A note on the uniqueness of certain types of differential-difference polynomials
UDC 517.9 We study the uniqueness problems of certain types of differential-difference polynomials sharing a small function.In this paper, we not only solve the open problem occurred in [A. Banerjee, S. Majumder, On the uniqueness of certain types of differential-difference polynomials, Anal. Math.,...
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| author | Majumder, S. Saha, S. Majumder, Sujoy Saha, Somnath Majumder, S. Saha, S. |
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We study the uniqueness problems of certain types of differential-difference polynomials sharing a small function.In this paper, we not only solve the open problem occurred in [A. Banerjee, S. Majumder, On the uniqueness of certain types of differential-difference polynomials, Anal. Math., 43, № 3, 415-444 (2017)], but also present our main results in a more generalized way.
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DOI: 10.37863/umzh.v73i5.379
UDC 517.9
S. Majumder (Raiganj Univ., West Bengal, India),
S. Saha (Mehendipara Jr. High School, Dist.-Dakshin Dinajpur, West Bengal, India)
A NOTE ON THE UNIQUENESS OF CERTAIN TYPES
OF DIFFERENTIAL-DIFFERENCE POLYNOMIALS
ПРО УНIКАЛЬНIСТЬ ДЕЯКИХ ТИПIВ
ДИФЕРЕНЦIАЛЬНО-РIЗНИЦЕВИХ ПОЛIНОМIВ
We study the uniqueness problems of certain types of differential-difference polynomials sharing a small function. In this
paper, we not only solve the open problem occurred in [A. Banerjee, S. Majumder, On the uniqueness of certain types of
differential-difference polynomials, Anal. Math., 43, № 3, 415 – 444 (2017)], but also present our main results in a more
generalized way.
Вивчається можливiсть розв’язання задач єдиностi для деяких типiв диференцiально-рiзницевих полiномiв, якi
мають спiльну малу функцiю. У цiй роботi не лише наведено розв’язок вiдкритої задачi з [A. Banerjee, S. Majumder,
On the uniqueness of certain types of differential-difference polynomials, Anal. Math., 43, № 3, 415 – 444 (2017)], а й
запропоновано бiльш загальний вигляд отриманого основного результату.
1. Introduction, definitions and results. In this paper by meromorphic functions we shall al-
ways mean meromorphic functions in the complex plane. We adopt the standard notations of value
distribution theory (see [8]). For a non-constant meromorphic function f, we denote by T (r, f)
the Nevanlinna characteristic of f and by S(r, f) any quantity satisfying S(r, f) = o\{ T (r, f)\} as
r \rightarrow \infty possibly outside a set of finite linear measure. A meromorphic function a is called a small
function of f, if T (r, a) = S(r, f). We denote by S(f) the set of all small functions of f. Also we
denote by \rho (f) the order of f.
Let f and g be two non-constant meromorphic functions. Let a \in S(f) \cap S(g). We say that f
and g share a counting multiplicities (CM) if f(z) - a(z) and g(z) - a(z) have the same zeros with
the same multiplicities and we say that f and g share a ignoring multiplicities (IM) if we do not
consider the multiplicities.
Let f be a transcendental meromorphic function and n \in \BbbN . Many authors have investigated the
value distributions of fn(z)f \prime (z). In 1959, W. K. Hayman (see [7], Corollary of Theorem 9) proved
the following theorem.
Theorem A [7]. Let f be a transcendental meromorphic function and n \in \BbbN such that n \geq 3.
Then fn(z)f \prime (z) = 1 has infinitely many solutions.
The case n = 2 was settled by Mues [13] in 1979. Bergweiler and Eremenko [3] showed that
f(z)f \prime (z) - 1 has infinitely many zeros. For an analog of the above results Laine and Yang [11]
investigated the value distribution of difference products of entire functions in the following manner.
Theorem B [11]. Let f be a transcendental entire function of finite order and c \in \BbbC \setminus \{ 0\} . Then
for n \in \BbbN \setminus \{ 1\} , fn(z)f(z + c) assumes every a \in \BbbC \setminus \{ 0\} infinitely often.
In 2010, Zhang [19] considered zeros of one certain type of difference polynomials that was not
studied previously and obtained the following theorem.
c\bigcirc S. MAJUMDER, S. SAHA, 2021
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 5 679
680 S. MAJUMDER, S. SAHA
Theorem C [19]. Let f be a transcendental entire function of finite order, \alpha ( \not \equiv 0) \in S(f),
c \in \BbbC \setminus \{ 0\} and n \in \BbbN . If n \geq 2, then fn(z)(f(z) - 1)f(z + c) - \alpha (z) has infinitely many zeros.
In 2012, Chen and Chen [5] further extended Theorem C as follows.
Theorem D [5]. Let f be a transcendental entire function of finite order, \alpha ( \not \equiv 0) \in S(f),
cj \in \BbbC and d,m, n, \nu j \in \BbbN , where j = 1, 2, . . . , d. If n \geq 2, then fn(z)(fm(z) - 1)
\prod d
j=1
(f(z +
+ cj))
\nu j - \alpha (z) has infinitely many zeros.
Chen and Chen [5] also found the uniqueness result corresponding to Theorem D. In 2014, Zhang
and Yi [18] treat the above investigations into a different way that was not dealt earlier. They paid
their attention to the kth derivative of more generalized difference expression and obtained a series
of results as follows.
Theorem E [18]. Let f be a transcendental entire function of finite order, \alpha ( \not \equiv 0) \in S(f),
cj \in \BbbC be distinct and d, m, n, \nu j \in \BbbN \cup \{ 0\} , j = 1, 2, . . . , d. If n \geq k + 2, then the differential-
difference polynomial
\biggl(
fn(z)(fm(z) - 1)
\prod d
j=1
(f(z + cj))
\nu j
\biggr) (k)
- \alpha (z) has infinitely many zeros.
Theorem F [18]. Let f be a transcendental entire function of finite order, \alpha ( \not \equiv 0) \in S(f),
cj \in \BbbC be distinct and d, m, n, \nu j \in \BbbN \cup \{ 0\} , j = 1, 2, . . . , d. If one of the following conditions
holds:
(i) n \geq k + 2, when m \leq k + 1;
(ii) n \geq 2k - m+ 3, when m > k + 1,
then the differential-difference polynomial
\Bigl(
fn(z)(f(z) - 1)m
\prod d
j=1
(f(z + cj))
\nu j
\Bigr) (k)
- \alpha (z) has
infinitely many zeros.
Theorem G [18]. Let f and g be two transcendental entire functions of finite order, \alpha (\not \equiv 0) \in
\in S(f) \cap S(g), cj \in \BbbC be distinct and d, m, n, \nu j \in \BbbN \cup \{ 0\} , where j = 1, 2, . . . , d and \sigma =
=
\sum d
j=1
\nu j . If n \geq 2k+m+\sigma +5 and
\Bigl(
fn(z)(fm(z) - 1)
\prod d
j=1
(f(z+cj))
\nu j
\Bigr) (k)
,
\Bigl(
gn(z)(gm(z) -
- 1)
\prod d
j=1
(g(z + cj))
\nu j
\Bigr) (k)
share \alpha (z) CM, then f \equiv tg, where tm = tn+\sigma = 1.
