Research Article | Open Access

Nan Li, Lianzhong Yang, "Value Distribution of Certain Type of Difference Polynomials", *Abstract and Applied Analysis*, vol. 2014, Article ID 278786, 6 pages, 2014. https://doi.org/10.1155/2014/278786

# Value Distribution of Certain Type of Difference Polynomials

**Academic Editor:**Guozhen Lu

#### Abstract

We investigate the value distribution of difference product , for and , respectively, where is a transcendental entire function of finite order and are constants satisfying .

#### 1. Introduction

In this paper, we assume that the reader is familiar with the basic notions of Nevanlinnaās value distribution theory (see [1ā3]). The notation is defined to be any quantity satisfying as , possibly outside a set of finite linear measures. In addition, we use the notation to denote the order of growth of the meromorphic function and to denote the exponent of convergence of zeros of .

Hayman proved the following theorem in [4].

Theorem 1. *Let be a transcendental integral function and let be an integer; then assumes all values except possibly zero infinitely often.*

Clunie proved that if , then Theorem 1 remains valid.

Recently, many papers (see [5ā17]) focus on complex difference. They obtain many new results on difference using the value distribution theory of meromorphic functions.

In [12], Laine and Yang found a difference analogue of Haymanās result as follows.

Theorem 2. *Let be a transcendental entire function of finite order and a nonzero complex constant. Then for , assumes every nonzero value infinitely often.*

Liu and Yang [14] proved the following theorem.

Theorem 3. *Let be a transcendental entire function of finite order and let be a nonzero complex constant, . Then for , has infinitely many zeros, where is a polynomial in .*

Chen [6] proved the following theorem.

Theorem 4. *Let be a transcendental entire function of finite order and let be a constant satisfying . Set where , and is an integer. Then the following statements hold.*(i)*If satisfies or has infinitely many zeros, then has infinitely many zeros.*(ii)*If has only finitely many zeros and , then has only finitely many zeros.*

It is natural to ask what condition will guarantee that assumes every nonzero and zero value infinitely often, where is a linear th order difference operator with varying shifts, operating on a transcendental entire function of finite order.

In this paper, we consider the above question for and , respectively, and obtain the following results.

Theorem 5. *Let be a transcendental entire function of finite order and let , āā be constant satisfying and when . Set , where are integers. Then the following statements hold.*(i)*If satisfies or has infinitely many zeros, then has infinitely many zeros.*(ii)*If has only finitely many zeros and , then has only finitely many zeros.*(iii)* has infinitely many zeros, and , where is a small function of .*

*Remark 6. *The result of Theorem 5 may be false if . For example, if , we have that (where is a constant satisfying ) has no zero, but has infinitely many zeros. This also shows that the restriction in Theorem 5(ii) is sharp. The following example shows that the assumption in Theorem 5(i) cannot be deleted. In fact, let ; we have .

By (i) and (iii) of Theorem 5, we can easily obtain the following corollary.

Corollary 7. *Let be a transcendental entire function of finite order and let , āā be constants satisfying and when . Set , where are integers. If or has infinitely many zeros, then takes every value infinitely often.*

Theorem 8. *Let be a finite-order transcendental entire function with a finite Borel exceptional value , and let be constants satisfying where . Set . Then the following statements hold. *(i)* takes every nonzero value infinitely often and satisfies .*(ii)*If , then has no finite Borel exceptional value.*(iii)*If , then is also the Borel exceptional value of . So that has no nonzero finite Borel exceptional value.*

Theorem 9. *Let be a transcendental entire function of finite order and let be constants satisfying . Set .**If there exists an infinite sequence satisfying , then takes every value (including ) infinitely often.*

Theorem 10. *Let be a transcendental entire function ofāāfinite order and let be distinct constants satisfying . Set , where is an integer.*(i)*If has only finitely many zeros and or has infinitely many zeros, then has infinitely many zeros.*(ii)*If has only finitely many zeros and , then has only finitely many zeros.*

*Example 11. *An entire function satisfies Theorem 8 (iii), it has Borel exceptional value , and let , , , , , and . Then
has also the Borel exceptional value since .

Simultaneously, also satisfies Theorem 10(i), although has no zero, we can also get has infinitely many zeros since .

*Example 12. *An entire function satisfies Theorem 8(ii), it has Borel exceptional value , and let , , , , , and . Then
has no finite Borel exceptional value.

#### 2. Some Lemmas

Lemma 13 (see [9]). *Let be a meromorphic function of finite order, , . Then
**
for all outside an exceptional set ofāāfinite logarithmic measures.*

Lemma 14 (see [7]). *Let be a nonconstant, finite-order meromorphic solution of
**
where , are difference polynomials in with meromorphic coefficients , and let . If the degree of as a polynomial in and its shifts is at most , then
**
for all outside an exceptional set of finite logarithmic measures.*

Lemma 15 (see [3]). *Let be meromorphic functions, and let be entire functions that satisfy the following:*(i)*;
*(ii)*when , is not a constant;*(iii)*when , ,
**where is of finite linear measure or finite logarithmic measure. Then .*

Lemma 16. *Let be a transcendental entire function of finite order and let be constants satisfying . Then is transcendental.*

*Proof. *If , then which contradicts our condition . Now we suppose that
where is a polynomial. Applying Lemma 14 to (8), we obtain that
Thus by (8), (9), and the first fundamental theorem of Nevanlinna theory, we obtain that
Since , this is a contradiction. Hence is a transcendental entire function.

