A Note on Poly-Bernoulli Polynomials Arising from Umbral Calculus

Adv. Studies Theor. Phys., Vol. 7, 2013, no. 15, 731 - 744 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/10.12988/astp.2013.3667

A Note on Poly-Bernoulli Polynomials Arising

from Umbral Calculus

Dae San Kim

Department of Mathematics, Sogang University Seoul 121-742, Republic of Korea [email protected]

Taekyun Kim

Department of Mathematics, Kwangwoon University Seoul 139-701, Republic of Korea [email protected]

Sang-Hun Lee

Division of General Education Kwangwoon University Seoul 139-701, Republic of Korea [email protected]

Copyright c 2013 Dae San Kim et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

In this paper, we give some recurrence formula and new and interesting 732 Dae San Kim, Taekyun Kim and Sang-Hun Lee

identities for the poly-Bernoulli numbers and polynomials which are derived from umbral calculus.

1 Introduction

The classical polylogarithmic function Lis(x) are

∞ k x ∈ Lis(x)= s ,s Z, (see [3, 5]). (1) k=1 k

In [5], poly-Bernoulli polynomials are deﬁned by the generating function to be

−t ∞ n Lik (1 − e ) (k) t ext = eB (x)t = B(k)(x) , (see [3, 5]), (2) − −t n 1 e n=0 n! (k) n (k) with the usual convention about replacing B (x) by Bn (x). As is well known, the Bernoulli polynomials of order r are deﬁned by the generating function to be

r ∞ t tn ext = B(r)(x) , (see [7, 9]). (3) t − n e 1 n=0 n!

(r) In the special case, r =1,Bn (x)=Bn(x) is called the n-th ordinary Bernoulli (r) polynomial. Here we denote higher-order Bernoulli polynomials as Bn to avoid conﬂict of notations. (k) (k) If x = 0, then Bn (0) = Bn is called the n-th poly-Bernoulli number. From (2), we note that n n (k) n (k) l n (k) n−l Bn (x)= Bn−lx = Bl x . (4) l=0 l l=0 l

Let F be the set of all formal power series in the variable t over C as follows: ∞ tk F ∈ = f(t)= ak ak C , (5) k=0 k! and let P = C[x] and P∗ denote the vector space of all linear functionals on P. L|p(x) denotes the acition of linear functional L on the polynomial p(x), and it is well known that the vector space oprations on P∗ are deﬁned Poly-Bernoulli polynomials arising from umbral calculus 733 by L + M|p(x) = L|p(x) + M|p(x), cL|p(x) = c L|p(x), where c is a complex constant (see [6, 9]). n For f(t) ∈F, let f(t) x = an. Then, by (5), we easily get k n t x = n!δn,k, (n, k ≥ 0), (see [1, 4, 6, 9, 10]), (6) where δn,k is the Kronecker’s symbol. ∞ k tk Let us assume that fL(t)= k=0 L x k! . Then, by (6), we see that n n fL(t) x = L x . That is, fL(t)=L. Additionally, the map L −→ fL(t)is a vector space isomorphism from P∗ onto F. Henceforth, F denotes both the algebra of the formal power series in t and the vector space of all linear functionals on P, and so an element f(t)ofF will be thought as a formal power series and a linear functional. F is called the umbral algebra. The umbral calculus is the study of umbral algebra. The order O (f(t)) of the power series f(t) = 0 is the smallest integer for which ak does not vanish. If O(f(t)) = 0, then f(t) is called an invertible series. If O(f(t)) = 1, then f(t) is called a delta series. For f(t),g(t) ∈F, we have

f(t)g(t)|p(x) = f(t)|g(t)p(x) = g(t)|f(t)p(x) . (7)

Let f(t) ∈F and p(x) ∈ P. Then we have ∞ ∞ tk xk f(t)= f(t)xk ,p(x)= tkp(x) , (see [6, 9]). (8) k=0 k! k=0 k! From (8), we can easily derive k ∞ l d p(x) t p(x) p(k)(x)= = xl−k. (9) k − dx l=k (l k)! Thus, by (8) and (9), we get p(k)(0) = tkp(x) = 1p(k)(x) . (10)

