Available online at www.tjnsa.com
J. Nonlinear Sci. Appl. 9 (2016), 2086–2098
Research Article
Identities involving degenerate Euler numbers and
polynomials arising from non-linear differential
equations
Taekyun Kima,b,∗, Dae San Kimc
a
Department of Mathematics, College of Science, Tianjin Polytechnic University, Tianjin City, 300387, China.
b
Department of Mathematics, Kwangwoon University, Seoul 139-701, Republic of Korea.
c
Department of Mathematics, Sogang University, Seoul 121-742, Republic of Korea.
Communicated by S.-H. Rim
Abstract
The purpose of this paper is to construct some new non-linear differential equations and investigate
the solutions of these non-linear differential equations. In addition, we give some new identities involving
degenerate Euler numbers and polynomials arising from those non-linear differential equations. c 2016 All
rights reserved.
Keywords: Degenerate Euler numbers, degenerate Euler polynomials, non-linear differential equation,
degenerate Bernoulli numbers, degenerate Bernoulli polynomials.
2010 MSC: 05A19, 11B83, 34A34.
1. Introduction
As is well known, the Euler polynomials of order r(∈ N) are defined by the generating function
(r)
2
t
e +1
r
ext =
∞
X
n=0
En(r) (x)
tn
,
n!
(see [1–15]) .
(r)
When x = 0, En = En (0) are called the higher-order Euler numbers.
∗
Corresponding author
Email addresses: tkkim@kw.ac.kr (Taekyun Kim), dskim@sogang.ac.kr (Dae San Kim)
Received 2015-12-29
(1.1)
T. Kim, D. S. Kim, J. Nonlinear Sci. Appl. 9 (2016), 2086–2098
2087
(1)
In particular, r = 1, En (x) = En (x) are called ordinary Euler polynomials.
In [2, 3], L. Carlitz considered the degenerate Euler polynomials which are given by the generating
function
!r
∞
X
x
2
tn
(r)
λ =
(1
+
λt)
(1.2)
En,λ (x) .
1
n!
(1 + λt) λ + 1
n=0
(r)
(r)
When x = 0, En,λ = En,λ (0) are called the higher-order degenerate Euler numbers.
(1)
(1)
In particular, for r = 1, En,λ = En,λ (0) and En,λ (x) = En,λ (x) are respectively called the degenerate
Euler numbers and the degenerate Euler polynomials.
From (1.2), we have
!r
∞
X
x
tn
2
(r)
(1 + λt) λ
En,λ (x)
=
(1.3)
1
n!
λ + 1
(1
+
λt)
n=0
!
∞
n X
X
n
tn
(r)
,
=
(x | λ)n−l El
n!
l
n=0
l=0
where (x | λ)n = x (x − λ) · · · (x − (n − 1) λ).
Thus, by (1.3), we get
(r)
En,λ (x)
=
n X
n
l=0
l
(r)
(x | λ)n−l El ,
(n ≥ 0) ,
(see [3, 10, 12]) .
(1.4)
In [9, 11], Kim and Kim, and Kim developed some new methods for obtaining identities related to
