IJMC: INTERNATIONAL JOURNAL OF MATHEMATICAL COMBINATORICS (2010), Vol 2 #A07-14
arXiv:1212.3989v1 [math.NT] 17 Dec 2012
GENERALIZATIONS OF POLY-BERNOULLI NUMBERS AND POLYNOMIALS
Hassan Jolany and M.R.Darafsheh, R.Eizadi Alikelaye School of Mathematics,college of science,university of Tehran, Tehran, Iran
[email protected]
Received April 12, 2010: , Accepted: May 28, 2010 Keywords:Poly-Bernoulli polynomials,Euler polynomials
,generalized Euler polynomials,generalized poly-Bernoulli polynomials
Abstract (k)
The Concepts of poly-Bernoulli numbers Bn , poly-Bernoulli polynomials Bnk (t) and the (k) (k) generalized poly-bernoulli numbers Bn (a, b) are generalized to Bn (t, a, b, c) which is called the generalized poly-Bernoulli polynomials depending on real parameters a,b,c .Some prop(k) (k) erties of these polynomials and some relationships between Bnk , Bn (t) , Bn (a, b) and (k) Bn (t, a, b, c) are established.
1. Introduction In this paper we shall develop a number of generalizations of the poly-Bernoulli numbers and polynomials , and obtain some results about these generalizations.They are fundamental objects in the theory of special functions. Euler numbers are denoted with Bk and are the coefficients of Taylor expansion of the t function et −1 ,as following; ∞ X t tk B = . k et − 1 k=0 k! The Euler polynomials En (x) are expressed in the following series ∞
X tk 2ext = E (x) . k et + 1 k=0 k! for more details, see [1]-[4].
2 In [10], Q.M.Luo, F.Oi and L.Debnath defined the generalization of Euler polynomials Ek (x, a, b, c) which are expressed in the following series : ∞
X 2cxt tk = Ek (x, a, b, c) . bt + at k! k=0 where a, b, c ∈ Z + .They proved that I) for a = 1 and b = c = e k X k Ej (x), Ek (x + 1) = j
(1)
Ek (x + 1) + Ek (x) = 2xk ,
(2)
Ek (x + 1, 1, b, b) + Ek (x, 1, b, b) = 2xk (ln b)k .
(3)
j=0
and
II) for a = 1 and b = c ,
In[5], Kaneko introduced and studied poly-Bernoulli numbers which generalize the clas(k) sical Bernoulli numbers. Poly-Bernoulli numbers Bn with k ∈ Z and n ∈ N , appear in the following power series: ∞
Lik (1 − e−x ) X (k) tn = Bn , 1 − e−x n! n=0
(∗)
where k ∈ Z and ∞ X zm Lik (z) = . mk m=1
|z| < 1
so for k ≤ 1, Li1 (z) = − ln(1 − z), Li0 (z) =
z z , Li−1 = , ... . 1−z (1 − z)2
Moreover when k ≥ 1, the left hand side of (∗) can be wrriten in the form of ”interated integrals”
3
1 e t e −1 t
Z
0
t
1 t e −1
Z
=
t
0
1 ... t e −1
∞ X
Bn(k)
n=0
Z
t
0
et
t dtdt...dt −1
tn . n!
(1)
In the special case, one can see Bn = Bn . (k)
Definition 1 The poly-Bernoulli polynomials,Bn (t), are appeared in the expansion of as following, ∞ (k) Lik (1 − e−x ) xt X Bn (t) n e = x 1 − e−x n! n=0
Lik (1−e−x ) xt e 1−e−x
(4)
for more details, see [6] − [11].
