About this Journal Submit a Manuscript Table of Contents
International Journal of Mathematics and Mathematical Sciences
Volume 2010 (2010), Article ID 507454, 11 pages
http://dx.doi.org/10.1155/2010/507454
Research Article

Unicity of Meromorphic Function Sharing One Small Function with Its Derivative

1Department of Mathematics, Shandong University, Jinan, Shandong 250100, China
2Department of Mathematics, Xinxiang University, Xinxiang, Henan 453002, China

Received 26 October 2009; Revised 2 February 2010; Accepted 3 February 2010

Academic Editor: Stanisława R. Kanas

Copyright © 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.

Abstract

We deal with the problem of uniqueness of a meromorphic function sharing one small function with its k's derivative and obtain some results.

1. 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 𝑓𝑏 and 𝑔𝑏 have same zeros counting multiplicities (ignoring multiplicities).

Moreover, we use the following notations.

Let 𝑘 be a positive integer. We denote by 𝑁𝑘)(𝑟,1/(𝑓𝑎)) the counting function for the zeros of 𝑓𝑎 with multiplicity 𝑘 and by 𝑁𝑘)(𝑟,1/(𝑓𝑎)) the corresponding one for which the multiplicity is not counted. Let 𝑁(𝑘(𝑟,1/(𝑓𝑎)) be the counting function for the zeros of 𝑓𝑎 with multiplicity 𝑘, and let 𝑁(𝑘(𝑟,1/(𝑓𝑎)) be the corresponding one for which the multiplicity is not counted. Set 𝑁𝑘(𝑟,1/(𝑓𝑎))=𝑁(𝑟1/(𝑓𝑎))+𝑁(2(𝑟,1/(𝑓𝑎))++𝑁(𝑘(𝑟,1/(𝑓𝑎)). And we define

𝛿𝑝(𝑎,𝑓)=1limsup𝑟+𝑁𝑝(𝑟,1/(𝑓𝑎)).𝑇(𝑟,𝑓)(1.1) Obviously, 1Θ(𝑎,𝑓)𝛿𝑝(𝑎,𝑓)𝛿(𝑎,𝑓)0. For more details, reader can see [1, 2].

Brück (see [3]) considered the uniqueness problems of an entire function sharing one value with its derivative and proved the following result.

Theorem 1 A. Let 𝑓be nonconstant entire function. If 𝑓 and 𝑓 share the value 1 CM and if 𝑁(𝑟,1/𝑓)=𝑆(𝑟,𝑓), then (𝑓1)/(𝑓1)𝑐 for some constant 𝑐{0}.

Yang [4], Zhang [5], and Yu [6] extended Theorem A and obtained many excellent results.

Theorem 1 B (see[5]). Let𝑓 be a nonconstant meromorphic function and, let 𝑘 be a positive integer. Suppose that 𝑓 and𝑓(𝑘) share 1 CM and 2𝑁(𝑟,𝑓)+𝑁1𝑟,𝑓1+𝑁𝑟,𝑓(𝑘)<(𝜆+𝑜(1))𝑇𝑟,𝑓(𝑘),(1.2) for 𝑟𝐼, where 𝐼 is a set of infinite linear measure and 𝜆 satisfies 0<𝜆<1, then (𝑓(𝑘)1)/(𝑓1)𝑐 for some constant 𝑐{0}.

Theorem 1 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)𝑓𝑎 and 𝑓(𝑘)𝑎 share the value 0 CM,(3)4𝛿(0,𝑓)+2(𝑘+8)Θ(,𝑓)>2𝑘+19, then 𝑓𝑓(𝑘), 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 1 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 𝑓𝑎 and 𝑓(𝑘)𝑎 share 0 IM and 4𝑁(𝑟,𝑓)+3𝑁21𝑟,𝑓(𝑘)+2𝑁1𝑟,(𝑓/𝑎)<(𝜆+𝑜(1))𝑇𝑟,𝑓(𝑘),(1.3) for 0<𝜆<1,𝑟𝐼, and 𝐼 is a set of infinite linear measure. Then (𝑓(𝑘)𝑎)(𝑓𝑎)𝑐 for some constant 𝑐{0}.

