(1) The numbers b(k)n =b(k)n (0) are the Bernoulli numbers of the second kind and of order k

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 [email protected]

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 of the second kind and general order k, b^(k)n (x), in terms of those of first order, b_n(x) =b⁽¹⁾n (x).

1. Introduction and Results

The Bernoulli polynomials b^(k)n (x) of the second kind and of order k, for any integerk, may be defined by (see [2,4,10])

✓ t log(1 +t)

◆k

(1 +t)^x= X1 n=0

b^(k)_n (x)tⁿ, |t|<1. (1)

The numbers b^(k)n =b^(k)n (0) are the Bernoulli numbers of the second kind and of order k; b⁽¹⁾n = bn, b⁽¹⁾n (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

b^(k)_n (x) = X

v1,···,vk2N0

v1+···+vk=n

b_v1(x/k)b_v2(x/k)· · ·b_vk(x/k), (2)

b^(k)_n = X

v1,···,vk2N0

v1+···+vk=n

bv1bv2· · ·bvk, (3)

1Corresponding author.

and

b^(k)_n (x) =b^(k)₀

✓x n

◆ +b^(k)₁

✓ x n 1

◆

+· · ·+b^(k)_{n 1}

✓x 1

◆

+b^(k)_n . (4) wheren2N0=N[{0},Nbeing the set of positive integers.

The numbersbn satisfy the recurrence relation (see [1, 5]) b0= 1,

Xn

j=0

( 1)^j

n j+ 1bj= 0 (n 1), (5)

so we findb₁=¹₂, b₂= ₁₂¹, b₃=₂₄¹, b₄= ₇₂₀¹⁹, b₅=₁₆₀³ , b₆= ₆₀₄₈₀⁸⁶³ .

The numbersb_n satisfy many interesting relations. For example (see [1,5,8]) bn=

Z ₁

0

✓x n

◆

dx, 1 log 2 = X1 n=1

|bn|

n+ 1, = X1 n=1

|bn| n , 1 +

X1 n=1

|bn|Hn

n =⇡² 6,

(6) where ^x_n =^{x(x 1)(x} 2)···(x n+1)

n! , is the Euler constant andHn=P_n

k=1 1 k is the n-th harmonic number.

The Bernoulli polynomialsBn^(k)(x) of orderk, for any integerk, may be defined by (see [3,4,6,7,10])

✓ t e^t 1

◆k

e^xt= X1 n=0

B_n^(k)(x)tⁿ

n!, |t|<2⇡. (7) The numbersBn^(k)=Bn^(k)(0) are the Bernoulli numbers of orderk, andBn⁽¹⁾=B_n are the ordinary Bernoulli numbers. The numbersBn⁽ⁿ⁾are called N¨orlund numbers (see [3]), or Cauchy numbers of the second kind (see [8, 13]). N¨orlund found the exponential generating function (see [10, p.150])

t

(1 +t) log(1 +t) = X1 n=0

B_n⁽ⁿ⁾tⁿ

n! (|t|<1). (8) These numbers b_n, b^(k)n , B⁽ⁿ⁾n and B^(k)n satisfy various identities. For example (see [1-4])

n!b_n=B_n⁽ⁿ⁾+nB_n⁽ⁿ₁¹⁾, B⁽ⁿ⁾_n =n!

Xn j=0

( 1)^{n j}b_j, and n!b^(k)_n = k

k nB^{(n k)}_n . (9) The paper’s central result is a formula expressing the Bernoulli polynomials of the second kind and general orderk,b^(k)_n (x), in terms of those of first order,b_n(x) = b⁽¹⁾n (x). That is, we shall prove the following main conclusion.

Theorem. Let n, k2Nandn k 1. Then ( 1)^{k 1}(k 1)!(n k)!b^(k)_n (x) =

k 1

X

j=0

(n 1 j)!

⇥ X

v1,···,vk j2N0

v1+···+vk j=j

(n j 1 x)^v1(n j 2 x)^v2· · ·(n k x)^{vk j}b_{n j}(x). (10)

By takingx= 0 in Equation (10), we can deduce the following.