Theorem H [18]. Under the same situation of Theorem G if n \geq 4k - m + \sigma + 9 and\Bigl(
fn(z)(f(z) - 1)m
\prod d
j=1
(f(z+ cj))
\nu j
\Bigr) (k)
,
\Bigl(
gn(z)(g(z) - 1)m
\prod d
j=1
(g(z+ cj))
\nu j
\Bigr) (k)
share \alpha (z)
CM, then f \equiv g.
In 2017, with the notion of weighted sharing as introduced in [10], Banerjee and Majumder [1]
rectified the errors occurred in Theorems G and H and generalised the results as follows.
Theorem I [1]. Let f and g be two transcendental entire functions of finite order, cj \in \BbbC , j =
= 1, 2, . . . , s, be distinct and let a( \not \equiv 0,\infty ) \in S(f)\cap S(g) with finitely many zeros. Let m, n, \mu j \in
\in \BbbN , j = 1, 2, . . . , s, such that n > 2k+m+\sigma +4, where \sigma =
\sum s
j=1
\mu j and P (\omega ) =
\sum m
j=0
aj\omega
j
be a polynomial, where a0(\not = 0), a1, . . . , am(\not = 0) \in \BbbC . If
\Bigl(
fn(z)P (f(z))
\prod s
j=1
(f(z+cj))
\mu j
\Bigr) (k)
-
- a(z) and
\Bigl(
gn(z)P (g(z))
\prod s
j=1
(g(z + cj))
\mu j
\Bigr) (k)
- a(z) share (0, 2), then:
(I) when P (\omega ) =
\sum m
j=0
aj\omega
j is a non-zero polynomial, one of the following two cases holds:
(I1) f \equiv tg, t \in \BbbC \setminus \{ 0\} such that td = 1, where d is the GCD of the elements of J, J = \{ k \in I :
ak \not = 0\} and I = \{ n+ \sigma , n+ \sigma + 1, . . . , n+ \sigma +m\} ,
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 5
A NOTE ON THE UNIQUENESS OF CERTAIN TYPES . . . 681
(I2) fn(z)P (f(z))
\prod s
j=1
(f(z + cj))
\mu j \equiv gn(z)P (g(z))
\prod s
j=1
(g(z + cj))
\mu j ;
(II) when P (\omega ) = \omega m - 1 and n \geq \sigma +2s+3, then f \equiv tg, t \in \BbbC \setminus \{ 0\} such that tm = tn+\sigma = 1;
(III) when P (\omega ) = (\omega - 1)m(m \geq 2), one of the following two cases holds:
(III1) f \equiv g,
(III2) fn(z)(f(z) - 1)m
\prod s
j=1
(f(z + cj))
\mu j \equiv gn(z)(g(z) - 1)m
\prod s
j=1
(g(z + cj))
\mu j .
In the same paper, Banerjee and Majumder [1] emerged the following question as an open
problem.
Question 1. Whether Theorem I can be obtained for any small function a \in S(f) \cap S(g)?
Our first objective to write this paper is to solve Question 1. Throughout this paper we use \scrP (\omega )
as follows:
\scrP (\omega ) = am\omega m + am - 1\omega
m - 1 + . . .+ a1\omega + a0, (1.1)
where ai \in S(f) \cap S(g) for i = 0, 1, 2, . . . ,m such that a0 \not \equiv 0, am \not \equiv 0.
Let c \in \BbbC such that \scrP (c) \not \equiv 0 and let \omega 1 = \omega - c. Then \scrP (\omega ) = \scrP (\omega 1 + c) = \scrP 1(\omega 1), say,
where \scrP 1(\omega 1) is of the form
\scrP 1(\omega 1) = bm\omega m
1 + bm - 1\omega
m - 1
1 + . . .+ b1\omega 1 + b0, (1.2)
bi \in S(f) \cap S(g) for i = 0, 1, 2, . . . ,m such that b0 \equiv amcm + am - 1c
m - 1 + . . . + a1c + a0 \not \equiv 0,
bm \equiv am \not \equiv 0. Throughout this paper we use \scrP 1(\omega 1) defined as in (1.2).
Our second objective to write this paper is to solve the following questions.
Question 2. Is Theorem I hold for \scrP (\omega ) instead of P (\omega )?
Question 3. Can one deduce more generalized result in which Theorem I will be included?
In 2017, with the notion of weakly weighted sharing and relaxed weighted sharing as introduced
in [12] and [2], respectively, Sahoo and Karmakar [15] obtained the following results.
Theorem J [15]. Let f and g be two transcendental entire functions of finite order and \alpha ( \not \equiv
\not \equiv 0) \in S(f) \cap S(g). Suppose that \eta \in \BbbC \setminus \{ 0\} , k \in \BbbN \cup \{ 0\} and m,n(> k) \in \BbbN satisfying
n \geq 2k + m + 6, when m \leq k + 1, and n \geq 4k - m + 10, when m > k + 1. If (fn(z)(f(z) -
- 1)mf(z+\eta ))(k) and (gn(z)(g(z) - 1)mg(z+\eta ))(k) share ``(\alpha (z), 2)"" and if f, g have no 1-points
with multiplicity less than or equal to k/m, when m \leq k, then either f \equiv g or f and g satisfy the
equation R(f, g) \equiv 0, where R(\omega 1, \omega 2) is given by
R(\omega 1, \omega 2) = \omega n
1 (\omega 1 - 1)m\omega 1(z + \eta ) - \omega n
2 (\omega 2 - 1)m\omega 2(z + \eta ).
Theorem K [15]. Let f and g be two transcendental entire functions of finite order and \alpha (\not \equiv
\not \equiv 0) \in S(f)\cap S(g). Suppose that \eta \in \BbbC \setminus \{ 0\} , k \in \BbbN \cup \{ 0\} and m,n(> k) \in \BbbN satisfying n \geq 3k+
+2m+8, when m \leq k+1, and n \geq 6k - m+13, when m > k+1. If (fn(z)(f(z) - 1)mf(z+\eta ))(k)
and (gn(z)(g(z) - 1)mg(z + \eta ))(k) share (\alpha (z), 2)\ast and if f, g have no 1-points with multiplicity
less than or equal to k/m, when m \leq k, then the conclusions of Theorem J hold.
Now our third objective to write this paper is to solve the following question.
Question 4. Can one remove the condition “f, g have no 1-points with multiplicity less than or
equal to k/m, when m \leq k” in Theorems J and K?
In this paper, taking the possible answers of the above questions into back ground we obtain main
results as follows.
