Notes on the uniqueness and value sharing for meromorphic functions concerning differential polynomials
We study the problem of uniqueness of meromorphic functions concerning differential polynomials, and obtain some results. These results improve the results obtained earlier by Li [J. Sichuan Univ. (Natural Science Edition), 45, 21–24 (2008)] and Dyavanal [J. Math. Anal. Appl., 374, 335–345 (2011)]....
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irk-123456789-1661182020-02-19T01:27:10Z Notes on the uniqueness and value sharing for meromorphic functions concerning differential polynomials Li, Jin-Dong Статті We study the problem of uniqueness of meromorphic functions concerning differential polynomials, and obtain some results. These results improve the results obtained earlier by Li [J. Sichuan Univ. (Natural Science Edition), 45, 21–24 (2008)] and Dyavanal [J. Math. Anal. Appl., 374, 335–345 (2011)]. Вивчається проблема єдиності мероморфних Функцій щодо диференціальних поліномів, отримано дєякі результати. Ці результати покращують результати, що отримані раніше в роботах Лі [J. Sichuan Univ. (Natural Science Edition). -2008. - 45. - P. 21 -24] та Д'яванала [J. Math. Anal. and Appl. - 2011. - 374. - P. 335-345]. 2014 Article Notes on the uniqueness and value sharing for meromorphic functions concerning differential polynomials / Jin-Dong Li // Український математичний журнал. — 2014. — Т. 66, № 10. — С. 1357–1366. — Бібліогр.: 9 назв. — англ. 1027-3190 http://dspace.nbuv.gov.ua/handle/123456789/166118 517.9 en Український математичний журнал Інститут математики НАН України |
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Статті Статті Li, Jin-Dong Notes on the uniqueness and value sharing for meromorphic functions concerning differential polynomials Український математичний журнал |
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We study the problem of uniqueness of meromorphic functions concerning differential polynomials, and obtain some results. These results improve the results obtained earlier by Li [J. Sichuan Univ. (Natural Science Edition), 45, 21–24 (2008)] and Dyavanal [J. Math. Anal. Appl., 374, 335–345 (2011)]. |
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Notes on the uniqueness and value sharing for meromorphic functions concerning differential polynomials |
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Notes on the uniqueness and value sharing for meromorphic functions concerning differential polynomials |
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Notes on the uniqueness and value sharing for meromorphic functions concerning differential polynomials |
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notes on the uniqueness and value sharing for meromorphic functions concerning differential polynomials |
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Інститут математики НАН України |
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Notes on the uniqueness and value sharing for meromorphic functions concerning differential polynomials / Jin-Dong Li // Український математичний журнал. — 2014. — Т. 66, № 10. — С. 1357–1366. — Бібліогр.: 9 назв. — англ. |
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Український математичний журнал |
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AT lijindong notesontheuniquenessandvaluesharingformeromorphicfunctionsconcerningdifferentialpolynomials |
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2025-07-14T20:46:59Z |
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UDC 517.9
Jin-Dong Li (College Management Sci., Chengdu Univ. Technology, China)
NOTES ON UNIQUENESS AND VALUE SHARING OF MEROMORPHIC
FUNCTIONS CONCERNING DIFFERENTIAL POLYNOMIALS*
ПРО ЄДИНIСТЬ ТА ПОДIЛ ЗНАЧЕНЬ ДЛЯ МЕРОМОРФНИХ ФУНКЦIЙ
ЩОДО ДИФЕРЕНЦIАЛЬНИХ ПОЛIНОМIВ
We study the problem of uniqueness of meromorphic functions concerning differential polynomials, and obtain some
results. The results improve earlier results by Li [J. Sichuan Univ. (Natural Science Edition). – 2008. – 45. – P. 21 – 24]
and Dyavanal [J. Math. Anal. and Appl. – 2011. – 374. – P. 335 – 345].
Вивчається проблема єдиностi мероморфних функцiй щодо диференцiальних полiномiв, отримано деякi результати.
