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# 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.

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

## 1 Introduction

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

$tetext=∑n=0∞Bn(x)tnn!,(see[1−21]).$(1)

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

$Bn(x)=∑i=0nnlBlxn−1,(n≥0),$(2)

and

$B0=1,Bn(1)−Bn=δ1,n,(n∈N),(see[1,19]),$(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

$t(1+λt)1λ−1(1+λt)xλ=∑n=0∞βn,λ(x)tnn!.$(4)

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

$tet−1ext=limλ→0⁡t(1+λt)1λ−1(1+λt)xλ=∑n=0∞limλ→0βn,λ(x)tnn!$(5)

Thus, by (5), we get

$Bn(x)=limλ→0βn,λ(x),(see[3,15]).$(6)

By (4), we get

$∑n=0∞(βm,λ(1)−βm,λ)tnn!=t,$(7)

and

$βn,λ(x)=∑l=0nnlβl,λλn−1xλn−1.$(8)

From (7), we have

$βn,λ1−βn,λ=δ1,n,(n≥0),β0,λ=1.$(9)

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

$log1+λt1λ1+λt1λ−11+λtxλ=∑n=0∞bn,λ(x)tnn!.$(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

$tet−1ext=limλ→0⁡log1+λt1λ1+λt1λ1+λtxλ=∑n=0∞limλ→0bn,λ(x)tnn!.$(11)

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

$Bn(x)=limλ→0bn,λ(x),(n≥0).$

The classical polylogarithm function Lik (x) is defined by

$Lik(t)=∑n=1∞tnnk,k∈Z,(see[10,11]).$(12)

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

$Lik1−e−t1−e−text=∑n=0∞Bn(k)(x)tnn!,(see[9,10,12]).$(13)

When k = 1, we have

$∑n=0∞Bn(1)(x)tnn!=t1−e−text=tet−1e(x+1)t=∑n=0∞Bn(x+1)tnn!.$(14)

$Bn(1)(x)=Bn(x+1),n≥0.$

By (14), we easily get

Let x = 0. Then ${B}_{n}^{\left(k\right)}={B}_{n}^{\left(k\right)}\left(0\right)$ 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

$Lik1−1+λt−1λ1−1+λt−1λ1+λtxλ=∑n=0∞βn,λ(k)(x)tnn!.$(15)

When $x=0,{\beta }_{n,\lambda }^{\left(k\right)}={\beta }_{n,\lambda }^{\left(k\right)}\left(0\right)$ are called the fully degenerate poly-Bernoulli numbers. From (13) and (15), we have

$Lik1−e−t1−e−text=limλ→0⁡Lik1−1+λt−1λ1−1+λt−1λ1+λtxλ=∑n=0∞limλ→0⁡βn,λ(k)(x)tnn!.$(16)

Thus, we get

$limλ→0⁡βn,λ(k)(x)=Bn(k)(x),n≥0.$(17)

By (15), we get

$∑n=0∞βn,λ(k)(x)tnn!=Lik1−1+λt−1λ1−1+λt−1λ1+λtxλ=∑n=0∞∑l=0nnlβl,λ(k)xλn−1λn−1tnn!.$(18)

Thus, from (18), we have

$βn,λ(k)x+y=∑l=0nnlyλn−1λn−1βl,λ(k)(x),n≥0,$(19)

and

$βn,λ(k)(x)=∑l=0nnlxλn−1λn−1βl,λ(k).$

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

For k ∈ ℤ, ≥ 0, we have

$βn,λ(k)(x+y)=∑l=0nnlyλn−1λn−1βl,λ(k)(x),n≥0,$(20)

and

$limλ→0⁡βn,λ(k)(x)=Bn(k)(x),$

where $\left(x{\right)}_{n}=x\left(x-1\right)\cdots \left(x-n+1\right)=\sum _{l=0}^{n}{S}_{1}\left(n,l\right){x}^{l}.$.

