Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2013, Article ID 459763, 6 pages
http://dx.doi.org/10.1155/2013/459763
Research Article
Some Identities on the High-Order 𝑞-Euler Numbers and
Polynomials with Weight 0
Jongsung Choi,1 Hyun-Mee Kim,2 and Young-Hee Kim1
1
2
Division of General Education-Mathematics, Kwangwoon University, Seoul 139-701, Republic of Korea
Department of General Education-Mathematics, Kookmin University, Seoul 136-702, Republic of Korea
Correspondence should be addressed to Young-Hee Kim; yhkim@kw.ac.kr
Received 7 February 2013; Accepted 2 April 2013
Academic Editor: Chun-Gang Zhu
Copyright © 2013 Jongsung Choi et al. This is an open access article distributed under the Creative Commons Attribution License,
which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
We construct the 𝑁th order nonlinear ordinary differential equation related to the generating function of 𝑞-Euler numbers with
weight 0. From this, we derive some identities on 𝑞-Euler numbers and polynomials of higher order with weight 0.
1. Introduction
As a well-known definition, the Euler polynomial 𝐸𝑛 (𝑥) is
given by
𝑡𝑛
2 𝑥𝑡 ∞
=
𝐸
𝑒
.
∑
(𝑥)
𝑛
𝑒𝑡 + 1
𝑛!
𝑛=0
(1)
In the special case, 𝑥 = 0, 𝐸𝑛 (0) = 𝐸𝑛 is the 𝑛th Euler number.
From (1), we note that
𝐸0 = 1,
(𝐸 + 1)𝑛 + 𝐸𝑛 = 0,
if 𝑛 > 0,
(2)
with the usual convention of replacing 𝐸𝑛 by 𝐸𝑛 (see [1–16]).
In the viewpoint of the 𝑞-extension of (1) and (2), let us
consider the following 𝑞-Euler number and polynomial:
∞
𝑛
2
𝑥𝑡
̃𝑛,𝑞 (𝑥) 𝑡 ,
𝐸
=
𝑒
∑
𝑞𝑒𝑡 + 1
𝑛!
𝑛=0
𝐸̃0,𝑞 =
2
,
1+𝑞
𝑛
𝑞(𝐸̃𝑞 + 1) + 𝐸̃𝑛,𝑞 = 0,
(3)
if 𝑛 > 0,
(4)
with the usual convention of replacing 𝐸̃𝑞𝑛 by 𝐸̃𝑛,𝑞 .
Equation (3) is called the generating function of 𝑞-Euler
polynomial with weight 0. In the case 𝑥 = 0, 𝐸̃𝑛,𝑞 (0) = 𝐸̃𝑛,𝑞 is
the 𝑛th 𝑞-Euler number with weight 0 (see [5, 11]).
Throughout this paper, let 𝑞 be a complex number with
|𝑞| < 1. As 𝑞 → 1, we obtain (1) and (2) from (3) and (4).
The generating function of Eulerian polynomial 𝐻𝑛 (𝑥 |
𝑢) is defined by
𝑡𝑛
1 − 𝑢 𝑥𝑡 ∞
|
𝑢)
=
𝐻
𝑒
,
∑
(𝑥
𝑛
𝑒𝑡 − 𝑢
𝑛!
𝑛=0
(5)
where 𝑢 ∈ C with 𝑢 ≠ 1. In the special case, 𝑥 = 0, 𝐻𝑛 (0 |
𝑢) = 𝐻𝑛 (𝑢) is called the 𝑛th Eulerian number (see [1–3]).
Sometimes that is called the 𝑛th Frobenius-Euler number (see
[9–11, 15]).
From (1) and (5), we note that 𝐻𝑛 (𝑥 | −1) = 𝐸𝑛 (𝑥). From
(5), we have
𝐻0 (𝑢) = 1,
𝐻𝑛 (1 | 𝑢) − 𝑢𝐻𝑛 (𝑢) = (1 − 𝑢) 𝛿0,𝑛 ,
(6)
where 𝛿𝑛,𝑘 is Kronecker symbol (see [9–11]).
