AM  Vol.8 No.8 , August 2017
Some Uniqueness Results of Q-Shift Difference Polynomials Involving Sharing Functions
ABSTRACT
In this paper, we mainly study the uniqueness of specific q-shift difference polynomials and of meromorphic functions, which share a common small function and get the corresponding results. In addition, we also investigate the problem of value distribution on q-shift difference polynomials of entire functions.

*Supported by the National Natural Science Foundation of China (No.11371139).

1. Introduction

In recent years, many Scholars have been interested in value distribution of difference operators of meromorphic functions (see [1] - [6] ). Furthermore, a large number of papers have studied and obtained the uniqueness results of difference polynomials of meromorphic functions, their shifts and difference operators (see [7] - [12] ). Our purpose in the paper is to study the value distribution for q-shift polynomials of transcendental meromorphic with zero order, and some results about entire functions.

For a meromorphic function f , we always assume that f is meromorphic in the complex plane . We use standard notations of the Nevanlinna Value Distribution Theory (see [13] ), such as m ( r , f ) , N ( r , f ) , N ¯ ( r , f ) , T ( r , f ) ,

S ( r , f ) , and define N 2 ( r , 1 f ) as the counting function of zero of f , such

that simple zero is counted once and multiple zeros are counted twice. We denote any quantity by S ( r , f ) , if it satisfies S ( r , f ) = o ( T ( r , f ) ) , as r outside of a possible exceptional set of r with finite logarithmic measure. In addition, the notation ρ ( f ) is the order of growth of f . Let meromorphic function α be a common small function of f ( z ) and g ( z ) , suppose that f ( z ) α ( z ) and g ( z ) α ( z ) have the same zeros counting multiplicities (ignoring multiplicities), then we say that f and g share α ( z ) CM(IM).

In this paper, we define a q-shift difference product of meromorphic function f ( z ) as follows.

F ( z ) = f n ( z ) j = 1 d f ( q j z + c j ) v j (1)

F 1 ( z ) = P n ( f ( z ) ) j = 1 d f ( q j z + c j ) v j (2)

where c j ( c j 0 , j = 1 , 2 , 3 , , d ) are distinct constants, q j ( j = 1 , 2 , , d ) be non-zero finite complex constants, let P n ( z ) = α n z n + α n 1 z n 1 + + α 1 z + α 0 be a non-zero polynomial, where α n ( 0 ) , α n 1 , , α 0 are small functions of f . Let n , d , v j ( j = 1 , 2 , , d ) are positive integers and σ = v 1 + v 2 + + v d .

Recently, Liu et al. [14] have considered and proved the uniqueness of q-shift difference polynomials of meromorphic functions.

Theorem A. Let f ( z ) and g ( z ) be two transcendental meromorphic functions with ρ ( f ) = ρ ( g ) = 0 . Let q and η be two non-zero finite complex constants. If f n ( z ) f ( q z + η ) and g n ( z ) g ( q z + η ) share 1 CM, then either f ( z ) = t g ( z ) or f ( z ) g ( z ) = t , where n ( N * ) 14 satisfying t n + 1 = 1 .

Theorem B. Let f ( z ) and g ( z ) be two transcendental meromorphic functions with ρ ( f ) = ρ ( g ) = 0 . Let q and η be two non-zero finite complex constants. If f n ( z ) f ( q z + η ) and g n ( z ) g ( q z + η ) share 1 IM, then either f ( z ) = t g ( z ) or f ( z ) g ( z ) = t , where n ( N * ) 26 satisfying t n + 1 = 1 .

First, we will prove the following theorems on value sharing results of q-shift difference polynomials extend the Theorem A, B, as follows:

Theorem 1.1. Let f ( z ) and g ( z ) be two transcendental meromorphic functions with ρ ( f ) = ρ ( g ) = 0 , and let α ( z ) ( 0 ) be a common small function of f ( z ) and g ( z ) . If F ( z ) and G ( z ) share α ( z ) CM, then f ( z ) = t g ( z ) , where n 4 m i n ( 2 d , σ ) + σ + 9 satisfying t n + σ = 1 .

