Available online at www.tjnsa.com
J. Nonlinear Sci. Appl. 9 (2016), 860–869
Research Article
Bernoulli polynomials of the second kind and their
identities arising from umbral calculus
Taekyun Kima,b , Dae San Kimc,∗, Dmitry V. Dolgyd , Jong-Jin Seoe
a
Department of Mathematics, College of Science, Tianjin Polytechnic University, Tianjin City, 300387, China.
Department of Mathematics, Kwangwoon University, Seoul 139-701, S. Korea.
c
Department of Mathematics, Sogang University, Seoul 121-742, Republic of Korea.
d
School of Natural Sciences, Far Eastern Federal University, 690950 Vladivostok, Russia.
e
Department of Applied Mathematics, Pukyong National University, Pusan 608-739, S. Korea.
b
Communicated by S.-H. Rim
Abstract
In this paper, we study the Bernoulli polynomials of the second kind with umbral calculus viewpoint and
c
derive various identities involving those polynomials by using umbral calculus. 2016
All rights reserved.
Keywords: Bernoulli polynomial of the second kind, umbral calculus.
2010 MSC: 05A40, 11B68, 11B83.
1. Introduction and preliminaries
As is well known, the ordinary Bernoulli polynomials are defined by the generating function to be
∞
X
tn
t
xt
e
=
B
(x)
,
n
et − 1
n!
(see [2, 10, 15, 17]) .
(1.1)
n=0
When x = 0, Bn = Bn (0) are called the Bernoulli numbers. The Bernoulli polynomials of the second kind
are given by the generating function to be
∞
X
tn
t
(1 + t)x =
bn (x) ,
log (1 + t)
n!
(see [19, 21, 22]) .
n=0
∗
Corresponding author
Email addresses: tkkim@kw.ac.kr, kimtk2015@gmail.com (Taekyun Kim), dskim@sogang.ac.kr (Dae San Kim),
d_dol@mail.ru (Dmitry V. Dolgy ), seo2011@pknu.ac.kr (Jong-Jin Seo)
Received 2015-10-08
(1.2)
T. Kim, D. S. Kim, D. V. Dolgy, J.-J. Seo, J. Nonlinear Sci. Appl. 9 (2016), 860–869
861
When x = 0, bn = bn (0) are called the Bernoulli numbers of the second kind.
The first few Bernoulli numbers bn of the second kind are
1
1
1
19
3
b0 = 1, b1 = , b2 = − , b3 = , b4 = −
, b5 =
,··· .
2
12
24
720
160
By (1.2), we easily get
bn (x) =
n X
n
bl (x)n−l ,
l
(1.3)
l=0
where (x)n = x (x − 1) · · · (x − n + 1), (n ≥ 0), and
bn (x) = Bn(n) (x + 1) ,
(see [21, 22]) ,
(1.4)
(n ≥ 0) .
(1.5)
(α)
where Bn (x) are the Bernoulli polynomials of order α.
The stirling number of the second kind is given by
xn =
n
X
S2 (n, l) (x)l ,
l=0
The Stirling number of the first kind is defined by
(x)n = x (x − 1) · · · (x − n + 1) =
n
X
S1 (n, l) xl ,
(n ≥ 0) .
(1.6)
l=0
Let C be the complex number field and let F be the set of all formal power series in the variable t:
(
)
∞
X
tk ak ak ∈ C .
F = f (t) =
k! (1.7)
k=0
Let us assume that P is the algebra of polynomials in the variable x over C and P∗ is the vector space of all
linear functionals on P. h L| p (x)i denotes the action of the linear functional L on a polynomial p (x). For
f (t) ∈ F, we define the continuous linear functional f (t) on P by
hf (t) |xn i = an ,
Thus, by (1.7) and (1.8), we get
D E
tk xn = n!δn,k ,
(n ≥ 0) ,
(n, k ≥ 0) ,
(see [21]) .
(see [1–22]) ,
(1.8)
(1.9)
where δn,k is the Kronecker’s symbol.
∞ L xk
P
h | i k
n
n
For fL (t) =
k! t , we have hfL (t) |x i = hL |x i , (n ≥ 0). Thus, we see that fL (t) = L. The
k=0
map L 7→ fL (t) is a vector space isomorphism from P∗ onto F. Henceforth, F is thought of as both a formal
power series and a linear functional. We call F the umbral algebra. The umbral calculus is the study of
umbral algebra. The order o (f (t)) of the non-zero power series f (t) is the smallest integer k for which
the coefficient of tk does not vanish (see [8, 21]). If o (f (t)) = 1, then f (t) is called a delta series and if
o (f (t)) = 0, then f (t) is called an invertible series. For f (t) ,Dg (t) ∈ F with
o (fE(t)) = 1 and o (g (t)) = 0,
k
there exists a unique sequence sn (x) of polynomials such that g (t) f (t) sn (x) = n!δn,k , where n, k ≥ 0.
