Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2013, Article ID 871512, 6 pages
http://dx.doi.org/10.1155/2013/871512
Research Article
Umbral Calculus and the Frobenius-Euler Polynomials
Dae San Kim,1 Taekyun Kim,2 and Sang-Hun Lee3
1
Department of Mathematics, Sogang University, Seoul 121-742, Republic of Korea
Department of Mathematics, Kwangwoon University, Seoul 139-701, Republic of Korea
3
Division of General Education, Kwangwoon University, Seoul 139-701, Republic of Korea
2
Correspondence should be addressed to Taekyun Kim; tkkim@kw.ac.kr
Received 27 November 2012; Accepted 19 December 2012
Academic Editor: Juan J. Trujillo
Copyright © 2013 Dae San Kim 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 study some properties of umbral calculus related to the Appell sequence. From those properties, we derive new and interesting
identities of the Frobenius-Euler polynomials.
1. Introduction
Let C be the complex number field. For 𝜆 ∈ C with 𝜆 ≠ 1, the
Frobenius-Euler polynomials are defined by the generating
function to be
∞
𝑡𝑛
1 − 𝜆 𝑥𝑡
𝐻(𝑥|𝜆)𝑡
=
𝑒
=
𝐻
𝑒
,
|
𝜆)
∑
(𝑥
𝑛
𝑒𝑡 − 𝜆
𝑛!
𝑛=0
For 𝑟 ∈ Z+ , the Frobenius-Euler polynomials of order 𝑟
are defined by the generating function to be
(
1 − 𝜆 𝑟 𝑥𝑡
1−𝜆
1 − 𝜆 𝑥𝑡
)𝑒 =( 𝑡
) × ⋅⋅⋅ × ( 𝑡
)𝑒
𝑒𝑡 − 𝜆
𝑒 −𝜆
𝑒 −𝜆
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
𝑟-times
∞
(1)
= ∑ 𝐻𝑛(𝑟) (𝑥 | 𝜆)
𝑛=0
(see [1–5]) with the usual convention about replacing 𝐻𝑛 (𝑥 |
𝜆) by 𝐻𝑛 (𝑥 | 𝜆).
In the special case, 𝑥 = 0, 𝐻𝑛 (0 | 𝜆) = 𝐻𝑛 (𝜆) are called
the 𝑛th Frobenius-Euler numbers. By (1), we get
𝑛
𝑛
(𝑟)
𝐻𝑛(𝑟) (𝑥 | 𝜆) = ∑ ( ) 𝐻𝑛−𝑙
(𝜆) 𝑥𝑙 ,
𝑙
(2)
𝑙=0
𝑙=0
(see [6–9]) with the usual convention about replacing 𝐻𝑛 (𝜆)
by 𝐻𝑛 (𝜆).
Thus, from (1) and (2), we note that
(𝐻 (𝜆) + 1)𝑛 − 𝜆𝐻𝑛 (𝜆) = (1 − 𝜆) 𝛿0,𝑛 ,
where 𝛿𝑛,𝑘 is the kronecker symbol (see [1, 10, 11]).
𝑡
.
𝑛!
In the special case, 𝑥 = 0, 𝐻𝑛(𝑟) (0 | 𝜆) = 𝐻𝑛(𝑟) (𝜆) are called the
𝑛th Frobenius-Euler numbers of order 𝑟 (see [1, 10]).
From (4), we can derive the following equation:
𝑛
𝑛
𝐻𝑛 (𝑥 | 𝜆) = ∑ ( ) 𝐻𝑛−𝑙 (𝜆) 𝑥𝑙 = (𝐻 (𝜆) + 𝑥)𝑛 ,
𝑙
(4)
𝑛
(3)
𝐻𝑛(𝑟)
(𝜆) =
∑
𝑙1 +⋅⋅⋅+𝑙𝑟 =𝑛
𝑛
(
) 𝐻𝑙1 (𝜆) ⋅ ⋅ ⋅ 𝐻𝑙𝑟 (𝜆) .
𝑙1 , . . . , 𝑙𝑟
(5)
By (5), we see that 𝐻𝑛(𝑟) (𝑥 | 𝜆) is a monic polynomial of degree
𝑛 with coefficients in Q(𝜆).