Theorem 1.2. Let f ( z ) and g ( z ) be two transcendental meromorphic functions with ρ ( f ) = ρ ( g ) = 0 , and let α ( z ) ( 0 ) be a common small function of f ( z ) and g ( z ) . If F ( z ) and G ( z ) share α ( z ) IM, then f ( z ) = t g ( z ) , where n 4 m i n ( 2 d , σ ) + σ + 6 d + 15 satisfying t n + σ = 1 .

Liu et al. [14] also considered some properties of q-shift difference poly- nomials of entire functions, as follow:

Theorem C. Let f ( z ) and g ( z ) be two transcendental entire functions with ρ ( f ) = ρ ( g ) = 0 , and let q and η are two non-zero finite complex constants, and let P n ( z ) = α n z n + α n 1 z n 1 + + α 1 z + α 0 be a non-zero polynomial, where α n ( 0 ) , α n 1 , , α 0 , are constants, and let m be the number of the distinct zero of P n ( z ) . If P n ( f ( z ) ) f ( q z + η ) and P n ( g ( z ) ) g ( q z + η ) share 1 CM, then only one of the following two cases holds:

a) f ( z ) = t g ( z ) , where n > 2 m + 1 , and k is greatest common divisor of ( λ 0 , λ 1 , , λ n ) , satisfying t k = 1 . When α i = 0 , then λ i = n + 1 , otherwise λ i = i + 1 . i = 0 , 1 , , n .

b) f ( z ) and g ( z ) satisfy a algebraic equation Q ( f ( z ) , g ( z ) ) = 0 , where

Q ( w 1 , w 2 ) = P n ( w 1 ) w 1 ( q z + c ) P n ( w 2 ) w 2 ( q z + c ) (3)

Next, it is easy to derive that P n ( f ( z ) ) f ( q z + η ) in Theorem C can be replaced by P n ( f ( z ) ) j = 1 d f ( q j z + c j ) v j , as follows

Theorem 1.3. Let f ( z ) and g ( z ) be two transcendental entire functions with ρ ( f ) = ρ ( g ) = 0 , and let α ( z ) be a common small function of f ( z ) and g ( z ) , and let k be the number of distinct zeros of P n ( z ) . If F 1 ( z ) and G 1 ( z ) share α ( z ) CM, then only one of the following results holds:

a) f ( z ) = t g ( z ) for a constant t such that t m = 1 , where n > 2 k + 2 d + σ and m is greatest common divisor of ( n + σ , n + σ 1, , n + σ i , , σ + 1 ) , α n i 0 , i = 0 , 1 , , n 1 .

b) f ( z ) and g ( z ) satisfy a algebraic equation Q ( f , g ) 0 , where

Q ( w 1 , w 2 ) = P n ( w 1 ) j = 1 d w 1 ( q j z + c j ) v j P n ( w 2 ) j = 1 d w 2 ( q j z + c j ) v j . (4)

2. Some Lemmas

Lemma 2.1. (see [15] ) Let n ( 1 ) be a positive integer, and let f ( z ) be a transcendental meromorphic function, and let α i ( i = 0 , 1 , , n ) be small meromorphic functions of f . If

P n ( f ( z ) ) = α n f n ( z ) + α n 1 f n 1 ( z ) + + α 1 f ( z ) + α 0 , (5)

then

T ( r , P n ( f ( z ) ) ) = n T ( r , f ( z ) ) + S ( r , f ( z ) ) (6)

Lemma 2.2. (see [9] ) Let q and η be two non-zero finite complex numbers, and let f ( z ) be a nonconstant meromorphic function with ρ ( f ) = 0 , then

m ( r , f ( q z + η ) f ( z ) ) = S ( r , f ) . (7)

on a set of logarithmic density 1.

