Available online at www.tjnsa.com
J. Nonlinear Sci. Appl. 9 (2016), 3098–3111
Research Article
Differential equations associated with λ-Changhee
polynomials
Taekyun Kima , Dae San Kimb , Jong-Jin Seoc,∗, Hyuck-In Kwona
a
Department of Mathematics, Kwangwoon University, Seoul 139-701, Republic of Korea.
b
Department of Mathematics, Sogang University, Seoul 121-742, Republic of Korea.
c
Department of Applied Mathematics, Pukyong National University, Busan, Republic of Korea.
Communicated by S.-H. Rim
Abstract
In this paper, we study linear differential equations arising from λ-Changhee polynomials (or called
degenerate Changhee polynomials) and give some explicit and new identities for the λ-Changhee polynomials
c
associated with linear differential equations. 2016
All rights reserved.
Keywords: λ-Changhee polynomials, differential equations.
2010 MSC: 05A19, 11B37, 34A30.
1. Introduction
For N ∈ N, we define the generalized harmonic numbers as follows:
HN,0 = 1,
HN,1 = HN = 1 +
HN,j =
∗
for all N,
1
1
+ ··· + ,
2
N
(see [7]) ,
HN −1,j−1 HN −2,j−1
Hj−1,j−1
+
+ ··· +
,
N
N −1
j
(1.1)
(2 ≤ j ≤ N ) .
Corresponding author
Email addresses: tkkim@kw.ac.kr (Taekyun Kim), dskim@sogang.ac.kr (Dae San Kim), seo2011@pknu.ac.kr (Jong-Jin
Seo), sura@kw.ac.kr (Hyuck-In Kwon)
Received 2016-02-12
T. Kim, D. S. Kim, J.-J. Seo, H.-I. Kwon, J. Nonlinear Sci. Appl. 9 (2016), 3098–3111
3099
For k ∈ N and N, j ∈ N ∪ {0}, we define the generalized Changhee power sums Sk,j (N ) as follows:
Sk,0 (N ) = (N + 1)k ,
Sk,j (N ) =
N
X
(1.2)
(j ≥ 1) ,
Sk,j−1 (l) ,
(see [7, 9]) .
(1.3)
l=0
In particular, for k = 1, we also define S1,−1 (N ) = 1.
As is well known, the Euler polynomials are defined by the generating function
∞
X
2
tn
xt
e
=
E
(x)
,
n
et + 1
n!
(see [1, 4, 11, 12]) .
(1.4)
n=0
With the viewpoint of deformed Euler polynomials, the Changhee polynomials are defined by the generating function
∞
X
2
tn
(t + 1)x =
(1.5)
Chn (x) , (see [3, 5, 6]) .
t+2
n!
n=0
From (1.4) and (1.5), we note that
∞
X
En (x)
n=0
∞
X
m
1
tn
=
Chm (x)
et − 1
n!
m!
=
m=0
∞
X
n
X
n=0
m=0
!
Chm (x) S2 (n, m)
tn
,
n!
(1.6)
where S2 (n, m) are the Stirling numbers of the second kind. Thus, by (1.6), we get
En (x) =
n
X
Chm (x) S2 (n, m)
(n ≥ 0) .
m=0
The Stirling numbers of the first kind S1 (n, l) appear in the expansion of the falling factorial
(x)0 = 1,
(x)n = x (x − 1) · · · (x − n + 1)
n
X
=
S1 (n, l) xl , (n ≥ 1) .
(1.7)
l=0
From (1.7), we note that the generating function of the Stirling numbers of the first kind is given by
(log (1 + t))n = n!
∞
X
S1 (m, n)
m=n
tm
,
m!
(see [8, 10, 13]) .
By (1.1), we easily get
∞
X
∞
X
tn
1
Chn (x)
=
Em (x)
(log (1 + t))m
n!
m!
n=0
m=0
!
∞
n
X
X
tm
=
S1 (n, m) Em (x)
.
m!
n=0
m=0
Thus, by (1.8), we have
Chn (x) =
n
X
m=0
S1 (n, m) Em (x) ,
(n ≥ 0) ,
(see [17, 18]) .
(1.8)
T. Kim, D. S. Kim, J.-J. Seo, H.-I. Kwon, J. Nonlinear Sci. Appl. 9 (2016), 3098–3111
3100
Recently, λ-Changhee polynomials (or called degenerate Changhee polynomials) are defined by the generating function
∞
2λ
log (1 + λt) x X
tn
1+
=
Chn,λ (x) , (see [14]) .