From (15), we can derive the following equation:

$∑n=0∞βn,λ(k)(x)tnn!=Lik1−1+λt−1λ1+λt1λ−11+λtx+1λ.$(21)

Thus, by (21), we get

$∑n=0∞βn,λ(k)−βn,λ(k)(−1)tnn!=Lik1−(1+λt)−1λ=∑m=1∞1−(1+λt)−1λmmk=∑m=0∞(−1)m+1(m+1)ke−1λlog(1+λt)−1m+1=∑m=0∞(−1)m+1(m+1)k(m+1)!∑l=m+1XS2(l,m+1)(−1)lλ−l(log(1+λt))ll!=∑l=1∞∑m=0l−1(−1)m+1(m+1)k(m+1)!S2(l,m+1)(−1)lλ−l∑n=l∞S1(n,l)λntnn!=∑n=1∞∑l=1n∑m=0l−1m!(m+1)(−1)l−m−1λn−lS2(l,m+1)S1(n,l)(m+1)ktnn!,$(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.

For k ∈ ℤ, n ≥ 1, we have

$βn,λ(k)−βn,λ(k)(−1)=∑l=1n∑m=0l−1m!(−1)l−m−1λn−1S2l,m+1S1n,l(m+1)k−1.$

From (12), we can easily derive the following equation:

$Lik′(t)=ddtLik(t)=1tLik−1(t).$(23)

Thus, by (23), the generating function of the fully degenerate poly-Bernoulli numbers is also written in terms of the following iterated integral:

$(1+λt)1λ(1+λt)1λ−1∫0t1(1+λt)1λ−1(1+λt)×∫0t1(1+λt)1λ−1(1+λt)⋯∫0tlog(1+λt)1λ(1+λt)1λ−1(1+λt)dt⋯dt⏟k−1times=∑n=0∞βn,λ(k)tnn!.$(24)

For k = 2, we have

$∑n=0∞βn,λ(2)tnn!=(1+λt)1λ(1+λt)1λ−1∫0tlog(1+λt)1λ(l+λt)1λ−1(1+λt)−λλdt=(1+λt)1λ(1+λt)1λ−1∑m=0∞bm,λ(−λ)1m!∫0ttmdt=t(1+λt)1λ−1(1+λt)1λ∑m=0∞bm,λ(−λ)(m+1)tmm!=∑n=0∞∑l=0nnlβl,λ(1)bn−l,λ(−λ)n−l+1tnn!.$(25)

Therefore, by (25), we obtain the following theorem.

For n ≥ 0, we have

$βn,λ(2)=∑l=0nnlβl,λ(1)bn−1,λ−λn−l+1.$

Note that

$Bn2=limλ→0⁡βn,λ(2)=∑l=0nnlBl(1)Bn−1n−l+1.$

From (15), we have

$∑n=0∞βn,λ(k)tnn!=Lik1−(1+λt)−1λ1−(1+λt)−1λ=∑m=0∞1(m+1)k1−(1+λt)−1λm=∑m=0∞(−1)m(m+1)ke−1λlog(1+λt)−1m=∑m=0∞(−1)m(m+1)km!∑l=m∞S2(l,m)−1λl(log(1+λt))ll!=∑l=0∞∑m=0l(−1)m+lm!(m+1)kS2(l,m)λ−l1l!(log(1+λt))l=∑l=0∞∑m=0l(−1)m+lm!(m+1)kS2(l,m)λ−l∑n=l∞S1(n,l)λntnn!=∑n=0∞∑l=0n∑m=0l(−1)m+lm!(m+1)kS2(l,m)S1(n,l)λn−ltnn!.$(26)

Therefore, by (26), we obtain the following theorem.