Bernoulli numbers of the second kind and Frobenius-Euler polynomials of higher order arising from certain
non-linear differential equations.
For example,
min{n,N −1}
(−1)N
X
(N +1−j)
(N − j)! (N − 1)!HN −1,N −1−j (n)j bn−j
j=0
(
Q
(N − l) ,
(−1)N N ! n−1
= Pn−N −1 nl=0
bn−l Ql+N
l=0
l=0 (n − l) ,
l n−l
if 0 ≤ n < N,
if n ≥ N + 1,
where HN,0 = 1, for all N (∈ N),
1
1
+ ··· + ,
2
N
HN −1,j−1 HN −2,j−1
H0,j−1
=
+
+ ··· +
, H0,j−1 = 0 (2 ≤ j ≤ N ) ,
N
N −1
1
= the nth Bernoulli numbers of the second kind with order k (see [9, 11]) .
HN,1 = HN = 1 +
HN,j
bn(k)
The rising factorial sequence is defined as
(x)n = x (x + 1) · · · (x + n − 1) =
n
X
|S1 (n, l)| xl ,
(n ≥ 0) ,
(1.5)
l=0
where |S1 (n, l)| are called the unsigned Stirling numbers of the first kind (see [1–9, 11–13]).
The purpose of this paper is to construct some new non-linear differential equations and investigate
the solutions of these non-linear differential equations. In addition, we give some new identities involving
degenerate Euler numbers and polynomials arising from those non-linear differential equations.
T. Kim, D. S. Kim, J. Nonlinear Sci. Appl. 9 (2016), 2086–2098
2088
2. Identities of degenerate Euler numbers and polynomials
1
Now, we construct the non-linear differential equations with the solution F (t) =
1
.
(1+λt) λ +1
Let
1
F = F (t) = F (t; λ) =
1
,
(2.1)
(1 + λt) λ + 1
and
F N = |F × F ×
{z· · · × F},
where N ∈ N.
(2.2)
N −times
From (2.1), we note that
F
(1)
dF
=
=
dt
=
1
− (1 + λt) λ
2
1
(1 + λt) λ + 1 (1 + λt)
(2.3)
(−1)
F − F2 .
1 + λt
Thus, by (2.3), we get
F (1) =
dF
(−1)
(t) =
F − F2 .
dt
1 + λt
(2.4)
From (2.4), we can derive
dF (1)
dt
(−1) (1)
(−1)2 λ
2
(1)
+
=
F
−
F
F
−
2F
F
(1 + λt)
(1 + λt)2
(−1)2 λ
(−1)
(−1)
(−1)
2
2
2
=
F −F +
F − F − 2F
F −F
(1 + λt) (1 + λt)
1 + λt
(1 + λt)2
(−1)2
(−1)3 2!
(−1)2 λ
2
2
2
3
F
−
F
+
F
−
F
+
=
2
2
2 F −F
(1 + λt)
(1 + λt)
(1 + λt)
2
3
(−1) (λ + 1)
(−1) 2!
2
2
3
=
F
−
F
+
F
−
F
,
(1 + λt)2
(1 + λt)2
F (2) =
(2.5)
and
dF (2)
dt
(−1)2 (λ + 1) (1)
λ (−1)3 2! (λ + 1)
2
(1)
=
F
−
F
+
F
−
2F
F
(1 + λt)3
(1 + λt)2
(−1)4 2!2λ
(−1)3 2! 2
3
(1)
2 (1)
+
F
−
F
+
2F
F
−
3F
F
(1 + λt)3
(1 + λt)2
(−1)4 (λ + 1) 2
(−1)4 2!2λ
(−1)3 (1 + λ) (2λ + 1)
2
2
3
=
F
−
F
+
F
−
F
+
F2 − F3
3
3
3
(1 + λt)
(1 + λt)
(1 + λt)
4
5
(−1) 2!2
(−1) 2!3
2
3
3
4
+
+
3 F −F
3 F −F
(1 + λt)
(1 + λt)
F (3) =
=
(−1)3 (1 + λ) (2λ + 1)
(−1)4 (2λ + 7) (λ + 1) 2 (−1)5 3! (λ + 2) 3
F
+
F +
F
(1 + λt)3
(1 + λt)3
(1 + λt)3
+
(−1)6 3! 4
F .
(1 + λt)3
(2.6)
T. Kim, D. S. Kim, J. Nonlinear Sci. Appl. 9 (2016), 2086–2098
