# Umbral Calculus and the Frobenius-Euler Polynomials

Authors: Dae San Kim; Taekyun Kim; Sang-Hun Lee
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We study some properties of umbral calculus related to the Appell sequence. From those properties, we derive new and interesting identities of the Frobenius-Euler polynomials.


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## Body

## 1. Introduction

LetC be the complex number field. For λ∈C with λ≠1, the Frobenius-Euler polynomials are defined by the generating function to be
(1)1-λet-λext=eH(x∣λ)t=∑n=0∞‍Hn(x∣λ)tnn!,
(see [1–5]) with the usual convention about replacing Hn(x∣λ) by Hn(x∣λ).In the special case,x=0,Hn(0∣λ)=Hn(λ) are called the nth Frobenius-Euler numbers. By (1), we get
(2)Hn(x∣λ)=∑l=0n‍(nl)Hn-l(λ)xl=(H(λ)+x)n,
(see [6–9]) with the usual convention about replacing Hn(λ) by Hn(λ).Thus, from (1) and (2), we note that
(3)(H(λ)+1)n-λHn(λ)=(1-λ)δ0,n,
where δn,k is the kronecker symbol (see [1, 10, 11]).Forr∈Z+, the Frobenius-Euler polynomials of order r are defined by the generating function to be
(4)(1-λet-λ)rext=(1-λet-λ)×⋯×(1-λet-λ)︸extr-times=∑n=0∞‍Hn(r)(x∣λ)tnn!.
In the special case, x=0,Hn(r)(0∣λ)=Hn(r)(λ) are called the nth Frobenius-Euler numbers of order r (see [1, 10]).From (4), we can derive the following equation:
(5)Hn(r)(x∣λ)=∑l=0n‍(nl)Hn-l(r)(λ)xl,Hn(r)(λ)=∑l1+⋯+lr=n‍(nl1,…,lr)Hl1(λ)⋯Hlr(λ).
By (5), we see that Hn(r)(x∣λ) is a monic polynomial of degree n with coefficients in Q(λ).Letℙ be the algebra of polynomials in the single variable x over C and let ℙ* be the vector space of all linear functionals on ℙ. As is known, 〈L∣p(x)〉 denotes the action of the linear functional L on a polynomial p(x) and we remind that the addition and scalar multiplication on ℙ* are, respectively, defined by
(6)〈L+M∣p(x)〉=〈L∣p(x)〉+〈M∣p(x)〉,〈cL∣p(x)〉=c〈L∣p(x)〉,
where c is a complex constant (see [3, 12]).LetF denote the algebra of formal power series:
(7)F={f(t)=∑k=0∞‍akk!tk∣ak∈C}
(see [3, 12]). The formal power series define a linear functional on ℙ by setting
(8)〈f(t)∣xn〉=an,∀n≥0.
Indeed, by (7) and (8), we get
(9)〈tk∣xn〉=n!δn,k(n,k≥0)
(see [3, 12]). This kind of algebra is called an umbral algebra.The orderO(f(t)) of a nonzero power series f(t) is the smallest integer k for which the coefficient of tk does not vanish. A series f(t) for which O(f(t))=1 is said to be an invertible series (see [2, 12]). For f(t),g(t)∈F, and p(x)∈ℙ, we have
(10)〈f(t)g(t)∣p(x)〉=〈f(t)∣g(t)p(x)〉=〈g(t)∣f(t)p(x)〉
(see [12]). One should keep in mind that each f(t)∈F plays three roles in the umbral calculus: a formal power series, a linear functional, and a linear operator. To illustrate this, let p(x)∈ℙ and f(t)=eyt∈F. As a linear functional, eyt satisfies 〈eyt∣p(x)〉=p(y). As a linear operator, eyt satisfies eytp(x)=p(x+y) (see [12]). Let sn(x) denote a polynomial in x with degree n. Let us assume that f(t) is a delta series and g(t) is an invertible series. Then there exists a unique sequence sn(x) of polynomials such that 〈g(t)f(t)k∣sn(x)〉=n!δn,k for all n,k≥0 (see [3, 12]). This sequence sn(x) is called the Sheffer sequence for (g(t),f(t)) which is denoted by sn(x)~(g(t),f(t)). If sn(x)~(1,f(t)), then sn(x) is called the associated sequence for f(t). If sn(x)~(g(t),t), then sn(x) is called the Appell sequence.Letsn(x)~(g(t),f(t)). Then we see that
(11)h(t)=∑k=0∞〈h(t)∣sk(x)〉k!g(t)f(t)k,h(t)∈F,p(x)=∑k=0∞〈g(t)f(t)k∣p(x)〉k!sk(x),p(x)∈ℙ,f(t)sn(x)=nsn-1(x),〈f(t)∣p(αx)〉=〈f(αt∣p(x)〉,(12)1g(f-(t))eyf-(t)=∑k=0∞sk(y)k!tk,∀y∈C,
where f-(t) is the compositional inverse of f(t) (see [3]). In this paper, we study some properties of umbral calculus related to the Appell sequence. For those properties, we derive new and interesting identities of the Frobenius-Euler polynomials.