The sequence sn (x) is called
sequence for (g (t) , f (t)) which is denoted by sn (x) ∼ (g (t) , f (t)).
yt the Sheffer
For p (x) ∈ P, we have e p (x) = p (y) and eyt p (x) = p (x + y). Let f (t) ∈ F and p (x) ∈ P. Then we
see that
∞ ∞ k X
X
t p (x) k
f (t) xk k
t , p (x) =
x .
(1.10)
f (t) =
k!
k!
k=0
Thus, by (1.10), we get
k=0
T. Kim, D. S. Kim, D. V. Dolgy, J.-J. Seo, J. Nonlinear Sci. Appl. 9 (2016), 860–869
D E
p(k) (0) = tk p (x) ,
D
E
1| p(k) (x) = p(k) (0) .
862
(1.11)
From (1.11), we have
tk p (x) = p(k) (x) =
dk p (x)
,
dxk
(k ≥ 0) .
For sn (x) ∼ (g (t) , f (t)), we have
n−1 E
dsn (x) X n D
=
f (t) xn−l sl (x) ,
dx
l
(n ≥ 1) ,
(1.12)
l=0
where f (t) is the compositional inverse of f (t) with f (f (t)) = f f (t) = t,
∞
X
1
tn
= exf (t) =
sn (x) ,
n!
g f (t)
n=0
f (t) sn (x) = nsn−1 (x) ,
(n ≥ 1) ,
for all x ∈ C,
sn (x + y) =
n X
n
j=0
j
sj (x) pn−j (y) ,
(1.13)
(1.14)
where pn (x) = g (t) sn (x),
h f (t)| xp (x)i = h ∂t f (t)| p (x)i ,
(1.15)
and
1
g 0 (t)
sn (x) , (n ≥ 0) , (see [1, 13, 16, 21]) .
sn+1 (x) = x −
0
g (t) f (t)
Assume that pn (x) ∼ 1, f (t) and qn (x) ∼ 1, g(t) . Then the transfer formula is given by
f (t) n −1
qn (x) = x
x pn (x) (n ≥ 1).
g(t)
(1.16)
(1.17)
For sn (x) ∼ (g (t) , f (t)), rn (x) ∼ (h (t) , l (t)), we have
sn (x) =
n
X
Cn,m rm (x) ,
(n ≥ 0) ,
(1.18)
m=0
where
Cn,m
1
=
m!
*
+
m m
h f (t)
l f (t)
,
x
g f (t)
(see [12, 21]) .
(1.19)
In this paper, we study the Bernoulli polynomials of the second kind with umbral calculus viewpoint and
derive various identities involving those polynomials by using umbral calculus.
2. Bernoulli polynomials of the second kind
For α ∈ N, the Bernoulli polynomials of the second kind with order α are defined by
α
∞
X
tn
t
x
(1 + t) =
b(α)
(x)
.
n
log (1 + t)
n!
(2.1)
n=0
(1)
(α)
(α)
Note that bn (x) = bn (x). When x = 0, bn = bn (0) are called the Bernoulli numbers of the second kind
with order α. Indeed, we note that
(n−α+1)
b(α)
(x + 1) .
n (x) = Bn
Let us consider the following two sheffer sequences :
T. Kim, D. S. Kim, D. V. Dolgy, J.-J. Seo, J. Nonlinear Sci. Appl. 9 (2016), 860–869
863
log (1 + t) α t
e −1
qn (x) ∼ 1,
t
and
(x)n ∼ 1, et − 1 .
Thus, by (1.17), we get
αn
t
qn (x) = x
x−1 (x)n
log (1 + t)
αn
t
(x − 1)n−1
=x
log (1 + t)
(αn)
= xbn−1 (x − 1) ,
That is,
(αn)
xbn−1 (x
(n ≥ 1) .
log (1 + t) α t
− 1) ∼ 1,
e −1 .
t
From (1.2) and (1.13), we have
bn (x) ∼
t
t
,e − 1 .
et − 1
(2.2)
By (2.2), we get
et
t
bn (x) ∼ 1, et − 1 ,
−1
(x)n ∼ 1, et − 1 .
(2.3)
Thus, we see that
n
et − 1
et − 1 X
bn (x) =
(x)n =
S1 (n, l) xl
t
t
(2.4)
l=0
n
X
S1 (n, l) l+1
x
= e −1
l+1
t
l=0
=
n
X
S1 (n, l) l=0
l+1
When x = 0, we have
bn =
(x + 1)l+1 − xl+1 .
n
X
S1 (n, l)
l=0
l+1
.
