# Differential Equation and Recursive Formulas of Sheffer Polynomial Sequences

Authors: Heekyung Youn; Yongzhi Yang
Journal: ISRN Discrete Mathematics
 (2011)

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## Abstract

We derive a differential equation and recursive formulas of Sheffer polynomial sequences utilizing matrix algebra. These formulas provide the defining characteristics
of, and the means to compute, the Sheffer polynomial sequences. The tools we use are
well-known Pascal functional and Wronskian matrices. The properties and the relationship
between the two matrices simplify the complexity of the generating functions of Sheffer
polynomial sequences. This work extends He and Ricci's work (2002) to a broader class of
polynomial sequences, from Appell to Sheffer, using a different method. The work is self-contained.


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## Body

## 1. Introduction

Sheffer polynomial sequences arise in numerous problems of applied mathematics, theoretical physics, approximation theory, and several other mathematical branches. In the past few decades, there has been a renewed interest in Sheffer polynomials. di Bucchianico recently summarized and documented more than five hundred old and new findings related to the study of Sheffer polynomial sequences in [1]. One aspect of such study is to find a differential equation and recursive formulas for Sheffer polynomial sequences. For instance, in [2], He and Ricci developed the differential equation and recursive formula for Appell polynomials, which is a subclass of Sheffer polynomial sequences. In this paper, we derive differential equation and recursive formulas for Sheffer polynomial sequences by using matrix algebra.The remainder of the paper is organized as follows. In Section2, we define Pascal functional and Wronskian matrices for analytic functions and derive their properties and the relationships between the functional matrices. In Section 3, we develop a differential equation for Sheffer polynomial sequences and present differential equations for some well-known Sheffer polynomials such as Laguerre, Lower factorial, Exponential, and Hermite polynomials. In Section 4, we discuss three recursive formulas for Sheffer polynomial sequences. An example will illustrate how these three forms of recursive formulas are useful in their own right.


## 2. Preliminaries

Let us start with the definitions of the generalized Pascal functional matrix of an analytic function [3] and the Wronskian matrix of several analytic functions. To avoid any unnecessary confusion, we use f(k) to stand for the kth order derivative of f and use fk to represent the kth power of f in the entire paper. In addition, f(0)=f and f0=1.Definition 2.1.
Letf(t) be an analytic function. The generalized Pascal functional matrix of f(t), denoted by 𝒫n[f(t)], is an (n+1) by (n+1) matrix and is defined as
(2.1)(Pn[f(t)])i,j={(ij)f(i-j)(t)ifi≥j,0otherwise,fori,j=0,1,2,…,n.Definition 2.2.
Thenth order Wronskian matrix of f1(t),f2(t),f3(t),…, and fm(t) is an (n+1) by m matrix and is defined as
(2.2)Wn[f1(t),f2(t),f3(t),…,fm(t)]=[f1(t)f2(t)f3(t)⋯fm(t)f1′(t)f2′(t)f3′(t)⋯fm′(t)⋮⋮⋮⋱⋮f1(n)(t)f2(n)(t)f3(n)(t)⋯fm(n)(t)].We study the Pascal functional and Wronskian matrices in a neighborhood oft=0. Hence, when we mention analytic, we mean analytic near t=0.Note 1.
Often, readers will encounter expressions such as𝒫n[f(x,t)]t=0 or 𝒲n[f(x,t)]t=0. In this context, the variable t is the working variable for the Pascal functional or the Wronkian matrix and the variable x is merely a parameter.
In the following, we list some properties and relationships between the Pascal functional and Wronskian matrices that will be the main tool for our work.Property 1.
(a)𝒫n[·] and 𝒲n[·] are linear, that is, for any constants a and b, and any analytic functions f(t) and g(t),
(2.3)Pn[af(t)+bg(t)]=aPn[f(t)]+bPn[g(t)],Wn[af(t)+bg(t)]=aWn[f(t)]+bWn[g(t)].
(b) For any analytic functionsf(t) and g(t),
(2.4)Pn[f(t)]Pn[g(t)]=Pn[g(t)]Pn[f(t)]=Pn[f(t)g(t)].
Furthermore, iff(t)≠0, then (𝒫n[f(t)])-1=𝒫n[f-1(t)], where f-1(t) denotes the multiplicative inverse of f(t).
(c) For any analytic functionsf(t) and g(t),
(2.5)Pn[f(t)]Wn[g(t)]=Pn[g(t)]Wn[f(t)]=Wn[(fg)(t)].
Furthermore, for any analytic functionsg(t), and f1(t),f2(t),…, and fm(t),
(2.6)Pn[g(t)]Wn[f1(t),f2(t),…,fm(t)]=Wn[(gf1)(t),(gf2)(t),…,(gfm)(t)].
(d) For any analytic functionsg(t) and f(t) with f(0)=0 and f′(0)≠0,
(2.7)Wn[g(f(t))]t=0=Wn[1,f(t),f2(t),f3(t),…,fn(t)]t=0Λn-1Wn[g(t)]t=0,
where Λn=diag[0!,1!,2!,…,n!]. The notational convention Λn will be used throughout this paper.Proof.
The proofs of Property1(a), 1(b), and 1(c) can be found in [4].
For Property1(d), let us express the functions f(t) and g(t) as series around t=0; f(t)=∑j=0∞fjtj and g(t)=∑j=0∞gjtj. Since f(0)=0 and f′(0)≠0, the leading term of fk(t) is tk. Therefore,
(2.8)(ddt)(k)g(f(t))|t=0=(ddt)(k)∑j=0∞gjfj(t)|t=0=(ddt)(k)∑j=0kgjfj(t)|t=0
because (d/dt)(k)fj(t)|t=0=0 for all j>k. By (2.8) and noting gk=g(k)(0)/k!, we have
(2.9)Wn[g(f(t))]t=0=[100⋯00f′(0)0⋯00f′′(0)(f2)′′(0)⋯0⋮⋮⋮⋱⋮0f(n)(0)(f2)(n)(0)⋯(fn)(n)(0)][g(0)(0)0!g(1)(0)1!g(2)(0)2!⋮g(n)(0)n!]=Wn[1,f(t),f2(t),f3(t),…,fn(t)]t=0Λn-1Wn[g(t)]t=0.
This completes the proof.


