ABSTRACT

In this paper, we mainly study the uniqueness of specific q-shift difference polynomials $f^n\left(z\right){\displaystyle \underset{j=1}{\overset{d}{\prod }}}f{\left(q_{j}z+c_j\right)}^{v_j}$ and $g^n\left(z\right){\displaystyle \underset{j=1}{\overset{d}{\prod }}}g{\left(q_{j}z+c_j\right)}^{v_j}$ 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 [1] - [6] ). 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 [7] - [12] ). 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 [13] ), such as $m\left(r\mathrm{,}f\right)$ , $N\left(r\mathrm{,}f\right)$ , $\bar{N}\left(r\mathrm{,}f\right)$ , $T\left(r\mathrm{,}f\right)$ ,

$S\left(r\mathrm{,}f\right)$ , and define $N_2\left(r\mathrm{,}\frac{1}{f}\right)$ 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 $S\left(r\mathrm{,}f\right)$ , if it satisfies $S\left(r\mathrm{,}f\right)=o\left(T\left(r\mathrm{,}f\right)\right)$ , 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.

$F\left(z\right)=f^n\left(z\right){\displaystyle \underset{j=1}{\overset{d}{\prod }}}f{\left(q_{j}z+c_j\right)}^{v_j}$ (1)

$F_1\left(z\right)=P_n\left(f\left(z\right)\right){\displaystyle \underset{j=1}{\overset{d}{\prod }}}f{\left(q_{j}z+c_j\right)}^{v_j}$ (2)

where $c_j\in ℂ$ $\left(c_j\ne 0,j=1,2,3,\cdots ,d\right)$ 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 ${\alpha }_n\left(\ne 0\right)\mathrm{,}{\alpha }_{n-1}\mathrm{,}\cdots ,{\alpha }_0$ 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. [14] 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 $n\left(\in N^{\mathrm{<mo>\; *\; </mo>}}\right)\ge 14$ 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 $n\left(\in N^{\mathrm{<mo>\; *\; </mo>}}\right)\ge 26$ 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 $\alpha \left(z\right)\left(\cancel{\equiv }0\right)$ 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 $n\ge 4min\left(2d\mathrm{,}\sigma \right)+\sigma +9$ 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 $\alpha \left(z\right)\left(\cancel{\equiv }0\right)$ 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 $n\ge 4min\left(2d\mathrm{,}\sigma \right)+\sigma +6d+15$ satisfying $t^{n+\sigma }=1$ .

Liu et al. [14] 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 ${\alpha }_n\left(\ne 0\right)\mathrm{,}{\alpha }_{n-1}\mathrm{,}\cdots ,{\alpha }_0$ , 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 $\left({\lambda }_0\mathrm{,}{\lambda }_1\mathrm{,}\cdots \mathrm{,}{\lambda }_n\right)$ , 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 $Q\left(f\left(z\right)\mathrm{,}g\left(z\right)\right)=0$ , where

$Q\left(w_1\mathrm{,}{w}_2\right)=P_n\left(w_1\right)w_1\left(qz+c\right)-P_n\left(w_2\right)w_2\left(qz+c\right)$ (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 $P_n\left(f\left(z\right)\right){\displaystyle \underset{j=1}{\overset{d}{\prod }}}f{\left(q_{j}z+c_j\right)}^{v_j}$ , 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 $\left(n+\sigma \mathrm{,}n+\sigma -\mathrm{1,}\cdots \mathrm{,}n+\sigma -i\mathrm{,}\cdots \mathrm{,}\sigma +1\right)$ , ${\alpha }_{n-i}\ne 0$ , $i=0,1,\cdots ,n-1$ .

