Volume 2013, Article ID 871512,6pages http://dx.doi.org/10.1155/2013/871512
Research Article
Umbral Calculus and the Frobenius-Euler Polynomials
Dae San Kim,
1Taekyun Kim,
2and Sang-Hun Lee
31Department of Mathematics, Sogang University, Seoul 121-742, Republic of Korea
2Department of Mathematics, Kwangwoon University, Seoul 139-701, Republic of Korea
3Division of General Education, Kwangwoon University, Seoul 139-701, Republic of Korea
Correspondence should be addressed to Taekyun Kim; [email protected] 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
LetC be the complex number field. For𝜆 ∈C with𝜆 ̸= 1, the Frobenius-Euler polynomials are defined by the generating function to be
1 − 𝜆
𝑒𝑡− 𝜆𝑒𝑥𝑡= 𝑒𝐻(𝑥|𝜆)𝑡=∑∞
𝑛=0
𝐻𝑛(𝑥 | 𝜆)𝑡𝑛
𝑛!, (1)
(see [1–5]) with the usual convention about replacing𝐻𝑛(𝑥 | 𝜆)by𝐻𝑛(𝑥 | 𝜆).
In the special case,𝑥 = 0, 𝐻𝑛(0 | 𝜆) = 𝐻𝑛(𝜆)are called the𝑛th Frobenius-Euler numbers. By (1), we get
𝐻𝑛(𝑥 | 𝜆) =∑𝑛
𝑙=0(𝑛𝑙)𝐻𝑛−𝑙(𝜆) 𝑥𝑙= (𝐻 (𝜆) + 𝑥)𝑛, (2) (see [6–9]) with the usual convention about replacing𝐻𝑛(𝜆) by𝐻𝑛(𝜆).
Thus, from (1) and (2), we note that
(𝐻 (𝜆) + 1)𝑛− 𝜆𝐻𝑛(𝜆) = (1 − 𝜆) 𝛿0,𝑛, (3)
where𝛿𝑛,𝑘is the kronecker symbol (see [1,10,11]).
For𝑟 ∈ Z+, the Frobenius-Euler polynomials of order𝑟 are defined by the generating function to be
(1 − 𝜆
𝑒𝑡− 𝜆)𝑟𝑒𝑥𝑡= (1 − 𝜆
𝑒𝑡− 𝜆) × ⋅ ⋅ ⋅ × (1 − 𝜆 𝑒𝑡− 𝜆)
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟𝑒𝑥𝑡
𝑟-times
=∑∞
𝑛=0𝐻𝑛(𝑟)(𝑥 | 𝜆)𝑡𝑛 𝑛!.
(4)
In the special case,𝑥 = 0, 𝐻𝑛(𝑟)(0 | 𝜆) = 𝐻𝑛(𝑟)(𝜆)are called the 𝑛th Frobenius-Euler numbers of order𝑟(see [1,10]).
From (4), we can derive the following equation:
𝐻𝑛(𝑟)(𝑥 | 𝜆) =∑𝑛
𝑙=0(𝑛𝑙)𝐻𝑛−𝑙(𝑟)(𝜆) 𝑥𝑙, 𝐻𝑛(𝑟)(𝜆) = ∑
𝑙1+⋅⋅⋅+𝑙𝑟=𝑛
( 𝑛
𝑙1, . . . , 𝑙𝑟) 𝐻𝑙1(𝜆) ⋅ ⋅ ⋅ 𝐻𝑙𝑟(𝜆) . (5)
By (5), we see that𝐻(𝑟)𝑛 (𝑥 | 𝜆)is a monic polynomial of degree 𝑛with coefficients inQ(𝜆).
