Poly-Bernoulli Polynomials Arising from Umbral Calculus

arXiv:1306.6697v1 [math.NT] 28 Jun 2013 ihteuulcneto bu replacing about convention usual the with h lsia oyoaihi function polylogarithmic classical The Introduction 1 nteseilcase, special the In oyoil eew eoehge-re enul oyoil as polynomials Bernoulli higher-order denote we Here polynomial. n[] oyBroliplnmasaedfie ytegnrtn funct generating the by defined are polynomials poly-Bernoulli [5], In si elkon h enul oyoil forder of polynomials Bernoulli the known, well is As eeaigfnto obe to function generating oyBroliplnmasaiigfo umbral from arising polynomials Poly-Bernoulli dniisfrtepl-enul ubr n polynomials and calculus. int numbers umbral and from new poly-Bernoulli derived and the formula recurrence for some identities give we paper, this In a a i,TeynKmadSn-u Lee Sang-Hun and Kim Taekyun Kim, San Dae Li k 1 (1 − − e − r e e Li t t 1, = − − t t s ) ( e 1 x xt = ) B r n ( = r e ) ( X xt k e ∞ x =1 B = ) = ( calculus k x ) k Abstract ( X n k x s ∞ =0 B ) s , t by n = B 1 ( Li x n ( X r n ∈ scle the called is ) ∞ ) s =0 ( ( Z x x B B ) , are ) n t n ( ( n k k ! se[3 (see ) ) , ( ( x x se[7 (see ) ) n t n n ! , , by 5]) n se[3 (see t riayBernoulli ordinary -th , r . 9]) B r endb the by defined are n ( . k ) ( , x hc are which 5]) ). eresting B , n ( r ) o obe to ion oavoid to (2) (3) (1) 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 n n B(k)(x)= B(k) xl = B(k)xn−l. (4) n l n−l l l l l X=0 X=0 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! ) k=0 X P P∗ and let = C[x] and denote the vector space of all linear functionals on P. hL|p(x)i 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 by hL + M|p(x)i = hL|p(x)i + hM|p(x)i, hcL|p(x)i = c hL|p(x)i, 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 f (t) xn = L xn . That is, f (t)= L. Additionally, the map L 7−→ f (t) is L L L a vector space isomorphism fromPP∗ 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) of F 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) =6 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

hf(t)g(t)|p(x)i = hf(t)|g(t)p(x)i = hg(t)|f(t)p(x)i . (7)

Let f(t) ∈F and p(x) ∈ P. Then we have

∞ tk ∞ xk f(t)= f(t) xk , p(x)= tk p(x) , (see [6, 9]). (8) k! k! k=0 k=0 X X

2 From (8), we can easily derive ∞ dkp(x) tl p(x) p(k)(x)= = xl−k. (9) dxk (l − k)! l=k X Thus, by (8) and (9), we get p(k)(0) = tk p(x) = 1 p(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), eyt p(x) = p(y). (12)

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 s (x) is called a Sheﬀer sequence for (g(t), f(t)) which is denoted by n 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

′ hf(t)|xp(x)i = h∂tf(t)|p(x)i = hf (t)|p(x)i , (see [9, 10]). (13)

Let sn(x) ∼ (g(t), f(t)). Then the following equations are known:

∞ ∞ k h(t) sk(x) g(t)f(t) p(x) h(t)= g(t)f(t)k, p(x)= s (x), (14) k! k! k k=0 k=0 X X where h(t) ∈F, p(x) ∈ P, ∞ 1 tk eyf¯(t) = s (y) , for all y ∈ C, (15) g f¯ t k k! ( ) k X=0 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 ∞ m t m! et − 1 = m! S (l, m) = S (l + m, m)tl+m. (17) 2 l! (l + m)! 2 l=m l=0 X X 3 Let sn(x) ∼ (g(t), t). Then the Appell identity is given by

n n n n s (x + y)= s (y)xn−k = s (y)xk, (see [6, 9]), (18) n k k k n−k k k X=0 X=0 and g′(t) s (x)= x − s (x), (see [6.9]). (19) n+1 g(t) n For sn(x) ∼ (g(t), f(t)), rn(x) ∼ (h(t),l(t)), we have

sn(x)= rm(x)cn,m, (20) m=0 X where 1 h f¯(t) m c l f¯ t xn , . n,m = ¯ ( ) (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 g t −t Let k( )= Lik(1−e ) . Then, by (2) and (15), we get

