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.
In recent years, many Scholars have been interested in value distribution of difference operators of meromorphic functions (see [
For a meromorphic function f , we always assume that f is meromorphic in the complex plane ℂ . We use standard notations of the Nevanlinna Value Distribution Theory (see [
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 ( 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. [
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. [
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)
Lemma 2.1. (see [
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 [
m ( r , f ( q z + η ) f ( z ) ) = S ( r , f ) . (7)
on a set of logarithmic density 1.
Lemma 2.3. (see [
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 [
(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 [
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 [
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.
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.
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.
We use the similar method as [
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)
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.
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.
Qian, X.X. and Ye, Y.S. (2017) Some Uniqueness Results of Q-Shift Difference Polynomials Involving Sharing Functions. Applied Mathematics, 8, 1117-1127. https://doi.org/10.4236/am.2017.88084