2089
Thus we are led to set
F (N ) =
(−1)N
N
+1
X
(1 + λt)N
i=1
where
F (N ) =
ai (N, λ) (−1)i−1 F i ,
(N ∈ N) ,
(2.7)
dN F
d
d
(t) =
× · · · × F (t).
dtN
dt
dt
|
{z
}
N −times
To determine the coefficients ai (N, λ) in (2.7), we take the derivative of (2.7) with respect to t as follows:
N +1
dF (N )
(−1)N +1 λN X
ai (N, λ) (−1)i−1 F i
=
N +1
dt
(1 + λt)
i=1
F (N +1) =
+
(−1)N
N
+1
X
(1 + λt)N
i=1
(2.8)
ai (N, λ) (−1)i−1 iF i−1 F (1) .
From (2.4) and (2.8), we have
F
(N +1)
=
N +1
(−1)N +1 λN X
(1 + λt)N +1
i=1
N
+1
X
N +1
+
(−1)
N +1
(1 + λt)
N +1
=
ai (N, λ) (−1)i−1 F i
(−1)
(1 + λt)N +1
+
N
+2
X
ai (N, λ) (−1)i−1 i F i − F i+1
i=1
(N +1
X
(λN ai (N, λ) + iai (N, λ)) (−1)i−1 F i
i=1
)
i−1
ai−1 (N, λ) (−1)
(i − 1) F
(2.9)
i
i=2
=
(−1)N +1
(1 + λt)N +1
+
N
+1
X
{(λN a1 (N, λ) + a1 (N, λ)) F
(λN ai (N, λ) + iai (N, λ) + (i − 1) ai−1 (N, λ)) (−1)i−1 F i
i=2
o
+aN +1 (N, λ) (−1)N +1 (N + 1) F N +2 .
By (2.7) and (2.9), we easily get
(−1)N +1 n
(1 + λt)N +1
+
N
+1
X
(λN a1 (N, λ) + a1 (N, λ)) F +aN +1 (N, λ) (−1)N +1 (N + 1) F N +2
(λN ai (N, λ) + iai (N, λ) + (i − 1) ai−1 (N, λ)) (−1)i−1 F i
o
(2.10)
i=2
=
(−1)N +1
(1 + λt)
N
+2
X
N +1
ai (N + 1, λ) (−1)i−1 F i .
i=1
By comparing the coefficients on both sides of (2.10), we get
a1 (N + 1, λ) = λN a1 (N, λ) + a1 (N, λ)
= (λN + 1) a1 (N, λ) ,
(2.11)
T. Kim, D. S. Kim, J. Nonlinear Sci. Appl. 9 (2016), 2086–2098
2090
aN +2 (N + 1, λ) = (N + 1) aN +1 (N, λ) ,
(2.12)
ai (N + 1, λ) = (λN + i) ai (N, λ) + (i − 1) ai−1 (N, λ) ,
(2.13)
and
where 2 ≤ i ≤ N + 1.
From (2.4) and (2.7), we have
(−1)
(−1)
F − F 2 = F (1) =
a1 (1, λ) F − a2 (1, λ) F 2 .
1 + λt
1 + λt
(2.14)
Thus, by (2.14), we get
a1 (1, λ) = 1,
and a2 (1, λ) = 1.
(2.15)
From (2.11) and (2.15), we can derive the following identities:
a1 (N + 1, λ) = (λN + 1) a1 (N, λ)
= (λN + 1) (λ (N − 1) + 1) a1 (N − 1, λ)
= (λN + 1) (λ (N − 1) + 1) (λ (N − 2) + 1) a1 (N − 2, λ)
..
.
(2.16)
= (λN + 1) (λ (N − 1) + 1) · · · (λ + 1) a1 (1, λ)
= (λN + 1) (λ (N − 1) + 1) · · · (λ + 1) · 1
1
N +1
=λ
,
λ N +1
and
aN +2 (N + 1, λ) = (N + 1) aN +1 (N, λ)
= (N + 1) N aN (N − 1, λ)
..
.
(2.17)
= (N + 1) N (N − 1) · · · 2a2 (1, λ)
= (N + 1)!.
We observe that
a1 (1, λ) = 1, a1 (2, λ) = (1 + λ) ,
a1 (3, λ) = (1 + λ) (1 + 2λ) ,
1
N
a1 (N, λ) = (1 + λ) (1 + 2λ) · · · (1 + (N − 1) λ) = λ
,
λ N
···
(2.18)
and
a2 (1, λ) = 1, a3 (2, λ) = 2!, a4 (3, λ) = 3!, . . . , aN +1 (N, λ) = N !.
That is, the matrix (ai (j, λ))1≤i≤N +1,1≤j≤N is given by
N

