The Sheffer A-Type 0 Orthogonal Polynomial Sequences and Related Results

Abstract

In this chapter, we present a rigorous development of I. M. Sheffer’s characterization of the A-Type 0 orthogonal polynomial sequences. We first develop the results that led to the main theorem that characterizes the general A-Type 0 polynomial sequences via a linear generating function. From there, we develop the additional theory that Sheffer utilized in order to determine which A-Type 0 polynomial sequences are also orthogonal. We then address Sheffer’s additional characterizations of B-Type and C-Type, as well as E.D. Rainville’s $\sigma$ -Type classification. Lastly, we cover J. Meixner’s approach to the same characterization problem studied by Sheffer and then discuss an extension of Meixner’s analysis by W.A. Al-Salam. Portions of the analysis addressed throughout this chapter are supplemented with informative concrete examples.

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In this chapter, we present a rigorous development of I. M. Sheffer’s characterization of the A-Type 0 orthogonal polynomial sequences. We first develop the results that led to the main theorem that characterizes the general A-Type 0 polynomial sequences via a linear generating function. From there, we develop the additional theory that Sheffer utilized in order to determine which A-Type 0 polynomial sequences are also orthogonal. We then address Sheffer’s additional characterizations of B-Type and C-Type, as well as E.D. Rainville’s $\sigma$ -Type classification. Lastly, we cover J. Meixner’s approach to the same characterization problem studied by Sheffer and then discuss an extension of Meixner’s analysis by W.A. Al-Salam. Portions of the analysis addressed throughout this chapter are supplemented with informative concrete examples.

1.1 Preliminaries

Throughout this chapter, we make use of each of the following definitions, terminologies and notations.

Definition 1.1.

We always assume that a set of polynomials ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ is such that each P _n (x) has degree exactly n, which we write as $\deg ({P}_{n}(x)) = n$.

Definition 1.2.

A set of polynomials ${\left \{{Q}_{n}(x)\right \}}_{n=0}^{\infty }$ is monic if ${Q}_{n}(x) - {x}^{n}$ is of degree at most n − 1 or equivalently if the leading coefficient of each Q _n (x) is unitary.

Definition 1.3.

We shall define a generating function for a polynomial sequence ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ as follows:

$$\sum\limits_{\Lambda }{\zeta }_{n}{P}_{n}(x){t}^{n} = F(x,t),$$

with $\Lambda \subseteq \left \{0,1,2,\ldots \right \}$ and ${\left \{{\zeta }_{n}\right \}}_{n=0}^{\infty }$ a sequence in n that is independent of x and t. Moreover, we say that the function F(x, t) generates the set ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$.

It is important to mention that a generating function need not converge, as in general, several relationships can be derived when F(x, t) is divergent.

Definition 1.4.

In this chapter, the term orthogonal polynomials refers to a set of polynomials ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ that satisfies one of the two weighted inner products below:

$$\begin{array}{rcl} & \mathrm{Continuous :} \left \langle {P}_{m}(x),{P}_{n}(x)\right \rangle ={ \int \nolimits \nolimits }_{{\Omega }_{1}}{P}_{m}(x){P}_{n}(x)w(x)dx = {\alpha }_{n}{\delta }_{m,n},&\end{array}$$

(1.1)

$$\begin{array}{rcl} & \mathrm{Discrete :} \left \langle {P}_{m}(x),{P}_{n}(x)\right \rangle =\sum\limits_{{\Omega }_{2}}{P}_{m}(x){P}_{n}(x)w(x) = {\beta }_{n}{\delta }_{m,n},&\end{array}$$

(1.2)

where ${\Omega }_{1} \subseteq \mathbb{R}$, ${\Omega }_{2} \subseteq \mathbb{W}$, ${\delta }_{m,n}$ denotes the Kronecker delta and w(x) > 0 is entitled the weight function.

For example, the Laguerre, Hermite, and Meixner–Pollaczek polynomials satisfy a continuous orthogonality relation of the form (1.1). On the other hand, the Charlier, Meixner, and Krawtchouk polynomials satisfy a discrete orthogonality relation of the form (1.2) (cf. [6]).

Now, it is well-known that a necessary and sufficient condition for a set of polynomials ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ to be orthogonal is that it satisfies a three-term recurrence relation (see [8]), which can be written in different (equivalent) forms. In particular, we utilize the following two forms in this chapter and adhere to the nomenclature used in [2].

Definition 1.5 (The Three-Term Recurrence Relations). 

It is a necessary and sufficient condition that an orthogonal polynomial sequence ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ satisfies an unrestricted three-term recurrence relation of the form

$$\begin{array}{rcl} & {P}_{n+1}(x) = ({A}_{n}x + {B}_{n}){P}_{n}(x) - {C}_{n}{P}_{n-1}(x), {A}_{n}{A}_{n-1}{C}_{n} > 0& \\ & \mathrm{where} {P}_{-1}(x) = 0 \mathrm{and} {P}_{0}(x) = 1. &\end{array}$$

(1.3)

If Q _n (x) represents the monic form of P _n (x), then it is a necessary and sufficient condition that ${\left \{{Q}_{n}(x)\right \}}_{n=0}^{\infty }$ satisfies the following monic three-term recurrence relation:

$$\begin{array}{rcl} & {Q}_{n+1}(x) = (x + {b}_{n}){Q}_{n}(x) - {c}_{n}{Q}_{n-1}(x), {c}_{n} > 0& \\ & \mathrm{where} {Q}_{-1}(x) = 0 \mathrm{and} {Q}_{0}(x) = 1. &\end{array}$$

(1.4)

We entitle the conditions ${A}_{n}{A}_{n-1}{C}_{n} > 0$ and c _n  > 0 above positivity conditions.

