# Uniqueness of entire functions concerning differential-difference
polynomials sharing small functions

Goutam Haldar  Goutam Haldar
Department of Mathematics, Malda College, Rabindra Avenue, Malda, West Bengal
732101, India<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

In this paper, we investigate the uniqueness problem of difference polynomials
$f^{n}(z)P(f(z))L_{c}(f)$ and $g^{n}(z)P(g(z))L_{c}(g)$, where
$L_{c}(f)=f(z+c)+c_{0}f(z)$, $P(z)$ is a polynomial with constant coefficients
of degree $m$ sharing a small function with the notions of weakly weighted
sharing and relaxed weighted sharing and obtained the corresponding results,
which improve and extend some recent results due to Sahoo and Biswas (Tamkang
J. Math., 49(2), 85–97 (2018)).

###### Key words and phrases:

Enitire function, difference operator, small function, weakly weighted
sharing, relaxed weighted sharing.

###### 2010 Mathematics Subject Classification:

30D35.

##  1\. Introduction

Let $f$ and $g$ be two non-constant meromorphic functions defined in the open
complex plane $\mathbb{C}$. If for some $a\in\mathbb{C}\cup\\{\infty\\}$, the
zero of $f-a$ and $g-a$ have the same locations as well as same
multiplicities, we say that $f$ and $g$ share the value $a$ CM (counting
multiplicities). If we do not consider the multiplicities, then $f$ and $g$
are said to share the value $a$ IM (ignoring multiplicities). We adopt the
standard notations of the Nevanlinna theory of meromorphic functions (see [5,
10, 20]). For a non-constant meromorphic function $f$, we denote by $T(r,f)$
the Nevanlinna characteristic function of $f$ and by $S(r,f)$ any quantity
satisfying $S(r,f)=o\\{T(r,f)\\}$ as $r\rightarrow\infty$ outside of an
exceptional set of finite linear measure. We say that $\alpha(z)$ is a small
function of $f$, if $\alpha(z)$ is a meromorphic function satisfying
$T(r,\alpha(z))=S(r,f)$. We denote by $E_{k)}(a,f)$ the set of all a-points of
$f$ with multiplicities not exceeding $k$, where an a-point is counted
according to its multiplicity. Also we denote by $\overline{E}_{k)}(a,f)$ the
set of distinct a-points of $f$ with multiplicities not exceeding $k$. In
addition, we need the following definitions.

######  Definition 1.1.

[6]Let $p$ be a positive integer and $a\in\mathbb{C}\cup\\{\infty\\}$.

  1. (i)

$N(r,a;f\mid\geq p)$ ($\overline{N}(r,a;f\mid\geq p)$)denotes the counting
function (reduced counting function) of those $a$-points of $f$ whose
multiplicities are not less than $p$.

  2. (ii)

$N(r,a;f\mid\leq p)$ ($\overline{N}(r,a;f\mid\leq p)$)denotes the counting
function (reduced counting function) of those $a$-points of $f$ whose
multiplicities are not greater than $p$.

######  Definition 1.2.

[7] Let $k$ be a positive integer or infinity. We denote by $N_{k}(r,a;f)$ the
counting function of a-points of $f$, where an a-point of multiplicity $m$ is
counted $m$ times if $m\leq k$ and $k$ times if $m>k$. Then

$\displaystyle N_{k}(r,a;f)=\overline{N}(r,a;f)+\overline{N}(r,a;f\mid\geq
2)+\ldots+\overline{N}(r,a;f\mid\geq k).$

Clearly, $N_{1}(r,a;f)=\overline{N}(r,a;f)$.

######  Definition 1.3.

Let $a\in\mathbb{C}\cup\\{\infty\\}$. We denote $N_{E}(r,a;f,g)$
$(N_{E}(r,a;f,g))$ by the counting function (reduced counting function) of all
common zeros of $f-a$ and $g-a$ with the same multiplicities and by
$N_{0}(r,a;f,g)$ $(N_{0}(r,a;f,g))$ the counting function (reduced counting
function) of all common zeros of $f-a$ and $g-a$ ignoring multiplicities. If

$\displaystyle\overline{N}(r,a;f)+\overline{N}(r,a;g)-2\overline{N}_{E}(r,a;f,g)=S(r,f)+S(r,g),$

then we say that $f$ and $g$ share the value a “CM”. If

$\displaystyle\overline{N}(r,a;f)+\overline{N}(r,a;g)-2\overline{N}_{0}(r,a;f,g)=S(r,f)+S(r,g),$

then we say that $f$ and $g$ share the value a “IM”. Definition 4

######  Definition 1.4.

[11] Let $f$ and $g$ share the value a “IM” and $k$ be a positive integer or
infinity. Then $\overline{N}^{E}_{k)}(r,a;f,g)$ denotes the reduced counting
function of those a-points of $f$ whose multiplicities are equal to the
corresponding a-points of $g$, and both of their multiplicities are not
greater than $k$. $\overline{N}^{0}_{(k}(r,a;f,g)$ denotes the reduced
counting function of those a-points of $f$ which are a-points of $g$, and both
of their multiplicities are not less than $k$.

######  Definition 1.5.

[11] Let $a\in\mathbb{C}\cup\\{\infty\\}$ and $k$ be a positive integer or
infinity. If

$\displaystyle\overline{N}(r,a;f\mid\leq
k)-\overline{N}^{E}_{k)}(r,a;f,g)=S(r,f),$
$\displaystyle\overline{N}(r,a;g\mid\leq
k)-\overline{N}^{E}_{k)}(r,a;f,g)=S(r,g),$
$\displaystyle\overline{N}(r,a;f\mid\geq
k+1)-\overline{N}_{0}^{(k+1}(r,a;f,g)=S(r,f),$
$\displaystyle\overline{N}(r,a;g\mid\geq
k+1)-\overline{N}_{0}^{(k+1}(r,a;f,g)=S(r,g),$

or if $k=0$ and

$\displaystyle\overline{N}(r,a;f)-\overline{N}_{0}(r,a;f,g)=S(r,f),$
$\displaystyle\overline{N}(r,a;g)-\overline{N}_{0}(r,a;f,g)=S(r,g),$

then we say that $f$ and $g$ share the value a weakly with weight $k$ and we
write $f$ and $g$ share $``(a,k)"$.

######  Definition 1.6.

[1] Let $k$ be a positive integer and for $a\in\mathbb{C}-\\{0\\}$,
$E_{k)}(a;f)=E_{k)}(a;g)$. Let $z_{0}$ be a zero of $f(z)-a$ of multiplicity
$p$ and a zero of $g(z)-a$ of multiplicity $q$. We denote by
$\overline{N}_{L}(r,a;f)(\overline{N}_{L}(r,a;g))$ the reduced counting
function of those $a$-points of $f$ and $g$ for which $p>q\geq k+1(q>p\geq
k+1)$, by $\overline{N}_{E}^{(k+1}(r,a;f)$ the reduced counting function of
those a-points of $f$ and $g$ for which $p=q\geq k+1$, by $\overline{N}_{f\geq
k+1}(r,a;f\mid g\neq a)$ the reduced counting functions of those a-points of
$f$ and $g$ for which $p\geq k+1$ and $q=0$.

######  Definition 1.7.

[1] Let $k$ be a positive integer and for$a\in\mathbb{C}-\\{0\\}$, let $f$,
$g$ share a “IM”. Let $z_{0}$ be a zero of $f(z)-a$ of multiplicity $p$ and a
zero of $g(z)-a$ of multiplicity $q$. We denote by $\overline{N}_{f\geq
k+1}(r,a;f\mid g=m)$ the reduced counting functions of those $a$-points of $f$
and $g$ for which $p\geq k+1$ and $q=m$. We can define
$\overline{N}_{L}(r,a;f)(\overline{N}_{L}(r,a;g))$ and
$\overline{N}_{E}^{(k+1}(r,a;f)$ in a similar manner as defined in the
previous definition.

######  Definition 1.8.

[8] Let $a,b\in\mathbb{C}\cup\\{\infty\\}$. We denote by $N(r,a;f\mid g=b)$
the counting function of those $a$-points of $f$, counted according to
multiplicity, which are $b$-points of $g$.

######  Definition 1.9.

[8] Let $a,b\in\mathbb{C}\cup\\{\infty\\}$. We denote by $N(r,a;f\mid g\neq
b)$ the counting function of those a-points of $f$, counted according to
multiplicity, which are not the b-points of $g$.

We define shift and difference operators of $f(z)$ by $f(z+c)$ and
$\Delta_{c}f(z)=f(z+c)-f(z)$, respectively. Note that
$\Delta_{c}^{n}f(z)=\Delta_{c}^{n-1}(\Delta_{c}f(z))$, where $c$ is a nonzero
complex number and $n\geq 2$ is a positive integer. For further generalization
of $\Delta_{c}f$, we now define the difference operator of an entire
(meromorphic) function $f$ as $L_{c}(f)=f(z+c)+c_{0}f(z)$, where $c_{0}$ is a
non-zero complex constant. Clearly, for the particular choice of the constant
$c_{0}=-1$, we get $L_{c}(f)=\Delta_{c}f.$

In 2007, Banerjee and Mukherjee [1] introduced a new type of sharing known as
relaxed weighted sharing which is weaker than weakly weighted sharing as
follows.