Corollary 1. Letn, k2Nandn k 1. Then ( 1)^{k 1}(k 1)!(n k)!b^(k)_n

=

k 1

X

j=0

(n 1 j)! X

v1,···,vk j2N0

v1+···+vk j=j

(n j 1)^v1(n j 2)^v2· · ·(n k)^{vk j}b_{n j}. (11)

Takingk= 2,3,4 in (10) and (11), we immediately deduce the following expres- sions for the first few higher order Bernoulli polynomials and numbers of the second kind:

b⁽²⁾_n (x) = (1 n)b_n(x) + (x+ 2 n)b_{n 1}(x) (n 1);

b⁽³⁾_n (x) = 1

2(n 1)(n 2)bn(x) +1

2(n 2)(2n 5 2x)bn 1(x) +1

2(n 3 x)²bn 2(x) (n 2);

b⁽⁴⁾_n (x) = 1

6(n 1)(n 2)(n 3)bn(x) 1

6(n 2)(n 3)(3n 9 3x)bn 1(x) 1

6(n 3) (n 3 x)²+ (n 3 x)(n 4 x) + (n 4 x)² bn 2(x) 1

6(n 4 x)³bn 3(x) (n 3);

b⁽²⁾_n = (1 n)b_n+ (2 n)b_{n 1} (n 1);

b⁽³⁾_n =1

2(n 1)(n 2)bn+1

2(n 2)(2n 5)bn 1+1

2(n 3)²bn 2 (n 2);

b⁽⁴⁾_n = 1

6(n 1)(n 2)(n 3)bn 1

6(n 2)(n 3)(3n 9)bn 1

1

6(n 3) (n 3)²+ (n 3)(n 4) + (n 4)² bn 2 1

6(n 4)³bn 3

(n 3).

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 k)! X

v1,···,vk2N0

v1+···+vk=n

Cv1Cv2· · ·Cvk

v₁!v₂!· · ·v_k!

=

k 1

X

j=0

X

v1,···,vk j2N0

v1+···+vk j=j

(n j 1)^v1(n j 2)^v2· · ·(n k)^{vk j}Cn j

n j. (12)

Corollary 2. Letn, k2N0 andm2N. Then ( 1)^kk!

Z m 0

b^(k)_n (x)dx=

n+1X

i=1

✓m i

◆Xk j=0

(n i j)!

(n i k)!

⇥ X

v1,···,vk+1 j2N0

v1+···+vk+1 j=j

(n i j)^v1(n i j 1)^v2· · ·(n i k)^{vk+1 j}bn+1 i j. (13)

By takingm= 1 in (13), we can deduce the following:

( 1)^kk!(n k 1)!

Z 1 0

b^(k)_n (x)dx= Xk j=0

(n 1 j)!

⇥ X

v1,···,vk+1 j2N0

v1+···+vk+1 j=j

(n j 1)^v1(n j 2)^v2· · ·(n k 1)^{vk+1 j}b_{n j}. (14)

Takingk= 0,1;m= 1,2,3 in (13) and noting thatb⁽⁰⁾n (x) = ^x_n , we have Z 1

0

✓x n

◆

dx=b_n, Z 2

0

✓x n

◆

dx= 2b_n+b_{n 1}, Z 3

0

✓x n

◆

dx= 3b_n+ 3b_{n 1}+b_{n 2}. and

Z 1 0

b_n(x)dx= (1 n)b_n+ (2 n)b_{n 1}, Z 2

0

b_n(x)dx= 2(1 n)b_n+ 3(2 n)b_{n 1}+ (3 n)b_{n 2}, Z 3

0

b_n(x)dx= 3(1 n)b_n+ 6(2 n)b_{n 1}+ 4(3 n)b_{n 2}+ (4 n)b_{n 3}.