Lastly, we mention that all of the power series in this chapter are formal power series, i.e., they may or may not converge. In [9], Sheffer used the symbol ‘$\cong$’ to denote formal series. For simplicity, we will use the equal sign throughout our present work and it will be tacitly assumed that each power series is nonetheless formal.

1.2 Sheffer’s Analysis of the Type 0 Polynomial Sequences

In this section, we discuss each of the theorems of I.M. Sheffer’s work [9] that were necessary in characterizing all of the Type 0 orthogonal sets. With respect to space constraints, we write each proof, and some examples as well, with as much detail as possible. To begin, we consider the very well-studied Appell polynomial sets ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$, which are defined as

$$\begin{array}{rcl} A(t){\mathrm{e}}^{xt} =\sum\limits_{n=0}^{\infty }{P}_{ n}(x){t}^{n}, A\left (t\right ) =\sum\limits_{n=0}^{\infty }{a}_{ n}{t}^{n}, {a}_{ 0} = 1.& &\end{array}$$

(1.5)

An example of an Appell set is ${\left \{{x}^{n}/n!\right \}}_{n=0}^{\infty }$, which is clear since

$${\mathrm{e}}^{xt} =\sum\limits_{n=0}^{\infty }\frac{{x}^{n}} {n!} {t}^{n}.$$

Now, we differentiate (1.5) with respect to x. The left-hand side becomes

$$\frac{\mathrm{d}} {\mathrm{d}x}\left [A(t){\mathrm{e}}^{xt}\right ] = tA(t){\mathrm{e}}^{xt} =\sum\limits_{n=0}^{\infty }{P}_{ n}(x){t}^{n+1} =\sum\limits_{n=1}^{\infty }{P}_{ n-1}(x){t}^{n}$$

and the right-hand side becomes

$$\sum\limits_{n=1}^{\infty }{P^{\prime}}_{ n}(x){t}^{n}.$$

Therefore, after comparing coefficients of t ⁿ in the results above, we achieve the equivalent characterization of Appell sets

$${P^{\prime}}_{n}(x) = {P}_{n-1}(x), n = 0,1,2,\ldots.$$

Next, we consider the set of Newton polynomials ${\left \{{N}_{n}(x)\right \}}_{n=0}^{\infty }$, which is not an Appell set:

$${N}_{0}(x) := 1, {N}_{n}(x) := \frac{x\left (x - 1\right )\cdots \left (x - n + 1\right )} {n!}, n = 1,2,\ldots.$$

For the difference operator defined by $\Delta f(x) := f(x + 1) - f(x)$, it can be shown that

$$\Delta {N}_{n}(x) = {N}_{n}(x + 1) - {N}_{n}(x) = {N}_{n-1}(x)$$

and

$${\left (1 + t\right )}^{x} ={ \mathrm{e}}^{x\ln (1+t)} =\sum\limits_{n=0}^{\infty }{N}_{ n}(x){t}^{n}.$$

We observe that the operator $\Delta$ functions as $\mathrm{d/d{\it { x}}}$ does on the Appell polynomials and that the generating function above is in a more general form than Eq. (1.5), i.e., the t in the exponent of Eq. (1.5) is replaced by $H(t) =\ln (t + 1)$. Due to this analysis, Sheffer was motivated to define a class of difference polynomial sets that satisfy

$$J\left [{P}_{n}\left (x\right )\right ] = {P}_{n-1}\left (x\right ), n = 0,1,2,\ldots $$

with J a general degree-lowering operator.

Thus, we now continue with a result regarding such a general degree-lowering operator J, which is an essential structure in all of the Type 0 analysis that follows.

Lemma 1.1.

Assume that J is a linear operator that acts on the set of monomials ${\left \{{x}^{n}\right \}}_{n=0}^{\infty }$ such that $\deg \left (J[{x}^{n}]\right ) \leq n$ . Then, J has the following structure:

$$\begin{array}{rcl} J[y(x)] =\sum\limits_{n=0}^{\infty }{L}_{ n}(x) \frac{{\mathrm{d}}^{n}} {\mathrm{d}{x}^{n}}y(x),& &\end{array}$$

(1.6)

which is valid for all polynomials y(x), with $\deg \left ({L}_{n}(x)\right ) \leq n$ .

Proof.

We first note that since J is assumed to be a linear operator that acts on the set of monomials ${\left \{{x}^{n}\right \}}_{n=0}^{\infty }$, it can act on any polynomial. Therefore, if we show that Eq. (1.6) holds for y(x) = x ⁿ, we have proven the theorem. Using the fact that

$$\frac{{\mathrm{d}}^{k}} {\mathrm{d}{x}^{k}}{x}^{n} = n(n - 1)(n - 2)\cdots (n - k + 1){x}^{n-k}$$

we can then recursively define L _n (x) by the following:

$$\begin{array}{rcl} J[{x}^{n}] =\sum\limits_{k=0}^{n}{L}_{ k}(x)[n(n - 1)(n - 2)\cdots (n - k + 1){x}^{n-k}], n = 0,1,2,\ldots.& &\end{array}$$

(1.7)

Since for each $n = 0,1,2,\ldots$ we assumed that $\deg \left (J[{x}^{n}]\right ) \leq n$, we must require that ${L}_{k}(x)[n(n - 1)(n - 2)\cdots (n - k + 1){x}^{n-k}]$ be of degree at most n for $k = 0,1,\ldots,n$. This will occur if and only if $\deg \left ({L}_{k}(x)\right ) \leq k$, since for any polynomials P _m (x) and Q _n (x), $\deg \left ({P}_{m}(x){Q}_{n}(x)\right ) =\deg \left ({P}_{m}(x)\right ) +\deg \left ({Q}_{n}(x)\right )$.