For 𝑁 ∈ N, the 𝑞-Euler polynomial of order 𝑁 is defined
by the generating function as follows:
2
2
) × ⋅⋅⋅ × ( 𝑡
)𝑒𝑥𝑡
𝐺𝑞𝑁 (𝑡, 𝑥) = ( 𝑡
𝑞𝑒 + 1
𝑞𝑒 + 1
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
𝑁-times
(7)
∞
𝑡𝑛
(𝑁)
= ∑ 𝐸̃𝑛,𝑞
(𝑥) .
𝑛!
𝑛=0
(𝑁)
(𝑁)
(0) = 𝐸̃𝑛,𝑞
is called the 𝑛th
In the special case, 𝑥 = 0, 𝐸̃𝑛,𝑞
𝑞-Euler number of order 𝑁 with weight 0 (see [5, 11]).
2
Abstract and Applied Analysis
In [9], Kim derived some identities between the sums
of products of Frobenius-Euler polynomials and FrobeniusEuler polynomials of higher order. The main idea is to construct nonlinear ordinary differential equations with respect
to 𝑡 which are closely related to the generating function
of Frobenius-Euler polynomial. In [3], Choi considered
nonlinear ordinary differential equations with respect to 𝑢
not 𝑡.
In this paper, we construct nonlinear ordinary differential
equations with respect to 𝑞. The purpose of this paper is to
give some new identities on the high order 𝑞-Euler numbers
and polynomials with weight 0 by using the differential
equations of 𝑞.
2. Construction of Nonlinear
Differential Equations
From (13) and (14), we get
𝑁!𝐺𝑁+1 = 𝑁 (𝑁 − 1)!𝐺𝑁
𝑁−1
+ ∑ 𝑘𝑎𝑘 (𝑁) 𝑞𝑘 𝐺(𝑘)
𝑘=0
𝑁
+ ∑ 𝑎𝑘−1 (𝑁) 𝑞𝑘 𝐺(𝑘) ,
𝑘=1
𝑘 (𝑘)
where 𝑁!𝐺𝑁+1 = ∑𝑁
𝑘=0 𝑎𝑘 (𝑁 + 1)𝑞 𝐺 .
By comparing coefficients on both sides of (15), we obtain
the following recurrence relations:
𝑎0 (𝑁 + 1) = 𝑁𝑎0 (𝑁) ,
𝑁-times
(8)
𝑎0 (𝑁 + 1) = 𝑁𝑎0 (𝑁)
for 𝑁 ∈ N.
= 𝑁 (𝑁 − 1) 𝑎0 (𝑁 − 1)
From (7) and (8), we note that
𝐺𝑞𝑁 (𝑡, 𝑥)
𝑁
𝑁
𝑁
𝑁 𝑥𝑡
= 2 𝐺 (𝑡, 𝑥) = 2 𝐺 𝑒 .
(1)
𝑞𝐺
𝑞𝑒𝑡 + 1 − 1
𝑑𝐺
𝐺 𝐺2
=− +
=−
,
2
𝑑𝑞
𝑞
𝑞
𝑞(𝑞𝑒𝑡 + 1)
(9)
By (10) and (13), we have
1
𝑞𝐺(1) + 𝐺 = 𝐺2 = ∑ 𝑎𝑘 (2) 𝑞𝑘 𝐺(𝑘)
𝑘=0
(10)
+𝐺=𝐺 .
From (18) and (19), we get
By differentiating (10) with respect to 𝑞, we get
(𝑁)
𝑁
(11)
𝑎0 (2) = 1,
𝑎1 (2) = 1,
where 𝐺 = 𝑑 𝐺/𝑑𝑞 .
By the derivative of (11) with respect to 𝑞, we have
𝑎𝑁 (𝑁 + 1) = 𝑎𝑁−1 (𝑁) = ⋅ ⋅ ⋅ = 𝑎1 (2) = 1.
(12)
𝑔 (𝑡, 𝑠) = ∑
(13)
𝑘=0
Let us consider the derivative of (13) with respect to 𝑞 to
find the coefficient 𝑎𝑘 (𝑁) in (13).