Proposition 1 (Kaneko theorem[6]) The Poly-Bernoulli numbers of non-negative index k ,satisfy the following
Bn(k)
= (−1)
n
n+1 (−1) X
m−1
(m − 1)! mk
m=1
n m−1
(5)
,
and for negative index k, we have min(n,k)
Bn(−k)
=
X
(j!)
j=0
where
n m
2
n+1 j+1
k+1 j+1
(6)
,
m (−1)m X m l ln m, n ≥ 0 (−1) = l m! l=0
(7)
(k)
Definition 2 Let a, b > 0 and a 6= b. The generalized poly-Bernoulli numbers Bn (a, b), (k) (k) the generalized poly-Bernoulli polynomials Bn (t, a, b) and the polynomial Bn (t, a, b, c) are appeared in the following series respectively; ∞
(k)
Lik (1 − (ab)−t ) X Bn (a, b) n 2π = t |t| < , bt − a−t n! | ln a + ln b| n=0
(8)
4 ∞
(k)
2π Lik (1 − (ab)−t ) xt X Bn (x, a, b) n e = t |t| < , t −t b −a n! | ln a + ln b| n=0 ∞
(9)
(k)
Lik (1 − (ab)−t ) xt X Bn (x, a, b, c) n 2π c = t |t| < , t −t b −a n! | ln a + ln b| n=0
(10)
2. The main theorems We present some recurrence formulae for generalized poly-Bernoulli polynomials. Theorem 1 Let x ∈ R and n ≥ 0. For every positive real numbers a,b and c such that a 6= b and b > a , we have − ln b (k) (k) Bn (a, b) = Bn (ln a + ln b)n , (11) ln a + ln b j X j (k) (k) j−i i j−i Bj , (12) (−1) (ln a + ln b) (ln b) Bj (a, b) = i i=1
Bn(k) (x; a, b, c)
n X n (k) (ln c)n−l Bl (a, b)xn−l , = l
(13)
l=0
b Bn(k) (x + 1; a, b, c) = Bn(k) (x; ac, , c), c
(14)
Bn(k) (t) = Bn(k) (et+1 , e−t ),
(15)
Bn(k) (x, a, b, c) = (ln a + ln b)n Bn(k) (
− ln b + x ln c ). ln a + ln b
(16)
Proof:By applying definition 2, we have (11) ∞
(k)
Lik (1 − (ab)−t ) X Bn (a, b) n = t bt − a−t n! n=0 1 = t b
Lik (1 − e−t ln ab ) 1 − e−t ln ab
−t ln b
=e
Lik (1 − e−t ln ab ) 1 − e−t(ln ab)
5
∞ X
=
Bn(k)
n=0
− ln b tn (ln a + ln b)n ln a + ln b n!
Therefore Bn(k) (a, b)
=
Bn(k)
− ln b (ln a + ln b)n . ln a + ln b
(12) 1 Lik (1 − (ab)−t ) = t t −t b −a b
Lik (1 − (ab)−t ) 1 − e−t ln ab
=
j ∞ X X j=0
∞ X (ln b)k
=
k!
k=0
(−1)k tk
!
(ln a + ln b)n n Bn(k) t n! n=0
!
i
(k) (ln a
∞ X
+ ln b) (ln b)j−i tj i!(j − i)!
(−1)j−i Bi
i=0
So we have (k) Bj (a, b)
=
j X
(−1)
j−i
i
(ln a + ln b) (ln b)
j−i
i=0
(13)
j i
∞
tn Lik (1 − (ab)−t ) xt X (k) c = Bn (x, a, b, c) bt − a−t n! n=0
=
∞ X l=0
=
tl (k) Bl (a, b)
l!
!
∞ X (ln c)i ti i=0
i!
i
x
!
∞ X l X (ln c)l−i (k) Bi (a, b)xl−i tl i!(l − i)! l=0 i=0
! ∞ n X X tn n (k) (ln c)n−l Bl (a, b)xn−l = . l n! n=0
(14)
l=0
(k)
Bi .
!
