﻿ Some Uniqueness Results of Q-Shift Difference Polynomials Involving Sharing Functions

Applied Mathematics
Vol.08 No.08(2017), Article ID:78579,11 pages
10.4236/am.2017.88084

Some Uniqueness Results of Q-Shift Difference Polynomials Involving Sharing Functions*

Xuexue Qian, Yasheng Ye

Department of Mathematics, College of Sciences, University of Shanghai for Science and Technology, Shanghai, China    Received: July 26, 2017; Accepted: August 18, 2017; Published: August 21, 2017

ABSTRACT

In this paper, we mainly study the uniqueness of specific q-shift difference polynomials ${f}^{n}\left(z\right)\underset{j=1}{\overset{d}{\prod }}f{\left({q}_{j}z+{c}_{j}\right)}^{{v}_{j}}$ and ${g}^{n}\left(z\right)\underset{j=1}{\overset{d}{\prod }}g{\left({q}_{j}z+{c}_{j}\right)}^{{v}_{j}}$ 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.

Keywords:

Value Distribution, Meromorphic Functions, Difference Polynomials, Uniqueness *Supported by the National Natural Science Foundation of China (No.11371139).

1. Introduction

In recent years, many Scholars have been interested in value distribution of difference operators of meromorphic functions (see  -  ). Furthermore, a large number of papers have studied and obtained the uniqueness results of difference polynomials of meromorphic functions, their shifts and difference operators (see  -  ). Our purpose in the paper is to study the value distribution for q-shift polynomials of transcendental meromorphic with zero order, and some results about entire functions.

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  ), such as $m\left(r,f\right)$ , $N\left(r,f\right)$ , $\stackrel{¯}{N}\left(r,f\right)$ , $T\left(r,f\right)$ ,

$S\left(r,f\right)$ , and define ${N}_{2}\left(r,\frac{1}{f}\right)$ 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 $S\left(r,f\right)$ , if it satisfies $S\left(r,f\right)=o\left(T\left(r,f\right)\right)$ , as $r\to \infty$ outside of a possible exceptional set of r with finite logarithmic measure. In addition, the notation $\rho \left(f\right)$ is the order of growth of $f$ . Let meromorphic function $\alpha$ be a common small function of $f\left(z\right)$ and $g\left(z\right)$ , suppose that $f\left(z\right)-\alpha \left(z\right)$ and $g\left(z\right)-\alpha \left(z\right)$ have the same zeros counting multiplicities (ignoring multiplicities), then we say that $f$ and $g$ share $\alpha \left(z\right)$ CM(IM).

In this paper, we define a q-shift difference product of meromorphic function $f\left(z\right)$ as follows.

$F\left(z\right)={f}^{n}\left(z\right)\underset{j=1}{\overset{d}{\prod }}f{\left({q}_{j}z+{c}_{j}\right)}^{{v}_{j}}$ (1)

${F}_{1}\left(z\right)={P}_{n}\left(f\left(z\right)\right)\underset{j=1}{\overset{d}{\prod }}f{\left({q}_{j}z+{c}_{j}\right)}^{{v}_{j}}$ (2)

where ${c}_{j}\in ℂ$ $\left({c}_{j}\ne 0,j=1,2,3,\cdots \text{ },d\right)$ are distinct constants, ${q}_{j}\left(j=1,2,\cdots ,d\right)$ be non-zero finite complex constants, let ${P}_{n}\left(z\right)={\alpha }_{n}{z}^{n}+{\alpha }_{n-1}{z}^{n-1}+\cdots +{\alpha }_{1}z+{\alpha }_{0}$ be a non-zero polynomial, where ${\alpha }_{n}\left(\ne 0\right),{\alpha }_{n-1},\cdots ,{\alpha }_{0}$ are small functions of $f$ . Let $n,d,{v}_{j}\left(j=1,2,\cdots ,d\right)$ are positive integers and $\sigma ={v}_{1}+{v}_{2}+\cdots +{v}_{d}$ .

Recently, Liu et al.  have considered and proved the uniqueness of q-shift difference polynomials of meromorphic functions.

Theorem A. Let $f\left(z\right)$ and $g\left(z\right)$ be two transcendental meromorphic functions with $\rho \left(f\right)=\rho \left(g\right)=0$ . Let $q$ and $\eta$ be two non-zero finite complex constants. If ${f}^{n}\left(z\right)f\left(qz+\eta \right)$ and ${g}^{n}\left(z\right)g\left(qz+\eta \right)$ share 1 CM, then either $f\left(z\right)=tg\left(z\right)$ or $f\left(z\right)g\left(z\right)=t$ , where $n\left(\in {N}^{*}\right)\ge 14$ satisfying ${t}^{n+1}=1$ .