In Lemma 1.1, we determined the structure that J must adhere to in order for $\deg \left (J[{x}^{n}]\right ) \leq n$. Next, we determine the form that J must have in order for $\deg \left (J[{x}^{n}]\right ) = n - 1$. As we shall see, this will amount to restrictions on L _n (x) in Eq. (1.6). Also, in order to naturally generalize our degree-lowering operator J, we additionally require that $J\left [c\right ] = 0$ for all constants c, analogous to $\frac{\mathrm{d}} {\mathrm{d}x}\left [c\right ] = 0$.

Lemma 1.2.

Necessary and sufficient conditions for J as defined in Eq. (1.6)to exist such that $\deg \left (J[{x}^{n}]\right ) = n - 1$ are as follows:

$$\begin{array}{rcl}{ L}_{0}(x) = 0, {L}_{n}(x) = {l}_{n,0} + {l}_{n,1}x + \cdots + {l}_{n,n-1}{x}^{n-1}, n = 1,2,\ldots & &\end{array}$$

(1.8)

and

$$\begin{array}{rcl}{ \lambda }_{n} := n{l}_{1,0} + n(n - 1){l}_{2,1} + \cdots + n!{l}_{n,n-1}\neq 0, n = 1,2,\ldots.& &\end{array}$$

(1.9)

Proof.

$(\Rightarrow )$ We initially assume that J[1] = 0 and $\deg \left (J[{x}^{n}]\right ) = n - 1$ for $n = 1,2,\ldots$ and that L _n (x) takes on the form

$${L}_{n}(x) = {l}_{n,0} + {l}_{n,1}x + \cdots + {l}_{n,n}{x}^{n},$$

from which we show that Eq. (1.8) and Eq. (1.9) necessarily follow. We begin by finding the coefficients of x ⁿ and x ^n − 1 in Eq. (1.7). Namely, we analyze the summand in Eq. (1.7):

$$\begin{array}{rcl}{ L}_{k}(x)[n(n - 1)(n - 2)\cdots (n - k + 1){x}^{n-k}]& &\end{array}$$

(1.10)

for $k = 0,1,2,\ldots,n$ and determine the leading coefficient in each case and subsequently add the results, thus obtaining the coefficient of x ⁿ, which will be valid for $n = 0,1,2,\ldots$. We follow the same procedure for achieving the coefficient of x ^n − 1.

For k = 0, we observe that Eq. (1.10) becomes ${L}_{0}(x){x}^{n} = {l}_{0,0}{x}^{n}$, which clearly has a leading coefficient of l _0, 0. For k = 1, we see that Eq. (1.10) turns out to be ${L}_{1}(x)n{x}^{n-1} = ({l}_{1,0} + {l}_{1,1}x)n{x}^{n-1}$ and therefore, the leading coefficient is nl _1, 1. With k = 2, Eq. (1.10) becomes ${L}_{2}(x)n(n - 1){x}^{n-2} = ({l}_{2,0} + {l}_{2,1}x + {l}_{2,2}{x}^{2})n(n - 1){x}^{n-2}$, which yields n(n − 1)l _2, 2 as the leading coefficient. Continuing in this fashion, we realize that for k = n the leading coefficient of Eq. (1.10) is n! l _n, n . So, the coefficient of x ⁿ is

$$\begin{array}{rcl}{ l}_{0,0} + n{l}_{1,1} + n(n - 1){l}_{2,2} + \cdots + n!{l}_{n,n}, n = 0,1,2,\ldots.& &\end{array}$$

(1.11)

We next successively compare (1.11) against $J\left [{x}^{n}\right ]$ for $n = 0,1,2,\ldots$. For n = 0, Eq. (1.11) becomes l _0, 0 and it must be that l _0, 0 = 0, since J[1] = 0, and thus L ₀(x) = 0. With n = 1, Eq. (1.11) turns out to be l _1, 1, which must be equal to zero, as $J[x] =\mathrm{ const}$. Continuing in this manner, it follows that l _j, j  = 0 for $j = 0,1,2,\ldots$, thus establishing (1.8).

Then, using the same logic that was used to determine the coefficient of x ⁿ, we achieve the coefficient of x ^n − 1 in Eq. (1.10), which we call ${\lambda }_{n}$, to be

$${\lambda }_{n} := n{l}_{1,0} + n(n - 1){l}_{2,1} + \cdots + n!{l}_{n,n-1}, n = 0,1,2,\ldots.$$

Since we have already shown that the coefficient of x ⁿ is zero, in order to have $\deg \left (J[{x}^{n}]\right ) = n - 1$ we must also require that ${\lambda }_{n}\neq 0$, thus proving the necessity of the statement.

$(\Leftarrow )$ From substituting Eq. (1.8) with the restriction (1.9) into Eq. (1.7), the sufficiency of the statement is immediate.