## 3. Sheffer Polynomial Sequence and Its Differential Equation

Let us first define a Sheffer polynomial sequence by a pair of generation functions(g(t),f(t)) as often done [5].Definition 3.1.
Letg(t) be an invertible analytic function, that is, g(0)≠0, and f(t) be analytic function with f(0)=0 and let  f′(0)≠0 that admits compositional inverse. Let f¯(t) denote the compositional inverse of f(t). Then, {sn(x)} is the Sheffer polynomial sequence for (g(t),f(t)) if and only if
(3.1)1g(f¯(t))exf¯(t)=∑k=0∞sk(x)k!tk.Note 2.
Since(1/g(f¯(t)))exf¯(t) is analytic, by Taylor’s Theorem,
(3.2)sk(x)=(ddt)(k)1g(f̅(t))exf̅(t)|t=0.The family of Sheffer polynomial sequences contains two simpler subclasses of polynomial sequences, Appell and associated polynomial sequences [5]. An Appell polynomial sequence is a Sheffer polynomial sequence where f(t)=t. Hence, we say {an(x)} is the Appell polynomial sequence for g(t) if and only if {an(x)} is the Sheffer polynomial sequence for (g(t),t). Associated polynomial sequence is a Sheffer polynomial sequence where g(t)=1. Hence, we say {qn(x)} is the associated polynomial sequence for f(t) if and only if {qn(x)} is the Sheffer polynomial sequence for (1,f(t)).We define the Sheffer polynomial sequence in vector form to utilize Wronskian matrices.Definition 3.2.
The Sheffer vector for(g(t),f(t)), denoted by S¯n(x), is defined as
(3.3)S¯n(x)=[s0(x)s1(x)s2(x)⋯sn(x)]T,
where {sn(x)} is the Sheffer polynomial sequence for (g(t),f(t)).
As noted in Note2, the Sheffer vector can be expressed as
(3.4)S¯n(x)=[s0(x)s1(x)s2(x)⋯sn(x)]T=Wn[1g(f¯(t))exf¯(t)]t=0.
In order to derive the differential equation for Sheffer polynomial sequence, we develop the following lemma.Lemma 3.3.
Let{sn(x)} be the Sheffer polynomial sequence for (g(t),f(t)). Then,
(3.5)Wn[s0(x),s1(x),…,sn(x)]TΛn-1=Wn[1,f̅(t),f̅2(t),…,f̅n(t)]t=0Λn-1Pn[1g(t)]t=0Pn[ext]t=0.Proof.
Using (3.4) and Property 1(d), we have
(3.6)S¯n(x)=Wn[exf̅(t)g(f̅(t))]t=0=Wn[1,f̅(t),f̅2(t),…,f̅n(t)]t=0Λn-1Wn[extg(t)]t=0.
Using Property1(c) and noting 𝒲n[ext]t=0=[1xx2⋯xn]T, we obtain
(3.7)S¯n(x)=Wn[1,f̅(t),f̅2(t),…,f̅n(t)]t=0Λn-1Pn[1g(t)]t=0[1xx2⋯xn]T.
Taking thekth order derivative with respect to x on both sides of (3.7) and dividing by k! yields
(3.8)1k![s0(k)(x)s1(k)(x)s2(k)(x)⋯sn(k)(x)]T=Wn[1,f¯(t),f¯2(t),…,f¯n(t)]t=0Λn-1Pn[1g(t)]t=0×[0⋯01(k+1k)x(k+2k)x2⋯(nk)xn-k]T.
The left-hand of (3.8) is the kth column of 𝒲n[s0(x),s1(x),…,sn(x)]TΛn-1, and the right-hand of (3.8) is the kth column of Wn[1,f̅(t),f̅2(t),…,f̅n(t)]t=0Λn-1𝒫n[1/g(t)]t=0𝒫n[ext]t=0.After the introduction of definitions and the lemma, we are ready to develop differential equations for Sheffer polynomials.Theorem 3.4.
Let{sn(x)} be the Sheffer polynomial sequence for (g(t),f(t)). Then, it satisfies the following differential equation:
(3.9)∑k=1n(βkx+αk)sn(k)(x)k!-nsn(x)=0,
where βk=(f(t)/f′(t))(k)|t=0 and αk=(-g′(t)f(t)/g(t)f′(t))(k)|t=0.Proof.
Let us consider𝒲n[t(d/dt)(exf¯(t)/g(f¯(t)))]t=0. On one hand, by Property 1(c), we have
(3.10)Wn[tddt(exf¯(t)g(f¯(t)))]t=0=Pn[t]t=0Wn[ddt(exf¯(t)g(f¯(t)))]t=0=[0000⋯0001000⋯0000200⋯0000030⋯000⋮⋮⋮⋱⋱⋮⋮⋮⋱⋱0000⋯n-1000000⋯0n0][s1(x)s2(x)s3(x)⋮sn(x)sn+1(x)].
On the other hand, by Properties1(c) and 1(d) and Lemma 3.3, we have(3.11)Wn[tddt(exf¯(t)g(f¯(t)))]t=0=Wn[(x-g′(f¯(t))g(f¯(t)))tf′(f¯(t))exf¯(t)g(f¯(t))]t=0=Wn[1,f¯(t),…,f¯n(t)]t=0Λn-1Wn[(x-g′(t)g(t))f(t)f′(t)extg(t)]t=0=Wn[1,f¯(t),…,f¯n(t)]t=0Λn-1Pn[1g(t)]t=0Pn[ext]t=0×Wn[xf(t)f′(t)-g′(t)f(t)g(t)f′(t)]t=0=Wn[s0(x),s1(x),…,sn(x)]TΛn-1Wn[xf(t)f′(t)-g′(t)f(t)g(t)f′(t)]t=0.