LetPbe the algebra of polynomials in the single variable 𝑥overC and letP∗be the vector space of all linear functionals onP. As is known, ⟨𝐿 | 𝑝(𝑥)⟩denotes the action of the linear functional𝐿on a polynomial𝑝(𝑥)and we remind that
the addition and scalar multiplication onP∗are, respectively, defined by
⟨𝐿 + 𝑀 | 𝑝 (𝑥)⟩ = ⟨𝐿 | 𝑝 (𝑥)⟩ + ⟨𝑀 | 𝑝 (𝑥)⟩,
⟨𝑐𝐿 | 𝑝 (𝑥)⟩ = 𝑐⟨𝐿 | 𝑝 (𝑥)⟩, (6) where𝑐is a complex constant (see [3,12]).
LetF denote the algebra of formal power series:
F= {𝑓 (𝑡) =∑∞
𝑘=0
𝑎𝑘
𝑘!𝑡𝑘| 𝑎𝑘∈C} (7) (see [3,12]). The formal power series define a linear functional onPby setting
⟨𝑓 (𝑡) | 𝑥𝑛⟩ = 𝑎𝑛, ∀𝑛 ≥ 0. (8) Indeed, by (7) and (8), we get
⟨𝑡𝑘| 𝑥𝑛⟩ = 𝑛!𝛿𝑛,𝑘 (𝑛, 𝑘 ≥ 0) (9) (see [3,12]). This kind of algebra is called an umbral algebra.
The order𝑂(𝑓(𝑡))of a nonzero power series𝑓(𝑡)is the smallest integer𝑘 for which the coefficient of 𝑡𝑘 does not vanish. A series𝑓(𝑡)for which𝑂(𝑓(𝑡)) = 1is said to be an invertible series (see [2,12]). For𝑓(𝑡), 𝑔(𝑡) ∈F, and𝑝(𝑥) ∈P, we have
⟨𝑓 (𝑡) 𝑔 (𝑡) | 𝑝 (𝑥)⟩ = ⟨ 𝑓 (𝑡) | 𝑔 (𝑡) 𝑝 (𝑥)⟩
= ⟨ 𝑔 (𝑡) | 𝑓 (𝑡) 𝑝 (𝑥)⟩ (10) (see [12]). One should keep in mind that each𝑓(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) ∈ Pand𝑓(𝑡) = 𝑒𝑦𝑡 ∈ F. As a linear functional,𝑒𝑦𝑡 satisfies⟨𝑒𝑦𝑡| 𝑝(𝑥)⟩ = 𝑝(𝑦). As a linear operator,𝑒𝑦𝑡satisfies 𝑒𝑦𝑡𝑝(𝑥) = 𝑝(𝑥 + 𝑦)(see [12]). Let𝑠𝑛(𝑥)denote a polynomial in𝑥with degree𝑛. Let us assume that𝑓(𝑡)is a delta series and 𝑔(𝑡)is an invertible series. Then there exists a unique sequence𝑠𝑛(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 by𝑠𝑛(𝑥) ∼ (𝑔(𝑡), 𝑓(𝑡)). If𝑠𝑛(𝑥) ∼ (1, 𝑓(𝑡)), then𝑠𝑛(𝑥)is called the associated sequence for𝑓(𝑡). If 𝑠𝑛(𝑥) ∼ (𝑔(𝑡), 𝑡), then 𝑠𝑛(𝑥)is called the Appell sequence.
Let𝑠𝑛(𝑥) ∼ (𝑔(𝑡), 𝑓(𝑡)). Then we see that ℎ (𝑡) =∑∞
𝑘=0
⟨ℎ (𝑡) | 𝑠𝑘(𝑥)⟩
𝑘! 𝑔 (𝑡) 𝑓(𝑡)𝑘, ℎ (𝑡) ∈F, 𝑝 (𝑥) =∑∞
𝑘=0
⟨𝑔 (𝑡) 𝑓(𝑡)𝑘| 𝑝 (𝑥)⟩
𝑘! 𝑠𝑘(𝑥) , 𝑝 (𝑥) ∈P, 𝑓 (𝑡) 𝑠𝑛(𝑥) = 𝑛𝑠𝑛−1(𝑥) ,
⟨𝑓 (𝑡) | 𝑝 (𝛼𝑥)⟩ = ⟨ 𝑓 (𝛼𝑡 | 𝑝 (𝑥)⟩ ,
(11)
1
𝑔 (𝑓 (𝑡))𝑒𝑦𝑓(𝑡)=∑∞
𝑘=0
𝑠𝑘(𝑦)
𝑘! 𝑡𝑘, ∀𝑦 ∈C, (12)
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.