2λ + log (1 + λt)
λ
n!
n=0
When x = 0, Chn,λ = Chn,λ (0) are called λ-Changhee numbers (or called degenerate Changhee numbers).
In [7], Kim-Kim gave some explicit and new identities for the Bernoulli numbers of the second kind
arising from nonlinear differential equations. It is known that some interesting identities and properties of
the Frobenius-Euler polynomials are also derived from the non-linear differential equations (see [9, 12]).
Recently, several authors have studied some interesting properties for the Changhee numbers and polynomials (see [1–19]).
In this paper, we develop some new method for obtaining identities related to λ-Changhee polynomials
arising from linear differential equations. From our study, we derive some explicit and new identities for the
λ-Changhee polynomials.
2. Some identities for the λ-Changhee polynomials arising from linear differential equations
First, we introduce lemma for the generalized Changhee power sum Sk,j (N ).
Lemma 2.1. For 2 ≤ r ≤ N and 1 ≤ i ≤ r − 1, we have
S1,i−1 (r − 1 − i) + S1,i−2 (r − i) = S1,i−1 (r − i) .
(A)
Proof. From (1.2) and (1.3), we have
S1,i−1 (r − 1 − i) + S1,i−2 (r − i) =
r−1−i
X
S1,i−2 (l) + S1,i−2 (r − i)
l=0
=
r−i
X
S1,i−2 (l)
l=0
= S1,i−1 (r − i) .
Let
F = F (t; x, λ) =
x
2λ
1 + λ−1 log (1 + λt) .
2λ + log (1 + λt)
(2.1)
Then, by (2.1), we get
F (1) =
d
F (t; x, λ)
dt
= λ (1 + λt)
F (2) =
−1
−1
− (2λ + log (1 + λt))
−1
+ x (λ + log (1 + λt))
(2.2)
F.
dF (1)
dt
n
= λ2 (1 + λt)−2 (2λ + log (1 + λt))−1 − x (λ + log (1 + λt))−1
+2 (2λ + log (1 + λt))−2 − 2x (2λ + log (1 + λt))−1 (λ + log (1 + λt))−1
o
+ (x)2 (λ + log (1 + λt))−2 F.
(2.3)
T. Kim, D. S. Kim, J.-J. Seo, H.-I. Kwon, J. Nonlinear Sci. Appl. 9 (2016), 3098–3111
3101
So, we are led to put
N
d
(N )
F
=
F (t; x, λ)
dt

N
= λ (1 + λt)
−N
(2.4)

(λ)
ai,j (N, x) (2λ
X

−i
+ log (1 + λt))
−j 
(λ + log (1 + λt))
F,
1≤i+j≤N
where N = 1, 2, . . . , and the sum is over all nonnegative integers i, j with 1 ≤ i + j ≤ N .
On the one hand, by (2.4), we get
dF (N )
dt
N +1
=λ
(1 + λt)−(N +1)




X
(λ)
× (−N )
ai,j (N, x) (2λ + log (1 + λt))−i (λ + log (1 + λt))−j


1≤i+j≤N

i≥0,j≥0
X
(λ)
−
iai−1,j (N, x) (2λ + log (1 + λt))−i (λ + log (1 + λt))−j
F (N +1) =
2≤i+j≤N +1
i≥1, j≥0
(λ)
(x − j + 1) ai,j−1 (N, x) (2λ + log (1 + λt))−i
X
+
2≤i+j≤N +1
i≥0,j≥1
o
× (λ + log (1 + λt))−j F.
On the other hand, by replacing N by N + 1 in (2.4), we have
F (N +1) = λN +1 (1 + λt)−(N +1)
X
(λ)
×
ai,j (N + 1, x) (2λ + log (1 + λt))−i
1≤i+j≤N +1
i,j≥0
× (λ + log (1 + λt))−j F.
Let i + j = r. Then 1 ≤ r ≤ N + 1. Comparing the terms with r = 1, we get
(λ)
(λ)
(λ)
(λ)
a1,0 (N + 1, x) = −N a1,0 (N, x)
(2.5)
a0,1 (N + 1, x) = −N a0,1 (N, x) .