By (10), we get
𝑑
𝑞 ((𝑁 − 1)! 𝐺𝑁) = 𝑁!𝐺𝑁−1 𝑞𝐺(1)
𝑑𝑞
= 𝑁!𝐺𝑁+1 − 𝑁 (𝑁 − 1)!𝐺𝑁.
∑ 𝑎𝑘 (𝑁)
𝑁≥1 0≤𝑘≤𝑁−1
𝑁−1
= 𝑁!𝐺𝑁−1 (−𝐺 + 𝐺2 )
(20)
𝑡𝑁 𝑘
𝑠 ,
𝑁!
(22)
where |𝑡| < 1 (see [9]).
From (17) and (22), we have
∑
∑ 𝑎𝑘+1 (𝑁 + 1)
𝑁≥1 0≤𝑘≤𝑁−1
= ∑
(14)
(21)
To find 𝑎𝑘 (𝑁) in (13) from (17), we set
Continuing this process, we get
(𝑁 − 1)!𝐺𝑁 = ∑ 𝑎𝑘 (𝑁) 𝑞𝑘 𝐺(𝑘) .
𝑎0 (𝑁) = (𝑁 − 1)!.
From the second part of (16), we have
𝑁
𝑞3 𝐺(3) + 9𝑞2 𝐺(2) + 18𝑞𝐺(1) + 3!𝐺 = 3!𝐺4 .
(19)
= 𝑎0 (2) 𝐺 + 𝑎1 (2) 𝑞𝐺(1) .
2
𝑞2 𝐺(2) + 4𝑞𝐺(1) + 2𝐺 = 2!𝐺3 ,
(18)
= ⋅ ⋅ ⋅ = 𝑁!𝑎0 (2) .
By differentiating (8) with respect to 𝑞, we get
𝐺(1) =
(17)
for 1 ≤ 𝑘 ≤ 𝑁 − 1 and 𝑎𝑘 (𝑁) = 0.
From the first part of (16), we have
1
,
𝑡
𝑞𝑒 + 1
⏟⏟𝑒𝑥𝑡
𝐺⏟⏟⏟⏟⏟⏟⏟
× ⋅⏟⏟⏟
⋅ ⋅⏟⏟⏟⏟⏟⏟⏟
×𝐺
𝐺𝑁 (𝑡, 𝑥) = ⏟⏟
𝑎𝑁 (𝑁 + 1) = 𝑎𝑁−1 (𝑁) , (16)
𝑎𝑘 (𝑁 + 1) = 𝑁𝑎𝑘 (𝑁) + 𝑘𝑎𝑘 (𝑁) + 𝑎𝑘−1 (𝑁) ,
We define
𝐺 = 𝐺 (𝑞) =
(15)
∑ 𝑁𝑎𝑘−1 (𝑁)
𝑁≥1 0≤𝑘≤𝑁−1
+ ∑
𝑡𝑁 𝑘
𝑠
𝑁!
∑
𝑁≥1 0≤𝑘≤𝑁−1
𝑡𝑁 𝑘
𝑠
𝑁!
(𝑘 + 1) 𝑎𝑘+1 (𝑁)
(23)
𝑡𝑁 𝑘
𝑠 + 𝑔 (𝑡, 𝑠) .
𝑁!
Abstract and Applied Analysis
3
From the left hand side of (23), we have
∑ 𝑎𝑘+1 (𝑁 + 1)
∑
𝑁≥1 0≤𝑘≤𝑁−1
We consider Cauchy problem for the following first-order
quasilinear partial differential equation:
𝑡𝑁 𝑘
𝑠
𝑁!
𝑃 (𝑥, 𝑦, 𝑧) 𝑧𝑥 + 𝑄 (𝑥, 𝑦, 𝑧) 𝑧𝑦
= 𝑅 (𝑥, 𝑦, 𝑧) ,
𝑡𝑁−1 𝑘
1
= ∑ ∑ 𝑎𝑘 (𝑁)
𝑠
𝑠 𝑁≥2 1≤𝑘≤𝑁−1
(𝑁 − 1)!
