2012
DOI: 10.1007/s11766-012-2795-x
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Some results on zeros and uniqueness of difference-differential polynomials

Abstract: We consider the zeros distributions on the derivatives of difference polynomials of meromorphic functions, and present some results which can be seen as the discrete analogues of Hayman conjecture [8], also partly answer the question given in [18, P448]. We also investigate the uniqueness problems of difference-differential polynomials of entire functions sharing one common value. These theorems improve the results of Luo and Lin [18] and some results of present authors [15].

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Cited by 22 publications
(11 citation statements)
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“…In 2012 Liu-Liu and Cao [9] first considered the value distribution of the differentialdifference counterpart of the above theorems and obtained the following result.…”
Section: Introduction Definitions and Resultsmentioning
confidence: 99%
“…In 2012 Liu-Liu and Cao [9] first considered the value distribution of the differentialdifference counterpart of the above theorems and obtained the following result.…”
Section: Introduction Definitions and Resultsmentioning
confidence: 99%
“…Recently, the topic of difference equation and difference product in the complex plane C has attracted many mathematicians, many papers have focused on value distribution of differences and differences operator analogues of Nevanlinna theory (including [3,6,8,7,17,23]), and many people dealt with the uniqueness of differences and difference polynomials of meromorphic function and obtained some interesting results ( [11,12,20,21,22,27,28]).…”
Section: Introduction and Main Resultsmentioning
confidence: 99%
“…. , n. Since g is an entire function and a n ̸ = 0, we can deduce from (20) that T (r, g) = m(r, g) = S(r, g), a contradiction. Then (19) holds.…”
Section: Subcase 22 Suppose That H Is Not a Constant Then We Claimmentioning
confidence: 90%
“…In view of Lemma 10, since ρ(h(z)) < 1, the degree of C 1 n q(z) n−1 q(z + c) + q(z) n e αc must be smaller than the degree of q(z) n . Hence, we arrive at the equality e αc = −C 1 n (16) provided that q(z) is not a constant. By using the same arguments as above, we also get D 2 (z) ≡ 0, that is C 2 n q(z) n−2 q(z + c)h(z) 2 + C 1 n q(z) n−1 h(z + c)e αc h(z) ≡ 0.…”
Section: Proof (Proof Of Theorem 8) Letmentioning
confidence: 94%