Цi результати покращують результати, що отриманi ранiше в роботах Лi [J. Sichuan Univ. (Natural Science Edition). –
2008. – 45. – P. 21 – 24] та Д’яванала [J. Math. Anal. and Appl. – 2011. – 374. – P. 335 – 345].
1. Introduction and results. Let f be a nonconstant meromorphic function defined in the whole
complex plane. It is assumed that the reader is familiar with the notations of the Nevanlinna theory
such as T (r, f), m(r, f), N(r, f), S(r, f) and so on, and these can be found, for instance in [5, 7].
Let f and g be two nonconstant meromorphic functions. If for a ∈ C = C∪{∞}, f−a and g−a
have the same set of zeros with the same multiplicities we say that f and g share the value a CM
(counting multiplicities), and if we do not consider the multiplicities then f and g are said to share
the value a IM (ignoring multiplicities). When f and g share the value 1 IM, Let z0 be a 1-points of
f of order p, a 1-points of g of order q, we denote by N11
(
r,
1
f − 1
)
the counting function of those
1-points of f and g where p = q = 1 and NL
(
r,
1
f − 1
)
is the counting function of those 1-points
of both f and g where p > q. In the same way, we can define N11
(
r,
1
g − 1
)
and NL
(
r,
1
g − 1
)
.
For any constant a, we define
Θ(a, f) = 1− lim
r→∞
N
(
r,
1
f − a
)
T (r, f)
.
Let f be a nonconstant meromorphic function. Let a be a finite complex number, and k be a
positive integer, we denote by Nk)
(
r,
1
f − a
) (
or Nk)
(
r,
1
f − a
))
the counting function for zeros
of f − a with multiplicity ≤ k (ignoring multiplicities), and by N(k
(
r,
1
f − a
) (
or N (k
(
r,
1
f − a
))
the counting function for zeros of f − a with multiplicity at least k (ignoring multiplicities). Set
Nk
(
r,
1
f − a
)
= N
(
r,
1
f − a
)
+N (2
(
r,
1
f − a
)
+ . . .+N (k
(
r,
1
f − a
)
.
We further define
* This work was supported by the Opening Fund of Geomathematics Key Laboratory of Sichuan Province of China
(No.scsxdz2011008).
c© JIN-DONG LI, 2014
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 10 1357
1358 JIN-DONG LI
δk(a, f) = 1− lim
r→∞
Nk
(
r,
1
f − a
)
T (r, f)
.
Hayman [2] and Clunie [1] proved the following result.
Theorem A. Let f(z) be a transcendental entire function, n ≥ 1 be a positive integer, then
fnf ′ = 1 has infinitely many solutions.
In 1997, Yang and Hua [6] obtained a unicity theorem corresponding to the above result and
proved the following result.
Theorem B. Let f(z) and g(z) be two transcendental entire functions, n ≥ 6 be a positive
integer. If fnf ′ and gng′ share 1 CM, then either f = tg for a constant t such that tn+1 = 1, or
f(z) = c2e
−cz, g(z) = c1e
cz, where c, c1 and c2 are three constants satisfying (c1c2)
n+1c2 = −1.
Theorem C. Let f(z) and g(z) be two nonconstant meromorphic functions, n ≥ 11 be a positive
integer. If fnf ′ and gng′ share 1 CM, then either f = tg for a constant t such that tn+1 = 1, or
f(z) = c2e
−cz, g(z) = c1e
cz, where c, c1 and c2 are three constants satisfying (c1c2)
n+1c2 = −1.
Recently, R. S. Dyavanal [4] improve above results and obtain the following results.
Theorem D. Let f(z) and g(z) be two nonconstant meromorphic functions, whose zeros and
poles are of multiplicities at least s, where s is a positive integer. Let n ≥ 2 be a positive integer
satisfying (n + 1)s ≥ 12. If fnf ′ and gng′ share 1 CM, then either f = tg for a constant t such
that tn+1 = 1, or f(z) = c2e
−cz, g(z) = c1e
cz, where c, c1 and c2 are three constants satisfying
(c1c2)
n+1c2 = −1.