Theorem B. Let $f\left(z\right)$ and $g\left(z\right)$ be two transcendental meromorphic functions with $\rho \left(f\right)=\rho \left(g\right)=0$ . Let $q$ and $\eta$ be two non-zero finite complex constants. If ${f}^{n}\left(z\right)f\left(qz+\eta \right)$ and ${g}^{n}\left(z\right)g\left(qz+\eta \right)$ share 1 IM, then either $f\left(z\right)=tg\left(z\right)$ or $f\left(z\right)g\left(z\right)=t$ , where $n\left(\in {N}^{*}\right)\ge 26$ 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 $f\left(z\right)$ and $g\left(z\right)$ be two transcendental meromorphic functions with $\rho \left(f\right)=\rho \left(g\right)=0$ , and let $\alpha \left(z\right)\left(\overline{)\equiv }0\right)$ be a common small function of $f\left(z\right)$ and $g\left(z\right)$ . If $F\left(z\right)$ and $G\left(z\right)$ share $\alpha \left(z\right)$ CM, then $f\left(z\right)=tg\left(z\right)$ , where $n\ge 4min\left(2d,\sigma \right)+\sigma +9$ satisfying ${t}^{n+\sigma }=1$ .

Theorem 1.2. Let $f\left(z\right)$ and $g\left(z\right)$ be two transcendental meromorphic functions with $\rho \left(f\right)=\rho \left(g\right)=0$ , and let $\alpha \left(z\right)\left(\overline{)\equiv }0\right)$ be a common small function of $f\left(z\right)$ and $g\left(z\right)$ . If $F\left(z\right)$ and $G\left(z\right)$ share $\alpha \left(z\right)$ IM, then $f\left(z\right)=tg\left(z\right)$ , where $n\ge 4min\left(2d,\sigma \right)+\sigma +6d+15$ satisfying ${t}^{n+\sigma }=1$ .

Liu et al.  also considered some properties of q-shift difference poly- nomials of entire functions, as follow:

Theorem C. Let $f\left(z\right)$ and $g\left(z\right)$ be two transcendental entire functions with $\rho \left(f\right)=\rho \left(g\right)=0$ , and let $q$ and $\eta$ are two non-zero finite complex constants, and let ${P}_{n}\left(z\right)={\alpha }_{n}{z}^{n}+{\alpha }_{n-1}{z}^{n-1}+\cdots +{\alpha }_{1}z+{\alpha }_{0}$ be a non-zero polynomial, where ${\alpha }_{n}\left(\ne 0\right),{\alpha }_{n-1},\cdots ,{\alpha }_{0}$ , are constants, and let m be the number of the distinct zero of ${P}_{n}\left(z\right)$ . If ${P}_{n}\left(f\left(z\right)\right)f\left(qz+\eta \right)$ and ${P}_{n}\left(g\left(z\right)\right)g\left(qz+\eta \right)$ share 1 CM, then only one of the following two cases holds:

a) $f\left(z\right)=tg\left(z\right)$ , where $n>2m+1$ , and $k$ is greatest common divisor of $\left({\lambda }_{0},{\lambda }_{1},\cdots ,{\lambda }_{n}\right)$ , satisfying ${t}^{k}=1$ . When ${\alpha }_{i}=0$ , then ${\lambda }_{i}=n+1$ , otherwise ${\lambda }_{i}=i+1$ . $i=0,1,\cdots ,n$ .

b) $f\left(z\right)$ and $g\left(z\right)$ satisfy a algebraic equation $Q\left(f\left(z\right),g\left(z\right)\right)=0$ , where

$Q\left({w}_{1},{w}_{2}\right)={P}_{n}\left({w}_{1}\right){w}_{1}\left(qz+c\right)-{P}_{n}\left({w}_{2}\right){w}_{2}\left(qz+c\right)$ (3)

Next, it is easy to derive that ${P}_{n}\left(f\left(z\right)\right)f\left(qz+\eta \right)$ in Theorem C can be replaced by ${P}_{n}\left(f\left(z\right)\right)\underset{j=1}{\overset{d}{\prod }}f{\left({q}_{j}z+{c}_{j}\right)}^{{v}_{j}}$ , as follows

Theorem 1.3. Let $f\left(z\right)$ and $g\left(z\right)$ be two transcendental entire functions with $\rho \left(f\right)=\rho \left(g\right)=0$ , and let $\alpha \left(z\right)$ be a common small function of $f\left(z\right)$ and $g\left(z\right)$ , and let $k$ be the number of distinct zeros of ${P}_{n}\left(z\right)$ . If ${F}_{1}\left(z\right)$ and ${G}_{1}\left(z\right)$ share $\alpha \left(z\right)$ CM, then only one of the following results holds:

a) $f\left(z\right)=tg\left(z\right)$ for a constant $t$ such that ${t}^{m}=1$ , where $n>2k+2d+\sigma$ and $m$ is greatest common divisor of $\left(n+\sigma ,n+\sigma -1,\cdots ,n+\sigma -i,\cdots ,\sigma +1\right)$ , ${\alpha }_{n-i}\ne 0$ , $i=0,1,\cdots ,n-1$ .

