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Meromorphic functions that share a polynomial with their difference operators
Advances in Difference Equations volume 2018, Article number: 194 (2018)
Abstract
In this paper, we prove the following result: Let f be a nonconstant meromorphic function of finite order, p be a nonconstant polynomial, and c be a nonzero constant. If f, \(\Delta _{c}f\), and \(\Delta_{c}^{n}f\) (\(n\ge 2\)) share ∞ and p CM, then \(f\equiv \Delta_{c}f\). Our result provides a difference analogue of the result of Chang and Fang in 2004 (Complex Var. Theory Appl. 49(12):871–895, 2004).
1 Introduction and main results
In this paper, we use the base notations of the Nevanlinna theory of meromorphic functions which are defined as follows [9, 18, 19].
Let f be a meromorphic function. Throughout this paper, a meromorphic function always means meromorphic in the whole complex plane.
Definition 1
\(m(r,f)\) is the average of the positive logarithm of \(\vert f(z) \vert \) on the circle \(\vert z \vert =r\).
Definition 2
where \(n(t,f)\) (\(\overline{n}(t,f)\)) denotes the number of poles of f in the disc \(\vert z \vert \le t\), multiples poles are counted according to their multiplicities (ignore multiplicity). \(n(0,f)\) (\(\overline{n}(0,f)\)) denotes the multiplicity of poles of f at the origin (ignore multiplicity).
\(N(r,f)\) is called the counting function of poles of f, and \(\overline{N}(r,f)\) is called the reduced counting function of poles of f.
Definition 3
\(T(r,f)\) is called the characteristic function of f. It plays a cardinal role in the whole theory of meromorphic functions.
Definition 4
Let f be a meromorphic function. The order of growth of f is defined as follows:
If \(\rho (f)< \infty \), then we say that f is a meromorphic function of finite order.
Definition 5
Let a, f be two meromorphic functions. If \(T(r,a)=S(r,f)\), where \(S(r,f)=o(T(r,f))\), as \(r\to \infty \) outside of a possible exceptional set of finite logarithmic measure. Then we say that a is a small function of f. And we use \(S(f)\) to denote the family of all small functions with respect to f.
Definition 6
Let f and g be two meromorphic functions, and p be a polynomial. We say that f and g share p CM, provided that \(f(z)-p(z)\) and \(g(z)-p(z)\) have the same zeros counting multiplicity. And if f and g have the same poles counting multiplicity, then we say that f and g share ∞ CM.
In this paper, we also use some known properties of the characteristic function \(T(r,f)\) as follows [9, 18, 19].
Property 1
Let \(f_{j}\) (\(j=1,2,\ldots,q\)) be q meromorphic functions in \(\vert z \vert < R\) and \(0< r< R\). Then
hold for \(1\le r< R\).
Property 2
Suppose that f is meromorphic in \(\vert z \vert < R\) (\(R\le \infty \)) and a is any complex number. Then, for \(0< r< R\), we have
Property 2 is the first fundamental theorem.
Property 3
Suppose that f is a nonconstant meromorphic function and \(a_{1}, a _{2},\ldots, a_{n}\) are \(n\ge 3\) distinct values in the extended complex plane. Then
Property 3 is the second fundamental theorem. For more properties about \(T(r,f)\), please see [9, 18, 19].
For a meromorphic function \(f(z)\), we define its shift by \(f_{c}(z) = f(z + c)\) and its difference operators by
In [10] the following result was proved.
Theorem 1
Let f be a nonconstant meromorphic function, and a be a nonzero finite complex number. If f, \(f'\), and \(f''\) share a CM, then \(f\equiv f'\).
In 2001, Li and Yang [12] considered the case when f, \(f'\), and \(f^{(n)}\) share one value.
Theorem 2
Let f be an entire function, a be a finite nonzero constant, and \({n\ge 2}\) be a positive integer. If f, \(f'\), and \(f^{(n)}\) share a CM, then f assumes the form
where b, w are two nonzero constants satisfying \(w^{n-1}=1\).
Remark 1
It is easy to see that the functions in ( 1.1 ) really share value a, since when \(b\neq 0\) and \(w^{n-1}=1\), from \(f^{(j)}(z)=a\), \(j=0,1,n\), it follows that \(bwe^{wz}=a\) for each \(j=0,1,n\). So, the functions \(f^{(j)}-a\), \(j=0, 1, n\), have the same zeros counting multiplicity.
