Research Article -Bernoulli, -Euler, and -Genocchi...
Transcript of Research Article -Bernoulli, -Euler, and -Genocchi...
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Research ArticleOn a Class of π-Bernoulli, π-Euler, and π-Genocchi Polynomials
N. I. Mahmudov and M. Momenzadeh
Eastern Mediterranean University, Gazimagusa, Northern Cyprus, Turkey
Correspondence should be addressed to M. Momenzadeh; [email protected]
Received 20 January 2014; Revised 3 April 2014; Accepted 3 April 2014; Published 28 April 2014
Academic Editor: Sofiya Ostrovska
Copyright Β© 2014 N. I. Mahmudov and M. Momenzadeh. This is an open access article distributed under the Creative CommonsAttribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work isproperly cited.
The main purpose of this paper is to introduce and investigate a class of π-Bernoulli, π-Euler, and π-Genocchi polynomials. Theπ-analogues of well-known formulas are derived. In addition, the π-analogue of the Srivastava-PinteΜr theorem is obtained. Somenew identities, involving π-polynomials, are proved.
1. Introduction
Throughout this paper, we always make use of the classicaldefinition of quantum concepts as follows.
The π-shifted factorial is defined by
(π; π)0= 1, (π; π)
π=
πβ1
β
π=0
(1 β πππ) , π β N,
(π; π)β=
β
β
π=0
(1 β πππ) ,
π < 1, π β C.
(1)
It is known that
(π; π)π=
π
β
π=0
[π
π]
π
π(1/2)π(πβ1)
(β1)πππ. (2)
The π-numbers and π-factorial are defined by
[π]π =1 β ππ
1 β π, (π ΜΈ= 1, π β C) ;
[0]π! = 1, [π]π! = [π]π[π β 1]π!.
(3)
The π-polynomial coefficient is defined by
[π
π]
π
=(π; π)π
(π; π)πβπ(π; π)π
, (π β©½ π, π, π β N) . (4)
In the standard approach to the π-calculus two exponen-tial functions are used, these π-exponential functions andimproved type π-exponential function (see [1]) are defined asfollows:
ππ(π§) =
β
β
π=0
π§π
[π]π!=
β
β
π=0
1
(1 β (1 β π) πππ§),
0 <π < 1, |π§| <
11 β π
,
πΈπ (π§) = π1/π (π§) =
β
β
π=0
π(1/2)π(πβ1)
π§π
[π]π!
=
β
β
π=0
(1 + (1 β π) πππ§) , 0 <
π < 1, π§ β C,
Eπ(π§) = π
π(π§
2)πΈπ(π§
2) =
β
β
π=0
(β1, π)π
2π
π§π
[π]π!
=
β
β
π=0
(1 + (1 β π) ππ(π§/2))
(1 β (1 β π) ππ (π§/2)), 0 < π < 1, |π§|<
2
1 β π.
(5)
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014, Article ID 696454, 10 pageshttp://dx.doi.org/10.1155/2014/696454
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2 Abstract and Applied Analysis
The form of improved type of π-exponential function Eπ(π§)
motivated us to define a new π-addition and π-subtraction as
(π₯ βππ¦)π
:=
π
β
π=0
[π
π]
π
(β1, π)π(β1, π)
πβπ
2ππ₯ππ¦πβπ,
π = 0, 1, 2, . . . ,
(π₯βππ¦)π
:=
π
β
π=0
[π
π]
π
(β1, π)π(β1, π)
πβπ
2ππ₯π(βπ¦)πβπ,
π = 0, 1, 2, . . . .
(6)
It follows that
Eπ (π‘π₯)Eπ (π‘π¦) =
β
β
π=0
(π₯ βππ¦)π π‘π
[π]π!. (7)
The Bernoulli numbers {π΅π}πβ₯0
are rational numbers in asequence defined by the binomial recursion formula:
π
β
π=0
(π
π)π΅πβ π΅π= {
1, π = 1,
0, π > 1,(8)
or equivalently, the generating function
β
β
π=0
π΅π
π‘π
π!=
π‘
ππ‘ β 1. (9)
π-Analogues of the Bernoulli numbers were first studiedby Carlitz [2] in the middle of the last century when heintroduced a new sequence {π½
π}πβ©Ύ0
:
π
β
π=0
(π
π)π½πππ+1
β π½π= {
1, π = 1,
0, π > 1.(10)
Here and in the remainder of the paper, for the parameter πwe make the assumption that |π| < 1. Clearly we recover (8)if we let π β 1 in (10). The π-binomial formula is known as
(1 β π)π
π= (π; π)
π=
πβ1
β
π=0
(1 β πππ)
=
π
β
π=0
[π
π]
π
π(1/2)π(πβ1)
(β1)πππ.