Comparing the terms with i + j = r (2 ≤ r ≤ N ),
r
X
(λ)
ai,r−i (N + 1, x) (2λ + log (1 + λt))−i (λ + log (1 + λt))i−r
i=0
= −N
−
+
r
X
(λ)
ai,r−i (N, x) (2λ + log (1 + λt))−i (λ + log (1 + λt))i−r
i=0
r
X
(2.6)
(λ)
iai−1,r−i (N, x) (2λ
i=1
r−1
X
i=0
(λ)
−i
+ log (1 + λt))
i−r
(λ + log (1 + λt))
(x + i − r + 1) ai,r−i−1 (N, x) (2λ + log (1 + λt))−i (λ + log (1 + λt))i−r .
T. Kim, D. S. Kim, J.-J. Seo, H.-I. Kwon, J. Nonlinear Sci. Appl. 9 (2016), 3098–3111
3102
Thus, by (2.6), we get
(λ)
(λ)
(λ)
(λ)
(λ)
a0,r (N + 1, x) = −N a0,r (N, x) + (x − r + 1) a0,r−1 (N, x) ,
(λ)
ar,0 (N + 1, x) = −N ar,0 (N, x) − rar−1,0 (N, x) ,
(2.7)
(2.8)
and
(λ)
(λ)
(λ)
ai,r−i (N + 1, x) = −N ai,r−i (N, x) − iai−1,r−i (N, x)
(λ)
(2.9)
+ (x + i − r + 1) ai,r−i−1 (N, x) ,
where 1 ≤ i ≤ r − 1.
Comparing the terms with i + j = N + 1, we get
N
+1
X
(λ)
ai,N +1−i (N + 1, x) (2λ + log (1 + λt))−i (λ + log (1 + λt))i−(N +1)
i=0
=−
+
N
+1
X
i=1
N
X
(λ)
iai−1,N +1−i (N, x) (2λ + log (1 + λt))−i (λ + log (1 + λt))i−(N +1)
(2.10)
(λ)
(x + i − N ) ai,N −i (N, x) (2λ + log (1 + λt))−i (λ + log (1 + λt))i−(N +1) .
i=0
From (2.10), we note that
(λ)
(λ)
a0,N +1 (N + 1, x) = (x − N ) a0,N (N, x) ,
(λ)
(2.11)
(λ)
aN +1,0 (N + 1, x) = − (N + 1) aN,0 (N, x) ,
and
(λ)
(λ)
(λ)
ai,N +1−i (N + 1, x) = −iai−1,N +1−i (N, x) + (x + i − N ) ai,N −i (N, x) ,
(2.12)
where 1 ≤ i ≤ N .
From (2.2) and (2.4), we have
λ (1 + λt)−1 − (2λ + log (1 + λt))−1 + x (λ + log (1 + λt))−1 F
= F (1)
(λ)
= λ (1 + λt)−1 a1,0 (1, x) (2λ + log (1 + λt))−1
(λ)
+a0,1 (1, x) (λ + log (1 + λt))−1 F.
By comparing the coefficients on both sides of (2.13), we have
(λ)
a1,0 (1, x) = −1,
(λ)
a0,1 (1, x) = x.
From (2.5), we note that
(λ)
a1,0 (N + 1, x) = (−1)N +1 N !,
(λ)
a0,1 (N + 1, x) = (−1)N N !x.
By (2.11), we easily get
(λ)
aN +1,0 (N + 1, x) = (−1)N +1 (N + 1)!,
(λ)
a0,N +1 (N + 1, x) = (x)N +1 .
(2.13)
T. Kim, D. S. Kim, J.-J. Seo, H.-I. Kwon, J. Nonlinear Sci. Appl. 9 (2016), 3098–3111
For i = 1 in (2.12), we have
(λ)
a1,N (N + 1, x)
(λ)
(λ)
= −a0,N (N, x) + (x + 1 − N ) a1,N −1 (N, x)
(λ)
= − (x)N + (x + 1 − N ) a1,N −1 (N, x)
(λ)
= − (x)N + (x + 1 − N ) − (x)N −1 + (x + 2 − N ) a1,N −2 (N − 1, x)
(λ)
= −2 (x)N + (x + 1 − N ) (x + 2 − N ) a1,N −2 (N − 1, x)
..
.
(λ)
= −N (x)N + (x + 1 − N ) (x + 2 − N ) · · · (x + N − N ) a1,0 (1, x)
= − (N + 1) (x)N
= −S1,0 (N ) (x)N .
