Some identities for the product of two Bernoulli and Euler polynomials

Kim Dae, Kim Taekyun, Lee Sang-Hun, Kim, Kim Young-Hee

Description

Let â„™ n be the space of polynomials of degree less than or equal to n . In this article, using the Bernoulli basis { B 0 ( x ), . . . , B n ( x )} for â„™ n consisting of Bernoulli polynomials, we investigate some new and interesting identities and formulae for the product of two Bernoulli and Euler polynomials like Carlitz did. Let â„™ n be the space of polynomials of degree less than or equal to n . In this article, using the Bernoulli basis { B 0 ( x ), . . . , B n ( x )} for â„™ n consisting of Bernoulli polynomials, we investigate some new and interesting identities and formulae for the product of two Bernoulli and Euler polynomials like Carlitz did.

Informations

• Publié par: biomed
• Publié le: 01 janvier 2012
• Nombre de lectures: 2
• Langue: English

Extrait

Kim et al. Advances in Difference Equations 2012, 2012:95
http://www.advancesindifferenceequations.com/content/2012/1/95
RESEARCH Open Access
Some identities for the product of two Bernoulli
and Euler polynomials
1 2* 3 3Dae San Kim , Taekyun Kim , Sang-Hun Lee and Young-Hee Kim
* Correspondence: Abstract
[e-mail]
2Department of Mathematics, Let ℙ be the space of polynomials of degree less than or equal to n. In this article,n
Kwangwoon University, Seoul 139- using the Bernoulli basis {B (x),..., B (x)} for ℙ consisting of Bernoulli polynomials,0 n n701, Republic of Korea
we investigate some new and interesting identities and formulae for the product ofFull list of author information is
available at the end of the article two Bernoulli and Euler polynomials like Carlitz did.
1 Introduction
The Bernoulli and Euler polynomials are defined by means of
∞ ∞n n t t 2 txt xte = B (x) , e = E (x) . (1)n nt te −1 n! e +1 n!
n=0 n=0
In the special case, x=0, B (0) = B and E (0) = E are called the n-th Bernoulli andn n n n
Euler numbers (see [1-17]).
From (1), we note that
n n n nn−l n−lB (x)= Bx ,E (x)= Ex . (2)n l n ll l
l=0 l=0
For n ≥ 0, we have
d d
B (x)= nB (x), E (x)= nE (x), (3)n n−1 n n−1
dx dx
(see [7,8]).
By (1), we get the following recurrence for the Bernoulli and the Euler numbers:
B =1,B (1) −B = δ and E =1,E (1)+E =2δ , (4)0 n n 1,n 0 n n 0,n
where δ is the Kronecker symbol (see [1-17]).k, n
Thus, from (3) and (4), we have
1 1
δ 2E0,n n+1
B (x)dx = E (x)dx = − (5), .n n
n+1 n+1
0 0
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provided the original work is properly cited.Kim et al. Advances in Difference Equations 2012, 2012:95 Page 2 of 14
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It is known [12] that
A 1
x x 1−m1 1−mnB ...B dx = a ...a B (x)...B (x)dx, (6)m m m m1 n n 1 n1a a1 n
0 0
where a , a,..., a are positive integers that are relatively prime in pairs A = a a ... a .1 2 n 1 2 n
For n = 2, there is the formula
1
Bp+qp+1B (x)B (x)dx=(−1) ,p q (7)p+q
0
q
where p + q ≥ 2 (see [3,4]). In [3,4], we can find the following formula for a product
of two Bernoulli polynomials:
B B (x) Bm n 2r m+n−2r m+nm+1 B (x)B (x)= n+ m +(−1) ,form+n ≥ 2.m n 2r 2r m+n −2r m+n (8)
r
n
Assume m, n, p ≥ 1. Then, by (7) and (8), we get
1 m n (m+n −2r −1)!p+1B (x)B (x)B (x)dx=(−1) p! n+ m B B , (9)m n p 2r m+n+p−2r
2r 2r (m+n+p −2r)!
r
0
(see [4]).
In [8], it is known that for nÎ ℤ ,+
n n
B (x)= B E (x)n k n−k (10)k
k=0
k=1
and
n En l+1
E (x)= −2 B (x).n n−l (11)
l l+1
l=0
i
Let ℙ ={∑ax|a Î ℚ} be the space of polynomials of degree less than or equal ton i i i
n. In this article, using the Bernoulli basis {B (x),..., B (x)} for ℙ consisting of Ber-0 n n
noulli polynomials, we investigate some new and interesting identities and formulae
for the product of two Bernoulli and Euler polynomials like Carlitz did.
2 Bernoulli identities arising from Bernoulli basis polynomials
From (1), we note that
∞t n1 t(e −1) 1 txt xte = e = (B (x+1) −B (x))n ntt e −1 t n!
n=0
∞ n1 t
= (B (x+1) −B (x)) (12)n n
t n!
n=1
∞ n B (x+1) −B (x) tn+1 n+1
= .
n+1 n!
n=0


