We affirm that n = 1, otherwise, let n \geq 2. If k = 1, from D1(z) \equiv 0,
then we get
B\prime
1(z) + \alpha szs - 1B1(z) = 0,
which implies the nontrivial solution B(z) of the above first order differential equation satisfying
\rho (B(z)) = s, which is a contradiction with \rho (B(z)) < s, thus, B1(z) \equiv 0. If k = 2, let g(z) =
= B\prime
1(z) + \alpha szs - 1B1(z), then we have g\prime (z) + \alpha szs - 1g(z) = 0, which also implies \rho (g(z)) = s,
a contradiction with \rho (g(z)) = \rho (B(z)) < s. Using this method for any positive integer k, we can
get B1(z) \equiv 0, that is
C1
nq(z)
n - 1q(z + c)h(z) + q(z)nh(z + c)e\alpha (C
1
s z
s - 1c+...+Cs - 1
s zcs - 1+cs) \equiv 0.
From Lemma 2.10, we get s = 1, which implies that
C1
nq(z)
n - 1q(z + c)h(z) + q(z)nh(z + c)e\alpha c \equiv 0. (5.9)
From Lemma 2.11, since \rho (h(z)) < 1, then the degree of C1
nq(z)
n - 1q(z + c) + q(z)ne\alpha c must be
less than the degree of q(z)n. Thus, we get
e\alpha c = - C1
n (5.10)
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80 K. LIU, T. B. CAO, X. L. LIU
provided that q(z) is not a constant. Using the same discussions as above, we also have B2(z) \equiv 0,
that is,
C2
nq(z)
n - 2q(z + c)h(z)2 + C1
nq(z)
n - 1h(z + c)e\alpha ch(z) \equiv 0.
Similar as the above, if q(z) is not a constant, we obtain
C1
ne
\alpha c = - C2
n. (5.11)
From (5.10) and (5.11), we get n = 0, a contradiction. Thus, we have n = 1. From (5.7), Lemmas 2.9
and 2.8, we obtain B1(z) \equiv 0. Then (5.9) is the same to (5.6). Thus, we get f(z) = q(z)+Aq(z)e\alpha z,
where e\alpha c = - 1.
Case 2.2. If p(z) - [q(z)nq(z + c)](k) \not \equiv 0, combining \rho (Dj(z)) < s, \rho (A(z)) < s with (5.8)
and Lemma 2.8, we get
[Dn+1(z) - A(z)]e(n+1)\alpha zs = p(z) - [q(z)nq(z + c)](k)
or
Dj(z)e
j\alpha zs = p(z) - [q(z)nq(z + c)](k),
which are impossible. Thus [f(z)nf(z + c)](k) - p(z) has infinitely many zeros.
Theorem 1.3 is proved.
6. Proof of Theorem 1.6. Similar as the beginning of proof of Theorem 1.3, f(z) also can
be written as f(z) = q(z) + h(z)e\alpha z
s
, where \alpha is a nonzero constant and h(z) is a nonzero entire
function with \lambda (h) \leq \rho (h) < \rho (f) = s. From (5.1), we have
f(z + c) - f(z) = q(z + c) - q(z) + (h1(z) - h(z))e\alpha z
s
:= q1(z) + h2(z)e
\alpha zs .
Suppose that [f(z)n(f(z + c) - f(z))](k) - p(z) has finitely many zeros, then from Hadamard
factorization theorem, we obtain\bigl[
f(z)n(f(z + c) - f(z))
\bigr] (k) - p(z) = B(z)e\gamma z
s
,
where B(z) is an entire function of finitely many zeros with order \rho (B) < s and \gamma is a nonzero
constant.
Case 1. If k = 0. From Theorem C, we get n = 1, hence,
f(z)
\bigl[
f(z + c) - f(z)
\bigr]
- p(z) = B(z)e\gamma z
s
. (6.1)
Substituting f(z) = q(z) + h(z)e\alpha z
s
into (6.1), we have
h(z)(h1(z) - h(z))e2\alpha z
s
+ (h(z)(q(z + c) - 2q(z)) + q(z)h1(z))e
\alpha zs =
= p(z) -
\bigl[
q(z)(q(z + c) - q(z))
\bigr]
+B(z)e\gamma z
s
. (6.2)
Case 1.1: p(z) - q(z)(q(z+c) - q(z)) \equiv 0. If h1(z) \equiv h(z), then \alpha = \gamma follows from the above
equation. From (5.2) and Lemma 2.10, we get s = 1, thus h(z) = e\alpha ch(z + c). From Lemma 2.6,
we get h(z) is a constant and e\alpha c = 1. Thus, f(z) = q(z) + he\alpha z, where e\alpha c = 1.
If h1(z) \not \equiv h(z), then \gamma = 2\alpha . From Lemma 2.9, we obtain
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 1
THE PROPERTIES ON DIFFERENTIAL-DIFFERENCE POLYNOMIALS 81
h(z)
\bigl[
q(z + c) - 2q(z)
\bigr]
+ q(z)h(z + c)e\alpha (C
1
s z
s - 1c+C2
s z
s - 2c2+...+Cs - 1
s zcs - 1+cs) = 0.
From Lemma 2.10, we get s = 1, thus,
h(z)
\bigl[
q(z + c) - 2q(z)
\bigr]
+ q(z)h(z + c)e\alpha c = 0.