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 5
682 S. MAJUMDER, S. SAHA
Theorem 1. Let f and g be two transcendental entire functions of finite order, c \in \BbbC , cj \in \BbbC ,
j = 1, 2, . . . , s, be distinct and let a(\not \equiv 0,\infty ) \in S(f) \cap S(g). Let k, m \in \BbbN \cup \{ 0\} , n, \sigma \in \BbbN ,
\mu j \in \BbbN \cup \{ 0\} , j = 1, 2, . . . , s, such that n \geq k + 1 and \sigma =
\sum s
j=1
\mu j . Suppose that
\left( (f(z) - c)n\scrP (f(z))
s\prod
j=1
(f(z + cj) - c)\mu j
\right) (k) - a(z)
and \left( (g(z) - c)n\scrP (g(z))
s\prod
j=1
(g(z + cj) - c)\mu j
\right) (k) - a(z)
share (0, 2), where \scrP (\omega ) is defined as in (1.1). Now:
(I) when \scrP (\omega ) \not \equiv (\omega - c)m - \beta , (\omega - c - \beta )m(m \geq 2), where \beta \in S(f) \cap S(g) and n \geq
\geq 2k +m+ \sigma + 5, then one of the following two cases holds:
(I1) f - c \equiv t(g - c), t \in \BbbC \setminus \{ 0\} such that td = 1, where d is the GCD of the elements of J,
J = \{ k \in I : bk \not = 0\} and I = \{ 0, 1, . . . ,m\} ;
(I2) (f(z) - c)n\scrP (f(z))
\prod s
j=1
(f(z+ cj) - c)\mu j \equiv (g(z) - c)n\scrP (g(z))
\prod s
j=1
(g(z+ cj) - c)\mu j ;
(II) when \scrP (\omega ) = (\omega - c)m - \beta , where \beta \in S(f)\cap S(g) and n \geq \mathrm{m}\mathrm{a}\mathrm{x}\{ 2k+m+\sigma +5, \sigma +2s+3\} ,
then f - c \equiv t(g - c), t \in \BbbC \setminus \{ 0\} such that tm = tn+\sigma = 1;
(III) when \scrP (\omega ) = (\omega - c - \beta )m, m \geq 2, where \beta \in S(f) \cap S(g) and
n \geq
\left\{ 2k +m+ \sigma + 5, if m \leq k + 1,
4k - m+ \sigma + 9, if m > k + 1,
then one of the following two cases holds:
(III1) f \equiv g,
(III2) (f(z) - c)n(f(z) - c - \beta (z))m
\prod s
j=1
(f(z + cj) - c)\mu j \equiv (g(z) - c)n(g(z) - c -
- \beta (z))m
\prod s
j=1
(g(z + cj) - c)\mu j .
Corollary 1. Let f and g be two transcendental entire functions of finite order, c \in \BbbC , cj \in \BbbC ,
j = 1, 2, . . . , s, be distinct, a( \not \equiv 0,\infty ) \in S(f)\cap S(g) and k,m \in \BbbN \cup \{ 0\} , n, \sigma \in \BbbN , \mu j \in \BbbN \cup \{ 0\} ,
j = 1, 2, . . . , s, such that n \geq k + 1 and \sigma =
\sum s
j=1
\mu j . Suppose that
\left( (f(z) - c)n\scrP (f(z))
s\prod
j=1
(f(z + cj) - c)\mu j
\right) (k) - a(z)
and \left( (g(z) - c)n\scrP (g(z))
s\prod
j=1
(g(z + cj) - c)\mu j
\right) (k) - a(z)
share ``(0, 2)"", where \scrP (\omega ) = (\omega - c - \beta )m and \beta \in S(f) \cap S(g). Now when
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 5
A NOTE ON THE UNIQUENESS OF CERTAIN TYPES . . . 683
n \geq
\left\{ 2k +m+ \sigma + 5, if m \leq k + 1,
4k - m+ \sigma + 9, if m > k + 1,
then one of the conclusions (III1) and (III2) of Theorem 1 holds.
Corollary 2. Let f and g be two transcendental entire functions of finite order, c \in \BbbC , cj \in \BbbC ,
j = 1, 2, . . . , s, be distinct, a( \not \equiv 0,\infty ) \in S(f)\cap S(g) and k, m \in \BbbN \cup \{ 0\} , n, \sigma \in \BbbN , \mu j \in \BbbN \cup \{ 0\} ,
j = 1, 2, . . . , s, such that n \geq k + 1 and \sigma =
\sum s
j=1
\mu j . Suppose that
\left( (f(z) - c)n\scrP (f(z))
s\prod
j=1
(f(z + cj) - c)\mu j
\right) (k) - a(z)
and \left( (g(z) - c)n\scrP (g(z))
s\prod
j=1
(g(z + cj) - c)\mu j
\right) (k) - a(z)
share (0, 2)\ast , where \scrP (\omega ) = (\omega - c - \beta )m and \beta \in S(f) \cap S(g). Now when
n \geq
\left\{ 3k + 2m+ 2\sigma + 6, if m \leq k + 1,
6k - m+ 2\sigma + 11, if m > k + 1,
then one of the conclusions (III1) and (III2) of Theorem 1 holds.
2. Lemmas. Let F and G be two non-constant meromorphic functions. Henceforth we shall
denote by H the function
H =
\biggl(
F \prime \prime
F \prime - 2F \prime
F - 1
\biggr)
-
\biggl(
G\prime \prime
G\prime - 2G\prime
G - 1
\biggr)
. (2.1)
Lemma 1 [16]. Let f be a non-constant meromorphic function and let an( \not \equiv 0), an - 1, . . . , a0 \in
\in S(f). Then T
\Bigl(
r,
\sum n
i=0
aif
i
\Bigr)
= nT (r, f) + S(r, f).
Lemma 2 [4]. Let f be a meromorphic function of finite order \rho and c \in \BbbC \setminus \{ 0\} be fixed. Then
for each \varepsilon > 0, we have
m
\biggl(
r,
f(z + c)
f(z)
\biggr)
+m
\biggl(
r,
f(z)
f(z + c)
\biggr)
= O
\bigl(
r\rho - 1+\varepsilon
\bigr)
.
The following lemma has little modifications of the original version (Theorem 2.1 of [4]).
Lemma 3. Let f be a transcendental meromorphic function of finite order, c \in \BbbC \setminus \{ 0\} be fixed.
Then T (r, f(z + c)) = T (r, f) + S(r, f).
Lemma 4 [6]. Let f be a non-constant meromorphic function of finite order and c \in \BbbC . Then
N(r, 0; f(z + c)) \leq N(r, 0; f(z)) + S(r, f) and N(r,\infty ; f(z + c)) \leq N(r,\infty ; f) + S(r, f).
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 5
684 S. MAJUMDER, S. SAHA
Lemma 5 ([8], Lemma 3.5). Suppose that F is meromorphic in a domain D and set f =
F \prime
F
.