Let P be the algebra of polynomials in the single variable
𝑥 over C and let P∗ be the vector space of all linear functionals
on P. As is known, ⟨𝐿 | 𝑝(𝑥)⟩ denotes the action of the
linear functional 𝐿 on a polynomial 𝑝(𝑥) and we remind that
2
Abstract and Applied Analysis
the addition and scalar multiplication on P∗ are, respectively,
defined by
⟨𝐿 + 𝑀 | 𝑝 (𝑥)⟩ = ⟨𝐿 | 𝑝 (𝑥)⟩ + ⟨𝑀 | 𝑝 (𝑥)⟩,
⟨𝑐𝐿 | 𝑝 (𝑥)⟩ = 𝑐⟨𝐿 | 𝑝 (𝑥)⟩,
(6)
where 𝑐 is a complex constant (see [3, 12]).
Let F denote the algebra of formal power series:
2. The Frobenius-Euler Polynomials and
Umbral Calculus
∞
𝑎𝑘 𝑘
𝑡 | 𝑎𝑘 ∈ C}
𝑘!
𝑘=0
F = {𝑓 (𝑡) = ∑
(7)
(see [3, 12]). The formal power series define a linear functional
on P by setting
⟨𝑓 (𝑡) | 𝑥𝑛 ⟩ = 𝑎𝑛 ,
∀𝑛 ≥ 0.
where 𝑓(𝑡) is the compositional inverse of 𝑓(𝑡) (see [3]).
In this paper, we study some properties of umbral calculus
related to the Appell sequence. For those properties, we
derive new and interesting identities of the Frobenius-Euler
polynomials.
(8)
By (4) and (12), we see that
𝐻𝑛(𝑟) (𝑥 | 𝜆) ∼ ((
𝑟
Indeed, by (7) and (8), we get
(𝑛, 𝑘 ≥ 0)
(9)
(see [3, 12]). This kind of algebra is called an umbral algebra.
The order 𝑂(𝑓(𝑡)) of a nonzero power series 𝑓(𝑡) is the
smallest integer 𝑘 for which the coefficient of 𝑡𝑘 does not
vanish. A series 𝑓(𝑡) for which 𝑂(𝑓(𝑡)) = 1 is said to be an
invertible series (see [2, 12]). For 𝑓(𝑡), 𝑔(𝑡) ∈ F, and 𝑝(𝑥) ∈ P,
we have
⟨𝑓 (𝑡) 𝑔 (𝑡) | 𝑝 (𝑥)⟩ = ⟨𝑓 (𝑡) | 𝑔 (𝑡) 𝑝 (𝑥)⟩
= ⟨𝑔 (𝑡) | 𝑓 (𝑡) 𝑝 (𝑥)⟩
∞
⟨ℎ (𝑡) | 𝑠𝑘 (𝑥)⟩
𝑔 (𝑡) 𝑓(𝑡)𝑘 ,
𝑘!
𝑘=0
ℎ (𝑡) = ∑
∞
𝑝 (𝑥) = ∑
⟨𝑔 (𝑡) 𝑓(𝑡)𝑘 | 𝑝 (𝑥)⟩
𝑘!
𝑘=0
𝑠𝑘 (𝑥) ,
⟨𝑓 (𝑡) | 𝑝 (𝛼𝑥)⟩ = ⟨𝑓 (𝛼𝑡 | 𝑝 (𝑥)⟩ ,
𝑔 (𝑓 (𝑡))
∞
𝑠𝑘 (𝑦) 𝑘
𝑡 ,
𝑘!
𝑘=0
𝑒𝑦𝑓(𝑡) = ∑
∀𝑦 ∈ C,
(12)
(14)
(15)
Then it is an (𝑛 + 1)-dimensional vector space over Q(𝜆).
So we see that {𝐻0(𝑟) (𝑥 | 𝜆), 𝐻1(𝑟) (𝑥 | 𝜆), . . . , 𝐻𝑛(𝑟) (𝑥 | 𝜆)} is
a basis for P𝑛 (𝜆). For 𝑝(𝑥) ∈ P𝑛 (𝜆), let
𝑛
𝑝 (𝑥) = ∑ 𝐶𝑘 𝐻𝑘(𝑟) (𝑥 | 𝜆) ,
(𝑛 ≥ 0) .
(16)
𝑘=0
Then, by (13), (14), and (16), we get
𝑟
𝑒𝑡 − 𝜆 𝑘
⟨(
) 𝑡 | 𝑝 (𝑥)⟩
1−𝜆
𝑟
𝑛
= ∑𝐶𝑙 ⟨(
𝑙=0
𝑒𝑡 − 𝜆 𝑘
) 𝑡 | 𝐻𝑙(𝑟) (𝑥 | 𝜆)⟩
1−𝜆
(17)
𝑛
= ∑𝐶𝑙 𝑙!𝛿𝑙,𝑘 = 𝑘!𝐶𝑘 .