2. The Frobenius-Euler Polynomials and Umbral Calculus
By (4) and (12), we see that
𝐻𝑛(𝑟)(𝑥 | 𝜆) ∼ ((𝑒𝑡− 𝜆
1 − 𝜆)𝑟, 𝑡) . (13) Thus, by (13), we get
⟨(𝑒𝑡− 𝜆
1 − 𝜆)𝑟𝑡𝑘| 𝐻𝑛(𝑟)(𝑥 | 𝜆)⟩ = 𝑛!𝛿𝑛,𝑘. (14) Let
P𝑛(𝜆) = {𝑝 (𝑥) ∈Q(𝜆) [𝑥] |deg𝑝 (𝑥) ≤ 𝑛} . (15) Then it is an(𝑛 + 1)-dimensional vector space overQ(𝜆).
So we see that{𝐻0(𝑟)(𝑥 | 𝜆), 𝐻1(𝑟)(𝑥 | 𝜆), . . . , 𝐻𝑛(𝑟)(𝑥 | 𝜆)}is a basis forP𝑛(𝜆). For𝑝(𝑥) ∈P𝑛(𝜆), let
𝑝 (𝑥) =∑𝑛
𝑘=0
𝐶𝑘𝐻𝑘(𝑟)(𝑥 | 𝜆) , (𝑛 ≥ 0) . (16) Then, by (13), (14), and (16), we get
⟨(𝑒𝑡− 𝜆
1 − 𝜆)𝑟𝑡𝑘| 𝑝 (𝑥)⟩
=∑𝑛
𝑙=0
𝐶𝑙⟨(𝑒𝑡− 𝜆 1 − 𝜆)
𝑟
𝑡𝑘| 𝐻𝑙(𝑟)(𝑥 | 𝜆)⟩
=∑𝑛
𝑙=0
𝐶𝑙𝑙!𝛿𝑙,𝑘= 𝑘!𝐶𝑘.
(17)
From (17), we have 𝐶𝑘= 1
𝑘!⟨(𝑒𝑡− 𝜆 1 − 𝜆)
𝑟
𝑡𝑘| 𝑝 (𝑥)⟩
= 1
𝑘!⟨(𝑒𝑡− 𝜆
1 − 𝜆)𝑟 | 𝐷𝑘𝑝 (𝑥)⟩
= 1
𝑘!(1 − 𝜆)𝑟
∑𝑟
𝑗=0(𝑟𝑗)(−𝜆)𝑟−𝑗⟨𝑒𝑗𝑡| 𝐷𝑘𝑝 (𝑥)⟩
= 1
𝑘!(1 − 𝜆)𝑟
∑𝑟
𝑗=0(𝑟𝑗)(−𝜆)𝑟−𝑗⟨𝑡0| 𝑒𝑗𝑡𝐷𝑘𝑝 (𝑥)⟩
= 1
𝑘!(1 − 𝜆)𝑟
∑𝑟
𝑗=0(𝑟𝑗)(−𝜆)𝑟−𝑗⟨𝑡0| 𝐷𝑘𝑝 (𝑥 + 𝑗)⟩ . (18)
Therefore, by (16) and (18), we obtain the following theorem.
Theorem 1. For𝑝(𝑥) ∈P𝑛(𝜆), let 𝑝 (𝑥) =∑𝑛
𝑘=0
𝐶𝑘𝐻𝑘(𝑟)(𝑥) . (19) Then one has
𝐶𝑘= 1 𝑘!(1 − 𝜆)𝑟
∑𝑟
𝑗=0(𝑟𝑗)(−𝜆)𝑟−𝑗𝐷𝑘𝑝 (𝑗) , (20) where𝐷𝑝(𝑥) = 𝑑𝑝(𝑥)/𝑑𝑥.