(11)
The above π-standard notation can be found in [3].Carlitz has introduced the π-Bernoulli numbers and poly-
nomials in [2]. Srivastava and PinteΜr proved some relationsand theorems between the Bernoulli polynomials and Eulerpolynomials in [4]. They also gave some generalizationsof these polynomials. In [4β16], the authors investigatedsome properties of the π-Euler polynomials and π-Genocchipolynomials. They gave some recurrence relations. In [17],Cenkci et al. gave the π-extension of Genocchi numbers ina different manner. In [18], Kim gave a new concept for theπ-Genocchi numbers and polynomials. In [19], Simsek et al.investigated the π-Genocchi zeta function and π-function by
using generating functions andMellin transformation.Thereare numerous recent studies on this subject by, among manyother authors, Cigler [20], Cenkci et al. [17, 21], Choi et al.[22], Cheon [23], Luo and Srivastava [8β10], Srivastava et al.[4, 24], Nalci and Pashaev [25] Gaboury and Kurt, [26], Kimet al. [27], and Kurt [28].
We first give the definitions of the π-numbers and π-polynomials. It should be mentioned that the definition of π-Bernoulli numbers in Definition 1 can be found in [25].
Definition 1. Let π β C, 0 < |π| < 1. The π-Bernoulli numbersbπ,π
and polynomials Bπ,π(π₯, π¦) are defined by means of the
generating functions:
BΜ (π‘) :=π‘ππ(βπ‘/2)
ππ (π‘/2) β ππ (βπ‘/2)
=π‘
Eπ (π‘) β 1
=
β
β
π=0
bπ,π
π‘π
[π]π!, |π‘| < 2π,
π‘
Eπ (π‘) β 1
Eπ(π‘π₯)E
π(π‘π¦)
=
β
β
π=0
Bπ,π(π₯, π¦)
π‘π
[π]π!, |π‘| < 2π.
(12)
Definition 2. Let π β C, 0 < |π| < 1. The π-Euler numberseπ,π
and polynomials Eπ,π(π₯, π¦) are defined by means of the
generating functions:
EΜ (π‘) :=2ππ(βπ‘/2)
ππ (π‘/2) + ππ (βπ‘/2)
=2
Eπ (π‘) + 1
=
β
β
π=0
eπ,π
π‘π
[π]π!, |π‘| < π,
2
Eπ(π‘) + 1
Eπ(π‘π₯)E
π(π‘π¦)
=
β
β
π=0
Eπ,π(π₯, π¦)
π‘π
[π]π!, |π‘| < π.
(13)
Definition 3. Let π β C, 0 < |π| < 1.The π-Genocchi numbersgπ,π
and polynomials Gπ,π(π₯, π¦) are defined by means of the
generating functions:
GΜ (π‘) :=2π‘ππ(βπ‘/2)
ππ(π‘/2) + π
π(βπ‘/2)
=2π‘
Eπ(π‘) + 1
=
β
β
π=0
gπ,π
π‘π
[π]π!, |π‘| < π,
2π‘
Eπ(π‘) + 1
Eπ (π‘π₯)Eπ (π‘π¦)
=
β
β
π=0
Gπ,π(π₯, π¦)
π‘π
[π]π!, |π‘| < π.
(14)
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Abstract and Applied Analysis 3
Note that Cigler [20] defined π-Genocchi numbers as
π‘ππ (π‘) + ππ (βπ‘)
ππ(π‘) + π
π(βπ‘)
=
β
β
π=0
π2π,π
(β1)πβ1(βπ; π)
2πβ1π‘2π
[2π]π!. (15)
Then comparing gπ,π
with ππ,π, we see that
(β1)πβ122π+1
g2π+2,π
= (βπ; π)2π+1
π2π+2,π
. (16)
Definition 4. Let π β C, 0 < |π| < 1. The π-tangent numbersTπ,π
are defined by means of the generating functions:
tanhππ‘ = βπ tan
π (ππ‘) =ππ (π‘) β ππ (βπ‘)
ππ(π‘) + π
π(βπ‘)
=Eπ (2π‘) β 1
Eπ(2π‘) + 1
=
β
β
π=1
T2π+1,π
(β1)ππ‘2π+1
[2π + 1]π!.
(17)
It is obvious that, by letting π tend to 1 from the left side,we lead to the classic definition of these polynomials:
bπ,π:= Bπ,π (0) , lim
πβ1β
Bπ,π (π₯) = π΅π (π₯) ,
limπβ1
β
Bπ,π(π₯, π¦) = π΅
π(π₯ + π¦) , lim
πβ1β
bπ,π= π΅π,
eπ,π:= Eπ,π (0) , lim
πβ1β
Eπ,π (π₯) = πΈπ (π₯) ,
limπβ1
β
Eπ,π(π₯, π¦) = πΈ
π(π₯ + π¦) , lim
πβ1β
eπ,π= πΈπ,
gπ,π:= Gπ,π(0) , lim
πβ1β
Gπ,π(π₯) = πΊ
π(π₯) ,
limπβ1
β
Gπ,π(π₯, π¦) = πΊ
π(π₯ + π¦) lim
πβ1β
gπ,π= πΊπ.