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

$Q\left(w_1,w_2\right)=P_n\left(w_1\right){\displaystyle \underset{j=1}{\overset{d}{\prod }}}{w}_1{\left(q_{j}z+c_j\right)}^{v_j}-P_n\left(w_2\right){\displaystyle \underset{j=1}{\overset{d}{\prod }}}{w}_2{\left(q_{j}z+c_j\right)}^{v_j}.$ (4)

2. Some Lemmas

Lemma 2.1. (see [15] ) 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

$P_n\left(f\left(z\right)\right)={\alpha }_n{f}^n\left(z\right)+{\alpha }_{n-1}{f}^{n-1}\left(z\right)+\cdots +{\alpha }_{1}f\left(z\right)+{\alpha }_0\mathrm{,}$ (5)

then

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

Lemma 2.2. (see [9] ) 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

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

on a set of logarithmic density 1.

Lemma 2.3. (see [12] ) 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 $L\cancel{\equiv }0$ , then

$\begin{array}{c}T\left(r\mathrm{,}f\right)+T\left(r\mathrm{,}g\right)\\ \le 2\left(N_2\left(r\mathrm{,}f\right)+N_2\left(r\mathrm{,}g\right)+N_2\left(r\mathrm{,}\frac{1}{f}\right)+N_2\left(r\mathrm{,}\frac{1}{g}\right)\right)\\ +3\left(\bar{N}\left(r\mathrm{,}f\right)+\bar{N}\left(r\mathrm{,}g\right)+\bar{N}\left(r\mathrm{,}\frac{1}{f}\right)+\bar{N}\left(r\mathrm{,}\frac{1}{g}\right)\right)+S\left(r\mathrm{,}f\right)+S\left(r\mathrm{,}g\right)\end{array}$ (9)

Lemma 2.4. (see [16] ) 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:

$\begin{array}{l}\text{(a)}\max \left\{T\left(r,f\right),T\left(r,g\right)\right\}\\ \le N_2\left(r,f\right)+N_2\left(r,g\right)+N_2\left(r,\frac{1}{f}\right)+N_2\left(r,\frac{1}{g}\right)+S\left(r,f\right)+S\left(r,g\right)\\ \text{(b)}f\equiv g;\\ \text{(c)}fg\equiv 1.\end{array}$ (10)

Lemma 2.5. (see [14] ) 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

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

on a set of logarithmic density 1.

Lemma 2.6. (see [14] ) Let $q$ and $\eta$ be two non-zero finite complex constants, and let $f$ be a nonconstant meromorphic function of zero order, then

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

$\left(n-\sigma \right)T\left(r\mathrm{,}f\right)+S\left(r\mathrm{,}f\right)\le T\left(r\mathrm{,}{F}_1\right)\le \left(n+\sigma \right)T\left(r\mathrm{,}f\right)+S\left(r\mathrm{,}f\right)$ (13)

Proof. Combining Lemma 2.1 with Lemma 2.5, we obtain

$\begin{array}{c}T\left(r,F_1\right)\le T\left(r,P_n\left(f\left(z\right)\right)\right)+T\left(r,{\displaystyle \underset{j=1}{\overset{d}{\prod }}}f{\left(q_{j}z+c_j\right)}^{v_j}\right)+S\left(r,f\right)\\ \le nT\left(r,f\left(z\right)\right)+{\displaystyle \underset{j=1}{\overset{d}{\sum }}}T\left(r,f{\left(q_{j}z+c_j\right)}^{v_j}\right)+S\left(r,f\right)\\ \le \left(n+\sigma \right)T\left(r,f\left(z\right)\right)+S\left(r,f\right)\end{array}$ (14)

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

$\begin{array}{l}\left(n+\sigma \right)T\left(r,f\left(z\right)\right)\le T\left(r,P_n\left(f\left(z\right)\right)f^{\sigma }\right)+S\left(r,f\right)\\ =m\left(r,P_n\left(f\left(z\right)\right)f^{\sigma }\right)+N\left(r,P_n\left(f\left(z\right)\right)f^{\sigma }\right)+S\left(r,f\right)\\ \le m\left(r,\frac{F_1\left(z\right)f^{\sigma }}{{\displaystyle \underset{j=1}{\overset{d}{\prod }}}f{\left(q_{j}z+c_j\right)}^{v_j}}\right)+N\left(r,\frac{F_1\left(z\right)f^{\sigma }}{{\displaystyle \underset{j=1}{\overset{d}{\prod }}}f{\left(q_{j}z+c_j\right)}^{v_j}}\right)+S\left(r,f\right)\\ \le m\left(r,F_1\right)+N\left(r,F_1\right)+T\left(r,\frac{f^{\sigma }}{{\displaystyle \underset{j=1}{\overset{d}{\prod }}}f{\left(q_{j}z+c_j\right)}^{v_j}}\right)+S\left(r,f\right)\\ \le T\left(r,F_1\right)+2\sigma T\left(r,f\right)+S\left(r,f\right)\end{array}$ (15)