FromTheorem 1, we note that 𝑝 (𝑥) = 1
(1 − 𝜆)𝑟
⋅∑𝑛
𝑘=0
{{ {
∑𝑟 𝑗=0
1
𝑘!(𝑟𝑗)(−𝜆)𝑟−𝑗𝐷𝑘𝑝 (𝑗)} }}
𝐻𝑘(𝑟)(𝑥 | 𝜆) . (21) Let us consider the operator ̃Δ𝜆with̃Δ𝜆𝑓(𝑥) = 𝑓(𝑥 + 1) − 𝜆𝑓(𝑥)and let𝐽𝜆= (1/(1 − 𝜆))̃Δ𝜆. Then we have
𝐽𝜆(𝑓) (𝑥) = 1
1 − 𝜆{𝑓 (𝑥 + 1) − 𝜆𝑓 (𝑥)} . (22) Thus, by (22), we get
𝐽𝜆(𝐻𝑛(𝑟)(𝑥 | 𝜆)) = 1
1 − 𝜆{𝐻𝑛(𝑟)(𝑥 + 1 | 𝜆) − 𝜆𝐻𝑛(𝑟)(𝑥 | 𝜆)} . (23) From (4), we can derive
∑∞ 𝑛=0
{𝐻𝑛(𝑟)(𝑥 + 1 | 𝜆) − 𝜆𝐻𝑛(𝑟)(𝑥 | 𝜆)}𝑡𝑛 𝑛!
= (1 − 𝜆
𝑒𝑡− 𝜆)𝑟𝑒(𝑥+1)𝑡− 𝜆(1 − 𝜆 𝑒𝑡− 𝜆)𝑟𝑒𝑥𝑡
= (1 − 𝜆
𝑒𝑡− 𝜆)𝑟𝑒𝑥𝑡(𝑒𝑡− 𝜆) = (1 − 𝜆) (1 − 𝜆 𝑒𝑡− 𝜆)𝑟−1𝑒𝑥𝑡
= (1 − 𝜆)∑∞
𝑛=0𝐻𝑛(𝑟−1)(𝑥 | 𝜆)𝑡𝑛 𝑛!.
(24) By (23) and (24), we get
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) = 𝐽𝜆𝑟𝐻𝑛(0)(𝑥 | 𝜆) = 𝐻𝑛(−𝑟)(𝑥 | 𝜆) = 𝐽𝜆2𝑟𝐻𝑛(𝑟)(𝑥 | 𝜆) .
(26)
For𝑠 ∈Z+, from (25), we have
𝐽𝜆𝑠(𝐻𝑛(𝑟)(𝑥 | 𝜆)) = 𝐻𝑛(𝑟−𝑠)(𝑥 | 𝜆) . (27) On the other hand, by (12), (13), and (25),
𝐽𝜆𝑠(𝐻𝑛(𝑟)(𝑥 | 𝜆)) = (𝑒𝑡− 𝜆 1 − 𝜆)
𝑠
(𝐻𝑛(𝑟)(𝑥 | 𝜆))
= 1
(1 − 𝜆)𝑠((1 − 𝜆) +∑∞
𝑘=1
𝑡𝑘 𝑘!)
𝑠
⋅ (𝐻𝑛(𝑟)(𝑥 | 𝜆)) .