(18)
Here π΅π(π₯), πΈ
π(π₯), and πΊ
π(π₯) denote the classical Bernoulli,
Euler, and Genocchi polynomials, respectively, which aredefined by
π‘
ππ‘ β 1ππ‘π₯=
β
β
π=0
π΅π(π₯)
π‘π
π!,
2
ππ‘ + 1ππ‘π₯=
β
β
π=0
πΈπ(π₯)
π‘π
π!,
2π‘
ππ‘ + 1ππ‘π₯=
β
β
π=0
πΊπ (π₯)
π‘π
π!.
(19)
The aim of the present paper is to obtain some resultsfor the above newly defined π-polynomials. It should bementioned that π-Bernoulli and π-Euler polynomials in ourdefinitions are polynomials of π₯ and π¦ and when π¦ = 0,they are polynomials of π₯. First advantage of this approach isthat for π β 1β, B
π,π(π₯, π¦) (E
π,π(π₯, π¦), G
π,π(π₯, π¦)) becomes
the classical BernoulliBπ(π₯ + π¦) (EulerE
π(π₯ + π¦), Genocchi
Gπ,π(π₯, π¦)) polynomial and wemay obtain the π-analogues of
well-known results, for example, Srivastava and PinteΜr [11],Cheon [23], and so forth. Second advantage is that, similarto the classical case, odd numbered terms of the Bernoullinumbers b
π,πand the Genocchi numbers g
π,πare zero, and
even numbered terms of the Euler numbers eπ,π
are zero.
2. Preliminary Results
In this section we will provide some basic formulae for theπ-Bernoulli, π-Euler, and π-Genocchi numbers and polyno-mials in order to obtain the main results of this paper in thenext section.
Lemma 5. The π-Bernoulli numbers bπ,π
satisfy the followingπ-binomial recurrence:
π
β
π=0
[π
π]
π
(β1, π)πβπ
2πβπbπ,πβ bπ,π= {
1, π = 1,
0, π > 1.(20)
Proof. By a simple multiplication of (8) we see that
BΜ (π‘)Eπ(π‘) = π‘ + BΜ (π‘) . (21)
Soβ
β
π=0
π
β
π=0
[π
π]
π
(β1, π)πβπ
2πβπbπ,π
π‘π
[π]π!= π‘ +
β
β
π=0
bπ,π
π‘π
[π]π!. (22)
The statement follows by comparing π‘π coefficients.
We use this formula to calculate the first few bπ,π:
b0,π= 1,
b1,π= β
1
2,
b2,π=1
4
π (π + 1)
π2 + π + 1=π[2]π
4[3]π
,
b3,π= 0.
(23)
The similar property can be proved for π-Euler numbers
π
β
π=0
[π
π]
π
(β1, π)πβπ
2πβπeπ,π+ eπ,π
= {2, π = 0,
0, π > 0.(24)
and π-Genocchi numbers
π
β
π=0
[π
π]
π
(β1, π)πβπ
2πβπgπ,π+ gπ,π
= {2, π = 1,
0, π > 1.(25)
Using the above recurrence formulae we calculate the firstfew eπ,π
and gπ,π
terms as well:
e0,π= 1, g
0,π= 0,
e1,π= β
1
2, g
1,π= 1,
e2,π= 0, g
2,π= β
[2]π
2= β
π + 1
2,
e3,π=[3]π[2]π β [4]π
8=π (1 + π)
8, g
3,π= 0.
(26)
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4 Abstract and Applied Analysis
Remark 6. The first advantage of the new π-numbersbπ,π, eπ,π, and g
π,πis that similar to classical case odd num-
bered terms of the Bernoulli numbers bπ,π
and the Genocchinumbers g
π,πare zero, and even numbered terms of the Euler
numbers eπ,π
are zero.
Next lemma gives the relationship between π-Genocchinumbers and π-Tangent numbers.
Lemma 7. For any π β N, we have
T2π+1,π
= g2π+2,π
(β1)πβ122π+1
[2π + 2]π
. (27)
Proof. First we recall the definition of π-trigonometric func-tions:
cosππ‘ =
ππ(ππ‘) + π
π(βππ‘)
2, sin
ππ‘ =
ππ(ππ‘) β π
π(βππ‘)
2π,
π tanππ‘ =
ππ(ππ‘) β π
π(βππ‘)
ππ (ππ‘) + ππ (βππ‘)
, cotππ‘ = π
ππ(ππ‘) + π
π(βππ‘)
ππ (ππ‘) β ππ (βππ‘)
.
(28)
Now by choosing π§ = 2ππ‘ in BΜ(π§), we get
BΜ (2ππ‘) =2ππ‘
Eπ(2ππ‘) β 1
=π‘ππ(βππ‘)
sinππ‘
=
β
β
π=0
bπ,π
(2ππ‘)π
[π]π!. (29)
It follows that
BΜ (2ππ‘) =π‘ππ(βππ‘)
sinππ‘
=π‘
sinππ‘(cosππ‘ β π sin
ππ‘) = π‘ cot
ππ‘ β ππ‘
= b0,π+ 2ππ‘b
1,π+
β
β
π=2
bπ,π
(2ππ‘)π
[π]π!