######  Definition 1.10.

[1] We denote by $N(r,a;f\mid=p;g\mid=q)$ the reduced counting function of
common a-points of $f$ and $g$ with multiplicities $p$ and $q$, respectively.

######  Definition 1.11.

[1] Let $a\in\mathbb{C}\cup\\{\infty\\}$ and $k$ be a positive integer or
infinity. Suppose that $f$ and $g$ share the value a “IM”. If for $p\neq q$,

$\displaystyle\sum_{p,q\leq k}N(r,a;f\mid=p;g\mid=q)=S(r),$

then we say that $f$ and $g$ share the value a with weight $k$ in a relaxed
manner and we write $f$ and $g$ share $(a,k)^{*}$

Let $P(z)=a_{m}z^{m}+a_{m-1}z^{m-1}+\ldots+a_{0}$ be a nonzero polynomial of
degree $m$, where $a_{m}(\neq 0),a_{m-1},\ldots,a_{0}(\neq 0)$ are complex
constants and $m$ is a positive integer.

In 1959, Hayman [4] proved the following result.

######  Theorem A.

[4] Let $f$ be a transcendental entire function and let $n(\geq 1)$ be an
integer. Then $f^{n}f^{\prime}=1$ has infinitely many solutions.

Regarding uniqueness of the above theorem, Yang and Hua [19], in 1997 obtained
the following result.

######  Theorem B.

[19] Let $f$ and $g$ be two non-constant entire functions, $n\geq 6$ a
positive integer. If $f^{n}f^{\prime}$ and $g^{n}g^{\prime}$ share $1$ CM,
then either $f(z)=c_{1}e^{cz}$, $g(z)=c_{2}e^{-cz}$, where $c_{1}$, $c_{2}$
and $c$ are three constants satisfying $(c_{1}c_{2})^{n+1}c^{2}=-1$ or
$f\equiv tg$ for a constant $t$ satisfying $t^{n+1}=1$.

In 2002, Fang and Fang [3] extends Theorem B in the following manner

######  Theorem C.

[3] Let $f$ and $g$ be two non-constant entire functions, and let $n(\geq 8)$
be an integer. If $f^{n}(f-1)f^{\prime}$ and $g^{n}(g-1)g^{\prime}$ share 1
CM, then $f\equiv g$.

In 2004, Lin and Yi [11] extended Theorem C by replacing value sharing with
fixed point sharing and obtained the following result.

######  Theorem D.

[11] Let $f$ and $g$ be two transcendental entire functions, and let $n(\geq
7)$ be an integer. If $f^{n}(f-1)f^{\prime}$ and $g^{n}(g-1)g^{\prime}$ share
$z$ CM, then $f\equiv g$.

In 2010, Zhang [21] obtained the following result by replacing $f^{\prime}(z)$
with $f(z+c)$, where $c$ is a non-zero constant.

######  Theorem E.

[21] Let $f$ and $g$ be two transcendental entire functions of finite order,
and $\alpha(z)(\not\equiv 0,\infty)$ be a small function with respect to both
$f$ and $g$. Suppose that $c$ is a nonzero complex constant and $n\geq 7$ be
an integer. If $f^{n}(z)(f(z)-1)f(z+c)$ and $g^{n}(z)(g(z)-1)g(z+c)$ share
$\alpha(z)$ CM, then $f(z)\equiv g(z)$.

Recently, Meng [13] improved Theorem E by relaxing the nature of sharing the
small function and obtained following results.

######  Theorem F.

[13] Let $f$ and $g$ be two transcendental entire functions of finite order,
and $\alpha(z)(\not\equiv 0,\infty)$ be a small function with respect to both
$f$ and $g$. Suppose that $c$ is a nonzero complex constant and $n\geq 7$ is
an integer. If $f^{n}(z)(f(z)-1)f(z+c)$ and $g^{n}(z)(g(z)-1)g(z+c)$ share
$``(\alpha(z),2)"$, then $f(z)\equiv g(z)$.

######  Theorem G.

[13] Let $f$ and $g$ be two transcendental entire functions of finite order,
and $\alpha(z)(\not\equiv 0,\infty)$ be a small function with respect to both
$f$ and $g$. Suppose that $c$ is a nonzero complex constant and $n\geq 10$ is
an integer. If $f^{n}(z)(f(z)-1)f(z+c)$ and $g^{n}(z)(g(z)-1)g(z+c)$ share
$(\alpha(z),2)^{*}$, then $f(z)\equiv g(z)$.

######  Theorem H.

[13] Let $f(z)$ and $g(z)$ be two transcendental entire functions of finite
order, and $\alpha(z)(\not\equiv 0,\infty)$ be a small function with respect
to both f $(z)$ and $g(z)$. Suppose that $c$ is a nonzero complex constant and
$n\geq 16$ is an integer. If
$\overline{E}_{2)}(\alpha(z),f^{n}(z)(f(z)-1)f(z+c))=\overline{E}_{2)}(\alpha(z),g^{n}(z)(g(z)-1)g(z+c))$,
then $f(z)\equiv g(z)$.

In 2018, Sahoo and Biswas [16] further extended Theorems E–H in the following.

######  Theorem I.

[16] Let $f(z)$ and $g(z)$ be two transcendental entire functions of finite
order, and $\alpha(z)(\not\equiv 0,\infty)$ be a small function with respect
to both $f(z)$ and $g(z)$ with finitely many zeros. Suppose that $c$ is a non-
zero complex constant, $n$, $k(\geq 0)$ and $m(\geq 1)$ are integers such that
$n\geq 2k+m+6$. If $(f^{n}(z)(f^{m}(z)-1)f(z+c))^{(k)}$ and
$(g^{n}(z)(g^{m}(z)-1)g(z+c))^{(k)}$ share $``(\alpha(z),2)"$, then
$f(z)\equiv tg(z)$, where $t^{m}=1$.

######  Theorem J.

[16] Let $f(z)$ and $g(z)$ be two transcendental entire functions of finite
order, and $\alpha(z)(\not\equiv 0,\infty)$ be a small function with respect
to both $f(z)$ and $g(z)$ with finitely many zeros. Suppose that $c$ is a non-
zero complex constant, $n$, $k(\geq 0)$ and $m(\geq 1)$ are integers such that
$n\geq 3k+2m+8$. If $(f^{n}(z)(f^{m}(z)-1)f(z+c))^{(k)}$ and
$(g^{n}(z)(g^{m}(z)-1)g(z+c))^{(k)}$ share $(\alpha(z),2)^{*}$, then
$f(z)\equiv tg(z)$, where $t^{m}=1$.

######  Theorem K.

[16] Let $f(z)$ and $g(z)$ be two transcendental entire functions of finite
order, and $\alpha(z)(\not\equiv 0,\infty)$ be a small function with respect
to both $f(z)$ and $g(z)$ with finitely many zeros. Suppose that $c$ is a non-
zero complex constant, $n$, $k(\geq 0)$ and $m(\geq 1)$ are integers such that
$n\geq 5k+4m+12$. If
$\overline{E}_{2)}(\alpha(z),(f^{n}(z)(f^{m}(z)-1)f(z+c))^{(k)})=\overline{E}_{2)}(\alpha(z),(g^{n}(z)(g^{m}(z)-1)g(z+c))^{(k)})$,
then $f(z)\equiv tg(z)$, where $t^{m}=1$.

Since no attempts, till now, have so far been made by any researchers
investigating the uniqueness problem by replacing $f(z+c)$ with $L_{c}(f)$ in
all the above-stated Theorems F–K, it is, therefore, inevitable to ask the
following question.

######  Question 1.1.

What can be said about the uniqueness of $f$ and $g$ if one replace
$f^{n}(f^{m}-1)f(z+c)$ by the difference polynomial $f(z)^{n}P(f(z))L_{c}(f)$
in Theorems F–K?

In this paper, we paid our attention to the above question and proved the
following three theorems that improve and extend Theorems I–K, respectively.
Indeed, the following theorems are the main results of the paper.

######  Theorem 1.1.

Let $f(z)$ and $g(z)$ be two transcendental entire functions of finite order,
$\alpha(z)(\not\equiv 0,\infty)$ be a small function with respect to both
$f(z)$ and $g(z)$. Suppose $c$ be a non-zero complex constant, $n$, $k(\geq
0)$, $m(\geq k+1)$ are integers such that $n\geq 2k+m+6$. If
$(f(z)^{n}P(f(z))L_{c}(f))^{(k)}$ and $(g(z)^{n}P(g(z))L_{c}(g))^{(k)}$ share
$``(\alpha(z),2)"$, then one of the following two conclusions can be realized.

