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formerly Central European Journal of Mathematics

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Volume 14, Issue 1 (Jan 2016)

Issues

Fully degenerate poly-Bernoulli numbers and polynomials

Taekyun Kim
  • Corresponding author
  • Department of Mathematics, College of Science, Tianjin Polytechnic University, Tianjin City, 300387, China and Department of Mathematics, Kwangwoon University, Seoul 139-701, Korea (Republic of)
  • Email:
/ Dae San Kim
  • Department of Mathematics, Sogang University, Seoul 121-742, Korea (Republic of)
  • Email:
/ Jong-Jin Seo
  • Department of Applied Mathematics, Pukyong Natioanl University, Pusan, Korea (Republic of)
  • Email:
Published Online: 2016-07-26 | DOI: https://doi.org/10.1515/math-2016-0048

Abstract

In this paper, we introduce the new fully degenerate poly-Bernoulli numbers and polynomials and inverstigate some properties of these polynomials and numbers. From our properties, we derive some identities for the fully degenerate poly-Bernoulli numbers and polynomials.

Keywords: Fully degenerate poly-Bernoulli polynomial; Fully degenerate poly-Bernoulli number; Umbral calculus

MSC 2010: 11B75; 11B83; 05A19; 05A40

1 Introduction

It is well known that the Bernoulli polynomials are defined by the generating function

tetext=n=0Bn(x)tnn!,(see[121]).(1)

When x = 0, Bn = Bn (0) are called the Bernoulli numbers. From (1), we note that

Bn(x)=i=0nnlBlxn1,(n0),(2)

and

B0=1,Bn(1)Bn=δ1,n,(nN),(see[1,19]),(3)

where δn k is the Kronecker’s symbol.

In [3], L. Carlitz considered the degenerate Bernoulli polynomials which are given by the generating function

t(1+λt)1λ1(1+λt)xλ=n=0βn,λ(x)tnn!.(4)

When x = 0, βn, λ = βn, λ (0) are called the degenerate Bernoulli numbers. From (1) and (4), we note that

tet1ext=limλ0t(1+λt)1λ1(1+λt)xλ=n=0limλ0βn,λ(x)tnn!(5)

Thus, by (5), we get

Bn(x)=limλ0βn,λ(x),(see[3,15]).(6)

By (4), we get

n=0(βm,λ(1)βm,λ)tnn!=t,(7)

and

βn,λ(x)=l=0nnlβl,λλn1xλn1.(8)

From (7), we have

βn,λ1βn,λ=δ1,n,(n0),β0,λ=1.(9)

Now, we consider the degenerate Bernoulli polynomials which are different from the degenerate Bernoulli polynomials of L. Carlitz as follows:

log1+λt1λ1+λt1λ11+λtxλ=n=0bn,λ(x)tnn!.(10)

When x = 0, bn, λ = bn, λ (0) are called the degenerate Bernoulli numbers.

Note. The degenerate Bernoulli polynomials are also called Daehee polynomials with λ-parameter (see [13]).

From (10), we note that

tet1ext=limλ0log1+λt1λ1+λt1λ1+λtxλ=n=0limλ0bn,λ(x)tnn!.(11)

By (1) and (11), we see that

Bn(x)=limλ0bn,λ(x),(n0).

The classical polylogarithm function Lik (x) is defined by

Lik(t)=n=1tnnk,kZ,(see[10,11]).(12)

It is known that the poly-Bernoulli polynomials are defined by the generating function

Lik1et1etext=n=0Bn(k)(x)tnn!,(see[9,10,12]).(13)

When k = 1, we have

n=0Bn(1)(x)tnn!=t1etext=tet1e(x+1)t=n=0Bn(x+1)tnn!.(14)

Bn(1)(x)=Bn(x+1),n0.

By (14), we easily get

Let x = 0. Then Bn(k)=Bn(k)(0) are called the poly-Bernoulli numbers.

In this paper, we introduce the new fully degenerate poly-Bernoulli numbers and polynomials and inverstigate some properties of these polynomials and numbers. From our investigation, we derive some identities for the fully degenerate poly-Bernoulli numbers and polynomials.