(28)
𝑧 (𝑥0 (𝑡) , 𝑦0 (𝑡)) = 𝑧0 (𝑡) ,
𝑡 ∈ 𝐼,
𝑡𝑁−1 𝑘
𝑡𝑁−1
1
𝑠 − 𝑎0 (𝑁)
)
= ∑ ( ∑ 𝑎𝑘 (𝑁)
𝑠 𝑁≥2 0≤𝑘≤𝑁−1
(𝑁 − 1)!
(𝑁 − 1)!
where 𝐼 is some interval.
We know that (28) has a unique solution under some
conditions as follows.
=
𝑡𝑁−1 𝑘
1
𝑠 − 𝑎0 (1) − ∑ 𝑡𝑁−1 )
( ∑ ∑ 𝑎𝑘 (𝑁)
𝑠 𝑁≥1 0≤𝑘≤𝑁−1
(𝑁 − 1)!
𝑁≥2
Theorem A (see [17, page 65]). Suppose that 𝑃, 𝑄, and 𝑅 are
of class 𝐶1 in a domain Ω of R3 containing the point (𝑥0 , 𝑦0 , 𝑧0 )
and suppose that
=
1
1
(𝑔𝑡 +
),
𝑠
𝑡−1
(24)
where 𝑔𝑡 = 𝜕𝑔/𝜕𝑡. From the first term of the right hand side
of (23), we have
∑ 𝑁𝑎𝑘+1 (𝑁)
∑
𝑁≥1 0≤𝑘≤𝑁−1
𝑃 (𝑥0 , 𝑦0 , 𝑧0 )
𝑑𝑦0 (𝑡0 )
𝑑𝑥 (𝑡 )
− 𝑄 (𝑥0 , 𝑦0 , 𝑧0 ) 0 0 ≠ 0.
𝑑𝑡
𝑑𝑡
Then in a neighborhood 𝑈 of (𝑥0 , 𝑦0 ) there exists a unique
solution of (28) at every point of initial curve contained in 𝑈.
Since (27) satisfies (29) and regularity conditions, there
exists a unique solution of (27).
It is customary to write (27) in the form
𝑁
𝑡 𝑘
𝑠
𝑁!
𝑡𝑁−1 𝑘
𝑡
𝑠
= ∑ ∑ 𝑎𝑘 (𝑁)
𝑠 𝑁≥1 1≤𝑘≤𝑁−1
(𝑁 − 1)!
=
𝑁−1
𝑡
𝑡
( ∑ ∑ 𝑎 (𝑁)
𝑠𝑘
𝑠 𝑁≥1 0≤𝑘≤𝑁−1 𝑘
(𝑁 − 1)!
(25)
𝑡
1
(𝑔 +
).
𝑠 𝑡 𝑡−1
∑
𝑁≥1 0≤𝑘≤𝑁−1
(𝑘 + 1) 𝑎𝑘+1 (𝑁)
= ∑ ∑ 𝑘𝑎𝑘 (𝑁)
𝑁≥1 1≤𝑘≤𝑁
(26)
𝑠 ∈ R.
1−𝑡
.
𝑠
(32)
𝑑𝑔
1
= −𝑔 − .
𝑑𝑠
𝑠
(33)
By the integrating factor method, we have
𝐸𝑖 (𝑠) = ∫
𝑠
(34)
−∞
𝑒𝑟
𝑑𝑟
𝑟
∞
𝑠𝑛
,
𝑛=1 𝑛 ⋅ 𝑛!
= 𝛾 + ln |𝑠| + ∑
where 𝑔𝑠 = 𝜕𝑔/𝜕𝑠.
From (22)–(26), we obtain the following initial value
problem quasilinear first-order partial differential equation:
𝑔 (0, 𝑠) = 0,
𝑔 = 0.
The exponential integral 𝐸𝑖 (𝑠) is defined by
𝑡𝑁 𝑘−1
= ∑ ∑ 𝑘𝑎𝑘 (𝑁)
𝑠 = 𝑔𝑠 ,
𝑁!
