Advances in Pure Mathematics
Vol.3 No.1(2013), Article ID:27361,6 pages DOI:10.4236/apm.2013.31012
The Zeros of a Certain Homogeneous Difference Polynomials of Meromorphic Functions*
1Department of Mathematics, Southwest University of Science and Technology, Mianyang, China
2Department of Material Science and Engineer, Southwest University of Science and Technology, Mianyang, China
Email: luqiankuo1965@hotmail.com, liaoql@swust.edu.cn
Received October 17, 2012; revised November 26, 2012; accepted December 4, 2012
Keywords: Meromorphic Functions; Zeros; Homogeneous Difference Polynomials
ABSTRACT
Let be a function transcendental and meromorphic in the plane of growth order less than 1. This paper focuses on discussing and estimating the number of the zeros of a certain homogeneous difference polynomials of degree k in
, and obtains that this certain homogeneous difference polynomials has infinitely many zeros.
1. Introduction and the Main Result
Let be a function transcendental and meromorphic in the plane. In what follow, we denote the convergence exponent of the zeros of
by
, the growth order of
by
, and the lower order of
by
.
Following Whittaker [1], define the forward differences to be k times iteration of the difference operator
, that is,
(1.1)
Recently, a number of papers research on complex difference equations and differences analogues of Nevanlinna’s theory [2-6]. Bergweiler and Langley [7] firstly investigated the existence of zeros of, and obtained a result as follow.
Theorem 1.1. Let f be a function transcendental and meromorphic of lower order in the plane. Let
be such that at most finitely many poles
of
satisfy
. Then
has infinitely many zeros.
In 2008, Z. X. Chen and K. H. Shon [8].
Theorem 1.2. Let and f be a function transcendental and meromorphic of lower order
in the plane. Let
and a set
consist of all poles of
, such that
at most except finitely many exceptions. Then has infinitely many zeros.
In 2009, Z. X. Chen and K. H. Shon [9] continue to investigate the existence of the zeros of the difference polynomials defined as follows
(1.2)
(1.3)
and obtained two results.
Theorem 1.3. Let f be a function transcendental and meromorphic of growth order, and
be two complex numbers, such that
, and
. If
has at most finitely many poles
satisfying
, then
has infinitely many zeros, and
.
In particular, if has at most finitely many zeros
satisfying
, then
has also infinitely many zeros, and
.
Theorem 1.4. Let satisfy the conditions in Theorem 1.3, If
has at most finitely many poles
satisfying
then
has infinitely many zeros, and
.
In particular, if has at most finitely many zeros
such that
, then
has also infinitely many zeros, and
.
It is not difficult to understand that defined by (1.2) is more general difference polynomials than
or
and Theorem 1.3 extends Theorem 1.1. Therefore, we pose naturally one question whether more general difference polynomials than
defined by (1.3) has also infinitely many zeros. In this paper, we focus on research a certain homogeneous difference polynomials and affirm to answer this problem.
Theorem 1.5. Suppose that k is a positive integer,. Let
be a function transcendental and meromorphic of growth order
, and there exists k complex numbers
such that
. If
has at most finitely many poles bj
satisfying
Then has infinitely many zeros, and
.
In particular, if has at most finitely many zeros
satisfying
, then
has also infinitely many zeros, and
.
2. Lemmas
Lemma 2.1. (see [7]) Let f be a function transcendental and meromorphic in the plane of growth order less than 1, and. Then there exists an
E such that
, (2.1)
as, uniformly in
for
.
Lemma 2.2. (see [7]) Let be a function transcendental and meromorphic in the plane of lower order
. Then there exists arbitrarily large R with the following properties. First,
. (2.2)
Second, there exists a set of linear measure
, such that for
,
(2.3)
on.
Lemma 2.3. Let be a function transcendental and meromorphic in the plane with growth order
. Supposed that
. If the homogeneous difference polynomials
or quotient of difference polynomials
is rational functions, then has at most finite many poles.
Proof. Without loss of generality, we assume that c1 = 1. Because that the homogeneous difference polynomials is rational, there exists a rational functions
such that
. (2.4)
Set
and
.
So there exists no poles of in the domain
and
.
Now we complete the proof of the conclusion that has at most finite.
Now we complete the proof of the conclusion that has at most finite many poles. Suppose not, there exists one domain Dj, for example D1, in which
has infinitely many poles. We assume that the set
consists of all poles of
in D1 and
and divide it into two cases:
Case 1. There exists, such that for an arbitrary
, there does not exist
such that
, that is, for an arbitrary
, we have
. In fact, since
and
, this case appears whenever
for every
. Thereforewe know
and that there exists a unbounded subsequence set
in which every
is the poles of
. Hence we know that there are at least one in these signs
, which takes every positive integer, for instance, m1 takes every positive integer.
Thus, , which contradicts the hypothesis of Lemma 2.3.
Case 2. There exists, such that for every
, there exists
, such that
. From
and
, we have that
. As the set A is infinite and B has only a finite elementary, there exists
, satisfying
(2.5)
By putting in order again, we have the following express
and
where
.