Hence, from (10), we have dkp(x) tkp(x)=p(k)(x)= , (see [1, 6, 9]). (11) dxk It is easy to show that eytp(x)=p(x + y), eytp(x) = p(y). (12) 734 Dae San Kim, Taekyun Kim and Sang-Hun Lee

Let O(f(t)) = 1 and O(g(t )) = 0. Then there exists a unique sequence sn(x) k of polynomials such that g(t)f(t) sn(x) = n!δn,k, for n, k ≥ 0. The sequence sn(x) is called a Sheﬀer sequence for (g(t),f(t)) which is denoted by sn(x) ∼ (g(t),f(t)). The Sheﬀer sequence sn(x) for (g(t),t) is called the Appell sequence for g(t). For p(x) ∈ P, f(t) ∈F, we have

f(t)|xp(x) = ∂tf(t)|p(x) = f (t)|p(x) , (see [9, 10]). (13)

Let sn(x) ∼ (g(t),f(t)). Then the following equations are known: ∞ ∞ k h(t) sk(x) k g(t)f(t) p(x) h(t)= g(t)f(t) ,p(x)= sk(x), (14) k=0 k! k=0 k! where h(t) ∈F, p(x) ∈ P,

∞ k 1 yf¯(t) t ∈ ¯ e = sk(y) , for all y C, (15) g f(t) k=0 k! where f¯(t) is the compositional inverse for f(t) with f¯(f(t)) = t, and

f(t)sn(x)=nsn−1(x). (16)

As is well known, the Stirling numbers of the second kind are also deﬁned by the generating function to be

∞ l ∞ t m t m! l+m e − 1 = m! S2(l, m) = S2(l + m, m)t . (17) l=m l! l=0 (l + m)!

Let sn(x) ∼ (g(t),t). Then the Appell identity is given by n n n n−k n k sn(x + y)= sk(y)x = sn−k(y)x , (see [6, 9]), (18) k=0 k k=0 k and g(t) sn+1(x)= x − sn(x), (see [6.9]). (19) g(t)

For sn(x) ∼ (g(t),f(t)), rn(x) ∼ (h(t),l(t)), we have n sn(x)= rm(x)cn,m, (20) m=0 Poly-Bernoulli polynomials arising from umbral calculus 735 where ¯ 1 h f(t) m n cn,m = l f¯(t) x , (see [9]). (21) m! g f¯(t)

The equations (20) and (21) are important in deriving our main results of this paper. In this paper, we give some recurrence formula and new and interesting identities for the poly-Bernoulli numbers and polynomials which are derive from umbral calculus.

2 Poly-Bernoulli numbers and polynomials

1−e−t Let gk(t)= −t . Then, by (2) and (15), we get Lik(1−e )

(k) Bn (x) ∼ (gk(t),t) . (22)

(k) That is, poly-Bernoulli polynomial Bn (x) is an Appell sequence. By (1), we easily get

d 1 (k) d (k) (k) Lik(x)= Lik−1(x),tB(x)= B (x)=nB (x). (23) dx x n dx n n−1

From (2) and (15), we have

−t (k) 1 n Lik (1 − e ) n Bn (x)= x = −t x . (24) gk(t) 1 − e

Let k ∈ Z and n ≥ 0. Then we have

−t ∞ −t m−1 Lik (1 − e ) (1 − e ) B(k)(x)= xn = xn (25) n 1 − e−t mk m=1 ∞ ∞ m 1 − −t m n 1 − j m −jt n = k 1 e x = k ( 1) e x m=0 (m +1) m=0 (m +1) j=0 j n m 1 − j m − n = k ( 1) (x j) . m=0 (m +1) j=0 j 736 Dae San Kim, Taekyun Kim and Sang-Hun Lee