Lemma 17 (see [17]). *Let be a nonconstant finite-order meromorphic function and let be an arbitrary complex number. Then
*

#### 3. Proof of Theorems 5 and 10

*Proof of Theorem 5. *(i) If has infinitely many zeros, then has infinitely many zeros since is an entire function and .

Now we suppose that has only finitely many zeros and . Thus since is transcendental, can be written as follows:
where , are polynomials, . Thus
Now we suppose that has only finitely many zeros. By Lemma 16, we see that is transcendental. So can be written as
where , are polynomials, . Set
where are constants and . Thus
where are constants. Since and
we see that are not constants.*Case **1*. If for any , are not constants, then by Lemma 15 and (14), we see that
which is a contradiction.*Case **2*. If there exists a satisfying where is a constant, then by (14), we have
By (19), Lemma 15, and , we obtain that
which is also a contradiction. Hence, has infinitely many zeros.

(ii) Suppose that has only finitely many zeros and . Then can be written as
where is a polynomial and , are constants. Thus
By the condition , we see that .

Hence has only finitely many zeros.

(iii) Case 1. . From , we get . If has only finitely zeros, then can be written as
where is a polynomial. By using a similar method as in the proof of Lemma 16, we get a contradiction. Thus has infinitely many zeros.

Case 2. . Suppose on contrary to the assertion that . If , that is, . By using a similar method as in the proof of Lemma 16, we get a contradiction. So we have . Thus, by Hadamard's theorem, can be written as
where is a polynomial and , are the canonical products formed by zeros and poles of , respectively, such that
Since , we get that
We set ; then from (25) and (26), we get

Differentiating (24) and eliminating , we get
where
*Case **2.1*. . Then from (28), we have
By integrating, we have
where is a nonzero constant. From (24) and (31), we have
By using a similar method as in the proof of Lemma 16, we get a contradiction.*Case **2.2*. . Let
Then from (28), we have
From Lemma 13 and Lemma 14, we have
Now for any given , we obtain from Lemma 17 and (27) that
It follows from (35) and (36) that
We obtain from the definition of that
Thus from (38) and (39), we have
Note that a zero of which is not a pole of is a pole of with the multiplicity at most 1, so from (34) and (27) we get that, for sufficiently small,
Hence from (33) and the above formula, we have
It follows from (37) and (42) that
Therefore, from (40) and (43), we have
which contradicts the assumption that is a transcendental entire function of finite order . This completes the proof of Theorem 5.

By using the same methods as in the proof of Theorem 5 (i) and (ii), we complete the proof of Theorem 10.

#### 4. Proof of Theorem 8

*Proof. *Firstly, we prove (ii) and (iii). (ii) Suppose that is the Borel exceptional value of . Then can be written as follows:
where is a positive integer, is a constant, and is an entire function satisfying
Thus
where is an entire function satisfying . So by using , we have
Since , we see that
By (48) and (49), we see that
If has the Borel exceptional value , then
where is a constant and is an entire function satisfying
By (48) and (51), we have
*Case **1*. If and , then by Lemma 15 and (53), we can obtain that
This contradicts with (49).*Case **2*. If or , then using the same method as above, we can also obtain a contradiction. Hence has no Borel exceptional value.

(iii) Suppose that is the Borel exceptional value of . Using the same method as above, we obtain
From (49) and
we see that is the finite Borel exceptional value of . Thus, has no nonzero finite Borel exceptional value.

Finally, we prove (i). By the assertion of (ii) and (iii), we see that if has the finite Borel exceptional value, then any nonzero finite value must not be the Borel exceptional value of . Hence takes the value infinitely often. By (50), we obtain .

#### 5. Proof of Theorem 9

*Proof. *Clearly, if , then has infinitely many zeros since is an entire function and has infinitely many zeros.

Now we suppose that . Suppose that has only finitely many zeros. Then can be written as follows:
where , are polynomials. By Lemma 16, we see that , . Differentiating (57) and eliminating , we obtain
Since there exists an infinite sequence satisfying , we see that there is a sufficiently large point such that and , at the same time.

From observation, we have the following: has a simple pole at and has pole at of multiplicity at least 2. This shows that (58) is a contradiction. Hence takes every value infinitely often.

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

The authors would like to thank the referee for the valuable suggestions to the present paper. This work was supported by the NNSF of China (no. 11171013 and no. 11371225).

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#### Copyright

Copyright © 2014 Nan Li and Lianzhong Yang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.