𝑙=0
From (17), we have
𝑟
𝐶𝑘 =
1
𝑒𝑡 − 𝜆 𝑘
⟨(
) 𝑡 | 𝑝 (𝑥)⟩
𝑘!
1−𝜆
𝑟
=
1
𝑒𝑡 − 𝜆
⟨(
) | 𝐷𝑘 𝑝 (𝑥)⟩
𝑘!
1−𝜆
=
𝑟
1
𝑟
𝑟−𝑗
⟨𝑒𝑗𝑡 | 𝐷𝑘 𝑝 (𝑥)⟩
𝑟 ∑ (𝑗) (−𝜆)
𝑘!(1 − 𝜆) 𝑗=0
=
𝑟
1
𝑟
𝑟−𝑗
⟨𝑡0 | 𝑒𝑗𝑡 𝐷𝑘 𝑝 (𝑥)⟩
𝑟 ∑ (𝑗) (−𝜆)
𝑘!(1 − 𝜆) 𝑗=0
=
𝑟
1
𝑟
( ) (−𝜆)𝑟−𝑗 ⟨𝑡0 | 𝐷𝑘 𝑝 (𝑥 + 𝑗)⟩ .
∑
𝑘!(1 − 𝜆)𝑟 𝑗=0 𝑗
(11)
𝑓 (𝑡) 𝑠𝑛 (𝑥) = 𝑛𝑠𝑛−1 (𝑥) ,
1
P𝑛 (𝜆) = {𝑝 (𝑥) ∈ Q (𝜆) [𝑥] | deg 𝑝 (𝑥) ≤ 𝑛} .
ℎ (𝑡) ∈ F,
𝑝 (𝑥) ∈ P,
𝑒𝑡 − 𝜆 𝑘
) 𝑡 | 𝐻𝑛(𝑟) (𝑥 | 𝜆)⟩ = 𝑛!𝛿𝑛,𝑘 .
1−𝜆
Let
(10)
(see [12]). One should keep in mind that each 𝑓(𝑡) ∈ F plays
three roles in the umbral calculus: a formal power series, a
linear functional, and a linear operator. To illustrate this, let
𝑝(𝑥) ∈ P and 𝑓(𝑡) = 𝑒𝑦𝑡 ∈ F. As a linear functional, 𝑒𝑦𝑡
satisfies ⟨𝑒𝑦𝑡 | 𝑝(𝑥)⟩ = 𝑝(𝑦). As a linear operator, 𝑒𝑦𝑡 satisfies
𝑒𝑦𝑡 𝑝(𝑥) = 𝑝(𝑥 + 𝑦) (see [12]). Let 𝑠𝑛 (𝑥) denote a polynomial
in 𝑥 with degree 𝑛. Let us assume that 𝑓(𝑡) is a delta series
and 𝑔(𝑡) is an invertible series. Then there exists a unique
sequence 𝑠𝑛 (𝑥) of polynomials such that ⟨𝑔(𝑡)𝑓(𝑡)𝑘 | 𝑠𝑛 (𝑥)⟩ =
𝑛!𝛿𝑛,𝑘 for all 𝑛, 𝑘 ≥ 0 (see [3, 12]). This sequence 𝑠𝑛 (𝑥) is
called the Sheffer sequence for (𝑔(𝑡), 𝑓(𝑡)) which is denoted
by 𝑠𝑛 (𝑥) ∼ (𝑔(𝑡), 𝑓(𝑡)). If 𝑠𝑛 (𝑥) ∼ (1, 𝑓(𝑡)), then 𝑠𝑛 (𝑥) is called
the associated sequence for 𝑓(𝑡). If 𝑠𝑛 (𝑥) ∼ (𝑔(𝑡), 𝑡), then
𝑠𝑛 (𝑥) is called the Appell sequence.
Let 𝑠𝑛 (𝑥) ∼ (𝑔(𝑡), 𝑓(𝑡)). Then we see that
(13)
Thus, by (13), we get
⟨(
⟨𝑡𝑘 | 𝑥𝑛 ⟩ = 𝑛!𝛿𝑛,𝑘
𝑟
𝑒𝑡 − 𝜆
) , 𝑡) .
1−𝜆
(18)
Therefore, by (16) and (18), we obtain the following theorem.
Abstract and Applied Analysis
3
Theorem 1. For 𝑝(𝑥) ∈ P𝑛 (𝜆), let
For 𝑠 ∈ Z+ , from (25), we have
𝐽𝜆𝑠 (𝐻𝑛(𝑟) (𝑥 | 𝜆)) = 𝐻𝑛(𝑟−𝑠) (𝑥 | 𝜆) .