  1. _(a)_

$f(z)\equiv tg(z)$, where $t$ is a constant such that $t^{d}=1$,
$d=\text{gcd}(\lambda_{0},\lambda_{1},\ldots,\lambda_{m})$, where
$\lambda_{j}$’s are defined by

$\displaystyle\lambda_{j}=\begin{cases}n+1+j,\;\;\;\text{if}\;a_{j}\neq 0\\\
n+1+m,\;\text{if}\;a_{j}=0,\end{cases}j=0,1,\ldots,m.$

  2. _(b)_

$f$ and $g$ satisfy the algebraic equation $R(w_{1},w_{2})=0$, where
$R(w_{1},w_{2})$ is given by
$R(w_{1},w_{2})=w_{1}^{n}P(w_{1})L_{c}(w_{1})-w_{2}^{n}P(w_{2})L_{c}(w_{2}).$

######  Theorem 1.2.

Let $f(z)$ and $g(z)$ be two transcendental entire functions of finite order,
$\alpha(z)(\not\equiv 0,\infty)$ be a small function with respect to both
$f(z)$ and $g(z)$. Suppose $c$ be a non-zero complex constant, $n$, $k(\geq
0)$, $m(\geq k+1)$ are integers such that $n\geq 3k+2m+8$. If
$(f(z)^{n}P(f(z))L_{c}(f))^{(k)}$ and $(g(z)^{n}P(g(z))L_{c}(g))^{(k)}$ share
$(\alpha(z),2)^{*}$, then the conclusions of Theorem 1.1 holds.

######  Theorem 1.3.

Let $f(z)$ and $g(z)$ be two transcendental entire functions of finite order,
$\alpha(z)(\not\equiv 0,\infty)$ be a small function with respect to both
$f(z)$ and $g(z)$. Suppose $c$ be a non-zero complex constant, $n$, $k(\geq
0)$, $m(\geq k+1)$ are integers such that $n\geq 9+(7k+5m)/2$. If
$E_{2)}(\alpha(z),(f^{n}P(f(z))L_{c}(f))^{(k)})=E_{2)}(\alpha(z),(g^{n}(z)P(g(z))L_{c}(g))^{(k)})$,
then the conclusions of Theorem 1.1 holds.

######  Remark 1.1.

Clearly, for the particular choice of $c_{0}=0$, $L_{c}(f)$ becomes $f(z+c)$,
and therefore Theorem 1.1 coincides with Theorem I and Theorem 1.2 with
Theorem J.

######  Remark 1.2.

In Theorem K, if one replace
$\overline{E}_{2)}(\alpha(z),f(z))=\overline{E}_{2)}(\alpha(z),g(z))$ by
$E_{2)}(\alpha(z),f(z))=E_{2)}(\alpha(z),g(z))$, for any two non-constant
meromorphic functions $f$ and $g$, then the lower bound of $n$ can
significantly reduced. Regarding this observation, we proved Theorem 1.3.

Since for a particular choice of $c_{0}=-1$, $L_{c}(f)=\Delta_{c}f$, we
observe the following corollaries.

######  Corollary 1.1.

Let $f(z)$ and $g(z)$ be two transcendental entire functions of finite order,
$\alpha(z)(\not\equiv 0,\infty)$ be a small function with respect to both
$f(z)$ and $g(z)$. Suppose $c$ be a non-zero complex constant, $n$, $k(\geq
0)$, $m(\geq k+1)$ are integers such that $n\geq 2k+m+6$. If
$(f(z)^{n}P(f(z))\Delta_{c}f)^{(k)}$ and $(g(z)^{n}P(g(z))\Delta_{c}g)^{(k)}$
share $``(\alpha(z),2)"$, then conclusion of Theorem 1.2 holds.

######  Corollary 1.2.

Let $f(z)$ and $g(z)$ be two transcendental entire functions of finite order,
$\alpha(z)(\not\equiv 0,\infty)$ be a small function with respect to both
$f(z)$ and $g(z)$. Suppose $c$ be a non-zero complex constant, $n$, $k(\geq
0)$, $m(\geq k+1)$ are integers such that $n\geq 3k+2m+8$. If
$(f(z)^{n}P(f(z))\Delta_{c}f)^{(k)}$ and $(g(z)^{n}P(g(z))\Delta_{c}g)^{(k)}$
share $(\alpha(z),2)^{*}$, then the conclusions of Theorem 1.1 holds.

######  Corollary 1.3.

Let $f(z)$ and $g(z)$ be two transcendental entire functions of finite order,
$\alpha(z)(\not\equiv 0,\infty)$ be a small function with respect to both
$f(z)$ and $g(z)$. Suppose $c$ be a non-zero complex constant, $n$, $k(\geq
0)$, $m(\geq k+1)$ are integers such that $n\geq 9+(7k+5m)/2$. If
$E_{2)}(\alpha(z),(f^{n}P(f(z))\Delta_{c}f)^{(k)})=E_{2)}(\alpha(z),(g^{n}(z)P(g(z))\Delta_{c}g)^{(k)})$,
then the conclusions of Theorem 1.1 holds.

##  2\. Some Lemmas

We now prove several lemmas which will play key roles in proving the main
results of the paper. Let $\mathcal{F}$ and $\mathcal{G}$ be two non-constant
meromorphic functions. Henceforth we shall denote by $\mathcal{H}$ the
following function

$H=\left(\frac{\;\;F^{\prime\prime}}{F^{\prime}}-\frac{2F^{\prime}}{F-1}\right)-\left(\frac{\;\;G^{\prime\prime}}{G^{\prime}}-\frac{2G^{\prime}}{G-1}\right).$
(2.1)

######  Lemma 2.1.

[2] Let $f(z)$ be a meromorphic function of finite order $\rho$, and let $c$
be a fixed non-zero complex constant. Then for each $\epsilon>0$, we have

$\displaystyle T(r,f(z+c))=T(r,f)+O(r^{\rho-1+\epsilon})+O{\log r}.$

######  Lemma 2.2.

[2] Let $f(z)$ be a meromorphic function of finite order $\rho$ and let $c$ be
a non-zero complex number. Then for each $\epsilon>0$, we have

$\displaystyle
m\left(r,\frac{f(z+c)}{f(z)}\right)+m\left(r,\frac{f(z)}{f(z+c)}\right)=O(r^{\rho-1+\epsilon}).$

######  Lemma 2.3.

[14] Let $f$ be a non-constant meromorphic function and let
$\mathcal{R}(f)=\sum\limits_{i=0}^{n}a_{i}f^{i}/\sum\limits_{j=0}^{m}b_{j}f^{j}$
be an irreducible rational function in $f$ with constant coefficients
$\\{a_{i}\\}$ and $\\{b_{j}\\}$ where $a_{n}\not=0$ and $b_{m}\not=0$. Then

$T(r,\mathcal{R}(f))=d\;T(r,f)+S(r,f),$

where $d=\max\\{n,m\\}$.

######  Lemma 2.4.

[9] If $N(r,0;f^{(k)}\mid f\not=0)$ denotes the counting function of those
zeros of $f^{(k)}$ which are not the zeros of $f$, where a zero of $f^{(k)}$
is counted according to its multiplicity then

$N\left(r,0;f^{(k)}\mid f\not=0\right)\leq
k\overline{N}(r,\infty;f)+N\left(r,0;f\mid<k\right)+k\overline{N}\left(r,0f\mid\geq
k\right)+S(r,f).$

######  Lemma 2.5.

Let $F=f(z)^{n}(z)P(f(z))L_{c}(f)$, where $f(z)$ is an entire function of
finite order, and $f(z)$, $f(z+c)$ share $0$ CM. Then

$\displaystyle T(r,F)=(n+m+1)T(r,f)+S(r,f).$

###### Proof.

Keeping in view of Lemmas 2.1 and 2.3, we have

$\displaystyle T(r,F)$ $\displaystyle=$ $\displaystyle
T(r,f(z)^{n}P(f(z))L_{c}(f))=m(r,f^{n}P(f)L_{c}(f))$ $\displaystyle\leq$
$\displaystyle m(r,f(z)^{n}P(f(z)))+m(r,L_{c}(f))+S(r,f)$ $\displaystyle\leq$
$\displaystyle
T(f(z)^{n}P(f(z)))+m\left(r,\frac{L_{c}(f)}{f(z)}\right)+m(r,f(z))+S(r,f).$

i.e.,

$\displaystyle T(r,F)\leq(n+m+1)T(r,f)+S(r,f).$

Since $f(z)$ and $f(z+c)$ share $0$ CM, we must have
$N\left(r,\infty;\frac{L_{c}(f)}{f(z)}\right)=S(r,f)$. So, keeping in view of
Lemmas 2.2 and 2.3, we have

$\displaystyle(n+m+1)T(r,f)=T(r,f(z)^{n+1}P(f(z)))=m(r,f(z)^{n+1}P(f(z)))$
$\displaystyle=$ $\displaystyle m\left(r,F\frac{f(z)}{L_{c}(f)}\right)\leq
m(r,F)+m\left(r,\frac{f(z)}{L_{c}(f)}\right)+S(r,f)$ $\displaystyle\leq$
$\displaystyle
T(r,F)+T\left(r,\frac{L_{c}(f)}{f(z)}\right)+S(r,f)=T(r,F)+N\left(r,\frac{L_{c}(f)}{f(z)}\right)$
$\displaystyle+m\left(r,\frac{L_{c}(f)}{f(z)}\right)+S(r,f)=T(r,F)+S(r,f).$