 1 (1 + λ) (1 + λ)(1 + 2λ)

 1!
×
×


2!
×

N + 1

3!




0
· · · λN
···
···
···
..
.
1
λ N
×
×
×
×
N!













(2.19)
T. Kim, D. S. Kim, J. Nonlinear Sci. Appl. 9 (2016), 2086–2098
2091
From (2.13), we have
a2 (N + 1, λ) = (λN + 2) a2 (N, λ) + a1 (N, λ)
= (λN + 2) {(λ (N − 1) + 2) a2 (N − 1, λ) + a1 (N − 1, λ)} + a1 (N, λ)
= (λN + 2) (λ (N − 1) + 2) a2 (N − 1, λ) + (λN + 2) a1 (N − 1, λ) + a1 (N, λ)
= a1 (N, λ) + (λN + 2) a1 (N − 1, λ) + (λN + 2) (λ (N − 1) + 2) a1 (N − 2, λ)
+ (λN + 2) (λ (N − 1) + 2) (λ (N − 2) + 2) a2 (N − 2, λ)
..
.
= a1 (N, λ) +
N
−1
X
mY
1 −1
m1 =1
l=0
!
(λ (N − l) + 2) a1 (N − m1 , λ)
+ (λN + 2) (λ (N − 1) + 2) · · · (λ + 2) · 1
N
−1
X
1
2
1
N
m1
N −m1
=λ
+
λ
+ N − m1 + 1
λ
λ N
λ
λ N −m1
m1
m1 =1
2
+λN
+1
λ
N
N
X
2
1
N −m1
m1
+ N − m1 + 1
λ
=
λ
λ
λ
m1
N −m1
(2.20)
m1 =0
= λN
N X
2
1
+ N − m1 + 1
,
λ
λ
m1
N −m1
m1 =0
and
a3 (N + 1, λ) = (λN + 3) a3 (N, λ) + 2!a2 (N, λ)
= 2!a2 (N, λ) + (λN + 3) {(λ (N − 1) + 3) a3 (N − 1, λ) + 2a2 (N − 1, λ)}
= 2!a2 (N, λ) + 2! (λN + 3) a2 (N − 1, λ)
+ (λN + 3) (λ (N − 1) + 3) a3 (N − 1, λ)
= 2!a2 (N, λ) + 2! (λN + 3) a2 (N − 1, λ)
+ 2! (λN + 3) (λ (N − 1) + 3) a2 (N − 2, λ)
+ (λN + 3) (λ (N − 1) + 3) (λ (N − 2) + 3) a3 (N − 2, λ)
..
.
= 2!a2 (N, λ) + 2!
N
−2
X
mY
2 −1
m2 =1
l=0
!
(λ (N − l) + 3) a2 (N − m2 , λ)
+2! (λN + 3) (λ (N − 1) + 3) · · · (3λ + 3) (2λ + 3)
N
−2
X
3
m2
= 2!a2 (N, λ) + 2!
λ
N − m2 + 1 +
a2 (N − m2 , λ)
λ m2
m2 =1
+2! (λN + 3) (λ (N − 1) + 3) · · · (3λ + 3) (2λ + 3)
N
−2
X
3
= 2!a2 (N, λ) + 2!
λm2 N − m2 + 1 +
a2 (N − m2 , λ)
λ m2
m2 =1
3
N −1
+2!λ
+2
λ
N −1
(2.21)
T. Kim, D. S. Kim, J. Nonlinear Sci. Appl. 9 (2016), 2086–2098
= 2!
N
−1
X
λ
m2
m2 =0
3
N − m2 + 1 +
λ
2092
a2 (N − m2 , λ)
m2
N
−1 N −m
X
X2 −1 3
= 2!λ
N − m2 + 1 +
λ m2
m2 =0 m1 =0
2
1
× N − m2 − m1 +
.
λ m1 λ N −m2 −m1 −1
N −1
From (2.13), we note that
a4 (N + 1, λ) = (λN + 4) a4 (N, λ) + 3a3 (N, λ) .
(2.22)
Thus, by (2.21) and (2.22), we get
a4 (N + 1, λ) = 3a3 (N, λ) + (λN + 4) {(λ (N − 1) + 4) a4 (N − 1, λ) + 3a3 (N − 1, λ)}
= 3a3 (N, λ) + 3 (λN + 4) a3 (N − 1, λ)
+ (λN + 4) (λ (N − 1) + 4) a4 (N − 1, λ)
= 3a3 (N, λ) + 3 (λN + 4) a3 (N − 1, λ)
+3 (λN + 4) (λ (N − 1) + 4) a3 (N − 2, λ)
+ (λN + 4) (λ (N − 1) + 4) (λ (N − 2) + 4) a4 (N − 2, λ)
..
.
= 3a3 (N, λ) + 3
N
−3
X
mY
3 −1
m3 =1
l=0
!
(λ (N − l) + 4) a3 (N − m3 , λ)
+3! (λN + 4) (λ (N − 1) + 4) · · · (3λ + 4)
N
−3
X
4
+ N − m3 + 1
a3 (N − m3 , λ)
= 3a3 (N, λ) + 3
λm3
λ
m
3
m3 =1
4
N −2
+3!λ
+3
λ