2 Fully degenerate poly-Bernoulli polynomials

For k ∈ ℤ, we define the fully degenerate poly-Bernoulli polynomials which are given by the generating function

Lik11+λt1λ11+λt1λ1+λtxλ=n=0βn,λ(k)(x)tnn!.(15)

When x=0,βn,λ(k)=βn,λ(k)(0) are called the fully degenerate poly-Bernoulli numbers. From (13) and (15), we have

Lik1et1etext=limλ0Lik11+λt1λ11+λt1λ1+λtxλ=n=0limλ0βn,λ(k)(x)tnn!.(16)

Thus, we get

limλ0βn,λ(k)(x)=Bn(k)(x),n0.(17)

By (15), we get

n=0βn,λ(k)(x)tnn!=Lik11+λt1λ11+λt1λ1+λtxλ=n=0l=0nnlβl,λ(k)xλn1λn1tnn!.(18)

Thus, from (18), we have

βn,λ(k)x+y=l=0nnlyλn1λn1βl,λ(k)(x),n0,(19)

and

βn,λ(k)(x)=l=0nnlxλn1λn1βl,λ(k).

Therefore, by (17) and (19), we obtain the following theorem.

Theorem 2.1: For k ∈ ℤ, ≥ 0, we haveβn,λ(k)(x+y)=l=0nnlyλn1λn1βl,λ(k)(x),n0,(20)andlimλ0βn,λ(k)(x)=Bn(k)(x),where (x)n=x(x1)(xn+1)=l=0nS1(n,l)xl..From (15), we can derive the following equation:n=0βn,λ(k)(x)tnn!=Lik11+λt1λ1+λt1λ11+λtx+1λ.(21)Thus, by (21), we getn=0βn,λ(k)βn,λ(k)(1)tnn!=Lik1(1+λt)1λ=m=11(1+λt)1λmmk=m=0(1)m+1(m+1)ke1λlog(1+λt)1m+1=m=0(1)m+1(m+1)k(m+1)!l=m+1XS2(l,m+1)(1)lλl(log(1+λt))ll!=l=1m=0l1(1)m+1(m+1)k(m+1)!S2(l,m+1)(1)lλln=lS1(n,l)λntnn!=n=1l=1nm=0l1m!(m+1)(1)lm1λnlS2(l,m+1)S1(n,l)(m+1)ktnn!,(22)where S2 (n, l) and S1 (n, l) are the Stirling numbers of the second kind and of the first kind, respectively. Therefore, by (22), we obtain the following theorem.

Theorem 2.2: For k ∈ ℤ, n ≥ 1, we haveβn,λ(k)βn,λ(k)(1)=l=1nm=0l1m!(1)lm1λn1S2l,m+1S1n,l(m+1)k1.From (12), we can easily derive the following equation:Lik(t)=ddtLik(t)=1tLik1(t).(23)Thus, by (23), the generating function of the fully degenerate poly-Bernoulli numbers is also written in terms of the following iterated integral:(1+λt)1λ(1+λt)1λ10t1(1+λt)1λ1(1+λt)×0t1(1+λt)1λ1(1+λt)0tlog(1+λt)1λ(1+λt)1λ1(1+λt)dtdtk1times=n=0βn,λ(k)tnn!.(24)For k = 2, we haven=0βn,λ(2)tnn!=(1+λt)1λ(1+λt)1λ10tlog(1+λt)1λ(l+λt)1λ1(1+λt)λλdt=(1+λt)1λ(1+λt)1λ1m=0bm,λ(λ)1m!0ttmdt=t(1+λt)1λ1(1+λt)1λm=0bm,λ(λ)(m+1)tmm!=n=0l=0nnlβl,λ(1)bnl,λ(λ)nl+1tnn!.(25)Therefore, by (25), we obtain the following theorem.