In 2004, Chang and Fang [1] considered the case when f, \(f'\), and \(f^{(n)}\) share a small function.
Theorem 3
Let f be an entire function, a be a nonzero small function of f, and \(n\ge 2\) be a positive integer. If f, \(f'\), and \(f^{(n)}\) share a CM, then \(f\equiv f'\).
Recently, value distribution in difference analogue of meromorphic functions has become a subject of some interest, see, e.g., [2–8, 11].
In 2012 and 2014, Chen et al. [2, 3] considered difference analogue of Theorem 1 and Theorem 2, and established the following result.
Theorem 4
Let f be a nonconstant entire function of finite order, and a (≢0) \(\in S(f)\) be a periodic entire function with period c. If f, \(\Delta_{c}f\), and \(\Delta_{c}^{n}f\) (\(n\ge 2\)) share a CM, then \(\Delta_{c}f\equiv \Delta_{c}^{n}f\).
For other related results, the reader is referred to the references due to Latreuch, El Farissi, Belaïdi [11], El Farissi, Latreuch, Asiri [5], El Farissi, Latreuch, Belaïdi and Asiri [6].
Remark 2
There are examples in [3] which show that the conclusion \(\Delta_{c}f\equiv \Delta_{c}^{n}f\) in Theorem 4 cannot be replaced by \(f\equiv \Delta_{c}f\), and the condition \(a(z)\not \equiv 0\) is necessary.
By Theorems 3 and 4, it is natural to ask: Can we provide a difference analogue of Theorem 3? Or, can we delete the condition that ‘\(a(z)\) is a periodic entire function with period c’ in Theorem 4?
In this paper, we study the problem and prove the following result.
Theorem 5
Let f be a nonconstant meromorphic function of finite order, and p be a nonconstant polynomial. If f, \(\Delta_{c}f\), and \(\Delta_{c} ^{n}f\) (\(n\ge 2\)) share p and ∞ CM, then \(f\equiv \Delta_{c}f\).
If f is an entire function, then f, \(\Delta_{c}f\), and \(\Delta_{c}^{n}f\) have no poles, obviously f, \(\Delta_{c}f\) and \(\Delta_{c}^{n}f\) share ∞ CM. By Theorem 5, we consequently get the following result.
Corollary 1
Let f be a nonconstant entire function of finite order, and \(n \ge 2\) be a positive integer. If f, \(\Delta_{c}f\), and \(\Delta_{c} ^{n}f\) share z CM, then \(f\equiv \Delta_{c}f\).
Example 1
Let A, a, b, c be four finite nonzero complex numbers satisfying \(a\neq b\), n (≥2) \(\in \mathbb{N}\) satisfying \([e^{Ac}-1]^{n-1}=1\), \(e^{Ac}-1=\frac{a}{a-b}\), and \(g(z)\) be a periodic entire function with period c, and let \(f(z)=g(z)e^{Az}+b\). By simple calculation, we obtain
It is easy to see that f, \(\Delta_{c}f\), and \(\Delta_{c} ^{n}f\) (\(n\ge 2\)) share a CM, and \(f\neq \Delta_{c}f\) when \(e^{Ac} \neq 2\). This example shows that ‘\(p(z)\) cannot be a constant’ in Theorem 5.
Example 2
Let A, b, c be three nonzero finite complex numbers satisfying \(e^{Ac}=1\), and \(f(z)=e^{Az}+b\), \(p(z)=b\). It is easy to see that f, \(\Delta_{c}f\), and \(\Delta_{c}^{n}f\) share \(p(z)\) CM. But \(\Delta_{c}f\equiv 0 \not \equiv f\). This example also shows that ‘\(p(z)\) cannot be a constant’ in Theorem 5.
Example 3
Let A, c be two nonzero finite complex numbers satisfying \(e^{Ac}=2\) and \(f(z)=e^{Az}\cot (\frac{\pi z}{c})\). By simple calculation, we obtain
Obviously, for any polynomial p, f, \(\Delta_{c}f\), and \(\Delta_{c}^{n}f\) share p and ∞ CM. This example satisfies Theorem 5.
