The value distribution and uniqueness of one certain type of differential-difference polynomials

Acta Mathematica Scientia 2014,34B(3):719–728 http://actams.wipm.ac.cn

THE VALUE DISTRIBUTION AND UNIQUENESS OF ONE CERTAIN TYPE OF DIFFERENTIAL-DIFFERENCE POLYNOMIALS∗

܎Œ)

Keyu ZHANG (

†

Department of Mathematics, Qilu Normal University, Jinan 250013, China Department of Mathematics, Shandong University, Jinan 251000, China E-mail : [email protected]; keyu [email protected]

¤öÊ)

Hongxun YI (

Department of Mathematics, Shandong University, Jinan 251000, China E-mail : [email protected]; [email protected] Abstract In this article, we investigate the distribution of the zeros and uniqueness of differential-difference polynomials G(z) = (f n (f m (z) − 1)

d Y

f (z + cj )vj )(k) − α(z),

d Y

f (z + cj )vj )(k) − α(z),

j=1

H(z) = (f n (f (z) − 1)m

j=1

where f is transcendental entire function of finite order, cj (j = 1, 2, · · · , d), n, m, d, and vj (j = 1, 2, · · · , d) are integers, and obtain some theorems, which extended and improved many previous results. Key words

Meromorphic; uniqueness; value distribution; differential-difference

2010 MR Subject Classification

1

30D35

Introduction and Main Results

In this article, we assume that reader is familiar with the standard notations and results such as the proximity functions m(r, f ), counting function N (r, f ), characteristic function T (r, f ), the first and second main theorems, lemma on the logarithmic derivatives of Nevanlinna theory, see [1–3]. S 1 Let p be a positive integer and a ∈ C {∞}, then we denote by Np (r, f −a ) the counting function of the zeros of f − a, where an m-fold zero is counted m times if m ≤ p and p times if m > p. We also need the following definition: Let f be a nonconstant meromorphic function, we ∗ Received

November 16, 2012; revised April 23, 2013. This article is supported by National Natural Science Foundation of China (11171184). † Corresponding author

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define difference operators as ∆η f (z) = f (z + η) − f (z), ∆nη f (z) = ∆n−1 (∆η f (z)), where η is a η nonzero complex number and n ≥ 2 is a positive integer. If η = 1, we denote ∆η f (z) = ∆f (z). The difference logarithmic derivative lemma, given by Chiang and Feng [4], Halburd and Korhonen [5], plays an important part in considering the difference analogues of Nevanlinna theory. With the development of difference analogue of Nevanlinna theory, many authors paid their attention to the zero distribution of difference polynomials [6–16]. Liu and Yang [9] also considered the zeros of f n (z)f (z + c) − p(z) and f n △c f , where p(z) is a nonzero polynomial, and obtain the following theorem: Theorem A Let f be a transcendental entire function of finite order and p(z) be a nonzero polynomial. If n ≥ 2, then f n (z)f (z + c) − p(z) has infinitely many zeros. If f is not a periodic function with period c and n ≥ 2, then, △c f = f (z + c) − f (z) has infinitely many zeros. In 2010, Zhang [17] considered zeros of one certain type of difference polynomials and obtained the following theorem. Theorem B Let f be transcendental entire function of finite order, α(z) 6≡ 0 be a small function with respect to f (z), cj (j = 1, 2, · · · , d), c be nonzero complex constant, and n be an integer. If n ≥ 2, then, f n (z)(f (z) − 1)f (z + c) − α(z) has infinitely many zeros. In 2012, Chen and Chen [18] considered zeros of one certain type of difference polynomials and obtained the following theorem. Theorem C Let f be transcendental entire function of finite order, α(z) 6≡ 0 be a small function with respect to f (z), cj (j = 1, 2, · · · , d), n, m, d, and vj (j = 1, 2, · · · , d) be integers. If d Q n ≥ 2, then, f n (f m (z) − 1) f (z + cj )vj − α(z) has infinitely many zeros. j=1

Theorem D Let f and g be two transcendental entire functions of finite order, α(z) 6≡ 0 be a common small function with respect to f and g, c be nonzero finite complex numbers. If d Q n ≥ m + 8σ, n, m, d, and vj (j = 1, 2, · · · , d) are integers, and f n (f m (z) − 1) f (z + cj )vj and j=1

g n (g m (z) − 1)

d Q

g(z + cj )vj share α(z) CM, then, f = tg, where tm = tn+σ = 1.