= 1 β ππ‘ +
β
β
π=2
bπ,π
(2ππ‘)π
[π]π!.
(30)
Since the function π‘ cotππ‘ is even in the above sum odd
coefficients b2π+1,π
, π = 1, 2, . . ., are zero, and we get
π‘ cotππ‘ = 1 +
β
β
π=2
bπ,π
(2ππ‘)π
[π]π!= 1 +
β
β
π=1
bπ,π
(2ππ‘)2π
[2π]π!. (31)
By choosing π§ = 2ππ‘ in GΜ(π§), we get
GΜ (2ππ‘) =4ππ‘
Eπ(2ππ‘) + 1
=2ππ‘ππ (βππ‘)
cosππ‘
=
β
β
π=0
gπ,π
(2ππ‘)π
[π]π!,
GΜ (2ππ‘) =4ππ‘
Eπ (2ππ‘) + 1
=2ππ‘ππ (βππ‘)
cosππ‘
=2ππ‘
cosππ‘(cosππ‘ β π sin
ππ‘)
= 2ππ‘ + 2π‘ tanππ‘ = g0,π+ 2ππ‘g
1,π+
β
β
π=2
gπ,π
(2ππ‘)π
[π]π!
= 2ππ‘ +
β
β
π=2
gπ,π
(2ππ‘)π
[π]π!.
(32)
It follows that
2π‘ tanππ‘ =
β
β
π=1
g2π,π
(2ππ‘)2π
[2π]π!,
tanππ‘ =
β
β
π=1
g2π,π
(β1)π(2π‘)2πβ1
[2π]π!,
tanhππ‘ = βπ tan
π(ππ‘) = βπ
β
β
π=1
g2π,π
(β1)π(2ππ‘)2πβ1
[2π]π!
= β
β
β
π=1
g2π,π
(2π‘)2πβ1
[2π]π!= β
β
β
π=1
g2π+2,π
(2π‘)2π+1
[2π + 2]π!,
(33)
Thus
tanhππ‘ = βπtan
π (ππ‘) =ππ(π‘) β π
π(βπ‘)
ππ(π‘) + π
π(βπ‘)
=Eπ(2π‘) β 1
Eπ(2π‘) + 1
=
β
β
π=1
T2π+1,π
(β1)ππ‘2π+1
[2π + 1]π!,
T2π+1,π
= g2π+2,π
(β1)πβ122π+1
[2π + 2]π
.
(34)
The following result is a π-analogue of the additiontheorem, for the classical Bernoulli, Euler, and Genocchipolynomials.
Lemma 8 (addition theorems). For all π₯, π¦ β C we have
Bπ,π(π₯, π¦) =
π
β
π=0
[π
π]
π
bπ,π(π₯ βππ¦)πβπ
,
Bπ,π(π₯, π¦) =
π
β
π=0
[π
π]
π
(β1, π)πβπ
2πβπBπ,π (π₯) π¦
πβπ,
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Abstract and Applied Analysis 5
Eπ,π(π₯, π¦) =
π
β
π=0
[π
π]
π
eπ,π(π₯ βππ¦)πβπ
,
Eπ,π(π₯, π¦) =
π
β
π=0
[π
π]
π
(β1, π)πβπ
2πβπEπ,π(π₯) π¦πβπ,
Gπ,π(π₯, π¦) =
π
β
π=0
[π
π]
π
gπ,π(π₯ βππ¦)πβπ
,
Gπ,π(π₯, π¦) =
π
β
π=0
π
β
π=0
[π
π]
π
(β1, π)πβπ
2πβπGπ,π(π₯) π¦πβπ.
(35)
Proof. We prove only the first formula. It is a consequence ofthe following identity:
β
β
π=0
Bπ,π(π₯, π¦)
π‘π
[π]π!=
π‘
Eπ(π‘) β 1
Eπ(π‘π₯)E
π(π‘π¦)
=
β
β
π=0
bπ,π
π‘π
[π]π!
β
β
π=0
(π₯ βππ¦)π π‘π
[π]π!
=
β
β
π=0
π
β
π=0
[π
π]
π
bπ,π(π₯ βππ¦)πβπ π‘π
[π]π!.