Theorem 2.3: For n ≥ 0, we haveβn,λ(2)=l=0nnlβl,λ(1)bn1,λλnl+1.Note thatBn2=limλ0βn,λ(2)=l=0nnlBl(1)Bn1nl+1.From (15), we haven=0βn,λ(k)tnn!=Lik1(1+λt)1λ1(1+λt)1λ=m=01(m+1)k1(1+λt)1λm=m=0(1)m(m+1)ke1λlog(1+λt)1m=m=0(1)m(m+1)km!l=mS2(l,m)1λl(log(1+λt))ll!=l=0m=0l(1)m+lm!(m+1)kS2(l,m)λl1l!(log(1+λt))l=l=0m=0l(1)m+lm!(m+1)kS2(l,m)λln=lS1(n,l)λntnn!=n=0l=0nm=0l(1)m+lm!(m+1)kS2(l,m)S1(n,l)λnltnn!.(26)Therefore, by (26), we obtain the following theorem.

Theorem 2.4: For n ≥ 0, we haveβn,λ(k)=l=0nm=0l1m+lm!m+1kS2l,mS1n,lλnl.Note thatBn(k)=limλ0βn,λ(k)=m=0n1m+nm!m+1kS2n,m.From (23), we haveddtLik11+λt1λ=111+λt1λ1+λt1λ1Lik111+λt1λ=1+λt1λ1n=0βn,λ(k1)tnn!.(27)On the other hand,ddtLik1(1+λt)1λ=ddt1(1+λt)1λ11(1+λt)1λLik1(1+λt)1λ)=(1+λt)1λ111(1+λt)1λLik1(1+λt)1λ+1(1+λt)1λddtn=0βn,λ(k)tnn!=(1+λt)1λ1n=0βn,λ(k)tnn!+1(1+λt)1λn=1βn,λ(k)tn1(n1)!.(28)By (27) and (28), we getn=0βn,λ(k1)tnn!=n=0βn,λ(k)tnn!+(1+λt)(1+λt)1λ1n=1βn,λ(k)tn1(n1)!=n=0βn,λ(k)tnn!+(1+λt)1λ1m=0βm+1,λ(k)tmm!+λ(1+λt)1λ1m=0βm,λ(k)mtmm!=n=0βn,λ(k)tnn!l=11λlλttll!m=0βm+1,λ(k)tmm!+λl=11λlλttll!m=0βm,λ(k)mtmm!=n=0βn,λ(k)tnn!+n=1m=0n11λnmλnmn!(nm)!m!βm+1,λ(k)tnn!+λn=1m=0n11λnmλnmmn!(nm)!m!βm,λ(k)tnn!=n=0βn,λ(k)tnn!+n=1m=0n1nm1λnmλnmβm+1,λ(k)tnn!+λn=1m=0n1nm1λnmλnmmβm,λ(k)tnn!,(29)where(1λ)n=1λ1λ11λn+1=l=0nS1nn,lλl,n0.Thus, by (29), we haveβn,λk1)=βn,λk)+m=0n1nm(1λ)nmλnmβm+1,λk+λm=0n1nm(1λ)nmλnmmβm,λk=n+1βn,λk)+m=1n1nm1(1λ)nm+1λnm+1βm,λk+λm=0n1nm(1λ)nmλnmmβm;λ,kn1.(30)Therefore, by (30), we obtain the following theorem

Theorem 2.5: For n ≥ 1, we haveβn,λk)=1n+1βn,λk1)m=1n1nm1βm,λk)(1λ)nm+1λnm+1λm=0n1nm(1λ)nmλnmmβm;λk.Note thatBn(k)=limλ0βn,λ(k)=1n+1Bn(k1)m=1n1nm1Bm(k).Now, we observe thatn=0(1(1+λt)1λ)n(n+1)k=n=1(1(1+λt)1λ)nnk11(1+λt)1λ=11(1+λt)1λLik(11+λt)1λ=n=0Xβn,λk)tnn!.(31)By (31), we getk=0n=0βn,λ(k)xnn!ykk!=k=0m=0(1(1+λt)1λ)m(m+1)kykk!=m=01(1+λx)1λmk=0(m+1)kykk!=m=0(1(1+λx)1λ)mem+1)y=j=0(1)j(e1λlog(1+λx)1)jej+1)y=j=0(1)jj!m=jS2m,j(1)mλm(log1+λx)mm!e(j+1)y=m=0j=0m(1)j+mj!S2m,jλmn=mS1n,mλnxnn!e(j+1)y=n=0m=0nj=0m(1)j+mj!S2m;jλnmS1n,mej+1)yxnn!=k=0n=0m=0nj=0m(1)j+mj!λnmS2m,jS1n,m(j+1)kxnn!ykk!.(32)Therefore, by (32), we obtain the following theorem.

