RESEARCH Open Access
Some results on difference polynomials sharing
values
Yong Liu
1,2
, XiaoGuang Qi
3*
and Hongxun Yi
1
* Correspondence:
[email protected]
3
Department of Mathematics, Jinan
University, Jinan 250022,
Shandong, P. R. China
Full list of author information is
available at the end of the article
Abstract
This article is devoted to studying uniqueness of difference polynomials sharing
values. The results improve those given by Liu and Yang and Heittokangas et al.
1 Introduction and main results
In this ar ticle, we shall assume that the reader is familiar with the fundamental results
and the standard notations of the Nevanlinna theory (e.g., see [1-3]). In addition, w e
will use the notations l(f) to denote the exponent of convergence of zero sequences of
meromorphic function f(z); s(f) to denote the order of f (z). We say that meromorphic
functions f and g shareafinitevalueaCMwhen f - a and g - a havethesamezeros
with the same multiplicities. For a non-zero constant c,theforwarddifference

n+1
c
f (z)=

f
share a finite value a CM, n is a positive integer, and c is a fixed constant, then

n
c
f − a
f − a
= τ
for some non-zero constant τ.
Heittokangas et al. [9], prove the following result which is a shifted analogue of
Brück conjecture valid for meromorphic functions.
Theorem B. Let f be a meromorphic function of order of growth s(f) <2, and let c Î
C. If f(z) and f(z + c) share the values a Î C and ∞ CM, then
f (z + c) − a
f (z) − a
= τ
Liu et al. Advances in Difference Equations 2012, 2012:1
http://www.advancesindifferenceequations.com/content/2012/1/1
© 2012 Liu et al; licensee Springer. This is an Open Access article distributed und er the t erms of the Creative Commons Attributi on
License (http://creativecommons.org/licenses/by/ 2.0), which permits unrestricted use, distribution, a nd reproduction in any medium,
provided the original work is properly cited.
for some constant τ.
Here, we also study the shift analo gue of Br ück conjecture, and obtain the results as
follows.
Theorem 1.1. Let f(z) be a non-constant entire function, s(f) <1 or 1 < s(f) <2 and l
(f) < s(f)=s. Set L
1
(f)=a
n
(z) f(z + n)+a

n-1
(z) f(z + n-1) + + a
1
(z) f(z +1)+e
z
f(z),a
j
(z)(1 ≤ j ≤
n) are entire functions with s(a
j
) <1 and a
n
(z) ≢ 0. If f and L
2
(f ) share 0 CM, then
L
2
(f )=h(z)f ,
where h(z) is an entire function of order no less than 1.
Theorem 1.3. Let f(z) be a non-constant entire function, s(f) <1 or 1 < s(f) <2, l(f) <
s(f). Set L
3
(f)=a
n
(z) f(z + n)+a
n-1
(z) f(z + n-1) + + a
1
(z) f(z +1)+a
0

Applying Lemma 2.1 to
1
f
, it is easy to see that for any given ε >0, there is a set E
2
⊂
(1, ∞) of finite logarithmic measure, such that
exp{−r
η+∈
}≤ |f (z)|≤exp{r
η+ε
},
holds for |z|=r ∉ [0, 1] ∪ E
2
, r ® ∞.
Lemma 2.2. [11]Let
Q(z)=b
n
z
n
+ b
n−1
z
n−1
+ ···+ b
0
,
Liu et al. Advances in Difference Equations 2012, 2012:1
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2n
− ε(j =0,1, ,2n − 1).
Then there exists a positive number R = R(ε) such that for |z|=r>R,
Re{Q(z)} >α
n
(1 − ε)sin(nε)r
n
if z Î S
j
where j is even; while
Re{Q(z)} < −α
n
(1 − ε)sin(nε)r
n
if z Î S
j
where j is odd.
Now for any given θ Î [0, 2π),if
θ = −
θ
n
n
+(2j − 1)
π
2n
,(j = 0, 1, , 2n-1), then we
take ε sufficiently small, there is some S
j
, j Î {0, 1, ,2n-1} such that θ Î S
j

,
holds for r ∉ E ∪ [0, 1], where n
z,h
is an integer depending on both z and h, b =max
{s - 2, 2l -2}if l <1 and b = max{s - 2, l - 1} if l ≥ 1 and l = max{l’, l’’}.
Lemma 2.4.[2]Let f(z) be an entire function of order s, then
σ = lim sup
r→∞
log ν(r)
log r
where ν(r) be the central index of f.
Lemma 2.5. [2,13,14]Let f be a transcendental entire function, let
0 <δ<
1
4
and z be
such that |z|=r and that
|f (z)| > M(r, g)ν(r, g)
−
1
4
+δ
holds. Then there exists a set F ⊂ R
+
of finite logarithmic measure, i.e.,

F
dt
t
< ∞

,apointrange
{z
k
= r
k
e
iω
k
}
such that |f(z
k
)| ≥ BM(r
k
, f
), ω
k
Î [0, 2π), lim
k®∞
ω
k
= ω
0
Î [0, 2π), r
k
∉ E ∪ [0, 1], r
k
® ∞, for any given ε >0,
we have
r
σ −ε

2
,
σ −1−α
2
}
, there exists a set E
1
⊂ (1, ∞)of|z|=r of finite
logarithmic measure, so that
f (z + j)
f (z)
= exp

j
f

(z)
f (z)
+ o(r
σ (f)−1−ε
)

, j =1,2, , n
(3:2)
holds for r ∉ E
1
∪ 0[1].
By Lemma 2.5, there exists a set E
2
⊂ (0, ∞) of finite logarithmic measure, such that