(28)
Thus, by (28), we get
𝐽𝜆𝑠(𝐻𝑛(𝑟)(𝑥 | 𝜆))
= ∑𝑠
𝑚=0
(𝑚𝑠) (1 − 𝜆)𝑚
∑∞ 𝑙=𝑚
( ∑
𝑘1+⋅⋅⋅+𝑘𝑚=𝑙 𝑘𝑗≥1
1
𝑘1! ⋅ ⋅ ⋅ 𝑘𝑚!) 𝑡𝑙(𝐻𝑛(𝑟)(𝑥 | 𝜆))
= ∑𝑠
𝑚=0
(𝑚𝑠) (1 − 𝜆)𝑚
∑∞ 𝑙=𝑚
1
𝑙!( ∑
𝑘1+⋅⋅⋅+𝑘𝑚=𝑙 𝑘𝑗≥1
( 𝑙
𝑘1, . . . , 𝑘𝑚) 𝐷𝑙)
⋅ 𝐻𝑛(𝑟)(𝑥 | 𝜆)
=min{𝑠,𝑛}∑
𝑚=0
(𝑚𝑠) (1 − 𝜆)𝑚
∑𝑛
𝑙=𝑚(𝑛𝑙) ∑𝑘
1+⋅⋅⋅+𝑘𝑚=𝑙 𝑘𝑗≥1
( 𝑙
𝑘1, . . . , 𝑘𝑚) 𝐻𝑛−𝑙(𝑟)(𝑥 | 𝜆)
=min{𝑠,𝑛}∑
𝑙=0
{{ {{ {{ {
(𝑛𝑙)∑𝑙
𝑚=0
(𝑚𝑠) (1 − 𝜆)𝑚
⋅ ∑
𝑘1+⋅⋅⋅+𝑘𝑚=𝑙 𝑘𝑗≥1
( 𝑙
𝑘1, . . . , 𝑘𝑚) }} }} }} }
𝐻𝑛−𝑙(𝑟)(𝑥 | 𝜆)
+ ∑𝑛
𝑙=min{𝑠,𝑛}+1
{{ {{ {{ {
(𝑛𝑙)min{𝑠,𝑛}∑
𝑚=0
(𝑚𝑠) (1 − 𝜆)𝑚
⋅ ∑
𝑘1+⋅⋅⋅+𝑘𝑚=𝑙 𝑘𝑗≥1
( 𝑙
𝑘1, . . . , 𝑘𝑚) }} }} }} }
𝐻𝑛−𝑙(𝑟)(𝑥 | 𝜆) .
(29) Therefore, by (27) and (29), we obtain the following theorem.
Theorem 2. For any𝑟, 𝑠 ≥ 0, one has 𝐻𝑛(𝑟−𝑠)(𝑥 | 𝜆)
=min{𝑠,𝑛}∑
𝑙=0
{{ {{ {{ {
(𝑛𝑙)∑𝑙
𝑚=0
(𝑚𝑠) (1 − 𝜆)𝑚 ∑
𝑘1+⋅⋅⋅+𝑘𝑚=𝑙 𝑘𝑗≥1
( 𝑙
𝑘1, . . . , 𝑘𝑚) }} }} }} }
⋅ 𝐻𝑛−𝑙(𝑟)(𝑥 | 𝜆)
+ ∑𝑛
𝑙=min{𝑠,𝑛}+1
{{ {{ {{ {
(𝑛𝑙)min{𝑠,𝑛}∑
𝑚=0
(𝑚𝑠) (1 − 𝜆)𝑚
⋅ ∑
𝑘1+⋅⋅⋅+𝑘𝑚=𝑙 𝑘𝑗≥1
( 𝑙
𝑘1, . . . , 𝑘𝑚) }} }} }} }
𝐻𝑛−𝑙(𝑟)(𝑥 | 𝜆) .
(30) Let us take𝑠 = 𝑟−1 (𝑟 ≥ 1)inTheorem 2. Then we obtain the following corollary.
Corollary 3. For𝑛 ≥ 0, 𝑟 ≥ 1, one has 𝐻𝑛(𝑥 | 𝜆)
=min{𝑟−1,𝑛}∑
𝑙=0
{{ {{ {{ {
(𝑛𝑙)∑𝑙
𝑚=0
(𝑟−1𝑚 ) (1 − 𝜆)𝑚 ∑
𝑘1+⋅⋅⋅+𝑘𝑚=𝑙 𝑘𝑗≥1
( 𝑙
𝑘1, . . . , 𝑘𝑚) }} }} }} }
⋅ 𝐻𝑛−𝑙(𝑟)(𝑥 | 𝜆)
+ ∑𝑛
𝑙=min{𝑟−1,𝑛}+1
{{ {{ {{ {
(𝑛𝑙)min{𝑟−1,𝑛}∑
𝑚=0
(𝑟−1𝑚 ) (1 − 𝜆)𝑚
⋅ ∑
𝑘1+⋅⋅⋅+𝑘𝑚=𝑙 𝑘𝑗≥1
( 𝑙
𝑘1, . . . , 𝑘𝑚) }} }} }} }
𝐻𝑛−𝑙(𝑟)(𝑥 | 𝜆) .