Theorem 2.6: For k ∈ ℤ and n ≥ 0, we haveβn,λ(k)=m=0nj=0m(1)j+mj!λnm(j+1)kS2m,jS1n,m.Note thatBn(k)=limλ0βn,λ(k)=j=0n(1)j+nj!(j+1)ks2(n,j).From Theorem 2.1, we haveddxβn,λ(k)(x)=l=0nlnβl,λ(k)ddxi=0nl1(xiλ)=l=0nlnβl,λ(k)j=0nl11(xjλ)i=0nl1(xiλ).The generalized falling factorial (x | λ)n is given by(x|λ)n=x(xλ)(x2λ)(x(n1)λ),(n0).(33)As is well known, the Bernoulli numbers of the second kind are defined by the generating funcdtiontlog(1+t)=n=0bntnn!,(see[20]).(34)We observe that01(1+λt)xλdx=n=0λn01xλndxtnn!=n=001x|λndxtnn!.(35)On the other hand,01(1+λt)xλdx=λlog(1+λt)(1+λt)1λ1=λtlog1+λt(1+λt)1λ1t=m=0bmλmtmm!l=0(1|λ)l+1l+1tll!=n=0l=0n(1|λ)l+1l+1λnlbnlnltnn!.(36)From (35) and (36), we have01(x|λ)ndx=l=0nnlλnlbn1(1|λ)l+1l+1,n0.(37)By Theorem 2.1, we get01βn,λ(k)xdx=l=0nnlβl,λ(k)01xλnlλnldx=l=0nnlβnl,λ(k)01(x|λ)ldx=l=0nm=0llmλlmblm(1|λ)m+1m+1nlβnl,λ(k).

3 Further remarks

Let ℂ be complex number field and let F be the set of all formal power series in the variable t over ℂ with

F=f(t)=k=oaktkk!akC.(38)

Let ℙ be the algebra of polynomials in a single variable x over ℂ and let ℙ* be the vector space of all linear functionals on ℙ. The action of linear functional L ∈ ℙ* on a polynomial p (x) is denoted by 〈L| p (x)〉, and linearly extended as

cL+cL|p(x)=cL|p(x)+cL|p(x),

where c, c′ ∈ ℂ.

For f(t)=k=0aktkk!, we define a linear functional on ℙ by setting

f(t)|xn=anforalln0.(39)

Thus, by (39), we get

tkxn=n!δn,k,(n,k0),(see[4,16,20]).(40)

For fL(t)=k=0L|xktkk!, by (40), we get 〈fL (i)| xn〉 = 〈L| xn〉. In addition, the mapping LfL (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 (i) of F can be regarded as both a formal power series and a linear functional. We refer to F umbral algebra. The umbral calculus is the study of umbral algebra (see [5, 15, 20]). The order o (f (i)) of the non-zero power series f (i) is the smallest integer k for which the coefficient of tk does not vanish.

If o (f (t)) = 1(respectively, o (f (t)) = 0), then f (t) is called a delta (respectively, an invertible) series (see [20]). For o (f (t)) = 1 and o (g (t)) = 0, there exists a unique sequence sn (x) of polynomials such that 〈g(t) f (t)k| sn (x〉 = n!δn, k, (n, k ≥ 0).

The sequence sn (x) is called the Sheffer sequence for (g (t), f (t)), and we write sn (x) ~ (g (t), f (t)) (see [20]).

Let f (t) ∈ F and p (x) ∈ ℙ. Then, by (40), we get

f(t)=k=of(t)xktkk!,p(x)=k=otkp(x)xkk!.(41)

From (41), we have

p(k)(0)=tkp(x)=1p(k)(x),(42)

where pk(x)=dkdxkp(x), (see [11, 14, 20]).