Lemma 2.3. (see [12] ) Let f ( z ) and g ( z ) be two non-constant meromorphic functions. Let f ( z ) and g ( z ) share 1 IM and

L = f f 2 f f 1 g g + 2 g g 1 (8)

If L 0 , then

T ( r , f ) + T ( r , g ) 2 ( N 2 ( r , f ) + N 2 ( r , g ) + N 2 ( r , 1 f ) + N 2 ( r , 1 g ) ) + 3 ( N ¯ ( r , f ) + N ¯ ( r , g ) + N ¯ ( r , 1 f ) + N ¯ ( r , 1 g ) ) + S ( r , f ) + S ( r , g ) (9)

Lemma 2.4. (see [16] ) Let f and g be two non-constant meromorphic functions. If f and g share 1 CM, then only one of the following results holds:

(a) max { T ( r , f ) , T ( r , g ) } N 2 ( r , f ) + N 2 ( r , g ) + N 2 ( r , 1 f ) + N 2 ( r , 1 g ) + S ( r , f ) + S ( r , g ) (b) f g ; (c) f g 1. (10)

Lemma 2.5. (see [14] ) Let q and η be two non-zero finite complex constants, and let f be a non-constant meromorphic function with ρ ( f ) = 0 , then

T ( r , f ( q z + η ) ) T ( r , f ( z ) ) + S ( r , f ) (11)

on a set of logarithmic density 1.

Lemma 2.6. (see [14] ) Let q and η be two non-zero finite complex constants, and let f be a nonconstant meromorphic function of zero order, then

N ¯ ( r , f ( q z + η ) ) N ¯ ( r , f ( z ) ) + S ( r , f ) N ¯ ( r , 1 f ( q z + η ) ) N ¯ ( r , 1 f ( z ) ) + S ( r , f ) N ( r , f ( q z + η ) ) N ( r , f ( z ) ) + S ( r , f ) N ( r , 1 f ( q z + η ) ) N ( r , 1 f ( z ) ) + S ( r , f ) . (12)

Lemma 2.7. Let f ( z ) be a non-constant meromorphic function of zero order, and F 1 ( z ) be defined as in (2). Then

( n σ ) T ( r , f ) + S ( r , f ) T ( r , F 1 ) ( n + σ ) T ( r , f ) + S ( r , f ) (13)

Proof. Combining Lemma 2.1 with Lemma 2.5, we obtain

T ( r , F 1 ) T ( r , P n ( f ( z ) ) ) + T ( r , j = 1 d f ( q j z + c j ) v j ) + S ( r , f ) n T ( r , f ( z ) ) + j = 1 d T ( r , f ( q j z + c j ) v j ) + S ( r , f ) ( n + σ ) T ( r , f ( z ) ) + S ( r , f ) (14)

In addition, by Lemma 2.1 and Lemma 2.5, we also get

( n + σ ) T ( r , f ( z ) ) T ( r , P n ( f ( z ) ) f σ ) + S ( r , f ) = m ( r , P n ( f ( z ) ) f σ ) + N ( r , P n ( f ( z ) ) f σ ) + S ( r , f ) m ( r , F 1 ( z ) f σ j = 1 d f ( q j z + c j ) v j ) + N ( r , F 1 ( z ) f σ j = 1 d f ( q j z + c j ) v j ) + S ( r , f ) m ( r , F 1 ) + N ( r , F 1 ) + T ( r , f σ j = 1 d f ( q j z + c j ) v j ) + S ( r , f ) T ( r , F 1 ) + 2 σ T ( r , f ) + S ( r , f ) (15)

which is equivalent to

( n σ ) T ( r , f ) + S ( r , f ) T ( r , F 1 ) (16)

Therefore, we get Lemma 2.7.

Lemma 2.8. Let f ( z ) be an entire function with ρ ( f ) = 0 , and F 1 ( z ) be stated as in (2). Then

T ( r , F 1 ) = ( n + σ ) T ( r , f ) + S ( r , f ) (17)

Proof. Using the same method as the Lemma 2.7, we can easily to prove.

3. Proof of Theorem

3.1. Proof of Theorem 1.1

Set F * ( z ) = F ( z ) α ( z ) , G * ( z ) = G ( z ) α ( z ) , than F * ( z ) and G * ( z ) share 1 CM.