From the above two inequalities, we must have

$\displaystyle T(r,F)=(n+m+1)T(r,f)+S(r,f).$

∎

######  Lemma 2.6.

[18] Let $f(z)$ and $g(z)$ be two non-constant meromorphic functions. Then

$\displaystyle
N\left(r,\infty;\frac{f}{g}\right)-N\left(r,\infty;\frac{g}{f}\right)=N(r,\infty;f)+N(r,0;g)-N(r,\infty;g)+N(r,0;f).$

######  Lemma 2.7.

Let $f(z)$ be a transcendental entire function of finite order,
$c\in\mathbb{C}$–$\\{0\\}$ be finite complex constants and $n\in\mathbb{N}$.
Let $F(z)=f(z)^{n}P(f(z))L_{c}(f)$, where $L_{c}(f)\not\equiv 0$. Then we have

$\displaystyle(n+m)T(r,f)\leq T(r,F)-N(r,0;L_{c}(f))+S(r,f).$

###### Proof.

Using Lemmas 2.2 and 2.6, we get

$\displaystyle m(f(z)^{n+1}P(f(z)))=m\left(r,\frac{f(z)F}{L_{c}(f)}\right)\leq
m(r,F)+m\left(r,\frac{f(z)}{L_{c}(f)}\right)+S(r,f)$ $\displaystyle\leq$
$\displaystyle
m(r,F)+T\left(r,\frac{f(z)}{L_{c}(f)}\right)-N\left(r,\infty;\frac{f(z)}{L_{c}(f)}\right)+S(r,f)$
$\displaystyle\leq$ $\displaystyle
m(r,F)+T\left(r,\frac{L_{c}(f)}{f(z)}\right)-N\left(r,\infty;\frac{f(z)}{L_{c}(f)}\right)+S(r,f)$
$\displaystyle\leq$ $\displaystyle
m(r,F)+N\left(r,\infty;\frac{L_{c}(f)}{f(z)}\right)+m\left(r,\frac{L_{c}(f)}{f(z)}\right)-N\left(r,\infty;\frac{f(z)}{L_{c}(f)}\right)+S(r,f)$
$\displaystyle\leq$ $\displaystyle m(r,F)+N(r,0;f)-N(r,0;L_{c}(f))+S(r,f).$

i.e.,

$\displaystyle m(f(z)^{n+1}P(f(z)))\leq T(r,F)+T(r,f)-N(r,0;L_{c}(f))+S(r,f).$

By Lemma 2.3, we get

$\displaystyle(n+m+1)T(r,f)=m(r,f^{n+1}P(f))\leq
T(r,F)+T(r,f)-N(r,0;L_{c}(f))+S(r,f).$

i.e.,

$\displaystyle(n+m)T(r,f)\leq T(r,F)-N(r,0;L_{c}(f))+S(r,f).$

∎

######  Lemma 2.8.

[1] Let $F$ and $G$ be two non-constant meromorphic functions that share
$(1,2)^{*}$. Then

$\displaystyle\overline{N}_{L}(r,1;F)+\overline{N}_{F\geq 3}(r,1;g|=1)$
$\displaystyle\leq$
$\displaystyle\overline{N}(r,0;F)+N(r,\infty;F)-\sum_{p=3}^{\infty}\overline{N}\left(r,0;\frac{F^{\prime}}{F}\mid\geq
p\right)-\overline{N}_{0}^{2}(r,0;F^{\prime})+S(r),$

where by $N_{0}^{2}(r,0;F^{\prime})$ is the counting function of those zeros
of $F^{\prime}$ which are not the zeros of $F(F-1)$, where each simple zero is
counted once and all other zeros are counted two times.

######  Lemma 2.9.

Let $F$ and $G$ be two non-constant meromorphic functions such that
$E_{2)}(1,F)=E_{2)}(1,G)$ and $H\not\equiv 0$. Then

$\displaystyle N(r,\infty;H)$ $\displaystyle\leq$
$\displaystyle\overline{N}(r,0;F\mid\geq 2)+\overline{N}(r,0;G\mid\geq
2)+\overline{N}_{L}(r,1;F)+\overline{N}_{L}(r,1;G)+\overline{N}(r,\infty;F\mid\geq
2)$ $\displaystyle+\overline{N}(r,\infty;G\mid\geq 2)+\overline{N}_{F\geq
3}(r,1;F\mid G\neq 1)+\overline{N}_{G\geq 3}(r,1;G\mid F\neq 1)$
$\displaystyle+\overline{N}_{0}(r,0;F^{\prime})+\overline{N}_{0}(r,0;G^{\prime})+S(r,F)+S(r,G).$

###### Proof.

It can be easily verified that all possible poles of $H$ occur at (i) multiple
zeros of $F$ and $G$, (ii) multiple poles of $F$ and $G$, (iii) the common
zeros of $F-1$ and $G-1$ whose multiplicities are different, (iii) those
$1$-points of $F(G)$ which are not the $1$-points of $F(G)$, (iv) zeros of
$F^{\prime}$ which are not the zeros of $F(F-1)$, (v) zeros of $G^{\prime}$
which are not zeros of $G(G-1)$. Since all the poles of $H$ are simple the
lemma follows from above. This proves the lemma. ∎

######  Lemma 2.10.

[1] If $f$, $g$ be share $``(1,1)"$ and $H\not\equiv 0$, then

$\displaystyle N(r,1;f\mid\leq 1)\leq N(r,0;H)+S(r,f)\leq
N(r,\infty;H)+S(r,f)+S(r,g).$

######  Lemma 2.11.

[1] If $f$, $g$ be two non-constant meromorphic functions such that
$E_{1)}(1;f)=E_{1)}(1;g)$ and $H\not\equiv 0$, then

$\displaystyle N(r,1;f\mid\leq 1)\leq N(r,0;H)\leq
N(r,\infty;H)+S(r,f)+S(r,g).$

######  Lemma 2.12.

[1] If $f$, $g$ be share $(1,1)^{*}$ and $H\not\equiv 0$, then

$\displaystyle N^{E}(r,1;f,g\mid\leq 1)\leq N(r,0;H)\leq
N(r,\infty;H)+S(r,f)+S(r,g).$

######  Lemma 2.13.

[1] If $f$, $g$ be share $(1,1)^{*}$ and $H\not\equiv 0$, then

$\displaystyle N(r,\infty;H)$ $\displaystyle\leq$
$\displaystyle\overline{N}(r,0;f\mid\geq 2)+\overline{N}(r,0;g\mid\geq
2)+\overline{N}(r,\infty;f\mid\geq 2)+\overline{N}_{*}(r,1;f,g)$
$\displaystyle+\overline{N}(r,\infty;g\mid\geq
2)+\overline{N}_{0}(r,0;f^{\prime})+\overline{N}_{0}(r,0;g^{\prime})+S(r,f)+S(r,g),$

where $\overline{N}_{0}(r,0;f^{\prime})$ is the reduced counting function of
those zeros of $f^{\prime}$ which are not the zeros of $f(f-1)$ and
$\overline{N}_{0}(r,0;g^{\prime})$ is similarly defined.

######  Lemma 2.14.

[1] Let $E_{2)}(1;f)=E_{2)}(1;g)$. Then

$\displaystyle\overline{N}_{f\geq 3}(r,1;f\mid g\neq 1)$ $\displaystyle\leq$
$\displaystyle\frac{1}{2}\overline{N}(r,0;f)+\frac{1}{2}\overline{N}(r,\infty;f)-\frac{1}{2}\sum_{p=3}^{\infty}\overline{N}\left(r,0;\frac{f^{\prime}}{F}\mid\geq
p\right)-\frac{1}{2}\overline{N}_{0}^{2}(r,0;f^{\prime})+S(r),$

######  Lemma 2.15.

[22] Let $f$ be a non-constant meromorphic function, and $p$, $k$ be positive
integers. Then

$\displaystyle N_{p}(r,0;f^{(k)})\leq
T(r,f^{(k)}\textasciiacute-T(r,f)+N_{p+k}(r,0;f)+S(r,f),$ $\displaystyle
N_{p}(r,0;f^{(k)})\leq k\overline{N}(r,\infty;f)+N_{p+k}(r,0;f)+S(r,f).$

##  3\. Proofs of the theorems

###### Proof of Theorem 1.1.

Let $F=\displaystyle\frac{F_{1}^{(k)}}{\alpha(z)}$ and
$G=\displaystyle\frac{G_{1}^{(k)}}{\alpha(z)}$, where
$F_{1}=f^{n}(z)P(f(z))L_{c}(f)$ and $G_{1}=(g^{n}(z)P(g(z))L_{c}(g))$. Then
$F$ and $G$ are two transcendental meromorphic functions that share “$(1,2)$”
except the zeros and poles of $\alpha(z)$. We consider the following two
cases.