In this article, we will pay our attention to the value sharing of 𝑓 and [𝑓𝑛](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 [𝑓𝑛](𝑘) share 𝑎(𝑧) IM and 4𝑁(𝑟,𝑓)+2𝑁1𝑟,(𝑓/𝑎)+2𝑁21𝑟,(𝑓𝑛)(𝑘)+𝑁1𝑟,(𝑓𝑛)(𝑘)(𝜆+𝑜(1))𝑇𝑟,(𝑓𝑛)(𝑘),(1.4) or 𝑓 and [𝑓𝑛](𝑘) share 𝑎(𝑧) CM and 2𝑁(𝑟,𝑓)+𝑁1𝑟,(𝑓/𝑎)+𝑁21𝑟,(𝑓𝑛)(𝑘)(𝜆+𝑜(1))𝑇𝑟,(𝑓𝑛)(𝑘),(1.5) for 0<𝜆<1, 𝑟𝐼, and 𝐼 is a set of infinite linear measure, then (𝑓𝑎)([𝑓𝑛](𝑘)𝑎)𝑐, for some constant 𝑐{0}.

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 [𝑓𝑛](𝑘) share 𝑎(𝑧) IM and (2𝑘+6)Θ(,𝑓)+3Θ(0,𝑓)+2𝛿𝑘+2(0,𝑓)>2𝑘+10,(1.6) or 𝑓 and [𝑓𝑛](𝑘) share 𝑎(𝑧) CM and (𝑘+3)Θ(,𝑓)+𝛿2(0,𝑓)+𝛿𝑘+2(0,𝑓)>𝑘+4,(1.7) then 𝑓(𝑓𝑛)(𝑘).

Clearly, Theorem 1.1 improves and extends Theorems B and D, while 1.2 improves and extends Theorem C.

2. 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 ; assume, that 𝐹 and 𝐺 share 1 IM; let 𝑧0 be a zero of 𝐹1 with multiplicity 𝑝 and a zero of 𝐺1 with multiplicity 𝑞. We denote by𝑁𝐸1)(𝑟,1/𝐹1)the counting function of the zeros of 𝐹1 where 𝑝=𝑞=1 and by𝑁𝐸(2(𝑟,1/𝐹1)the counting function of zeros of 𝐹1 where 𝑝=𝑞2. We denotes by 𝑁𝐿(𝑟,1/𝐹1) the counting function of the zeros of 𝐹1 where 𝑝>𝑞1; each point is counted according to its multiplicity, and 𝑁𝐿(𝑟,1/𝐹1) denote its reduced form. In the same way, we can define𝑁𝐸1)(𝑟,1/𝐺1), 𝑁𝐸(2(𝑟,1/𝐺1),𝑁𝐿(𝑟,1/𝐺1), and so on.

Definition 2.2. In this paper 𝑁0(𝑟,1/𝐹) denotes the counting function of the zeros of𝐹 which are not the zeros of 𝐹 and𝐹1, and𝑁0(𝑟,1/𝐹) denotes its reduced form. In the same way, we can define 𝑁0(𝑟,1/𝐺) and 𝑁0(𝑟,1/𝐺).