For n ≥ 0, we have

$βn,λ(k)=∑l=0n∑m=0l−1m+lm!m+1kS2l,mS1n,lλn−l.$

Note that

$Bn(k)=limλ→0⁡βn,λ(k)=∑m=0n−1m+nm!m+1kS2n,m.$

From (23), we have

$ddtLik1−1+λt−1λ=11−1+λt−1λ1+λt−1λ−1Lik−11−1+λt−1λ=1+λt−1λ−1∑n=0∞βn,λ(k−1)tnn!.$(27)

On the other hand,

$ddtLik1−(1+λt)−1λ=ddt1−(1+λt)−1λ11−(1+λt)−1λLik1−(1+λt)−1λ)=(1+λt)−1λ−111−(1+λt)−1λLik1−(1+λt)−1λ+1−(1+λt)−1λddt∑n=0∞⁡βn,λ(k)tnn!=(1+λt)−1λ−1∑n=0∞⁡βn,λ(k)tnn!+1−(1+λt)−1λ∑n=1∞⁡βn,λ(k)tn−1(n−1)!.$(28)

By (27) and (28), we get

$∑n=0∞βn,λ(k−1)tnn!=∑n=0∞βn,λ(k)tnn!+(1+λt)(1+λt)1λ−1∑n=1∞βn,λ(k)tn−1(n−1)!=∑n=0∞βn,λ(k)tnn!+(1+λt)1λ−1∑m=0∞βm+1,λ(k)tmm!+λ(1+λt)1λ−1∑m=0∞βm,λ(k)mtmm!=∑n=0∞βn,λ(k)tnn!∑l=1∞1λlλttll!∑m=0∞βm+1,λ(k)tmm!+λ∑l=1∞1λlλttll!∑m=0∞βm,λ(k)mtmm!=∑n=0∞βn,λ(k)tnn!+∑n=1∞∑m=0n−11λn−mλn−mn!(n−m)!m!βm+1,λ(k)tnn!+λ∑n=1∞∑m=0n−11λn−mλn−mm⋅n!(n−m)!m!βm,λ(k)tnn!=∑n=0∞βn,λ(k)tnn!+∑n=1∞∑m=0n−1nm1λn−mλn−mβm+1,λ(k)tnn!+λ∑n=1∞∑m=0n−1nm1λn−mλn−mmβm,λ(k)tnn!,$(29)

where

$(1λ)n=1λ1λ−1⋯1λ−n+1=∑l=0nS1nn,lλ−l,n≥0.$

Thus, by (29), we have

$βn,λk−1)=βn,λk)+∑m=0n−1nm(1λ)n−mλn−mβm+1,λk+λ∑m=0n−1nm(1λ)n−mλn−mmβm,λk=n+1βn,λk)+∑m=1n−1nm−1(1λ)n−m+1λn−m+1βm,λk+λ∑m=0n−1nm(1λ)n−mλn−mmβm;λ,kn≥1.$(30)

Therefore, by (30), we obtain the following theorem

For n ≥ 1, we have

$βn,λk)=1n+1βn,λk−1)−∑m=1n−1nm−1βm,λk)(1λ)n−m+1λn−m+1−λ∑m=0n−1nm(1λ)n−mλn−mmβm;λk.$

Note that

$Bn(k)=limλ→0βn,λ(k)=1n+1Bn(k−1)−∑m=1n−1nm−1Bm(k).$

Now, we observe that

$∑n=0∞(1−(1+λt)−1λ)n(n+1)−k=∑n=1∞(1−(1+λt)−1λ)nnk11−(1+λt)−1λ=11−(1+λt)−1λLik(1−1+λt)−1λ=∑n=0Xβn,λk)tnn!.$(31)

By (31), we get

$∑k=0∞∑n=0∞βn,λ(−k)xnn!ykk!=∑k=0∞∑m=0∞(1−(1+λt)−1λ)m(m+1)kykk!=∑m=0∞1−(1+λx)−1λm∑k=0∞(m+1)kykk!=∑m=0∞(1−(1+λx)−1λ)mem+1)y=∑j=0∞(−1)j(e−1λlog(1+λx)−1)jej+1)y=∑j=0∞(−1)jj!∑m=j∞S2m,j(−1)mλ−m(log1+λx)mm!e(j+1)y=∑m=0∞∑j=0m(−1)j+mj!S2m,jλ−m∑n=m∞S1n,mλnxnn!e(j+1)y=∑n=0∞∑m=0n∑j=0m(−1)j+mj!S2m;jλn−mS1n,mej+1)yxnn!=∑k=0∞∑n=0∞∑m=0n∑j=0m(−1)j+mj!λn−mS2m,jS1n,m(j+1)kxnn!ykk!.$(32)