By (17) and (25), we get

n ∞ (k) 1 − a m! a+m n Bn (x)= k ( 1) S2(a + m, m)t x (26) m=0 (m +1) a=0 (a + m)! n n−m 1 − a m! n−a−m = k ( 1) S2(a + m, m)(n)a+mx m=0 (m +1) a=0 (a + m)! n n−l − n−m−l ( 1) n − l = k m!S2(n l, m) x , l=0 m=0 (m +1) l where (a)n = a(a − 1)(a − 2) ···(a − n + 1). Now, we use the well-known transfer formula for Appell sequences (see equation (19)). By (19) and (22), we get (k) gk(t) (k) Bn+1(x)= x − Bn (x), (27) gk(t) where gk(t) −t −t = (log gk(t)) = log 1 − e − log Lik 1 − e (28) gk(t) −t − −t e − Lik−1 (1 e ) = − −t 1 − −t 1 e Lik (1 e ) −t −t 1 Lik (1 − e ) − Lik−1 (1 − e ) = t −t . e − 1 Lik (1 − e )

From (27) and (28), we have

(k) (k) gk(t) (k) Bn+1(x)=xBn (x) − Bn (x) (29) gk(t) −t −t t Lik (1 − e ) − Lik−1 (1 − e ) = xB(k)(x) − xn. n et − 1 t (1 − e−t)

Here, we note that −t −t ∞ Lik (1 − e ) − Lik−1 (1 − e ) 1 1 m−1 = − 1 − e−t (30) 1 − e−t mk mk−1 m=2 1 1 = − t + ··· 2k 2k−1 Poly-Bernoulli polynomials arising from umbral calculus 737 is a delta series. For any delta series f(t), we observe that

f(t) xn+1 xn = f(t) . (31) t n +1 By (29), (30) and (31), we get −t −t (k) t 1 Lik (1 − e ) − Lik−1 (1 − e ) B (x)=xB(k)(x) − xn+1 n+1 n et − 1 n +1 1 − e−t ∞ (k) 1 Bl l (k) (k−1) = xBn (x) − t Bn+1(x) − Bn+1 (x) n +1 l=0 l! n+1 (k) 1 n +1 (k) (k−1) = xBn (x) − Bl Bn+1−l(x) − Bn+1−l(x) . (32) n +1 l=0 l

Therefore, by (32), we obtain the following theorem.

Theorem 1. For k ∈ Z, n ≥ 0, we have

n+1 (k) (k) 1 n +1 (k) (k−1) Bn+1(x)=xBn (x) − Bl Bn+1−l(x) − Bn+1−l(x) , n +1 l=0 l where Bn is the n-th ordinary Bernoulli number.

It is easy to show that n n (k) n (k) l+1 n (k) l txBn (x)=t Bn−lx = Bn−l(l +1)x (33) l=0 l l=0 l n−1 n n − 1 (k) l n (k) n = nx Bn−1−lx + Bn−lx l=0 l l=0 l (k) (k) = nxBn−1(x)+Bn (x).

Applying to t on both sides of Theorem 1, by (33), we get n (k) (k) (k) n (k) (k−1) (n +1)Bn (x)=nxBn−1(x)+Bn (x) − Bn−l Bl (x) − Bl (x) . l=0 l (34)

Therefore, by (34), we obtain the following corollary. 738 Dae San Kim, Taekyun Kim and Sang-Hun Lee