j=1

In this article, we investigate the following difference polynomial: (f n (f m (z) − 1)

d Y

f (z + cj )vj )(k) and (f n (f (z) − 1)m

j=1

d Y

f (z + cj )vj )(k) ,

j=1

where f is transcendental entire function of finite order, cj (j = 1, 2, · · · , d), n, m, d, and vj (j = d P 1, 2, · · · , d) are nonnegative integers, and σ = vj . j=1

Theorem 1 Let f be transcendental entire function of finite order, α(z) 6≡ 0 be a small function with respect to f , cj (j = 1, 2, · · · , d) be distinct finite complex numbers, and n, m, d, and vj (j = 1, 2, · · · , d) be nonnegative integers. If n ≥ k + 2, then, the differential-difference d Q polynomial (f n (f m (z) − 1) f (z + cj )vj )(k) − α(z) has infinitely many zeros. j=1

Remark 1 If k = 0, we can easily get Theorem C. Theorem 2 Let f be transcendental entire function of finite order, α(z) 6≡ 0 be a small function with respect to f , cj (j = 1, 2, · · · , d) be distinct finite complex numbers, and n, m, d, and vj (j = 1, 2, · · · , d) are nonnegative integers. If one of the following conditions holds:

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(i) n ≥ k + 2 when m ≤ k + 1, (ii) n ≥ 2k − m + 3 when m > k + 1. Then, the differential-difference polynomial (f n (f (z)−1)m

d Q

f (z+cj )vj )(k) −α(z) has infinitely

j=1

many zeros. Theorem 3 Let f and g be transcendental entire functions of finite order, α(z) 6≡ 0 be a common small function with respect to f and g, cj (j = 1, 2, · · · , d) be distinct finite complex numbers, and n, m, d, and vj (j = 1, 2, · · · , d) are nonnegative integers. If n ≥ 2k + m + σ + 5, d Q and the differential-difference polynomial (f n (f m (z) − 1) f (z + cj )vj )(k) and (g n (g m (z) − j=1

1)

d Q

vj (k)

g(z + cj ) )

share α(z) CM, then, f = tg, where t

m

= tn+σ = 1.

j=1

Theorem 4 Let f and g be transcendental entire functions of finite order, α(z) 6≡ 0 be a common small function with respect to f and g, cj (j = 1, 2, · · · , d) be distinct finite complex numbers, and n, m, d, and vj (j = 1, 2, · · · , d) are nonnegative integers. If n ≥ 4k − m + σ + 9, d Q and the differential-difference polynomial (f n (f (z) − 1)m f (z + cj )vj )(k) and (g n (g(z) − j=1

1)m

d Q

g(z + cj )vj )(k) share α(z) CM, then, f ≡ g.

j=1

2

Some Lemmas

Lemma 1 (see [1]) Let f and g be two transcendental meromorphic function, n ≥ 1, n ∈ Z. Let αi , i = 1, 2, · · · , n be meromorphic functions and T (r, αi ) = S(r, f ). Let P (f ) = an f n + an−1 f n−1 + ... + a1 f, an 6= 0. Then, T (r, P (f )) = nT (r, f ) + S(r, f ). Lemma 2 (see [4])

(2.1)

Let f be transcendental meromorphic function of finite order, then, T (r, f (z + c)) = T (r, f ) + S(r, f ).

(2.2)

Lemma 3 (see [18]) Let f be entire function of finite order and F = f n (f m (z) − 1)

d Y

f (z + cj )vj .

j=1

Then, T (r, F ) = (n + m + σ)T (r, f ) + S(r, f ), σ =

d X

vj .

(2.3)

j=1

Lemma 4 Let f be entire function of finite order and F = f n (f (z) − 1)m

d Q

f (z + cj )vj .

j=1

Then, T (r, F ) = (n + m + σ)T (r, f ) + S(r, f ), σ =

d X j=1

vj .

(2.4)

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Applying the same method of Lemma 3, we can easily proof it.

Lemma 5 (see [1]) fuction of f . Then,

Let f be nonconstant meromorphic function, αi (i = 1, 2, 3) be small

T (r, f ) ≤

3 X

N (r,

i=1

Lemma 6 (see [19]) Then, Np (r,



1 ) + S(r, f ). f − ai

(2.5)

Letf be nonconstant meromorphic function, k, p be positive integer.