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, k2N0 andm2N. Then b^(k+1)_n (x) = n k

k b^(k)_n (x) +n k 1 x

k b^(k)_{n 1}(x). (15)

Proof. By (1), we have X1 n=0

(n k)b^(k)_n (x)t^{n k 1}= d dt

✓ 1 log(1 +t)

◆k

(1 +t)^x

= k

✓ 1 log(1 +t)

◆k+1

(1 +t)^{x 1}+x

✓ 1 log(1 +t)

◆k

(1 +t)^{x 1}, (16) i.e.,

X1 n=1

(n k 1)b^(k)_{n 1}(x)t^{n k 1}

= kt

✓ 1 log(1 +t)

◆k+1

(1 +t)^{x 1}+xt

✓ 1 log(1 +t)

◆k

(1 +t)^{x 1}. (17) By (16) and (17), we have

X1 n=0

(n k)b^(k)_n (x)t^{n k 1}+ X1 n=1

(n k 1)b^(k)_{n 1}(x)t^{n k 1}

= k

✓ 1 log(1 +t)

◆k+1

(1 +t)^x+x

✓ 1 log(1 +t)

◆k

(1 +t)^x

= k

X1 n=0

b^(k+1)_n (x)t^{n k 1}+x X1 n=0

b^(k)_n (x)t^{n k}. (18)

Comparing the coefficient oft^{n k 1} on both sides of (18), we get b^(k+1)_n (x) = n k

k b^(k)_n (x) +n k 1 x

k b^(k)_{n 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]).

Proof of theorem. First note that (10) holds fork= 1,2, by (15). Now suppose (10) is true for some natural numberkand all n k 1. By superposition of (15), we have

( 1)^kk!(n k 1)!b^(k+1)n (x)

= ( 1)^{k 1}(k 1)!(n k)!b^(k)n (x) + ( 1)^{k 1}(k 1)!(n k 1)!(n k 1 x)b^(k)_{n 1}(x)

=

k 1

X

j=0

(n 1 j)!

⇥ X

v₁,···,v_{k j}2N0 v₁+···+v_{k j}=j

(n j 1 x)^v1(n j 2 x)^v2· · ·(n k x)^{vk j}bn j(x)

+(n k 1 x)

k 1

X

j=0

(n 2 j)!

⇥ X

v1,···,v_{k j}2N0 v₁+···+vk j=j

(n j 2 x)^v1(n j 3 x)^v2· · ·(n k 1 x)^{vk j}bn 1 j(x)

=

k 1

X

j=0

(n 1 j)!

⇥ X

v₁,···,v_{k j}2N0 v₁+···+v_{k j}=j

(n j 1 x)^v1(n j 2 x)^v2· · ·(n k x)^{vk j}bn j(x)

+(n k 1 x)

Xk

j=1

(n 1 j)!

⇥ X

v₁,···,v_{k+1 j}2N0 v₁+···+vk+1 j=j 1

(n j 1 x)^v1(n j 2 x)^v2· · ·(n k 1 x)^{vk+1 j}bn j(x)

= Xk j=0

(n 1 j)!

⇥ X

v₁,···,v_{k+1 j}2N0 v₁+···+v_{k+1 j}=j

(n j 1 x)^v1(n j 2 x)^v2· · ·(n k 1 x)^{vk+1 j}bn j(x),

which shows that (10) is also true for the natural numberk+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 _dx^d b^(k)n (x) =b^(k_{n 1}¹⁾(x) (see [11]),

we have ( 1)^kk!

Z m 0

b^(k)_n (x)dx= ( 1)^kk!⇣

b^(k+1)_n+1 (m) b^(k+1)_n+1 ⌘

= ( 1)^kk!

n+1X

i=1

✓m i

◆ b^(k+1)_{n+1 i}

=

n+1X

i=1

✓m i

◆Xk j=0

(n i j)!

(n i k)!

⇥ X

v1,···,vk+1 j2N0

v1+···+vk+1 j=j

(n i j)^v1(n i j 1)^v2· · ·(n i k)^{vk+1 j}b_{n+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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