For i = 2 in (2.12), we note that
(λ)
a2,N −1 (N + 1, x)
(λ)
= −2a1,N −1 (N, x)
(λ)
+ (x + 2 − N ) a2,N −2 (N, x)
= (−1)2 2N (x)N −1
n
o
(λ)
+ (x + 2 − N ) (−1)2 2 (N − 1) (x)N −2 + (x + 3 − N ) a2,N −3 (N − 1, x)
(λ)
= (−1)2 2 {N + (N − 1)} (x)N −1 + (x + 2 − N ) (x + 3 − N ) a2,N −3 (N − 1, x)
..
.
= (−1)2 2 {N + (N − 1) + (N − 2) + · · · + 2} (x)N −1
(λ)
+ (x + 2 − N ) (x + 3 − N ) · · · xa2,0 (2, x)
= (−1)2 2 {N + (N − 1) + · · · + 2 + 1} (x)N −1
= (−1)2 2!S1,1 (N − 1) (x)N −1 .
Let i = 3 in (2.12). Then we have
(λ)
a3,N −2 (N + 1, x)
(λ)
(λ)
= −3a2,N −2 (N, x) + (x + 3 − N ) a3,N −3 (N, x)
(λ)
= (−1)3 3!S1,1 (N − 2) (x)N −2 + (x + 3 − N ) a3,N −3 (N, x)
..
.
= (−1)3 3! {S1,1 (N − 2) + · · · + S1,1 (1)} (x)N −2
(λ)
+ (x + 3 − N ) (x + 4 − N ) · · · xa3,0 (3, x)
= (−1)3 3! {S1,1 (N − 2) + · · · + S1,1 (1) + S1,1 (0)} (x)N −2
= (−1)3 3!S1,2 (N − 2) (x)N −2 .
Continuing this process, we get
(λ)
ai,N +1−i (N + 1, x) = (−1)i i!S1,i−1 (N − i + 1) (x)N −i+1 ,
(1 ≤ i ≤ N ) .
3103
T. Kim, D. S. Kim, J.-J. Seo, H.-I. Kwon, J. Nonlinear Sci. Appl. 9 (2016), 3098–3111
3104
Let 2 ≤ r ≤ N . Then, by (2.7), we get
(λ)
(λ)
(λ)
a0,r (N + 1, x) = (x − r + 1) a0,r−1 (N, x) − N a0,r (N, x)
(λ)
= (x − r + 1) a0,r−1 (N, x)
n
o
(λ)
(λ)
−N (x − r + 1) a0,r−1 (N − 1, x) − (N − 1) a0,r (N − 1, x)
n
o
(λ)
(λ)
= (x − r + 1) a0,r−1 (N, x) − N a0,r−1 (N − 1, x)
(λ)
+ (−1)2 N (N − 1) a0,r (N − 1, x)
(2.14)
..
.
= (x − r + 1)
N
−r
X
(λ)
(−1)i (N )i a0,r−1 (N − i, x)
i=0
N −r+1
(λ)
N (N − 1) · · · ra0,r (r, x)
+ (−1)
= (x − r + 1)
NX
−r+1
(λ)
(−1)i (N )i a0,r−1 (N − i, x) .
i=0
(λ)
Now, we give an explicit expression for a0,r (N + 1, x) (2 ≤ r ≤ N ).
For r = 2 in (2.14), we have
(λ)
a0,2 (N + 1, x) = (x − 1)
N
−1
X
(λ)
(−1)i (N )i a0,1 (N − i, x)
i=0
= (x)2 (−1)N −1
N
−1
X
(N )i (N − i − 1)!
i=0
N
−1
X
= (x)2 (−1)N −1 N !
= (x)2 (−1)N −1 N !
i=0
N
X
i=1
N −1
= (x)2 (−1)
1
N −i
1
i
N !HN,1 .
Let us consider r = 3 in (2.14). From (2.15), we note that
(λ)
a0,3 (N
+ 1, x) = (x − 2)
= (x − 2)
N
−2
X
i=0
N
−2
X
(λ)
(−1)i (N )i a0,2 (N − i, x)
(−1)i (N )i (x)2 (−1)N −i−2 (N − i − 1)!HN −i−1
i=0
= (x)3 (−1)N −2 N !