In Examples 1 and 2 , we have \(\Delta_{c}f\equiv tf+a(1-t)\) and \(f(z)=e^{Az+B}+a\), respectively, when f, \(\Delta_{c}f\), and \(\Delta_{c}^{n}f\) (\(n\ge 2\)) share a nonzero constant a CM. Hence we posed the following problem.
Problem 1
Assume that f is a nonconstant entire function of finite order, a is a nonzero constant, and that f, \(\Delta_{c}f\), and \(\Delta_{c}^{n}f\) (\(n\ge 2\)) share a CM. Whether or not, one of the following two cases occurs:
-
(1)
\(\Delta_{c}f\equiv tf+a(1-t)\), where t is a constant satisfying \(t^{n-1}=1\),
-
(2)
\(f(z)=e^{Az+B}+a\), where A (≠0), B are two constants satisfying \(e^{Ac}=1\).
2 Some lemmas
Lemma 1
Let f be a meromorphic function of finite order, and c be a nonzero complex constant. Then
Lemma 2
Let \(c\in \mathbb{C}\), k be a positive integer, and f be a meromorphic function of finite order. Then
Lemma 3
Let \(n\ge 2\) be a positive integer. Suppose that \(f_{i}(z)\) (\(i = 1,2,\ldots,n\)) are meromorphic functions and \(g_{i}(z)\) (\(i = 1,2,\ldots,n\)) are entire functions satisfying
-
(i)
\(\sum_{i=1}^{n}f_{i}(z)e^{g_{i}(z)}\equiv 0\),
-
(ii)
the orders of \(f_{i}\) are less than those of \(e^{g_{k}-g_{l}}\) for \(1\le i \le n\), \(1\le k < l \le n\).
Then \(f_{i}(z)\equiv 0\) (\(i=1,2,\ldots,n\)).
The following lemma is well known.
Lemma 4
Let the function f satisfy the following difference equation:
in the complex plane.
Then the following formula holds:
for every \(w\in \mathbb{C}\) and \(k\in \mathbb{N}^{+}\).
Formula (2.1) has many applications. For example, many solvable difference equations are essentially solved by using it (see [15–17]), and by using such obtained formulas the behavior of their solution can be studied (see, for example, recent papers [13, 14]; see also many related references therein). As another simple application, by using a linear change of variables, the following corollary is obtained:
Corollary 2
Let the function f satisfy the following difference equation:
in the complex plane.
Then the following formula holds:
for every \(w\in \mathbb{C}\) and \(k\in \mathbb{N}^{+}\).
From the ideas of Chang and Fang [1] and Chen and Li [3], we prove the following lemma.
Lemma 5
Let f be a nonconstant meromorphic function of finite order, p (≢0) be a polynomial, and \(n\ge 2\) be an integer. Suppose that
where α and β are two polynomials, and that
Then \(\Delta_{c}f\equiv tf+b(1-t)\), where t, b are constants satisfying \(t^{n-1}=1\) and \(b\neq 0\). Moreover, if \(t\neq 1\), then \(p(z)\equiv b\).
Proof
Firstly, we prove that f cannot be a rational function. Otherwise, suppose that \(f(z)=P(z)/Q(z)\), where \(P(z)\) and \(Q(z)\) are two co-prime polynomials. It follows from (2.2) that \(f(z)\) and \(\Delta_{c}f(z)\) share ∞ CM. We claim that \(Q(z)\) is a constant. Otherwise, suppose that there exists \(z_{0}\) such that \(Q(z_{0}+c)=0\). Since \(f(z)\) and \(\Delta_{c}f(z)\) share ∞ CM, and
We deduce that all zeros of \(Q(z+c)\) must be the zeros of \(Q(z)\). Otherwise, suppose that there exists \(z_{1}\) such that \(Q(z_{1}+c)=0\) but \(Q(z_{1})\neq 0\), then it follows from (2.4) that \(z_{1}+c\) is a pole of \(\Delta_{c}f\) but not the pole of f, which contradicts with \(f(z)\) and \(\Delta_{c}f(z)\) share ∞ CM. Then we get
This implies that \(Q(z)\) has infinitely many zeros, which is a contradiction. Thus, the claim is proved.