Theorem E. Let f(z) and g(z) be two transcendental entire functions, whose zeros are of
multiplicities at least s, where s is a positive integer. Let n be a positive integer satisfying (n+1)s ≥
≥ 7. If fnf ′ and gng′ share 1 CM, then either f = tg for a constant t such that tn+1 = 1, or
f(z) = c2e
−cz, g(z) = c1e
cz, where c, c1 and c2 are three constants satisfying (c1c2)
n+1c2 = −1.
Remark 1.1. If s = 1 in Theorem D and Theorem E, respectively, then Theorem D and Theo-
rem E reduces to Theorem B and Theorem C, respectively.
Naturally, one can pose the following question: what can be stated if CM is replaced with IM in
the above results.
In 2008, Li [3] prove the following result.
Theorem F. Let f(z) and g(z) be two nonconstant meromorphic functions, n ≥ 23 be a positive
integer. If fnf ′ and gng′ share 1 IM, then either f = tg for a constant t such that tn+1 = 1, or
f(z) = c2e
−cz, g(z) = c1e
cz, where c, c1 and c2 are three constants satisfying (c1c2)
n+1c2 = −1.
In this paper, we shall generalize and improve the above the results and obtain the following two
theorems.
Theorem 1.1. Let f(z) and g(z) be two nonconstant meromorphic functions, whose zeros and
poles are of multiplicities at least s, where s is a positive integer. Let n ≥ 2 be a positive integer
satisfying (n + 1)s ≥ 24. If fnf ′ and gng′ share 1 IM, then either f = tg for a constant t such
that tn+1 = 1, or f(z) = c2e
−cz, g(z) = c1e
cz, where c, c1 and c2 are three constants satisfying
(c1c2)
n+1c2 = −1.
Remark 1.2. If s = 1 in Theorem 1.1, then Theorem 1.1 improves Theorem F.
Remark 1.3. Giving specific values for s in Theorem 1.1, we can get the following interesting
cases:
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 10
NOTES ON UNIQUENESS AND VALUE SHARING OF MEROMORPHIC FUNCTIONS. . . 1359
(i) if s = 1, then n ≥ 23,
(ii) if s = 2, then n ≥ 11,
(iii) if s = 3, then n ≥ 7,
(iv) if s ≥ 4, then n ≥ 5.
We can conclude that f and g have zeros and poles of higher order multiplicity, then we can
reduce the value of n.
Theorem 1.2. Let f(z) and g(z) be two transcendental entire functions, whose zeros are of
multiplicities at least s, where s is a positive integer. Let n be a positive integer satisfying (n+1)s ≥
≥ 13. If fnf ′ and gng′ share 1 IM, then either f = tg for a constant t such that tn+1 = 1, or
f(z) = c2e
−cz, g(z) = c1e
cz, where c, c1 and c2 are three constants satisfying (c1c2)
n+1c2 = −1.
If s = 1 in Theorem 1.2, then Theorem 1.2 reduces to the following result.
Corollary 1.1. Let f(z) and g(z) be two transcendental entire functions, and let n be a positive
integer satisfying n ≥ 12. If fnf ′ and gng′ share 1 IM, then either f = tg for a constant t such
that tn+1 = 1, or f(z) = c2e
−cz, g(z) = c1e
cz, where c, c1 and c2 are three constants satisfying
(c1c2)
n+1c2 = −1.
2. Some lemmas. For the proof of our result, we need the following lemmas.
Lemma 2.1 (see [2]). Let f be nonconstant meromorphic function, a0, a1, . . . , an be finite com-
plex numbers such that an 6= 0.Then
T (r, anf
n + an−1f
n−1 + . . .+ a0) = nT (r, f) + S(r, f).
Lemma 2.2 (see [2]). Let f(z) be a nonconstant meromorphic function, k be a positive integer,
and let c be a nonzero finite complex number. Then
T (r, f) ≤ N(r, f) +N
(
r,
1
f
)
+N
(
r,
1
f (k) − c
)
−N
(
r,
1
f (k+1)
)
+ S(r, f) ≤
≤ N(r, f) +Nk+1
(
r,
1
f
)
+N
(
r,
1
f (k) − c
)
−N0
(
r,
1
f (k+1)
)
+ S(r, f).