1 1 ) ≤ T (r, f (k) ) − T (r, f ) + Nk+p (r, ) + S(r, f ), f f (k)

(2.6)

Lemma 7 (see [20]) Let H=



2F ′ F ′′ − ′ F F −1



−



 2G′ G′′ − , G′ G−1

(2.7)

where F and G are nonconstant meromorphic functions. If F and G share 1 CM and H 6= 0, then, 1 1 ) + N2 (r, ) F G +N2 (r, F ) + N2 (r, G)) + S(r, F ) + S(r, G). T (r, F ) + T (r, G) ≤ 2(N2 (r,

(2.8)

Lemma 8 Let f and g are nonconstant entire functions, n, m, k be positive integers, and let F = (f n (f m (z) − 1)

d Y

f (z + cj )vj )(k) , G = (g n (g m (z) − 1)

j=1

d Y

g(z + cj )vj )(k) .

j=1

If there exist nonzero constants a1 ,a2 such that N (r,

1 1 1 1 ) = N (r, ), N (r, ) = N (r, ), F − a1 G G − a2 F

(2.9)

then n ≤ 2k + 2 + m + σ. Proof By the second fundamental theorem, we have 1 1 ) + N (r, ) + S(r, F ) F F − a1 1 1 ≤ N (r, ) + N (r, ) + S(r, F ). F G 1 1 ≤ N1 (r, ) + N1 (r, ) + S(r, F ). F G By Lemmas 3 and 6, we have T (r, F ) ≤ N (r,

T (r, F ) ≤ T (r, F ) − T (r, f n (f m (z) − 1)

d Y

f (z + cj )vj )

j=1

+T (r, G) − T (r, g n (g m (z) − 1)

d Y

g(z + cj )vj )

j=1

+Nk+1 r,

1 f n (f m (z) − 1)

d Q

j=1

f (z + cj )vj

!

(2.10)

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1 r, d Q g n (g m (z) − 1) g(z + cj )vj

+Nk+1

!

+ S(r, f ) + S(r, g).

(2.11)

j=1

Hence, 1

(n + m + σ)T (r, f ) ≤ Nk+1 r,

f n (f m (z) − 1)

d Q

f (z + cj )vj

!

j=1

+Nk+1

1 r, d Q g n (g m (z) − 1) g(z + cj )vj

!

+ S(r, f ) + S(r, g)

j=1

≤ (k + 1 + m + σ)(T (r, f ) + T (r, g)) + S(r, f ) + S(r, g).

(2.12)

By the same method, we have (n + m + σ)T (r, g) ≤ (k + 1 + m + σ)(T (r, f ) + T (r, g)) + S(r, f ) + S(r, g).

(2.13)

(n − 2k − 2 − m − σ)(T (r, f ) + T (r, g)) ≤ S(r, f ) + S(r, g).

(2.14)

Hence,

The proof of Lemma 8 is completed.



Lemma 9 (see [18]) Suppose that f and g are transcendental entire function of finite order, cj (j = 1, 2, · · · , d) are distinct finite complex numbers, and n, m, d, vj (j = 1, 2, · · · , d) are integers. If n > m + 5σ, we have f n (f m (z) − 1)

d Y

f (z + cj )vj = g n (g m (z) − 1)

j=1

d Y

g(z + cj )vj .

j=1

Then, f = tg, where tm = tn+σ = 1.

3

Proofs of the Theorems Let F = f n (f m (z) − 1)

Proof of Theorem 1

d Q

f (z + cj )vj . From Lemma 3, we know

j=1

that (2.3) holds. Which means that F is also a transcendental entire function. Suppose that F (k) − α(z) has only finitely many zeros, then, we obtain N (r,

1 ) = O(log r) = S(r, f ). F (k) − α(z)

(3.1)

From Lemma 5, we have T (r, F (k) ) ≤ N (r,

1 F (k)

) + N (r,

F (k)

1 1 ) + S(r, f ) = N1 (r, (k) ) + S(r, f ). − α(z) F

(3.2)

By applying Lemma 6 to the right side of (3.2), we have T (r, F ) ≤ Nk+1 (r,

1 ) + S(r, f ). F

From (2.3) and (3.3), we have (n + m + σ)T (r, f ) + S(r, f ) = T (r, F ) ≤ Nk+1 (r,

1 ) + S(r, f ) F

(3.3)

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1 ) + mT (r, f ) + σT (r, f ) + S(r, f ) f ≤ (k + 1 + m + σ)T (r, f ) + S(r, f ). ≤ (k + 1)N (r,

(3.4)

From (3.3)and (3.4), we have (n − k − 1)T (r, f ) ≤ S(r, f ). Which contradicts assumption n ≥ k + 2. The proof of Theorem 1 is completed.



Applying the same method of Theorem 1, we have

Proof of Theorem 2

(n + m + σ)T (r, f ) + S(r, f ) = T (r, F ) ≤ Nk+1 (r,

1 ) + S(r, f ). F

(3.5)

If m ≤ k + 1, Nk+1 (r,

1 1 ) ≤ (k + 1)N (r, ) + mT (r, f ) + σT (r, f ) + S(r, f ) F f ≤ (k + 1 + m + σ)T (r, f ) + S(r, f ).