## 2. The Frobenius-Euler Polynomials and Umbral Calculus

By (4) and (12), we see that
(13)Hn(r)(x∣λ)~((et-λ1-λ)r,t).
Thus, by (13), we get
(14)〈(et-λ1-λ)rtk∣Hn(r)(x∣λ)〉=n!δn,k.
Let
(15)ℙn(λ)={p(x)∈Q(λ)[x]∣degp(x)≤n}.
Then it is an (n+1)-dimensional vector space over Q(λ).So we see that{H0(r)(x∣λ),H1(r)(x∣λ),…,Hn(r)(x∣λ)} is a basis for ℙn(λ). For p(x)∈ℙn(λ), let
(16)p(x)=∑k=0n‍CkHk(r)(x∣λ),(n≥0).
Then, by (13), (14), and (16), we get
(17)〈(et-λ1-λ)rtk∣p(x)〉=∑l=0n‍Cl〈(et-λ1-λ)rtk∣Hl(r)(x∣λ)〉=∑l=0n‍Cll!δl,k=k!Ck.
From (17), we have
(18)Ck=1k!〈(et-λ1-λ)rtk∣p(x)〉=1k!〈(et-λ1-λ)r∣Dkp(x)〉=1k!(1-λ)r∑j=0r‍(rj)(-λ)r-j〈ejt∣Dkp(x)〉=1k!(1-λ)r∑j=0r‍(rj)(-λ)r-j〈t0∣ejtDkp(x)〉=1k!(1-λ)r∑j=0r‍(rj)(-λ)r-j〈t0∣Dkp(x+j)〉.
Therefore, by (16) and (18), we obtain the following theorem.Theorem 1.
Forp(x)∈ℙn(λ), let
(19)p(x)=∑k=0n‍CkHk(r)(x).
Then one has
(20)Ck=1k!(1-λ)r∑j=0r‍(rj)(-λ)r-jDkp(j),
where Dp(x)=dp(x)/dx.From Theorem1, we note that
(21)p(x)=1(1-λ)r·∑k=0n‍{∑j=0r‍1k!(rj)(-λ)r-jDkp(j)}Hk(r)(x∣λ).
Let us consider the operator Δ~λ with Δ~λf(x)=f(x+1)-λf(x) and let Jλ=(1/(1-λ))Δ~λ. Then we have
(22)Jλ(f)(x)=11-λ{f(x+1)-λf(x)}.
Thus, by (22), we get
(23)Jλ(Hn(r)(x∣λ))=11-λ{Hn(r)(x+1∣λ)-λHn(r)(x∣λ)}.
From (4), we can derive
(24)∑n=0∞‍{Hn(r)(x+1∣λ)-λHn(r)(x∣λ)}tnn!=(1-λet-λ)re(x+1)t-λ(1-λet-λ)rext=(1-λet-λ)rext(et-λ)=(1-λ)(1-λet-λ)r-1ext=(1-λ)∑n=0∞‍Hn(r-1)(x∣λ)tnn!.
By (23) and (24), we get
(25)Jλ(Hn(r)(x∣λ))=Hn(r-1)(x∣λ).
From (25), we have
(26)Jλr(Hn(r)(x∣λ))=Jλr-1(Hn(r-1)(x∣λ))=⋯=Hn(0)(x∣λ)=xn,Jλr(xn)=JλrHn(0)(x∣λ)=Hn(-r)(x∣λ)=Jλ2rHn(r)(x∣λ).
For s∈Z+, from (25), we have
(27)Jλs(Hn(r)(x∣λ))=Hn(r-s)(x∣λ).
On the other hand, by (12), (13), and (25),
(28)Jλs(Hn(r)(x∣λ))=(et-λ1-λ)s(Hn(r)(x∣λ))=1(1-λ)s((1-λ)+∑k=1∞‍tkk!)s·(Hn(r)(x∣λ)).