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

4 Let k ∈ Z and n ≥ 0. Then we have

∞ Li (1 − e−t) (1 − e−t)m−1 B(k)(x)= k xn = xn (25) n 1 − e−t mk m=1 ∞ X ∞ m 1 m 1 m = 1 − e−t xn = (−1)j e−jtxn (m + 1)k (m + 1)k j m=0 m=0 j=0 Xn m X X 1 m = (−1)j (x − j)n. (m + 1)k j m=0 j=0 X X By (17) and (25), we get

n ∞ 1 m! B(k)(x)= (−1)a S (a + m, m)ta+mxn (26) n (m + 1)k (a + m)! 2 m=0 a=0 Xn Xn−m 1 m! = (−1)a S (a + m, m)(n) xn−a−m (m + 1)k (a + m)! 2 a+m m=0 a=0 Xn n−l X (−1)n−m−l n = m!S (n − l, m) xl, (m + 1)k l 2 l (m=0 ) X=0 X 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

g′ (t) B(k) (x)= x − k B(k)(x), (27) n+1 g (t) n k 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 Li (1 − e ) − Li − (1 − e ) = k k 1 . et − 1 Li (1 − e−t) k

5 From (27) and (28), we have

′ (k) (k) gk(t) (k) Bn+1(x)= xBn (x) − Bn (x) (29) gk(t) −t −t t Li (1 − e ) − Li − (1 − e ) = xB(k)(x) − k k 1 xn. n et − 1 t (1 − e−t) Here, we note that

−t −t ∞ Li (1 − e ) − Li − (1 − e ) 1 1 m−1 k k 1 = − 1 − e−t (30) 1 − e−t mk mk−1 m=2 X 1 1 = − t + ··· 2k 2k−1 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 1 Li (1 − e−t) − Li (1 − e−t) B(k) (x)= xB(k)(x) − k k−1 xn+1 n+1 n et − 1 n +1 1 − e−t ∞ 1 B = xB(k)(x) − l tl B(k) (x) − B(k−1)(x) n n +1 l! n+1 n+1 l=0 n o Xn 1 +1 n +1 = xB(k)(x) − B B(k) (x) − B(k−1) (x) . (32) n n +1 l l n+1−l n+1−l l X=0 n o Therefore, by (32), we obtain the following theorem.

Theorem 1. For k ∈ Z, n ≥ 0, we have

n 1 +1 n +1 B(k) (x)= xB(k)(x) − B B(k) (x) − B(k−1) (x) , n+1 n n +1 l l n+1−l n+1−l l X=0 n o where Bn is the n-th ordinary Bernoulli number.

6 It is easy to show that

n n n n txB(k)(x)= t B(k) xl+1 = B(k) (l + 1)xl (33) n l n−l l n−l l=0 l=0 Xn− X n 1 n − 1 n = nx B(k) xl + B(k) xn l n−1−l l n−l l l X=0 X=0 (k) (k) = nxBn−1(x)+ Bn (x).

Applying to t on both sides of Theorem 1, by (33), we get

n n (n +1)B(k)(x)= nxB(k) (x)+ B(k)(x) − B B(k)(x) − B(k−1)(x) . n n−1 n l n−l l l l=0 n o X (34) Therefore, by (34), we obtain the following corollary.