(3.6)

From (3.5) and (3.6), we have (n − k − 1)T (r, f ) ≤ S(r, f ). Which contradicts assumption n ≥ k + 2. If m > k + 1, Nk+1 (r,

1 1 ) ≤ (k + 1)T (r, ) + (k + 1)T (r, f ) + σT (r, f ) + S(r, f ) F f ≤ (2k + 2 + σ)T (r, f ) + S(r, f ).

(3.7)

From (3.5) and (3.7), we have (n + m − 2k − 2)T (r, f ) ≤ S(r, f ). Which contradicts assumption n ≥ 2k − m + 3. The proof of Theorem 2 is completed. Let

Proof of Theorem 3 (f n (f m (z) − 1) F =



d Q

f (z + cj )vj )(k)

j=1

α(z)

(g n (g m (z) − 1) , G=

d Q

g(z + cj )vj )(k)

j=1

α(z)

.

(3.8)

Then, F and G are transcendental meromorphic functions that share 1 CM. Let H be given by (2.7), If H 6= 0, by Lemma 7 we know that T (r, F ) + T (r, G) 1 1 ≤ 2(N2 (r, ) + N2 (r, ) + N2 (r, F ) + N2 (r, G)) + S(r, F ) + S(r, G). F G From Lemma 6 and Lemma 3, we have ! 1 1 N2 (r, ) ≤ N2 r, + S(r, f ) d Q F v n m (k) j (f (f (z) − 1) f (z + cj ) ) j=1

≤ T (r, (f n (f m (z) − 1)

d Y

j=1

f (z + cj )vj )(k) ) − (n + m + σ)T (r, f )

(3.9)

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+Nk+2 r, f n (f m (z)

d Q

− 1)

f (z + cj

)vj

!

+ S(r, f )

!

+ S(r, f ).

j=1

= T (r, F ) − (n + m + σ)T (r, f ) 1

+Nk+2 r, f n (f m (z)

− 1)

d Q

f (z + cj

)vj

(3.10)

j=1

Similarly, we have N2 (r,

1 ) ≤ T (r, G) − (n + m + σ)T (r, g) G ! 1 + Nk+2 r, + S(r, g) d Q g n (g m (z) − 1) g(z + cj )vj

(3.11)

j=1

1

1 N2 (r, ) ≤ Nk+2 r, F

f n (f m (z)

− 1)

d Q

f (z + cj

)vj

!

+ S(r, f ),

(3.12)

!

+ S(r, f ).

(3.13)

j=1

and 1 N2 (r, ) ≤ Nk+2 r, G

1 d Q g n (g m (z) − 1) g(z + cj )vj j=1

Combining (3.10), (3.11), (3.12), and (3.13), we have (n + m + σ)(T (r, f ) + T (r, g)) 1 1 ≤ T (r, F ) + T (r, G) − N2 (r, ) − N2 (r, ) + Nk+2 r, F G

1 f n (f m (z) − 1)

d Q

f (z + cj )vj

!

j=1

+Nk+2

1 r, d Q g n (g m (z) − 1) g(z + cj )vj

!

j=1

1

≤ 2Nk+2 r,

f n (f m (z) − 1)

d Q

f (z + cj )vj

!

j=1

+2Nk+2

1 r, d Q g n (g m (z) − 1) g(z + cj )vj

!

+ S(r, f ) + S(r, g)

j=1

1 1 ) + N (r, )) + (2m + 2σ)(T (r, f ) + T (r, g)) + S(r, f ) + S(r, g). f g Which contradicts the assumption that n ≥ 2k + m + σ + 5. Therefore, H ≡ 0. Integrating twice, we deduce from (2.3) that ≤ 2(k + 2)(N (r,

1 A = + B, F −1 G−1

(3.14)

(3.15)

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where A 6= 0 and B are constants. By (3.15), we have F =

(B − A)F + (A − B − 1) (B + 1)G + (A − B − 1) , G= . BG + (A − B) BF − (B + 1)

(3.16)

Case 1: B 6= 0, −1. From (3.16), we have N (r,

1 ) = N (r, G). F − B+1 B

(3.17)

By the second fundamental theorem, we have 1 1 ) + N (r, ) + S(r, f ) F F − B+1 B 1 1 ≤ N (r, ) + S(r, f ) = N1 (r, ) + S(r, f ). F F Applying the same method with the proof of (3.10), we obtain T (r, F ) ≤ N (r,

N1 (r,

(3.18)

1 ) ≤ T (r, F ) − (n + m + σ)T (r, f ) F 1

+Nk+1 r,

f n (f m (z) − 1)

d Q

f (z + cj )vj

!