Case 1: Suppose $H\not\equiv 0$. Since $F$ and $G$ share $``(1,2)"$, it
follows that $F$ and $G$ share $(1,1)^{*}$. Keeping in view of Lemmas 2.10 and
2.13, we see that

$\displaystyle\overline{N}(r,1;F)$ $\displaystyle=$ $\displaystyle
N(r,1;F\mid\leq 1)+\overline{N}(r,1;F\mid\geq 2)\leq
N(r,\infty;H)+\overline{N}(r,1;F\mid\geq 2)$ (3.1) $\displaystyle\leq$
$\displaystyle\overline{N}(r,0;F\mid\geq 2)+\overline{N}(r,0;G\mid\geq
2)+\overline{N}_{*}(r,1;F,G)+\overline{N}(r,1;F\mid\geq 2)$
$\displaystyle\overline{N}_{0}(r,0;F^{\prime})+\overline{N}_{0}(r,0;G^{\prime})+S(r,F)+S(r,G).$

Since $F$, $G$ share $``(1,2)"$, we must have $\overline{N}_{F\geq
2}(r,1;F\mid G\neq 1)=S(r,F)$ and $\overline{N}(r,1;F\mid\geq
2,G\mid=1)=S(r,F)$. Therefore, keeping in view of the above observation and
Lemma 2.4, we get

$\displaystyle\overline{N}_{0}(r,0;G^{\prime})+\overline{N}(r,1;F\mid\geq
2)+\overline{N}_{*}(r,1;F,G)$ (3.2) $\displaystyle\leq$
$\displaystyle\overline{N}_{0}(r,0;G^{\prime})+\overline{N}(r,1;F\mid\geq
3)+\overline{N}_{F\geq 2}(r,1;F\mid G\neq 1)+\overline{N}(r,1;F\mid\geq
2,G\mid=1)$ $\displaystyle+\overline{N}(r,1;F\mid\geq 2,G\mid\geq 2)+S(r,G)$
$\displaystyle\leq$
$\displaystyle\overline{N}_{0}(r,0;G^{\prime})+\overline{N}(r,1;G\mid\geq
3)+\overline{N}(r,1;G\mid\geq 2)+S(r,F)+S(r,G)$ $\displaystyle\leq$
$\displaystyle N(r,0;G^{\prime}\mid G\neq 0)\leq\overline{N}(r,0;G)+S(r,G).$

Hence using (3.1), (3.2), Lemmas 2.2, 2.7 and 2.15, we get from second
fundamental theorem that

$\displaystyle(n+m)T(r,f)\leq T(r,F_{1})-N(r,0;L_{c}(f))+S(r,f)$
$\displaystyle\leq$ $\displaystyle
T(r,F)+N_{k+2}(r,0;F_{1})-N_{2}(r,0;F)-N(r,0;L_{c}(f))+S(r,f)$
$\displaystyle\leq$
$\displaystyle\overline{N}(r,0;F)+\overline{N}(r,1;F)+\overline{N}(r,\infty;F)+N_{k+2}(r,0;F_{1})-N_{2}(r,0;F)$
$\displaystyle-N(r,0;L_{c}(f))-\overline{N}_{0}(r,0;F^{\prime})+S(r,f)$
$\displaystyle\leq$
$\displaystyle\overline{N}(r,0;F)+\overline{N}(r,0;F\mid\geq
2)+\overline{N}(r,0;G\mid\geq 2)+\overline{N}(r,1;F\mid\geq
2)+\overline{N}_{L}(r,1;F)$
$\displaystyle+\overline{N}_{L}(r,1;G)+\overline{N}_{0}(r,0;G^{\prime})+N_{k+2}(r,0;F_{1})-N_{2}(r,0;F)-N(r,0;L_{c}(f))$
$\displaystyle\leq$ $\displaystyle
N_{k+2}(r,0;F_{1})+N_{2}(r,0;G)-N(r,0;L_{c}(f))+S(r,f)+S(r,g)$
$\displaystyle\leq$ $\displaystyle
N_{k+2}(r,0;f(z)^{n}P(f(z))L_{c}(f))+N_{k+2}(r,0;g(z)^{n}P(g(z))L_{c}(g))$
$\displaystyle-N(r,0;L_{c}(f))+S(r,f)+S(r,g)$ $\displaystyle\leq$
$\displaystyle
N_{k+2}(r,0;f^{n}(z))+N_{k+2}(r,0;P(f))+N_{k+2}(r.0;g^{n}(z))+N_{k+2}(r,0;P(g))$
$\displaystyle+N(r,0;L_{c}(g))+S(r,f)+S(r,g)$ $\displaystyle\leq$
$\displaystyle(k+2)\overline{N}(r,0;f)+N(r,0;P(f))+(k+2)\overline{N}(r,0;g)+N(r,0;P(g))$
$\displaystyle+N(r,0;L_{c}(f))+S(r,f)+S(r,g)$ $\displaystyle\leq$
$\displaystyle(k+m+1)(T(r,f)+T(r,g))+T(r,L_{c}(g))+S(r,f)+S(r,g).$

i.e.,

$\displaystyle(n+m)T(r,f)$ (3.3) $\displaystyle\leq$
$\displaystyle(k+m+1)(T(r,f)+T(r,g))+m(r,L_{c}(g))+S(r,f)+S(r,g)$
$\displaystyle\leq$
$\displaystyle(k+m+1)(T(r,f)+T(r,g))+m\left(r,\frac{L_{c}(g)}{g}\right)+m(r,g)+S(r,f)+S(r,g)$
$\displaystyle\leq$
$\displaystyle(k+m+1)(T(r,f)+T(r,g))+T(r,g)+S(r,f)+S(r,g).$

In a similarly, we get

$\displaystyle(n+m)T(r,g)\leq(k+m+1)(T(r,f)+T(r,g))+T(r,f)+S(r,f)+S(r,g).$
(3.4)

Combining (3) and (3.4), we get

$\displaystyle(n-2k-m-5)(T(r,f)+T(r,g))\leq S(r,f)+S(r,g),$

which is a contradicts with $n\geq 2k+m+6$.

Case 2: Suppose $H\equiv 0$. Then by integration we get

$\displaystyle F=\frac{AG+B}{CG+D},$ (3.5)

where $A,\;B,\;C,\;D$ are complex constant satisfying $AD-BC\neq 0$.

Subcase 2.1: Suppose $AC\neq 0$. Then $F-\frac{A}{C}=\frac{-(AD-
BC)}{C(CG+D)}\neq 0.$ So $F$ omits the value $\frac{A}{C}.$

Therefore, by Lemma 2.7 and the Second Fundamental Theorem of Nevalinna, we
get

$\displaystyle(n+m)T(r,f)\leq
T(r,f(z)^{n}P(f(z))L_{c}(f))-N(r,0;L_{c}(f))+S(r,f)$ $\displaystyle\leq$
$\displaystyle T(r,F_{1})-N(r,0;L_{c}(f))+S(r,f)\leq
T(r,F)+N_{k+1}(r,0;F_{1})-\overline{N}(r,0;F)$
$\displaystyle-N(r,0;L_{c}(f))+S(r,f)\leq\overline{N}(r,0;F)+\overline{N}(r,\infty;F)+\overline{N}\left(r,\frac{A}{C};F\right)$
$\displaystyle-N(r,0;L_{c}(f))+N_{k+1}(r,0;F_{1})-N(r,0;F)+S(r,f)$
$\displaystyle\leq$ $\displaystyle
N_{k+1}(r,0;f^{n}P(f)L_{c}(f))-N(r,0;L_{c}(f))+S(r,f)\leq(k+1)\overline{N}(r,0;f)$
$\displaystyle+N(r,0;P(f))+S(r,f)\leq(k+m+1)T(r,f)+S(r,f),$

which is a contradicts with $n\geq 2k+m+6$.

Subcase 2.2: Suppose $AC=0$. Since $AD-BC\neq 0$, $A$ and $C$ both can not be
simultaneously zero.