Due to Lemma 1.2, we can now modify the structure of Eq. (1.6), since our primary concern is when $\deg \left (J[{x}^{n}]\right ) = n - 1$. We have

$$\begin{array}{rcl} J[y(x)] =\sum\limits_{n=1}^{\infty }[{l}_{ n,0} + {l}_{n,1}x + \cdots + {l}_{n,n-1}{x}^{n-1}] \frac{{\mathrm{d}}^{n}} {\mathrm{d}{x}^{n}}y(x), {\lambda }_{n}\neq 0, n = 1,2,\ldots.& &\end{array}$$

(1.12)

The summation above starts at 1 via L ₀(x) = 0.

Next, given a set of polynomials $\mathcal{S} ={ \left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$, we want to determine how many operators J exist, such that J transforms each polynomial ${P}_{k}(x) \in \mathcal{S}$ to the polynomial immediately preceding it in the sequence, i.e., to ${P}_{k-1}(x) \in \mathcal{S}$. As it turns out, there is exactly one such operator.

Theorem 1.1.

For a given polynomial set $\mathcal{S} ={ \left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ , there exists a unique operator J such that

$$\begin{array}{rcl} J[{P}_{n}(x)] = {P}_{n-1}(x), n = 1,2,\ldots & &\end{array}$$

(1.13)

with J[P ₀ (x)] := 0.

Proof.

To show the existence and uniqueness of J, we substitute $y(x) = {P}_{n}(x) \in \mathcal{S}$ into Eq. (1.12), which yields

$$\begin{array}{rcl} J[{P}_{n}(x)] =\sum\limits_{k=1}^{n}[{l}_{ k,0} + {l}_{k,1}x + \cdots + {l}_{k,k-1}{x}^{k-1}] \frac{{\mathrm{d}}^{k}} {\mathrm{d}{x}^{k}}{P}_{n}(x).& & \\ \end{array}$$

Moreover, from Eq. (1.13), we require

$$\begin{array}{rcl} \sum\limits_{k=1}^{n}[{l}_{ k,0} + {l}_{k,1}x + \cdots + {l}_{k,k-1}{x}^{k-1}] \frac{{\mathrm{d}}^{k}} {\mathrm{d}{x}^{k}}{P}_{n}(x) = {P}_{n-1}(x).& &\end{array}$$

(1.14)

Therefore, upon successively comparing the coefficients of Eq. (1.14) for $n = 1,2,\ldots$ we see that the sequence $\left \{{l}_{i,j}\right \}$ is uniquely determined, thus establishing the uniqueness of J given $\mathcal{S}$.

We say that the set ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ corresponds to the operator J if Eq. (1.13) is satisfied.

Example 1.1.

To concretely demonstrate how the sequence $\left \{{l}_{i,j}\right \}$ is uniquely constructed, we consider J as in Eq. (1.12) acting on the Appell set

$$\mathcal{S} ={ \left \{{x}^{n}/n!\right \}}_{ n=0}^{\infty }.$$

First, for n = 0, we see that J[1] = 0 gives us no information. For n = 1, we require J[x] = 1. Therefore,

$$\begin{array}{rcl} J[x] =\sum\limits_{k=1}^{1}[{l}_{ k,0} + {l}_{k,1}x + \cdots + {l}_{k,k-1}{x}^{k-1}] \frac{{\mathrm{d}}^{k}} {\mathrm{d}{x}^{k}}[x] = {l}_{1,0} \cdot 1,& & \\ \end{array}$$

which implies that l _1, 0 = 1. Then, for n = 2, we must have J[x ² ∕ 2! ] = x and thus

$$\begin{array}{rcl} J\left [\frac{{x}^{2}} {2!} \right ]& =\sum\limits_{k=1}^{2}[{l}_{k,0} + {l}_{k,1}x + \cdots + {l}_{k,k-1}{x}^{k-1}] \frac{{\mathrm{d}}^{k}} {\mathrm{d}{x}^{k}}\left [\frac{{x}^{2}} {2!} \right ]& \\ & = {l}_{1,0} \cdot x + ({l}_{2,0} + {l}_{2,1}x) \cdot 1 & \\ & = {l}_{2,0} + (1 + {l}_{2,1})x. & \\ \end{array}$$

Therefore, ${l}_{2,0} = {l}_{2,1} = 0$. Next, for n = 3, we see that $J[{x}^{3}/3!] = {x}^{2}/2!$ must hold. Hence,

$$\begin{array}{rcl} J\left [\frac{{x}^{3}} {3!} \right ]& =\sum\limits_{k=1}^{3}[{l}_{k,0} + {l}_{k,1}x + \cdots + {l}_{k,k-1}{x}^{k-1}] \frac{{\mathrm{d}}^{k}} {\mathrm{d}{x}^{k}}\left [\frac{{x}^{3}} {3!} \right ] & \\ & = {l}_{1,0} \cdot \frac{{x}^{2}} {2!} + ({l}_{2,0} + {l}_{2,1}x) \cdot x + ({l}_{3,0} + {l}_{3,1}x + {l}_{3,2}{x}^{2}) \cdot 1& \\ & = {l}_{3,0} + {l}_{3,1}x + \left ( \frac{1} {2!} + {l}_{3,2}\right ){x}^{2}. & \\ \end{array}$$

Therefore, it must be that ${l}_{3,0} = {l}_{3,1} = {l}_{3,2} = 0$.

In fact, continuing in this fashion, the interested reader can readily show that all of the l-values in this sequence will be uniquely determined to be zero, except l _1, 0 = 1. This is certainly clear since

$$\frac{\mathrm{d}} {\mathrm{d}x}\left [\frac{{x}^{n}} {n!} \right ] = \frac{{x}^{n-1}} {(n - 1)!}.$$

We next prove the converse of Theorem 1.1.