Then, for n \in \BbbN ,
F (n)
F
= fn +
n(n - 1)
2
fn - 2f \prime +Anf
n - 3f \prime \prime +Bnf
n - 4(f \prime )2 + Pn - 3(f),
where An =
1
6
n(n - 1)(n - 2), Bn =
1
8
n(n - 1)(n - 2)(n - 3) and Pn - 3(f) is a differential
polynomial with constant coefficients, which vanishes identically for n \leq 3 and has degree n - 3
when n > 3.
Lemma 6. Let f be a transcendental meromorphic function of finite order such that N(r,\infty ; f) =
= S(r, f) and cj \in \BbbC , m \in \BbbN \cup \{ 0\} , n, \sigma \in \BbbN , \mu j \in \BbbN \cup \{ 0\} , j = 1, 2, . . . , s. Then, for each
\varepsilon > 0, we have
T
\left( r, fn(z)\scrP (f(z))
s\prod
j=1
(f(z + cj))
\mu j
\right) = T
\bigl(
r, fn+\sigma (z)\scrP (f(z))
\bigr)
+ S(r, f).
Proof of lemma follows from the proof of Lemma 6 [1].
Lemma 7. Let f and g be two transcendental entire functions of finite order, c, cj \in \BbbC , and k,
m \in \BbbN \cup \{ 0\} , n, \sigma \in \BbbN , \mu j \in \BbbN \cup \{ 0\} , j = 1, 2, . . . , s, such that n \geq k + 1. Suppose that
F (z) =
\Bigl(
(f(z) - c)n\scrP (f(z))
\prod s
j=1
(f(z + cj) - c)\mu j
\Bigr) (k)
\alpha (z)
,
G(z) =
\Bigl(
(g(z) - c)n\scrP (g(z))
\prod s
j=1
(g(z + cj) - c)\mu j
\Bigr) (k)
\alpha (z)
,
where \alpha \in S(f) \cap S(g) and H \equiv 0. If one of the following conditions holds:
(1) \scrP (\omega ) \not \equiv (\omega - c - \beta )m, where \beta \in S(f) \cap S(g) and n \geq 2k +m+ \sigma + 5,
(2) \scrP (\omega ) \equiv (\omega - c - \beta )m, where \beta \in S(f) \cap S(g) and n \geq 2k +m+ \sigma + 5 when m \leq k + 1
and n \geq 4k - m+ \sigma + 9 when m > k + 1,
then one of the following two cases holds:
(i)
\Bigl(
(f(z) - c)n\scrP (f(z))
\prod s
j=1
(f(z + cj) - c)\mu j
\Bigr) (k)
\times
\times
\Bigl(
(g(z) - c)n\scrP (g(z))
\prod s
j=1
(g(z + cj) - c)\mu j
\Bigr) (k)
\equiv \alpha 2(z),
(ii) (f(z) - c)n\scrP (f(z))
\prod s
j=1
(f(z+ cj) - c)\mu j \equiv (g(z) - c)n\scrP (g(z))
\prod s
j=1
(g(z+ cj) - c)\mu j .
Proof. Note that when \scrP (\omega ) = (\omega - c - \beta )m, where \beta \in S(f) \cap S(g) and m > k + 1. Then
Nk+1
\left( r, 0; (f(z) - c)n\scrP (f(z))
s\prod
j=1
(f(z + cj) - c)\mu j
\right) +
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A NOTE ON THE UNIQUENESS OF CERTAIN TYPES . . . 685
= Nk+1
\left( r, 0; fn
1 (z)\scrP 1(f1(z))
s\prod
j=1
(f1(z + cj))
\mu j
\right) =
= Nk+1
\left( r, 0; fn
1 (z)(f1(z) - \beta (z))m
s\prod
j=1
(f1(z + cj))
\mu j
\right) =
= Nk+1(r, 0; f
n
1 ) +Nk+1(r, 0; (f1(z) - \beta (z))m) +Nk+1
\left( r, 0;
s\prod
j=1
(f1(z + cj))
\mu j
\right) \leq
\leq 2(k + 1)T (r, f1) +N
\left( r, 0;
s\prod
j=1
(f1(z + cj))
\mu j
\right) + S(r, f1),
where f1 = f - c. Similar expression holds for (g(z) - c)n\scrP (g(z))
\prod s
j=1
(g(z + cj) - c)\mu j . We
omit the detail proof, since proof of lemma follows from the proof of Lemma 7 [1].
Lemma 8 ([1], Lemma 8). Let f be a transcendental meromorphic function of finite order and
cj \in \BbbC , j = 1, 2, . . . , s. Suppose that n, \sigma \in \BbbN and \mu j \in \BbbN \cup \{ 0\} , j = 1, 2, . . . , s. Let \Phi (z) =
= fn(z)
\prod s
j=1
(f(z + cj))
\mu j . Then (n - \sigma ) T (r, f) \leq T (r,\Phi ) + S(r, f).
Lemma 9. Let f and g be transcendental entire functions of finite order, \alpha ( \not \equiv 0,\infty ) \in S(f)\cap
\cap S(g), c, cj \in \BbbC and n, m, s, \sigma \in \BbbN , k, \mu j \in \BbbN \cup \{ 0\} , j = 1, 2, . . . , s. If n \geq k + 1, then\left( (f(z) - c)n\scrP (f(z))
s\prod
j=1
(f(z + cj) - c)\mu j
\right) (k)\left( (g(z) - c)n\scrP (g(z))
s\prod
j=1
(g(z + cj) - c)\mu j
\right) (k) \not \equiv
\not \equiv \alpha 2(z).
Proof. Suppose on contrary that\left( (f(z) - c)n\scrP (f(z))
s\prod
j=1
(f(z + cj) - c)\mu j
\right) (k)\left( (g(z) - c)n\scrP (g(z))
s\prod
j=1
(g(z + cj) - c)\mu j
\right) (k) \equiv
\equiv \alpha 2(z).
Then\left( fn
1 (z)\scrP 1(f1(z))
s\prod
j=1
(f1(z + cj))
\mu j
\right) (k)\left( gn1 (z)\scrP 1(g1(z))
s\prod
j=1
(g1(z + cj))
\mu j
\right) (k) \equiv \alpha 2(z), (2.2)
where f1 = f - c and g1 = g - c. Note that S(r, f) = S(r, f1) and S(r, g) = S(r, g1). Since
n \geq k + 1, from (2.2), we have
N(r, 0; f1) \leq N(r, 0;\alpha 2) = S(r, f1). (2.3)
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686 S. MAJUMDER, S. SAHA
Since f and g are transcendental entire functions of finite order and so are f1 and g1. Therefore we
can take f1(z) = \gamma (z)e\delta (z) and g1(z) = \eta (z)e\zeta (z), where \gamma (z)(\not \equiv 0), \eta (z)(\not \equiv 0) are entire functions
such that N(r, 0; \gamma ) = S(r, f1), N(r, 0; \eta ) = S(r, f1) and \delta (z), \zeta (z) are non-zero polynomials. We
now consider following two cases.