b) $f\left(z\right)$ and $g\left(z\right)$ satisfy a algebraic equation $Q\left(f,g\right)\equiv 0$ , where

$Q\left({w}_{1},{w}_{2}\right)={P}_{n}\left({w}_{1}\right)\underset{j=1}{\overset{d}{\prod }}{w}_{1}{\left({q}_{j}z+{c}_{j}\right)}^{{v}_{j}}-{P}_{n}\left({w}_{2}\right)\underset{j=1}{\overset{d}{\prod }}{w}_{2}{\left({q}_{j}z+{c}_{j}\right)}^{{v}_{j}}.$ (4)

2. Some Lemmas

Lemma 2.1. (see  ) Let $n\left(\ge 1\right)$ be a positive integer, and let $f\left(z\right)$ be a transcendental meromorphic function, and let ${\alpha }_{i}\left(i=0,1,\cdots ,n\right)$ be small meromorphic functions of $f$ . If

${P}_{n}\left(f\left(z\right)\right)={\alpha }_{n}{f}^{n}\left(z\right)+{\alpha }_{n-1}{f}^{n-1}\left(z\right)+\cdots +{\alpha }_{1}f\left(z\right)+{\alpha }_{0},$ (5)

then

$T\left(r,{P}_{n}\left(f\left(z\right)\right)\right)=nT\left(r,f\left(z\right)\right)+S\left(r,f\left(z\right)\right)$ (6)

Lemma 2.2. (see  ) Let $q$ and $\eta$ be two non-zero finite complex numbers, and let $f\left(z\right)$ be a nonconstant meromorphic function with $\rho \left(f\right)=0$ , then

$m\left(r,\frac{f\left(qz+\eta \right)}{f\left(z\right)}\right)=S\left(r,f\right).$ (7)

on a set of logarithmic density 1.

Lemma 2.3. (see  ) Let $f\left(z\right)$ and $g\left(z\right)$ be two non-constant meromorphic functions. Let $f\left(z\right)$ and $g\left(z\right)$ share 1 IM and

$L=\frac{{f}^{″}}{{f}^{\prime }}-2\frac{{f}^{\prime }}{f-1}-\frac{{g}^{″}}{{g}^{\prime }}+2\frac{{g}^{\prime }}{g-1}$ (8)

If $L\overline{)\equiv }0$ , then

$\begin{array}{c}T\left(r,f\right)+T\left(r,g\right)\\ \le 2\left({N}_{2}\left(r,f\right)+{N}_{2}\left(r,g\right)+{N}_{2}\left(r,\frac{1}{f}\right)+{N}_{2}\left(r,\frac{1}{g}\right)\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+3\left(\stackrel{¯}{N}\left(r,f\right)+\stackrel{¯}{N}\left(r,g\right)+\stackrel{¯}{N}\left(r,\frac{1}{f}\right)+\stackrel{¯}{N}\left(r,\frac{1}{g}\right)\right)+S\left(r,f\right)+S\left(r,g\right)\end{array}$ (9)

Lemma 2.4. (see  ) Let $f$ and $g$ be two non-constant meromorphic functions. If $f$ and $g$ share 1 CM, then only one of the following results holds:

$\begin{array}{l}\text{(a)}\text{\hspace{0.17em}}\mathrm{max}\left\{T\left(r,f\right),T\left(r,g\right)\right\}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\le {N}_{2}\left(r,f\right)+{N}_{2}\left(r,g\right)+{N}_{2}\left(r,\frac{1}{f}\right)+{N}_{2}\left(r,\frac{1}{g}\right)+S\left(r,f\right)+S\left(r,g\right)\\ \text{(b)}\text{\hspace{0.17em}}f\equiv g;\\ \text{(c)}\text{\hspace{0.17em}}fg\equiv 1.\end{array}$ (10)

Lemma 2.5. (see  ) Let $q$ and $\eta$ be two non-zero finite complex constants, and let $f$ be a non-constant meromorphic function with $\rho \left(f\right)=0$ , then

$T\left(r,f\left(qz+\eta \right)\right)\le T\left(r,f\left(z\right)\right)+S\left(r,f\right)$ (11)

on a set of logarithmic density 1.