Then f is a nonconstant polynomial, suppose that
Then we get \(\Delta_{c}f(z)=f(z+c)-f(z)\), and obviously, \(\deg \Delta_{c}^{n}f(z)\le \deg \Delta_{c}f(z) < \deg f(z)\). Then it follows from (2.2) that \(\alpha (z)\), \(\beta (z)\) are constants, and we let \(e^{\alpha (z)}=a\), \(e^{\beta (z)}=b\). So we have
Then we get \(\deg f=k\le \deg p=m\) since \(\deg \Delta_{c}^{n}f(z) \le \deg \Delta_{c}f(z) < \deg f(z)\). If \(k< m\), then we get \(a=b=1\). This implies \(f\equiv \Delta_{c}f(z) \equiv \Delta_{c}^{n}f\), which contradicts with \(\deg \Delta_{c}f(z) < \deg f(z)\). If \(k= m\) then we get \(a=b=b_{k}/(a_{k}-b_{k})\), \(\Delta_{c}f(z) \equiv \Delta_{c}^{n}f\), and hence \(f(z)\) is a constant, which is a contradiction.
Hence, f is a transcendental meromorphic function. Thus \(T(r,p)=S(r,f)\).
Next, we consider two cases.
Case 1. \(\beta (z)\) is a nonconstant polynomial. It follows from the second equation in (2.2) that \(\Delta_{c}f(z)=e^{\beta (z)}(f(z)-p(z))+p(z)\), and that
where \(a_{1}(z)=e^{\beta (z)}+1\), \(b_{1}(z)=p(z)[1-e^{\beta (z)}]\).
By Corollary 2 , it is easy to get, for any \(k\in \mathbb{N^{+}}\),
where
It follows from (2.5) and (2.6) that
where
In particular, \(\lambda_{1,t}=(-1)^{n-1-t}C_{n-1}^{t}\), which implies
By ( 2.3 ), ( 2.8 ), and Lemma 1 , it is easy to get
Since \(\beta (z)\) is a nonconstant polynomial, we have that
where \(l_{i}\) (\(0\le i \le m \)) are constants satisfying \(l_{m}\neq 0\) and \(m\ge 1\). Obviously, for any \(j\in \{0, 1,\ldots, n-1\}\), we have
where \(P_{i,j}(z)\) are polynomials with degree less than m for \(i\in \{1, 2,\ldots,n\}\), \(j\in \{0,1,\ldots, {C_{n}^{i}-1}\}\).
It follows from the first equation in (2.2) that
From (2.7) and (2.13), we have
From ( 2.3 ), ( 2.10 ) and since \(T(r,p)=S(r,f)\), we have that
If \(\mu_{n}(z)-e^{\alpha (z)}\not \equiv 0\), by ( 2.14 ), ( 2.15 ), Property 1 , and Property 2 , we obtain
which is a contradiction.
Hence \(\mu_{n}(z)-e^{\alpha (z)}\equiv 0\). Combining this with (2.12), we get
Next, we consider three subcases.
Case 1.1. \(\deg \alpha (z)> m\). Then, for any \(1\le i \le n\), \(1\le k< j \le n\), we have
Since \(P_{i,j}(z)\) are polynomials with degree less than m for \(i\in \{1, 2,\ldots,n\}\), \(j\in \{0,1,\ldots, C_{n}^{i}-1\}\), then for \(i=1,2,\ldots,n-1\),
By ( 2.16 ) and using Lemma 3, we obtain \(e^{P_{n,0}}\equiv 0\), which is a contradiction.
Case 1.2. \(\deg \alpha (z)< m\). Then, for any \(1\le i \le n\), \(1\le k< j \le n\), we have
Since \(P_{i,j}(z)\) are polynomials with degree less than m for \(i\in \{1, 2,\ldots,n\}\), \(j\in \{0,1,\ldots, C_{n}^{i}-1\}\), then for \(i=1,2,\ldots,n-1\),
By ( 2.16 ) and using Lemma 3, we obtain \(e^{P_{n,0}}\equiv 0\), which is a contradiction.
Case 1.3. \(\deg \alpha (z)= m\). Set \(\alpha (z)=dz^{m}+\alpha^{*}(z)\), where \(d\neq 0\) and \(\deg \alpha^{*}(z)< m\). Rewrite (2.16) as
If \(d\neq jl_{m}\), for any \(j=1,2,\ldots,n\), we have
Combining this with (2.17), by using Lemma 3, we get a contradiction.