Thus by Nevanlinna second fundamental theory, Lemma 2.5 and Lemma 2.7, we have

( n σ ) T ( r , f ) + S ( r , f ) T ( r , F * ( z ) ) N ¯ ( r , F * ( z ) ) + N ¯ ( r , 1 F * ( z ) ) + N ¯ ( r , 1 F * ( z ) 1 ) + S ( r , F * ( z ) ) N ¯ ( r , f n ) + N ¯ ( r , j = 1 d f ( q j z + c j ) v j ) + N ¯ ( r , 1 f n ) + N ¯ ( r , 1 j = 1 d f ( q j z + c j ) v j ) + N ¯ ( r , 1 G * ( z ) 1 ) + S ( r , f ) ( 2 d + 2 ) T ( r , f ) + ( n + σ ) T ( r , g ) + S ( r , g ) + S ( r , f ) (18)

Then

( n 2 d σ 2 ) T ( r , f ) ( n + σ ) T ( r , g ) + S ( r , g ) + S ( r , f ) (19)

Similarly,

( n 2 d σ 2 ) T ( r , g ) ( n + σ ) T ( r , f ) + S ( r , f ) + S ( r , g ) (20)

It follows that S ( r , f ) = S ( r , g ) .

Then by Lemma 2.4, we consider three subcases.

Case 1. Suppose that max { T ( r , F * ( z ) ) , T ( r , G * ( z ) ) } N 2 ( r , F * ( z ) ) + N 2 ( r , 1 F * ( z ) ) + N 2 ( r , G * ( z ) ) + N 2 ( r , 1 G * ( z ) ) + S ( r , F * ( z ) ) + S ( r , G * ( z ) ) holds.

Through simple calculation, we have

N 2 ( r , F * ( z ) ) N 2 ( r , f n ) + N 2 ( r , j = 1 d f ( q j z + c j ) v j ) { 2 + min ( 2 d , σ ) } T ( r , f ) + S ( r , f ) (21)

In the same way,

N 2 ( r , 1 F * ( z ) ) { 2 + min ( 2 d , σ ) } T ( r , f ) + S ( r , f ) N 2 ( r , G * ( z ) ) { 2 + min ( 2 d , σ ) } T ( r , g ) + S ( r , g ) N 2 ( r , 1 G * ( z ) ) { 2 + min ( 2 d , σ ) } T ( r , g ) + S ( r , g ) (22)

Combining Lemma 2.4, Lemma 2.7, Equations ((21) and (22)), we obtain that

( n σ ) ( T ( r , f ) + T ( r , g ) ) T ( r , F * ( z ) ) + T ( r , G * ( z ) ) 2 N 2 ( r , F * ( z ) ) + 2 N 2 ( r , 1 F * ( z ) ) + 2 N 2 ( r , G * ( z ) ) + 2 N 2 ( r , 1 G * ( z ) ) + S ( r , F * ( z ) ) + S ( r , G * ( z ) ) 4 [ 2 + min ( 2 d , σ ) ] ( T ( r , f ) + T ( r , g ) ) + S ( r , f ) + S ( r , g ) (23)

Then

( n σ 8 4 m i n ( 2 d , σ ) ) ( T ( r , f ) + T ( r , g ) ) S ( r , f ) (24)

Which is impossible, since n 4 m i n ( 2 d , σ ) + σ + 9 .

Case 2. Suppose that F * ( z ) G * ( z ) holds, we obtain

f n ( z ) j = 1 d f ( q j z + c j ) v j = g n ( z ) j = 1 d g ( q j z + c j ) v j . (25)

We assume that h ( z ) : = f ( z ) g ( z ) . If h ( z ) C (constant), then f = t g , and by substituting f = t g into (25), we obtain that

g n j = 1 d g ( q j z + c j ) v j [ t n + σ 1 ] = 0. (26)

Since g is a transcendental meromorphic function, than g n j = 1 d g ( q j z + c j ) v j 0 . It follows that t n + σ = 1 .

Suppose that h ( z ) C (constant), then using (25), we deduce that h n ( z ) = j = 1 d 1 h ( q j z + c j ) v j ,

So

n T ( r , h ( z ) ) = T ( r , j = 1 d 1 h ( q j z + c j ) v j ) σ T ( r , h ( z ) ) + S ( r , h ( z ) ) (27)

We get a contradiction, since n 4 m i n ( 2 d , σ ) + σ + 9 .