Lemma 2.6. (see  ) Let $q$ and $\eta$ be two non-zero finite complex constants, and let $f$ be a nonconstant meromorphic function of zero order, then

$\begin{array}{l}\stackrel{¯}{N}\left(r,f\left(qz+\eta \right)\right)\le \stackrel{¯}{N}\left(r,f\left(z\right)\right)+S\left(r,f\right)\\ \stackrel{¯}{N}\left(r,\frac{1}{f\left(qz+\eta \right)}\right)\le \stackrel{¯}{N}\left(r,\frac{1}{f\left(z\right)}\right)+S\left(r,f\right)\\ N\left(r,f\left(qz+\eta \right)\right)\le N\left(r,f\left(z\right)\right)+S\left(r,f\right)\\ N\left(r,\frac{1}{f\left(qz+\eta \right)}\right)\le N\left(r,\frac{1}{f\left(z\right)}\right)+S\left(r,f\right).\end{array}$ (12)

Lemma 2.7. Let $f\left(z\right)$ be a non-constant meromorphic function of zero order, and ${F}_{1}\left(z\right)$ be defined as in (2). Then

$\left(n-\sigma \right)T\left(r,f\right)+S\left(r,f\right)\le T\left(r,{F}_{1}\right)\le \left(n+\sigma \right)T\left(r,f\right)+S\left(r,f\right)$ (13)

Proof. Combining Lemma 2.1 with Lemma 2.5, we obtain

$\begin{array}{c}T\left(r,{F}_{1}\right)\le T\left(r,{P}_{n}\left(f\left(z\right)\right)\right)+T\left(r,\underset{j=1}{\overset{d}{\prod }}f{\left({q}_{j}z+{c}_{j}\right)}^{{v}_{j}}\right)+S\left(r,f\right)\\ \le nT\left(r,f\left(z\right)\right)+\underset{j=1}{\overset{d}{\sum }}\text{ }\text{ }T\left(r,f{\left({q}_{j}z+{c}_{j}\right)}^{{v}_{j}}\right)+S\left(r,f\right)\\ \le \left(n+\sigma \right)T\left(r,f\left(z\right)\right)+S\left(r,f\right)\end{array}$ (14)

In addition, by Lemma 2.1 and Lemma 2.5, we also get

$\begin{array}{l}\left(n+\sigma \right)T\left(r,f\left(z\right)\right)\le T\left(r,{P}_{n}\left(f\left(z\right)\right){f}^{\sigma }\right)+S\left(r,f\right)\\ =m\left(r,{P}_{n}\left(f\left(z\right)\right){f}^{\sigma }\right)+N\left(r,{P}_{n}\left(f\left(z\right)\right){f}^{\sigma }\right)+S\left(r,f\right)\\ \le m\left(r,\frac{{F}_{1}\left(z\right){f}^{\sigma }}{\underset{j=1}{\overset{d}{\prod }}f{\left({q}_{j}z+{c}_{j}\right)}^{{v}_{j}}}\right)+N\left(r,\frac{{F}_{1}\left(z\right){f}^{\sigma }}{\underset{j=1}{\overset{d}{\prod }}f{\left({q}_{j}z+{c}_{j}\right)}^{{v}_{j}}}\right)+S\left(r,f\right)\\ \le m\left(r,{F}_{1}\right)+N\left(r,{F}_{1}\right)+T\left(r,\frac{{f}^{\sigma }}{\underset{j=1}{\overset{d}{\prod }}f{\left({q}_{j}z+{c}_{j}\right)}^{{v}_{j}}}\right)+S\left(r,f\right)\\ \le T\left(r,{F}_{1}\right)+2\sigma T\left(r,f\right)+S\left(r,f\right)\end{array}$ (15)

which is equivalent to

$\left(n-\sigma \right)T\left(r,f\right)+S\left(r,f\right)\le T\left(r,{F}_{1}\right)$ (16)

Therefore, we get Lemma 2.7.

Lemma 2.8. Let $f\left(z\right)$ be an entire function with $\rho \left(f\right)=0$ , and ${F}_{1}\left(z\right)$ be stated as in (2). Then

$T\left(r,{F}_{1}\right)=\left(n+\sigma \right)T\left(r,f\right)+S\left(r,f\right)$ (17)

Proof. Using the same method as the Lemma 2.7, we can easily to prove.

3. Proof of Theorem

3.1. Proof of Theorem 1.1

Set ${F}^{*}\left(z\right)=\frac{F\left(z\right)}{\alpha \left(z\right)}$ , ${G}^{*}\left(z\right)=\frac{G\left(z\right)}{\alpha \left(z\right)}$ , than ${F}^{*}\left(z\right)$ and ${G}^{*}\left(z\right)$ share 1 CM.