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

For k ∈ ℤ and n ≥ 0, we have

$βn,λ(−k)=∑m=0n∑j=0m(−1)j+mj!λn−m(j+1)kS2m,jS1n,m.$

Note that

$Bn(−k)=limλ→0βn,λ(−k)=∑j=0n(−1)j+nj!(j+1)ks2(n,j).$

From Theorem 2.1, we have

$ddxβn,λ(k)(x)=∑l=0nlnβl,λ(k)ddx∏i=0n−l−1(x−iλ)=∑l=0nlnβl,λ(k)∑j=0n−l−11(x−jλ)∏i=0n−l−1(x−iλ).$

The generalized falling factorial (x | λ)n is given by

$(x|λ)n=x(x−λ)(x−2λ)⋯(x−(n−1)λ),(n≥0).$(33)

As is well known, the Bernoulli numbers of the second kind are defined by the generating funcdtion

$tlog(1+t)=∑n=0∞bntnn!,(see[20]).$(34)

We observe that

$∫01(1+λt)xλdx=∑n=0∞λn∫01xλndxtnn!=∑n=0∞∫01x|λndxtnn!.$(35)

On the other hand,

$∫01(1+λt)xλdx=λlog(1+λt)(1+λt)1λ−1=λtlog1+λt(1+λt)1λ−1t=∑m=0∞bmλmtmm!∑l=0∞(1|λ)l+1l+1tll!=∑n=0∞∑l=0n(1|λ)l+1l+1λn−lbn−lnltnn!.$(36)

From (35) and (36), we have

$∫01(x|λ)ndx=∑l=0nnlλn−lbn−1(1|λ)l+1l+1,n≥0.$(37)

By Theorem 2.1, we get

$∫01βn,λ(k)xdx=∑l=0nnlβl,λ(k)∫01xλn−lλn−ldx=∑l=0nnlβn−l,λ(k)∫01(x|λ)ldx=∑l=0n∑m=0llmλl−mbl−m(1|λ)m+1m+1nlβn−l,λ(k).$

## 3 Further remarks

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

$F=f(t)=∑k=o∞aktkk!ak∈C.$(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

$cL+c′L′|p(x)=cL|p(x)+c′L′|p(x),$

where c, c′ ∈ ℂ.

For $f\left(t\right)=\sum _{k=0}^{\mathrm{\infty }}{a}_{k}\frac{{t}^{k}}{k!}$, we define a linear functional on ℙ by setting

$f(t)|xn=anforalln≥0.$(39)

Thus, by (39), we get

$tkxn=n!δn,k,(n,k≥0),(see[4,16,20]).$(40)

For ${f}_{L}\left(t\right)=\sum _{k=0}^{\mathrm{\infty }}⟨L|{x}^{k}⟩\frac{{t}^{k}}{k!}$, by (40), we get 〈fL (i)| xn〉 = 〈L| xn〉. In addition, the mapping LfL (t) is a vector space isomorphism from ℙ* onto $\mathcal{F}$ . Henceforth, $\mathcal{F}$ 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 $\mathcal{F}$ can be regarded as both a formal power series and a linear functional. We refer to $\mathcal{F}$ 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) ∈ $\mathcal{F}$ and p (x) ∈ ℙ. Then, by (40), we get

$f(t)=∑k=o∞f(t)xktkk!,p(x)=∑k=o∞tkp(x)xkk!.$(41)

From (41), we have

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

where ${p}^{k}\left(x\right)=\frac{{d}^{k}}{d{x}^{k}}p\left(x\right)$, (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