Corollary 2. For k ∈ Z and n ≥ 1, we have

n−2 (k) 1 (k) n (k) (n + 1)B (x) − n x + B (x)+ B − B (x) n 2 n−1 l n l l l X=0 n n = B B(k−1)(x). l n−l l l X=0 From (2) and (6), we note that

−t −t (k) Lik (1 − e ) yt n Lik (1 − e ) yt n−1 Bn (y)= −t e x = −t e xx (35) * 1 − e + * 1 − e +

−t Lik (1 − e ) yt n−1 = ∂t −t e x * 1 − e + −t −t Lik (1 − e ) yt n−1 Lik (1 − e ) yt n−1 = ∂t −t e x + y −t e x * 1 − e + * 1 − e + −t −t Lik−1 (1 − e ) − Lik (1 − e ) (y−1)t n−1 (k ) = 2 e x + yBn− 1(y). * (1 − e−t) +

7 Now, we observe that −t −t ∞ −t m −t m Li − (1 − e ) − Li (1 − e ) 1 (1 − e ) (1 − e ) k 1 k = − (1 − e−t)2 (1 − e−t)2 mk−1 mk m=1 X (36) ∞ 1 1 m− = − 1 − e−t 2 mk−1 mk m=2 X∞ 1 1 m = − 1 − e−t . (m + 2)k−1 (m + 2)k m=0 X 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 = − 1 − e e x (m + 2)k−1 (m + 2)k m=0 Xn−1 1 1 m = − 1 − e−t (x + y − 1)n−1 (m + 2)k−1 (m + 2)k m=0 Xn−1 n−1 1 1 n − 1 m = − (y − 1)n−1−a 1 − e−t xa . (m + 2)k−1 (m + 2)k a m=0 a=0 X X 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 n − 1 1 1 = (−1)a+m m! − a (m + 2)k−1 (m + 2)k m=0 a=0 X X n−1−a × S2(a, m)(y − 1) n− n− 1 1 n − 1 1 1 = (−1)n−1−l+m m! − l (m + 2)k−1 (m + 2)k m=0 l=0 X X l × S2(n − 1 − l, m)(y − 1) . Therefore, by (35) and (39), we obtain the following theorem.

8 Theorem 3. For k ∈ Z and n ≥ 1, we have

n− 1 n − 1 B(k)(x)= xB(k) (x)+ (−1)n−1−l n n−1 l l=0 n− X 1 (m + 1)! × (−1)m S (n − 1 − l, m) (x − 1)l. (m + 2)k 2 (m=0 ) X −t n+1 Now, we try to compute Lik (1 − e ) x in two ways. On the one hand,

−t Lik (1 − e ) Li e−t xn+1 e−t xn+1 k 1 − = 1 − −t (40) * 1 − e + −t Lik (1 − e ) −t n+1 = −t 1 − e x * 1 − e +

−t Lik (1 − e ) n+1 n+1 = −t x − (x − 1) * 1 − e +

n −t n +1 n−m Lik (1 − e ) m = (− 1) 1 −t x m * 1 − e + m=0 Xn n +1 n−m (k) = (−1) Bm . m m=0 X On the other hand,

t −t n+1 −s ′ n+1 Lik 1 − e x = Lik 1 − e ds x (41) * 0 + Z t −s Lik −1 (1 − e ) e−s ds xn+1 = −s * 0 1 − e + Z t ∞ a ∞ (k−1) (−s) B m = sm ds xn+1 a! m! * 0 a=0 ! m=0 ! + Z X X ∞ l l 1 t = (−1)l−mB(k−1) slds xn+1 m m l! * l m=0 ! 0 + X=0 X Z n l (k−1) n l Bm n = (−1)l−m (n + 1)!δ = (−1)n−mB(k−1). m (l + 1)! n+1,l+1 m m l m=0 m=0 X=0 X X

9 By (40) and (41), we get

n n n n +1 (−1)n−m B(k−1) = (−1)n−mB(k). (42) m m m m m=0 m=0 X X By (2) and (3), we see that

1 − e−t et − 1 r B(k)(x) ∼ , t , B(r)(x) ∼ , t , r ≥ 0. (43) n Li (1 − e−t) n t k From (20), (21) and (43), we have

n (k) B(r) Bn (x)= Cn,m m (x), (44) m=0 X where r et−1 1 t m n C = −t t x (45) n,m 1−e m! * −t + Lik(1−e ) −t t r 1 Lik (1 − e ) e − 1 tmxn = −t m! * 1 − e t + −t t r n Lik (1 − e ) e − 1 xn−m . = −t m * 1 − e t +