Next we present some lemmas which will be needed in the sequel. Let 𝐹,𝐺 be two non-constant meromorphic functions defined in . We shall denote by 𝐻 the following function:

𝐹𝐻=𝐹𝐹2𝐺𝐹1𝐺𝐺2𝐺1.(2.1)

Lemma 2.3 (see[2]). Let 𝐹,𝐺 be two nonconstant meromorphic functions defined in . If 𝐹 and 𝐺 are sharing 1 IM, then 𝑁(𝑟,𝐻)𝑁(𝑟,𝐹)+𝑁(21𝑟,𝐹+𝑁(21𝑟,𝐺+𝑁𝐿1𝑟,+𝐹1𝑁𝐿1𝑟,+𝐺1𝑁01𝑟,𝐹+𝑁01𝑟,𝐺+𝑆(𝑟,𝑓).(2.2) If 𝐹 and 𝐺 are sharing 1 CM, then 𝑁(𝑟,𝐻)𝑁(𝑟,𝐹)+𝑁(21𝑟,𝐹+𝑁(21𝑟,𝐺+𝑁01𝑟,𝐹+𝑁01𝑟,𝐺+𝑆(𝑟,𝑓).(2.3)

Lemma 2.4 (see[1]). Let 𝑓 be a meromorphic function and 𝑎 is a finite complex number. Then (i)𝑇(𝑟,1/(𝑓𝑎))=𝑇(𝑟,𝑓)+𝑂(1),(ii)𝑚(𝑟,𝑓(𝑘)/𝑓(𝑙))=𝑆(𝑟,𝑓) for 𝑘>𝑙0,(iii)𝑇(𝑟,𝑓)𝑁(𝑟,𝑓)+𝑁(𝑟,1/(𝑓𝑎1(𝑧)))+𝑁(𝑟,1/(𝑓𝑎2(𝑧)))+𝑆(𝑟,𝑓),where 𝑎1(𝑧)𝑎2(𝑧) are two meromorphic functions such that 𝑇(𝑟,𝑎𝑖)=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 𝑁𝑝1𝑟,𝑓(𝑘)𝑁𝑝+𝑘1𝑟,𝑓+𝑘𝑁(𝑟,𝑓)+𝑆(𝑟,𝑓).(2.4)

Lemma 2.6 (see[9]). Let 𝑓 be a non-constant meromorphic function and let 𝑛 be a positive integer. 𝑃(𝑓)=𝑎𝑛𝑓𝑛+𝑎𝑛1𝑓𝑛1++𝑎1𝑓 where 𝑎𝑖 are meromorphic functions such that 𝑇(𝑟,𝑎𝑖)=𝑆(𝑟,𝑓)(𝑖=1,2,,𝑛),  and 𝑎𝑛0. Then 𝑇(𝑟,𝑃(𝑓))=𝑛𝑇(𝑟,𝑓)+𝑆(𝑟,𝑓).(2.5)

3. Proof of Theorem 1.1

Let 𝐹=𝑓(𝑧)/𝑎(𝑧),𝐺=(𝑓𝑛(𝑧))(𝑘)/𝑎(𝑧), then

𝐹1=𝑓(𝑧)𝑎(𝑧),𝑎(𝑧)𝐺1=(𝑓𝑛(𝑧))(𝑘)𝑎(𝑧).𝑎(𝑧)(3.1) From the definitions of 𝐹,𝐺 and recalling that 𝐹 and 𝐺 share value 1 IM(CM), we get

𝑁𝐸1)1𝑟,𝐹1=𝑁𝐸1)1𝑟,𝐺1+𝑆(𝑟,𝑓),𝑁𝐸(21𝑟,=𝐹1𝑁𝐸(21𝑟,𝐺1+𝑆(𝑟,𝑓),(3.2)𝑁𝐿1𝑟,𝐹1𝑁1𝑟,𝐹+𝑁(𝑟,𝐹)+𝑆(𝑟,𝐹),(3.3)𝑁1𝑟,=𝐹1𝑁1𝑟,𝐺1+𝑆(𝑟,𝐹)=𝑁𝐸1)1𝑟,+𝐹1𝑁𝐸(21𝑟,+𝐹1𝑁𝐿1𝑟,+𝐹1𝑁𝐿1𝑟,𝐺1+𝑆(𝑟,𝑓).(3.4)

We will distinguish two cases below.