N
−2
X
i=0
N −2
= (x)3 (−1)
HN −i−1
N −i
N !HN,2 .
For r = 4 in (2.14), we have
(λ)
a0,4 (N
+ 1, x) = (x − 3)
N
−3
X
i=0
(λ)
(−1)i (N )i a0,3 (N − i, x)
(2.15)
T. Kim, D. S. Kim, J.-J. Seo, H.-I. Kwon, J. Nonlinear Sci. Appl. 9 (2016), 3098–3111
= (x − 3)
N
−3
X
3105
(−1)i (N )i (x)3 (−1)N −i−3 (N − i − 1)!HN −i−1,2
i=0
= (x)4 (−1)N −3 N !
N
−3
X
i=0
N −3
= (x)4 (−1)
N!
HN −i−1,2
N −i
HN −1,2 HN −2,2
H2,2
+
+ ··· +
N
N −1
3
= (x)4 (−1)N −3 N !HN,3 .
Continuing this process, we get
(λ)
a0,r (N + 1, x) = (x)r (−1)N −r+1 N !HN,r−1 .
For 2 ≤ r ≤ N , by (2.8), we get
(λ)
(λ)
(λ)
ar,0 (N + 1, x) = −rar−1,0 (N, x) − N ar,0 (N, x)
..
.
= −r
= −r
= −r
N
−r
X
i=0
N
−r
X
(λ)
(λ)
(−1)i (N )i ar−1,0 (N − i, x) + (−1)N −r+1 N (N − 1) · · · rar,0 (r, x)
(2.16)
(λ)
(−1)i (N )i ar−1,0 (N − i, x) + (−1)N −r+1 (N )N −r+1 (−1)r r!
i=0
NX
−r+1
(λ)
(−1)i (N )i ar−1,0 (N − i, x) .
i=0
Let r = 2 in (2.16). Then, we have
(λ)
a2,0 (N
+ 1, x) = −2
N
−1
X
(λ)
(−1)i (N )i a1,0 (N − i, x)
i=0
= 2 (−1)N +1
N
+1
X
(N )i (N − i − 1)!
i=0
N
−1
X
= 2 (−1)N +1 N !
= 2 (−1)N +1 N !
i=0
N
X
i=1
N +1
= 2 (−1)
1
N −i
(2.17)
1
i
N !HN,1 .
For r = 3 in (2.16), by (2.17), we get
(λ)
a3,0 (N
+ 1, x) = −3
= −3
N
−2
X
i=0
N
−2
X
(λ)
(−1)i (N )i a2,0 (N − i, x)
(−1)i (N )i 2 (−1)N −i (N − i − 1)!HN −i−1,1
i=0
= 3! (−1)N +1 N !
N
−2
X
i=0
N +1
= 3! (−1)
HN −i−1,1
N −i
N !HN,2 .
(2.18)
T. Kim, D. S. Kim, J.-J. Seo, H.-I. Kwon, J. Nonlinear Sci. Appl. 9 (2016), 3098–3111
3106
From (2.18), by r = 4 in (2.16), we note that
(λ)
a4,0 (N
+ 1, x) = −4
N
−3
X
(λ)
(−1)i (N )i a3,0 (N − i, x)
i=0
= (−1)N +1 4!N !
N
−3
X
i=0
N +1
= (−1)
HN −i−1,2
N −i
4!N !HN,3 .
Continuing this process, we get
(λ)
ar,0 (N + 1, x) = (−1)N +1 N !r!HN,r−1 ,
(2 ≤ r ≤ N ) .
Let 2 ≤ r ≤ N , 1 ≤ i ≤ r − 1. Then, by (2.9), we get
(λ)
(λ)
(λ)
(λ)
ai,r−i (N + 1, x) = (x + i − r + 1) ai,r−i−1 (N, x) − iai−1,r−i (N, x) − N ai,r−i (N, x)
n
o
(λ)
(λ)
= (x + i − r + 1) ai,r−i−1 (N, x) − N ai,r−i−1 (N − 1, x)
n
o
(λ)
(λ)
−i ai−1,r−i (N, x) − N ai−1,r−i (N − 1, x)
(λ)
+ (−1)2 N (N − 1) ai,r−i (N − 1, x)
..
.