Theorem 1.2.

Associated to each operator J are infinitely many sets ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ such that Eq. (1.13)holds. More specifically, exactly one of these sets ${\left \{{B}_{n}(x)\right \}}_{n=0}^{\infty }$ , entitled the basic set , is such that

$${B}_{0}(x) = 1 and {B}_{n}(0) = 0, n = 0,1,2,\ldots.$$

Proof.

Let Q _m (x) be a polynomial such that $\deg ({Q}_{m}) = m$. Then, via Eq. (1.12), we can construct a polynomial, say P _m + 1(x), such that $J[{P}_{m+1}(x)] = {Q}_{m}(x)$, where $\deg ({P}_{m+1}(x)) = m + 1$. However, since $J[\mathrm{const}] = 0$, the polynomial ${P}_{m+1}(x)$ is unique only up to an additive constant. This proves the infinitude of sets ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ that correspond to a given J.

By assigning B ₀(x) : = 1 and assuming B _n (0) = 0 for n > 0, one can successively and uniquely determine the set ${\left \{{B}_{n}(x)\right \}}_{n=0}^{\infty }$ such that $J[{B}_{n}(x)] = {B}_{n-1}(x)$ and $\deg ({B}_{n}(x)) = n$.

The next result states that for ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ to be a set corresponding to J, it must be expressed as a linear combination of polynomials from the basic set. However, the scalers in this linear combination appear in a special way and play a key role in the later characterizations.

Theorem 1.3.

A necessary and sufficient condition that ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ be a set corresponding to J is that there exist a sequence of constants $\left \{{a}_{k}\right \}$ such that

$$\begin{array}{rcl}{ P}_{n}(x) = {a}_{0}{B}_{n}(x) + {a}_{1}{B}_{n-1}(x) + \cdots + {a}_{n}{B}_{0}(x), {a}_{0}\neq 0.& &\end{array}$$

(1.15)

Proof.

$(\Rightarrow )$ Assume that ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ satisfies (1.15). Therefore, $\deg ({P}_{n}(x)) = n$ and by the linearity of J we have

$$J[{P}_{n}(x)] = J\left [\sum\limits_{i=0}^{n}{a}_{ i}{B}_{n-i}(x)\right ] =\sum\limits_{i=0}^{n}{a}_{ i}J[{B}_{n-i}(x)] =\sum\limits_{i=0}^{n-1}{a}_{ i}{B}_{n-i-1}(x) = {P}_{n-1}(x),$$

which follows since $J[{B}_{0}(x)] = 0$ and J[B _n (x)] = B _n − 1(x) for n > 0 via Theorem 1.2. Thus J[P _n (x)] = P _n − 1(x).

$(\Leftarrow )$ We now assume that ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ corresponds to J. Since $\deg ({B}_{n}(x)) = n$, given P _n (x), there must exist constants ${\left \{{a}_{n,k}\right \}}_{k=0}^{n}$ such that

$${P}_{n}(x) = {a}_{n,0}{B}_{n}(x) + {a}_{n,1}{B}_{n-1}(x) + \cdots + {a}_{n,n}{B}_{0}(x), {a}_{n,0}\neq 0.$$

We act on this relation with J and see that the left-hand side becomes

$$J[{P}_{n}(x)] = {P}_{n-1}(x) = {a}_{n-1,0}{B}_{n-1}(x)+{a}_{n-1,1}{B}_{n-2}(x)+\cdots +{a}_{n-1,n-1}{B}_{0}(x)$$

and since $J[{B}_{0}(x)] = 0$, we see that the right-hand side turns into

$${a}_{n,0}{B}_{n-1}(x) + {a}_{n,1}{B}_{n-2}(x) + \cdots + {a}_{n,n-1}{B}_{0}(x).$$

Therefore, from comparing coefficients of the results directly above, we infer that

$${a}_{n,k} = {a}_{n-1,k}, k = 0,1,\ldots,n - 1.$$

Next, we momentarily fix k. Then, the relation immediately above implies that for all $n \geq k$ each a _n, k is equal to a _{n − 1, k} . Thus, the first index in the series $\left \{{a}_{n,k}\right \}$ is superfluous and thus can be omitted. We conclude that $\left \{{a}_{k}\right \}$ exist such that Eq. (1.15) is satisfied.

It may at first appear counterintuitive that the elements of the sequence $\left \{{a}_{k}\right \}$ appear as they do in Eq. (1.15). However, the proof of Theorem 1.3 shows why this is the case. For emphasis, consider expressing a polynomial P _n (x) corresponding to J as a linear combination of basic polynomials in the following “natural” way:

$${P}_{n}(x) = {a}_{n}{B}_{n}(x) + {a}_{n-1}{B}_{n-1}(x) + \cdots + {a}_{0}{B}_{0}(x).$$

Then,

$$J[{P}_{n}(x)] = J\left [\sum\limits_{i=0}^{n}{a}_{ i}{B}_{i}(x)\right ] =\sum\limits_{i=0}^{n}{a}_{ i}J[{B}_{i}(x)] =\sum\limits_{i=0}^{n}{a}_{ i}{B}_{i-1}(x)\neq {P}_{n-1}(x).$$

We next wish to determine what conditions are needed for a set ${\left \{{Q}_{n}(x)\right \}}_{n=0}^{\infty }$ to correspond to J, given that a set ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ corresponds to J. As it turns out, Q _n (x) must be written as a linear combination of ${\left \{{P}_{k}(x)\right \}}_{k=0}^{n}$.