6
Lik (1 − (ab)−t ) (x+1)t Lik (1 − (ab)−t ) xt t c = c .c bt − a−t bt − a−t
∞
Lik (1 − (ab)−t ) xt X (k) b tn = c = B (x; ac, , c) . n b t −t c n! − (ac) n=0 c
(15)
Lik (1 − e−x ) xt Lik (1 − e−x ) Lik (1 − e−x ) e = = 1 − e−x e−xt − e−x−xt (e−t )x − (e1+t )−x so, Bn(k) (t) = Bn(k) (et+1 , e−t ). (16)we can write ∞ X
Bn(k) (x, a, b, c)
n=0
t(− ln b+x ln c)
=e
Lik (1 − (ab)−t ) xt 1 Lik (1 − (ab)−t ) xt tn = c = c n! bt − a−t bt (1 − (ab)−t )
Lik (1 − e−t ln ab ) 1 − e−t(ln ab)
=
∞ X
(ln a + ln b)
Bn(k)
n=0
so Bn(k) (x, a, b, c)
n
= (ln a + ln b)
n
Bn(k)
− ln b + x ln c ln a + ln b
− ln b + x ln c ln a + ln b
tn n!
.
Theorem 2 Let x ∈ R , n ≥ 0. For every positive real numbers a,b such that a 6= b and b > a > 0, we have
Bn(k) (x+y, a, b, c)
n ∞ X X n n (k) n−l (k) n−l (ln c)n−l Bl (y, a, b, c)xn−l. (ln c) Bl (x; a, b, c)y = = l l l=0
l=0
(17)
Proof: We have ∞
Lik (1 − (ab)−t ) (x+y)t X (k) tn c = B (x + y; a, b, c) n bt − a−t n! n=0
7
Lik (1 − (ab)−t ) xt yt = c .c = bt − a−t
∞ X
tn Bn(k) (x; a, b, c) n! n=0
!
∞ X y i (ln c)i i=0
i!
ti
!
! ∞ n X X tn n (k) y n−l(ln c)n−l Bl (x, a, b, c) = l n! n=0
so we get
l=0
Lik (1 − (ab)t ) (x+y)t Lik (1 − (ab)−t yt xt c = c c bt − a−t bt − a−t ! ∞ n X X tn n n−l n−l (k) . x (ln c) Bl (y, a, b, c) = l n! n=0
l=0
Theorem 3 Let x ∈ R and n ≥ 0.For every positive real numbers a,b and c such that a 6= b and b > a > 0, we have
Bn(k) (x; a, b, c)
n X − ln b n n−l (k) (ln a + ln b)l xn−l , (ln c) Bl = l ln a + ln b
(18)
l=0
Bn(k) (x; a, b, c)
=
n X l X
(−1)
l−j
l=0 j=0
n l
l j
(k)
(ln c)n−l (ln b)l−j (ln a + ln b)j Bj xn−k .
(19)
Proof: (18)By applying Theorems 1 and 2 we know ,
Bn(k) (x; a, b, c)
n X n (k) (ln c)n−l Bl (a, b)xn−l = l l=0
and Bn(k) (a, b) = Bn(k) (
−lnb )(ln a + ln b)n ln a + ln b
The relation (18)will follow if we combine these formulae. (19) proof is similar to(18). Now,we give some results about derivatives and integrals of the generalized poly-Bernoulli polynomials in the following theorem.