N −2
N
−2
X
4
m3
=3
λ
+ N − m3 + 1
a3 (N − m3 , λ)
λ
m3
(2.23)
m3 =0
= 3!
N
−2
X
λ
m3
m3 =0
4
+ N − m3 + 1
λ
λN −m3 −2
m3
N −m
3 −m2 −2 X3 −2 N −mX
3
×
N − m3 − m2 +
λ m2
m2 =0
m1 =0
2
1
× N − m3 − m2 − m1 − 1 +
.
λ m1 λ N −m3 −m2 −m1 −2
By (2.23), we see that
a4 (N + 1, λ) = 3!λN −2
N
−2 N −m
3 −m2 −2 X
X3 −2 N −mX
m3 =0
m2 =0
× N − m3 − m2 +
m1 =0
3
λ
4
+ N − m3 + 1
λ
m3
(2.24)
m2
2
× N − m3 − m2 − m1 − 1 +
λ
1
.
λ N −m3 −m2 −m1 −2
m1
T. Kim, D. S. Kim, J. Nonlinear Sci. Appl. 9 (2016), 2086–2098
2093
Continuing this process, we get
N −mi−1 −···−m2 −i+2 NX
−i+2 N −mX
i−1 −i+2
i
···
ai (N + 1, λ) = (i − 1)!λ
+ N − mi−1 + 1
λ
mi−1
m1 =0
mi−2 =0
mi−1 =0
2
i−1
(2.25)
· · · N − mi−1 − · · · − m1 − i + 3 +
× N − mi−1 − mi−2 +
λ
λ
mi−2
m1
1
×
.
λ N −mi−1 −mi−2 −···−m1 −i+2
N −i+2
X
Therefore, by (2.7) and (2.25), we obtain the following theorem.
Theorem 2.1. For N ∈ N, let us consider the following non-linear differential equation with respect to t:
F
(N )
=
N
+1
X
(−1)N
N
(1 + λt)
ai (N, λ) (−1)i−1 F i ,
(2.26)
i=1
where
ai (N, λ) = (i − 1)!λ
N −i+1
NX
−i+1 N −mX
i−1 −i+1
mi−1 =0
N −mi−1 −···−m2 −i+1 X
···
mi−2 =0
m1 =0
i
N − mi−1 +
λ
mi−1
i−1
2
× N − mi−1 − mi−2 − 1 +
· · · N − mi−1 − · · · − m1 − i + 2 +
λ
λ m1
mi−2
1
.
×
λ N −mi−1 −mi−2 −···−m1 −i+1
Then F = F (t) =
1
is a solution of (2.26).
1
(1+λt) λ +1
Now, we observe that
F (N )
1 dN
=
2 dtN
=
!
2
1
(1 + λt) λ + 1
∞
X
tm
Em,λ
m!
1 dN
2 dtN
m=0
∞
1 X
m (m − 1) · · · (m − N + 1) m−N
=
Em,λ
t
2
m!
=
1
2
m=N
∞
X
Em+N,λ
m=0
(2.27)
tm
.
m!
Thus, by (2.27), we get
(1 + λt)N F (N )
!
!
∞ ∞
X
tm
N l l
1X
=
Em+N,λ
λt
l
2
m!
m=0
l=0
!
∞
n
1X X N
tn
(n)l λl En−l+N,λ
,
=
2
l
n!
n=0
where (x)n = x (x − 1) · · · (x − n + 1), (n ≥ 0) .
l=0
(2.28)
T. Kim, D. S. Kim, J. Nonlinear Sci. Appl. 9 (2016), 2086–2098
2094
From (1.2), we have
1
F = i
2
!
2
i
× ··· ×
1
(1 + λt) λ + 1
|
1
2i
=
1
(1 + λt) λ + 1
∞
X
!
2
{z
}
i−times
(i)
En,λ
n=0
tn
n!
(2.29)
.
Therefore, by Theorem 2.1, (2.28), and (2.29), we obtain the following theorem.
Theorem 2.2. For n ≥ 0, N ∈ N, we have
n X
N
(n)l λl En−l+N,λ
l
l=0
=
N
+1
X
(i − 1)!λN −i+1
NX
−i+1 N −mX
i−1 −i+1
mi−1 =0
i=1
N −mi−1 −···−m2 −i+1 X
···
N − mi−1 +
m1 =0
mi−2 =0
i
λ
mi−1
i−1
···
× N − mi−1 − mi−2 − 1 +
λ
mi−1
2
1
× N − mi−1 − mi−2 − · · · − m1 − i + 2 +
λ m1 λ N −mi−1 −···−m1 −i+1
1 (i)
E ,
2i−1 n,λ
× (−1)N +i−1
where (x)n = x (x − 1) · · · (x − n + 1).
Let
1
F (t) =
1
.
(2.30)
(1 + λt) λ − 1
Then, by (2.30), we get
F (1)