= (x + i − r + 1)
N
−r
X
(λ)
(−1)s (N )s ai,r−i−1 (N − s, x)
s=0
−i
N
−r
X
s
(−1)
(2.19)
(λ)
(N )s ai−1,r−i (N
− s, x)
s=0
(λ)
+ (−1)N −r+1 (N )N −r+1 ai,r−i (r, x)
= (x + i − r + 1)
N
−r
X
(λ)
(−1)s (N )s ai,r−i−1 (N − s, x)
s=0
−i
N
−r
X
(λ)
(−1)s (N )s ai−1,r−i (N − s, x)
s=0
+ (−1)N −r+1 (N )N −r+1 (−1)i i!S1,i−1 (r − i) (x)r−i .
Let r = 2 in (2.19). Then i = 1. From (2.19), we note that
(λ)
a1,1 (N + 1, x) = x
N
−2
X
−
(λ)
(−1)s (N )s a1,0 (N − s, x)
s=0
N
−2
X
(λ)
(−1)s (N )s a0,1 (N − s, x)
s=0
+ (−1)N −1 N ! (−1) S1,0 (1) x
=x
N
−2
X
−
(−1)s (N )s (−1)N −s (N − s − 1)!
s=0
N
−2
X
s=0
(−1)s (N )s (−1)N −s−1 (N − s − 1)!x + 2 (−1)N N !x
T. Kim, D. S. Kim, J.-J. Seo, H.-I. Kwon, J. Nonlinear Sci. Appl. 9 (2016), 3098–3111
N
−2
X
N
= 2x (−1)
(N )s (N − s − 1)! + 2 (−1)N N !x
s=0
N
−2
X
= 2x (−1)N N !
s=0
N
3107
= 2 (−1) N !x
1
+ 2 (−1)N N !x
N −s
1
1 1
1
+
+ ··· + +
N
N −1
2 1
= 2 (−1)N N !xHN .
Let r = 3. Then 1 ≤ i ≤ 2. From (2.19), we note that
(λ)
ai,3−i (N + 1, x) = (x + i − 2)
N
−3
X
(λ)
(−1)s (N )s ai,2−i (N − s, x)
s=0
−i
N
−3
X
(2.20)
(λ)
(−1)s (N )s ai−1,3−i (N − s, x)
s=0
+ (−1)N −2 (N )N −2 (−1)i i!S1,i−1 (3 − i) (x)3−i .
For i = 1 in (2.20), we have
(λ)
a1,2 (N + 1, x)
= (x − 1)
N
−3
X
(λ)
(−1)s (N )s a1,1 (N − s, x)
s=0
−
N
−3
X
(λ)
(−1)s (N )s a0,2 (N − s, x)
s=0
+ (−1)N −2 (N )N −2 (−1) S1,0 (2) (x)2
= (x − 1)
N
−3
X
(−1)s (N )s 2 (−1)N −s−1 (N − s − 1)!HN −s−1,1 x
(2.21)
s=0
−
N
−3
X
(−1)s (N )s (x)2 (−1)N −s−2 (N − s − 1)!HN −s−1,1
s=0
+3 (−1)N −1 (N )N −2 (x)2
= 3 (x)2 (−1)N −1 N !
N
−3
X
s=0
N −1
= 3 (x)2 (−1)
N!
HN −s−1,1
1
+ 3 (x)2 (−1)N −1 N !
N −s
2
HN −1,1 HN −2,1
H2,1 H1,1
+
+ ··· +
+
N
N −1
3
2
= 3 (x)2 (−1)N −1 N !HN,2 .
Let i = 2 in (2.20). From (2.21), we note that
(λ)
a2,1 (N + 1, x) = x
N
−3
X
(λ)
(−1)s (N )s a2,0 (N − s, x)
s=0
N
−3
X
−2
s=0
(λ)
(−1)s (N )s a1,1 (N − s, x)
T. Kim, D. S. Kim, J.-J. Seo, H.-I. Kwon, J. Nonlinear Sci. Appl. 9 (2016), 3098–3111
3108
+ (−1)N −2 (N )N −2 (−1)2 2!S1,1 (1) x
=x
N
−3
X
(−1)s (N )s 2 (−1)N −s (N − s − 1)!HN −s−1,1
s=0
N
−3
X
−2
(−1)s (N )s 2 (−1)N −s−1 (N − s − 1)!HN −s−1,1 x
s=0
+ (−1)N −2 (N )N −2 (−1)2 3!x
N
−3
X
HN −s−1,1
+ 6x (−1)N (N )N −2
N −s
s=0
HN −1,1
H2,1 H1,1
N
= 6x (−1) N !