Thus by Nevanlinna second fundamental theory, Lemma 2.5 and Lemma 2.7, we have

$\begin{array}{l}\left(n-\sigma \right)T\left(r,f\right)+S\left(r,f\right)\le T\left(r,{F}^{*}\left(z\right)\right)\\ \le \stackrel{¯}{N}\left(r,{F}^{*}\left(z\right)\right)+\stackrel{¯}{N}\left(r,\frac{1}{{F}^{*}\left(z\right)}\right)+\stackrel{¯}{N}\left(r,\frac{1}{{F}^{*}\left(z\right)-1}\right)+S\left(r,{F}^{*}\left(z\right)\right)\\ \le \stackrel{¯}{N}\left(r,{f}^{n}\right)+\stackrel{¯}{N}\left(r,\underset{j=1}{\overset{d}{\prod }}f{\left({q}_{j}z+{c}_{j}\right)}^{{v}_{j}}\right)+\stackrel{¯}{N}\left(r,\frac{1}{{f}^{n}}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\stackrel{¯}{N}\left(r,\frac{1}{\underset{j=1}{\overset{d}{\prod }}f{\left({q}_{j}z+{c}_{j}\right)}^{{v}_{j}}}\right)+\stackrel{¯}{N}\left(r,\frac{1}{{G}^{*}\left(z\right)-1}\right)+S\left(r,f\right)\\ \le \left(2d+2\right)T\left(r,f\right)+\left(n+\sigma \right)T\left(r,g\right)+S\left(r,g\right)+S\left(r,f\right)\end{array}$ (18)

Then

$\left(n-2d-\sigma -2\right)T\left(r,f\right)\le \left(n+\sigma \right)T\left(r,g\right)+S\left(r,g\right)+S\left(r,f\right)$ (19)

Similarly,

$\left(n-2d-\sigma -2\right)T\left(r,g\right)\le \left(n+\sigma \right)T\left(r,f\right)+S\left(r,f\right)+S\left(r,g\right)$ (20)

It follows that $S\left(r,f\right)=S\left(r,g\right)$ .

Then by Lemma 2.4, we consider three subcases.

Case 1. Suppose that $\begin{array}{l}\mathrm{max}\left\{T\left(r,{F}^{*}\left(z\right)\right),T\left(r,{G}^{*}\left(z\right)\right)\right\}\le {N}_{2}\left(r,{F}^{*}\left(z\right)\right)+{N}_{2}\left(r,\frac{1}{{F}^{*}\left(z\right)}\right)+{N}_{2}\left(r,{G}^{*}\left(z\right)\right)\\ +{N}_{2}\left(r,\frac{1}{{G}^{*}\left(z\right)}\right)+S\left(r,{F}^{*}\left(z\right)\right)+S\left(r,{G}^{*}\left(z\right)\right)\end{array}$ holds.

Through simple calculation, we have

$\begin{array}{l}{N}_{2}\left(r,{F}^{*}\left(z\right)\right)\le {N}_{2}\left(r,{f}^{n}\right)+{N}_{2}\left(r,\underset{j=1}{\overset{d}{\prod }}f{\left({q}_{j}z+{c}_{j}\right)}^{{v}_{j}}\right)\\ \le \left\{2+\mathrm{min}\left(2d,\sigma \right)\right\}T\left(r,f\right)+S\left(r,f\right)\end{array}$ (21)

In the same way,

$\begin{array}{l}{N}_{2}\left(r,\frac{1}{{F}^{*}\left(z\right)}\right)\le \left\{2+\mathrm{min}\left(2d,\sigma \right)\right\}T\left(r,f\right)+S\left(r,f\right)\\ {N}_{2}\left(r,{G}^{*}\left(z\right)\right)\le \left\{2+\mathrm{min}\left(2d,\sigma \right)\right\}T\left(r,g\right)+S\left(r,g\right)\\ {N}_{2}\left(r,\frac{1}{{G}^{*}\left(z\right)}\right)\le \left\{2+\mathrm{min}\left(2d,\sigma \right)\right\}T\left(r,g\right)+S\left(r,g\right)\end{array}$ (22)

Combining Lemma 2.4, Lemma 2.7, Equations ((21) and (22)), we obtain that

$\begin{array}{l}\left(n-\sigma \right)\left(T\left(r,f\right)+T\left(r,g\right)\right)\le T\left(r,{F}^{*}\left(z\right)\right)+T\left(r,{G}^{*}\left(z\right)\right)\\ \le 2{N}_{2}\left(r,{F}^{*}\left(z\right)\right)+2{N}_{2}\left(r,\frac{1}{{F}^{*}\left(z\right)}\right)+2{N}_{2}\left(r,{G}^{*}\left(z\right)\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+2{N}_{2}\left(r,\frac{1}{{G}^{*}\left(z\right)}\right)+S\left(r,{F}^{*}\left(z\right)\right)+S\left(r,{G}^{*}\left(z\right)\right)\\ \le 4\left[2+\mathrm{min}\left(2d,\sigma \right)\right]\left(T\left(r,f\right)+T\left(r,g\right)\right)+S\left(r,f\right)+S\left(r,g\right)\end{array}$ (23)

Then

$\left(n-\sigma -8-4min\left(2d,\sigma \right)\right)\left(T\left(r,f\right)+T\left(r,g\right)\right)\le S\left(r,f\right)$ (24)

Which is impossible, since $n\ge 4min\left(2d,\sigma \right)+\sigma +9$ .