Case 1. Suppose that k \in \BbbN . Let
Fi(z) = bi(z)f
n+i
1 (z)
s\prod
j=1
(f1(z + cj))
\mu j =
= bi(z)\gamma
n+i(z)
s\prod
j=1
(\gamma (z + cj))
\mu j e(n+i)\delta (z)+
\sum s
j=1 \mu j\delta (z+cj) = P1i(z)e
P2i(z), (2.4)
where P1i(z) = bi\gamma
n+i(z)
\prod s
j=1 (\gamma (z + cj))
\mu j and P2i(z) = (n + i)\delta (z) +
\sum s
j=1
\mu j\delta (z + cj)
for i = 0, 1, 2, . . . ,m. Let J1 = \{ j \in I1 : bj(z) \not \equiv 0\} , where I1 = \{ 0, 1, . . . ,m\} . Note that
N(r,\infty ;Fi) = S(r, f1) for i \in J1. Using (2.3) and Lemma 4, we obtain N(r, 0, P1i) = S(r, f1) and
N(r,\infty , P1i) = S(r, f1) for i \in J1. By Lemmas 1 and 6, we have T (r, Fi) = (n+ i+ \sigma )T (r, f1)+
+S(r, f1) and so S(r, Fi) = S(r, f1) for i \in J1. Note that \gamma (z) \not \equiv 0 and so
\prod s
j=1
(\gamma (z + cj))
\mu j \not \equiv
\not \equiv 0. Therefore, we have P1i(z) \not \equiv 0 for i \in J1. Let
hi =
F \prime
i
Fi
=
P \prime
1i
P1i
+ P \prime
2i
for i \in J1. Clearly,
T (r, hi) = N
\biggl(
r,
F \prime
i
Fi
\biggr)
+m
\biggl(
r,
F \prime
i
Fi
\biggr)
= N(r,\infty ;Fi) +N(r, 0;Fi) + S(r, Fi) =
= S(r, f1) + S(r, Fi) = S(r, f1) (2.5)
for i \in J1. By using (2.5), we obtain
T
\bigl(
r, h
(p)
i
\bigr)
\leq (p+ 1)T (r, hi) + S(r, hi) = S(r, f1), (2.6)
where p \in \BbbN \cup \{ 0\} and i \in J1. From (2.6) and Lemma 1, we get
T
\bigl(
r,
\bigl(
h
(p)
i
\bigr) q\bigr)
= q T
\bigl(
r, h
(p)
i
\bigr)
+ S(r, hi) = S(r, f1), (2.7)
where q \in \BbbN \cup \{ 0\} and i \in J1. By Lemma 5, we have F
(k)
i = QiFi, i.e.,\bigl(
Fi(z)
\bigr) (k)
= Qi(z)P1i(z)e
P2i(z),
where Qi = hki +
k(k - 1)
2
hk - 2
i h\prime i +Akh
k - 3
i h\prime \prime i +Bkh
k - 4
i (h\prime i)
2 +Pk - 3(hi) and i \in J1. Since f(z)
is a transcendental entire function, it follows that Fi(z) is also a transcendental entire function for
i \in J1. Consequently, Qi(z) \not \equiv 0 for i \in J1. Then, from (2.5) and (2.7), it follows that
T (r,Qi) = T
\biggl(
r, hki +
k(k - 1)
2
hk - 2
i h
\prime
i +Akh
k - 3
i h
\prime \prime
i +Bkh
k - 4
i (h\prime i)
2 + Pk - 3(hi)
\biggr)
\leq
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A NOTE ON THE UNIQUENESS OF CERTAIN TYPES . . . 687
\leq T (r, hki ) + T (r, hk - 2
i ) + T (r, h
\prime
i) + T (r, hk - 3
i ) + T (r, h
\prime \prime
i )+
+T (r, hk - 4
i ) + T (r, (h\prime i)
2)) + T (r, Pk - 3(hi)) + S(r, f1) = S(r, f1)
for i \in J1. Note that, for i \in J1,\left( fn
1 (z)\scrP 1(f1(z))
s\prod
j=1
(f1(z + cj))
\mu j
\right) (k) =
=
m\sum
i=0
\left( bif
n+i
1 (z)
s\prod
j=1
(f1(z + cj))
\mu j
\right) (k) =
=
m\sum
i=0
Qi(z)P1i(z)e
P2i(z) =
= \gamma n(z)
s\prod
j=1
(\gamma (z + cj))
\mu j en\delta (z)+
\sum s
j=1 \mu j\delta (z+cj)
m\sum
i=0
biQi(z)\gamma
i(z)ei\delta (z) =
= \gamma n(z)
s\prod
j=1
(\gamma (z + cj))
\mu j en\delta (z)+
\sum s
j=1 \mu j\delta (z+cj)
m\sum
i=0
biQi(z)f
i
1(z). (2.8)
Also, from (2.2), we have
\Bigl(
fn
1 (z)\scrP 1(f1(z))
\prod s
j=1
(f1(z + cj))
\mu j
\Bigr) (k)
\not \equiv 0 and so
N
\left( r, 0;
\left( fn
1 (z)\scrP 1(f1(z))
s\prod
j=1
(f1(z + cj))
\mu j
\right) (k)
\right) \leq N(r, 0;\alpha 2) \leq S(r, f1).
From (2.8), we obtain
N
\Biggl(
r, 0;
m\sum
i=0
biQif
i
1
\Biggr)
= S(r, f1). (2.9)
Since bi, Qi \in S(f1), from Lemma 1, we get m T (r, f1) = T
\Bigl(
r,
\sum m
i=0
biQif
i
1
\Bigr)
+ S(r, f1). This
shows that S(r, f1) = S
\Bigl(
r,
\sum m
i=0
biQif
i
1
\Bigr)
. Similarly, we have S(r, f1) = S
\Bigl(
r,
\sum m
i=1
biQif
i
1
\Bigr)
.
Now we claim that
\sum m
i=1
biQif
i
1 is not a rational function. If possible suppose that
\sum m
i=1
biQif
i
1
is a rational function. Since bi, Qi \in S(f1), we obtain
m T (r, f1) = T
\Biggl(
r,
m\sum
i=1
biQif
i
1
\Biggr)
+ S(r, f1) = O(\mathrm{l}\mathrm{o}\mathrm{g} r) + S(r, f1) = S(r, f1),
which is not possible. Hence,
\sum m
i=1
biQif
i
1 is a transcendental meromorphic function such that
N
\Biggl(
r,\infty ,
m\sum
i=1
biQif
i
1
\Biggr)
= S(r, f1). (2.10)
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688 S. MAJUMDER, S. SAHA
Note that T (r, b0Q0) = S(r, f1) = S
\Bigl(
r,
\sum m
i=1
biQif
i
1
\Bigr)
, which shows that b0Q0 is a small function
of
\sum m
i=1
biQif
i
1. Then, using (2.3), (2.9), (2.10) and Lemma 1, we get from the second fundamental
theorem for small functions (see [17]) that
m T (r, f1) = T
\Biggl(
r,
m\sum
i=1
biQif
i
1
\Biggr)
+ S(r, f1) \leq
\leq N
\Biggl(
r, 0;
m\sum
i=1
biQif
i
1
\Biggr)
+N
\Biggl(
r,\infty ;
m\sum
i=1
biQif
i
1
\Biggr)
+N
\Biggl(
r, 0;
m\sum
i=0
biQif
i
1
\Biggr)
+ S(r, f1) \leq
\leq N(r, 0; f1) +N
\Biggl(
r, 0;
m\sum
i=1
biQif
i - 1
1
\Biggr)
+ S(r, f1) \leq
\leq T
\Biggl(
r,
m\sum
i=1
biQif
i - 1
1
\Biggr)
+ S(r, f1) = (m - 1) T (r, f1) + S(r, f1),
which is not possible.