Here N0
(
r,
1
f (k+1)
)
is the counting function which only counts those points such that f (k+1) = 0
but f
(
f (k) − c
)
6= 0.
Lemma 2.3 (see [3]). Let f(z) be a transcendental meromorphic function, and let a1(z), a2(z)
be two meromorphic functions such that T (r, ai) = S(r, f), i = 1, 2. Then
T (r, f) ≤ N(r, f) +N
(
r,
1
f − a1
)
+N
(
r,
1
f − a2
)
+ S(r, f).
Lemma 2.4 (see [8]). Let f be a nonconstant meromorphic function, k, p be two positve integers,
then
Np
(
r,
1
f (k)
)
≤ Np+k
(
r,
1
f
)
+ kN(r, f) + S(r, f) ≤
≤ (p+ k)N
(
r,
1
f
)
+ kN(r, f) + S(r, f).
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 10
1360 JIN-DONG LI
Clearly N
(
r,
1
f (k)
)
= N1
(
r,
1
f (k)
)
.
Lemma 2.5. Let f(z) and g(z) be two meromorphic functions, and let k be a positive integer.
If f (k) and g(k) share the value 1 IM and
∆ = (2k + 3)Θ(∞, f) + (2k + 4)Θ(∞, g) + (k + 2)Θ(0, f) + (2k + 3)Θ(0, g)+
+δk+1(0, f) + δk+1(0, g) > 7k + 13, (2.1)
then either f (k)g(k) ≡ 1 or f ≡ g.
Proof. Let
h(z) =
f (k+2)(z)
f (k+1)(z)
− 2
f (k+1)(z)
f (k)(z)− 1
− g(k+2)(z)
g(k+1)(z)
+ 2
g(k+1)(z)
g(k)(z)− 1
. (2.2)
If z0 is a common simple 1-point of f (k) and g(k), substituting their Taylor series at z0 into (2.2),
we see that z0 is a zero of h(z). Thus, we have
N11
(
r,
1
f (k) − 1
)
= N11
(
r,
1
g(k) − 1
)
≤ N
(
r,
1
h
)
≤ T (r, h) +O(1) ≤
≤ N(r, h) + S(r, f) + S(r, g). (2.3)
By our assumptions, h(z) have poles only at zeros of f (k+1) and g(k+1) and poles of f and
g, and those 1-points of f (k) and g(k) whose multiplicities are distinct from the multiplicities of
corresponding 1-points of g(k) and f (k) respectively. Thus, we deduce from (2.2) that
N(r, h) ≤ N(r, f) +N(r, g) +N
(
r,
1
f
)
+N
(
r,
1
g
)
+N0
(
r,
1
f (k+1)
)
+
+N0
(
r,
1
g(k+1)
)
+NL
(
r,
1
f (k) − 1
)
+NL
(
r,
1
g(k) − 1
)
(2.4)
here N0
(
r,
1
f (k+1)
)
has the same meaning as in Lemma 2.2.