8 Theorem 4 Let x ∈ R.If a, b and c > 0 , a 6= b and b > a > 0 ,For any non-negative integer l and real numbers α and β we have (k)
∂ l Bn (x, a, b, c) n! (k) = (ln c)l Bn−l (x, a, b, c) l ∂x (n − l)! Z
β
Bn(k) (x, a, b, c)dx =
α
(20)
1 (k) (k) [Bn+1 (β, a, b, c) − Bn+1 (α, a, b, c)] (n + 1) ln c
(21)
Proof: By applying (20) and (21) can be proved by induction on n. In [9], GI-Sang Cheon investigated the classical relationship between Bernoulli and Euler polynomials, in this paper we study the relationship between the generalized poly-Bernoulli and Euler polynomials. Theorem 5 For b > 0 we have n 1X n (k1 ) [Bn(k1 ) (y, 1, b, b) + Bn(k1 ) (y + 1, 1, b, b)]En−k (x, 1, b, b). Bn (x + y, 1, b, b) = k 2 k=0 Proof:We know Bn(k1 ) (x
∞ X n (k ) (ln b)n−k Bk 1 (y; 1, b, b)xn−k + y, 1, b, b) = k k=0
and Ek (x + y, 1, b, b) + Ek (x, 1, b, b) = 2xk (ln b)k (k )
So, We obtain Bn 1 (x + y, 1, b, b) = n
1X 2 k=0
n k
n
1X = 2 k=0
(ln b)
n k
n−k
(k ) Bk 1 (y; 1, b, b)
(k )
Bk 1 (y; 1, b, b)
n
1X = 2 k=0
"
n k
1 (En−k (x, 1, b, b) + En−k (x + 1, 1, b, b)) (ln b)n−k
!# n−k X n−k Ej (x, 1, b, b) En−k (x, 1, b, b) + j j=0
(k )
Bk 1 (y; 1, b, b)En−k (x, 1, b, b)
9 n
1X + 2 j=0 =
1 2
1 2
n j
n−j X n−j (k ) Bk 1 (y, 1, b, b) Ej (x; 1, b, b) k k=0
n X k=0
n X j=0
n k
n j
n k
(k )
Bk 1 (y; 1, b, b)En−k (x, 1, b, b)+ (k )
1 Bn−j (y + 1; 1, b, b)Ej (x, 1, b, b)
So we have n
Bn(k1 ) (x
1X + y, 1, b, b) = 2 k=0
[Bn(k1 ) (y, 1, b, b) + Bn(k1 ) (y + 1, 1, b, b)]En−k (x, 1, b, b).
Corollary 1 In theorem 5, if we set k1 = 1 and b = e, we obtain Bn (x) =
n X
(k=0),(k6=1)
n k
Bk En−k (x).
For more detail see [7]. Acknowledgement:The authors would like to thank the referee for has valuable comments concerning theorem 5.
References [1] T.Arakawa and M.Kaneko:On poly-Bernoulli numbers,Comment.Math.univ.st.Pauli48, (1999),159-167 [2] B.N. Oue and F.Qi:Generalization of Bernoulli polynomials ,internat.j.math.ed.sci.tech 33, (2002),No,3,428-431 [3] M.-S.Kim and T.Kim:An explicit formula on the generalized Bernoulli number with order n ,Indian.j.pure and applied math 31, (2000),1455-1466 [4] Hassan Jolany and M.R.Darafsheh:Some another remarks on the generalization of Bernoulli and Euler polynomials,Scientia magna Vol5,NO 3 [5] M.Kaneko:poly-Bernoulli numbers ,journal de theorides numbers de bordeaux. 9, (1997),221-228
10 [6] Y.Hamahata,H.Masubuch:Special Multi-Poly- Bernoulli numbers ,Journal of Integer sequences vol 10, (2007) [7] H.M.Srivastava and A.Pinter:Remarks on some relationships Between the Bernoulli and Euler polynomials,Applied .Math.Letter 17, (2004)375-380 [8] Chad Brewbaker :A combinatorial Interpretion of the poly-Bernoulli numbers and two fermat analogues, Integers journal 8, (2008) [9] GI-Sang Cheon:A note on the Bernoulli and Euler polynomials ,Applied.Math.Letter 16, (2003),365-368 [10] Q.M.Luo,F.Oi and L.Debnath Generalization of Euler numbers and polynomials,Int.J.Math.Sci (2003)3893-3901 [11] Y.Hamahata,H.Masubuchi Recurrence bers,Integers journal 7(2007)
formulae
for
Multi-poly-Bernoulli
num-
[12] Hassan Jolany, Explicit formula for generalization of poly-Bernoulli numbers and polynomials with a,b,c parameters. http://arxiv.org/pdf/1109.1387v1.pdf