By (1.12), we get
n−1
E
X n D
d
bn (x) =
log (1 + t)| xn−l bl (x)
dx
l
l=0
* ∞
+
n−1
X n X
(−1)m−1 m n−l
=
t x
bl (x)
m
l
m=1
l=0
n−1
X n
=
(n − l − 1)! (−1)n−l−1 bl (x)
l
=
l=0
n−1
X
l=0
n!
(−1)n−l−1 bl (x) ,
l!(n − l)
Therefore, by (2.5), we obtain the following lemma.
(n ≥ 1) .
(2.5)
T. Kim, D. S. Kim, D. V. Dolgy, J.-J. Seo, J. Nonlinear Sci. Appl. 9 (2016), 860–869
864
Lemma 1. For n ≥ 1, we have
n−1
X
d
n!
bn (x) =
(−1)n−l−1 bl (x) .
dx
l!(n − l)
l=0
From (1.9), we have
(1 + t) xn
bn (y) =
*
+
X
m
∞
t
t
n
=
(y)m
x
log (1 + t) m!
m=0
n
X
n−m
n
t
x
=
(y)m
log (1 + t) m
m=0
n
X
n
=
(y)m
bn−m .
m
t
log (1 + t)
y
(2.6)
m=0
Therefore, by (2.6), we obtain the following proposition.
Proposition 2. For n ≥ 0, we have
n X
n
bn (x) =
bn−m (x)m
m
m=0
n
X
n
x
=
m!
bn−m .
m m
m=0
By (1.2), we get
n
X
t
bn (x) =
(x)n =
S1 (n, l)
log (1 + t)
l=0
t
log (1 + t)
xl
(2.7)
n
X
l
X
bm m l
=
S1 (n, l)
t x
m!
m=0
l=0
n
l X
X
l
=
S1 (n, l)
bm xl−m
m
m=0
l=0
n
l
XX
l
=
S1 (n, l)
bl−m xm .
m
l=0 m=0
By (1.14), we get
n X
n
bn (x + y) =
bj (x) (y)n−j .
j
(2.8)
j=0
Let
Pn = { p (x) ∈ C [x]| deg p (x) ≤ n} ,
(n ≥ 0) .
Then it is an (n + 1)-dimensional vector space over C. Now, we consider the polynomial p (x) in Pn which
is given by
n
X
p (x) =
Cm bm (x) .
(2.9)
m=0
Thus, by (2.9), we get
T. Kim, D. S. Kim, D. V. Dolgy, J.-J. Seo, J. Nonlinear Sci. Appl. 9 (2016), 860–869
X
n
m m t
t
t
t
e − 1 p (x) =
e − 1 bl (x)
Cl
et − 1
et − 1
865
(2.10)
l=0
=
n
X
Cl l!δm,l = m!Cm .
l=0
From (2.10), we have
1
=
m!
Cm
t
t
e −1
m e − 1 p (x) .
t
(2.11)
Therefore, by (2.11), we obtain the following theorem.
Theorem 3. Let p (x) ∈ Pn with
n
X
p (x) =
Cm bm (x) .
m=0
Then, we have
1
=
m!
Cm
t
t
e −1
m e − 1 p (x) .
t
For example, let us take p (x) = Bn (x) ∈ Pn . Then, we have
Bn (x) =
n
X
Cm bm (x) ,
(2.12)
m=0
where
Cm
m t
1
t
e − 1 Bn (x)
=
m!
et − 1
n
X
n
t
Bn−l (x)
=
S2 (l, m)
et − 1 l
l=m
X
n−l
n
X
k
n−l
t
n
x
=
S2 (l, m)
Bn−l−k
l
k
et − 1 k=0
l=m
n
n−l
XX
n n−l
S2 (l, m)
Bn−l−k Bk .
=
l
k
(2.13)
l=m k=0
Therefore, by (2.12) and (2.13), we obtain the following theorem.
Theorem 4. For n ≥ 0, we have
Bn (x) =
n
X
m=0
(
n X
n−l X
n n−l
S2 (l, m) Bn−l−k Bk
l
k
)
bm (x) .
l=m k=0
Remark. From (2.13), for m ≥ 1, we have
m t
1
t
Cm =
e − 1 Bn (x)
m!
et − 1
E
m−1 1 D t
=
e −1
tBn (x)
m! D
E
m−1 n
et − 1
=
Bn−1 (x)
m!
(2.14)
T. Kim, D. S. Kim, D. V. Dolgy, J.-J. Seo, J. Nonlinear Sci. Appl. 9 (2016), 860–869
866
n−1
E
X
n
1 D l =
(m − 1)!
t Bn−1 (x)
S2 (l, m − 1)
m!
l!
l=m−1
n−1
n X
n−1
=
S2 (l, m − 1)
Bn−1−l .
m
l
l=m−1
Therefore, by (2.12) and (2.14), we get
(
)
n
n−1
n X
X
n X
n−1
n
Bn (x) =
S2 (l, m − 1)
Bn−1−l bm (x) +
Bn−k Bk .
m
l
k
m=1
l=m−1
k=0
The classical polylogarithm function is given by
Lik (x) =
∞
X
xn
n=1
nk
(k ∈ Z, x > 0) .