By Lemma 2.2, we have
T (r, f) ≤ N(r, f) +Nk+1
(
r,
1
f
)
+N
(
r,
1
f (k) − c
)
−N0
(
r,
1
f (k+1)
)
+ S(r, f), (2.5)
T (r, g) ≤ N(r, g) +Nk+1
(
r,
1
g
)
+N
(
r,
1
g(k) − c
)
−N0
(
r,
1
g(k+1)
)
+ S(r, g). (2.6)
Since f (k) and g(k) share the value 1 IM, we obtain
N
(
r,
1
f (k) − 1
)
+N
(
r,
1
g(k) − 1
)
≤
≤ N11
(
r,
1
f (k) − 1
)
+NL
(
r,
1
g(k) − 1
)
+N
(
r,
1
f (k) − 1
)
≤
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 10
NOTES ON UNIQUENESS AND VALUE SHARING OF MEROMORPHIC FUNCTIONS. . . 1361
≤ N11
(
r,
1
f (k) − 1
)
+NL
(
r,
1
g(k) − 1
)
+ T (r, f (k)) +O(1) ≤
≤ N11
(
r,
1
f (k) − 1
)
+NL
(
r,
1
g(k) − 1
)
+m(r, f)+
+m
(
r,
f (k)
f
)
+N(r, f) + kN(r, f) + S(r, f) ≤
≤ N11
(
r,
1
f (k) − 1
)
+NL
(
r,
1
g(k) − 1
)
+ T (r, f) + kN(r, f) + S(r, f). (2.7)
Noting that, by Lemma 2.4, we get
N
(
r,
1
f (k)
)
= N1
(
r,
1
f (k)
)
≤ N1+k
(
r,
1
f
)
+ kN(r, f) + S(r, f) ≤
≤ (k + 1)N
(
r,
1
f
)
+ kN(r, f) + S(r, f),
(2.8)
NL
(
r,
1
f (k) − 1
)
≤ N
(
r,
1
f (k) − 1
)
−N
(
r,
1
f (k) − 1
)
≤ N
(
r,
f (k)
f (k+1)
)
≤
≤ N
(
r,
f (k+1)
f (k)
)
+ S(r, f) ≤ N(r, f) +N
(
r,
1
f (k)
)
+ S(r, f).
So, we have
NL
(
r,
1
f (k) − 1
)
≤ (k + 1)N(r, f) + (k + 1)N
(
r,
1
f
)
+ S(r, f). (2.9)
Similarly
NL
(
r,
1
g(k) − 1
)
≤ (k + 1)N(r, g) + (k + 1)N
(
r,
1
g
)
+ S(r, g). (2.10)
We obtain from (2.3) – (2.10) that
T (r, g) ≤ (2k + 3)N(r, f) + (2k + 4)N(r, g) + (k + 2)N
(
r,
1
f
)
+
+(2k + 3)N
(
r,
1
g
)
+Nk+1
(
r,
1
f
)
+Nk+1
(
r,
1
g
)
+ S(r, f) + S(r, g).
Without loss of generality, we suppose that there exists a set I with infinite measure such that
T (r, f) ≤ T (r, g) for r ∈ I. Hence
T (r, g) ≤
{[
(7k + 14)− (2k + 3)Θ(∞, f)− (2k + 4)Θ(∞, g)−
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 10
1362 JIN-DONG LI
−(k + 2)Θ(0, f)− (2k + 3)Θ(0, g)− δk+1(0, f)− δk+1(0, g)
]
+ ε
}
T (r, g) + S(r, g) (2.11)
for r ∈ I and 0 < ε < ∆− (7k + 13). Thus, we obtain from (2.1) and (2.11) that T (r, g) ≤ S(r, g)
for r ∈ I, a contradiction.
Hence, we get h(z) ≡ 0; that is
f (k+2)(z)
f (k+1)(z)
− 2
f (k+1)(z)
f (k)(z)− 1
=
g(k+2)(z)
g(k+1)(z)
− 2
g(k+1)(z)
g(k)(z)− 1
.
By solving this equation, we obtain
1
f (k) − 1
=
bg(k) + a− b
g(k) − 1
. (2.12)
Where a, b are two constants. Next, we consider three cases.
Case 1. b 6= 0 and a = b.
Subcase 1.1. b = −1. Then we deduce from (2.12) that f (k)(z)g(k)(z) ≡ 1.
Subcase 1.2. b 6= −1. Then we get from (2.12) that
1
f (k)
=
bg(k)
(1 + b)g(k) − 1
so
N
r, 1
g(k) − 1
1 + b
≤ N (r, 1
f (k)
)
. (2.13)
From (2.13) and (2.8), we get
N
r, 1
g(k) − 1
1 + b
≤ (k + 1)N
(
r,
1
f
)
+ kN(r, f) + S(r, f).