𝑁≥1 0≤𝑘≤𝑁−1
|𝑡| < 1,
(31)
𝑠 = 𝑝,
Since 𝑑𝑡/(𝑡 − 1) = 𝑑𝑠/𝑠 is separable, we get
𝑁
(𝑡 − 1) 𝑔𝑡 + 𝑠𝑔𝑠 = −𝑠𝑔 − 1,
𝑡 = 0,
𝑢2 (𝑡, 𝑠, 𝑔) = 𝑒𝑠 𝑔 + 𝐸𝑖 (𝑠) .
𝑡𝑁 𝑘
𝑠
𝑁!
𝑡 𝑘−1
𝑠
𝑁!
(30)
𝑢1 is a solution of partial differential equation of (27).
From (30), we get the linear equation
From the second term of the right hand side of (23), we have
∑
𝑑𝑔
𝑑𝑠
𝑑𝑡
=
=
,
𝑡−1
𝑠
−𝑠𝑔 − 1
𝑢1 (𝑡, 𝑠, 𝑔) =
𝑎 (𝑁) 𝑁−1
−∑ 0
𝑡 )
𝑁≥1 (𝑁 − 1)!
=
(29)
(27)
(35)
(𝑠 ∈ R, 𝑠 ≠ 0) ,
where 𝛾 is Euler constant.
𝑢2 is another solution of partial differential equation of
(27), and 𝑢1 and 𝑢2 are linearly independent.
From the parameterized initial conditions (31), (33), and
(34), we get
𝑢2 = 𝐸𝑖 (
1
),
𝑢1
𝑒𝑠 𝑔 + 𝐸𝑖 (𝑥) = 𝐸𝑖 (
𝑠
).
1−𝑡
(36)
4
Abstract and Applied Analysis
Thus, from (35) and (36), we obtain the following unique
solution of (27):
∞
𝑛
𝑛
𝑠
1
((
) − 1)) .
𝑛
⋅
𝑛!
1
−𝑡
𝑛=1
(37)
𝑔 (𝑡, 𝑠) = 𝑒−𝑠 (− ln |1 − 𝑡| + ∑
In the case of 𝑘 ≥ 𝑁 in (40), from (41), we get
𝑘
(−1)𝑘−𝑙
(−1)𝑘
𝑙+𝑁−1
+∑
(
)
𝑁
𝑁 ⋅ 𝑘! 𝑙=1 (𝑘 − 𝑙)!𝑙 ⋅ 𝑙!
= (−1)𝑘 (
Moreover, if we choose another initial condition
∞
𝑔 (𝑡, 0) = ∑ 𝑎0 (𝑁)
𝑁≥1
𝑁
∞
= (−1)𝑘
𝑁
𝑡
𝑡
= ∑
𝑁! 𝑁≥1 𝑁
(38)
𝑡𝑁)
𝑁≥1
𝑙1 +⋅⋅⋅+𝑙𝑛 =𝑁
+ ∑ (𝑁 − 1)!
𝑁≥1
(39)
= ∑
𝑁≥1
𝑡𝑁
(−1)𝑘 𝑘
𝑠 )(∑ )
𝑘!
𝑁≥1 𝑁
𝑘≥0
(−1)𝑘 𝑘
𝑠 )
𝑘!
𝑘≥0
𝐺𝑁 (𝑞) =
𝑠𝑛
𝑛+𝑁−1 𝑁
)𝑡 )
∑(
𝑁
𝑛≥1 𝑛 ⋅ 𝑛! 𝑁≥1
× (∑
𝑁≥1 1≤𝑘≤𝑁−1
where 𝐺(𝑘) = 𝑑𝑘 𝐺(𝑞) /𝑑𝑞𝑘 and 𝐺𝑁(𝑞) = ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
𝐺(𝑞) × ⋅ ⋅ ⋅ × 𝐺(𝑞).
𝑘−𝑙
(−1)
𝑙+𝑁−1
(
)) 𝑡𝑁𝑠𝑘
𝑁
(𝑘 − 𝑙)! 𝑙 ⋅ 𝑙!
𝑙=1
(∑
𝑘
(−1)𝑘−𝑙
𝑙+𝑁−1
(
)) 𝑡𝑁𝑠𝑘 .
𝑁
−
𝑙)!
𝑙
⋅
𝑙!