Case 2. Suppose that ${F}^{*}\left(z\right)\equiv {G}^{*}\left(z\right)$ holds, we obtain

$\begin{array}{l}{f}^{n}\left(z\right)\underset{j=1}{\overset{d}{\prod }}f{\left({q}_{j}z+{c}_{j}\right)}^{{v}_{j}}={g}^{n}\left(z\right)\underset{j=1}{\overset{d}{\prod }}g{\left({q}_{j}z+{c}_{j}\right)}^{{v}_{j}}.\hfill \end{array}$ (25)

We assume that $h\left(z\right):=\frac{f\left(z\right)}{g\left(z\right)}$ . If $h\left(z\right)\equiv C$ (constant), then $f=tg$ , and by substituting $f=tg$ into (25), we obtain that

$\begin{array}{l}{g}^{n}\underset{j=1}{\overset{d}{\prod }}g{\left({q}_{j}z+{c}_{j}\right)}^{{v}_{j}}\left[{t}^{n+\sigma }-1\right]=0.\hfill \end{array}$ (26)

Since $g$ is a transcendental meromorphic function, than ${g}^{n}\underset{j=1}{\overset{d}{\prod }}g{\left({q}_{j}z+{c}_{j}\right)}^{{v}_{j}}\overline{)\equiv }0$ . It follows that ${t}^{n+\sigma }=1$ .

Suppose that $h\left(z\right)\overline{)\equiv }C$ (constant), then using (25), we deduce that ${h}^{n}\left(z\right)=\underset{j=1}{\overset{d}{\prod }}\frac{1}{h{\left({q}_{j}z+{c}_{j}\right)}^{{v}_{j}}}$ ,

So

$nT\left(r,h\left(z\right)\right)=T\left(r,\underset{j=1}{\overset{d}{\prod }}\frac{1}{h{\left({q}_{j}z+{c}_{j}\right)}^{{v}_{j}}}\right)\le \sigma T\left(r,h\left(z\right)\right)+S\left(r,h\left(z\right)\right)$ (27)

We get a contradiction, since $\begin{array}{l}n\ge 4min\left(2d,\sigma \right)+\sigma +9\hfill \end{array}$ .

Case 3. Suppose that ${F}^{*}\left(z\right){G}^{*}\left(z\right)\equiv 1$ holds, then ${f}^{n}\left(z\right)\underset{j=1}{\overset{d}{\prod }}f{\left({q}_{j}z+{c}_{j}\right)}^{{v}_{j}}\cdot {g}^{n}\left(z\right)\underset{j=1}{\overset{d}{\prod }}g{\left({q}_{j}z+{c}_{j}\right)}^{{v}_{j}}={\alpha }^{2}\left(z\right).$

We define ${h}_{1}\left(z\right)=f\left(z\right)\cdot g\left(z\right)$ , we easily get ${h}_{1}^{n}\left(z\right)=\underset{j=1}{\overset{d}{\prod }}\frac{{\alpha }^{2}\left(z\right)}{{h}_{1}{\left({q}_{j}z+{c}_{j}\right)}^{{v}_{j}}}$ is non-constant, hence

$nT\left(r,{h}_{1}\left(z\right)\right)=T\left(r,\underset{j=1}{\overset{d}{\prod }}\frac{{\alpha }^{2}\left(z\right)}{{h}_{1}{\left({q}_{j}z+{c}_{j}\right)}^{{v}_{j}}}\right)\le \sigma T\left(r,{h}_{1}\left(z\right)\right)+S\left(r,{h}_{1}\left(z\right)\right)$ (28)

We get a contradiction, since $n\ge 4\mathrm{min}\left(2d,\sigma \right)+\sigma +9$ . This implies that ${h}_{1}\left(z\right)$ is a constant, which is impossible.

3.2. Proof of Theorem 1.2

Set ${F}^{*}\left(z\right)=\frac{F\left(z\right)}{\alpha \left(z\right)}$ , ${G}^{*}\left(z\right)=\frac{G\left(z\right)}{\alpha \left(z\right)}$ , So ${F}^{*}\left(z\right)$ and ${G}^{*}\left(z\right)$ share 1 IM.

Using the same arguments as in Theorem 1.1, we prove that (18)-(22) holds.

We can easily get

$\begin{array}{l}\stackrel{¯}{N}\left(r,{F}^{*}\left(z\right)\right)\le \left(1+d\right)T\left(r,f\right)+S\left(r,f\right)\\ \stackrel{¯}{N}\left(r,\frac{1}{{F}^{*}\left(z\right)}\right)\le \left(1+d\right)T\left(r,f\right)+S\left(r,f\right)\\ \stackrel{¯}{N}\left(r,{G}^{*}\left(z\right)\right)\le \left(1+d\right)T\left(r,g\right)+S\left(r,g\right)\\ \stackrel{¯}{N}\left(r,\frac{1}{{G}^{*}\left(z\right)}\right)\le \left(1+d\right)T\left(r,g\right)+S\left(r,g\right)\end{array}$ (29)