dF
(−1) 
=
=

dt
(1 + λt) 
=
F (2) =
(1 + λt)



1
λ
1
(1 + λt) λ − 1
2


(2.31)
(−1)
F + F2 ,
1 + λt
dF (1)
(−1)2 λ
(−1) (1)
2
(1)
=
F
+
F
+
F
+
2F
F
dt
1 + λt
(1 + λt)2
(−1)2 (λ + 1)
(−1)2 (λ + 3) 2
(−1)2 2 3
=
F
+
F
+
F
(1 + λt)2
(1 + λt)2
(1 + λt)2
(2.32)
and
dF (2)
dt
(−1)3 (λ + 1) (2λ + 1)
(−1)3 (2λ + 7) (λ + 1) 2
=
F
+
F
(1 + λt)3
(1 + λt)3
F (3) =
+
(−1)3 3! (λ + 2) 3
(−1)3 3! 4
F
+
F .
(1 + λt)3
(1 + λt)3
(2.33)
T. Kim, D. S. Kim, J. Nonlinear Sci. Appl. 9 (2016), 2086–2098
2095
So we are led to put
F (N ) =
N
+1
X
(−1)N
N
(1 + λt)
ai (N, λ) F i .
(2.34)
i=1
Thus, by (2.34), we get
dF (N )
dt
F (N +1) =
N +1
(−1)N +1 λN X
=
(1 + λt)N +1
+
(−1)N
(1 + λt)N
(−1)N +1
=
i=1
N
+1
X
ai (N, λ) iF i−1 F (1)
i=1
N
+1
X
N +1
(1 + λt)
N +1
+
(−1)
ai (N, λ) F i
(λN + i) ai (N, λ) F i
i=1
N
+2
X
(1 + λt)N +1
(2.35)
ai−1 (N, λ) (i − 1) F i .
i=2
From (2.34) and (2.35), we note that
F (N +1) =
(−1)N +1
(1 + λt)
+
N
+1
X
N +1
(λN + 1) a1 (N, λ) F + aN +1 (N, λ) (N + 1) F N +2
)
((λN + i) ai (N, λ) + (i − 1) ai−1 (N, λ)) F
i
(2.36)
i=2
=
(−1)N +1
N
+2
X
(1 + λt)N +1
i=1
ai (N + 1, λ) F i .
By comparing the coefficients on the both sides of (2.36), we get
a1 (N + 1, λ) = (λN + 1) a1 (N, λ) ,
aN +2 (N + 1, λ) = (N + 1) aN +1 (N, λ) ,
(2.37)
and
(λN + i) ai (N, λ) + (i − 1) ai−1 (N, λ) = ai (N + 1, λ) ,
(2 ≤ i ≤ N + 1) .
(2.38)
Also, we observe that
(−1) a1 (1, λ) F + a2 (1, λ) F 2
1 + λt
(−1)
=
F + F2 .
1 + λt
F (1) =
(2.39)
Thus, from (2.39), we get
a1 (1, λ) = 1,
and a2 (1, λ) = 1.
(2.40)
Therefore the relations in (2.37), (2.38), and (2.40) are the same as the ones in (2.11), (2.12), (2.13), and
(2.15). Hence, from (2.25), we obtain the following theorem.
Theorem 2.3. For N ∈ N, the following non-linear differential equation
F
has the solution F = F (t) =
(N )
1
1
(1+λt) λ −1
=
(−1)N
N
+1
X
(1 + λt)N
i=1
, where
ai (N, λ) F i
(2.41)
T. Kim, D. S. Kim, J. Nonlinear Sci. Appl. 9 (2016), 2086–2098
ai (N, λ) = (i − 1)!λ
N −i+1
NX
−i+1 N −mX
i−1 −i+1
N −mi−1 −···−m2 −i+1 X
···
mi−2 =0
mi−1 =0
2096
m1 =0
i
N − mi−1 +
λ
mi−1
i−1
2
× N − mi−1 − mi−2 − 1 +
· · · N − mi−1 − · · · − m1 − i + 2 +
λ
λ m1
mi−2
1
×
.
λ N −mi−1 −mi−2 −···−m1 −i+1
For r ∈ N, the degenerate Bernoulli polynomials of order r are defined by Carlitz as
!r
∞
X
x
tn
t
(r)
λ
βn,λ (x) , (see [3]) .
(1 + λt) =
1
n!
(1 + λt) λ − 1
n=0
(r)
(2.42)
(r)
When x = 0, βn,λ = βn,λ (0) are called the degenerate higher-order Bernoulli numbers. In particular, r = 1,
(1)
βn,λ = βn,λ are called the degenerate Bernoulli numbers. Note that β0,λ = 1.
We observe that
1
F = F (t) =
1
(1 + λt) λ − 1
∞
∞
tn X
tn−1 1
1X
βn,λ =
βn,λ
+
=
t
n!
n!
t
=
n=0
∞
X
βn+1,λ tn
n=0
(2.43)
n=1
1
+ .
n + 1 n!
t
Thus, by (2.43), we get
F
(N −1)
dN −1
= N −1
dt
1
(1 + λt) λ − 1
∞
X
βn+1,λ tn−N +1
(−1)N −1
+
(N − 1)!
n + 1 (n − N + 1)!
tN
=
=
!
1
n=N −1
∞
X
βn+N,λ tn
n=0
n + N n!
+
(2.44)
1
(−1)N −1 (N − 1)!.
tN
From (2.44), we have
tN F (N −1) =
=
∞
X
βn+1,λ
tn+1
+ (−1)N −1 (N − 1)!
n + 1 (n − N + 1)!
n=N −1
∞
X
βn,λ
n=N
tn
+ (−1)N −1 (N − 1)!.
n (n − N )!
(2.45)
Replacing N by N + 1, we get
∞
X
βn,λ
tn
+ (−1)N N ! (1 + λt)N
n (n − N − 1)!
n=N +1
!
n−N
X−1 N βn−l,λ
tn
λl
n (n − 1) · · · (n − l − N )
l n−l
n!
(1 + λt)N tN +1 F (N ) = (1 + λt)N
=
∞
X
n=N +1
l=0
N
+ (−1) N !
N
X
n=0
(N )n λn
tn
,
n!
(2.46)
T. Kim, D. S. Kim, J. Nonlinear Sci. Appl. 9 (2016), 2086–2098
2097
where (x)n = x (x − 1) · · · (x − n + 1).
From Theorem 2.3, we have
(1 + λt)N tN +1 F (N ) = (−1)N
N
+1
X
aj (N, λ) F j tN +1
j=1
N
= (−1)
N
+1
X
aj (N, λ)
j=1
N
= (−1)
N
X
!j
t
tN +1−j
1
λ
(1 + λt) − 1
aN +1−j (N, λ) t
j
∞
X
m
(N +1−j) t
βm,λ
(2.47)
m!
m=0
j=0