Kim et al. Advances in Difference Equations 2012, 2012:95 Page 3 of 14
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Thus, from (12), we have
n1 1 n+1nx = (B (x+1) −B (x)) = B (x). (13)n+1 n+1 lln+1 n+1
l=0
From (13), we note that {B (x), B (x),..., B (x)} spans ℙ . For p(x) Î ℙ ,let0 1 n n n
np(x)= a B (x) and g(x)= p(x+1)-p(x). Then we havek kk=0
n n
k−1g(x)= a (B (x+1) −B (x)) = ka x .k k k k (14)
k=0 k=0
From (14), we can derive the following Equation (15):
n
(r) k−r−1g (x)= k(k −1)...(k −r)a x , (15)k
k=r+1
rd g(x)(r)where and r=0,1,2,..., n. Let us take x = 0 in (15). Then we haveg (x)= rdx
(r) (16)g (0) = (r+1)!a .r+1
By (16), we get, for r=1,2,..., n,
(r−1)g (0) 1 (r−1) (r−1) (17)a = (1) −p (0)).= (pr
r! r!
n
Let 0= p(x)= a B (x). Then, from (17), we havek kk=0
1 1(r−1) (r−1) (r−1)a = g (0) = (p (1) −p (0)) = 0. (18)r
r! r!
From (18), we note that {B (x), B (x),..., B (x)} is a linearly independent set.0 1 n
Therefore, we obtain the following theorem.
Proposition 1 The set of Bernoulli polynomials {B (x), B (x),..., B (x)} is a basis forℙ .0 1 n n
Let us consider polynomial p(x)Î ℙ as a linear combination of Bernoulli basis poly-n
nomials with
p(x)= C B (x)+C B (x)+···+C B (x). (19)0 0 1 1 n n
We can write (19) as a dot product of two variables:
⎛ ⎞
C0
⎜ ⎟C1⎜ ⎟
p(x)=(B (x),B (x),...,B (x))⎜ ⎟. (20)0 1 n ..⎝ ⎠.
Cn
From (20), we can derive the following equation:
⎛ ⎞
⎛ ⎞1 b b ··· b12 13 1n+1 C0⎜ ⎟01 b ··· b23 2n+1⎜ ⎟⎜ ⎟C1⎜ ⎟⎜ ⎟00 1 ··· b3n+1⎜ ⎟⎜ ⎟2 n C2p(x)=(1,x,x ,...,x )⎜ ⎟⎜ ⎟, (21). . . ... . . . .⎜ ⎟⎜ ⎟... . . . .⎜ ⎟⎝ ⎠.⎝ ⎠00 0 ··· bnn+1
Cn
00 0 ··· 1Kim et al. Advances in Difference Equations 2012, 2012:95 Page 4 of 14
http://www.advancesindifferenceequations.com/content/2012/1/95
where b are the coefficients of the power basis that are used to determine theij
respective Bernoulli polynomials. It is easy to show that
1 1 3 12 3 2B (x)=1, B (x)= x − , B (x)= x −x+ , B (x)= x − x + x, ....0 1 2 3
2 6 2 2
In the quadratic case (n = 2), the matrix representation is
⎛ ⎞⎛ ⎞1 11 − C02 6
2 ⎝ ⎠⎝ ⎠p(x)=(1, x, x ) 01 −1 C . (22)1
00 1 C2