𝑛
𝑝 (𝑥) = ∑ 𝐶𝑘 𝐻𝑘(𝑟) (𝑥) .
(19)
On the other hand, by (12), (13), and (25),
𝑘=0
Then one has
𝐶𝑘 =
𝑟
1
𝑟
𝑟−𝑗 𝑘
𝐷 𝑝 (𝑗) ,
𝑟 ∑ (𝑗) (−𝜆)
𝑘!(1 − 𝜆) 𝑗=0
𝐽𝜆𝑠 (𝐻𝑛(𝑟) (𝑥 | 𝜆)) = (
(20)
𝑠
1
(1 − 𝜆)𝑟
Thus, by (28), we get
𝑛 { 𝑟
}
1 𝑟
⋅ ∑ { ∑ ( ) (−𝜆)𝑟−𝑗 𝐷𝑘 𝑝 (𝑗)} 𝐻𝑘(𝑟) (𝑥 | 𝜆) .
𝑗
𝑘!
𝑘=0 𝑗=0
{
}
(21)
𝐽𝜆𝑠 (𝐻𝑛(𝑟) (𝑥 | 𝜆))
̃ 𝜆 𝑓(𝑥) = 𝑓(𝑥 + 1) −
̃ 𝜆 with Δ
Let us consider the operator Δ
̃ 𝜆 . Then we have
𝜆𝑓(𝑥) and let 𝐽𝜆 = (1/(1 − 𝜆))Δ
1
{𝑓 (𝑥 + 1) − 𝜆𝑓 (𝑥)} .
1−𝜆
𝑠
( 𝑚𝑠 ) ∞
∑
𝑚∑(
𝑚=0 (1 − 𝜆) 𝑙=𝑚
𝑘 +⋅⋅⋅+𝑘
= ∑
1
(22)
( 𝑚𝑠 ) ∞ 1
∑
𝑚∑ (
𝑚=0 (1 − 𝜆) 𝑙=𝑚 𝑙!
𝑘 +⋅⋅⋅+𝑘
𝑠
1
1
{𝐻𝑛(𝑟) (𝑥 + 1 | 𝜆) − 𝜆𝐻𝑛(𝑟) (𝑥 | 𝜆)} .
1−𝜆
(23)
∞
∑ {𝐻𝑛(𝑟) (𝑥 + 1 | 𝜆) − 𝜆𝐻𝑛(𝑟) (𝑥 | 𝜆)}
𝑛=0
1 − 𝜆 𝑟 𝑥𝑡
1 − 𝜆 𝑟 (𝑥+1)𝑡
𝑒
−
𝜆(
)
)𝑒
𝑒𝑡 − 𝜆
𝑒𝑡 − 𝜆
=(
1 − 𝜆 𝑟 𝑥𝑡 𝑡
1 − 𝜆 𝑟−1 𝑥𝑡
) 𝑒 (𝑒 − 𝜆) = (1 − 𝜆) ( 𝑡
) 𝑒
𝑡
𝑒 −𝜆
𝑒 −𝜆
𝑛=0
min{𝑠,𝑛}
∑
𝑚=0
𝑡𝑛
𝑛!
=(
= (1 − 𝜆) ∑ 𝐻𝑛(𝑟−1) (𝑥 | 𝜆)
𝑙
(
) 𝐷𝑙 )
𝑘1 , . . . , 𝑘𝑚
=
( 𝑚𝑠 ) 𝑛 𝑛
∑( ) ∑
(1 − 𝜆)𝑚 𝑙=𝑚 𝑙 𝑘 +⋅⋅⋅+𝑘
1
𝑚 =𝑙
𝑙
(𝑟)
(
) 𝐻𝑛−𝑙
(𝑥 | 𝜆)
𝑘1 , . . . , 𝑘𝑚
𝑘𝑗 ≥1
min{𝑠,𝑛} {
{
{
𝑙
( 𝑚𝑠 )
𝑛
∑ {( ) ∑
{ 𝑙 𝑚=0 (1 − 𝜆)𝑚
𝑙=0 {
{
}
}
} (𝑟)
𝑙
⋅ ∑ (
)} 𝐻𝑛−𝑙
(𝑥 | 𝜆)
𝑘1 , . . . , 𝑘𝑚 }
}
𝑘1 +⋅⋅⋅+𝑘𝑚 =𝑙
𝑘𝑗 ≥1
}
𝑛
𝑡
.
𝑛!
