Let f(z) be a function transcendental and meromorphic in the plane of growth order less than 1. This paper focuses on discuss and estimate the number of the zeros of a certain homogeneous difference polynomials of degree k in f(z), and obtains that this certain homogeneous difference polynomials has infinitely many zeros.
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 [
Recently, a number of papers research on complex difference equations and differences analogues of Nevanlinna’s theory [2-6]. Bergweiler and Langley [
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 [
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 [
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.
Lemma 2.1. (see [
as, uniformly in for.
Lemma 2.2. (see [
Second, there exists a set of linear measure, such that for
,
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
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
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
By Lemma 2.3, has at most finite many poles. Again from Lemma 2.1, there exists such that as, we have
It follows that from (2.6) and (2.7)
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,
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)
Since and is transcendental, there exists a sequence, such that for arbitrary, we have that
Then, we induce that from (2.4) and (2.11)
Therefore, from (2.12) and (2.13) we have
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
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
We have that from and (2.16)
If is a poles of with multiplicity m, then must be a poles of with multiplicity, so that we denote by, that is,
where is a canonical product of distinct poles of. By (2.16), we obtain that
From (2.15) and (2.18), we deduce that
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.
From Lemma 2.2 we see that there exists a sufficiently large, a positive number such that
and there exists a set with linear measure, such that for any, we have the following equation
where satisfies the following express,
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
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
From (2.23)-(2.25), we know that there exists such that, , , , and have no zeros and poles on the circle. Therefore,
Applying Rouché’s Theorem to and, we obtain the following equation
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
Since the pole z0 of has multiplicity, we have the following equality
And obviously,
Substituting (2.28), (2.30) into (2.27), we obtain
If, then. Thus, we have that by (2.31)
then.
If, we have that from (2.31)
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,.