By (42), we easily get

tkp(x)=pk(x),eytp(x)=p(x+y),andeytp(x)=p(y).(43)

From (43), we have

eyt1tp(x)=xx+yp(u)du,eyt1p(x)=p(y)p(0).

Let f (t) be the linear functional such that

f(t)p(x)=oyp(u)du,(44)

for all polynomials p (x). Then it can be determined by (41) to be

f(t)=k=0f(t)|xkk!tk=k=0yk+1(k+1)!tk=1t(eyt1).(45)

Thus, for p (x) ∈ ℙ, we have

eyt1tp(x)=0yp(u)du.(46)

It is known that

sn(x)(g(t),f(t))1gf¯(t)exf¯(t)=k=0sk(x)tkk!,(xC)(47)

where f¯(t) is the compositional inverse of f (t) such that ff¯(t)=f¯(f(t))=t (see [11, 20]). From (15), we note that

βn,λ(k)(x)1etLik(1et),1λeλt1.(48)

That is,

n=0βn,λ(k)(x)tnn!=Lik1(1+λt)1λ1(1+λt)1λ(1+λt)xλ.

Thus, by (48),

1λ(eλt1)βn,λ(k)(x)=nβn1,λ(k)(x).(49)

On the other hand,

(eλt1)βn,λ(k)(x)=βn,λ(k)(x+λ)βn,λ(k)(x).(50)

Therefore, by (49) and (50), we obtain the following theorem.

Theorem 3.1: For n ∈ ℕ, we haveλβn1,λ(k)(x)=1nβn,λ(k)(x+λ)βn,λ(k)(x).By (46), we geteyt1tβn,λ(k)(x)=xx+yβn,λ(k)(u)du.(51)From (51), we haveeyt1tβn,λ(k)(x)=0yβn,λ(k)(u)du.(52)Thus, by (52), we getet1tβn,λ(k)(x)=01βn,λ(k)(u)du=l=0nm=0llmnlλlmblmβnl,λk(1|λ)m+1m+1.(53)Therefore, by (53), we obtain the following theorem.,

Theorem 3.2: For n ≥ 0 we haveet1tβn,λ(k)(x)=l=0nm=0llmnlλlmblmβnl,λk(1|λ)m+1m+1.Note thatet1tBn(k)(x)=limλ0et1tβn,λ(k)(x)=limλ001βn,λ(k)(u)du=l=0nnlBnlk)1l+1LetPn={pxCx|degpxn},n0.For p(x) ∈ ℙn with p(x)=m=0namβm,λ(k)(x), we have1etLik1et1λeλt1mpx=l=0nal1etLik1et1λeλt1mβl,λk)x(54)From (48), we note that1etLik1et1λeλt1mβl,λk)x=l!δl,m.(55)By (54) and (55), we getam=1m!1etLik1et1λeλt1mpx,(m0).(56)Therefore, by (56), we obtain the following theorem.

Theorem 3.3: For p (x) ∈ ℙn, we havepx=m=0namβm,λ(k)(x),wheream=1m!1etLik1et1λeλt1mpx.For example, let us take p(x)=βn(k)(x)(n0). Then, by Theorem 3.3, we haveBn(k)x=m=0namβm,λ(k)(x),(57)wheream=1m!1etLik1et1λeλt1mBn(k)x=1m!1etLik1et1λeλt1mLik1et1etxn=λmm!eλt1mxn=λml=mS2l,mλll!tlxn=λnmS2n,m.(58)From (57) and (58), we haveBn(k)x=m=0nλnmS2n,mβm,λ(k)x.(59)

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About the article

Received: 2015-05-26

Accepted: 2016-06-29

Published Online: 2016-07-26

Published in Print: 2016-01-01


Citation Information: Open Mathematics, ISSN (Online) 2391-5455, DOI: https://doi.org/10.1515/math-2016-0048. Export Citation

© 2016 Kim et al., published by De Gruyter Open. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License. (CC BY-NC-ND 3.0)

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