Thus,(3.12)Wn[tddt(exf¯(t)g(f¯(t)))]t=0=[s0(x)00⋯0s1(x)s1′(x)1!0⋯0s2(x)s2′(x)1!s2′′(x)2!⋯0⋮⋮⋮⋱⋮sn(x)sn′(x)1!sn′′(x)2!⋯sn(n)(x)n!][β0x+α0β1x+α1β2x+α2⋮βnx+αn].
Equating the last rows of (3.10) and (3.12), we get
(3.13)nsn(x)=∑k=0nsn(k)(x)k!(βkx+αk).
Sincef(0)=0, α0=β0=0, a rearrangement of (3.13) produces the desired result.The following corollaries are immediate consequences of Theorem3.4. When g(t)=1, αk=0 for all k>0, and we have a differential equation for associated polynomials.Corollary 3.5.
Let{qn(x)} be the associated polynomial sequence for f(t). Then, it satisfies the following differential equation:
(3.14)∑k=1nβkxqn(k)(x)k!-nqn(x)=0,
where βk=(f(t)/f′(t))(k)|t=0.Settingf(t)=t in Theorem 3.4, we get a differential equation for Appell polynomials.Corollary 3.6.
Let{an(x)} be the Appell polynomial sequence for g(t). Then, it satisfies the following differential equation:
(3.15)∑k=2nα¯k-1(k-1)!an(k)(x)+(x+α¯0)an′(x)-nan(x)=0,
where α¯k=(-g′(t)/g(t))(k)|t=0.Proof.
Sincef(t)=t, β1=1 and βk=0 for all k≠1. Furthermore,
(3.16)αk=(-tg′(t)g(t))(k)|t=0=k(-g′(t)g(t))(k-1)|t=0=kα¯k-1.
Hence, the above differential equation follows from Theorem3.4.Remark 3.7.
The above differential equation for Appell is equivalent to the one in Theorem  2.1 in [2].
Let us apply Theorem3.4 to derive differential equations of some well-known Sheffer polynomials. For the sake of brevity of context, we leave the detailed calculations of examples in the paper to interested readers.Example 3.8.
LetLn{a}(x) denote the Laguerre polynomial of order a, which is the Sheffer polynomial for (g(t),f(t))=((1-t)-a-1,t/(t-1)). Ln{a}(x)=∑k=0n(n+an-k)(n!/k!)(-x)k satisfies the following differential equation:
(3.17)x(Ln{a}(x))′′-(x-a-1)(Ln{a}(x))′+nLn{a}(x)=0.
In particular fora=0, Ln{0}(x)=Ln(x), generally known as the Laguerre polynomial, satisfies
(3.18)xLn′′(x)-(x-1)Ln′(x)+nLn(x)=0.Example 3.9.
Letpn(x) denote the Poisson-Charlier polynomial of order a, which is the Sheffer polynomial for (g(t),f(t))=(ea(et-1),a(et-1)). pn(x)=∑k=0n(nk)(-1)n-ka-k(x)k, where (x)k=x(x-1)(x-2)⋯(x-k+1) is lower factorial polynomial, satisfies the following differential equation:
(3.19)∑k=1n((-1)k+1x-a)pn(k)(x)k!-npn(x)=0.Example 3.10.
The Actuarial polynomial, denoted byAn(x), is the Sheffer polynomial for (g(t),f(t))=((1-t)-b,ln(1-t)). An(x)=∑k=0n(bk)(∑j=knS(n,j)(j)k(-x)j-k), where S(n,k) is the Stirling number of the second kind and (j)k is the lower factorial polynomial. Then,
(3.20)∑k=2n(xk(k-1)+bk)An(k)(x)-(x-b)An′(x)+nAn(x)=0,forn≥2.Example 3.11.
The Lower factorial polynomial is the associated polynomial sequence forf(t)=eat-1. Lower factorial polynomial wn(x)=(x/a)n=(x/a)(x/a-1)⋯(x/a-n+1) satisfies the following differential equation:
(3.21)x∑k=1n(-a)k-1k!wn(k)(x)-nwn(x)=0.Example 3.12.
The exponential polynomial is the associated polynomial sequence forf(t)=ln(1+t). Let ϕn(x) denote the exponential polynomial and ϕn(x)=∑k=0nS(n,k)xk, where S(n,k) is the Stirling number of the second kind. Then,
(3.22)x∑k=2n(-1)kϕn(k)(x)k(k-1)+xϕn′(x)-nϕn(x)=0,forn≥2.Example 3.13.
The Hermite polynomial of orderv is the Appell polynomial sequence for g(t)=evt2/2. Let us denote the Hermite polynomial of order v as Hn{v}(x) and Hn{v}(x)=∑k=0⌊n/2⌋(-v/2)k((n)2k/k!)xn-2k. Then
(3.23)v(Hn{v}(x))′′-x(Hn{v}(x))′+nHn{v}(x)=0.


