INTEGERS 13 (2013)
#A75
AN EXPLICIT FORMULA FOR HIGHER ORDER BERNOULLI
POLYNOMIALS OF THE SECOND KIND
Haiqing Wang
Dept. of Mathematics, Huizhou University, Huizhou, Guangdong, P. R. China
Guodong Liu1
Dept. of Mathematics, Huizhou University, Huizhou, Guangdong, P. R. China
gdliu@pub.huizhou.gd.cn
Received: 3/15/13, Accepted: 11/5/13, Published: 11/27/13
Abstract
In this paper, the authors establish a formula expressing the Bernoulli polynomials
(k)
of the second kind and general order k, bn (x), in terms of those of first order,
(1)
bn (x) = bn (x).
1. Introduction and Results
(k)
The Bernoulli polynomials bn (x) of the second kind and of order k, for any
integer k, may be defined by (see [2,4,10])
✓
t
log(1 + t)
(k)
◆k
(1 + t)x =
1
X
n
b(k)
n (x)t ,
n=0
|t| < 1.
(1)
(k)
The numbers bn = bn (0) are the Bernoulli numbers of the second kind and
(1)
(1)
of order k; bn = bn , bn (x) = bn (x) are the ordinary Bernoulli numbers and
polynomials of the second kind (see [1,4,5, 10-13]), and Cn = n!bn are the Cauchy
numbers of the first kind (see [8, 13]). By (1.1), we have
X
b(k)
bv1 (x/k)bv2 (x/k) · · · bvk (x/k),
(2)
n (x) =
v1 ,··· ,vk 2N0
v1 +···+vk =n
b(k)
n =
X
v1 ,··· ,vk 2N0
v1 +···+vk =n
1 Corresponding
author.
bv1 bv2 · · · bvk ,
(3)
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and
b(k)
n (x)
=
(k)
b0
✓ ◆
✓
◆
✓ ◆
x
x
x
(k)
(k)
+ b1
+ · · · + bn 1
+ b(k)
n .
n
n 1
1
(4)
where n 2 N0 = N [ {0}, N being the set of positive integers.
The numbers bn satisfy the recurrence relation (see [1, 5])
n
X
b0 = 1,
j=0
( 1)j
bj = 0 (n
n j+1
1),
(5)
1
1
19
3
863
so we find b1 = 12 , b2 = 12
, b3 = 24
, b4 = 720
, b5 = 160
, b6 = 60480
.
The numbers bn satisfy many interesting relations. For example (see [1,5,8])
bn =
Z
1
0
✓ ◆
x
dx,
n
1
X
|bn |
log 2 =
,
n
+1
n=1
1
X
|bn |Hn
⇡2
1+
=
,
n
6
n=1
(6)
Pn
is the Euler constant and Hn = k=1 k1 is the
1
1
X
|bn |
=
,
n
n=1
2)···(x n+1)
where nx = x(x 1)(x n!
,
n-th harmonic number.
(k)
The Bernoulli polynomials Bn (x) of order k, for any integer k, may be defined
by (see [3,4,6,7,10])
✓
t
et
1
(k)
◆k
ext =
1
X
Bn(k) (x)
n=0
tn
,
n!
|t| < 2⇡.
(7)
(k)
(1)
The numbers Bn = Bn (0) are the Bernoulli numbers of order k, and Bn = Bn
(n)
are the ordinary Bernoulli numbers. The numbers Bn are called Nörlund numbers
(see [3]), or Cauchy numbers of the second kind (see [8, 13]). Nörlund found the
exponential generating function (see [10, p.150])
1
X
t
tn
=
Bn(n)
(1 + t) log(1 + t) n=0
n!
(k)
(n)
These numbers bn , bn , Bn
(see [1-4])
n!bn =
Bn(n)
+
(n 1)
nBn 1 ,
Bn(n)
(k)
and Bn
(8)
satisfy various identities. For example
n
X
= n!
( 1)n
j=0
(|t| < 1).
j
bj ,
and n!b(k)
n =
k
k
n
Bn(n
k)
.
(9)
The paper’s central result is a formula expressing the Bernoulli polynomials of
(k)
the second kind and general order k, bn (x), in terms of those of first order, bn (x) =
(1)
bn (x). That is, we shall prove the following main conclusion.
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INTEGERS: 13 (2013)
Theorem. Let n, k 2 N and n
( 1)k
1
(k
k)!b(k)
n (x) =
1)!(n
1. Then
k
X1
(n
1
j)!
j=0
X
⇥
k
(n
j
1
x)v1 (n
j
v1 ,··· ,vk j 2N0
v1 +···+vk j =j
2
x)v2 · · · (n
k
x)vk j bn
j (x).