Subcase 2.2.1: Let $A\neq 0$ and $C=0$. Then (3.5) becomes $F=A_{1}G+B_{1}$,
where $A_{1}=A/D$ and $B_{1}=B/D$. If $f$ has no 1-point, then by Lemma 2.7
and the second fundamental theorem of Nevallina, we get

$\displaystyle(n+m)T(r,f)\leq
T(r,f(z)^{n}P(f(z))L_{c}(f))-N(r,0;L_{c}(f))+S(r,f)$ $\displaystyle\leq$
$\displaystyle T(r,F_{1})-N(r,0;L_{c}(f))+S(r,f)\leq
T(r,F)+N_{k+1}(r,0;F_{1})-\overline{N}(r,0;F)$
$\displaystyle-N(r,0;L_{c}(f))+S(r,f)\leq\overline{N}(r,0;F)+\overline{N}(r,\infty;F)+\overline{N}\left(r,1;F\right)$
$\displaystyle+N_{k+1}(r,0;F_{1})-\overline{N}(r,0;F)-N(r,0;L_{c}(f))+S(r,f)$
$\displaystyle\leq$
$\displaystyle\overline{N}_{k+1}(r,0;f^{n}P(f)L_{c}(f))-N(r,0;L_{c}(f))\leq(k+m+1)T(r,f)+S(r,f),$

which is a contradiction since $n\geq 2k+m+6$. Let $f$ has some 1-point. Then
$A_{1}+B_{1}=1$. Therefore, $F=A_{1}G+1-A_{1}$. If $A_{1}\neq 1$, then using
Lemmas 2.7, 2.5, 2.15 and the second fundamental theorem, we get

$\displaystyle(n+m)T(r,g)\leq T(r,g^{n}P(g)L_{c}(g))-N(r,0;L_{c}(g))+S(r,g)$
$\displaystyle\leq$ $\displaystyle T(r,G_{1})-N(r,0;L_{c}(g))+S(r,g)$
$\displaystyle\leq$ $\displaystyle
T(r,G)+N_{k+1}(r,0;G_{1})-\overline{N}(r,0;G)-N(r,0;L_{c}(g))+S(r,g)$
$\displaystyle\leq$
$\displaystyle\overline{N}(r,0;G)+\overline{N}(r,\infty;G)+\overline{N}\left(r,\frac{1-A_{1}}{A_{1}};G\right)+N_{k+1}(r,0;G_{1})-\overline{N}(r,0;G)$
$\displaystyle-N(r,0;L_{c}(g))+S(r,g)$ $\displaystyle\leq$ $\displaystyle
N_{k+1}(r,0;G_{1})+\overline{N}(r,0;F)-N(r,0;L_{c}(g))+S(r,g)$
$\displaystyle\leq$ $\displaystyle
N_{k+1}(r,0;F_{1})+N_{k+1}(r,0;G_{1})-N(r,0;L_{c}(g))+S(r,g)$
$\displaystyle\leq$
$\displaystyle(k+m+1)T(r,f)+T(r,L_{c}(f))+(k+m+1)T(r,g)+S(r,f)+S(r,g)$
$\displaystyle\leq$ $\displaystyle(2k+2m+3)T(r,g)+S(r,g).$

i.e.,

$\displaystyle(n-2k-m-3)T(r,g)\leq S(r,g),$

which is a contradiction since $n\geq 2k+m+6$. Hence $A_{1}=1$, and therefore
we have $F\equiv G.$ i.e.,

$\displaystyle(f(z)^{n}P(f(z))L_{c}(f))^{(k)}\equiv(g(z)^{n}P(g(z))L_{c}(g))^{(k)}.$

Integrating $k$ times we get

$\displaystyle f^{n}P(f)L_{c}(f)\equiv g^{n}P(g)L_{c}(g)+p(z),$ (3.6)

where $p(z)$ is a polynomial of degree at most $k-1$. Suppose $p(z)\not\equiv
0$. Then from (3.6), we have

$\displaystyle\frac{f^{n}P(f)L_{c}(f)}{p(z)}\equiv\frac{g^{n}P(g)L_{c}(g)}{p(z)}+1.$
(3.7)

Now, using Lemmas 2.2, 2.6 and the second fundamental theorem, we get

$\displaystyle(n+m)T(r,f)\leq T(r,f^{n}P(f)L_{c}(f))-N(r,0;L_{c}(f))+S(r,f)$
$\displaystyle\leq$ $\displaystyle
T(r,f^{n}P(f)L_{c}(f)/p(z))-N(r,0;L_{c}(f))+S(r,f)$ $\displaystyle\leq$
$\displaystyle\overline{N}\left(r,0;\frac{f^{n}P(f)L_{c}(f)}{p}\right)+\overline{N}\left(r,\infty;\frac{f^{n}P(f)L_{c}(f)}{p}\right)+\overline{N}\left(r,1;\frac{f^{n}P(f)L_{c}(f)}{p}\right)$
$\displaystyle-N(r,0;L_{c}(f))+S(r,f)$ $\displaystyle\leq$
$\displaystyle\overline{N}(r,0;f)+\overline{N}(r,0;P(f))+\overline{N}\left(r,0;\frac{g^{n}P(g)L_{c}(g)}{p}\right)+S(r,f)$
$\displaystyle\leq$
$\displaystyle\overline{N}(r,0;f)+\overline{N}(r,0;P(f))+\overline{N}(r,0;g)+\overline{N}(r,0;P(g))+\overline{N}(r,0;L_{c}(g))+S(r,f)$
$\displaystyle\leq$
$\displaystyle(m+1)T(r,f)+(m+1)T(r,g)+T(r,L_{c}(g))+S(r,f)+S(r,g)$
$\displaystyle\leq$
$\displaystyle(m+1)T(r,f)+(m+1)T(r,g)+m\left(r,\frac{L_{c}(g)}{g}\right)+m(r,g)+S(r,f)+S(r,g)$
$\displaystyle\leq$ $\displaystyle(m+1)T(r,f)+(m+2)T(r,g)+S(r,f)+S(r,g).$

Similarly, we get

$\displaystyle(n+m)T(r,g)\leq(m+1)T(r,g)+(m+2)T(r,f)+S(r,f)+S(r,g).$

Combining the above two equations, we get

$\displaystyle(n-m-3)(T(r,f)+T(r,g))\leq S(r,f)+S(r,g),$

which contradicts to the fact that $n\geq 2k+m+6$. Hence $p(z)\equiv 0$ and so
from (3.6), we obtain

$\displaystyle f^{n}P(f)L_{c}(f)\equiv g^{n}P(g)L_{c}(g).$

i.e.

$\displaystyle
f^{n}(a_{m}f^{m}+a_{m-1}f^{m-1}+\ldots+a_{1}f+a_{0})(f(z+c)+c_{0}f(z))$
$\displaystyle\equiv
g^{n}(a_{m}g^{m}+a_{m-1}g^{m-1}+\ldots+a_{1}g+a_{0})(g(z+c)+c_{0}g(z)).$ (3.8)

Let $h=f/g$. Then the above equation can be written as

$\displaystyle[a_{m}(h^{n+m}h(z+c)-1)g^{m}+a_{m-1}(h^{n+m-1}h(z+c)-1)g^{m-1}+\ldots$
$\displaystyle+a_{0}(h^{n}h(z+c)-1)]g(z+c)$ $\displaystyle\equiv-
c_{0}[a_{m}(h^{n+m+1}-1)g^{m}+a_{m-1}(h^{n+m}-1)g^{m-1}+\ldots+a_{0}(h^{n+1}-1)]g(z).$

If $h$ is constant, then the above equation can be written as

$\displaystyle[a_{m}(h^{n+m+1}-1)g^{m}+a_{m-1}(h^{n+m}-1)g^{m-1}+\ldots+a_{0}(h^{n+1}-1)]L_{c}(g)\equiv
0.$

Since $L_{c}(g)\not\equiv 0$, we must have

$\displaystyle
a_{m}(h^{n+m+1}-1)g^{m}+a_{m-1}(h^{n+m}-1)g^{m-1}+\ldots+a_{0}(h^{n+1}-1)=0.$

Then by a similar argument as in the Case 2 in the proof of Theorem 11 [17],
we obtain $f=tg$, where $t$ a constant such that $t^{d}=1$,
$d=\text{gcd}(\lambda_{0},\lambda_{1},\ldots,\lambda_{m})$, where
$\lambda_{j}$’s are defined by

$\displaystyle\lambda_{j}=\begin{cases}n+1+j,\;\;\;\text{if}\;a_{j}\neq 0\\\
n+1+m,\;\text{if}\;a_{j}=0,\end{cases}j=0,1,\ldots,m.$

If $h$ is not constant, then it follows that $f,\;g$ satisfy the algebraic
equation $R(w_{1},w_{2})=0$, where

$\displaystyle
R(w_{1},w_{2})=w_{1}^{n}P(w_{1})L_{c}(w_{1})-w_{2}^{n}P(w_{2})L_{c}(w_{2}).$

Subcase 2.2.2: Let $A=0$ and $C\neq 0$. Then (3.5) becomes

$\displaystyle F=\frac{1}{A_{2}G+B_{2}},$ (3.9)

where $A_{2}=C/B$ and $B_{2}=D/B$. If $F$ has no $1$-point, then by a similar
argument as done in Subcase 2.2.1, we can get a contradiction. Let $F$ has
some $1$-point. Then $A_{2}+B_{2}=1$. If $A_{2}\neq 1$, then (3.9) can be
written as

$\displaystyle F=\frac{1}{A_{2}G+1-A_{2}}.$ (3.10)

Since $F$ is entire and $A_{2}\neq 0$, $G$ omits the value $(1-A_{2})/A_{2}$.
Therefore, by Lemma 2.7 and the second fundamental theorem, we get