,
The poly-Bernoulli polynomials are defined by the generating function to be
∞
Lik 1 − et xt X (k)
tn
e
=
.
B
(x)
n
et − 1
n!
(2.15)
(2.16)
n=0
Thus, by (2.16), we see that
Bn(k) (x)
∼
et − 1
,t .
Lik (1 − e−t )
(2.17)
(k)
Let us take p (x) = Bn (x) ∈ Pn . Then we have
Bn(k) (x) =
n
X
Cm bm (x) ,
(2.18)
m=0
where
Cm
m (k)
1
t
t
=
e − 1 Bn (x)
m! et − 1
n
X
n
t (k)
B
=
S2 (l, m)
(x)
et − 1 n−l
l
l=m
X
n
n−l X
n
n−l
t j
(k)
=
x
S2 (l, m)
Bn−l−j
et − 1 l
j
j=0
l=m
=
n X
n−l X
l=m j=0
(k)
(2.19)
n n−l
(k)
S2 (l, m) Bn−l−j Bj ,
l
j
(k)
where Bn = Bn (0) are the poly-Bernoulli numbers. Therefore, by (2.18) and (2.19), we obtain the
following theorem.
Theorem 5. For n ≥ 0, we have


n X
n X
n−l 
X
n n−l
(k)
Bn(k) (x) =
S2 (l, m) Bn−j−l Bj bm (x) .


l
j
m=0
l=m j=0
Let us consider p (x) = xn ∈ Pn . Then, we have
T. Kim, D. S. Kim, D. V. Dolgy, J.-J. Seo, J. Nonlinear Sci. Appl. 9 (2016), 860–869
xn =
n
X
Cm bm (x) ,
867
(2.20)
m=0
where
m n
t
t
x
e
−
1
et − 1
n
t n−l
=
S2 (l, m)
x
l
et − 1 l=m
n
X
n
=
S2 (l, m)
Bn−l .
l
Cm =
1
m!
n
X
(2.21)
l=m
Thus, by (2.20) and (2.21), we get
n
X
xn =
m=0
(
n
X
l=m
)
n
S2 (l, m)
Bn−l bm (x) .
l
(2.22)
Let us consider the following two Sheffer sequences :
t
t
,e − 1 ,
bn (x) ∼
et − 1
and
Bn(k) (x) ∼
(2.23)
et − 1
,
t
.
Lik (1 − e−t )
Then, by (1.17) and (1.18), we get
Bn(k) (x) =
n
X
Cn,m bm (x) ,
(2.24)
m=0
where
Cn,m
+
m n
Lik 1 − e−t
t
t
e − 1 x
et − 1
et − 1
*
+
n
X
Lik 1 − e−t t
n
n−l
S2 (l, m)
=
x
t
e −1
l
et − 1
l=m
*
+
X
n
n−l X
Lik 1 − e−t j
n−l
n
=
S2 (l, m)
Bn−l−j
x
j
l
et − 1
1
=
m!
*
j=0
l=m
n X
n−l X
n n−l
(k)
=
S2 (l, m) Bn−l−j Bj .
l
j
l=m j=0
Therefore, by (2.24) and (2.25), we obtain the following theorem.
Theorem 6. For n ≥ 0, we have
Bn(k) (x) =

n X
n X
n−l X
n n−l
m=0

l=m j=0
l
Let us consider the following Sheffer sequences:
j
(k)
S2 (l, m) Bn−l−j Bj



bm (x) .
(2.25)
T. Kim, D. S. Kim, D. V. Dolgy, J.-J. Seo, J. Nonlinear Sci. Appl. 9 (2016), 860–869
868
t
t
bn (x) ∼
,e − 1 ,
et − 1
t
e −1
Bn (x) ∼
,t .
t
Then we have
bn (x) =
n
X
(2.26)
Cn,m Bm (x) ,
(2.27)
m=0
where
Cn,m
t
t
1
m n
(log (1 + t)) x
=
m! log (1 + t) log (1 + t)
*
+
2 n X
n
t
n−l
=
S1 (l, m)
x
l
log (1 + t) l=m
n X
n
(2)
=
S1 (l, m) bn−l ,
l
(2.28)
l=m
(2)
where bn are di-Bernoulli numbers of the second kind.
Therefore, by (2.27) and (2.28), we get
bn (x) =
n
X
m=0
n X
n
(2)
S1 (l, m) bn−l
l
!
Bm (x) .
(2.29)
l=m
Acknowledgement
This paper is supported by grant No. 14-11-00022 of Russian Scientific fund.
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