which is equivalent to

$\left(n-\sigma \right)T\left(r\mathrm{,}f\right)+S\left(r\mathrm{,}f\right)\le T\left(r\mathrm{,}{F}_1\right)$ (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

$T\left(r\mathrm{,}{F}_1\right)=\left(n+\sigma \right)T\left(r\mathrm{,}f\right)+S\left(r\mathrm{,}f\right)$ (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 $F^{*}\left(z\right)=\frac{F\left(z\right)}{\alpha \left(z\right)}$ , $G^{\mathrm{<mo>\; *\; </mo>}}\left(z\right)=\frac{G\left(z\right)}{\alpha \left(z\right)}$ , than $F^{\mathrm{<mo>\; *\; </mo>}}\left(z\right)$ and $G^{\mathrm{<mo>\; *\; </mo>}}\left(z\right)$ share 1 CM.

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

$\begin{array}{l}\left(n-\sigma \right)T\left(r,f\right)+S\left(r,f\right)\le T\left(r,F^{*}\left(z\right)\right)\\ \le \bar{N}\left(r,F^{*}\left(z\right)\right)+\bar{N}\left(r,\frac{1}{F^{*}\left(z\right)}\right)+\bar{N}\left(r,\frac{1}{F^{*}\left(z\right)-1}\right)+S\left(r,F^{*}\left(z\right)\right)\\ \le \bar{N}\left(r,f^n\right)+\bar{N}\left(r,{\displaystyle \underset{j=1}{\overset{d}{\prod }}}f{\left(q_{j}z+c_j\right)}^{v_j}\right)+\bar{N}\left(r,\frac{1}{f^n}\right)\\ +\bar{N}\left(r,\frac{1}{{\displaystyle \underset{j=1}{\overset{d}{\prod }}}f{\left(q_{j}z+c_j\right)}^{v_j}}\right)+\bar{N}\left(r,\frac{1}{G^{*}\left(z\right)-1}\right)+S\left(r,f\right)\\ \le \left(2d+2\right)T\left(r,f\right)+\left(n+\sigma \right)T\left(r,g\right)+S\left(r,g\right)+S\left(r,f\right)\end{array}$ (18)

Then

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

Similarly,

$\left(n-2d-\sigma -2\right)T\left(r\mathrm{,}g\right)\le \left(n+\sigma \right)T\left(r\mathrm{,}f\right)+S\left(r\mathrm{,}f\right)+S\left(r\mathrm{,}g\right)$ (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 $\begin{array}{l}\max \left\{T\left(r,F^{*}\left(z\right)\right),T\left(r,G^{*}\left(z\right)\right)\right\}\le N_2\left(r,F^{*}\left(z\right)\right)+N_2\left(r,\frac{1}{F^{*}\left(z\right)}\right)+N_2\left(r,G^{*}\left(z\right)\right)\\ +N_2\left(r,\frac{1}{G^{*}\left(z\right)}\right)+S\left(r,F^{*}\left(z\right)\right)+S\left(r,G^{*}\left(z\right)\right)\end{array}$ holds.