Case 1 (𝐻0). From (2.1) it is easy to see that 𝑚(𝑟,𝐻)=𝑆(𝑟,𝑓).Subcase 1.1. Suppose that 𝑓 and (𝑓𝑛)(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 𝑁(𝑟,𝐹)=𝑁(𝑟,𝑓)+𝑆(𝑟,𝑓),𝑁(𝑟,𝐺)=𝑁(𝑟,𝑓)+𝑆(𝑟,𝑓).(3.5)
Let 𝑧0 be a simple zero of 𝐹1 and 𝐺1, but 𝑎(𝑧0)0,. Through a simple calculation we know that 𝑧0 is a zero of 𝐻, so 𝑁𝐸1)1𝑟,1𝐹1𝑁𝑟,𝐻+𝑆(𝑟,𝑓)𝑇(𝑟,𝐻)+𝑆(𝑟,𝑓)𝑁(𝑟,𝐻)+𝑆(𝑟,𝑓).(3.6)
From (3.4)–(3.6) and Lemma 2.3, we have 𝑁1𝑟,𝐺1𝑁(𝑟,𝐹)+2𝑁𝐿1𝑟,𝐹1+2𝑁𝐿1𝑟,+𝐺1𝑁(21𝑟,𝐹+𝑁(21𝑟,𝐺+𝑁𝐸(21𝑟,+𝐹1𝑁01𝑟,𝐹+𝑁01𝑟,𝐺+𝑆(𝑟,𝑓)𝑁(𝑟,𝑓)+2𝑁1𝑟,𝐹+2𝑁𝐿1𝑟,+𝐺1𝑁(21𝑟,𝐺+𝑁01𝑟,𝐺+𝑆(𝑟,𝑓).(3.7) It follows by the second fundamental theorem, (3.5), and (3.7) that 𝑇(𝑟,𝐺)𝑁(𝑟,𝐺)+𝑁1𝑟,𝐺+𝑁1𝑟,𝐺1𝑁01𝑟,𝐺+𝑆(𝑟,𝐺)2𝑁(𝑟,𝑓)+2𝑁1𝑟,𝐹+2𝑁1𝑟,𝐺+𝑁1𝑟,𝐺+𝑆(𝑟,𝑓).(3.8) By Lemma 2.5, we have 𝑇𝑟,(𝑓𝑛)(𝑘)4𝑁(𝑟,𝑓)+2𝑁1𝑟,(𝑓/𝑎)+2𝑁21𝑟,(𝑓𝑛)(𝑘)+𝑁1𝑟,(𝑓𝑛)(𝑘)+𝑆(𝑟,𝑓),(3.9) which contradicts (1.4).
Subcase 1.2. Suppose that 𝑓 and (𝑓𝑛)(𝑘) share 𝑎(𝑧) CM.
Let 𝑧0 be a simple zero of 𝐹1 and 𝐺1, but 𝑎(𝑧0)0,. By a simple calculation, we can still get 𝐻(𝑧0)=0. Therefore 𝑁1)1𝑟,1𝐹1𝑁𝑟,𝐻+𝑆(𝑟,𝑓)𝑁(𝑟,𝐻)+𝑆(𝑟,𝑓).(3.10)
Noting that 𝑁1)(𝑟,1/(𝐹1))=𝑁1)(𝑟,1/(𝐺1))+𝑆(𝑟,𝑓), by (3.4) and Lemma 2.3, we can deduce 𝑁1𝑟,𝐺1𝑁(𝑟,𝐹)+𝑁(21𝑟,𝐹+𝑁(21𝑟,𝐺+𝑁01𝑟,𝐹+𝑁01𝑟,𝐺+𝑁(21𝑟,𝐹1+𝑆(𝑟,𝑓).(3.11)
By the second fundamental theorem, (3.5), and (3.11), we have 𝑇(𝑟,𝐺)𝑁(𝑟,𝐺)+𝑁1𝑟,𝐺+𝑁1𝑟,𝐺1𝑁01𝑟,𝐺+𝑆(𝑟,𝐺)2𝑁(𝑟,𝑓)+𝑁21𝑟,𝐺+𝑁1𝑟,𝐹+𝑆(𝑟,𝑓).(3.12)
Taking into account (3.1), we have 𝑇𝑟,(𝑓𝑛)(𝑘)2𝑁(𝑟,𝑓)+𝑁1𝑟,(𝑓/𝑎)+𝑁21𝑟,(𝑓𝑛)(𝑘)+𝑆(𝑟,𝑓).(3.13)
This contradicts (1.5).