Case 2. Suppose that k = 0. Then, from (2.2), we get
fn
1 (z)\scrP 1(f1(z))
s\prod
j=1
(f1(z + cj))
\mu jgn1 (z)\scrP 1(g1(z))
s\prod
j=1
(g1(z + cj))
\mu j \equiv \alpha 2(z). (2.11)
Now, from (2.11), we have
N(r, 0;\scrP 1(f1)) \leq N(r, 0;\alpha 2) = S(r, f1), i.e., N
\Biggl(
r, 0;
m\sum
i=0
bif
i
1
\Biggr)
= S(r, f1). (2.12)
One can easily prove that
\sum m
i=1
bif
i
1 is a transcendental meromorphic function such that
N
\Biggl(
r,\infty ;
m\sum
i=0
bif
i
1
\Biggr)
= S(r, f1) (2.13)
and b0 is a small function of
\sum m
i=1
bif
i
1. Now, by using (2.12), (2.13) and Lemma 1, we get, from
the second fundamental theorem for small functions (see [17]),
m T (r, f1) = T
\Biggl(
r,
m\sum
i=1
bif
i
1
\Biggr)
+ S(r, f1) \leq
\leq N
\Biggl(
r, 0;
m\sum
i=1
bif
i
1
\Biggr)
+N
\Biggl(
r,\infty ;
m\sum
i=1
bif
i
1
\Biggr)
+N
\Biggl(
r, 0;
m\sum
i=0
bif
i
1
\Biggr)
+ S(r, f1) \leq
\leq N(r, 0; f1) +N
\Biggl(
r, 0;
m\sum
i=1
bif
i - 1
1
\Biggr)
+ S(r, f1) \leq
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A NOTE ON THE UNIQUENESS OF CERTAIN TYPES . . . 689
\leq T
\Biggl(
r,
m\sum
i=1
bif
i - 1
1
\Biggr)
+ S(r, f1) = (m - 1) T (r, f1) + S(r, f1),
which is not possible.
Lemma 9 is proved.
Lemma 10. Let f and g be two transcendental entire functions of finite order, c, cj \in \BbbC , and
m \in \BbbN \cup \{ 0\} , n, \sigma \in \BbbN , \mu j \in \BbbN \cup \{ 0\} , j = 1, 2, . . . , s. Suppose that
(f(z) - c)n\scrP (f(z))
s\prod
j=1
(f(z + cj) - c)\mu j \equiv (g(z) - c)n\scrP (g(z))
s\prod
j=1
(g(z + cj) - c)\mu j .
Now:
(I) when \scrP (\omega ) \not \equiv (\omega - c)m - \beta , (\omega - c - \beta )m, m \geq 2, where \beta \in S(f) \cap S(g), then one of
the following two cases holds:
(I1) f - c \equiv t(g - c), t \in \BbbC \setminus \{ 0\} such that td = 1, where d is the GCD of the elements of J,
J = \{ k \in I : bk \not = 0\} and I = \{ 0, 1, . . . ,m\} ;
(I2) (f(z) - c)n\scrP (f(z))
\prod s
j=1
(f(z+ cj) - c)\mu j \equiv (g(z) - c)n\scrP (g(z))
\prod s
j=1
(g(z+ cj) - c)\mu j ;
(II) when \scrP (\omega ) = (\omega - c)m - \beta , where \beta \in S(f)\cap S(g) and n > \sigma +2s+2, then f - c \equiv t(g - c),
t \in \BbbC \setminus \{ 0\} such that tm = tn+\sigma = 1;
(III) when \scrP (\omega ) = (\omega - c - \beta )m, m \geq 2, where \beta \in S(f) \cap S(g), then one of the following
two cases holds:
(III1) f \equiv g,
(III2) (f(z) - c)n(f(z) - c - \beta (z))m
\prod s
j=1
(f(z + cj) - c)\mu j \equiv (g(z) - c)n(g(z) - c -
- \beta (z))m
\prod s
j=1
(g(z + cj) - c)\mu j .
Proof. Suppose that
(f(z) - c)n\scrP (f(z))
s\prod
j=1
(f(z + cj) - c)\mu j \equiv (g(z) - c)n\scrP (g(z))
s\prod
j=1
(g(z + cj) - c)\mu j , (2.14)
i.e.,
fn
1 (z)\scrP 1(f1(z))
s\prod
j=1
(f1(z + cj))
\mu j \equiv gn1 (z)\scrP 1(g1(z))
s\prod
j=1
(g1(z + cj))
\mu j , (2.15)
where f1(z) = f(z) - c and g1(z) = g(z) - c. Now, from (2.15), Lemmas 1 and 6, we have
T (r, f1) + S(r, f1) = T (r, g1) + S(r, g1) and so S(r, f1) = S(r, g1). We consider the following
cases.
Case 1. Suppose \scrP (\omega ) \not \equiv (\omega - c)m - \beta or (\omega - c - \beta )m, m \geq 2, where \beta \in S(f) \cap S(g).
Let h =
f1
g1
. If h is a constant, by putting f1 = hg1 in (2.15), we get
bmgm1 (hn+m+\sigma - 1) + bm - 1g
m - 1
1 (hn+m+\sigma - 1 - 1) + . . .+ b1g1(h
n+\sigma +1 - 1) + b0(h
n+\sigma - 1) \equiv 0,
which implies that hd = 1, where d is the GCD of the elements of J, J = \{ k \in I : bk \not = 0\} and
I = \{ 0, 1, . . . ,m\} . Otherwise, by Lemma 1, we have T (r, g1) = S(r, g1), which is impossible.
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690 S. MAJUMDER, S. SAHA
Thus, f1 \equiv tg1, i.e., f - c = t(g - c), t \in \BbbC \setminus \{ 0\} such that td = 1, where d is the GCD of the
elements of J, J = \{ k \in I : bk \not = 0\} and I = \{ 0, 1, . . . ,m\} .
If h is not a constant, then we know by (2.14) that
(f(z) - c)n\scrP (f(z))
s\prod
j=1
(f(z + cj) - c)\mu j \equiv (g(z) - c)n\scrP (g(z))
s\prod
j=1
(g(z + cj) - c)\mu j .