Through simple calculation, we have

$\begin{array}{l}{N}_2\left(r,F^{*}\left(z\right)\right)\le N_2\left(r,f^n\right)+N_2\left(r,{\displaystyle \underset{j=1}{\overset{d}{\prod }}}f{\left(q_{j}z+c_j\right)}^{v_j}\right)\\ \le \left\{2+\min \left(2d,\sigma \right)\right\}T\left(r,f\right)+S\left(r,f\right)\end{array}$ (21)

In the same way,

$\begin{array}{l}{N}_2\left(r,\frac{1}{F^{*}\left(z\right)}\right)\le \left\{2+\min \left(2d,\sigma \right)\right\}T\left(r,f\right)+S\left(r,f\right)\\ N_2\left(r,G^{*}\left(z\right)\right)\le \left\{2+\min \left(2d,\sigma \right)\right\}T\left(r,g\right)+S\left(r,g\right)\\ N_2\left(r,\frac{1}{G^{*}\left(z\right)}\right)\le \left\{2+\min \left(2d,\sigma \right)\right\}T\left(r,g\right)+S\left(r,g\right)\end{array}$ (22)

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

$\begin{array}{l}\left(n-\sigma \right)\left(T\left(r,f\right)+T\left(r,g\right)\right)\le T\left(r,F^{*}\left(z\right)\right)+T\left(r,G^{*}\left(z\right)\right)\\ \le 2N_2\left(r,F^{*}\left(z\right)\right)+2N_2\left(r,\frac{1}{F^{*}\left(z\right)}\right)+2N_2\left(r,G^{*}\left(z\right)\right)\\ +2N_2\left(r,\frac{1}{G^{*}\left(z\right)}\right)+S\left(r,F^{*}\left(z\right)\right)+S\left(r,G^{*}\left(z\right)\right)\\ \le 4\left[2+\min \left(2d,\sigma \right)\right]\left(T\left(r,f\right)+T\left(r,g\right)\right)+S\left(r,f\right)+S\left(r,g\right)\end{array}$ (23)

Then

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

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

Case 2. Suppose that $F^{\mathrm{<mo>\; *\; </mo>}}\left(z\right)\equiv G^{\mathrm{<mo>\; *\; </mo>}}\left(z\right)$ holds, we obtain

$\begin{array}{l}{f}^n\left(z\right){\displaystyle \underset{j=1}{\overset{d}{\prod }}}f{\left(q_{j}z+c_j\right)}^{v_j}=g^n\left(z\right){\displaystyle \underset{j=1}{\overset{d}{\prod }}}g{\left(q_{j}z+c_j\right)}^{v_j}.\hfill \end{array}$ (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

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

Since $g$ is a transcendental meromorphic function, than $g^n{\displaystyle \underset{j=1}{\overset{d}{\prod }}}g{\left(q_{j}z+c_j\right)}^{v_j}\cancel{\equiv }0$ . It follows that $t^{n+\sigma }=1$ .

Suppose that $h\left(z\right)\cancel{\equiv }C$ (constant), then using (25), we deduce that $h^n\left(z\right)={\displaystyle \underset{j=1}{\overset{d}{\prod }}}\frac{1}{h{\left(q_{j}z+c_j\right)}^{v_j}}$ ,

So

$nT\left(r,h\left(z\right)\right)=T\left(r,{\displaystyle \underset{j=1}{\overset{d}{\prod }}}\frac{1}{h{\left(q_{j}z+c_j\right)}^{v_j}}\right)\le \sigma T\left(r,h\left(z\right)\right)+S\left(r,h\left(z\right)\right)$ (27)

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

Case 3. Suppose that $F^{\mathrm{<mo>\; *\; </mo>}}\left(z\right)G^{\mathrm{<mo>\; *\; </mo>}}\left(z\right)\equiv 1$ holds, then $f^n\left(z\right){\displaystyle \underset{j=1}{\overset{d}{\prod }}}f{\left(q_{j}z+c_j\right)}^{v_j}\cdot g^n\left(z\right){\displaystyle \underset{j=1}{\overset{d}{\prod }}}g{\left(q_{j}z+c_j\right)}^{v_j}={\alpha }^2\left(z\right).$

We define $h_1\left(z\right)=f\left(z\right)\cdot g\left(z\right)$ , we easily get $h_1^n\left(z\right)={\displaystyle \underset{j=1}{\overset{d}{\prod }}}\frac{{\alpha }^2\left(z\right)}{h_1{\left(q_{j}z+c_j\right)}^{v_j}}$ is non-constant, hence

$nT\left(r,h_1\left(z\right)\right)=T\left(r,{\displaystyle \underset{j=1}{\overset{d}{\prod }}}\frac{{\alpha }^2\left(z\right)}{h_1{\left(q_{j}z+c_j\right)}^{v_j}}\right)\le \sigma T\left(r,h_1\left(z\right)\right)+S\left(r,h_1\left(z\right)\right)$ (28)

We get a contradiction, since $n\ge 4\min \left(2d,\sigma \right)+\sigma +9$ . This implies that $h_1\left(z\right)$ is a constant, which is impossible.