Case 2 (𝐻0). Integration yields 1𝐴𝐹1𝐺1+𝐵,(3.14) where 𝐴,𝐵 are constants and 𝐴0. 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 𝐵0, then we can rewrite (3.14) as 1𝐹1𝐵(𝐺1+𝐴/𝐵).𝐺1(3.16) So 𝑁1𝑟,=𝐺1+𝐴/𝐵𝑁(𝑟,𝐹)=𝑆(𝑟,𝑓).(3.17) If 𝐴𝐵, then by Lemma 2.4 and (3.17) we have 𝑇(𝑟,𝐺)𝑁(𝑟,𝐺)+𝑁1𝑟,𝐺+𝑁1𝑟,𝐺1+𝐴/𝐵+𝑆(𝑟,𝑓)𝑁1𝑟,𝐺+𝑆(𝑟,𝑓)𝑇(𝑟,𝐺)+𝑆(𝑟,𝑓).(3.18) Hence 𝑇(𝑟,𝐺)=𝑁1𝑟,𝐺+𝑆(𝑟,𝑓),(3.19) that is, 𝑇𝑟,(𝑓𝑛)(𝑘)=𝑁1𝑟,(𝑓𝑛)(𝑘)+𝑆(𝑟,𝑓).(3.20) This is a contradiction with (1.4) and (1.5). If 𝐴=𝐵, then from (3.14) we get 1/(𝐹1)=𝐴𝐺/(𝐺1). We rewrite it as 𝑎2𝑓𝑛(𝐴𝑓𝑎𝑎𝐴)(𝑓𝑛)(𝑘)𝑓𝑛.(3.21) So by Lemmas 2.4 and 2.6 and (3.15), we have (𝑛+1)𝑇(𝑟,𝑓)=𝑇𝑟,(𝑓𝑛)(𝑘)𝑓𝑛1+𝑆(𝑟,𝑓)𝑛𝑁𝑟,𝑓+𝑘𝑁(𝑟,𝑓)+𝑆(𝑟,𝑓)𝑛𝑇(𝑟,𝑓)+𝑆(𝑟,𝑓).(3.22) This implies that 𝑇(𝑟,𝑓)=𝑆(𝑟,𝑓), since 𝑛1. This is impossible. Hence our claim is right. So (𝐺1)/(𝐹1)=𝐴. Theorem 1.1 is, thus, completely proved.