2
∪E
3
and
G = {−
ϕ
n
n
+(2j − 1)
π
2n
|j =0,1}∪{
π
2
,
3π
2
}
. By Lemma 2.6, there
exist a positive number
B ∈ [
3
4
,1]
, a point range
{z
k
= r
k
e

σ (f)+ε
k
(3:5)
By (3.1)-(3.3), we have that
a
n
exp

n(1 + o(1))
ν(r
k
, f )
z
k

+ ···+ a
1
exp

(1 + o(1))
ν(r
k
, f )
z
k
} + a
0
= e
Q(z)


0
Î S
0
; (ii) θ
0
Î S
1
.
Case (i). θ
0
Î S
1
.SinceS
j
is an opened set and lim
k®∞
θ
k
= θ
0
,thereisaK>0 such
that θ
k
Î S
j
when k>K. By Lemma 2.2, we have
Re{Q(r
k
e
iθ

, f )
z
k





≤ 3




a
n
exp

n(1 + o(1))
ν(r
k
, f )
z
k

+ ···+ a
1
exp{(1 + o(1))
ν(r
k
, f )




a
n
a
0
exp

n(1 + o(1))
ν(r
k
, f )
z
k

+ ···+
a
1
a
0
exp

(1 + o(1))
ν(r
k
, f )
z
k


k
exp {r
α+ε
k
},
(3:9)
which implies that 1 <0, r ® ∞, a contradiction.
Case (ii). θ
0
Î S
0
.SinceS
0
is an opened set and lim
k®∞
θ
k
= θ
0
,thereisK>0such
that θ
k
Î S
j
when k>K. By Lemma 2.2, we have
Re{Q(r
k
e
iθ
k

|
= |e
Q(z)
|≥e
ηr
k
.
(3:11)
From (3.11), we get that s(f) ≥ 2, a contradiction. Theorem 1.1 is thus proved.
4 Proof of Theorem 1.2
Under the hypothesis of Theorem 1.2, see [3], it is easy to get that
L
2
(f )
f
= e
Q(z)
,
(4:1)
where Q(z) is an entire function. For Q(z), we discuss the following two cases.
Case (1): Q(z) is a polynomial with deg Q = n ≥ 1. Then Theorem 1.2 is proved.
Case (2): Q(z) is a constant. Using the similar reasoning as in the proof of Theore m
1.1, we get that
Liu et al. Advances in Difference Equations 2012, 2012:1
http://www.advancesindifferenceequations.com/content/2012/1/1
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a
n
exp






a
n
exp

n(1 + o(1))
ν(r
k
, f )
z
k

+ ···+ a
1
exp

(1 + o(1))
ν(r
k
, f )
z
k






, f )
z
k
− r
α+ε
k

≤ 2




a
n
exp

n(1 + o(1))
ν(r
k
, f )
z
k

+ ···+ a
1
exp

(1 + o(1))
ν(r
k

then s(F)=s(f). Substituting F(z)=f(z) -p(z) into (5.1), we obtain
a
n
(z)F ( z + n)+a
n−1
(z)F ( z + n − 1) + ···+ a
1
(z)F ( z +1)
F(z)
+ a
0
(z)+
b(z)
F(z)
= e
Q(z)
,
(5:2)
where b(z)=a
n
(z) P(z + n)+ + a
1
(z) P (z +1)+a
0
(z) p(z) is a polynomial. We dis-
cuss the following two cases.
Case 1. Q(z) is a complex constant. Then Theorem 1.3 holds.
Case 2. Q(z) is a polynomial with deg Q = 1. By Lemma 2.3 and l(f) < s(f)=s,for
any given
0 <ε<min{

By Lemma 2.5, there exists a set E
2
⊂ (0, ∞) of finite logarithmic measure, such that
f

(z)
f (z)
=(1+o(1))
ν(r, f )
z
,
(5:4)
holds for |z|=r ∉ E
2
∪ [0, 1], where z is chosen as in Lemma 2.5.
Liu et al. Advances in Difference Equations 2012, 2012:1
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Set E = E
1
∪ E
2
and
G = {−
ϕ
n
n
+(2j − 1)
π
2n

k®∞
θ
k
= θ
0
Î [0, 2π)\G, r
k
∉ E ∪ 0[1], r
k
® ∞, for any given ε
>0, as r
k
® ∞, we have
r
σ (f )−ε
k
<ν(r
k
, f ) < r
σ (f )+ε
k
.
(5:5)
Since F is a transcendental entire function and |f(z
k
)| ≥ BM (r
k
, f), we obtain
b(z
k


+ a
0
+ o(1) = e
Q(z)
.
(5:7)
Using similar p roof as in proof of Theorem 1.1, we ca n get a contradiction. Hence,
Theorem 1.3 holds.
6 Proof of Theorem 1.4
Using similar proof as in proof of Theorem 1.1, we can get Theorem 1.4 holds.
Acknowledgements
The authors thank the referee for his/her valuable suggestions to improve the present article. This research was partly
supported by the NNSF of China (No. 11171184), the NSF of Shangdong Province, China (No. Z2008A01) and
Shandong University graduate student independent innovation fund (yzc11024).
Author details
1
Department of Mathematics, Shandong University, Jinan 250100, Shandong, P. R. China
2
Department of Physics and
Mathematics, Joensuu Campus, University of Eastern Finland, P.O. Box 111, Joensuu FI-80101, Finland
3
Department of
Mathematics, Jinan University, Jinan 250022, Shandong, P. R. China
Author's contributions
YL completed the main part of this article, YL, XQ and HX corrected the main theorems. All authors read and
approved the final manuscript.
Competing interests
The authors declare that they have no competing interests.
Received: 9 June 2011 Accepted: 5 January 2012 Published: 5 January 2012

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