Thus, by (28), we get
(29)Jλs(Hn(r)(x∣λ))=∑m=0s‍(sm)(1-λ)m∑l=m∞(∑k1+⋯+km=lkj≥1‍1k1!⋯km!)tl(Hn(r)(x∣λ))=∑m=0s‍(sm)(1-λ)m∑l=m∞‍1l!(∑k1+⋯+km=lkj≥1‍(lk1,…,km)Dl)·Hn(r)(x∣λ)=∑m=0min{s,n}‍(sm)(1-λ)m∑l=mn‍(nl)∑k1+⋯+km=lkj≥1‍(lk1,…,km)Hn-l(r)(x∣λ)=∑l=0min{s,n}‍{(nl)∑m=0l(sm)(1-λ)m∑l=0min{s,n}‍·∑k1+⋯+km=lkj≥1(lk1,…,km)}Hn-l(r)(x∣λ)+∑l=min{s,n}+1n{(nl)∑m=0min{s,n}‍(sm)(1-λ)m∑m=0l=min{s,n}+1‍·∑k1+⋯+km=lkj≥1(lk1,…,km)}Hn-l(r)(x∣λ).
Therefore, by (27) and (29), we obtain the following theorem.Theorem 2.
For anyr,s≥0, one has
(30)Hn(r-s)(x∣λ)=∑l=0min{s,n}‍{(nl)∑m=0l‍(sm)(1-λ)m∑k1+⋯+km=lkj≥1‍(lk1,…,km)}·Hn-l(r)(x∣λ)+∑l=min{s,n}+1n{(nl)∑m=0min{s,n}‍(sm)(1-λ)mbbbbbbbbb·∑k1+⋯+km=lkj≥1‍(lk1,…,km)}Hn-l(r)(x∣λ).Let us takes=r-1(r≥1) in Theorem 2. Then we obtain the following corollary.Corollary 3.
Forn≥0,r≥1, one has
(31)Hn(x∣λ)=∑l=0min{r-1,n}‍{(nl)∑m=0l‍(r-1m)(1-λ)m∑k1+⋯+km=lkj≥1‍(lk1,…,km)}·Hn-l(r)(x∣λ)+∑l=min{r-1,n}+1n‍{(nl)∑m=0min{r-1,n}‍(r-1m)(1-λ)m·∑k1+⋯+km=lkj≥1‍(lk1,…,km)}Hn-l(r)(x∣λ).Let us takes=r(r≥1) in Theorem 2. Then we obtain the following corollary.Corollary 4.
Forn≥0,r≥1, one has
(32)xn=∑l=0min{r,n}‍{(nl)∑m=0l‍(rm)(1-λ)m∑k1+⋯+km=lkj≥1‍(lk1,…,km)}·Hn-l(r)(x∣λ)+∑l=min{r,n}+1n‍{(nl)∑m=0min{r,n}‍(rm)(1-λ)m∑l=min{r,n}+1n‍·∑k1+⋯+km=lkj≥1‍(lk1,…,km)}Hn-l(r)(x∣λ).Now, we define the analogue of Stirling numbers of the second kind as follows:(33)Sλ(n,k)=1k!∑j=0k‍(kj)(-λ)k-jjn,(n,k≥0).
Note that S1(n,k)=S(n,k) is the Stirling number of the second kind.From the definition ofΔ~λ, we have
(34)Δ~λnf(0)=∑k=0n‍(nk)(-λ)n-kf(k).
By (33) and (34), we get
(35)Sλ(n,k)=1k!Δ~λk0n,(n,k≥0).
Let us take s=2r. Then we have