(24)
By (23) and (24), we get
𝑚 =𝑙
𝑘𝑗 ≥1
⋅ 𝐻𝑛(𝑟) (𝑥 | 𝜆)
=
From (4), we can derive
∞
𝑚
𝑘𝑗 ≥1
1
) 𝑡𝑙 (𝐻𝑛(𝑟) (𝑥 | 𝜆))
𝑘
!
⋅
⋅
⋅
𝑘
!
1
𝑚
=𝑙
= ∑
Thus, by (22), we get
𝐽𝜆 (𝐻𝑛(𝑟) (𝑥 | 𝜆)) =
(28)
⋅ (𝐻𝑛(𝑟) (𝑥 | 𝜆)) .
From Theorem 1, we note that
𝐽𝜆 (𝑓) (𝑥) =
𝑠
𝑒𝑡 − 𝜆
) (𝐻𝑛(𝑟) (𝑥 | 𝜆))
1−𝜆
∞ 𝑘
𝑡
1
((1
−
𝜆)
+
)
=
∑
𝑠
𝑘!
(1 − 𝜆)
𝑘=1
where 𝐷𝑝(𝑥) = 𝑑𝑝(𝑥)/𝑑𝑥.
𝑝 (𝑥) =
(27)
{
{
{ 𝑛 min{𝑠,𝑛} ( 𝑚𝑠 )
+
( ) ∑
∑
{
{ 𝑙 𝑚=0 (1 − 𝜆)𝑚
𝑙=min{𝑠,𝑛}+1 {
{
𝑛
𝐽𝜆 (𝐻𝑛(𝑟) (𝑥 | 𝜆)) = 𝐻𝑛(𝑟−1) (𝑥 | 𝜆) .
(25)
From (25), we have
𝐽𝜆𝑟 (𝐻𝑛(𝑟) (𝑥 | 𝜆)) = 𝐽𝜆𝑟−1 (𝐻𝑛(𝑟−1) (𝑥 | 𝜆))
= ⋅ ⋅ ⋅ = 𝐻𝑛(0) (𝑥 | 𝜆) = 𝑥𝑛 ,
𝐽𝜆𝑟 (𝑥𝑛 ) = 𝐽𝜆𝑟 𝐻𝑛(0) (𝑥 | 𝜆) = 𝐻𝑛(−𝑟) (𝑥 | 𝜆) = 𝐽𝜆2𝑟 𝐻𝑛(𝑟) (𝑥 | 𝜆) .
(26)
}
}
} (𝑟)
𝑙
⋅
)} 𝐻𝑛−𝑙
∑ (
(𝑥 | 𝜆) .
𝑘1 , . . . , 𝑘𝑚 }
}
𝑘1 +⋅⋅⋅+𝑘𝑚 =𝑙
𝑘𝑗 ≥1
}
(29)
Therefore, by (27) and (29), we obtain the following theorem.
4
Abstract and Applied Analysis
Theorem 2. For any 𝑟, 𝑠 ≥ 0, one has
{
{
{ 𝑛 min{𝑟,𝑛} ( 𝑚𝑟 )
( ) ∑
+
∑
{
{ 𝑙 𝑚=0 (1 − 𝜆)𝑚
𝑙=min{𝑟,𝑛}+1 {
{
𝑛
𝐻𝑛(𝑟−𝑠) (𝑥 | 𝜆)
=
min{𝑠,𝑛} {
{
{
}
𝑙
}
}
( 𝑚𝑠 )
𝑙
𝑛
)}
∑ {( ) ∑
∑ (
𝑚
,
.
.
.
,
𝑘
𝑘
𝑙
{
1
𝑚 }
}
𝑚=0 (1 − 𝜆) 𝑘1 +⋅⋅⋅+𝑘𝑚 =𝑙
𝑙=0 {
𝑘
≥1
𝑗
{
}
}
}
} (𝑟)
𝑙
⋅ ∑ (
)} 𝐻𝑛−𝑙
(𝑥 | 𝜆) .
𝑘1 , . . . , 𝑘𝑚 }
}
𝑘1 +⋅⋅⋅+𝑘𝑚 =𝑙
𝑘𝑗 ≥1
}
(32)
(𝑟)
⋅ 𝐻𝑛−𝑙
(𝑥 | 𝜆)
{
{
{ 𝑛 min{𝑠,𝑛} ( 𝑚𝑠 )
+
( ) ∑
∑
{
{ 𝑙 𝑚=0 (1 − 𝜆)𝑚
𝑙=min{𝑠,𝑛}+1 {
{
𝑛
}
}
} (𝑟)
𝑙
⋅ ∑ (
)} 𝐻𝑛−𝑙
(𝑥 | 𝜆) .