In the cubic case (n = 3), the matrix representation is
⎛ ⎞⎛ ⎞1 11 − 0 C02 6
1⎜ ⎟⎜ ⎟01 −1 C2 3 1⎜ 2 ⎟⎜ ⎟p(x)=(1,x,x ,x ) . (23)3⎝ ⎠⎝ ⎠00 1 − C22
00 0 1 C3
In many applications of Bernoulli polynomials, a matrix formulation for the Bernoulli
polynomials seems to be useful.
There are many ways of obtaining polynomial identities in general. Here, in
Theorems 2-9, we use the Bernoulli basis in order to express certain polynomials as linear
combinations of that basis and hence to get some new and interesting polynomial
identities.
1Let . Then, by integration by parts, we getI = B (x)B (x)dxform,n ∈ Zm,n m n +0
Bm+nm+nI = I =0, I =(−1) ,(m, n ≥ 2).0,n m,0 m,n
m+n (24)
m
For nÎ ℤ with n ≥ 2, let us consider the following polynomials in ℙ :+ n
n
p(x)= B (x)B (x) ∈ P .k n−k n (25)
k=0
Then, from (25), we have
n(n+1)!(r)p (x)= B (x)B (x),k−r n−k (26)
(n −r+1)!
k=r
where r=0,1,2,... n.
By Proposition 1, we see that p(x) can be written as
n
p(x)= a B (x).k k (27)
k=0
From (25) and (27), we note that
1 n n−1 k−1 n (−1) (1+(−1) ) 2
a = p(t)dt = I = B = B = B .0 k,n−k n n n
n n+2 n+2
k=0 k=10 kKim et al. Advances in Difference Equations 2012, 2012:95 Page 5 of 14
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By (18) and (26), we get
1 (r) (r)a = (p (1) −p (0))r+1
(r+1)!
n(n+1)!
= (B (1)B (1) −B B )k−r n−k k−r n−k
(r+1)!(n −r+1)!
k=r
n1 n+r
= {(δ +B )(δ +B ) −B B }1,k−r k−r 1,n−k n−k k−r n−k (28)r+1n+2
k=r

1 n+2
= (B +B + δ )n−r−1 n−r−1 r,n−2
r+1n+2
⎧
⎨ n+22 B if r = n −2.n−r−1n+2= r+1
⎩
0if r = n −2.
Therefore, by (25), (27) and (28), we obtain the following theorem.
Theorem 2 For nÎ ℤ with n ≥ 2, we have+
n n−2 2 n+2
B (x)B (x)= B B (x)+(n+1)B (x).k n−k n−k k n
kn+2
k=0 k=0
For nÎ ℤ with n ≥ 2, let us take polynomial p(x)in ℙ as follows:+ n
n 1
p(x)= (x)B (x) ∈ P .Bk n−k n (29)
k!(n −k)!
k=0
From Proposition 1, we note that p(x) is given by means of Bernoulli basis
polynomials:
n
p(x)= a B (x) ∈ P .k k n (30)
k=0
By (24), (29) and (30), we get
1 n n−1 k−1 1 2I (−1)0,n
a = p(t)dt = I = + B0 k,n−k n
k!(n −k)! n! n
k=0 k=1 k!(n −k)!0
k (31)
n−1 nB B (1+(−1) ) Bn n nk−1
= (−