(36)
In particular, setting π¦ = 0 in (35), we get the followingformulae for π-Bernoulli, π-Euler and π-Genocchi polynomi-als, respectively:
Bπ,π(π₯) =
π
β
π=0
[π
π]
π
(β1, π)πβπ
2πβπbπ,ππ₯πβπ,
Eπ,π(π₯) =
π
β
π=0
[π
π]
π
(β1, π)πβπ
2πβπeπ,ππ₯πβπ,
(37)
Gπ,π (π₯) =
π
β
π=0
[π
π]
π
(β1, π)πβπ
2πβπgπ,ππ₯πβπ. (38)
Setting π¦ = 1 in (35), we get
Bπ,π(π₯, 1) =
π
β
π=0
[π
π]
π
(β1, π)πβπ
2πβπBπ,π(π₯) ,
Eπ,π(π₯, 1) =
π
β
π=0
[π
π]
π
(β1, π)πβπ
2πβπEπ,π(π₯) ,
Gπ,π(π₯, 1) =
π
β
π=0
[π
π]
π
(β1, π)πβπ
2πβπGπ,π(π₯) .
(39)
Clearly (39) is π-analogues of
π΅π (π₯ + 1) =
π
β
π=0
(π
π)π΅π (π₯) ,
πΈπ(π₯ + 1) =
π
β
π=0
(π
π)πΈπ(π₯) ,
πΊπ(π₯ + 1) =
π
β
π=0
(π
π)πΊπ(π₯) ,
(40)
respectively.
Lemma 9. The odd coefficients of the π-Bernoulli numbers,except the first one, are zero. That means b
π,π= 0 where
π = 2π + 1 (π β N).
Proof. It follows from the fact that the function
π (π‘) =
β
β
π=0
bπ,π
π‘π
[π]π!β b1,ππ‘
=π‘
Eπ (π‘) β 1
+π‘
2=π‘
2(Eπ (π‘) + 1
Eπ (π‘) β 1
) ,
(41)
By using π-derivative we obtain the next lemma.
Lemma 10. One has
π·π,π₯Bπ,π (π₯) = [π]π
Bπβ1,π (π₯) +Bπβ1,π (ππ₯)
2,
π·π,π₯Eπ,π(π₯) = [π]π
Eπβ1,π
(π₯) + Eπβ1,π
(ππ₯)
2,
π·π,π₯Gπ,π (π₯) = [π]π
Gπβ1,π
(π₯) +Gπβ1,π
(ππ₯)
2.
(42)
Lemma 11 (difference equations). One has
Bπ,π(π₯, 1) βB
π,π(π₯) =
(β1; π)πβ1
2πβ1[π]ππ₯πβ1, π β₯ 1, (43)
Eπ,π (π₯, 1) + Eπ,π (π₯) = 2
(β1; π)π
2ππ₯π, π β₯ 0, (44)
Gπ,π(π₯, 1) +G
π,π(π₯) = 2
(β1; π)πβ1
2πβ1[π]ππ₯πβ1, π β₯ 1. (45)
Proof. Weprove the identity for the π-Bernoulli polynomials.From the identity
π‘Eπ(π‘)
Eπ(π‘) β 1
Eπ (π‘π₯) = π‘Eπ (π‘π₯) +
π‘
Eπ(π‘) β 1
Eπ (π‘π₯) , (46)
it follows thatβ
β
π=0
π
β
π=0
[π
π]
π
(β1, π)πβπ
2πβπBπ,π (π₯)
π‘π
[π]π!
=
β
β
π=0
(β1, π)π
2ππ₯π π‘π+1
[π]π!+
β
β
π=0
Bπ,π(π₯)
π‘π
[π]π!.
(47)
-
6 Abstract and Applied Analysis
From (43) and (37), (44) and (38), we obtain the followingformulae.
Lemma 12. One has
π₯π=
2π
(β1; π)π[π]π
π
β
π=0
[π + 1
π]
π
(β1; π)π+1βπ
2π+1βπBπ,π(π₯) ,
π₯π=
2πβ1
(β1; π)π
(
π
β
π=0
[π
π]
π
(β1; π)πβπ
2πβπEπ,π (π₯) + Eπ,π (π₯)) ,
π₯π=
2πβ1
(β1; π)π[π + 1]π
Γ (
π+1
β
π=0
[π + 1
π]
π
(β1; π)π+1βπ
2π+1βπGπ,π(π₯) +G
π+1,π(π₯)) .
(48)
The above formulae are π-analogues of the followingfamiliar expansions:
π₯π=
1
π + 1
π
β
π=0
(π + 1
π)π΅π(π₯) ,
π₯π=1
2[
π
β
π=0
(π
π)πΈπ (π₯) + πΈπ (π₯)] ,
π₯π=
1
2 (π + 1)[
π+1
β
π=0
(π + 1
π)πΈπ(π₯) + πΈ
π+1(π₯)] ,
(49)
respectively.
Lemma 13. The following identities hold true:
π
β
π=0
[π
π]
π
(β1, π)πβπ
2πβπBπ,π(π₯, π¦) βB
π,π(π₯, π¦)
= [π]π(π₯ βπ π¦)πβ1
,
π
β
π=0
[π
π]
π
(β1, π)πβπ
2πβπEπ,π(π₯, π¦) + E
π,π(π₯, π¦) = 2(π₯ β
ππ¦)π
,
π
β
π=0
[π
π]
π
(β1, π)πβπ
2πβπGπ,π(π₯, π¦) +G
π,π(π₯, π¦)
= 2[π]π(π₯ βπ π¦)πβ1
.