By Lemma 2.2, we have
T (r, g) ≤ N(r, g) +Nk+1
(
r,
1
g
)
+N
r, 1
g(k) − 1
b+ 1
−N0
(
r,
1
g(k+1)
)
≤
≤ N(r, g) +Nk+1
(
r,
1
g
)
+ kN(r, f) + (k + 1)N
(
r,
1
f
)
+ S(r, f) + S(r, g) ≤
≤ (2k + 3)N(r, f) + (2k + 4)N(r, g) + (k + 2)N
(
r,
1
f
)
+
+(2k + 3)N
(
r,
1
g
)
+Nk+1
(
r,
1
f
)
+Nk+1
(
r,
1
g
)
+ S(r, f) + S(r, g).
That is T (r, g) ≤ (7k + 14−∆)T (r, g) + S(r, g) for r ∈ I.
Thus, by (2.1), we obtain that T (r, g) ≤ S(r, g) for r ∈ I, a contradiction.
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NOTES ON UNIQUENESS AND VALUE SHARING OF MEROMORPHIC FUNCTIONS. . . 1363
Case 2. b 6= 0 and a 6= b.
Subcase 2.1. b = −1. Then we obtain from (2.12) that
f (k) =
a
−g(k) + a+ 1
.
Therefore
N
(
r,
a
−g(k) + a+ 1
)
= N(r, f (k)) = N(r, f).
By Lemma 2.2, we have
T (r, g) ≤ N(r, g) +Nk+1
(
r,
1
g
)
+N
(
r,
1
g(k) − (a+ 1)
)
−N0
(
r,
1
g(k+1)
)
+ S(r, g) ≤
≤ N(r, g) +Nk+1
(
r,
1
g
)
+N(r, f) + S(r, f) + S(r, g) ≤
≤ (2k + 3)N(r, f) + (2k + 4)N(r, g) + (k + 2)N
(
r,
1
f
)
+
+(2k + 3)N
(
r,
1
g
)
+Nk+1
(
r,
1
f
)
+Nk+1
(
r,
1
g
)
+ S(r, f) + S(r, g).
Using the argument as in case 1, we get a contradiction.
Subcase 2.2. b 6= −1. Then we get from (2.12) that
f (k) −
(
1 +
1
b
)
=
−a
b2
(
g(k) +
a− b
b
).
Therefore
N
r, 1
g(k) +
a− b
b
= N
(
r, f (k) −
(
1 +
1
b
))
= N(r, f).
By Lemma 2.2, we get
T (r, g) ≤ N(r, g) +Nk+1
(
r,
1
g
)
+N
r, 1
g(k) +
a− b
b
−N0
(
r,
1
g(k+1)
)
+ S(r, g) ≤
≤ N(r, g) +Nk+1
(
r,
1
g
)
+N(r, f) + S(r, f) + S(r, g) ≤
≤ (2k + 3)N(r, f) + (2k + 4)N(r, g) + (k + 2)N
(
r,
1
f
)
+
+(2k + 3)N
(
r,
1
g
)
+Nk+1
(
r,
1
f
)
+Nk+1
(
r,
1
g
)
+ S(r, f) + S(r, g).
Using the argument as in case 1, we get a contradiction.
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1364 JIN-DONG LI
Case 3. b = 0. From (2.12), we obtain
f =
1
a
g + P (z), (2.14)
where P (z) is a polynomial. If P (z) 6= 0, then by Lemma 2.3, we have
T (r, f) ≤ N(r, f) +N
(
r,
1
f
)
+N
(
r,
1
f − P
)
+ S(r, f) ≤
≤ N(r, f) +N
(
r,
1
f
)
+N
(
r,
1
g
)
+ S(r, f). (2.15)
From (2.14), we obtain T (r, f) = T (r, g) + S(r, f). Hence, substituting this into (15), we get
T (r, f) ≤
{
3−
[
Θ(∞, f) + Θ(0, f) + Θ(0, g)
]
+ ε
}
T (r, f) + S(r, f),
where
0 < ε < 1− δk+1(0, f) + 1− δk+1(0, g) + (2k + 2)
[
1−Θ(∞, f)
]
+
+(2k + 4)
[
1−Θ(∞, g)
]
+
[
1−Θ(0, f)
]
+ 2
[
1−Θ(0, g)
]
.
Therefore T (r, f) ≤ [7k + 14−∆]T (r, f) + S(r, f).
That is
[
∆− (7k + 13)
]
T (r, f) < S(r, f).