$eyt−1tp(x)=∫xx+yp(u)du,eyt−1p(x)=p(y)−p(0).$

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

$f(t)=∑k=0∞f(t)|xkk!tk=∑k=0∞yk+1(k+1)!tk=1t(eyt−1).$(45)

Thus, for p (x) ∈ ℙ, we have

$eyt−1tp(x)=∫0yp(u)du.$(46)

It is known that

$sn(x)∼(g(t),f(t))⟺1gf¯(t)exf¯(t)=∑k=0∞sk(x)tkk!,(x∈C)$(47)

where $\overline{f}\left(t\right)$ is the compositional inverse of f (t) such that $f\left(\overline{f}\left(t\right)\right)=\overline{f}\left(f\left(t\right)\right)=t$ (see [11, 20]). From (15), we note that

$βn,λ(k)(x)∼1−e−tLik(1−e−t),1λeλt−1.$(48)

That is,

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

Thus, by (48),

$1λ(eλt−1)βn,λ(k)(x)=nβn−1,λ(k)(x).$(49)

On the other hand,

$(eλt−1)βn,λ(k)(x)=βn,λ(k)(x+λ)−βn,λ(k)(x).$(50)

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

For n ∈ ℕ, we have

$λβn−1,λ(k)(x)=1nβn,λ(k)(x+λ)−βn,λ(k)(x).$

By (46), we get

$eyt−1tβn,λ(k)(x)=∫xx+yβn,λ(k)(u)du.$(51)

From (51), we have

$eyt−1tβn,λ(k)(x)=∫0yβn,λ(k)(u)du.$(52)

Thus, by (52), we get

$et−1tβn,λ(k)(x)=∫01βn,λ(k)(u)du=∑l=0n∑m=0llmnlλl−mbl−mβn−l,λk(1|λ)m+1m+1.$(53)

Therefore, by (53), we obtain the following theorem.,

For n ≥ 0 we have

$et−1tβn,λ(k)(x)=∑l=0n∑m=0llmnlλl−mbl−mβn−l,λk(1|λ)m+1m+1.$

Note that

$et−1tBn(k)(x)=limλ→0et−1tβn,λ(k)(x)=limλ→0∫01βn,λ(k)(u)du=∑l=0nnlBn−lk)1l+1$

Let

$Pn={px∈Cx|degpx≤n},n≥0.$

For p(x) ∈ ℙn with $p\left(x\right)=\sum _{m=0}^{n}{a}_{m}{\beta }_{m,\lambda }^{\left(k\right)}\left(x\right)$, we have

$1−e−tLik1−e−t1λeλt−1mpx=∑l=0nal1−e−tLik1−e−t1λeλt−1mβl,λk)x$(54)

From (48), we note that

$1−e−tLik1−e−t1λeλt−1mβl,λk)x=l!δl,m.$(55)

By (54) and (55), we get

$am=1m!1−e−tLik1−e−t1λeλt−1mpx,(m≥0).$(56)

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

For p (x) ∈ ℙn, we have

$px=∑m=0namβm,λ(k)(x),$

where

$am=1m!1−e−tLik1−e−t1λeλt−1mpx.$

For example, let us take $p\left(x\right)={\beta }_{n}^{\left(k\right)}\left(x\right)\left(n\ge 0\right)$. Then, by Theorem 3.3, we have

$Bn(k)x=∑m=0namβm,λ(k)(x),$(57)

where

$am=1m!1−e−tLik1−e−t1λeλt−1mBn(k)x=1m!1−e−tLik1−e−t1λeλt−1mLik1−e−t1−e−txn=λ−mm!eλt−1mxn=λ−m∑l=m∞S2l,mλll!tlxn=λn−mS2n,m.$(58)

From (57) and (58), we have

$Bn(k)x=∑m=0nλn−mS2n,mβm,λ(k)x.$(59)

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Accepted: 2016-06-29

Published Online: 2016-07-26

Published in Print: 2016-01-01

Citation Information: Open Mathematics, ISSN (Online) 2391-5455,

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