Unicity of Meromorphic Function Sharing One Small Function with Its Derivative

Copyright q 2010 Ang Chen et al. 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. We deal with the problem of uniqueness of a meromorphic function sharing one small function with its k's derivative and obtain some results.


Introduction and Main Results
In this article, a meromorphic function means meromorphic in the open complex plane.We assume that the reader is familiar with the Nevanlinna theory of meromorphic functions and the standard notations such as T r, f , m r, f , N r, f , N r, f , and so on.
Let f and g be two nonconstant meromorphic functions; a meromorphic function a z / ≡ ∞ is called a small functions with respect to f provided that T r, a S r, f .Note that the set of all small function of f is a field.Let b z be a small function with respect to f and g.We say that f and g share b z CM IM provided that f − b and g − b have same zeros counting multiplicities ignoring multiplicities .
Moreover, we use the following notations.
Let k be a positive integer.We denote by N k r, 1/ f − a the counting function for the zeros of f − a with multiplicity ≤ k and by N k r, 1/ f − a the corresponding one for which the multiplicity is not counted.Let N k r, 1/ f − a be the counting function for the zeros of f − a with multiplicity ≥ k, and let N k r, 1/ f − a be the corresponding one for which the multiplicity is not counted.Set

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Obviously, 1 ≥ Θ a, f ≥ δ p a, f ≥ δ a, f ≥ 0. For more details, reader can see 1, 2 .Br ück see 3 considered the uniqueness problems of an entire function sharing one value with its derivative and proved the following result.
Yang 4 , Zhang 5 , and Yu 6 extended Theorem A and obtained many excellent results.
Theorem B see 5 .Letf be a nonconstant meromorphic function and, let k be a positive integer.Suppose that f and f k share 1 CM and for r ∈ I, where I is a set of infinite linear measure and λ satisfies 0 Theorem C see 6 .Let f be a nonconstant, nonentire meromorphic function and a z / ≡ 0, ∞ be a small function with respect tof.If 1 f and a z have no common poles, 2 f − a and f k − a share the value 0 CM, where k is a positive integer.
In the same paper, Yu 6 posed four open questions.Lahiri and Sarkar 7 and Zhang 8 studied the problem of a meromorphic or an entire function sharing one small function with its derivative with weighted shared method and obtained the following result, which answered the open questions posed by Yu 6 .Theorem D see 8 .Let f be a non-constant meromorphic function and, let k be a positive integer.Also let a z / ≡ 0, ∞ be a meromorphic function such that T r, a S r, f .Suppose that f − a and for 0 < λ < 1, r ∈ I, and I is a set of infinite linear measure.Then In this article, we will pay our attention to the value sharing of f and f n k that share a small function and obtain the following results, which are the improvements and complements of the above theorems.Theorem 1.1.Let k ≥1 , n ≥1 be integers and let f be a non-constant meromorphic function.Also let a z / ≡ 0, ∞ be a small function with respect to f.If f and f n k share a z IM and Theorem 1.2.Let k ≥1 , n ≥1 be integers and f be a non-constant meromorphic function.Also let a z / ≡ 0, ∞ be a small function with respect to f.If f and f n k share a z IM and or f and f n k share a z CM and Clearly, Theorem 1.1 improves and extends Theorems B and D, while 1.2 improves and extends Theorem C.

Some Lemmas
In this section, first of all, we give some definitions which will be used in the whole paper.Definition 2.1.Let F and G be two meromorphic functions defined in C; assume, that F and G share 1 IM; let z 0 be a zero of F − 1 with multiplicity p and a zero of G − 1 with multiplicity q.We denote byN 1 E r, 1/F − 1 the counting function of the zeros of F − 1 where p q 1 and byN 2 E r, 1/F − 1 the counting function of zeros of F − 1 where p q ≥ 2. We denotes by N L r, 1/F − 1 the counting function of the zeros of F − 1 where p > q ≥ 1; each point is counted according to its multiplicity, and N L r, 1/F − 1 denote its reduced form.In the same way, we can defineN and so on.International Journal of Mathematics and Mathematical Sciences Definition 2.2.In this paper N 0 r, 1/F denotes the counting function of the zeros ofF which are not the zeros of F and F − 1, and N 0 r, 1/F denotes its reduced form.In the same way, we can define N 0 r, 1/G and N 0 r, 1/G .Next we present some lemmas which will be needed in the sequel.Let F, G be two nonconstant meromorphic functions defined in C. We shall denote by H the following function:

2.2
If F and G are sharing 1 CM, then Lemma 2.4 see 1 .Let f be a meromorphic function and a is a finite complex number.Then where a 1 z a 2 z are two meromorphic functions such that T r, a i S r, f , i 1, 2 .
Lemma 2.5 see 7 .Let f be a non-constant meromorphic function, and k, p are two positive integers.Then Lemma 2.6 see 9 .Let f be a non-constant meromorphic function and let n be a positive integer.
where a i are meromorphic functions such that T r, a i S r, f i 1, 2, . . ., n , and a n / ≡ 0. Then T r, P f nT r, f S r, f .2.5

Proof of Theorem 1.1
Let F f z /a z , G f n z k /a z , then

3.1
From the definitions of F, G and recalling that F and G share value 1 IM CM , we get

3.4
We will distinguish two cases below.
Case 1 H / ≡ 0 .From 2.1 it is easy to see that m r, H S r, f .Subcase 1.1.Suppose that f and f n k share a z IM.According to 3.1 , F and G share 1 IM except the zeros and poles of a z .By 3.1 , we have Let z 0 be a simple zero of F −1 and G−1, but a z 0 / 0, ∞.Through a simple calculation we know that z 0 is a zero of H, so From 3.4 -3.6 and Lemma 2.3, we have

3.7
It follows by the second fundamental theorem, 3.5 , and 3.7 that
Subcase 1.2.Suppose that f and f n k share a z CM.Let z 0 be a simple zero of F − 1 and G − 1, but a z 0 / 0, ∞.By a simple calculation, we can still get H z 0 0. Therefore

3.11
International Journal of Mathematics and Mathematical Sciences 7 By the second fundamental theorem, 3.5 , and 3.11 , we have

3.12
Taking into account 3.1 , we have
Case 2 H ≡ 0 .Integration yields where A, B are constants and A / 0. It is easy to see that F and G share 1 CM.Now we claim that B 0. If N r, f / S r, f , then by 3.14 we get B 0. So our claim holds.Hence we can assume that N r, f S r, f .

3.15
If B / 0, then we can rewrite 3.14 as

3.17
If A / B, then by Lemma 2.4 and 3.17 we have

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Hence This is a contradiction with 1.4 and 1.5 .If A B, then from 3.14 we get 1/ F − 1 AG/ G − 1 .We rewrite it as

3.21
So by Lemmas 2.4 and 2.6 and 3.15 , we have

3.22
This implies that T r, f S r, f , since n ≥ 1.This is impossible.Hence our claim is right.So G − 1 / F − 1 A. Theorem 1.1 is, thus, completely proved.

Proof of Theorem 1.2
The proof is similar to the proof of Theorem 1.1.Let F and G be defined as in Theorem 1.1; hence, we have 3.1 -3.5 .We still distinguish two cases.
Case 1. H / ≡ 0 Subcase 1.1.Suppose that f and f n k share a z IM, then we can still get 3.6 and 3.7 .
Then by the second fundamental theorem, Lemma 2.3, and 3.5 we have

4.1
International Journal of Mathematics and Mathematical Sciences 9 Applying Lemma 2.5 to the above inequality and noticing the definition of F, G, we get

4.2
This implies that This contradicts 1.6 .
Subcase 1.2.Suppose that f and f n k share a z CM.Similarly as above, we can easily obtain

4.4
So by the second fundamental theorem, 4.4 , and using Lemma 2.5 again, we have

4.5
This implies that This contradicts 1.7 .
Case 2 H ≡ 0 .Similarly, we can also get 3.14 .Next we claim that B 0. If N r, f / S r, f , then it follows that B 0 from 3.14 .Hence, we may assume that 3.15 holds.If B / 0 and B / − 1, then

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Again by second fundamental theorem and 4.4 we have Then we have T r, f N r, 1/f , and it follows that Θ 0, f 0 and from 3.15 we have Θ ∞, f 1; then with 1.6 and 1.7 we may deduce δ k 2 0, f > 1.It is impossible, and we can assume that B −1; thus, we can get which with 3.15 may deduce N k 2 r, 1/f T r, f S r, f ; so δ k 2 o, f 0, which with Θ ∞, f 1 and 1.6 may deduce Θ 0, f > 1, which is impossible.Hence our claim holds.Next we will prove that A 1. From 3.17 we have that is, Θ ∞, f > 1, which is also a contradiction.Hence A 1 and f ≡ f n k .Now Theorem 1.2 has been completely proved.

Theorem A .
Let fbe nonconstant entire function.If f and f share the value 1 CM and if

1 f n k − a A 1 N
by 4.11 , then we have f • f n k ≡ a 2 , which with the above equality may lead to T r, f S r, f , which is impossible.If A / − 1, then by second fundamental theorem, Lemma 2.5, 3.15 , and 4.11 we have T r, f n k ≤ N r,