(𝑘
𝑙=1
(40)
+ ∑ ∑ (∑
Then 𝐺(𝑞) = 1/(𝑞𝑒𝑡 + 1) is a solution of (45).
It is known that
𝑁-times
Let us define 𝐺(𝑘) (𝑡, 𝑥) = 𝐺(𝑘) (𝑞)𝑒𝑥𝑡 . Then we obtain the
following corollary.
Corollary 2. For 𝑁 ∈ N, one considers
𝐺𝑁 (𝑡, 𝑥) =
2
𝑁−1
1
𝑁−1
) 𝑞𝑘 𝐺(𝑘) (𝑡, 𝑥)
∑ (𝑁 − 𝑘 − 1)!(
𝑘
(𝑁 − 1)! 𝑘=0
𝑁−1
𝑙 −𝑁
𝑙+𝑁−1
( ),
)=
(−1)𝑙 (
𝑁
𝑁 𝑙
𝑘 𝑁
𝑘+𝑁
).
∑( )( ) = (
𝑙
𝑙
𝑘
(45)
1 𝑁 − 1 𝑘 (𝑘)
= ∑ (
)𝑞 𝐺 ,
𝑘
𝑘!
𝑘=0
(−1)𝑘 𝑁 𝑘
(−1)𝑘 𝑁 𝑘
𝑡 𝑠 + ∑ ∑
𝑡 𝑠
𝑁≥1 0≤𝑘≤𝑁−1 𝑁 ⋅ 𝑘!
𝑁≥1 𝑘≥𝑁 𝑁 ⋅ 𝑘!
𝑘
2
𝑁−1
1
𝑁−1
) 𝑞𝑘 𝐺(𝑘)
∑ (𝑁 − 𝑘 − 1)!(
𝑘
(𝑁 − 1)! 𝑘=0
𝑁−1
∑
𝑙=0
(44)
Theorem 1. For 𝑞 ∈ C with |𝑞| < 1 and 𝑁 ∈ N, one can
consider the following nonlinear (𝑁 − 1)th order ordinary
differential equation with respect to 𝑞:
+ (∑
𝑘
2
𝑁 − 1 𝑡𝑁 𝑘
𝑠 ,
)
(𝑁 − 𝑘 − 1)!(
𝑘
𝑁!
0≤𝑘≤𝑁−1
∑
Therefore, by (13) and (44), we obtain the following theorem.
𝑔 (𝑡, 𝑠) = (∑
𝑁≥1 𝑘≥𝑁
(43)
2
By (37) and (39), we get
∑
𝑡𝑁
𝑁!
𝑁−1
𝑎𝑘 (𝑁) = (𝑁 − 𝑘 − 1)!(
).
𝑘
𝑁≥1
+ ∑
(𝑁 − 1)! 𝑘 − 𝑁 𝑡𝑁 𝑘
(
𝑠
)
𝑘
𝑘!
𝑁!
𝑘 𝑁−1
where ( 𝑘−𝑁
𝑘 ) = (−1) ( 𝑘 ). Thus, by (22) and (43), we get
𝑛+𝑁−1 𝑁
= ∑(
)𝑡 .
𝑁
= ∑
∑ (−1)𝑘
𝑁≥1 1≤𝑘≤𝑁−1
𝑛-times
∑
1
𝑘−𝑁
(1 + (
) − 1) = 0.
𝑘
𝑁 ⋅ 𝑘!
𝑔 (𝑡, 𝑠) = ∑
1 𝑛
) − 1 = ( ∑ 𝑡𝑙1 ) × ⋅ ⋅ ⋅ × ( ∑ 𝑡𝑙𝑛 ) − 1
1−𝑡
𝑙1 ≥0
𝑙𝑛 ≥0
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
= ∑(
(42)
By (40) and (41), we get
from (20) and (22), then (37) satisfies it.
We note that
(
1
1 𝑘 𝑘 −𝑁
+
∑ ( ) ( ))
𝑙
𝑁 ⋅ 𝑘! 𝑁 ⋅ 𝑘! 𝑙=1 𝑙
1 𝑁 − 1 𝑘 (𝑘)
(
) 𝑞 𝐺 (𝑡, 𝑥) .