Case 2. Suppose \scrP (\omega ) = (\omega - c)m - \beta , where \beta \in S(f) \cap S(g). Clearly \beta \in S(f1) \cap S(g1).
Then, from (2.15), we have
fn
1 (z)(f
m
1 (z) - \beta (z))
s\prod
j=1
(f1(z + cj))
\mu j \equiv gn1 (z)(g
m
1 (z) - \beta (z))
s\prod
j=1
(g1(z + cj))
\mu j . (2.16)
Let h =
f1
g1
. Clearly, from (2.16), we get
gm1 (z)
\left( hn+m(z)
s\prod
j=1
(h(z + cj))
\mu j - 1
\right) \equiv \beta (z)
\left( hn(z)
s\prod
j=1
(h(z + cj))
\mu j - 1
\right) . (2.17)
First we suppose that h is non-constant. We assert that both hn+m(z)
\prod s
j=1
(h(z + cj))
\mu j ( \not \equiv 0) and
hn(z)
\prod s
j=1
(h(z + cj))
\mu j ( \not \equiv 0) are non-constant. If not, let hn+m(z)
\prod s
j=1
(h(z + cj))
\mu j \equiv d1 \in
\in \BbbC \setminus \{ 0\} . Then we have
hn+m(z) \equiv d1\prod s
j=1
(h(z + cj))
\mu j
.
Now, by Lemmas 1, 2 and 4, we get
(n+m) T (r, h) = T (r, hn+m) + S(r, h) = T
\left( r,
d1\prod s
j=1
(h(z + cj))
\mu j
\right) + S(r, h) \leq
\leq
s\sum
j=1
\mu jN(r, 0;h(z + cj)) +
s\sum
j=1
\mu j m
\biggl(
r,
1
h(z + cj)
\biggr)
+ S(r, h) \leq
\leq
s\sum
j=1
\mu j N(r, 0;h(z)) +
s\sum
j=1
\mu j m
\biggl(
r,
1
h(z)
\biggr)
+ S(r, h) \leq
\leq \sigma T (r, h) + S(r, h),
which is a contradiction. Similarly we can prove that hn(z)
\prod s
j=1
(h(z + cj))
\mu j is non-constant.
Thus, from (2.17), we have
fm
1 (z) \equiv \beta (z)hm(z)
hn(z)
\prod s
j=1
(h(z + cj))
\mu j - 1
hn+m(z)
\prod s
j=1
(h(z + cj))
\mu j - 1
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A NOTE ON THE UNIQUENESS OF CERTAIN TYPES . . . 691
and
gm1 (z) \equiv \beta (z)
hn(z)
\prod s
j=1
(h(z + cj))
\mu j - 1
hn+m(z)
\prod s
j=1
(h(z + cj))
\mu j - 1
. (2.18)
First we claim that h is a transcendental meromorphic function. If not, suppose that h is a rational
function. Then, from (2.18) and Lemma 1, we have m T (r, g1) = S(r, g1), which is impossible.
Hence, h is a transcendental meromorphic function. Note that T (r, h) \leq T (r, f1) + T (r, g1) =
= 2 T (r, f1) + S(r, f1) and so T (r, h) = O(T (r, f1)). By Lemmas 1 and 3, we obtain
T
\left( r, hn(z)
s\prod
j=1
(h(z + cj))
\mu j - 1
\right) \leq T
\left( r, hn(z)
s\prod
j=1
(h(z + cj))
\mu j
\right) +O(1) \leq
\leq T (r, hn) + T
\left( r,
s\prod
j=1
(h(z + cj))
\mu j
\right) +O(1) \leq
\leq T (r, hn) +
s\sum
j=1
\mu j T (r, h(z + cj)) +O(1) =
= T (r, hn) +
s\sum
j=1
\mu j T (r, h) + S(r, h) =
= (n+ \sigma ) T (r, h) + S(r, h).
Similarly, we have T (r, hn+m(z)
\prod s
j=1
(h(z + cj))
\mu j - 1) \leq (n+m+ \sigma ) T (r, h) + S(r, h). Now,
from (2.18) and Lemma 1, we get m T (r, g1) \leq (2n + m + 2\sigma )T (r, h) + S(r, h) + S(r, g1), i.e.,
T (r, g1) = O(T (r, h)). This shows that S(r, g1) = S(r, h) and so \beta \in S(h). Let z0 be a zero of
hn+m(z)
\prod s
j=1
(h(z + cj))
\mu j - 1 such that \beta (z0) \not = 0,\infty . Since g1 is an entire function, it follows
that z0 is also a zero of hn(z)
\prod s
j=1
(h(z + cj))
\mu j - 1. Then clearly hm(z0) - 1 = 0 and so
N
\Bigl(
r, 1;hn+m
\prod s
j=1
(h(z + cj))
\mu j
\Bigr)
\leq N(r, 1;hm) \leq m T (r, h) + S(r, h).
So, in view of Lemmas 1, 4, and 8 and the second fundamental theorem, we get
(n+m - \sigma ) T (r, h) =
= T
\left( r, hn+m(z)
s\prod
j=1
(h(z + cj))
\mu j
\right) + S(r, h) \leq
\leq N
\left( r, 0;hn+m
s\prod
j=1
(h(z + cj))
\mu j
\right) +N
\left( r,\infty ;hn+m
s\prod
j=1
(h(z + cj))
\mu j
\right) +
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 5
692 S. MAJUMDER, S. SAHA
+N
\left( r, 1;hn+m
s\prod
j=1
(h(z + cj))
\mu j
\right) + S(r, h) \leq
\leq N(r, 0;h) +
s\sum
j=1
N(r, 0;h(z + cj)) +N(r,\infty ;h)+
+
s\sum
j=1
N(r,\infty ;h(z + cj)) +m T (r, h) + S(r, h) \leq
\leq N(r, 0;h) +
s\sum
j=1
N(r, 0;h(z)) +N(r,\infty ;h) +
s\sum
j=1
N(r,\infty ;h(z)) +m T (r, h) + S(r, h) \leq
\leq (m+ 2s+ 2) T (r, h) + S(r, h),
which contradicts with n > \sigma + 2s + 2. Hence, h is a constant. Since g1 is transcendental entire
function, from (2.17), we have
hn+m(z)
s\prod
j=1
(h(z + cj))
\mu j - 1 \equiv 0 \Leftarrow \Rightarrow hn(z)
s\prod
j=1
(h(z + cj))
\mu j - 1 \equiv 0
and so hm(z) = 1 and hn+\sigma = 1. Thus, f \equiv tg, t \in \BbbC \setminus \{ 0\} such that tm = tn+\sigma = 1.