(31) Let us take𝑠 = 𝑟 (𝑟 ≥ 1)inTheorem 2. Then we obtain the following corollary.
Corollary 4. For𝑛 ≥ 0, 𝑟 ≥ 1, one has
𝑥𝑛 =min{𝑟,𝑛}∑
𝑙=0
{{ {{ {{ {
(𝑛𝑙)∑𝑙
𝑚=0
(𝑚𝑟) (1 − 𝜆)𝑚 ∑
𝑘1+⋅⋅⋅+𝑘𝑚=𝑙 𝑘𝑗≥1
( 𝑙
𝑘1, . . . , 𝑘𝑚) }} }} }} }
⋅ 𝐻𝑛−𝑙(𝑟)(𝑥 | 𝜆)
+ ∑𝑛
𝑙=min{𝑟,𝑛}+1
{{ {{ {{ {
(𝑛𝑙)min{𝑟,𝑛}∑
𝑚=0
(𝑚𝑟) (1 − 𝜆)𝑚
⋅ ∑
𝑘1+⋅⋅⋅+𝑘𝑚=𝑙 𝑘𝑗≥1
( 𝑙
𝑘1, . . . , 𝑘𝑚) }} }} }} }
𝐻𝑛−𝑙(𝑟)(𝑥 | 𝜆) .
(32) Now, we define the analogue of Stirling numbers of the second kind as follows:
𝑆𝜆(𝑛, 𝑘) = 1 𝑘!
∑𝑘
𝑗=0(𝑘𝑗)(−𝜆)𝑘−𝑗𝑗𝑛, (𝑛, 𝑘 ≥ 0) . (33) Note that 𝑆1(𝑛, 𝑘) = 𝑆(𝑛, 𝑘) is the Stirling number of the second kind.
From the definition of̃Δ𝜆, we have
̃Δ𝑛𝜆𝑓 (0) =∑𝑛
𝑘=0(𝑛𝑘)(−𝜆)𝑛−𝑘𝑓 (𝑘) . (34) By (33) and (34), we get
𝑆𝜆(𝑛, 𝑘) = 1
𝑘!̃Δ𝑘𝜆0𝑛, (𝑛, 𝑘 ≥ 0) . (35) Let us take𝑠 = 2𝑟. Then we have
𝐽𝜆𝑟𝑥𝑛
= 𝐻𝑛(−𝑟)(𝑥 | 𝜆)
=min{2𝑟,𝑛}∑
𝑙=0
{{ {{ {{ {
(𝑛𝑙)∑𝑙
𝑚=0
(2𝑟𝑚) (1 − 𝜆)𝑚 ∑
𝑘1+⋅⋅⋅+𝑘𝑚=𝑙 𝑘𝑗≥1
( 𝑙
𝑘1, . . . , 𝑘𝑚) }} }} }} }
⋅ 𝐻𝑛−𝑙(𝑟)(𝑥 | 𝜆)
+ ∑𝑛
𝑙=min{2𝑟,𝑛}+1
{{ {{ {{ {
(𝑛𝑙)min{2𝑟,𝑛}∑
𝑚=0
(2𝑟𝑚) (1 − 𝜆)𝑚
⋅ ∑
𝑘1+⋅⋅⋅+𝑘𝑚=𝑙 𝑘𝑗≥1
( 𝑙
𝑘1, . . . , 𝑘𝑚) }} }} }} }
𝐻𝑛−𝑙(𝑟)(𝑥 | 𝜆) ,
𝐽𝜆𝑟𝑥𝑛= ( 1
1 − 𝜆̃Δ𝜆)𝑟𝑥𝑛
= 1
(1 − 𝜆)𝑟
∑𝑟
𝑗=0(𝑟𝑗)(−𝜆)𝑟−𝑗(𝑥 + 𝑗)𝑛.