(50)
Proof. We prove the identity for the π-Bernoulli polynomi-als. From the identity
π‘Eπ(π‘)
Eπ(π‘) β 1
Eπ (π‘π₯)Eπ (π‘π¦)
= π‘Eπ(π‘π₯)E
π(π‘π¦) +
π‘
Eπ (π‘) β 1
Eπ(π‘π₯)E
π(π‘π¦) ,
(51)
it follows thatβ
β
π=0
π
β
π=0
[π
π]
π
(β1, π)πβπ
2πβπBπ,π(π₯, π¦)
π‘π
[π]π!
=
β
β
π=0
π
β
π=0
[π
π]
π
(β1, π)π(β1, π)
πβπ
2ππ₯ππ¦πβπ π‘π+1
[π]π!
+
β
β
π=0
Bπ,π(π₯, π¦)
π‘π
[π]π!.
(52)
3. Some New Formulae
The classical Cayley transformation π§ βCay(π§, π) := (1 +ππ§)/(1βππ§)motivated us to derive the formula forE
π(ππ‘). In
addition, by substituting Cay(π§, (π β 1)/2) in the generatingformula we have
BΜπ(ππ‘) BΜ
π (π‘) = (BΜq (ππ‘) β πBΜπ (π‘) (1 + (1 β π)π‘
2))
Γ1
1 β πΓ
2
Eπ (π‘) + 1
.
(53)
The right hand side can be presented by π-Euler numbers.Now equating coefficients of π‘π we get the following identity.In the case that π = 0, we find the first improved π-Eulernumber which is exactly 1.
Proposition 14. For all π β₯ 1,π
β
π=0
[π
π]
π
Bπ,πBπβπ,π
ππ
= βπ
π
β
π=0
[π
π]
π
(β1, π)πβπ
2πβπBπ,πEπβπ,π[π β 1]π
βπ
2
πβ1
β
π=0
[π β 1
π]
π
(β1, π)πβπβ1
2πβπβ1Bπ,πEπβπβ1,π[π]π.
(54)
Let us take a π-derivative from the generating function,after simplifying the equation, by knowing the quotient rulefor quantum derivative, and also using
Eπ(ππ‘) =
1 β (1 β π) (π‘/2)
1 + (1 β π) (π‘/2)Eπ(π‘) ,
π·π(Eq (π‘)) =
Eπ(ππ‘) +E
π (π‘)
2,
(55)
one has
BΜπ(ππ‘) BΜ
π(π‘) =
2 + (1 β π) π‘
2Eπ (π‘) (π β 1)
(πBΜπ(π‘) β BΜ
π(ππ‘)) . (56)
It is clear that Eβ1π(π‘) = E
π(βπ‘). Now, by equating
coefficients of π‘π we obtain the following identity.
-
Abstract and Applied Analysis 7
Proposition 15. For all π β₯ 1,
2π
β
π=0
[2π
π]
π
Bπ,πB2πβπ,π
ππ
= βπ
2π
β
π=0
[2π
π]
π
(β1, π)2πβπ
22πβπBπ,π[π β 1]π(β1)
π
+π (1 β π)
2
2πβ1
β
π=0
[2π β 1
π]
π
(β1, π)2πβ1βπ
22πβ1βπBπ,π
Γ [π β 1]π(β1)π,
2π+1
β
π=0
[2π + 1
π]
π
Bπ,πB2πβπ+1,π
ππ
= π
2π+1
β
π=0
[2π + 1
π]
π
(β1, π)2π+1βπ
22π+1βπBπ,π[π β 1]π(β1)
π
βπ (1 β π)
2
2π
β
π=0
[2π
π]
π
(β1, π)2πβπ
22πβπBπ,π[π β 1]π(β1)
π.
(57)
We may also derive a differential equation for BΜπ(π‘). If
we differentiate both sides of the generating function withrespect to π‘, after a little calculation we find that
π
ππ‘BΜπ(π‘)
= BΜπ (π‘) (
1
π‘β(1 β π)E
π(π‘)
Eπ(π‘) β 1
(
β
β
π=0
4ππ
4 β (1 β π)2π2π
)) .
(58)
If we differentiate BΜπ(π‘) with respect to π, we obtain,
instead,
π
ππBΜπ(π‘) = βBΜ
2
π(π‘)Eπ(π‘)
β
β
π=0
4π‘ (πππβ1
β (π + 1) ππ)
4 β (1 β π)2π2π
. (59)
Again, using the generating function and combining thiswith the derivative we get the partial differential equation.
Proposition 16. Consider the following:
π
ππ‘BΜπ(π‘) β
π
ππBΜπ(π‘)
=BΜπ(π‘)
π‘+
BΜ2π(π‘)Eπ(π‘)
π‘
Γ
β
β
π=0
4π‘ (πππβ1
β (π + 1) ππ) β ππ(1 β π)
4 β (1 β π)2π2π
.