If \(d= jl_{m}\), for some \(j=1,2,\ldots,n-1\), without loss of generality, we assume that \(j=1\), then (2.17) can be rewritten as
And then, by using the same argument as above, we get a contradiction.
Hence, \(d= nl_{m}\). Rewrite (2.17) as
Using the same argument as in Case 1.1 and using Lemma 3, we obtain
Then, by (2.9) and (2.18), we get
If \(m\ge 2\), then for any \(t=0,1,\ldots,n-1\), we have
where \(q_{t}(z)\) are polynomials with \(\deg q_{t}(z)< m-1\).
From (2.19) with (2.20), we get
Using the same argument as Case 1.1, we obtain a contradiction.
Hence \(m=1\). Thus \(\beta (z)=l_{1}z+l_{0}\), where \(l_{1}\neq 0\). Then, for any \(n\ge 1\), we deduce that
Hence, it follows from (2.19) that \((e^{l_{1}c}-1)^{n-1}\equiv 0\), which yields \(e^{l_{1}c}=1\). Then, for any \(t\in \mathbb{N^{+}}\), we have
By the second equation in (2.2) and (2.21), we get
By mathematical induction, it is easy to get, for any integer \(t\ge 2\),
where \(b_{1}(z)=p(z)\), \(b_{t}(z)=p(z)e^{(t-1)\beta (z)}+\Delta_{c}b _{t-1}=\sum_{i=0}^{t-1}e^{(t-1-i)\beta (z)}\Delta_{c}^{i}p(z)\).
Hence,
where \(b_{n}(z)=\sum_{i=0}^{n-1}e^{(n-1-i)\beta (z)}\Delta_{c}^{i}p(z)\).
From the first equation in (2.2) and (2.22), we have
If \(e^{\alpha (z)}-e^{n\beta (z)}\not \equiv 0\), then by ( 2.15 ), ( 2.23 ), Property 1 , and Property 2 , we have
which is a contradiction.
Hence \(e^{\alpha (z)}-e^{n\beta (z)}\equiv 0\). It follows from (2.22) and (2.23) that
That is,
If \(p(z)\) is a constant, as \(\Delta_{c}^{i}p(z)=0\) for any \(i\in \mathbb{N^{+}}\). It follows from (2.24) that
Hence, \(e^{(n-1)\beta (z)}\equiv 1\), which is a contradiction.
If \(p(z)\) is a nonconstant polynomial, then \(p(z)-\Delta_{c}^{i}p(z) \not \equiv 0\) for any \(i\in \mathbb{N^{+}}\). It follows from (2.24) that
Thus we have
a contradiction.
Case 2. \(\beta (z)=\beta \in \mathbb{C}\) is a constant. By the second equation in (2.2), we get
By mathematical induction, it is easy to get, for any integer \(t\ge 2\),
where \(b_{2}(z)=\Delta_{c}p(z)\), \(b_{t}(z)=\Delta_{c}p(z)e^{(t-1) \beta }+\Delta_{c}b_{t-1}=\sum_{i=1}^{t-1}\Delta_{c}^{i}p(z)e^{(t-1-i) \beta }\).
Hence,
where \(b_{n}(z)=\sum_{i=1}^{n-1}\Delta_{c}^{i}p(z)e^{(n-1-i)\beta }\).
Using the same argument as the above, it is easy to get \(e^{\alpha }=e ^{n\beta }\). Then it follows from (2.2) and \(e^{\alpha }=e^{n\beta }\) that
If \(\Delta_{c}f(z) \not \equiv \Delta_{c}^{n}f(z)\), it follows from (2.26) that \(e^{(n-1)\beta }\neq 1\). Combining (2.25) and (2.26), we have
If \(p(z)\) is a constant, then the left-hand side of equation (2.27) is equal to 0, and hence \(p(z)\equiv 0\), which is a contradiction.
If \(p(z)\) is a nonconstant polynomial, let \(d=\deg p(z)\ge 1\), then the left-hand side of equation (2.27) is a polynomial with degree less than d, but the right-hand side of the equation is a polynomial with degree d, which is a contradiction.
Hence \(\Delta_{c}f(z)\equiv \Delta_{c}^{n}f(z)\), and \(e^{(n-1)\beta }= 1\).