𝑘1 , . . . , 𝑘𝑚 }
}
𝑘1 +⋅⋅⋅+𝑘𝑚 =𝑙
𝑘𝑗 ≥1
}
(30)
Now, we define the analogue of Stirling numbers of the
second kind as follows:
𝑆𝜆 (𝑛, 𝑘) =
=
∑
𝑙=0
By (33) and (34), we get
{
{
{ 𝑛 min{𝑟−1,𝑛} ( 𝑟−1
𝑚 )
+
( ) ∑
∑
{
𝑙
{
(1
−
𝜆)𝑚
𝑚=0
𝑙=min{𝑟−1,𝑛}+1 {
{
𝑛
}
}
} (𝑟)
𝑙
⋅ ∑ (
)} 𝐻𝑛−𝑙
(𝑥 | 𝜆) .
𝑘1 , . . . , 𝑘𝑚 }
}
𝑘1 +⋅⋅⋅+𝑘𝑚 =𝑙
𝑘𝑗 ≥1
}
(31)
(35)
𝐽𝜆𝑟 𝑥𝑛
=
min{2𝑟,𝑛} {
{
{
}
𝑙
}
}
( 2𝑟
𝑙
𝑛
𝑚)
(
)
(
)
∑
∑
𝑚
{
}
,
.
.
.
,
𝑘
𝑘
𝑙
{
1
𝑚 }
{
}
𝑚=0 (1 − 𝜆) 𝑘1 +⋅⋅⋅+𝑘𝑚 =𝑙
𝑘𝑗 ≥1
{
}
∑
𝑙=0
(𝑟)
⋅ 𝐻𝑛−𝑙
(𝑥 | 𝜆)
{
{
{ 𝑛 min{2𝑟,𝑛} ( 2𝑟
𝑚)
+
( ) ∑
∑
𝑚
{
𝑙
{
𝑚=0 (1 − 𝜆)
𝑙=min{2𝑟,𝑛}+1 {
{
𝑛
}
}
} (𝑟)
𝑙
(
)} 𝐻𝑛−𝑙
(𝑥 | 𝜆) ,
𝑘1 , . . . , 𝑘𝑚 }
}
𝑘1 +⋅⋅⋅+𝑘𝑚 =𝑙
𝑘𝑗 ≥1
}
⋅
Corollary 4. For 𝑛 ≥ 0, 𝑟 ≥ 1, one has
}
}
}
𝑙
𝑛
(
)
∑ {( ) ∑
∑
𝑚
{ 𝑙 𝑚=0 (1 − 𝜆) 𝑘 +⋅⋅⋅+𝑘 =𝑙 𝑘1 , . . . , 𝑘𝑚 }
}
}
𝑙=0 {
1
𝑚
𝑘𝑗 ≥1
{
}
(𝑟)
⋅ 𝐻𝑛−𝑙
(𝑥 | 𝜆)
(𝑛, 𝑘 ≥ 0) .
Let us take 𝑠 = 2𝑟. Then we have
Let us take 𝑠 = 𝑟 (𝑟 ≥ 1) in Theorem 2. Then we obtain
the following corollary.
𝑥𝑛 =
1 ̃𝑘 𝑛
Δ 0 ,
𝑘! 𝜆
= 𝐻𝑛(−𝑟) (𝑥 | 𝜆)
(𝑟)
⋅ 𝐻𝑛−𝑙
(𝑥 | 𝜆)
min{𝑟,𝑛} {
{
{
(34)
𝑘=0
𝑆𝜆 (𝑛, 𝑘) =
}
𝑙
}
}
( 𝑟−1
𝑙
𝑛
𝑚 )
(
)
(
)
∑
∑
𝑚
{
}
,
.
.
.
,
𝑘
𝑘
𝑙
{
(1
−
𝜆)
1
𝑚 }
{
}
𝑚=0
𝑘1 +⋅⋅⋅+𝑘𝑚 =𝑙
𝑘
≥1
𝑗
{
}
(33)
𝑛
̃ 𝑛 𝑓 (0) = ∑ (𝑛) (−𝜆)𝑛−𝑘 𝑓 (𝑘) .
Δ
𝜆
𝑘
𝐻𝑛 (𝑥 | 𝜆)
min{𝑟−1,𝑛} {
{
{
(𝑛, 𝑘 ≥ 0) .
Note that 𝑆1 (𝑛, 𝑘) = 𝑆(𝑛, 𝑘) is the Stirling number of the
second kind.
̃ 𝜆 , we have
From the definition of Δ
Let us take 𝑠 = 𝑟−1 (𝑟 ≥ 1) in Theorem 2. Then we obtain
the following corollary.