(10)
By taking x = 0 in Equation (10), we can deduce the following.
Corollary 1. Let n, k 2 N and n
( 1)k
=
1
k
X1
(k
(n
j=0
1)!(n
1
j)!
k
1. Then
j
1)v1 (n
k)!b(k)
n
X
v1 ,··· ,vk j 2N0
v1 +···+vk j =j
(n
j
2)v2 · · · (n
k)vk j bn
j.
(11)
Taking k = 2, 3, 4 in (10) and (11), we immediately deduce the following expressions for the first few higher order Bernoulli polynomials and numbers of the second
kind:
b(2)
n)bn (x) + (x + 2 n)bn 1 (x) (n 1);
n (x) = (1
1
b(3)
1)(n 2)bn (x)
n (x) = (n
2
1
1
+ (n 2)(2n 5 2x)bn 1 (x) + (n 3 x)2 bn 2 (x) (n 2);
2
2
1
1
b(4)
(n 1)(n 2)(n 3)bn (x)
(n 2)(n 3)(3n 9 3x)bn 1 (x)
n (x) =
6
6
1
(n 3) (n 3 x)2 + (n 3 x)(n 4 x) + (n 4 x)2 bn 2 (x)
6
1
(n 4 x)3 bn 3 (x) (n 3);
6
b(2)
n)bn + (2 n)bn 1 (n 1);
n = (1
1
1
1
b(3)
1)(n 2)bn + (n 2)(2n 5)bn 1 + (n 3)2 bn 2 (n 2);
n = (n
2
2
2
1
1
(4)
bn =
(n 1)(n 2)(n 3)bn
(n 2)(n 3)(3n 9)bn 1
6
6
1
1
(n 3) (n 3)2 + (n 3)(n 4) + (n 4)2 bn 2
(n 4)3 bn 3
6
6
(n 3).
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By (11), (3), and noting that Cn = n!bn , we obtain an explicit formula for the
sum involving Cauchy numbers of the first kind:
( 1)k
1
(k
1)!(n
X
k)!
Cv1 Cv2 · · · Cvk
v1 !v2 ! · · · vk !
v1 ,··· ,vk 2N0
v1 +···+vk =n
=
k
X1
X
(n
1)v1 (n
j
2)v2 · · · (n
j
j=0 v1 ,··· ,vk j 2N0
v1 +···+vk j =j
k)vk
j
Cn j
. (12)
n j
Corollary 2. Let n, k 2 N0 and m 2 N. Then
( 1)k k!
Z
m
0
X
⇥
b(k)
n (x)dx =
n+1
X✓
i=1
(n
v1
i
◆ k
m X (n
i j=0 (n
j) (n
i
i
i
j)!
k)!
1)v2 · · · (n
j
v1 ,··· ,vk+1 j 2N0
v1 +···+vk+1 j =j
k)vk+1 j bn+1
i
i j.
(13)
By taking m = 1 in (13), we can deduce the following:
k
( 1) k!(n
k
1)!
Z
0
⇥
X
(n
j
1
b(k)
n (x)dx
k
X
=
(n
1
j)!
j=0
1)v1 (n
2)v2 · · · (n
j
v1 ,··· ,vk+1 j 2N0
v1 +···+vk+1 j =j
k
1)vk+1 j bn
(0)
Taking k = 0, 1; m = 1, 2, 3 in (13) and noting that bn (x) =
Z
0
1
✓ ◆
x
dx = bn ,
n
Z
0
2
✓ ◆
x
dx = 2bn + bn
n
Z
1,
3
0
x
n
j.
(14)
, we have
✓ ◆
x
dx = 3bn + 3bn
n
1
+ bn
and
Z
1
bn (x)dx = (1
n)bn + (2
n)bn
1,
0
Z
2
bn (x)dx = 2(1
n)bn + 3(2
n)bn
1
+ (3
bn (x)dx = 3(1
n)bn + 6(2
n)bn
1
+ 4(3
n)bn
2,
0
Z
0
3
n)bn
2
+ (4
n)bn
3.
2.
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2. Proof of Theorem
In this section, we shall complete the proof of the theorem. First, the following
lemma (see [2,10]) is crucial to the proof of the theorem. To be more self-contained,
we present a simpler proof here.
Lemma. Let n, k 2 N0 and m 2 N. Then
b(k+1)
(x) =
n
n
k (k)
n
bn (x) +
k
k
1
x
k
(k)
bn
1 (x).