Corollary 1.1.

Given that ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ is a set corresponding to J, a necessary and sufficient condition that ${\left \{{Q}_{n}(x)\right \}}_{n=0}^{\infty }$ also corresponds to J is that constants $\left \{{b}_{k}\right \}$ exist such that

$${Q}_{n}(x) = {b}_{0}{P}_{n}(x) + {b}_{1}{P}_{n-1}(x) + \cdots + {b}_{n}{P}_{0}(x), {b}_{0}\neq 0.$$

Proof.

The proof is similar to that of Theorem 1.3 and is left as an exercise for the reader.

In light of the preceding theorems, we now state the definition of Sheffer Type k.

Definition 1.6.

Let the set $\mathcal{S} :={ \left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ correspond to the unique operator J. Then, $\mathcal{S}$ is of Sheffer Type k, or simply Type k, if the coefficients ${\left \{{L}_{j}(x)\right \}}_{j=0}^{\infty }$ in Eq. (1.12) are such that $\deg ({L}_{j}(x)) \leq k$ for all j and there exists at least one ${L}_{i}(x) \in {\left \{{L}_{j}(x)\right \}}_{j=0}^{\infty }$ such that $\deg ({L}_{i}(x)) = k$. If ${\left \{{L}_{j}(x)\right \}}_{j=0}^{\infty }$ is unbounded, then $\mathcal{S}$ is of Type  $\infty$.

With this definition, we have the following result.

Theorem 1.4.

There exist infinitely many sets for each Sheffer Type k (k finite or infinite).

Proof.

We know from Theorem 1.2 that associated to each operator J are infinitely many sets ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ such that Eq. (1.13) holds. This result is entirely independent of the degrees of the coefficients ${\left \{{L}_{j}(x)\right \}}_{j=0}^{\infty }$ in Eq. (1.12) and therefore the Type. Hence, Theorem 1.2 holds for all ${\left \{{L}_{j}(x)\right \}}_{j=0}^{\infty }$, even if it is unbounded, so there are infinitely many sets of every Type (finite or infinite).

We now consider what effect replacing $\mathcal{S} :={ \left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ with ${\left \{{c}_{n}{P}_{n}(x)\right \}}_{n=0}^{\infty }$ has on the Type classification of $\mathcal{S}$. Assuming that $\mathcal{S}$ corresponds to J, we immediately observe that

$$J[{c}_{n}{P}_{n}(x)] = {c}_{n}J[{P}_{n}(x)] = {c}_{n}{P}_{n-1}(x)\neq {c}_{n-1}{P}_{n-1}(x).$$

This simple manipulation tells us that the Type is not necessarily preserved since we may need a new operator, say $\breve{J}$, such that $\breve{J}[{c}_{n}{P}_{n}(x)] = {c}_{n-1}{P}_{n-1}(x)$. We demonstrate this concretely in the following examples.

Example 1.2.

In Example 1.1, we analyzed the Appell set $\mathcal{S} :={ \left \{{x}^{n}/n!\right \}}_{n=0}^{\infty }$ and showed that l _1, 0 = 1 and that every other l-value was zero by utilizing (1.12). Moreover, we actually showed that L ₁(x) = 1 and L _j (x) ≡ 0 for $j = 2,3,\ldots$ and thus, that $\mathcal{S}$ is a Type 0 set, since $k = 0$ in Definition 1.6. We next consider the set $\breve{\mathcal{S}} :={ \left \{{c}_{n}{x}^{n}/n!\right \}}_{n=0}^{\infty }$, where ${c}_{i}\neq 0$ and each c _i is distinct.

For n = 1, we require J[c ₁ x] = c ₀. Therefore,

$$\begin{array}{rcl} J[{c}_{1}x] =\sum\limits_{k=1}^{1}[{l}_{ k,0} + {l}_{k,1}x + \cdots + {l}_{k,k-1}{x}^{k-1}] \frac{{\mathrm{d}}^{k}} {\mathrm{d}{x}^{k}}[{c}_{1}x] = {l}_{1,0} \cdot {c}_{1},& & \\ \end{array}$$

which implies that l _1, 0 = c ₀ ∕ c ₁ and therefore L ₁(x) = c ₀ ∕ c ₁. Then, for n = 2, we must have J[c ₂ x ² ∕ 2! ] = c ₁ x and thus

$$\begin{array}{rcl} J\left [\frac{{c}_{2}{x}^{2}} {2!} \right ]& =\sum\limits_{k=1}^{2}[{l}_{k,0} + {l}_{k,1}x + \cdots + {l}_{k,k-1}{x}^{k-1}] \frac{{\mathrm{d}}^{k}} {\mathrm{d}{x}^{k}}\left [\frac{{c}_{2}{x}^{2}} {2!} \right ]& \\ & = {l}_{1,0} \cdot {c}_{2}x + ({l}_{2,0} + {l}_{2,1}x) \cdot {c}_{2} & \\ & = {l}_{2,0}{c}_{2} + ({c}_{0}{c}_{2}/{c}_{1} + {l}_{2,1}{c}_{2})x. & \\ \end{array}$$

So, l _2, 0 = 0 and l _2, 1 = (c ₁ ² − c ₀ c ₂) ∕ (c ₁ c ₂) giving ${L}_{1}(x) = \frac{({c}_{1}^{2}-{c}_{ 0}{c}_{2})} {{c}_{1}{c}_{2}} x$. Therefore, we already see that $\breve{\mathcal{S}}$ is not a Type 0 set. In fact, the interested reader can show that $\breve{\mathcal{S}}$ is actually Type $\infty$.