Case 3. Suppose \scrP (\omega ) = (\omega - c - \beta )m, m \geq 2, where \beta \in S(f) \cap S(g). Then, from (2.14),
we have
(f(z) - c)n(f(z) - c - \beta (z))m
s\prod
j
(f(z + cj) - c)\mu j \equiv
\equiv (g(z) - c)n(g(z) - c - \beta (z))m
s\prod
j=1
(g(z + cj) - c)\mu j , (2.19)
i.e.,
fn
1 (z)(f1(z) - \beta (z))m
s\prod
j
(f1(z + cj))
\mu j \equiv gn1 (z)(g1(z) - \beta (z))m
s\prod
j=1
(g1(z + cj))
\mu j . (2.20)
Let h =
f1
g1
. First we suppose that h is non-constant. Then we know from (2.19), that
(f(z) - c)n(f(z) - c - \beta (z))m
s\prod
j
(f(z + cj) - c)\mu j \equiv
\equiv (g(z) - c)n(g(z) - c - \beta (z))m
s\prod
j=1
(g(z + cj) - c)\mu j .
Next we suppose that h is constant. Then, from (2.20), we get
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 5
A NOTE ON THE UNIQUENESS OF CERTAIN TYPES . . . 693
fn
1 (z)
s\prod
j=1
(f1(z + cj))
\mu j
m\sum
i=0
mCm - i f
m - i
1 (z)\beta i(z) \equiv
\equiv gn1 (z)
s\prod
j=1
(g1(z + cj))
\mu j
m\sum
i=0
mCm - ig
m - i
1 (z)\beta i(z). (2.21)
Now substituting f1 = hg1 in (2.21), we get
m\sum
i=0
mCm - i \beta
igm - i
1 (z)(hn+m+\sigma - i(z) - 1) \equiv 0,
which implies that h = 1. Hence, f1 \equiv g1, i.e., f \equiv g.
Lemma 10 is proved.
3. Proof of Theorem 1. Let
F (z) =
\Bigl(
(f(z) - c)n\scrP (f(z))
\prod s
j=1
(f(z + cj) - c)\mu j
\Bigr) (k)
a(z)
=
=
\Bigl(
fn
1 (z)\scrP 1(f1(z))
\prod s
j=1
(f1(z + cj))
\mu j
\Bigr) (k)
a(z)
and
G(z) =
\Bigl(
(g(z) - c)n\scrP (g(z))
\prod s
j=1
(g(z + cj) - c)\mu j
\Bigr) (k)
a(z)
=
=
\Bigl(
gn1 (z)\scrP 1(g1(z))
\prod s
j=1
(g1(z + cj))
\mu j
\Bigr) (k)
a(z)
,
where f1 = f - c and g1 = g - c. Clearly F and G share (1, 2) except for the zeros and poles of
a. Note that when \scrP (\omega ) = (\omega - c - \beta )m, where \beta \in S(f) \cap S(g) and m > k + 1, then
Nk+2
\left( r, 0; (f(z) - c)n\scrP (f(z))
s\prod
j=1
(f(z + cj) - c)\mu j
\right) =
= Nk+2
\left( r, 0; fn
1 (z)\scrP 1(f1(z))
s\prod
j=1
(f1(z + cj))
\mu j
\right) =
= Nk+2
\left( r, 0; fn
1 (z)(f1(z) - \beta (z))m
s\prod
j=1
(f1(z + cj))
\mu j
\right) \leq
\leq 2(k + 2)T (r, f1) +N
\left( r, 0;
s\prod
j=1
(f1(z + cj))
\mu j
\right) + S(r, f1).
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 5
694 S. MAJUMDER, S. SAHA
Similar expression holds for (g(z) - c)n\scrP (g(z))
\prod s
j=1
(g(z+cj) - c)\mu j . So we omit the detail proof,
since when H \not \equiv 0 we follow the proof of Theorem 2 [1] while for H \equiv 0 we follow Lemmas 7, 9
and 10.
Theorem 1 is proved.
Proof of Corollary 1. When H \not \equiv 0 we follow the proof of Theorem 1 [14], while for H \equiv 0
we follow Lemmas 7, 9 and 10. So we omit the detail proof.
Proof of Corollary 2. When H \not \equiv 0 we follow the proof of Theorem 2 [14], while for H \equiv 0
we follow Lemmas 7, 9 and 10. So we omit the detail proof.
References
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Received 30.09.18,
after revision — 17.03.19
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 5
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| id | umjimathkievua-article-379 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:02:37Z |
| publishDate | 2021 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/56/4efbc578a2b47e19d194e233bb382e56.pdf |
| spelling | umjimathkievua-article-3792025-03-31T08:48:07Z A note on the uniqueness of certain types of differential-difference polynomials A NOTE ON THE UNIQUENESS OF CERTAIN TYPES OF DIFFERENTIAL-DIFFERENCE POLYNOMIALS A note on the uniqueness of certain types of differential-difference polynomials Majumder, S. Saha, S. Majumder, Sujoy Saha, Somnath Majumder, S. Saha, S. Meromorphic function difference polynomia uniqueness Meromorphic function difference polynomia uniqueness UDC 517.9 We study the uniqueness problems of certain types of differential-difference polynomials sharing a small function.In this paper, we not only solve the open problem occurred in [A. Banerjee, S. Majumder, On the uniqueness of certain types of differential-difference polynomials, Anal. Math., 43, № 3, 415-444 (2017)], but also present our main results in a more generalized way.   УДК 517.9 Про унiкальнiсть деяких типiв диференцiально-рiзницевих полiномiв Вивчається можливість розв'язання задач єдиності для деяких типів диференціально-різницевих поліномів, які мають спільну малу функцію. У цій роботі не лише наведено розв'язок відкритої задачі з [A. Banerjee, S. Majumder, On the uniqueness of certain types of differential-difference polynomials, Anal. Math., 43, № 3, 415-444 (2017)], а й запропоновано більш загальний вигляд отриманого основного результату. Institute of Mathematics, NAS of Ukraine 2021-05-24 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/379 10.37863/umzh.v73i5.379 Ukrains’kyi Matematychnyi Zhurnal; Vol. 73 No. 5 (2021); 679 - 694 Український математичний журнал; Том 73 № 5 (2021); 679 - 694 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/379/9019 Copyright (c) 2021 Sujoy Majumder, Somnath Saha |
| spellingShingle | Majumder, S. Saha, S. Majumder, Sujoy Saha, Somnath Majumder, S. Saha, S. A note on the uniqueness of certain types of differential-difference polynomials |
| title | A note on the uniqueness of certain types of differential-difference polynomials |
| title_alt | A NOTE ON THE UNIQUENESS OF CERTAIN TYPES OF DIFFERENTIAL-DIFFERENCE POLYNOMIALS A note on the uniqueness of certain types of differential-difference polynomials |
| title_full | A note on the uniqueness of certain types of differential-difference polynomials |
| title_fullStr | A note on the uniqueness of certain types of differential-difference polynomials |
| title_full_unstemmed | A note on the uniqueness of certain types of differential-difference polynomials |
| title_short | A note on the uniqueness of certain types of differential-difference polynomials |
| title_sort | note on the uniqueness of certain types of differential-difference polynomials |
| topic_facet | Meromorphic function difference polynomia uniqueness Meromorphic function difference polynomia uniqueness |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/379 |
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