+ ··· +
+
N
3
2
= 6x (−1)N N !
= 6x (−1)N N !HN,2 .
Therefore, we obtain the following theorem.
Theorem 2.2. Let 1 ≤ r ≤ N + 1, 0 ≤ i ≤ r. Then we have
(λ)
ai,r−i (N + 1, x) = (−1)N +1+i−r i!S1,i−1 (r − i) N !HN,r−1 (x)r−i .
Proof. We showed that it is true for r = 1 and r = N + 1. Assume that 2 ≤ r ≤ N . If i = 0 or i = r, then
it is also true. So we prove the assertion by induction on r when 2 ≤ r ≤ N , 1 ≤ i ≤ r − 1. For r = 2, i = 1,
we showed that
(λ)
a1,1 (N + 1, x) = 2 (−1)N N !HN,1 x.
Assume now that it is true for r − 1 (3 ≤ r ≤ N ).
From 2.19, we note that
(λ)
ai,r−i (N
+ 1, x) = (x + i − r + 1)
N
−r
X
(λ)
(−1)s (N )s ai,r−i−1 (N − s, x)
s=0
−i
N
−r
X
(λ)
(−1)s (N )s ai−1,r−i (N − s, x)
s=0
+ (−1)N −r+1 (N )N −r+1 (−1)i i!S1,i−1 (r − i) (x)r−i
= (x + i − r + 1)
N
−r
X
(−1)s (N )s (−1)N −s−1+i−r i!
s=0
(2.22)
×S1,i−1 (r − 1 − i) (N − s − 1)!HN −s−1,r−2 (x)r−1−i
−i
N
−r
X
(−1)s (N )s (−1)N −s+i−r (i − 1)!S1,i−2 (r − i)
s=0
× (N − s − 1)!HN −s−1,r−2 (x)r−i
+ (−1)N −r+1 (N )N −r+1 (−1)i i!S1,i−1 (r − i) (x)r−i
= (−1)N +1+i−r i!N ! (x)r−i
(
× (S1,i−1 (r − 1 − i) + S1,i−2 (r − i))
N
−r
X
s=0
HN −s−1,r−2 S1,i−1 (r − i)
+
N −s
(r − 1)!
)
.
T. Kim, D. S. Kim, J.-J. Seo, H.-I. Kwon, J. Nonlinear Sci. Appl. 9 (2016), 3098–3111
3109
By Lemma 2.1 and (2.22), we get
(λ)
ai,r−i (N + 1, x) = (−1)N +1+i−r i!S1,i−1 (r − i) N ! (x)r−i
HN −1,r−2
Hr−1,r−2 Hr−2,r−2
×
+ ··· +
+
N
r
r−1
= (−1)N +1+i−r i!S1,i−1 (r − i) N !HN,r−1 (x)r−i .
Therefore, we obtain the following theorem.
Theorem 2.3. The linear differential equations
N X
r
X
F (N ) = λN (1 + λt)−N ×
!
(λ)
ai,r−i (N, x) (2λ + log (1 + λt))−i (λ + log (1 + λt))−(r−i)
F
r=1 i=0
has a solution
x
F = F (t; x, λ) = 2λ (2λ + log (1 + λt))−1 1 + λ−1 log (1 + λt) ,
where, for 1 ≤ r ≤ N , 0 ≤ i ≤ r,
(λ)
ai,r−i (N, x) = (−1)N +i−r i!S1,i−1 (r − i) (N − 1)!HN −1,r−1 (x)r−i .
Recall that the λ-Changhee polynomials, Chn,λ (x), (n ≥ 0), are given by the generating function
F = F (t; x, λ)
2λ
=
2λ + log (1 + λt)
∞
X
tn
=
Chn,λ (x) .
n!
x
1
1 + log (1 + λt)
λ
(2.23)
n=0
Thus, by (2.23), we get
F
(N )
d N
=
F (t; x, λ)
dt
∞
X
tk
Chk+N,λ (x) .
=
k!
k=0
On the other hand, by Theorem 2.3, we get
F
(N )
N
r
∞
X
N + l − 1 l l X X (λ)
m i+m−1
=λ
(−1)
λt
ai,r−i (N, x)
(−1)
l
m
r=1 i=0
m=0
l=0
∞
X
r + n − i − 1 −(r−i)−n
= (2λ)−i−m (log (1 + λt))m
(−1)n
λ
(log (1 + λt))n
n
N
∞
X
l
n=0
×
=λ
∞
X
Chs,λ (x)
s=0
∞
X
N
l=0
ts
s!