Corollary 2. For k ∈ Z and n ≥ 1, we have n−2 (k) 1 (k) n (k) (n +1)Bn (x) − n x + Bn−1(x)+ Bn−lBl (x) 2 l=0 l n n (k−1) = Bn−lBl (x). l=0 l From (2) and (6), we note that −t −t Lik (1 − e ) Lik (1 − e ) B(k)(y)= eytxn = eytxxn−1 (35) n 1 − e−t 1 − e−t −t Lik (1 − e ) yt n−1 = ∂t e x 1 − e−t −t −t Lik (1 − e ) yt n−1 Lik (1 − e ) yt n−1 = ∂t e x + y e x 1 − e−t 1 − e−t −t −t Lik−1 (1 − e ) − Lik (1 − e ) (y−1)t n−1 (k) = 2 e x + yBn−1(y). (1 − e−t) Now, we observe that −t −t ∞ −t m −t m Lik−1 (1 − e ) − Lik (1 − e ) 1 (1 − e ) (1 − e ) = − − −t 2 − −t 2 k−1 k (1 e ) (1 e ) m=1 m m (36) ∞ 1 1 m−2 = − 1 − e−t mk−1 mk m=2 ∞ 1 − 1 − −t m = k−1 k 1 e . m=0 (m +2) (m +2) Thus, by (36), we get −t −t Lik−1 (1 − e ) − Lik (1 − e ) (y−1)t n−1 2 e x (37) (1 − e−t) ∞ 1 − 1 − −t m (y−1)t n−1 = k−1 k 1 e e x m=0 (m +2) (m +2) n−1 1 − 1 − −t m − n−1 = k−1 k 1 e (x + y 1) m=0 (m +2) (m +2) n−1 n−1 − 1 − 1 n 1 − n−1−a − −t m a = k−1 k (y 1) 1 e x . m=0 (m +2) (m +2) a=0 a Poly-Bernoulli polynomials arising from umbral calculus 739

From (6) and (7), we have −t m a a+m 1 − e x =(−1) m!S2(a, m). (38)

From (37) and (38), wehave −t −t Lik−1 (1 − e ) − Lik−1 (1 − e ) (y−1)t n−1 2 e x (39) (1 − e−t) n−1 n−1 − − a+m n 1 1 − 1 = ( 1) m! k−1 k m=0 a=0 a (m +2) (m +2) n−1−a × S2(a, m)(y − 1) n−1 n−1 − − n−1−l+m n 1 1 − 1 = ( 1) m! k−1 k m=0 l=0 l (m +2) (m +2) l × S2(n − 1 − l, m)(y − 1) .

Therefore, by (35) and (39), we obtain the following theorem.

Theorem 3. For k ∈ Z and n ≥ 1, we have n−1 (k) n − 1 B(k)(x)=xB (x)+ (−1)n−1−l n n−1 l l=0 n−1 × − m (m + 1)! − − − l ( 1) k S2(n 1 l, m) (x 1) . m=0 (m +2) −t n+1 Now, we try to compute Lik (1 − e ) x in two ways. On the one hand, − −t −t n+1 −t Lik (1 e ) n+1 Lik 1 − e x = 1 − e x (40) 1 − e−t −t Lik (1 − e ) = 1 − e−t xn+1 1 − e−t −t Lik (1 − e ) = xn+1 − (x − 1)n+1 1 − e−t n −t n +1 Lik (1 − e ) = (−1)n−m 1 xm − −t m=0 m 1 e n n +1 n−m (k) = (−1) Bm . m=0 m 740 Dae San Kim, Taekyun Kim and Sang-Hun Lee

On the other hand, t −t n+1 −s n+1 Lik 1 − e x = Lik 1 − e dsx (41) 0 t −s Lik−1 (1 − e ) −s n+1 = e −s ds x 0 1 − e t ∞ a ∞ (k−1) (−s) Bm = sm dsxn+1 0 a! m! a=0 m=0 ∞ l l 1 t − l−m (k−1) l n+1 = ( 1) Bm s dsx l=0 m=0 m l! 0 n l (k−1) n l l−m Bm n n−m (k−1) = (−1) (n + 1)!δn+1,l+1 = (−1) Bm . l=0 m=0 m (l + 1)! m=0 m By (40) and (41), we get n n n−m n (k−1) n +1 n−m (k) (−1) Bm = (−1) Bm . (42) m=0 m m=0 m By (2) and (3), we see that − −t t − r (k) ∼ 1 e B(r) ∼ e 1 ≥ Bn (x) −t ,t , n (x) ,t ,r 0. (43) Lik (1 − e ) t From (20), (21) and (43), we have n (k) (r) Bn (x)= Cn,mBm (x), (44) m=0 where r et−1 1 t m n Cn,m = 1−e−t t x (45) m! Li (1−e−t) k −t t r 1 Lik (1 − e ) e − 1 = tmxn m! 1 − e−t t −t t r n Lik (1 − e ) e − 1 = xn−m . m 1 − e−t t By (17), we easily get t r ∞ e − 1 r! l = S2(l + r, r)t . (46) t l=0 (l + r)! Poly-Bernoulli polynomials arising from umbral calculus 741