}
∞ min{n,N
 tn
X
X
n!
(N +1−j)
βn−j,λ
= (−1)N
aN +1−j (N, λ)
 n!

(n − j)!
n=0
j=0


min{n,N }
∞ 
 tn
X
X
(N +1−j)
=
(−1)N
aN +1−j (N, λ) n (n − 1) · · · (n − j + 1) βn−j,λ
.

 n!
n=0
j=0
Therefore, by (2.46) and (2.47), we obtain the following theorem.
Theorem 2.4. For n ≥ 0, we have
min{n,N }
X
N
(−1)
(N +1−j)
aN +1−j (N, λ) n (n − 1) · · · (n − j + 1) βn−j,λ
j=0
(
(−1)N N ! (N ) λn
= Pn−N −1 l N n βn−l,λ
λ l n−l n (n − 1) · · · (n − l − N )
l=0
if 0 ≤ n ≤ N,
if n ≥ N + 1,
where
ai (N, λ) = (i − 1)!λ
N −i+1
NX
−i+1 N −mX
i−1 −i+1
mi−1 =0
mi−2 =0
N −mi−1 −···−m2 −i+1 ···
X
m1 =0
i
N − mi−1 +
λ
mi−1
i−1
2
× N − mi−1 − mi−2 − 1 +
· · · N − mi−1 − · · · − m1 − i + 2 +
λ
λ m1
mi−2
1
×
.
λ N −mi−1 −mi−2 −···−m1 −i+1
Acknowledgements
The first author is appointed as a chair professor at Tian- jin Polytechnic University by Tianjin City in
China from August 2015 to August 2019.
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