(15)
Proof. By (1), we have
1
X
(n
n=0
=
n k 1
k)b(k)
n (x)t
k
✓
1
log(1 + t)
◆k+1
d
=
dt
✓
(1 + t)x
◆k
1
(1 + t)x
log(1 + t)
✓
◆k
1
1
+x
(1 + t)x
log(1 + t)
1
,
(16)
i.e.,
1
X
(n
n=1
=
kt
(k)
k
✓
1)bn
1
log(1 + t)
n k 1
1 (x)t
◆k+1
(1 + t)x
+ xt
✓
(n
k
1
1
log(1 + t)
◆k
(1 + t)x
1
.
(17)
By (16) and (17), we have
1
X
(n
n=0
=
=
n k 1
k)b(k)
n (x)t
+
1
X
(k)
1)bn
n k 1
1 (x)t
n=1
✓
◆k+1
✓
◆k
1
1
x
k
(1 + t) + x
(1 + t)x
log(1 + t)
log(1 + t)
1
1
X
X
n k 1
n k
k
b(k+1)
(x)t
+
x
b(k)
.
n
n (x)t
n=0
Comparing the coefficient of tn
b(k+1)
(x) =
n
(18)
n=0
n
k 1
on both sides of (18), we get
k (k)
n
b (x) +
k n
k
1
k
x
(k)
bn
1 (x).
This proves the lemma.
Now we complete the proof of the theorem by using mathematical induction and
the method of coefficients (see [9]).
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INTEGERS: 13 (2013)
Proof of theorem. First note that (10) holds for k = 1, 2, by (15). Now suppose (10)
is true for some natural number k and all n k 1. By superposition of (15), we
have
( 1)k k!(n
= ( 1)k
=
k
X1
1
(n
j=0
1)!b(k+1)
(x)
n
k
(k
1)!(n
1
1
x)v1 (n
j
2
x)v2 · · · (n
k
x)vk
(k
(n
j
1
v1 ,··· ,vk j 2N0
v1 +···+vk j =j
+(n
k
1
x)
k
X1
(n
2
k
X1
(n
j=0
⇥
(n
⇥
2
x)v1 (n
j
3
x)v2 · · · (n
k
1
1
k
X
j=0
X
(n
k
1
x)
⇥
X
x)bn
1 (x)
j
bn
j (x)
x)vk
j
1
x)v1 (n
j
2
x)v2 · · · (n
k
x)vk
j
bn
1 j (x)
k
X
(n
1
j
bn
j (x)
j)!
j=1
X
(n
(k)
1
j)!
(n
j
1
x)v1 (n
j
2
v1 ,··· ,vk+1 j 2N0
v1 +···+vk+1 j =j 1
=
k
j
v1 ,··· ,vk j 2N0
v1 +···+vk j =j
+(n
1)!(n
j)!
v1 ,··· ,vk j 2N0
v1 +···+vk j =j
=
k
j=0
X
⇥
1)!(n
j)!
X
⇥
k
k)!b(k)
n (x) + ( 1)
1
x)v2 · · · (n
k
x)vk+1
1
j
bn
j (x)
j)!
(n
j
1
x)v1 (n
j
2
v1 ,··· ,vk+1 j 2N0
v1 +···+vk+1 j =j
x)v2 · · · (n
k
1
x)vk+1
j
bn
j (x),
which shows that (10) is also true for the natural number k +1. The theorem follows
by induction.
2
Now we complete the proof of Corollary 2.
Proof of Corollary 2. By (11), (4), and noting that
d (k)
dx bn (x)
(k 1)
1 (x)
= bn
(see [11]),
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INTEGERS: 13 (2013)
we have
k
( 1) k!
Z
m
b(k)
n (x)dx
0
=
n+1
X✓
i=1
⇥
⇣
(k+1)
= ( 1) k! bn+1 (m)
k
◆ k
m X (n i j)!
i j=0 (n i k)!
X
(n i j)v1 (n
v1 ,··· ,vk+1 j 2N0
v1 +···+vk+1 j =j
i
j
(k+1)
bn+1
⌘
k
= ( 1) k!
1)v2 · · · (n
n+1
X✓
i=1
i
◆
m (k+1)
b
i n+1 i
k)vk+1 j bn+1
i j.
This completes the proof.
Acknowledgement. The authors would like to thank the referee for his/her very
useful suggestions. This work was supported by the Guangdong Provincial Natural
Science Foundation (No. 8151601501000002).
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(n)
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(x)
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(x)
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