## 4. Recurrence Relations for the Sheffer Polynomials

Finding recursive formulas is one of main interests on study of the Sheffer polynomial sequences. For instance, Lehmer in [6] developed six recursive relations for the Bernoulli polynomial sequence (one of Appell polynomial sequences). In this section, we derive three recursive formulas. The first formula expresses sn+1(x) in terms of sn(x) and its derivatives, and the second and third formulas express sn+1(x) in terms of sk(x) for k=0,1,…,n.


### 4.1. The First Recursive Formula for the Sheffer Polynomials

Theorem 4.1 (Recursive Formula I).
Let{sn(x)} denote the Sheffer polynomial sequence for (g(t),f(t)). Then, s0(x)=1/g(0) and
(4.1)sn+1(x)=∑k=0n(γkx+δk)sn(k)(x)k!,forn≥0,
where γk=(1/f′(t))(k)|t=0 and δk=(-g′(t)/g(t)f′(t))(k)|t=0.Proof.
The proof is similar to the proof of Theorem3.4. Let us consider 𝒲n[(d/dt)(exf¯(t)/g(f¯(t)))]t=0. On one hand,
(4.2)Wn[ddt(exf¯(t)g(f¯(t)))]t=0=[s1(x)s2(x)s3(x)⋯sn+1(x)]T.
(17
On the other hand, by Properties1(c) and 1(d) and Lemma 3.3, we have(4.3)Wn[ddt(exf¯(t)g(f¯(t)))]t=0=Wn[(x-g′(f¯(t))g(f¯(t)))1f′(f¯(t))exf¯(t)g(f¯(t))]t=0=Wn[1,f¯(t),…,f¯n(t)]t=0Λn-1Pn[1g(t)]t=0Pn[ext]t=0×Wn[x1f′(t)-g′(t)g(t)f′(t)]t=0=Wn[s0(x),s1(x),…,sn(x)]TΛn-1Wn[x1f′(t)-g′(t)g(t)f′(t)]t=0.
Thus,(4.4)Wn[ddt(exf¯(t)g(f¯(t)))]t=0=[s0(x)00⋯0s1(x)s1′(x)1!0⋯0s2(x)s2′(x)1!s2′′(x)2!⋯0⋮⋮⋮⋱⋮sn(x)sn′(x)1!sn′′(x)2!⋯sn(n)(x)n!][γ0x+δ0γ1x+δ1γ2x+δ2⋮γnx+δn].
Equating the last rows of (17) and (4.4), we get the desired result.Example 4.2.
ForAn(x)=∑k=0n(bk)(∑j=knS(n,j)(j)k(-x)j-k), the Actuarial polynomial,
(4.5)A0(x)=1,An+1(x)=(-x+b)An(x)+xAn′(x),forn≥0.Example 4.3.
LetLn(x) denote the Laguerre polynomial. Then,
(4.6)L0(x)=1,Ln+1(x)=(-x+1)Ln(x)+(2x-1)Ln′(x)-xLn′′(x),forn≥0.Example 4.4.
Hermite polynomial is defined as the Sheffer polynomial sequence for(g(t),f(t))=(et2/4,t/2) Let us denote the polynomial as Hn(x). Then,
(4.7)H0(x)=1,Hn+1(x)=2xHn(x)-Hn′(x),forn≥0.Corollary 4.5 (Recursive Formula I for associated polynomial sequences).
Let{qn(x)} denote the associated polynomial sequence for f(t). Then, q0(x)=1 and
(4.8)qn+1(x)=x∑k=0nγkk!qn(k)(x),forn≥0,
where γk=(1/f′(t))(k)|t=0.Proof.
It follows from Theorem4.1 since g(t)=1 and hence δk=0 for all k.Example 4.6.
For the exponential polynomialϕn(x),
(4.9)ϕ0(x)=1,ϕn+1(x)=x[ϕn(x)+ϕn′(x)],forn≥0.Corollary 4.7 (Recursive Formula I for Appell polynomial sequences).
Let{an(x)} denote the Appell polynomial sequence for g(t). Then, a0(x)=1/g(0) and
(4.10)an+1(x)=xan(x)+∑k=0nδkk!an(k)(x),forn≥0,
where δk=(-g′(t)/g(t))(k)|t=0.Proof.
Sincef(t)=t, δk=(-g′(t)/g(t)f′(t))(k)|t=0=(-g′(t)/g(t))(k)|t=0,γ0=1,  and γk=0 for all other k in Theorem 4.1.Example 4.8.
For the Hermite polynomial of orderv, denoted by Hn{v}(x),
(4.11)H0{v}(x)=1,Hn+1{v}(x)=xHn{v}(x)-v(Hn{v})′(x),forn≥0.Example 4.9.
Let us consider the Stirling polynomial sequenceSn(x), which is the Sheffer polynomial sequence for (g(t),f¯(t))=(e-t,ln(t/1-e-t)). To obtain the recursive formula for Sn(x) by Theorem 4.1, we have to find f(t), that is, solve the insolvable transcendental equation
(4.12)et=y1-e-y
for y.This example shows a type of Sheffer sequences for which Theorem4.1 fails to produce a recursive formula. This motivate us to develop other recursive formulas, which represent sn+1(x) in term of its previous terms sk(x) and the derivatives of f¯(t) and g(t).