Case 3. Suppose that F * ( z ) G * ( z ) 1 holds, then f n ( z ) j = 1 d f ( q j z + c j ) v j g n ( z ) j = 1 d g ( q j z + c j ) v j = α 2 ( z ) .

We define h 1 ( z ) = f ( z ) g ( z ) , we easily get h 1 n ( z ) = j = 1 d α 2 ( z ) h 1 ( q j z + c j ) v j is non-constant, hence

n T ( r , h 1 ( z ) ) = T ( r , j = 1 d α 2 ( z ) h 1 ( q j z + c j ) v j ) σ T ( r , h 1 ( z ) ) + S ( r , h 1 ( z ) ) (28)

We get a contradiction, since n 4 min ( 2 d , σ ) + σ + 9 . This implies that h 1 ( z ) is a constant, which is impossible.

3.2. Proof of Theorem 1.2

Set F * ( z ) = F ( z ) α ( z ) , G * ( z ) = G ( z ) α ( z ) , So F * ( z ) and G * ( z ) share 1 IM.

Using the same arguments as in Theorem 1.1, we prove that (18)-(22) holds.

We can easily get

N ¯ ( r , F * ( z ) ) ( 1 + d ) T ( r , f ) + S ( r , f ) N ¯ ( r , 1 F * ( z ) ) ( 1 + d ) T ( r , f ) + S ( r , f ) N ¯ ( r , G * ( z ) ) ( 1 + d ) T ( r , g ) + S ( r , g ) N ¯ ( r , 1 G * ( z ) ) ( 1 + d ) T ( r , g ) + S ( r , g ) (29)

Let

L ( z ) = F * ( z ) F * ( z ) 2 F * ( z ) F * ( z ) 1 G * ( z ) G * ( z ) + 2 G * ( z ) G * ( z ) 1 (30)

If L 0 , combining Lemma 2.3, (21), (22) with (29), we obtain

( n σ ) ( T ( r , f ) + T ( r , g ) ) T ( r , F * ( z ) ) + T ( r , G * ( z ) ) [ 14 + 6 d + 4 m i n ( 2 d , σ ) ] ( T ( r , f ) + T ( r , g ) ) + S ( r , f ) + S ( r , g ) (31)

Then,

( n σ 14 6 d 4 min ( 2 d , σ ) ) ( T ( r , f ) + T ( r , g ) ) S ( r , f ) + S ( r , g ) (32)

that is impossible, since n 4 m i n ( 2 d , σ ) + σ + 6 d + 15 . Hence, we get L 0 .

By integrating L twice, we obtain that

F * = ( b + 1 ) G * + ( a b 1 ) b G * + ( a b ) (33)

which yields T ( r , F * ) = T ( r , G * ) + O ( 1 ) . From Lemma 2.8, we deduced that T ( r , f ) = T ( r , g ) + S ( r , f ) . Next, we will consider the following three subcases.

Case 1. b 0 and b 1 . Suppose that a b 1 0 , by (33), we get

N ¯ ( r , 1 F * ) = N ¯ ( r , 1 G * a b 1 b + 1 ) (34)

Combining the second fundamental theory with Lemma 2.5, Lemma 2.7, (29), and (34), we have

( n σ ) T ( r , g ) T ( r , G * ( z ) ) + S ( r , g ) N ¯ ( r , G * ( z ) ) + N ¯ ( r , 1 G * ( z ) ) + N ¯ ( r , 1 G * a b 1 b + 1 ) + S ( r , g ) N ¯ ( r , G * ( z ) ) + N ¯ ( r , 1 G * ( z ) ) + N ¯ ( r , 1 F * ) + S ( r , g ) ( 2 + 2 d ) T ( r , g ) + ( 1 + d ) T ( r , f ) + S ( r , g ) ( 3 + 3 d ) T ( r , g ) + S ( r , g ) (35)

which is impossible, since n 4 m i n ( 2 d , σ ) + σ + 6 d + 15 . Therefore, a b 1 = 0 , so

F * = ( b + 1 ) G * b G * + 1 (36)