$\displaystyle(n+m)T(r,g)\leq T(r,g^{n}P(g)L_{c}(g))-N(r,0;L_{c}(g))+S(r,g)$
$\displaystyle\leq$ $\displaystyle T(r,G_{1})-N(r,0;L_{c}(g))+S(r,g)$
$\displaystyle\leq$ $\displaystyle
T(r,G)+N_{k+1}(r,0;G_{1})-\overline{N}(r,0;G)-N(r,0;L_{c}(g))+S(r,g)$
$\displaystyle\leq$
$\displaystyle\overline{N}(r,0;G)+\overline{N}(r,\infty;G)+\overline{N}\left(r,\frac{1-A_{2}}{A_{2}};G\right)+N_{k+1}(r,0;G_{1})-\overline{N}(r,0;G)$
$\displaystyle-N(r,0;\Delta_{c}g)+S(r,g).$

i.e.,

$\displaystyle(n+m)T(r,g)\leq
N_{k+1}(r,0;g^{n}P(g)L_{c}(g))-N(r,0;L_{c}(g))+S(r,g)$ $\displaystyle\leq$
$\displaystyle N_{k+1}(r,0;g^{n})+N_{k+1}(r,0;P(g))+S(r,g)$
$\displaystyle\leq$
$\displaystyle(k+1)\overline{N}(r,0;g)+N(r,0;P(g))+S(r,g)\leq(k+m+1)T(r,g)+S(r,g),$

which is a contradiction since $n\geq 2k+m+6$. Hence $A_{2}=1$. So, from
(3.10), we get $FG\equiv 1$. i.e.,

$\displaystyle(f(z)^{n}P(f(z))L_{c}(f))^{(k)}(g(z)^{n}P(g(z))L_{c}(g))^{(k)}\equiv\alpha^{2}(z).$
(3.11)

Let $u_{1},u_{2},\ldots,u_{t}$, $1\leq t\leq m$ be the distinct zeros of
$P(z)$. Since $m\geq k+1$, $a_{0}\neq 0$ and $f$ is entire, it is easily seen
from (3.11) that $f$ has atleast two finite Picard exceptional values, which
is not possible. Hence the proof is complete. ∎

###### Proof of Theorem 1.2.

Let $F$ and $G$ be defined as in Theorem 1.1. Then $F$ and $G$ are two
transcendental meromorphic functions that share $(1,2)^{*}$ except the zeros
and poles of $\alpha(z)$. We consider the following two cases.

Case 1: Suppose $H\not\equiv 0$. Since $F$ and $G$ share $(1,2)^{*}$, it
follows that $F$ and $G$ share $(1,1)^{*}$. Also we note that
$\overline{N}(r,1;F\mid=1,G\mid=0)=S(r,F)+S(r,G)$. Keeping in view of Lemmas
2.12 and 2.13, we see that

$\displaystyle\overline{N}(r,1;F)=N(r,1;F\mid\leq
1)+\overline{N}(r,1;F\mid\geq 2)$ (3.12) $\displaystyle\leq$
$\displaystyle\overline{N}(r,1;F\mid=1,G\mid=0)+\overline{N}^{E}(r,1;F,G\mid\leq
1)+\overline{N}(r,1;F\mid\geq 2)$ $\displaystyle\leq$ $\displaystyle
N(r,\infty;H)+\overline{N}(r,1;F\mid\geq 2)+S(r,F)+S(r,G)$ $\displaystyle\leq$
$\displaystyle\overline{N}(r,0;F\mid\geq 2)+\overline{N}(r,0;G\mid\geq
2)+\overline{N}_{*}(r,1;F,G)+\overline{N}(r,1;F\mid\geq 2)$
$\displaystyle\overline{N}_{0}(r,0;F^{\prime})+\overline{N}_{0}(r,0;G^{\prime})+S(r,F)+S(r,G).$

Since $F$, $G$ share $(1,2)^{*}$, we must have $\overline{N}_{F\geq
2}(r,1;F\mid G\neq 1)=S(r,F)+S(r,G)$,
$\overline{N}(r,1;F\mid=2,G\mid=1)=S(r,F)+S(r,G)$. Therefore, using Lemma 2.8,
we get

$\displaystyle\overline{N}(r,1;F\mid\geq 2)$ (3.13) $\displaystyle\leq$
$\displaystyle\overline{N}_{F\geq 2}(r,1;F\mid G\neq
1)+\overline{N}(r,1;F\mid\geq 2,G\mid=1)+\overline{N}(r,1;F\mid\geq
2,G\mid\geq 2)$ $\displaystyle\leq$ $\displaystyle\overline{N}_{F\geq
2}(r,1;F\mid G\neq 1)+\overline{N}(r,1;F\mid=2,G\mid=1)+\overline{N}_{F\geq
3}(r,1;G\mid=1)$ $\displaystyle+\overline{N}(r,1;G\mid\geq 2)+S(r,F)+S(r,G)$
$\displaystyle\leq$
$\displaystyle\overline{N}(r,0;F)+\overline{N}(r,1;G\mid\geq
2)+S(r,F)+S(r,G).$

Again using (3.13) and Lemma 2.4, we get

$\displaystyle\overline{N}_{0}(r,0;G^{\prime})+\overline{N}(r,1;F\mid\geq
2)+\overline{N}_{*}(r,1;F,G)$ (3.14) $\displaystyle\leq$
$\displaystyle\overline{N}_{0}(r,0;G^{\prime})+\overline{N}(r,1;G\mid\geq
2)+\overline{N}(r,1;G\mid\geq 3)+\overline{N}(r,0;F)+S(r,G)$
$\displaystyle\leq$
$\displaystyle\overline{N}_{0}(r,0;G^{\prime})+N(r,1;G)-\overline{N}(r,1;G)+\overline{N}(r,0;F)+S(r,F)+S(r,G)$
$\displaystyle\leq$ $\displaystyle N(r,0;G^{\prime}\mid G\neq
0)+\overline{N}(r,0;F)+S(r,F)+S(r,G)$ $\displaystyle\leq$
$\displaystyle\overline{N}(r,0;F)+\overline{N}(r,0;G)+S(r,F)+S(r,G).$

Hence using (3.12), (3.14), Lemmas 2.2 and 2.7, Second fundamental theorem of
Nevalinna, we get

$\displaystyle(n+m)T(r,f)\leq T(r,f^{n}P(f)L_{c}(f))-N(r,0;L_{c}(f))+S(r,f)$
$\displaystyle\leq$ $\displaystyle
T(r,F)+N_{k+2}(r,0;F_{1})-N_{2}(r,0;F)-N(r,0;L_{c}(f))+S(r,f).$

i.e.,

$\displaystyle(n+m)T(r,f)\leq\overline{N}(r,0;F)+\overline{N}(r,\infty;F)+\overline{N}(r,1;F)-\overline{N}(r,0;F^{\prime})$
(3.15) $\displaystyle+N_{k+2}(r,0;F_{1})-N_{2}(r,0;F)-N(r,0;L_{c}(f))+S(r,f)$
$\displaystyle\leq$ $\displaystyle
N_{2}(r,0;F)+N_{2}(r,0;G)+\overline{N}(r,0;F)+N_{k+2}(r,0;F_{1})-N_{2}(r,0;F)$
$\displaystyle-N(r,0;L_{c}(f))+S(r,f)+S(r,g)$ $\displaystyle\leq$
$\displaystyle
N_{k+2}(r,0;F_{1})+N_{k+2}(r,0;G_{1})+N_{k+1}(r,0;F_{1})-N(r,0;L_{c}(f))$
$\displaystyle+S(r,f)+S(r,g)\leq(k+2)(\overline{N}(r,0;f)+\overline{N}(r,0;g))+N(r,0;P(f))$
$\displaystyle+N(r,0;P(g))+N(r,0;P(f))+N(r,0;L_{c}(g))+(k+1)\overline{N}(r,0;f)$
$\displaystyle+N(r,0;L_{c}(f))+S(r,f)+S(r,g)\leq(k+m+2)(T(r,f)+T(r,g))$
$\displaystyle+(k+m+1)T(r,f)+T(r,L_{c}(f))+T(r,L_{c}(g))+S(r,f)+S(r,g)$
$\displaystyle\leq$
$\displaystyle(k+m+2)(T(r,f)+T(r,g))+(k+m+1)T(r,f)+m\left(r,\frac{L_{c}(f)}{f}\right)$
$\displaystyle+m(r,f)+m\left(r,\frac{L_{c}(g)}{g}\right)+m(r,g)+S(r,f)+S(r,g)$
$\displaystyle\leq$ $\displaystyle(k+m+3)(T(r,f)+T(r,g))+(k+m+1)T(r,f)$
$\displaystyle+S(r,f)+S(r,g).$

In a similar manner, we obtain

$\displaystyle(n+m)T(r,g)$ $\displaystyle\leq$
$\displaystyle(k+m+3)(T(r,f)+T(r,g))+(k+m+1)T(r,g)$ (3.16)
$\displaystyle+S(r,f)+S(r,g).$

Combining (3.15) and (3.16), we get

$\displaystyle(n-3k-2m+7)(T(r,f)+T(r,g))\leq S(r,f)+S(r,g),$

which is a contradicts to the fact that $n\geq 3k+2m+8$.