Corollary 3. For 𝑛 ≥ 0, 𝑟 ≥ 1, one has
1 𝑘 𝑘
∑ ( ) (−𝜆)𝑘−𝑗 𝑗𝑛 ,
𝑘! 𝑗=0 𝑗
𝑙
( 𝑚𝑟 )
𝐽𝜆𝑟 𝑥𝑛 = (
=
∑
1 ̃ 𝑟 𝑛
Δ )𝑥
1−𝜆 𝜆
𝑟
1
𝑟
𝑛
( ) (−𝜆)𝑟−𝑗 (𝑥 + 𝑗) .
∑
(1 − 𝜆)𝑟 𝑗=0 𝑗
(36)
Abstract and Applied Analysis
5
Let us consider 𝑠 = 2𝑟 − 1 in the identity of Theorem 2.
Then we have
By (36), we get
𝑟
1
𝑟
𝑛
𝑟−𝑗
(𝑥 + 𝑗)
𝑟 ∑ (𝑗) (−𝜆)
−
𝜆)
(1
𝑗=0
=
=
𝐽𝜆𝑟−1 𝑥𝑛
= 𝐻𝑛−(𝑟−1) (𝑥 | 𝜆)
1
̃ 𝑟 𝑥𝑛
Δ
(1 − 𝜆)𝑟 𝜆
=
min{2𝑟,𝑛} {
{
{
}
𝑙
}
}
( 2𝑟
𝑙
𝑛
𝑚)
(
)
(
)
∑
∑
𝑚
{
}
,
.
.
.
,
𝑘
𝑘
𝑙
{
1
𝑚 }
{
}
𝑚=0 (1 − 𝜆) 𝑘1 +⋅⋅⋅+𝑘𝑚 =𝑙
𝑘
≥1
𝑗
{
}
∑
𝑙=0
(𝑟)
⋅ 𝐻𝑛−𝑙
(𝑥 | 𝜆)
{
{
{ 𝑛 min{2𝑟−1,𝑛} ( 2𝑟−1
𝑚 )
+
( ) ∑
∑
{
𝑙
{
(1
−
𝜆)𝑚
𝑚=0
𝑙=min{2𝑟−1,𝑛}+1 {
{
𝑛
{
{
{ 𝑛 min{2𝑟,𝑛} ( 2𝑟
𝑚)
( ) ∑
𝑚
{
{ 𝑙
𝑚=0 (1 − 𝜆)
𝑙=min{2𝑟,𝑛}+1 {
{
𝑛
∑
}
}
}
} (𝑟)
𝑙
⋅ ∑ (
)} 𝐻𝑛−𝑙
(𝑥 | 𝜆) .
𝑘1 , . . . , 𝑘𝑚 }
}
𝑘1 +⋅⋅⋅+𝑘𝑚 =𝑙
}
𝑘𝑗 ≥1
}
(37)
Let us take 𝑥 = 0 in (37). Then we obtain the following
theorem.
Theorem 5. We have
𝑟!
𝑆 (𝑛, 𝑟)
(1 − 𝜆)𝑟 𝜆
=
}
}
}
𝑙
}
}
( 2𝑟
𝑙
𝑛
𝑚)
)}
( )∑
∑ (
𝑚
{
,
.
.
.
,
𝑘
𝑘
𝑙
{
−
𝜆)
(1
1
𝑚 }
}
{
𝑚=0
𝑘1 +⋅⋅⋅+𝑘𝑚 =𝑙
}
{
}
{
≥1
𝑘
𝑗
}
{
min{2𝑟,𝑛} {
{
{
𝑙=0
⋅
(𝑟)
𝐻𝑛−𝑙
+
(𝜆)
{
{
{ 𝑛 min{2𝑟,𝑛} ( 2𝑟
𝑚)
( ) ∑
{
𝑙
{
(1
−
𝜆)𝑚
𝑚=0
𝑙=min{2𝑟,𝑛}+1 {
{
𝑛
∑
𝑚=0
=
1
̃ 𝑟−1 𝑥𝑛 .
Δ
𝜆
(1 − 𝜆)𝑟−1
(39)
(𝑟 − 1)!
𝑆𝜆 (𝑛, 𝑟 − 1)
(1 − 𝜆)𝑟−1
=
̃ 𝑟−1 0𝑛
(𝑟 − 1)! Δ
𝜆
(1 − 𝜆)𝑟−1 (𝑟 − 1)!
min{2𝑟−1,𝑛} {
{
{
}
𝑙
}
}
( 2𝑟−1
𝑙
𝑛
𝑚 )
(
)
(
)
∑
∑
𝑚
{
}
,
.