### 4.2. The Second Recursive Formula for the Sheffer Polynomials

Theorem 4.10 (Recursive Formula II).
Let{sn(x)} be the Sheffer polynomial sequence for (g(t),f(t)). Then, s0(x)=1/g(0) and
(4.13)ϵ0sn+1(x)=xsn(x)+∑k=0n(nk)θksn-k(x)-∑k=1n(nk)ϵksn+1-k(x),forn≥0,
where ϵk=(f′(f¯(t)))(k)|t=0=(1/f̅′(t))(k)|t=0andθk=(-g′(f¯(t))/g(f¯(t)))(k)|t=0.Proof.
Let us consider𝒲n[f′(f¯(t))(d/dt)(exf¯(t)/g(f¯(t)))]t=0. On one hand, applying Property 1(c),(4.14)Wn[f′(f¯(t))ddt(exf¯(t)g(f¯(t)))]t=0=Pn[ddt(exf¯(t)g(f¯(t)))]t=0Wn[f′(f¯(t))]t=0=[s1(x)00⋯0s2(x)s1(x)0⋯0s3(x)(21)s2(x)s1(x)⋯0⋮⋮⋮⋱⋮sn+1(x)(n1)sn(x)(n2)sn-1(x)⋯s1(x)][ϵ0ϵ1ϵ2⋮ϵn].
On the other hand, by Properties1(a) and 1(c),
(4.15)Wn[f′(f¯(t))ddt(exf¯(t)g(f¯(t)))]t=0=Wn[(x-g′(f¯(t))g(f¯(t)))exf¯(t)g(f¯(t))]t=0=xWn[exf¯(t)g(f¯(t))]t=0-Pn[exf¯(t)g(f¯(t))]t=0Wn[g′(f¯(t))g(f¯(t))]t=0=x[s0(x)s1(x)s2(x)⋮sn(x)]+[s0(x)00⋯0s1(x)s0(x)0⋯0s2(x)(21)s1(x)s0(x)⋯0⋮⋮⋮⋱⋮sn(x)(n1)sn-1(x)(n2)sn-2(x)⋯s0(x)][θ0θ1θ2⋮θn].
Equating the last rows of (4.14) and (4.15) leads to
(4.16)∑k=0n(nk)sn+1-k(x)ϵk=xsn(x)+∑k=0n(nk)sn-k(x)θk.
A rearrangement of the above yields the desired result.Example 4.11.
LetLn{a}(x) be the Laguerre polynomial of order a, pn(x) the Poisson-Charlier polynomial of order a, and Mn(x) the Meixner polynomial of the first kind of order (b,c), which is the Sheffer polynomial sequence for (g(t),f(t))=(((1-c)/(1-cet))b,(1-et)/(c-1-et)). Then,
(4.17)L0{a}(x)=1,Ln+1{a}(x)=(2n+a+1-x)Ln{a}(x)-(n2+na)Ln-1{a}(x),forn≥0,p0(x)=1,apn+1(x)=(x-n-a)pn(x)-npn-1(x),forn≥0,M0(x)=1,cMn+1(x)=((c-1)x+n(c+1)+bc)Mn(x)-(n(n-1)+nb)Mn-1(x),forn≥0.Corollary 4.12 (Recursive Formula II for associated polynomial sequences).
Let{qn(x)} denote the associated polynomial sequence for f(t). Then, q0(x)=1 and
(4.18)ϵ0qn+1(x)=xqn(x)-∑k=1n(nk)ϵkqn+1-k(x),forn≥0,
where ϵk=(f′(f¯(t)))(k)|t=0.Example 4.13.
Letϕn(x) denote the exponential polynomial. Then,
(4.19)ϕ0(x)=1,ϕn+1(x)=xϕn(x)+∑k=0n-1(nk+1)(-1)kϕn-k(x),forn≥0.Corollary 4.14 (Recursive Formula II for Appell polynomial sequences).
Let{an(x)} denote the Appell polynomial sequence for g(t). Then, a0(x)=1/g(0) and
(4.20)an+1(x)=xan(x)+∑k=0n(nk)θkan-k(x),forn≥0,
where θk=(-g′(t)/g(t))(k)|t=0.Remark 4.15.
Theorem  2.2 and Theorem  2.4 in [2] are special cases of Corollary 4.14.Example 4.16.
LetHn{v}(x) denote the Hermite polynomial of order v. Then,
(4.21)H0{v}(x)=1,Hn+1{v}(x)=xHn{v}(x)-nvHn-1{v}(x),forn≥0.Here, we would like to revisit Example4.9. In order to obtain the recursive formula for the Stirling polynomial sequence {Sn(x)} by Theorem 4.10, we need to compute
(4.22)ϵk=(f′(f¯(t)))(k)|t=0=(1f̅′(t))(k)|t=0=(t(et-1)et-1-t)(k)|t=0.Equation  (4.22) seems difficult to evaluate and does not lead to any nice result. This is a motivation for us to develop yet another formula in Theorem 4.17.