(36)
By (36), we get 1 (1 − 𝜆)𝑟
∑𝑟
𝑗=0(𝑟𝑗)(−𝜆)𝑟−𝑗(𝑥 + 𝑗)𝑛
= 1
(1 − 𝜆)𝑟̃Δ𝑟𝜆𝑥𝑛
=min{2𝑟,𝑛}∑
𝑙=0
{{ {{ {{ {
(𝑛𝑙)∑𝑙
𝑚=0
(2𝑟𝑚) (1 − 𝜆)𝑚 ∑
𝑘1+⋅⋅⋅+𝑘𝑚=𝑙 𝑘𝑗≥1
( 𝑙
𝑘1, . . . , 𝑘𝑚) }} }} }} }
⋅ 𝐻𝑛−𝑙(𝑟)(𝑥 | 𝜆)
+ ∑𝑛
𝑙=min{2𝑟,𝑛}+1
{{ {{ {{ {
(𝑛𝑙)min{2𝑟,𝑛}∑
𝑚=0
(2𝑟𝑚) (1 − 𝜆)𝑚
⋅ ∑
𝑘1+⋅⋅⋅+𝑘𝑚=𝑙 𝑘𝑗≥1
( 𝑙
𝑘1, . . . , 𝑘𝑚) }} }} }} }} }
𝐻𝑛−𝑙(𝑟)(𝑥 | 𝜆) .
(37) Let us take 𝑥 = 0in (37). Then we obtain the following theorem.
Theorem 5. We have 𝑟!
(1 − 𝜆)𝑟𝑆𝜆(𝑛, 𝑟)
= 𝑟!
(1 − 𝜆)𝑟
̃Δ𝑟𝜆0𝑛 𝑟!
=min{2𝑟,𝑛}∑
𝑙=0
{{ {{ {{ {{ {{ {
(𝑛𝑙)∑𝑙
𝑚=0
(2𝑟𝑚) (1 − 𝜆)𝑚 ∑
𝑘1+⋅⋅⋅+𝑘𝑚=𝑙 𝑘𝑗≥1
( 𝑙
𝑘1, . . . , 𝑘𝑚) }} }} }} }} }} }
⋅ 𝐻𝑛−𝑙(𝑟)(𝜆)
+ ∑𝑛
𝑙=min{2𝑟,𝑛}+1
{{ {{ {{ {
(𝑛𝑙)min{2𝑟,𝑛}∑
𝑚=0
(2𝑟𝑚) (1 − 𝜆)𝑚
⋅ ∑
𝑘1+⋅⋅⋅+𝑘𝑚=𝑙 𝑘𝑗≥1
( 𝑙
𝑘1, . . . , 𝑘𝑚) }} }} }} }
𝐻𝑛−𝑙(𝑟)(𝜆)
=min{𝑟,𝑛}∑
𝑚=0
(𝑚𝑟) (1 − 𝜆)𝑚 ∑
𝑘1+⋅⋅⋅+𝑘𝑚=𝑛 𝑘𝑗≥1
( 𝑛
𝑘1, . . . , 𝑘𝑚) .
(38)
Let us consider𝑠 = 2𝑟 − 1in the identity ofTheorem 2.