Then, N ¯ ( r , 1 F * ) = N ¯ ( r , 1 G * + 1 / b ) . Similarly, we have

( n σ ) T ( r , g ) N ¯ ( r , G * ( z ) ) + N ¯ ( r , 1 G * ( z ) ) + N ¯ ( r , 1 G * + 1 / b ) + S ( r , g ) N ¯ ( r , G * ( z ) ) + N ¯ ( r , 1 G * ( z ) ) + N ¯ ( r , 1 F * ) + S ( r , g ) ( 2 + 2 d ) T ( r , g ) + ( 1 + d ) T ( r , f ) + S ( r , g ) ( 3 + 3 d ) T ( r , g ) + S ( r , g ) (37)

Which is impossible, since n 4 m i n ( 2 d , σ ) + σ + 6 d + 15 .

Case 2. If b = 0 and a = 1 , then F * G * obviously. From the proof of case 2 in theorem 1.1, we get f ( z ) = t g ( z ) , where t n + σ = 1 . Therefore, we consider b = 0 and a 1 . Then from (33), we obtain

F * = G * + a 1 a . (38)

Using the same discuss as Case 1, we get contradiction.

Case 3. If b = 1 and a = 1 , then F * G * 1 obviously. Thus from the proof of case 3 in theorem 1.1, we get a contradiction. Therefore, we consider b = 1 and a 1 . From (33), we get

F * = a a + 1 G * . (39)

Which is impossible, using the similar method as Case 1.

3.3. Proof of Theorem 1.3

We use the similar method as [14] . By the theorem condition that F 1 ( z ) α ( z ) and G 1 ( z ) α ( z ) share 0 CM, hence there exist an entire function u ( z ) , than

F 1 ( z ) α ( z ) G 1 ( z ) α ( z ) = e u ( z ) . (40)

Since ρ ( f ) = ρ ( g ) = 0 , than e u ( z ) η is a constant.

Rewriting (40)

G 1 ( z ) = F 1 ( z ) + ( η 1 ) α ( z ) (41)

If η 1 , we can use Nevanlinnas two fundamental theorems, Lemma 2.5 and Lemma 2.8 to get a contradiction, since n > σ + 2 k + 2 d .

So we get η = 1 . Rewriting (40)

P n ( f ( z ) ) j = 1 d f ( q j z + c j ) v j = P n ( g ( z ) ) j = 1 d g ( q j z + c j ) v j . (42)

Set h ( z ) : = f ( z ) g ( z ) , suppose that h ( z ) C (constant), then f = t g . Then we take f = t g into (42) and get

j = 1 d g ( q j z + c j ) v j [ α n g n ( t n + σ 1 ) + α n 1 g n 1 ( t n + σ 1 1 ) + + α 1 g ( t σ + 1 1 ) ] 0. (43)

where α n is a non-zero complex constant. And j = 1 d g ( q j z + c j ) v j 0 , since g

is transcendental meromorphic function. So h m = 1 , where m is greatest common divisor of ( n + σ , n + σ 1, , n + σ i , , σ + 1 ) , α n i 0 ( i = 0 , 1 , , n 1 ).

Suppose that h ( z ) C (constant), Equation (43) imply that f ( z ) and g ( z ) satisfy a algebraic equation Q ( f , g ) 0 , where

Q ( w 1 , w 2 ) = P n ( w 1 ) j = 1 d w 1 ( q j z + c j ) v j P n ( w 2 ) j = 1 d w 2 ( q j z + c j ) v j . (44)

4. Conclusion

In this paper, we obtain some important results about the uniqueness of specific q-shift difference polynomials of meromorphic functions by Nevanlinna and value distribution theory and extend previous results. In addition, we also investigate the problem of value distribution on q-shift difference polynomials of entire functions.

Acknowledgements

Sincere thanks to the members of Xuexue Qian and Yasheng YE for their professional performance, and special thanks to managing editor for a rare attitude of high quality.

Cite this paper
Qian, X. and Ye, Y. (2017) Some Uniqueness Results of Q-Shift Difference Polynomials Involving Sharing Functions. Applied Mathematics, 8, 1117-1127. doi: 10.4236/am.2017.88084.
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