.
.
,
𝑘
𝑘
𝑙
{
1
𝑚 }
{
}
𝑚=0 (1 − 𝜆) 𝑘1 +⋅⋅⋅+𝑘𝑚 =𝑙
𝑘𝑗 ≥1
{
}
∑
𝑙=0
(𝑟)
⋅ 𝐻𝑛−𝑙
(𝜆)
∑
min{𝑟,𝑛}
𝑟−1
1
𝑟−1
𝑛
) (−𝜆)𝑟−1−𝑗 (𝑥 + 𝑗)
∑(
𝑟−1
𝑗
(1 − 𝜆) 𝑗=0
=
{
{
{ 𝑛 min{2𝑟−1,𝑛} ( 2𝑟−1
𝑚 )
+
( ) ∑
∑
{
𝑙
{
−
(1 𝜆)𝑚
𝑚=0
𝑙=min{2𝑟−1,𝑛}+1 {
{
𝑛
}
}
} (𝑟)
𝑙
⋅ ∑ (
) } 𝐻𝑛−𝑙
(𝜆)
𝑘1 , . . . , 𝑘𝑚 }
}
𝑘1 +⋅⋅⋅+𝑘𝑚 =𝑙
𝑘𝑗 ≥1
}
=
=
Theorem 6. For 𝑛 ≥ 0 and 𝑟 ≥ 1, one has
̃ 𝑟 0𝑛
𝑟! Δ
𝜆
(1 − 𝜆)𝑟 𝑟!
∑
}
}
} (𝑟)
𝑙
⋅ ∑ (
)} 𝐻𝑛−𝑙
(𝑥 | 𝜆)
𝑘1 , . . . , 𝑘𝑚 }
}
𝑘1 +⋅⋅⋅+𝑘𝑚 =𝑙
𝑘𝑗 ≥1
}
Let us take 𝑥 = 0 in (39). Then we obtain the following
theorem.
{
{
=
}
𝑙
}
}
( 2𝑟−1
𝑙
𝑛
𝑚 )
(
)
(
)
∑
∑
𝑚
{
}
,
.
.
.
,
𝑘
𝑘
𝑙
{
1
𝑚 }
{
}
𝑚=0 (1 − 𝜆) 𝑘1 +⋅⋅⋅+𝑘𝑚 =𝑙
𝑘𝑗 ≥1
{
}
∑
𝑙=0
(𝑟)
⋅ 𝐻𝑛−𝑙
(𝑥 | 𝜆)
+
min{2𝑟−1,𝑛} {
{
{
( 𝑚𝑟 )
∑
(1 − 𝜆)𝑚 𝑘 +⋅⋅⋅+𝑘
1
𝑚 =𝑛
𝑘𝑗 ≥1
𝑛
(
).
𝑘1 , . . . , 𝑘𝑚
(38)
}
}
} (𝑟)
𝑙
⋅ ∑ (
)} 𝐻𝑛−𝑙
(𝜆) .
𝑘1 , . . . , 𝑘𝑚 }
}
𝑘1 +⋅⋅⋅+𝑘𝑚 =𝑙
𝑘𝑗 ≥1
}
(40)
6
Abstract and Applied Analysis
Remark 7. Note that
(𝑟 − 1)!
𝑆𝜆 (𝑛, 𝑟 − 1)
(1 − 𝜆)𝑟−1
=
min{𝑟,𝑛} {
{
{
}
𝑙
}
}
( 𝑚𝑟 )
𝑙
𝑛
(
)
∑ {( ) ∑
∑
𝑚
{ 𝑙 𝑚=0 (1 − 𝜆) 𝑘 +⋅⋅⋅+𝑘 =𝑙 𝑘1 , . . . , 𝑘𝑚 }
}
}
𝑙=0 {
1
𝑚
𝑘
≥1
𝑗
{
}
⋅ 𝐻𝑛−𝑙 (𝜆)
+
{
{
{ 𝑛 min{𝑟,𝑛} ( 𝑚𝑟 )
( ) ∑
{
{ 𝑙 𝑚=0 (1 − 𝜆)𝑚
𝑙=min{𝑟,𝑛}+1 {
{
𝑛
∑
}
}
}
𝑙
⋅ ∑ (
) } 𝐻𝑛−𝑙 (𝜆) .
𝑘1 , . . . , 𝑘𝑚 }
}
𝑘1 +⋅⋅⋅+𝑘𝑚 =𝑙
𝑘𝑗 ≥1
}
(41)
Acknowledgment
The authors would like to express their gratitude to the
referees for their valuable suggestions.
References
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