Example 1.3.

We next consider a very important Type 0 set, the importance of which will become most evident upon the completion of Sect. 1.3. This set is defined as ${\mathcal{H}}_{n}(x) := {2}^{-n}{H}_{n}(x)/n!$, where

$$\begin{array}{rcl}{ H}_{n}(x) := {2}^{n}n!\sum\limits_{k=0}^{\left \lfloor n/2\right \rfloor } \frac{{(-1)}^{k}{x}^{n-2k}} {{2}^{2k}k!(n - 2k)!}& & \\ \end{array}$$

are the Hermite polynomials. For convenience, we write out the first four polynomials from the set ${\left \{{\mathcal{H}}_{n}(x)\right \}}_{n=0}^{\infty }$:

$${\mathcal{H}}_{0}(x) = 1, {\mathcal{H}}_{1}(x) = x, {\mathcal{H}}_{2}(x) = \frac{1} {2}{x}^{2} -\frac{1} {4} \mathrm{and} {\mathcal{H}}_{3}(x) = \frac{1} {6}{x}^{3} -\frac{1} {4}x.$$

We initially see that $J[{\mathcal{H}}_{1}(x)] = {\mathcal{H}}_{0}(x)$ implies

$$J[{\mathcal{H}}_{1}(x)] =\sum\limits_{k=1}^{1}[{l}_{ k,0} + {l}_{k,1}x + \cdots + {l}_{k,k-1}{x}^{k-1}] \frac{{\mathrm{d}}^{k}} {\mathrm{d}{x}^{k}}\left [{\mathcal{H}}_{1}(x)\right ] = {l}_{1,0} = 1$$

and therefore, L ₁(x) = 1. Then, $J[{\mathcal{H}}_{2}(x)] = {\mathcal{H}}_{1}(x)$ yields

$$\begin{array}{rcl} J[{\mathcal{H}}_{2}(x)]& =\sum\limits_{k=1}^{2}[{l}_{k,0} + {l}_{k,1}x + \cdots + {l}_{k,k-1}{x}^{k-1}] \frac{{\mathrm{d}}^{k}} {\mathrm{d}{x}^{k}}\left [{\mathcal{H}}_{2}(x)\right ]& \\ & = {l}_{2,0} + (1 + {l}_{2,1}x)x = x & \\ \end{array}$$

and it must be that l _2, 0 = l _2, 1 = 0, i.e., L ₂(x) = 0. Continuing, one concludes that L ₁(x) = 1 and L _j (x) = 0 for j = 2, 3, … and thus ${\left \{{\mathcal{H}}_{n}(x)\right \}}_{n=0}^{\infty }$ is a Type 0 set. Moreover, writing (1.12) in the operator form

$$\begin{array}{rcl} J =\sum\limits_{k=1}^{\infty }{L}_{ k}(x) \frac{{\mathrm{d}}^{k}} {\mathrm{d}{x}^{k}},& &\end{array}$$

(1.16)

we find the unique operator J for our current set to be $J = \mathrm{d}/\mathrm{d}x$.

Example 1.4.

Now consider the set $\mathrm{{H}}_{n}(x) := {H}_{n}(x)/{(n!)}^{2}$, where ${\left \{{H}_{n}(x)\right \}}_{n=0}^{\infty }$ are the Hermite polynomials as defined in the last example. The first four polynomials from the set ${\left \{\mathrm{{H}}_{n}(x)\right \}}_{n=0}^{\infty }$ are

$$\mathrm{{H}}_{0}(x) = 1, \mathrm{{H}}_{1}(x) = 2x, \mathrm{{H}}_{2}(x) = {x}^{2} -\frac{1} {2} \mathrm{and} \mathrm{{H}}_{3}(x) = \frac{2} {9}{x}^{3} -\frac{1} {3}x.$$

Using the same procedure as in Examples 1.2 and 1.3, one can show that ${L}_{1}(x) = 1/2$, ${L}_{2}(x) = \frac{1} {2}x$, L ₃(x) =  − 1 ∕ 4, and L _j (x) = 0 for j = 4, 5, …. Thus, ${\left \{\mathrm{{H}}_{n}(x)\right \}}_{n=0}^{\infty }$ is a Type 1 set and Eq. (1.16) becomes

$$J = \frac{1} {2} \frac{\mathrm{d}} {\mathrm{d}x} + \frac{1} {2}x \frac{{\mathrm{d}}^{2}} {\mathrm{d}{x}^{2}} -\frac{1} {4} \frac{{\mathrm{d}}^{3}} {\mathrm{d}{x}^{3}}.$$

Next, we notice that if ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ is a Type 0 set, then each L _n (x) must be a constant and we can therefore restate Definition 1.6 specifically for Type 0 sets as follows.

Definition 1.7.

${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ is a Type 0 set if Eq. (1.13) holds with J defined by

$$\begin{array}{rcl} J[y(x)] :=\sum\limits_{n=1}^{\infty }{c}_{ n}{y}^{(n)}(x), {c}_{ 1}\neq 0.& &\end{array}$$

(1.17)

We emphasize that as we have seen in Examples 1.2–1.4, Eq. (1.17) may or may not terminate, i.e., it may be finite Type k or Type $\infty$. We also have the following definition.

Definition 1.8.