(60)
4. Explicit Relationship between theπ-Bernoulli and π-Euler Polynomials
In this section, we give some explicit relationship betweenthe π-Bernoulli and π-Euler polynomials.We also obtain newformulae and some special cases for them.These formulae areextensions of the formulae of Srivastava and PinteΜr, Cheon,and others.
We present natural π-extensions of themain results in thepapers [9, 11]; see Theorems 17 and 19.
Theorem 17. For π β N0, the following relationships hold true:
Bπ,π(π₯, π¦)
=1
2
π
β
π=0
[π
π]
π
ππβπ [
[
Bπ,π (π₯) +
π
β
π=0
[π
π]
π
(β1, π)πβπ
Bπ,π (π₯)
2πβπππβπ]
]
Γ Eπβπ,π
(ππ¦)
=1
2
π
β
π=0
[π
π]
π
ππβπ
[Bπ,π(π₯) +B
π,π(π₯,
1
π)]Eπβπ,π
(ππ¦) .
(61)
Proof. Using the following identity
π‘
Eπ(π‘) β 1
Eπ(π‘π₯)E
π(π‘π¦)
=π‘
Eπ(π‘) β 1
Eπ (π‘π₯)
β Eπ (π‘/π) + 1
2β
2
Eπ(π‘/π) + 1
Eπ(π‘
πππ¦)
(62)
we have
β
β
π=0
Bπ,π(π₯, π¦)
π‘π
[π]π!
=1
2
β
β
π=0
Eπ,π(ππ¦)
π‘π
ππ[π]π!
β
β
π=0
(β1, π)π
ππ2π
π‘π
[π]π!
β
β
π=0
Bπ,π(π₯)
π‘π
[π]π!
+1
2
β
β
π=0
Eπ,π(ππ¦)
π‘π
ππ[π]π!
β
β
π=0
Bπ,π (π₯)
π‘π
[π]π!
=: πΌ1+ πΌ2.
(63)
It is clear that
πΌ2=1
2
β
β
π=0
Eπ,π(ππ¦)
π‘π
ππ[π]π!
β
β
π=0
Bπ,π(π₯)
π‘π
[π]π!
=1
2
β
β
π=0
π
β
π=0
[π
π]
π
ππβπ
Bπ,π (π₯)Eπβπ,π (ππ¦)
π‘π
[π]π!.
(64)
-
8 Abstract and Applied Analysis
On the other hand
πΌ1=1
2
β
β
π=0
Eπ,π(ππ¦)
π‘π
ππ[π]π!
Γ
β
β
π=0
π
β
π=0
[π
π]
π
Bπ,π (π₯)
(β1, π)πβπ
ππβπ2πβπ
π‘π
[π]π!
=1
2
β
β
π=0
π
β
π=0
[π
π]
π
Eπβπ,π
(ππ¦)
Γ
π
β
π=0
[π
π]
π
Bπ,π(π₯) (β1, π)
πβπ
ππβπππβπ2πβπ
π‘π
[π]π!.
(65)
Thereforeβ
β
π=0
Bπ,π(π₯, π¦)
π‘π
[π]π!
=1
2
β
β
π=0
π
β
π=0
[π
π]
π
ππβπ
Γ [
[
Bπ,π (π₯) +
π
β
π=0
[π
π]
π
(β1, π)πβπ
Bπ,π(π₯)
2πβπππβπ]
]
Γ Eπβπ,π
(ππ¦)π‘π
[π]π!.
(66)
It remains to equate the coefficients of π‘π.
Next we discuss some special cases of Theorem 17.
Corollary 18. For π β N0the following relationship holds true:
Bπ,π(π₯, π¦)
=
π
β
π=0
[π
π]
π
(Bπ,π (π₯) +
(β1; π)πβ1
2π[π]ππ₯πβ1)Eπβπ,π
(π¦) .
(67)
The formula (67) is a π-extension of the Cheonβs mainresult [23].
Theorem 19. For π β N0, the following relationships
Eπ,π(π₯, π¦)
=1
[π + 1]π
Γ
π+1
β
π=0
1
ππ+1βπ[π + 1
π]
π
Γ (
π
β
π=0
[π
π]
π
(β1, π)πβπ
ππβπ2πβπEπ,π(π¦) β E
π,π(π¦))
ΓBπ+1βπ,π (ππ₯)
(68)
hold true between the π-Bernoulli polynomials and π-Eulerpolynomials.
Proof. The proof is based on the following identity:2
Eπ(π‘) + 1
Eπ (π‘π₯)Eπ (π‘π¦)
=2
Eπ (π‘) + 1
Eπ(π‘π¦)
β Eπ(π‘/π) β 1
π‘β
π‘
Eπ(π‘/π) β 1
Eπ(π‘
πππ₯) .
(69)
Indeedβ
β
π=0
Eπ,π(π₯, π¦)
π‘π
[π]π!