### 4.3. The Third Recursive Formula for the Sheffer Polynomials

Theorem 4.17 (Recursive Formula III).
Let{sn(x)} denote the Sheffer polynomial sequence for (g(t),f(t)). Then, s0(x)=1/g(0) and
(4.23)sn+1(x)=∑k=0n(nk)(xαk+βk)sn-k(x),forn≥0,
where αk=(1/f′(f¯(t)))(k)|t=0=(f¯(t))(k+1)|t=0andβk=(-g′(f¯(t))/g(f¯(t))f′(f¯(t)))(k)|t=0.Proof.
We have(4.24)[s1(x)s2(x)s3(x)⋯sn+1(x)]T=Wn[ddt(exf¯(t)g(f¯(t)))]t=0.
Also,(4.25)Wn[ddt(exf¯(t)g(f¯(t)))]t=0=Wn[(x-g′(f¯(t))g(f¯(t)))1f′(f¯(t))exf¯(t)g(f¯(t))]t=0=Pn[(x-g′(f¯(t))g(f¯(t)))1f′(f¯(t))]t=0Wn[exf¯(t)g(f¯(t))]t=0=[xα0+β000⋯0xα1+β1xα0+β00⋯0xα2+β22(xα1+β1)xα0+β0⋯0⋮⋮⋮⋱⋮xαn+βn(n1)(xαn-1+βn-1)(n2)(xαn-2+βn-2)⋯xα0+β0][s0(x)s1(x)s2(x)⋮sn(x)].
Equating the last rows of (4.24) and (4.25), we get the desired result.Example 4.18.
LetAn(x) be the Actuarial polynomial. Then,
(4.26)A0(x)=1,An+1(x)=-x∑k=0n(nk)Ak(x)+bAn(x),forn≥0.Corollary 4.19 (Recursive Formula III for associated polynomial sequences).
Let{qn(x)} denote the associated polynomial sequence for f(t). Then, q0(x)=1 and
(4.27)qn+1(x)=x∑k=0n(nk)αkqn-k(x),forn≥0,
where αk=(1/f′(f¯(t)))(k)|t=0=(f¯(t))(k+1)|t=0.Example 4.20.
Letϕn(x) denote the exponential polynomial. By Corollary 4.19, we have the following well-known result
(4.28)ϕ0(x)=1,ϕn+1(x)=x∑k=0n(nk)ϕk(x),forn≥0.Finally, let us finish up Example4.9 and conclude this paper.To obtain the recursive formula for the Stirling polynomial sequence{Sn(x)}by Theorem 4.17, we need to compute(4.29)αk=(1f′(f¯(t)))(k)|t=0=(f¯(t))(k+1)|t=0=(1t-1et-1)(k)|t=0βk=(-g′(f¯(t))g(f¯(t))f′(f¯(t)))(k)|t=0=(1t-1et-1)(k)|t=0.
Noting t/(et-1) is the exponential generating function for Bernoulli number sequence {ℬn} in [7], we can easily compute (4.29) to get(4.30)αk=βk=(1t-1et-1)(k)|t=0(1t[1-tet-1])(k)|t=0=(-∑j=0∞Bj+1(j+1)!tj)(k)|t=0=-Bk+1k+1.Therefore, the recursive formula for the Stirling polynomialSn(x) is S0(x)=1 and(4.31)Sn+1(x)=-∑k=0n(nk)(x+1)Bk+1k+1Sn-k(x)=-(x+1)n+1∑k=0n(n+1k+1)Bk+1Sn-k(x).