𝑘
𝑘!
𝑘=0
= ∑
(41)
(46)
Then 𝐺(𝑡, 𝑥) = 𝑒𝑥𝑡 /(𝑞𝑒𝑡 + 1) is a solution of (46).
Abstract and Applied Analysis
5
∞
3. Identities on the High-Order
𝑞-Euler Numbers and
Polynomials with Weight 0
𝐺𝑁 (𝑞) =
1
2
2
( 𝑡
) × ⋅⋅⋅ × ( 𝑡
)
2𝑁 ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
𝑞𝑒 + 1
𝑞𝑒 + 1
𝑛=0
𝑙1 +⋅⋅⋅+𝑙𝑁 =𝑛
= ∑(
∑
𝑛=0
𝑙1 +⋅⋅⋅+𝑙𝑁 =𝑛
𝑡𝑛
𝑛!
𝑡𝑛
𝑛
(
) 𝐸̃𝑙1 ,𝑞 ⋅ ⋅ ⋅ 𝐸̃𝑙𝑁 ,𝑞 ) .
𝑙1 , . . . , 𝑙𝑁
𝑛!
Therefore, by (49) and (52), we obtain the following corollary.
1 ∞ (𝑁) 𝑡𝑛
= 𝑁 ∑ 𝐸̃𝑛,𝑞
,
2 𝑛=0
𝑛!
(47)
𝑡𝑛
1 2
1∞
= ∑ 𝐸̃𝑛,𝑞 .
𝑡
2 𝑞𝑒 + 1 2 𝑛=0
𝑛!
Corollary 5. For 𝑁 ∈ N and 𝑛 ∈ N ∪ {0}, one has
∑
𝑙1 +⋅⋅⋅+𝑙𝑁 =𝑛
𝑛
(
) 𝐸̃ ⋅ ⋅ ⋅ 𝐸̃𝑙𝑁 ,𝑞
𝑙1 , . . . , 𝑙𝑁 𝑙1 ,𝑞
𝑁−1
From (47), we note that
𝐺(𝑘) =
𝑙1 ! ⋅ ⋅ ⋅ 𝑙𝑁!
)
(52)
𝑁-times
𝐺 (𝑞) =
∑
∞
From (3), (7), and (8), we get
𝑛!𝐸̃𝑙1 ,𝑞 ⋅ ⋅ ⋅ 𝐸̃𝑙𝑁 ,𝑞
= ∑(
=2
𝑑𝑘 𝐺 (𝑞) 1 ∞ 𝑑 𝐸̃𝑛,𝑞 𝑡𝑛
= ∑
.
2 𝑛=0 𝑑𝑞𝑘 𝑛!
𝑑𝑞𝑘
𝑘
(48)
𝑘
1 𝑁 − 1 𝑘 𝑑 𝐸̃𝑛,𝑞
.
)𝑞
∑ (
𝑘
𝑘!
𝑑𝑞𝑘
𝑘=0
𝑁−1
(53)
Acknowledgment
Therefore, by (47), (48), and (45), we obtain the following
theorem.
The present research has been conducted by the research
grant of Kwangwoon University in 2013.
Theorem 3. For 𝑁 ∈ N and 𝑛 ∈ N ∪ {0}, one has
References
𝑘
𝑁−1
1 𝑁 − 1 𝑘 𝑑 𝐸̃𝑛,𝑞
(𝑁)
= 2𝑁−1 ∑ (
.
𝐸̃𝑛,𝑞
)𝑞
𝑘
𝑘!
𝑑𝑞𝑘
𝑘=0
(49)
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𝑘
∞ 𝑛 𝑛
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𝑘
𝑁−1
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2
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𝑞𝑒 + 1
𝑁-times
∞
= ( ∑ 𝐸̃𝑙1 ,𝑞
𝑙1 =0
∞
𝑡𝑙1
𝑡𝑙𝑁
) × ⋅ ⋅ ⋅ × ( ∑ 𝐸̃𝑙𝑁 ,𝑞 )
𝑙1 !
𝑙𝑁!
𝑙 =0
𝑁
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