=
β
β
π=0
Eπ,π(π¦)
π‘π
[π]π!
β
β
π=0
(β1, π)π
ππ2π
π‘πβ1
[π]π!
Γ
β
β
π=0
Bπ,π(ππ₯)
π‘π
ππ[π]π!
β
β
β
π=0
Eπ,π(π¦)
π‘πβ1
[π]π!
β
β
π=0
Bπ,π(ππ₯)
π‘π
ππ[π]π!
=: πΌ1β πΌ2.
(70)
It follows that
πΌ2=1
π‘
β
β
π=0
Eπ,π(π¦)
π‘π
[π]π!
β
β
π=0
Bπ,π(ππ₯)
π‘π
ππ[π]π!
=1
π‘
β
β
π=0
π
β
π=0
[π
π]
π
1
ππβπEπ,π(π¦)B
πβπ,π (ππ₯)π‘π
[π]π!
=
β
β
π=0
1
[π + 1]π
Γ
π+1
β
π=0
[π + 1
π]
π
1
ππ+1βπEπ,π(π¦)B
π+1βπ,π (ππ₯)π‘π
[π]π!,
πΌ1=1
π‘
β
β
π=0
Bπ,π(ππ₯)
π‘π
ππ[π]π!
Γ
β
β
π=0
π
β
π=0
[π
π]
π
(β1, π)πβπ
ππβπ2πβπEπ,π(π¦)
π‘π
[π]π!
=1
π‘
β
β
π=0
π
β
π=0
[π
π]
π
1
ππβπBπβπ,π
(ππ₯)
Γ
π
β
π=0
[π
π]
π
(β1, π)πβπ
ππβπ2πβπEπ,π(π¦)
π‘πβ1
[π]π!.
(71)
Next we give an interesting relationship between the π-Genocchi polynomials and the π-Bernoulli polynomials.
-
Abstract and Applied Analysis 9
Theorem 20. For π β N0, the following relationship
Gπ,π(π₯, π¦)
=1
[π + 1]π
Γ
π+1
β
π=0
1
ππβπ[π + 1
π]
π
Γ (
π
β
π=0
[π
π]
π
(β1, π)πβπ
ππβπ2πβπGπ,π(π₯) βG
π,π(π₯))
ΓBπ+1βπ,π
(ππ¦) ,
Bπ,π(π₯, π¦)
=1
2[π + 1]π
Γ
π+1
β
π=0
1
ππβπ[π + 1
π]
π
Γ (
π
β
π=0
[π
π]
π
(β1, π)πβπ
ππβπ2πβπBπ,π(π₯) +B
π,π(π₯))
ΓGπ+1βπ,π
(ππ¦)
(72)
holds true between the π-Genocchi and the π-Bernoulli polyno-mials.
Proof. Using the following identity
2π‘
Eπ(π‘) + 1
Eπ(π‘π₯)E
π(π‘π¦)
=2π‘
Eπ (π‘) + 1
Eπ(π‘π₯) β (E
π(π‘
π) β 1)
π
π‘
β π‘/π
Eπ(π‘/π) β 1
β Eπ(π‘
πππ¦)
(73)
we have
β
β
π=0
Gπ,π(π₯, π¦)
π‘π
[π]π!
=π
π‘
β
β
π=0
Gπ,π(π₯, π¦)
π‘π
[π]π!
Γ
β
β
π=0
(β1, π)π
ππ2π
π‘π
[π]π!
β
β
π=0
Bπ,π(ππ¦)
π‘π
ππ[π]π!
βπ
π‘
β
β
π=0
Gπ,π(π₯, π¦)
π‘π
[π]π!
β
β
π=0
Bπ,π(ππ¦)
π‘π
ππ[π]π!
=π
π‘
β
β
π=0
(
π
β
π=0
[π
π]
π
(β1, π)πβπ
ππβπ2πβπGπ,π (π₯) βGπ,π (π₯))
π‘π
[π]π!
Γ
β
β
π=0
Bπ,π(ππ¦)
π‘π
ππ[π]π!
=π
π‘
β
β
π=0
π
β
π=0
1
ππβπ[π
π]
π
Γ (
π
β
π=0
[π
π]
π
(β1, π)πβπ
ππβπ2πβπGπ,π(π₯) βG
π,π(π₯))
ΓBπβπ,π
(ππ¦)π‘π
[π]π!
=
β
β
π=0
1
[π + 1]π
π+1
β
π=0
1
ππβπ[π + 1
π]
π
Γ (
π
β
π=0
[π
π]
π
(β1, π)πβπ
ππβπ2πβπGπ,π(π₯) βG
π,π(π₯))
ΓBπ+1βπ,π
(ππ¦)π‘π
[π]π!.
(74)
The second identity can be proved in a like manner.
Conflict of Interests
The authors declare that there is no conflict of interests withany commercial identities regarding the publication of thispaper.
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Algebra
Discrete Dynamics in Nature and Society
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