Now set
where
Since are between
and
,
are between
and
, we know that
, that is,
. From
, and (2.4), we know that one of
and
is the pole of
. If
is the pole of
, then from the some argument above we have one of
is also the pole of
. If
is the pole of
, then one of
is also the pole of
. On the analogy of this, it is not difficult to find there exists at least one of
, for instance, we assume that is j, such that j takes all value of
. From
to
,
to
, and
to
, repeating above proceeding, we have
where
Therefore, we can see that there exist infinite many poles of whose expressions are as follows
where
in which we can find that one of takes every positive integer. Thus,
, which still contradict the hypothesis on the growth order of
in Lemma 2.3.
By the similar method to above, it is easy to prove that has at most finite many poles whenever quotient of difference polynomials
is rational functions.
Lemma 2.4. Let be a function transcendental and meromorphic in the plane with growth order
. Supposed that
, then the homogeneous difference polynomials
and
also are transcendental.
Proof. Suppose first that there exists a rational function, such that
(2.6)
By Lemma 2.3, has at most finite many poles. Again from Lemma 2.1, there exists
such that as
, we have
(2.7)
It follows that from (2.6) and (2.7)
(2.8)
We write for a polynomial formed by the pole of
, and
. So
is an entire function, and
. With the standard result in the Wiman-Valiron Theory, we know that there exists a subset
with finite logarithmic measure
, in which for an sufficiently large
the following equality holds
.
Thus,
(2.9)
where, and
, as
. Set
. Since
is
, we have that F1 also is of finite logarithmic measure. Therefore, for all z,
, and
we immediately deduce that from (2.8) and (2.9)
(2.10)
Since and
is transcendental, there exists a sequence
, such that for arbitrary
, we have that
(2.11)
. (2.12)
Then, we induce that from (2.4) and (2.11)
. (2.13)
Therefore, from (2.12) and (2.13) we have
(2.14)
By (2.10), (2.12), and (2.14), we deduce easily that, which contradicts the assumption on
, that is,
transcendental.
Lemma 2.5. Let be a function transcendental and meromorphic in the plane, whose growth orde
. Supposed that
, and
. Then
Proof. For of growth order
, from Hadamard’s factorization theorem we have
, (2.15)
where and
are respectively the canonical product of zeros and poles of
, satisfying
.
From (2.15), we have
Therefore, if, we deduce that
. For
, the following equations hold
(2.16)
We have that from and (2.16)
. (2.17)
If is a poles of
with multiplicity m, then
must be a poles of
with multiplicity
, so that we denote
by
, that is,
, (2.18)
where is a canonical product of distinct poles of
. By (2.16), we obtain that
. (2.19)
From (2.15) and (2.18), we deduce that
(2.20)
Thus, if z0 is the pole of (that is,
), then
,
, but
. Hence, we have that
is not the zero of
.
So that
and
This completes the proof of Lemma 2.5.
3. Proofs of Theorem 1.5
From Lemma 2.2 we see that there exists a sufficiently large, a positive number
such that
(2.21)
and there exists a set with linear measure
, such that for any
, we have the following equation
, (2.22)
where satisfies the following express,
, (2.23)
here.
On the other hand, under the condition of Theorem 1.5 and from Lemma 2.4 we know transcendental.
Suppose that E concludes all of zeros and poles of
, and
. Setting
Since the property of and
, we have that
is with finite logarithmic measure, and
has linear measure
for sufficiently large
.
We assume that is a set, such that
. (2.24)
Noting that there exists many points
at most from (2.22), at which
is not continuous, and also for any
holds for some
whenever
. Therefore,
has linear measure
, (2.25)
From (2.23)-(2.25), we know that there exists such that
,
,
,
, and
have no zeros and poles on the circle
. Therefore,
. (2.26)
Applying Rouché’s Theorem to and
, we obtain the following equation
. (2.27)
Without loss of generality, we may assume that
for all poles of
. From the assumption in Theorem 1.3, we know that there exists positive number
, which does not depend on R and r, such that if
is a pole of
with multiplicity
,
then by the expression of
and
,
we see that z0, are respectively the pole of
with multiplicity
. Therefore, we deduce that
. (2.28)
Since the pole z0 of has multiplicity
, we have the following equality
. (2.29)
And obviously,
. (2.30)
Substituting (2.28), (2.30) into (2.27), we obtain
(2.31)
If, then
. Thus, we have that by (2.31)
, (2.32)
then.
If, we have that from (2.31)
. (2.33)
By Lemma 2.5 and (2.33), we deduce that
.
In particular, if is the zero of
then z0 is, also the zero of
. On the other hand, if z1 is the zero of
, but not the zero of
, then z1 must be the zero of
, that is,
for some j. From the assumption in Theorem 1.5 that
has at most finitely many zeros
satisfying
, we have
Therefore,.
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NOTES
*Supported by China Industrial Technology Development Program (B3120110001) and Sichuan Provincial Science and Technology Department of Applied Foundation Project (07JY029-006).