4. 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. 𝐻0Subcase 1.1. Suppose that 𝑓 and (𝑓𝑛)(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 𝑇(𝑟,𝐹)𝑁(𝑟,𝐹)+𝑁1𝑟,𝐹+𝑁1𝑟,𝐹1𝑁01𝑟,𝐹+𝑆(𝑟,𝐹)2𝑁(𝑟,𝑓)+2𝑁1𝑟,𝐺+2𝑁1𝑟,𝐹+𝑁1𝑟,𝐹+𝑆(𝑟,𝑓).(4.1) Applying Lemma 2.5 to the above inequality and noticing the definition of 𝐹,𝐺, we get 𝑇(𝑟,𝑓)(2𝑘+6)𝑁(𝑟,𝑓)+3𝑁1𝑟,𝑓+2𝑁𝑘+2𝑁1𝑟,𝑓+𝑆(𝑟,𝑓)(2𝑘+6)(1Θ(,𝑓))+33Θ(0,𝑓)+22𝛿𝑘+2(0,𝑓)𝑇(𝑟,𝑓)+𝑆(𝑟,𝑓).(4.2) This implies that (2𝑘+6)Θ(,𝑓)+3Θ(0,𝑓)+2𝛿𝑘+2(0,𝑓)2𝑘+10.(4.3) This contradicts (1.6).Subcase 1.2. Suppose that 𝑓 and (𝑓𝑛)(𝑘) share 𝑎(𝑧) CM. Similarly as above, we can easily obtain 𝑁1)(𝑟,1/(𝐹1))=𝑁1)(𝑟,1/(𝐺1))+𝑆(𝑟,𝑓); by Lemma 2.3, we can deduce 𝑁1𝑟,𝐹1𝑁(𝑟,𝐹)+𝑁(21𝑟,𝐹+𝑁(21𝑟,𝐺+𝑁01𝑟,𝐹+𝑁01𝑟,𝐺+𝑁(21𝑟,𝐺1+𝑆(𝑟,𝑓).(4.4) So by the second fundamental theorem, (4.4), and using Lemma 2.5 again, we have 𝑇(𝑟,𝐹)𝑁(𝑟,𝐹)+𝑁1𝑟,𝐹+𝑁1𝑟,𝐹1𝑁01𝑟,𝐹+𝑆(𝑟,𝑓)2𝑁(𝑟,𝑓)+𝑁21𝑟,𝑓+𝑁1𝑟,𝐺+𝑆(𝑟,𝑓)(𝑘+5)(𝑘+3)Θ(,𝑓)𝛿2(0,𝑓)𝛿𝑘+2(0,𝑓)𝑇(𝑟,𝑓)+𝑆(𝑟,𝑓).(4.5) This implies that (𝑘+3)Θ(,𝑓)+𝛿2(0,𝑓)+𝛿𝑘+2(0,𝑓)𝑘+4.(4.6) This contradicts (1.7).

Case 2 (𝐻0). 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 𝐵0and𝐵1, then 𝐴𝐺1𝐵𝐹(𝐵+1),𝐹1(4.7) and so 𝑁(𝑟,𝐺)=𝑁1𝑟,𝐹(𝐵+1)/𝐵.(4.8)
Again by second fundamental theorem and (4.4) we have 𝑇(𝑟,𝐹)=𝑁1𝑟,𝐹+𝑆(𝑟,𝑓),(4.9) that is, 𝑇(𝑟,𝑓)𝑁1𝑟,𝑓+𝑆(𝑟,𝑓)𝑇(𝑟,𝑓)+𝑆(𝑟,𝑓).(4.10) 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 𝛿𝑘+2(0,𝑓)>1. It is impossible, and we can assume that 𝐵=1; thus, we can get (𝑓𝑛)(𝑘)𝑎1(𝐴+1)𝐴𝑎𝑓.(4.11) It shows that 𝑇(𝑟,𝑓)=𝑇(𝑟,(𝑓𝑛)(𝑘)).
If 𝐴=1, by (4.11), then we have 𝑓(𝑓𝑛)(𝑘)𝑎2, which with the above equality may lead to 𝑇(𝑟,𝑓)=𝑆(𝑟,𝑓), which is impossible. If 𝐴1, then by second fundamental theorem, Lemma 2.5, (3.15), and (4.11) we have 𝑇𝑟,(𝑓𝑛)(𝑘)𝑁1𝑟,(𝑓𝑛)(𝑘)+𝑎(𝐴+1)𝑁1𝑟,(𝑓𝑛)(𝑘)+𝑆(𝑟,𝑓),𝑘𝑁(𝑟,𝑓)+𝑁𝑘+21𝑟,𝑓+𝑆(𝑟,𝑓)𝑇(𝑟,𝑓)+𝑆(𝑟,𝑓),(4.12) which with (3.15) may deduce 𝑁𝑘+2(𝑟,1/𝑓)=𝑇(𝑟,𝑓)+𝑆(𝑟,𝑓); so 𝛿𝑘+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 𝐺1𝐴(𝐹1). Then 𝑁1𝑟,𝐺=𝑁1𝑟,.𝐹+1/𝐴1(4.13) If 𝐴1, then we have 𝑇(𝑟,𝐹)𝑁(𝑟,𝐹)+𝑁1𝑟,𝐹+𝑁1𝑟,𝐺+𝑆(𝑟,𝑓).(4.14) By Lemma 2.5, we get 𝑇(𝑟,𝑓)(𝑘+1)𝑁(𝑟,𝑓)+𝑁1𝑟,𝑓+𝑁𝑘+21𝑟,𝑓+𝑆(𝑟,𝑓).(4.15) It implies that (𝑘+1)Θ(,𝑓)+Θ(0,𝑓)+𝛿𝑘+2(0,𝑓)𝑘+2.(4.16) Combining (4.16) with (1.6) yields 2(𝑘+2)+Θ(0,𝑓)2(𝑘+3)Θ(,𝑓)+3Θ(0,𝑓)+2𝛿2+𝑘(0,𝑓)4Θ(,𝑓)>2𝑘+6,(4.17) that is, Θ(0,𝑓)>2. This is a contradiction.
Combining (4.16) with (1.7) yields 𝑘+2+2Θ(,𝑓)(𝑘+3)Θ(,𝑓)+Θ(0,𝑓)+𝛿𝑘+2(0,𝑓)>𝑘+4,(4.18) that is, Θ(,𝑓)>1, which is also a contradiction. Hence 𝐴=1 and 𝑓(𝑓𝑛)(𝑘). Now Theorem 1.2 has been completely proved.