Hence, by (2.1), we deduce that T (r, f) ≤ S(r, f) for r ∈ I, a contradiction.
Therefore, we deduce that P (z) ≡ 0, that is
f =
1
a
g. (2.16)
If a 6= 1, then f (k) and g(k) sharing the value 1 IM, we deduce from (2.16) that g(k) 6= 1. That is
N
(
r,
1
g(k) − 1
)
= 0.
Next, we can deduce a contradiction as in case 1. Thus, we get that a = 1, that is f ≡ g.
Lemma 2.5 is proved.
Lemma 2.6 (see [9]). Let f and g be two nonconstant entire functions, n ≥ 1. If fnf ′gng′ = 1,
then f(z) = c2e
−cz, g(z) = c1e
cz, where c, c1 and c2 are three constants satisfying (c1c2)
n+1c2 =
= −1.
3. Proof of Theorem 1.1. Let F =
fn+1
n+ 1
and G =
gn+1
n+ 1
. Then F ′ = fnf ′ and G′ = gng′
share the value 1 IM.
Consider
N
(
r,
1
F
)
= N
(
r,
1
fn+1
)
≤ 1
s(n+ 1)
N
(
r,
1
F
)
≤ 1
s(n+ 1)
[
T
(
r,
1
F
)
+O(1)
]
.
Therefore
Θ(0, F ) = 1− lim
r→∞
N
(
r,
1
F
)
T (r, F )
≥ 1− 1
s(n+ 1)
, (3.1)
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NOTES ON UNIQUENESS AND VALUE SHARING OF MEROMORPHIC FUNCTIONS. . . 1365
δk+1(0, F ) = 1− lim
r→∞
Nk+1
(
r,
1
F
)
T (r, F )
≥ 1− lim
r→∞
(k + 1)N
(
r,
1
F
)
T (r, F )
≥ 1− k + 1
s(n+ 1)
. (3.2)
Similarly
Θ(0, G) ≥ 1− 1
s(n+ 1)
, (3.3)
Θ(∞, F ) ≥ 1− 1
s(n+ 1)
, (3.4)
Θ(∞, G) ≥ 1− 1
s(n+ 1)
, (3.5)
δk+1(0, G) ≥ 1− k + 1
s(n+ 1)
. (3.6)
From (3.1) – (3.6), we get
∆ = (2k + 3)Θ(∞, f) + (2k + 4)Θ(∞, g) + (k + 2)Θ(0, f) + (2k + 3)Θ(0, g)+
+δk+1(0, f) + δk+1(0, g) ≥ 7k + 14− 9k + 14
s(n+ 1)
. (3.7)
Since s(n+ 1) ≥ 24, for k = 1, we obtain ∆ > 20 from (3.7). Hence by Lemma 2.5, we get either
F ′G′ ≡ 1 or F ≡ G.
Consider the case F ′G′ ≡ 1, that is
fnf ′gng′ ≡ 1. (3.8)
Suppose that f has a pole z0 (with order p ≥ s say). Then z0 is a zero of g (with order m ≥ s
say). By (3.8), we get
nm+m− 1 = np+ p+ 1.
That is, (m − p)(n + 1) = 2, which is impossible since n ≥ 2 and m, p are positive integers.
Therefore, we conclude that f and g are entire functions. From Lemma 2.6, we get f(z) = c2e
−cz,
g(z) = c1e
cz, where c, c1 and c2 are three constants satisfying (c1c2)
n+1c2 = −1.
Next we consider another case F ≡ G. This gives fn+1 = gn+1. So f = tg for a constant t such
that tn+1 = 1.
Theorem 1.1 is proved.
4. Proof of Theorem 1.2. Since f and g are entire functions, we have N(r, f) = N(r, g) = 0.
Proceeding as in the proof Theorem 1.1 and applying Lemma 2.5 we shall obtain that Theorem 1.2
holds.
5. One open question.
Question 1. Can the condition (n+ 1)s ≥ 24 in Theorem 1.1 be further relaxed?
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1366 JIN-DONG LI
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Received 11.10.12
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