If \(e^{\beta }\neq 1\) and \(p(z)\) is a nonconstant polynomial, then it follows from (2.25)–(2.26) that \(b_{n}(z)\equiv 0\). Thus
Let \(p(z)=a_{m}z^{m}+a_{m-1}z^{m-1}+\cdots +a_{0}\). It follows that \(\deg \Delta_{c}^{i}p(z)=m-i\) if \(m\ge i\). If \(m\ge 2\), then the left-hand side of (2.28) is a polynomial with degree \(m-1\ge 1\), which is a contradiction.
Hence \(m=1\), that is, \(p(z)=a_{1}z+a_{0}\). Thus \(\Delta_{c}p(z)=a_{1}c \neq 0\). It follows from (2.28) that \(a_{1}ce^{(n-2)\beta }=0\), which is a contradiction.
From the above discussion, we obtain that if \(e^{\beta }\neq 1\), then \(p(z)\) (≡b) is a nonzero constant, hence
where \(t=e^{\beta }\) satisfying \(t^{n-1}=1\).
Thus, Lemma 5 is proved. □
Lemma 6
(Hadamard’s factorization theorem [18])
Let f be an entire function of finite order \(\rho (f)\) with zeros \(\{z_{1}, z_{2},\ldots \}\subset \mathbb{C}\backslash \{0\}\) and a k-fold zero at the origin. Then
where α is the canonical product of f formed with the non-null zeros of f, and β is a polynomial of degree \(\le \rho (f)\).
3 Proof of Theorem 5
Proof
Since the order of f is finite, and f, \(\Delta_{c}f\), \(\Delta_{c}^{n}f\) share ∞ and \(p(z)\) CM, obviously \((\Delta _{c}^{n}f(z)-p(z))/(f(z)-p(z))\) and \((\Delta_{c}f(z)-p(z))/(f(z)-p(z))\) have no zeros and poles. By Lemmas 1 and 6 , we have
where \(\alpha (z)\) and \(\beta (z)\) are two polynomials with degree \(\le \rho (f)\).
Using the same discussion as in Lemma 5, we deduce that f cannot be a rational function. Hence, f is a transcendental meromorphic function, and \(T(r,p)=S(r,f)\).
Set \(F(z):=f(z)-p(z)\), then \(T(r,f)=T(r,F)+S(r,f)\) and \(T(r,p)=S(r,F)\).
Obviously, we have
Rewrite (3.1) as
Since \(p(z)\) is a nonconstant polynomial, it follows that \(\Delta_{c}^{n}p(z)-p(z)\not \equiv 0\) and \(\Delta_{c}p(z)-p(z) \not \equiv 0\). Set
Next, we consider two cases.
Case 1. \(\phi (z)\not \equiv 0\). Then, by \(T(r,p)=S(r,F)\), Lemma 1, and Lemma 2, we get
By (3.2)–(3.3), we can rewrite \(\phi (z)\) as
Since \(p(z)\) is a polynomial, we deduce that \(N(r,\phi )=S(r,F)\). Hence, we get
Since \(\phi (z)\not \equiv 0\), by (3.5) we have
Then by (3.6), (3.7), \(T(r,p)=S(r,F)\), Property 2, and Property 3, we have
Hence by ( 3.6 ), Property 1 , and the previous inequality, we get
Hence, by Lemma 5 and since \(p(z)\) is a nonconstant polynomial, we obtain \(f\equiv \Delta_{c}{f}\).
Case 2. \(\phi (z)\equiv 0\). That is,
By simple calculation, we can rewrite (3.10) as follows:
Since \(p(z)\) is a polynomial, it follows from (3.12) that \(e^{\alpha (z)-\beta (z)}\) is a constant. Suppose that \(e^{\alpha (z)- \beta (z)}=A\), then we get \(p(z)-\Delta_{c}^{n}p(z)=A(p(z)-\Delta_{c}p(z))\). It follows that \(A=1\) and \(p(z)\) is a constant, which is a contradiction.
This completes the proof of Theorem 5 . □
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Acknowledgements
Research supported by the NNSF of China (Grant No. 11371149; 11701188) and the Graduate Student Overseas Study Program from South China Agricultural University (Grant No. 2017LHPY003).
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Deng, B., Liu, D., Gu, Y. et al. Meromorphic functions that share a polynomial with their difference operators. Adv Differ Equ 2018, 194 (2018). https://doi.org/10.1186/s13662-018-1645-4
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DOI: https://doi.org/10.1186/s13662-018-1645-4