Acknowledgment

The authors would like to express their sincere thanks to the referee for helpful comments and suggestions.

References

  1. W. K. Hayman, Meromorphic Functions, Oxford Mathematical Monographs, Clarendon Press, Oxford, UK, 1964. View at MathSciNet
  2. C. C. Yang and H.-X. Yi, Uniqueness Theory of Meromorphic Functions, vol. 557 of Mathematics and Its Applications, Science Press, Beijing, China; Kluwer Academic, New York, NY, USA, 2003. View at MathSciNet
  3. R. Brück, “On entire functions which share one value CM with their first derivative,” Results in Mathematics, vol. 30, no. 1-2, pp. 21–24, 1996. View at Zentralblatt MATH · View at MathSciNet
  4. L. Z. Yang, “Solution of a differential equation and its applications,” Kodai Mathematical Journal, vol. 22, no. 3, pp. 458–464, 1999. View at Publisher · View at Google Scholar · View at Zentralblatt MATH · View at MathSciNet
  5. Q. C. Zhang, “The uniqueness of meromorphic functions with their derivatives,” Kodai Mathematical Journal, vol. 21, no. 2, pp. 179–184, 1998. View at Publisher · View at Google Scholar · View at Zentralblatt MATH · View at MathSciNet
  6. K.-W. Yu, “On entire and meromorphic functions that share small functions with their derivatives,” Journal of Inequalities in Pure and Applied Mathematics, vol. 4, no. 1, article 21, p. 7, 2003. View at Zentralblatt MATH · View at MathSciNet
  7. I. Lahiri and A. Sarkar, “Uniqueness of a meromorphic function and its derivative,” Journal of Inequalities in Pure and Applied Mathematics, vol. 5, no. 1, article 20, p. 9, 2004. View at Zentralblatt MATH · View at MathSciNet
  8. Q. C. Zhang, “Meromorphic function that shares one small function with its derivative,” Journal of Inequalities in Pure and Applied Mathematics, vol. 6, no. 4, article 116, p. 13, 2005. View at Zentralblatt MATH · View at MathSciNet
  9. C. C. Yang, “On deficiencies of differential polynomials. II,” Mathematische Zeitschrift, vol. 125, pp. 107–112, 1972. View at Publisher · View at Google Scholar · View at Zentralblatt MATH · View at MathSciNet