Thus, from (46), we have t r n−m e − 1 n−m r! n−m−l x = S2(l + r, r)(n − m)lx . (47) t l=0 (l + r)! By (45) and (47), we get n−m −t n r! Lik (1 − e ) − n−m−l Cn,m = S2(l + r, r)(n m)l − −t x (48) m l=0 (l + r)! 1 e n−m −t n r! 0Lik (1 − e ) n−m−l = S2(l + r, r)(n − m)l t x − −t m l=0 (l + r)! 1 e n−m n r! (k) = S2(l + r, r)(n − m)lBn−m−l. m l=0 (l + r)! Therefore, by (44) and (48), we obtain the following theorem.

Theorem 4. For k ∈ Z and r ∈ Z≥0, we have n n−m (k) n r!(n − m)l (k) (r) Bn (x)= S2(l + r, r)Bn−l−m Bn (x). m=0 m l=0 (l + r)!

For r ∈ Z≥0, the Euler polynomials of order r are deﬁned by the generating function to be r ∞ n 2 xt (r) t t e = En (x) , (see [2, 6, 8]). (49) e +1 n=0 n! By (2) and (49), we see that − −t t r (k) ∼ 1 e (r) ∼ e +1 Bn (x) −t ,t ,En (x) ,t . (50) Lik (1 − e ) 2 From (20), (21) and (50), we have n (k) (r) Bn (x)= Cn,mEm (x), (51) m=0 where −t t r 1 Lik (1 − e ) e +1 m n Cn,m = t x (52) m! 1 − e−t 2 n −t Lik (1 − e ) r = m et +1 xn−m 2r 1 − e−t n r −t r Lik (1 − e ) = m ejtxn−m r − −t 2 j=0 j 1 e 742 Dae San Kim, Taekyun Kim and Sang-Hun Lee

n r −t r Lik (1 − e ) m 0 n−m = r t − −t (x + j) 2 j=0 j 1 e n r m r (k) = r Bn−m(j). 2 j=0 j Therefore, by (51) and (52), we obtain the following theorem.

Theorem 5. For k ∈ Z and r ∈ Z≥0, we have n r (k) 1 n r (k) (r) Bn (x)= r Bn−m(j) En (x). 2 m=0 m j=0 j

Let λ ∈ C with λ =1.For r ∈ Z≥0, the Frobenius-Euler polynomials are also deﬁned by the generating function to be r ∞ 1 − λ tn ext = H(r)(x|λ) , (see [2, 6, 7]). (53) t − n e λ n=0 n! From (2), (15) and (53), we note that − −t t − r (k) ∼ 1 e (r) | ∼ e λ Bn (x) −t ,t ,Hn (x λ) ,t . (54) Lik (1 − e ) 1 − λ By (20), (21) and (54), we get n (k) (r) Bn (x)= Cn,mHm (x|λ), (55) m=0 where −t t r 1 Lik (1 − e ) e − λ m n Cn,m = t x (56) m! 1 − e−t 1 − λ n −t Lik (1 − e ) r = m et − λ xn−m (1 − λ)r 1 − e−t n r −t r Lik (1 − e ) = m (−λ)r−j ejtxn−m − r − −t (1 λ) j=0 j 1 e n r −t r Lik (1 − e ) = m (−λ)r−j t0 (x + j)n−m − r − −t (1 λ) j=0 j 1 e n r r (k) = m (−λ)r−jB (j). − r n−m (1 λ) j=0 j Therefore, by (55) and (56), we obtain the following theorem. Poly-Bernoulli polynomials arising from umbral calculus 743

Theorem 6. For k ∈ Z and r ∈ Z≥0, we have n r (k) 1 n r − r−j (k) (r) | Bn (x)= − r ( λ) Bn−m(j) Hm (x λ). (1 λ) m=0 m j=0 j

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Received: June 28, 2013