3.2. Proof of Theorem 1.2

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

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

We can easily get

$\begin{array}{l}\bar{N}\left(r,F^{*}\left(z\right)\right)\le \left(1+d\right)T\left(r,f\right)+S\left(r,f\right)\\ \bar{N}\left(r,\frac{1}{F^{*}\left(z\right)}\right)\le \left(1+d\right)T\left(r,f\right)+S\left(r,f\right)\\ \bar{N}\left(r,G^{*}\left(z\right)\right)\le \left(1+d\right)T\left(r,g\right)+S\left(r,g\right)\\ \bar{N}\left(r,\frac{1}{G^{*}\left(z\right)}\right)\le \left(1+d\right)T\left(r,g\right)+S\left(r,g\right)\end{array}$ (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 $L\cancel{\equiv }0$ , combining Lemma 2.3, (21), (22) with (29), we obtain

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

Then,

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

that is impossible, since $n\ge 4min\left(2d\mathrm{,}\sigma \right)+\sigma +6d+15$ . 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 $T\left(r\mathrm{,}{F}^{\mathrm{<mo>\; *\; </mo>}}\right)=T\left(r\mathrm{,}{G}^{\mathrm{<mo>\; *\; </mo>}}\right)+O\left(1\right)$ . From Lemma 2.8, we deduced that $T\left(r\mathrm{,}f\right)=T\left(r\mathrm{,}g\right)+S\left(r\mathrm{,}f\right)$ . 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

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

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

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

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

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

Then, $\bar{N}\left(r\mathrm{,}\frac{1}{F^{\mathrm{<mo>\; *\; </mo>}}}\right)=\bar{N}\left(r\mathrm{,}\frac{1}{G^{\mathrm{<mo>\; *\; </mo>}}+1/b}\right)$ . Similarly, we have

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

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

Case 2. If $b=0$ and $a=1$ , then $F^{\mathrm{<mo>\; *\; </mo>}}\equiv G^{\mathrm{<mo>\; *\; </mo>}}$ 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 $F^{\mathrm{<mo>\; *\; </mo>}}{G}^{\mathrm{<mo>\; *\; </mo>}}\equiv 1$ 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 [14] . 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

$\frac{F_1\left(z\right)-\alpha \left(z\right)}{G_1\left(z\right)-\alpha \left(z\right)}={\text{e}}^{u\left(z\right)}\mathrm{.}$ (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)

$\begin{array}{l}\begin{array}{l}{P}_n\left(f\left(z\right)\right){\displaystyle \underset{j=1}{\overset{d}{\prod }}}f{\left(q_{j}z+c_j\right)}^{v_j}\\ =P_n\left(g\left(z\right)\right){\displaystyle \underset{j=1}{\overset{d}{\prod }}}g{\left(q_{j}z+c_j\right)}^{v_j}.\end{array}\hfill \end{array}$ (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

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

where ${\alpha }_n$ is a non-zero complex constant. And ${\displaystyle \underset{j=1}{\overset{d}{\prod }}}g{\left(q_{j}z+c_j\right)}^{v_j}\cancel{\equiv }0$ , since $g$

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

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

$Q\left(w_1,w_2\right)=P_n\left(w_1\right){\displaystyle \underset{j=1}{\overset{d}{\prod }}}{w}_1{\left(q_{j}z+c_j\right)}^{v_j}-P_n\left(w_2\right){\displaystyle \underset{j=1}{\overset{d}{\prod }}}{w}_2{\left(q_{j}z+c_j\right)}^{v_j}.$ (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

References