## 4.1. The First Recursive Formula for the Sheffer Polynomials

Theorem 4.1 (Recursive Formula I).
Let{sn(x)} denote the Sheffer polynomial sequence for (g(t),f(t)). Then, s0(x)=1/g(0) and
(4.1)sn+1(x)=∑k=0n(γkx+δk)sn(k)(x)k!,forn≥0,
where γk=(1/f′(t))(k)|t=0 and δk=(-g′(t)/g(t)f′(t))(k)|t=0.Proof.
The proof is similar to the proof of Theorem3.4. Let us consider 𝒲n[(d/dt)(exf¯(t)/g(f¯(t)))]t=0. On one hand,
(4.2)Wn[ddt(exf¯(t)g(f¯(t)))]t=0=[s1(x)s2(x)s3(x)⋯sn+1(x)]T.
(17
On the other hand, by Properties1(c) and 1(d) and Lemma 3.3, we have(4.3)Wn[ddt(exf¯(t)g(f¯(t)))]t=0=Wn[(x-g′(f¯(t))g(f¯(t)))1f′(f¯(t))exf¯(t)g(f¯(t))]t=0=Wn[1,f¯(t),…,f¯n(t)]t=0Λn-1Pn[1g(t)]t=0Pn[ext]t=0×Wn[x1f′(t)-g′(t)g(t)f′(t)]t=0=Wn[s0(x),s1(x),…,sn(x)]TΛn-1Wn[x1f′(t)-g′(t)g(t)f′(t)]t=0.
Thus,(4.4)Wn[ddt(exf¯(t)g(f¯(t)))]t=0=[s0(x)00⋯0s1(x)s1′(x)1!0⋯0s2(x)s2′(x)1!s2′′(x)2!⋯0⋮⋮⋮⋱⋮sn(x)sn′(x)1!sn′′(x)2!⋯sn(n)(x)n!][γ0x+δ0γ1x+δ1γ2x+δ2⋮γnx+δn].
Equating the last rows of (17) and (4.4), we get the desired result.Example 4.2.
ForAn(x)=∑k=0n(bk)(∑j=knS(n,j)(j)k(-x)j-k), the Actuarial polynomial,
(4.5)A0(x)=1,An+1(x)=(-x+b)An(x)+xAn′(x),forn≥0.Example 4.3.
LetLn(x) denote the Laguerre polynomial. Then,
(4.6)L0(x)=1,Ln+1(x)=(-x+1)Ln(x)+(2x-1)Ln′(x)-xLn′′(x),forn≥0.Example 4.4.
Hermite polynomial is defined as the Sheffer polynomial sequence for(g(t),f(t))=(et2/4,t/2) Let us denote the polynomial as Hn(x). Then,
(4.7)H0(x)=1,Hn+1(x)=2xHn(x)-Hn′(x),forn≥0.Corollary 4.5 (Recursive Formula I for associated polynomial sequences).
Let{qn(x)} denote the associated polynomial sequence for f(t). Then, q0(x)=1 and
(4.8)qn+1(x)=x∑k=0nγkk!qn(k)(x),forn≥0,
where γk=(1/f′(t))(k)|t=0.Proof.
It follows from Theorem4.1 since g(t)=1 and hence δk=0 for all k.Example 4.6.
For the exponential polynomialϕn(x),
(4.9)ϕ0(x)=1,ϕn+1(x)=x[ϕn(x)+ϕn′(x)],forn≥0.Corollary 4.7 (Recursive Formula I for Appell polynomial sequences).
Let{an(x)} denote the Appell polynomial sequence for g(t). Then, a0(x)=1/g(0) and
(4.10)an+1(x)=xan(x)+∑k=0nδkk!an(k)(x),forn≥0,
where δk=(-g′(t)/g(t))(k)|t=0.Proof.
Sincef(t)=t, δk=(-g′(t)/g(t)f′(t))(k)|t=0=(-g′(t)/g(t))(k)|t=0,γ0=1,  and γk=0 for all other k in Theorem 4.1.Example 4.8.
For the Hermite polynomial of orderv, denoted by Hn{v}(x),
(4.11)H0{v}(x)=1,Hn+1{v}(x)=xHn{v}(x)-v(Hn{v})′(x),forn≥0.Example 4.9.
Let us consider the Stirling polynomial sequenceSn(x), which is the Sheffer polynomial sequence for (g(t),f¯(t))=(e-t,ln(t/1-e-t)). To obtain the recursive formula for Sn(x) by Theorem 4.1, we have to find f(t), that is, solve the insolvable transcendental equation
(4.12)et=y1-e-y
for y.This example shows a type of Sheffer sequences for which Theorem4.1 fails to produce a recursive formula. This motivate us to develop other recursive formulas, which represent sn+1(x) in term of its previous terms sk(x) and the derivatives of f¯(t) and g(t).


## 4.2. The Second Recursive Formula for the Sheffer Polynomials

Theorem 4.10 (Recursive Formula II).
Let{sn(x)} be the Sheffer polynomial sequence for (g(t),f(t)). Then, s0(x)=1/g(0) and
(4.13)ϵ0sn+1(x)=xsn(x)+∑k=0n(nk)θksn-k(x)-∑k=1n(nk)ϵksn+1-k(x),forn≥0,
where ϵk=(f′(f¯(t)))(k)|t=0=(1/f̅′(t))(k)|t=0andθk=(-g′(f¯(t))/g(f¯(t)))(k)|t=0.Proof.
Let us consider𝒲n[f′(f¯(t))(d/dt)(exf¯(t)/g(f¯(t)))]t=0. On one hand, applying Property 1(c),(4.14)Wn[f′(f¯(t))ddt(exf¯(t)g(f¯(t)))]t=0=Pn[ddt(exf¯(t)g(f¯(t)))]t=0Wn[f′(f¯(t))]t=0=[s1(x)00⋯0s2(x)s1(x)0⋯0s3(x)(21)s2(x)s1(x)⋯0⋮⋮⋮⋱⋮sn+1(x)(n1)sn(x)(n2)sn-1(x)⋯s1(x)][ϵ0ϵ1ϵ2⋮ϵn].
On the other hand, by Properties1(a) and 1(c),
(4.15)Wn[f′(f¯(t))ddt(exf¯(t)g(f¯(t)))]t=0=Wn[(x-g′(f¯(t))g(f¯(t)))exf¯(t)g(f¯(t))]t=0=xWn[exf¯(t)g(f¯(t))]t=0-Pn[exf¯(t)g(f¯(t))]t=0Wn[g′(f¯(t))g(f¯(t))]t=0=x[s0(x)s1(x)s2(x)⋮sn(x)]+[s0(x)00⋯0s1(x)s0(x)0⋯0s2(x)(21)s1(x)s0(x)⋯0⋮⋮⋮⋱⋮sn(x)(n1)sn-1(x)(n2)sn-2(x)⋯s0(x)][θ0θ1θ2⋮θn].
Equating the last rows of (4.14) and (4.15) leads to
(4.16)∑k=0n(nk)sn+1-k(x)ϵk=xsn(x)+∑k=0n(nk)sn-k(x)θk.
A rearrangement of the above yields the desired result.Example 4.11.
LetLn{a}(x) be the Laguerre polynomial of order a, pn(x) the Poisson-Charlier polynomial of order a, and Mn(x) the Meixner polynomial of the first kind of order (b,c), which is the Sheffer polynomial sequence for (g(t),f(t))=(((1-c)/(1-cet))b,(1-et)/(c-1-et)). Then,
(4.17)L0{a}(x)=1,Ln+1{a}(x)=(2n+a+1-x)Ln{a}(x)-(n2+na)Ln-1{a}(x),forn≥0,p0(x)=1,apn+1(x)=(x-n-a)pn(x)-npn-1(x),forn≥0,M0(x)=1,cMn+1(x)=((c-1)x+n(c+1)+bc)Mn(x)-(n(n-1)+nb)Mn-1(x),forn≥0.Corollary 4.12 (Recursive Formula II for associated polynomial sequences).
Let{qn(x)} denote the associated polynomial sequence for f(t). Then, q0(x)=1 and
(4.18)ϵ0qn+1(x)=xqn(x)-∑k=1n(nk)ϵkqn+1-k(x),forn≥0,
where ϵk=(f′(f¯(t)))(k)|t=0.Example 4.13.
Letϕn(x) denote the exponential polynomial. Then,
(4.19)ϕ0(x)=1,ϕn+1(x)=xϕn(x)+∑k=0n-1(nk+1)(-1)kϕn-k(x),forn≥0.Corollary 4.14 (Recursive Formula II for Appell polynomial sequences).
Let{an(x)} denote the Appell polynomial sequence for g(t). Then, a0(x)=1/g(0) and
(4.20)an+1(x)=xan(x)+∑k=0n(nk)θkan-k(x),forn≥0,
where θk=(-g′(t)/g(t))(k)|t=0.Remark 4.15.
Theorem  2.2 and Theorem  2.4 in [2] are special cases of Corollary 4.14.Example 4.16.
LetHn{v}(x) denote the Hermite polynomial of order v. Then,
(4.21)H0{v}(x)=1,Hn+1{v}(x)=xHn{v}(x)-nvHn-1{v}(x),forn≥0.Here, we would like to revisit Example4.9. In order to obtain the recursive formula for the Stirling polynomial sequence {Sn(x)} by Theorem 4.10, we need to compute
(4.22)ϵk=(f′(f¯(t)))(k)|t=0=(1f̅′(t))(k)|t=0=(t(et-1)et-1-t)(k)|t=0.Equation  (4.22) seems difficult to evaluate and does not lead to any nice result. This is a motivation for us to develop yet another formula in Theorem 4.17.