Then we have 𝐽𝜆𝑟−1𝑥𝑛
= 𝐻𝑛−(𝑟−1)(𝑥 | 𝜆)
=min{2𝑟−1,𝑛}
∑
𝑙=0
{{ {{ {{ {
(𝑛𝑙)∑𝑙
𝑚=0
(2𝑟−1𝑚 ) (1 − 𝜆)𝑚 ∑
𝑘1+⋅⋅⋅+𝑘𝑚=𝑙 𝑘𝑗≥1
( 𝑙
𝑘1, . . . , 𝑘𝑚) }} }} }} }
⋅ 𝐻𝑛−𝑙(𝑟)(𝑥 | 𝜆)
+ ∑𝑛
𝑙=min{2𝑟−1,𝑛}+1
{{ {{ {{ {
(𝑛𝑙)min{2𝑟−1,𝑛}
∑
𝑚=0
(2𝑟−1𝑚 ) (1 − 𝜆)𝑚
⋅ ∑
𝑘1+⋅⋅⋅+𝑘𝑚=𝑙 𝑘𝑗≥1
( 𝑙
𝑘1, . . . , 𝑘𝑚) }} }} }} }
𝐻𝑛−𝑙(𝑟)(𝑥 | 𝜆)
= 1
(1 − 𝜆)𝑟−1
𝑟−1∑
𝑗=0(𝑟 − 1𝑗 ) (−𝜆)𝑟−1−𝑗(𝑥 + 𝑗)𝑛
= 1
(1 − 𝜆)𝑟−1̃Δ𝑟−1𝜆 𝑥𝑛.
(39) Let us take 𝑥 = 0in (39). Then we obtain the following theorem.
Theorem 6. For𝑛 ≥ 0and𝑟 ≥ 1, one has (𝑟 − 1)!
(1 − 𝜆)𝑟−1𝑆𝜆(𝑛, 𝑟 − 1)
= (𝑟 − 1)!
(1 − 𝜆)𝑟−1
̃Δ𝑟−1𝜆 0𝑛 (𝑟 − 1)!
=min{2𝑟−1,𝑛}
∑
𝑙=0
{{ {{ {{ {
(𝑛𝑙)∑𝑙
𝑚=0
(2𝑟−1𝑚 ) (1 − 𝜆)𝑚 ∑
𝑘1+⋅⋅⋅+𝑘𝑚=𝑙 𝑘𝑗≥1
( 𝑙
𝑘1, . . . , 𝑘𝑚) }} }} }} }
⋅ 𝐻𝑛−𝑙(𝑟)(𝜆)
+ ∑𝑛
𝑙=min{2𝑟−1,𝑛}+1
{{ {{ {{ {
(𝑛𝑙)min{2𝑟−1,𝑛}
∑
𝑚=0
(2𝑟−1𝑚 ) (1 − 𝜆)𝑚
⋅ ∑
𝑘1+⋅⋅⋅+𝑘𝑚=𝑙 𝑘𝑗≥1
( 𝑙
𝑘1, . . . , 𝑘𝑚) }} }} }} }
𝐻𝑛−𝑙(𝑟)(𝜆) .
(40)
Remark 7. Note that (𝑟 − 1)!
(1 − 𝜆)𝑟−1𝑆𝜆(𝑛, 𝑟 − 1)
=min{𝑟,𝑛}∑
𝑙=0
{{ {{ {{ {
(𝑛𝑙)∑𝑙
𝑚=0
(𝑚𝑟) (1 − 𝜆)𝑚 ∑
𝑘1+⋅⋅⋅+𝑘𝑚=𝑙 𝑘𝑗≥1
( 𝑙
𝑘1, . . . , 𝑘𝑚) }} }} }} }
⋅ 𝐻𝑛−𝑙(𝜆)
+ ∑𝑛
𝑙=min{𝑟,𝑛}+1
{{ {{ {{ {
(𝑛𝑙)min{𝑟,𝑛}∑
𝑚=0
(𝑚𝑟) (1 − 𝜆)𝑚
⋅ ∑
𝑘1+⋅⋅⋅+𝑘𝑚=𝑙 𝑘𝑗≥1
( 𝑙
𝑘1, . . . , 𝑘𝑚) }} }} }} }
𝐻𝑛−𝑙(𝜆) .
(41)
Acknowledgment
The authors would like to express their gratitude to the referees for their valuable suggestions.
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