Let

$L\left(z\right)=\frac{{F}^{*}{}^{\prime \text{​}\prime }\left(z\right)}{{F}^{*}{}^{\prime }\left(z\right)}-2\frac{{F}^{*}{}^{\prime }\left(z\right)}{{F}^{*}\left(z\right)-1}-\frac{{G}^{*}{}^{\prime \text{​}\prime }\left(z\right)}{{G}^{*}{}^{\prime }\left(z\right)}+2\frac{{G}^{*}{}^{\prime }\left(z\right)}{{G}^{*}\left(z\right)-1}$ (30)

If $L\overline{)\equiv }0$ , combining Lemma 2.3, (21), (22) with (29), we obtain

$\begin{array}{l}\left(n-\sigma \right)\left(T\left(r,f\right)+T\left(r,g\right)\right)\le T\left(r,{F}^{*}\left(z\right)\right)+T\left(r,{G}^{*}\left(z\right)\right)\\ \le \left[14+6d+4min\left(2d,\sigma \right)\right]\left(T\left(r,f\right)+T\left(r,g\right)\right)+S\left(r,f\right)+S\left(r,g\right)\end{array}$ (31)

Then,

$\left(n-\sigma -14-6d-4\mathrm{min}\left(2d,\sigma \right)\right)\left(T\left(r,f\right)+T\left(r,g\right)\right)\le S\left(r,f\right)+S\left(r,g\right)$ (32)

that is impossible, since $n\ge 4min\left(2d,\sigma \right)+\sigma +6d+15$ . Hence, we get $L\equiv 0$ .

By integrating L twice, we obtain that

${F}^{*}=\frac{\left(b+1\right){G}^{*}+\left(a-b-1\right)}{b{G}^{*}+\left(a-b\right)}$ (33)

which yields $T\left(r,{F}^{*}\right)=T\left(r,{G}^{*}\right)+O\left(1\right)$ . From Lemma 2.8, we deduced that $T\left(r,f\right)=T\left(r,g\right)+S\left(r,f\right)$ . Next, we will consider the following three subcases.

Case 1. $b\ne 0$ and $b\ne -1$ . Suppose that $a-b-1\ne 0$ , by (33), we get

$\stackrel{¯}{N}\left(r,\frac{1}{{F}^{*}}\right)=\stackrel{¯}{N}\left(r,\frac{1}{{G}^{*}-\frac{a-b-1}{b+1}}\right)$ (34)

Combining the second fundamental theory with Lemma 2.5, Lemma 2.7, (29), and (34), we have

$\begin{array}{l}\left(n-\sigma \right)T\left(r,g\right)\le T\left(r,{G}^{*}\left(z\right)\right)+S\left(r,g\right)\\ \le \stackrel{¯}{N}\left(r,{G}^{*}\left(z\right)\right)+\stackrel{¯}{N}\left(r,\frac{1}{{G}^{*}\left(z\right)}\right)+\stackrel{¯}{N}\left(r,\frac{1}{{G}^{*}-\frac{a-b-1}{b+1}}\right)+S\left(r,g\right)\\ \le \stackrel{¯}{N}\left(r,{G}^{*}\left(z\right)\right)+\stackrel{¯}{N}\left(r,\frac{1}{{G}^{*}\left(z\right)}\right)+\stackrel{¯}{N}\left(r,\frac{1}{{F}^{*}}\right)+S\left(r,g\right)\\ \le \left(2+2d\right)T\left(r,g\right)+\left(1+d\right)T\left(r,f\right)+S\left(r,g\right)\\ \le \left(3+3d\right)T\left(r,g\right)+S\left(r,g\right)\end{array}$ (35)

which is impossible, since $n\ge 4min\left(2d,\sigma \right)+\sigma +6d+15$ . Therefore, $a-b-1=0$ , so

${F}^{*}=\frac{\left(b+1\right){G}^{*}}{b{G}^{*}+1}$ (36)

Then, $\stackrel{¯}{N}\left(r,\frac{1}{{F}^{*}}\right)=\stackrel{¯}{N}\left(r,\frac{1}{{G}^{*}+1/b}\right)$ . Similarly, we have

$\begin{array}{c}\left(n-\sigma \right)T\left(r,g\right)\le \stackrel{¯}{N}\left(r,{G}^{*}\left(z\right)\right)+\stackrel{¯}{N}\left(r,\frac{1}{{G}^{*}\left(z\right)}\right)+\stackrel{¯}{N}\left(r,\frac{1}{{G}^{*}+1/b}\right)+S\left(r,g\right)\\ \le \stackrel{¯}{N}\left(r,{G}^{*}\left(z\right)\right)+\stackrel{¯}{N}\left(r,\frac{1}{{G}^{*}\left(z\right)}\right)+\stackrel{¯}{N}\left(r,\frac{1}{{F}^{*}}\right)+S\left(r,g\right)\\ \le \left(2+2d\right)T\left(r,g\right)+\left(1+d\right)T\left(r,f\right)+S\left(r,g\right)\\ \le \left(3+3d\right)T\left(r,g\right)+S\left(r,g\right)\end{array}$ (37)

Which is impossible, since $n\ge 4min\left(2d,\sigma \right)+\sigma +6d+15$ .