(36)Jλrxn=Hn(-r)(x∣λ)=∑l=0min{2r,n}‍{(nl)∑m=0l‍(2rm)(1-λ)m∑k1+⋯+km=lkj≥1‍(lk1,…,km)}·Hn-l(r)(x∣λ)+∑l=min{2r,n}+1n‍{(nl)∑m=0min{2r,n}‍(2rm)(1-λ)m∑l=min{2r,n}+1n‍·∑k1+⋯+km=lkj≥1‍(lk1,…,km)}Hn-l(r)(x∣λ),Jλrxn=(11-λΔ~λ)rxnJλrxn=1(1-λ)r∑j=0r‍(rj)(-λ)r-j(x+j)n.
By (36), we get
(37)1(1-λ)r∑j=0r‍(rj)(-λ)r-j(x+j)n=1(1-λ)rΔ~λrxn=∑l=0min{2r,n}‍{(nl)∑m=0l‍(2rm)(1-λ)m∑k1+⋯+km=lkj≥1‍(lk1,…,km)}·Hn-l(r)(x∣λ)+∑l=min{2r,n}+1n‍{(nl)∑m=0min{2r,n}‍(2rm)(1-λ)m∑l=min{2r,n}+1n‍·∑k1+⋯+km=lkj≥1‍(lk1,…,km)}Hn-l(r)(x∣λ).
Let us take x=0 in (37). Then we obtain the following theorem.Theorem 5.
We have(38)r!(1-λ)rSλ(n,r)=r!(1-λ)rΔ~λr0nr!=∑l=0min{2r,n}‍{(nl)∑m=0l‍(2rm)(1-λ)m∑k1+⋯+km=lkj≥1‍(lk1,…,km)}·Hn-l(r)(λ)+∑l=min{2r,n}+1n{(nl)∑m=0min{2r,n}‍(2rm)(1-λ)m·∑k1+⋯+km=lkj≥1‍(lk1,…,km)}Hn-l(r)(λ)=∑m=0min{r,n}‍(rm)(1-λ)m∑k1+⋯+km=nkj≥1‍(nk1,…,km).Let us considers=2r-1 in the identity of Theorem 2. Then we have
(39)Jλr-1xn=Hn-(r-1)(x∣λ)=∑l=0min{2r-1,n}‍{(nl)∑m=0l‍(2r-1m)(1-λ)m∑k1+⋯+km=lkj≥1‍(lk1,…,km)}·Hn-l(r)(x∣λ)+∑l=min{2r-1,n}+1n‍{(nl)∑m=0min{2r-1,n}‍(2r-1m)(1-λ)m+∑l=min{2r-1,n}+1n‍·∑k1+⋯+km=lkj≥1‍(lk1,…,km)}Hn-l(r)(x∣λ)=1(1-λ)r-1∑j=0r-1‍(r-1j)(-λ)r-1-j(x+j)n=1(1-λ)r-1Δ~λr-1xn.
Let us take x=0 in (39). Then we obtain the following theorem.Theorem 6.
Forn≥0 and r≥1, one has
(40)(r-1)!(1-λ)r-1Sλ(n,r-1)=(r-1)!(1-λ)r-1Δ~λr-10n(r-1)!=∑l=0min{2r-1,n}‍{(nl)∑m=0l‍(2r-1m)(1-λ)m∑k1+⋯+km=lkj≥1‍(lk1,…,km)}·Hn-l(r)(λ)+∑l=min{2r-1,n}+1n‍{(nl)∑m=0min{2r-1,n}‍(2r-1m)(1-λ)m∑l=min{2r-1,n}+1n‍·∑k1+⋯+km=lkj≥1‍(lk1,…,km)}Hn-l(r)(λ).Remark 7.
Note that(41)(r-1)!(1-λ)r-1Sλ(n,r-1)=∑l=0min{r,n}‍{(nl)∑m=0l‍(rm)(1-λ)m∑k1+⋯+km=lkj≥1‍(lk1,…,km)}·Hn-l(λ)+∑l=min{r,n}+1n‍{(nl)∑m=0min{r,n}‍(rm)(1-λ)m·∑k1+⋯+km=lkj≥1‍(lk1,…,km)}Hn-l(λ).



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