Case 2: Let $H\equiv 0$. This case can be carried out similarly as done in
case 2 of the proof of Theorem 1.1. So, we omit the details. This proves
Theorem 1.2. ∎

###### Proof of Theorem 1.3.

Let $F$ and $G$ be defined as in Theorem 1.1. Then $F$ and $G$ are
transcendental meromorphic functions such that $E_{2)}(1,F)=E_{2)}(1,G)$
except the zeros and poles of $\alpha(z)$. Let us discuss the following two
cases.

Case 1: Let $H\not\equiv 0$. Since $E_{2)}(1,F)=E_{2)}(1,G)$, it follows that
$E_{1)}(1,F)=E_{1)}(1,G)$. Keeping in view of Lemmas 2.9, 2.11 and 2.14, we
see that

$\displaystyle\overline{N}(r,1;F)=N(r,1;F\mid\leq
1)+\overline{N}(r,1;F\mid\geq 2)$ $\displaystyle\leq$ $\displaystyle
N(r,H)+\overline{N}(r,1;F\mid=2)+\overline{N}_{F\geq 3}(r,1;F\mid G\neq
1)+\overline{N}(r,1;F\mid\geq 3,G\mid\geq 3)$ $\displaystyle\leq$
$\displaystyle
N(r,\infty;H)+\overline{N}(r,1;G\mid=2)+\overline{N}(r,1;G\mid\geq
3)+\overline{N}_{F\geq 3}(r,1;F\mid G\neq 1)$ $\displaystyle+S(r,F)+S(r,G)$
$\displaystyle\leq$ $\displaystyle N(r,\infty;H)+\overline{N}(r,1;G\mid\geq
2)+\overline{N}_{F\geq 3}(r,1;F\mid G\neq 1)+S(r,F)+S(r,G)$
$\displaystyle\leq$ $\displaystyle\overline{N}(r,0;F\mid\geq
2)+\overline{N}(r,0;G\mid\geq
2)+\overline{N}_{L}(r,1;F)+\overline{N}_{L}(r,1;G)+\overline{N}(r,1;G\mid\geq
2)$ $\displaystyle+2\overline{N}_{F\geq 3}(r,1;F\mid G\neq
1)+\overline{N}_{G\geq 3}(r,1;G\mid F\neq
1)+\overline{N}_{0}(r,0;F^{\prime})+\overline{N}_{0}(r,0;G^{\prime})$
$\displaystyle+S(r,F)+S(r,G).$

i.e.,

$\displaystyle\overline{N}(r,1;F)$ (3.17) $\displaystyle\leq$
$\displaystyle\overline{N}(r,0;F\mid\geq 2)+\overline{N}(r,0;G\mid\geq
2)+\overline{N}_{L}(r,1;F)+\overline{N}_{L}(r,1;G)+\overline{N}(r,1;G\mid\geq
2)$ $\displaystyle+2\overline{N}_{F\geq 3}(r,1;F\mid G\neq
1)+\overline{N}_{G\geq 3}(r,1;G\mid F\neq
1)+\overline{N}_{0}(r,0;F^{\prime})+\overline{N}_{0}(r,0;G^{\prime})$
$\displaystyle+S(r,F)+S(r,G)$ $\displaystyle\leq$
$\displaystyle\overline{N}(r,0;F\mid\geq 2)+\overline{N}(r,0;G\mid\geq
2)+\overline{N}_{L}(r,1;F)+\overline{N}_{L}(r,1;G)+\overline{N}(r,1;G\mid\geq
2)$ $\displaystyle+$
$\displaystyle\overline{N}(r,0;F)+\frac{1}{2}\overline{N}(r,0;G)+\overline{N}_{0}(r,0;F^{\prime})+\overline{N}_{0}(r,0;G^{\prime})+S(r,F)+S(r,G).$

Now using Lemma 2.4, we get

$\displaystyle\overline{N}_{0}(r,0;G^{\prime})+\overline{N}(r,1;G\mid\geq
2)+\overline{N}_{L}(r,1;F)+\overline{N}_{L}(r,1;G)$ (3.18) $\displaystyle\leq$
$\displaystyle\overline{N}_{0}(r,0;G^{\prime})+\overline{N}(r,1;G\mid\geq
2)+\overline{N}(r,1;G\mid\geq 3)+S(r,G)$ $\displaystyle\leq$
$\displaystyle\overline{N}_{0}(r,0;G^{\prime})+N(r,1;G)-\overline{N}(r,1;G)+S(r,F)+S(r,G)$
$\displaystyle\leq$ $\displaystyle N(r,0;G^{\prime}\mid G\neq
0)\leq\overline{N}(r,0;G)+S(r,F)+S(r,G).$

Therefore, using (3.17), (3.18), Lemmas 2.2 and 2.7, we get from Second
Fundamental Theorem that

$\displaystyle(n+m)T(r,f)\leq T(r,f^{n}P(f)L_{c}(f))-N(r,0;L_{c}(f))+S(r,f)$
(3.19) $\displaystyle\leq$ $\displaystyle
T(r,F)+N_{k+2}(r,0;F_{1})-N_{2}(r,0;F)-N(r,0;L_{c}(f))+S(r,f)$
$\displaystyle\leq$
$\displaystyle\overline{N}(r,0;F)+\overline{N}(r,\infty;F)+\overline{N}(r,1;F)-\overline{N}(r,0;F^{\prime})+N_{k+2}(r,0;F_{1})$
$\displaystyle-N_{2}(r,0;F)-N(r,0;L_{c}(f))+S(r,f)$ $\displaystyle\leq$
$\displaystyle
N_{2}(r,0;F)+N_{2}(r,0;G)+\overline{N}(r,0;F)+\frac{1}{2}\overline{N}(r,0;G)+N_{k+2}(r,0;F_{1})$
$\displaystyle-N_{2}(r,0;F)-N(r,0;L_{c}(f))+S(r,f)+S(r,g)$ $\displaystyle\leq$
$\displaystyle
N_{k+2}(r,0;F_{1})+N_{k+2}(r,0;G_{1})+N_{k+1}(r,0;F_{1})+\frac{1}{2}N_{k+1}(r,0;G_{1})$
$\displaystyle-N(r,0;L_{c}(f))+S(r,f)+S(r,g)$ $\displaystyle\leq$
$\displaystyle(k+2)(\overline{N}(r,0;f)+\overline{N}(r,0;g))+2N(r,0;P(f))+N(r,0;P(g))$
$\displaystyle+N(r,0;L_{c}(g))+(k+1)\overline{N}(r,0;f)+N(r,0;L_{c}(f))+\frac{1}{2}(k+1)\overline{N}(r,0;g)$
$\displaystyle+\frac{1}{2}N(r,0;P(g))+\frac{1}{2}N(r,0;L_{c}(g))+S(r,f)+S(r,g)$
$\displaystyle\leq$
$\displaystyle(k+m+2)(T(r,f)+T(r,g))+(k+m+1)T(r,f)+T(r,L_{c}(f))$
$\displaystyle+T(r,L_{c}(g))+\frac{1}{2}(k+m+1)T(r,g)+\frac{1}{2}T(r,L_{c}(g))+S(r,f)+S(r,g)$
$\displaystyle\leq$
$\displaystyle(k+m+2)(T(r,f)+T(r,g))+(k+m+1)T(r,f)+m\left(r,\frac{L_{c}(f)}{f}\right)$
$\displaystyle+m(r,f)+m\left(r,\frac{L_{c}(g)}{g}\right)+m(r,g)+\frac{1}{2}(k+m+1)T(r,g)+\frac{1}{2}m\left(r,\frac{L_{c}(g)}{g}\right)$
$\displaystyle+\frac{1}{2}m(r,g)+S(r,f)+S(r,g)$ $\displaystyle\leq$
$\displaystyle(k+m+3)(T(r,f)+T(r,g))+(k+m+1)T(r,f)+\frac{1}{2}(k+m+2)T(r,g)$
$\displaystyle+S(r,f)+S(r,g).$

Similarly, we obtain

$\displaystyle(n+m)T(r,g)$ $\displaystyle\leq$
$\displaystyle(k+m+3)(T(r,f)+T(r,g))+(k+m+1)T(r,g)$ (3.20)
$\displaystyle+\frac{1}{2}(k+m+2)T(r,f)+S(r,f)+S(r,g).$

Combining (3.19) and (3.20), we get

$\displaystyle\left(n-\frac{7k}{2}-\frac{5m}{2}-8\right)(T(r,f)+T(r,g))\leq
S(r,f)+S(r,g),$

which is not possible since
$n\geq\displaystyle\frac{7k}{2}+\displaystyle\frac{5m}{2}+9$.

Case 2: Let $H\equiv 0$. This case can be carried out similarly as done in
case 2 of the proof of Theorem 1.1. So, we omit the details. This proves
Theorem 1.3. ∎

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