Case 2. If $b=0$ and $a=1$ , then ${F}^{*}\equiv {G}^{*}$ obviously. From the proof of case 2 in theorem 1.1, we get $f\left(z\right)=tg\left(z\right)$ , where ${t}^{n+\sigma }=1$ . Therefore, we consider $b=0$ and $a\ne 1$ . Then from (33), we obtain

${F}^{*}=\frac{{G}^{*}+a-1}{a}.$ (38)

Using the same discuss as Case 1, we get contradiction.

Case 3. If $b=-1$ and $a=-1$ , then ${F}^{*}{G}^{*}\equiv 1$ obviously. Thus from the proof of case 3 in theorem 1.1, we get a contradiction. Therefore, we consider $b=-1$ and $a\ne -1$ . From (33), we get

${F}^{*}=\frac{a}{a+1-{G}^{*}}.$ (39)

Which is impossible, using the similar method as Case 1.

3.3. Proof of Theorem 1.3

We use the similar method as  . By the theorem condition that ${F}_{1}\left(z\right)-\alpha \left(z\right)$ and ${G}_{1}\left(z\right)-\alpha \left(z\right)$ share 0 CM, hence there exist an entire function $u\left(z\right)$ , than

$\frac{{F}_{1}\left(z\right)-\alpha \left(z\right)}{{G}_{1}\left(z\right)-\alpha \left(z\right)}={\text{e}}^{u\left(z\right)}.$ (40)

Since $\rho \left(f\right)=\rho \left(g\right)=0$ , than ${\text{e}}^{u\left(z\right)}\equiv \eta$ is a constant.

Rewriting (40)

${G}_{1}\left(z\right)={F}_{1}\left(z\right)+\left(\eta -1\right)\alpha \left(z\right)$ (41)

If $\eta \ne 1$ , we can use Nevanlinnas two fundamental theorems, Lemma 2.5 and Lemma 2.8 to get a contradiction, since $n>\sigma +2k+2d$ .

So we get $\eta =1$ . Rewriting (40)

$\begin{array}{l}\begin{array}{l}{P}_{n}\left(f\left(z\right)\right)\underset{j=1}{\overset{d}{\prod }}f{\left({q}_{j}z+{c}_{j}\right)}^{{v}_{j}}\\ ={P}_{n}\left(g\left(z\right)\right)\underset{j=1}{\overset{d}{\prod }}g{\left({q}_{j}z+{c}_{j}\right)}^{{v}_{j}}.\end{array}\hfill \end{array}$ (42)

Set $h\left(z\right):=\frac{f\left(z\right)}{g\left(z\right)}$ , suppose that $h\left(z\right)\equiv C$ (constant), then $f=tg$ . Then we take $f=tg$ into (42) and get

$\begin{array}{l}\underset{j=1}{\overset{d}{\prod }}g{\left({q}_{j}z+{c}_{j}\right)}^{{v}_{j}}\left[{\alpha }_{n}{g}^{n}\left({t}^{n+\sigma }-1\right)+{\alpha }_{n-1}{g}^{n-1}\left({t}^{n+\sigma -1}-1\right)+\cdots +{\alpha }_{1}g\left({t}^{\sigma +1}-1\right)\right]\equiv 0.\hfill \end{array}$ (43)

where ${\alpha }_{n}$ is a non-zero complex constant. And $\underset{j=1}{\overset{d}{\prod }}g{\left({q}_{j}z+{c}_{j}\right)}^{{v}_{j}}\overline{)\equiv }0$ , since $g$

is transcendental meromorphic function. So ${h}^{m}=1$ , where $m$ is greatest common divisor of $\left(n+\sigma ,n+\sigma -1,\cdots ,n+\sigma -i,\cdots ,\sigma +1\right)$ , ${\alpha }_{n-i}\ne 0$ ( $i=0,1,\cdots ,n-1$ ).

Suppose that $h\left(z\right)\overline{)\equiv }C$ (constant), Equation (43) imply that $f\left(z\right)$ and $g\left(z\right)$ satisfy a algebraic equation $Q\left(f,g\right)\equiv 0$ , where

$Q\left({w}_{1},{w}_{2}\right)={P}_{n}\left({w}_{1}\right)\underset{j=1}{\overset{d}{\prod }}{w}_{1}{\left({q}_{j}z+{c}_{j}\right)}^{{v}_{j}}-{P}_{n}\left({w}_{2}\right)\underset{j=1}{\overset{d}{\prod }}{w}_{2}{\left({q}_{j}z+{c}_{j}\right)}^{{v}_{j}}.$ (44)

4. Conclusion

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.

Acknowledgements

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.

Cite this paper

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

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