Since \rho (h) < 1, from Lemma 2.11, then q(z) must be a constant or e\alpha c = 1 when q(z) is not a
constant. If q(z) is a constant, from Lemma 2.6, then h(z) reduces a constant, thus f(z) is a periodic
function with period c, a contradiction. If e\alpha c = 1 and q(z) is not a constant, from (6.2), we have
h(z)(h(z + c) - h(z)) = B(z), combining with \rho (h) < 1 and h(z) is an entire function, we obtain
\rho (h(z + c) - h(z)) < 1, thus B(z) must have infinitely many zeros, a contradiction.
Case 1.2: p(z) - q(z)(q(z + c) - q(z)) \not \equiv 0, this case is similar as the Case 1.2 in the proof of
Theorem 1.3.
Case 2. If k \geq 1, we also can get a similar equation as (5.8),
E1(z)e
\alpha zs + . . .+ Ej(z)e
j\alpha zs + . . .+ En(z)e
n\alpha zsEn+1(z)e
(n+1)\alpha zs - B(z)e\beta z
s
=
= p(z) -
\bigl[
q(z)n(q(z + c) - q(z))
\bigr] (k)
,
where Ej are differential polynomials of h(z), h2(z), q(z) and q(z+c) and their powers derivatives
and \rho (Ej) < s. In the following, we also divided into two cases p(z) - [q(z)n(q(z+c) - q(z))](k) \equiv 0
and p(z) - [q(z)n(q(z + c) - q(z))](k) \not \equiv 0. Using the similar method as the Case 2 of the proof of
Theorem 1.3, then we can have the completed proof of Theorem 1.6.
7. Discussions. In the paper, we gave the example to show that [f(z)f(z + c)]\prime - \alpha (z) and
[f(z)(f(z + c) - f(z))]\prime - \alpha (z) can admit finitely many zeros in Remarks 1.1, 1.4, and 1.5. By
theorems of the paper, we see that if f be a transcendental entire function of finite order and n \geq 2,
k \geq 1, then [f(z)nf(z + c)](k) - \alpha (z) or [f(z)n(f(z + c) - f(z))](k) - \alpha (z) have infinitely many
zeros under some additional conditions, for example N
\biggl(
r,
1
F
\biggr)
= S(r, f) or f(z) has infinitely many
multiorder zeros or f(z) has a Borel exceptional polynomial. Unfortunately, we have no methods to
remove these additional conditions. For the further studying, we raise the following two questions,
where \alpha (z) is a nonzero small function with respect to f(z).
Question 1: Let f be a transcendental entire function of finite order and n \geq 2, k \geq 1. Can we
get [f(z)nf(z + c)](k) - \alpha (z) have infinitely many zeros?
Question 2: Let f be a transcendental entire function of finite order, not a periodic function
with period c and n \geq 2, k \geq 1. Can we get [f(z)n(f(z + c) - f(z))](k) - \alpha (z) have infinitely
many zeros?
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ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 1
|
| id | umjimathkievua-article-1676 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:10:22Z |
| publishDate | 2017 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/d2/4198c83a563d606d929eeca676c0f6d2.pdf |
| spelling | umjimathkievua-article-16762019-12-05T09:23:35Z The properties on differential-difference polynomials Властивостi диференцiально-рiзницевих полiномiв Cao, T. B. Liu, K. Liu, X. L. Цао, Т. Б. Лю, К. Лю, X. Л. The main aim of this paper is to improve some classical results on the distribution of zeros for differential polynomials and differential-difference polynomials. We present some results on the distribution of zeros of $[f(z)^nf(z + c)]^{(k)} \alpha (z)$ and $[f(z)^n(f(z + c) f(z))]^{(k)} \alpha (z)$ and give some examples to show that the results are best possible in a certain sense. Основною метою роботи є полiпшення деяких класичних результатiв про розподiли нулiв для диференцiальних та диференцiально-рiзницевих полiномiв. Нaведено деякi результати щодо розподiлiв нулiв для $[f(z)^nf(z + c)]^{(k)} \alpha (z)$ та $[f(z)^n(f(z + c) f(z))]^{(k)} \alpha (z)$, а також деякi приклади, якi демонструють, що отриманi результати є, в певному розумiннi, найкращими. Institute of Mathematics, NAS of Ukraine 2017-01-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/1676 Ukrains’kyi Matematychnyi Zhurnal; Vol. 69 No. 1 (2017); 71-82 Український математичний журнал; Том 69 № 1 (2017); 71-82 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/1676/658 Copyright (c) 2017 Cao T. B.; Liu K.; Liu X. L. |
| spellingShingle | Cao, T. B. Liu, K. Liu, X. L. Цао, Т. Б. Лю, К. Лю, X. Л. The properties on differential-difference polynomials |
| title | The properties on differential-difference polynomials |
| title_alt | Властивостi диференцiально-рiзницевих полiномiв |
| title_full | The properties on differential-difference polynomials |
| title_fullStr | The properties on differential-difference polynomials |
| title_full_unstemmed | The properties on differential-difference polynomials |
| title_short | The properties on differential-difference polynomials |
| title_sort | properties on differential-difference polynomials |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/1676 |
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