By (17), we easily get ∞ et − 1 r r! = S (l + r, r)tl. (46) t (l + r)! 2 l X=0 Thus, from (46), we have

n−m et − 1 r r! xn−m = S (l + r, r)(n − m) xn−m−l. (47) t (l + r)! 2 l l X=0

10 By (45) and (47), we get

n−m n r! Li (1 − e−t) C = S (l + r, r)(n − m) k xn−m−l (48) n,m m (l + r)! 2 l 1 − e−t l * + X=0 n−m − t n r! Lik (1 − e ) S l r, r n m t0 xn−m−l = 2( + )( − )l −t m (l + r)! * 1 − e + l=0 nX−m n r! k = S (l + r, r)(n − m) B( ) . m (l + r)! 2 l n−m−l l X=0 Therefore, by (44) and (48), we obtain the following theorem.

Theorem 4. For k ∈ Z and r ∈ Z≥0, we have n n−m n r!(n − m) B(k)(x)= l S (l + r, r)B(k) B(r)(x). n m (l + r)! 2 n−l−m n m=0 ( l ) X X=0 For r ∈ Z≥0, the Euler polynomials of order r are deﬁned by the generating function to be ∞ 2 r tn ext = E(r)(x) , (see [2, 6, 8]). (49) et +1 n n! n=0 X By (2) and (49), we see that 1 − e−t et +1 r B(k)(x) ∼ , t , E(r)(x) ∼ , t . (50) n Li (1 − e−t) n 2 k From (20), (21) and (50), we have n (k) (r) Bn (x)= Cn,mEm (x), (51) m=0 X where 1 Li (1 − e−t) et +1 r C k tmxn n,m = −t (52) m! * 1 − e 2 + n −t Lik (1 − e ) r m et xn−m = r −t +1 2 * 1 − e + n r −t r Lik (1 − e ) m ejtxn−m = r −t 2 j * 1 − e + j=0 X

11 r n r Li (1 − e−t) = m t0 k (x + j)n−m 2r j 1 − e−t j=0 * + X n r r (k) = m B (j). 2r j n−m j=0 X Therefore, by (51) and (52), we obtain the following theorem.

Theorem 5. For k ∈ Z and r ∈ Z≥0, we have n r 1 n r B(k)(x)= B(k) (j) E(r)(x). n 2r m j n−m n m=0 ( j=0 ) X X Let λ ∈ C with λ =6 1. For r ∈ Z≥0, the Frobenius-Euler polynomials are also deﬁned by the generating function to be ∞ 1 − λ r tn ext = H(r)(x|λ) , (see [2, 6, 7]). (53) et − λ n n! n=0 X From (2), (15) and (53), we note that 1 − e−t et − λ r B(k)(x) ∼ , t , H(r)(x|λ) ∼ , t . (54) n Li (1 − e−t) n 1 − λ k By (20), (21) and (54), we get n (k) (r) Bn (x)= Cn,mHm (x|λ), (55) m=0 X where 1 Li (1 − e−t) et − λ r C k tmxn n,m = −t (56) m! * 1 − e 1 − λ + n −t Lik (1 − e ) r m et λ xn−m = r −t − (1 − λ) * 1 − e + n r −t r Lik (1 − e ) = m (−λ)r− j ejtxn−m (1 − λ)r j 1 − e−t j=0 * + X r n r Li (1 − e− t) = m (−λ)r−j t0 k (x + j)n−m (1 − λ)r j 1 − e−t j=0 * + X n r r k = m (−λ)r−jB( ) (j). (1 − λ)r j n−m j=0 X

12 Therefore, by (55) and (56), we obtain the following theorem.

Theorem 6. For k ∈ Z and r ∈ Z≥0, we have

n r 1 n r B(k)(x)= (−λ)r−jB(k) (j) H(r)(x|λ). n (1 − λ)r m j n−m m m=0 ( j=0 ) X X References

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

Department of Mathematics, Kwangwoon University, Seoul 139-701, Republic of Korea e-mail: [email protected] Division of General Education, Kwangwoon University, Seoul 139-701, Repub- lic of Korea e-mail: [email protected]