+ S(r, f ).

(3.19)

j=1

Hence, (n + m + σ)T (r, f ) + S(r, f ) + T (r, F ) 1 ≤ (k + 1)N (r, ) + mT (r, f ) + σT (r, f ) + S(r, f ), f

(3.20)

which contradicts the assumption that n ≥ 2k + m + σ + 5. Case 2: B = 0. From (3.15), we have G + (A − 1) , G = AF − (A − 1). A

(3.21)

1 1 1 1 ) = N (r, ), N (r, ) = N (r, ). A−1 G F G + (A − 1) F− A

(3.22)

F = If A 6= 1, then, N (r,

By Lemma 8, we have n ≤ 2k + 2 + m + σ, which contradicts the assumption that n ≥ 2k + m + σ + 5. So, F = G; that is to say, ak−1 6= 0. By integration, we have (f n (f m (z) − 1)

d Y

f (z + cj )vj )(k−1) = (g n (g m (z) − 1)

j=1

d Y

g(z + cj )vj )(k−1) + ak−1 ,

j=1

where ak−1 is a constant. If ak−1 6= 0, then from Lemma 8, we can get n ≤ 2k + 4 + m + σ, which is a contradiction. Hence, ak−1 = 0. Repeating the same process k − 1 times, we obtain f n (f m (z) − 1)

d Y

f (z + cj )vj = g n (g m (z) − 1)

j=1

From Lemma 9, we know that (i) holds.

d Y

j=1

g(z + cj )vj .

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Case 3: B = −1, then, F =

(A + 1)F − A A , G= . −G + (A + 1) F

(3.23)

If A 6= −1, then, by the same reason as Case 1 and Case 2, we know that it is a contradiction, d Q so, A = −1. Because F G = α2 (z), that is to say, (f n (f m (z) − 1) f (z + cj )vj )(k) (g n (g m (z) − j=1

1)

d Q

g(z +cj )vj )(k) = α2 (z). Because n ≥ 2k+m+σ +5, then, f has no zeros. Let f (z) = eβ(z) ,

j=1

then,

n

m

(f (f (z) − 1)

d Y

vj (k)

f (z + cj ) )

= (e

nβ(z)

(e

mβ(z)

− 1)

j=1

= (enβ(z)

d Y

d Y

evj β(z+cj ) )(k)

j=1

evj β(z+cj ) )(k) (emβ(z) − 1))(k)

j=1

= (e

nβ(z)+γ(z)

(emβ(z) − 1))(k) = (e(n+m)β(z)+γ(z) − enβ(z)+γ(z) )(k)

= e(n+m)β(z)+γ(z) P1 (β(z), γ(z), β ′ (z)), γ ′ (z) · · · β (k) (z), γ (k) (z)) −enβ(z)+γ(z) P2 (β(z), γ(z), β ′ (z)), γ ′ (z) · · · β (k) (z), γ (k) (z)) = enβ(z)+γ(z) (P1 emβ(z) − P2 ), P where γ(z) = vj β(z + cj ). Obviously, P1 emβ(z) − P2 has infinite zeros, so it is impossible. The proof of Theorem 3 is completed.  Proof of Theorem 4 Applying the same method of Theorem 3, we can prove that f n (f m (z) − 1)

d Y

f (z + cj )vj = g n (g m (z) − 1)

j=1

or F G ≡ α2 (z). Let h(z) = where t is a constant.

f (z) g(z) .

d Y

g(z + cj )vj ,

j=1

Applying the same methods of Lemma 9, we obtain h(z) = t,

So, tn+σ g n (tg(z) − 1)m

d Y

(g(z + cj ))vj = g n (g(z) − 1)m

j=1

d Y

g(z + cj )vj ,

(3.24)

j=1

tn+σ (tg(z) − 1)m = (g(z) − 1)m ,

(3.25)

then, 1 m−1 m−1 1 m−1 g (z) · · · + (−1)m , tn+σ (tm g m (z) − Cm t g (z) · · · + (−1)m ) = g m (z) − Cm

(3.26)

tn+σ+m = tn+σ+m−1 = · · · = tn = 1 , so, we know t = 1, then f (z) ≡ g(z). d d Q Q If F G ≡ α2 (z), that is to say, (f n (f (z) − 1)m f (z + cj )vj )(k) (g n (g(z) − 1)m g(z + j=1

j=1

cj )vj )(k) = α2 (z). Applying the same methods of Theorem 3, we know that it is impossible too. The proof of Theorem 4 is completed. 

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