(−1)l (N + l − 1)l λl
tl
l!
(2.24)
T. Kim, D. S. Kim, J.-J. Seo, H.-I. Kwon, J. Nonlinear Sci. Appl. 9 (2016), 3098–3111
×
N X
r
X
r=1 i=0
×
∞
X
∞
X
(λ)
ai,N −i (N, x)
3110
(−1)m (i + m − 1)m (2λ)−i−m
m=0
∞
e+m
S1 (e + m, m) λ
te+m X
(−1)n (r + n − i − 1)n
(e + m)!
n=0
e=0
∞
X
×λi−r−n
S1 (f + n, n)
f =0
∞
×λf +n
ts
tf +n X
Chs,λ (x)
(f + n)!
s!
s=0
N X
r
X
= λN
(λ)
ai,r−i (N, x)
r=1 i=0
×
×
×
=
∞
X
n=0
∞
X
∞
X
(−1)m (i + m − 1)m (2λ)−i−m λm
m=0
∞
(−1)n (r + n − i − 1)n λi−r−n λn
tn X
tl
(−1)l (N + l − 1)l λl
n!
l!
l=0
S1 (e + m, m) λe
e=0
∞
X
∞
te X
e!m!
(e + m)! e!
S1 (f + n, n) λf
f =0
i−r−n n
×λ
λ
∞
tn X

n!

a=0
N X
r
X
l
(−1) λ
a−s
l+e+f +s=a
(λ)
ai,r−i (N, x) λ−r 2−i
r=1 i=0
a
l,e,f,s
e+m f +n
m
n
∞
X
ta
a!
X
k=0
× (r + n − i −
+ n − i − 1)n
X
S1 (e + m, m) S1 (f + n, n) Chs,λ (x))
=λ
f !n! tf
(f + n)! f !
ts
Chs,λ (x)
s!
s=0
N X
r
∞
X
X
(λ)
λN
ai,r−i (N, x)
(−1)m
r=1 i=0
m=0
∞
tm X
× (i + m − 1)m (2λ)−i−m λm
(−1)n (r
m!
n=0
N
tm
m!
(N + l − 1)l
k
m, n, a
m+n+a=k
a
l,e,f,s,
l a−s
1)n
(−1) λ
e+m f +n
m
n
l+e+f +s=a
m
1
−
(−1)n (i + m − 1)m
2
X
tk
× (N + l − 1)l S1 (e + m, m) S1 (f + n, n) Chs,λ (x))
k!
(
∞
N X
r
X
X
(λ)
=
λN
ai,r−i (N, x) λ−r 2−i
r=1 i=0
k=0
×
X
m+n+a=k
×
X
l+e+f +s=a
k
m, n, a
m
1
−
(−1)n (i + m − 1)m (r + n − i − 1)n
2
a
(−1)l λa−s
l,e,f,s
e+m f +n
m
n
(N + l − 1)l
×S1 (e + m, m) S1 (f + n, n) Chs,λ (x)}
tk
.
k!
(2.25)
T. Kim, D. S. Kim, J.-J. Seo, H.-I. Kwon, J. Nonlinear Sci. Appl. 9 (2016), 3098–3111
3111
Therefore, by comparing the coefficients on both sides of (2.24) and (2.25), we obtain the following
theorem.
Theorem 2.4. For N ∈ N, and k ∈ N ∪ {0}, we have
Chk+N,λ (x) = λ
N
N X
r
X
(λ)
ai,r−i (N, x) λ−r 2−i
r=1 i=0
X
m+n+a=k
k
m, n, a
m
1
× −
(−1)n (i + m − 1)m (r + n − i − 1)n
2
a
X
l,e,f,s
l a−s
×
(−1) λ
e+m f +n
l+e+f +s=a
m
n
× (N + l − 1)l S1 (e + m, m) S1 (f + n, n) Chs,λ (x) ,
where, for 1 ≤ r ≤ N , 0 ≤ i ≤ r,
(λ)
ai,r−i (N, x) = (−1)N +i−r i!S1,i−1 (r − i) (N − 1)!HN −1,r−1 (x)r−i .
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