## 4.3. The Third Recursive Formula for the Sheffer Polynomials

Theorem 4.17 (Recursive Formula III).
Let{sn(x)} denote the Sheffer polynomial sequence for (g(t),f(t)). Then, s0(x)=1/g(0) and
(4.23)sn+1(x)=∑k=0n(nk)(xαk+βk)sn-k(x),forn≥0,
where αk=(1/f′(f¯(t)))(k)|t=0=(f¯(t))(k+1)|t=0andβk=(-g′(f¯(t))/g(f¯(t))f′(f¯(t)))(k)|t=0.Proof.
We have(4.24)[s1(x)s2(x)s3(x)⋯sn+1(x)]T=Wn[ddt(exf¯(t)g(f¯(t)))]t=0.
Also,(4.25)Wn[ddt(exf¯(t)g(f¯(t)))]t=0=Wn[(x-g′(f¯(t))g(f¯(t)))1f′(f¯(t))exf¯(t)g(f¯(t))]t=0=Pn[(x-g′(f¯(t))g(f¯(t)))1f′(f¯(t))]t=0Wn[exf¯(t)g(f¯(t))]t=0=[xα0+β000⋯0xα1+β1xα0+β00⋯0xα2+β22(xα1+β1)xα0+β0⋯0⋮⋮⋮⋱⋮xαn+βn(n1)(xαn-1+βn-1)(n2)(xαn-2+βn-2)⋯xα0+β0][s0(x)s1(x)s2(x)⋮sn(x)].
Equating the last rows of (4.24) and (4.25), we get the desired result.Example 4.18.
LetAn(x) be the Actuarial polynomial. Then,
(4.26)A0(x)=1,An+1(x)=-x∑k=0n(nk)Ak(x)+bAn(x),forn≥0.Corollary 4.19 (Recursive Formula III for associated polynomial sequences).
Let{qn(x)} denote the associated polynomial sequence for f(t). Then, q0(x)=1 and
(4.27)qn+1(x)=x∑k=0n(nk)αkqn-k(x),forn≥0,
where αk=(1/f′(f¯(t)))(k)|t=0=(f¯(t))(k+1)|t=0.Example 4.20.
Letϕn(x) denote the exponential polynomial. By Corollary 4.19, we have the following well-known result
(4.28)ϕ0(x)=1,ϕn+1(x)=x∑k=0n(nk)ϕk(x),forn≥0.Finally, let us finish up Example4.9 and conclude this paper.To obtain the recursive formula for the Stirling polynomial sequence{Sn(x)}by Theorem 4.17, we need to compute(4.29)αk=(1f′(f¯(t)))(k)|t=0=(f¯(t))(k+1)|t=0=(1t-1et-1)(k)|t=0βk=(-g′(f¯(t))g(f¯(t))f′(f¯(t)))(k)|t=0=(1t-1et-1)(k)|t=0.
Noting t/(et-1) is the exponential generating function for Bernoulli number sequence {ℬn} in [7], we can easily compute (4.29) to get(4.30)αk=βk=(1t-1et-1)(k)|t=0(1t[1-tet-1])(k)|t=0=(-∑j=0∞Bj+1(j+1)!tj)(k)|t=0=-Bk+1k+1.Therefore, the recursive formula for the Stirling polynomialSn(x) is S0(x)=1 and(4.31)Sn+1(x)=-∑k=0n(nk)(x+1)Bk+1k+1Sn-k(x)=-(x+1)n+1∑k=0n(n+1k+1)Bk+1Sn-k(x).



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*Source: 476462-2011-11-03.xml*