Let J(t) be the formal power series

$$\begin{array}{rcl} J(t) :=\sum\limits_{n=1}^{\infty }{c}_{ n}{t}^{n}, {c}_{ 1}\neq 0,& & \\ \end{array}$$

which we entitle the generating function for J, with J as in Eq. (1.17).

Now, let the formal power series inverse of J(t) be

$$\begin{array}{rcl} H(t) :=\sum\limits_{n=1}^{\infty }{s}_{ n}{t}^{n}, {s}_{ 1} = {c}_{1}^{-1}\neq 0.& &\end{array}$$

(1.18)

This is a valid definition because if J(t) is formally substituted for t in (1.18) and the coefficients are collected to form a single power series in t, then the coefficient of each t ⁿ is a polynomial in ${c}_{1},{c}_{2},\ldots,{c}_{n},{s}_{1},{s}_{2},\ldots,{s}_{n}$. Therefore, we can choose each s _n recursively and uniquely as a function of ${c}_{1},{c}_{2},\ldots,{c}_{n},{s}_{1},{s}_{2},\ldots,{s}_{n-1}$ so the series has a single term t, i.e.,

$$J(H(t)) = H(J(t)) = t.$$

In considering $\mathrm{exp}\left (xH(t)\right )$, we see that each coefficient of t ⁿ in the formal power series expansion only comprises ${s}_{1},{s}_{2},\ldots,{s}_{n}$. Upon multiplying $\mathrm{exp}\left (xH(t)\right )$ by

$$A(t) :=\sum\limits_{n=0}^{\infty }{a}_{ n}{t}^{n}, {a}_{ 0}\neq 0,$$

we achieve a series in t where the coefficient of each t ⁿ involves elements of the sequences ${a}_{1},{a}_{2},\ldots,{a}_{n}$ and ${s}_{1},{s}_{2},\ldots,{s}_{n}$, such that each coefficient of t ⁿ is a polynomial in $x$ of degree exactly n, adhering to Definition 1.1. This leads to the main result of this section.

Theorem 1.5.

The set P _n (x) _n=0 ^∞ corresponds to the operator J and is of Sheffer Type 0 if and only if the sequence ${\left \{{a}_{n}\right \}}_{n=0}^{\infty }$ exists such that

$$\begin{array}{rcl} A(t){\mathrm{e}}^{xH(t)} =\sum\limits_{n=0}^{\infty }{P}_{ n}(x){t}^{n},& &\end{array}$$

(1.19)

where

$$\begin{array}{rcl} A(t) :=\sum\limits_{n=0}^{\infty }{a}_{ n}{t}^{n}, {a}_{ 0} = 1 \mathrm{and} H(t) :=\sum\limits_{n=1}^{\infty }{s}_{ n}{t}^{n}, {s}_{ 1} = 1.& &\end{array}$$

(1.20)

Proof.

We show that both necessity and sufficiency will follow if we prove that the basic set ${\left \{{B}_{n}(x)\right \}}_{n=0}^{\infty }$ corresponding to J in Eq. (1.13) has the following generating function:

$$\begin{array}{rcl}{ \mathrm{e}}^{xH(t)} =\sum\limits_{n=0}^{\infty }{B}_{ n}(x){t}^{n}.& &\end{array}$$

(1.21)

Since $\exp \left [xH(t)\right ] =\sum\limits_{n=0}^{\infty }[{H}^{n}(t){x}^{n}/n!]$, we let this expansion have the form

$$\begin{array}{rcl}{ \mathrm{e}}^{xH(t)} =\sum\limits_{n=0}^{\infty }{C}_{ n}(x){t}^{n}.& &\end{array}$$

(1.22)

Then, ${\left \{{C}_{n}(x)\right \}}_{n=0}^{\infty }$ must be such that $\deg ({C}_{n}(x)) = n$. We show that ${\left \{{C}_{n}(x)\right \}}_{n=0}^{\infty }$ is the basic set.

Upon setting x = 0 in Eq. (1.22) and comparing coefficients, we see immediately that C ₀(0) = 1, and therefore C ₀(x) = 1, and that C _n (0) = 0 for n > 0. Thus, ${\left \{{C}_{n}(x)\right \}}_{n=0}^{\infty }$ satisfies the initial conditions of the basic set. We clearly have J[C ₀(x)] = 0 and next show that $J[{C}_{n}(x)] = {C}_{n-1}(x)$ for $n = 1,2,3,\ldots$. We apply J to Eq. (1.22):

$$\begin{array}{rcl} J\left [\sum\limits_{n=0}^{\infty }{C}_{ n}(x){t}^{n}\right ]& =\sum\limits_{ n=0}^{\infty }J[{C}_{ n}(x)]{t}^{n} = J\left [{\mathrm{e}}^{xH(t)}\right ] =\sum\limits_{ n=0}^{\infty }{c}_{ n} \frac{{\mathrm{d}}^{n}} {\mathrm{d}{x}^{n}}\left [{\mathrm{e}}^{xH(t)}\right ] & \\ & = {c}_{1} \frac{\mathrm{d}} {\mathrm{d}x}\left [{\mathrm{e}}^{xH(t)}\right ] + {c}_{ 2} \frac{{\mathrm{d}}^{2}} {\mathrm{d}{x}^{2}} \left [{\mathrm{e}}^{xH(t)}\right ] + {c}_{3} \frac{{\mathrm{d}}^{3}} {\mathrm{d}{x}^{3}} \left [{\mathrm{e}}^{xH(t)}\right ] + \cdots & \\ & = \left ({c}_{1}H(t) + {c}_{2}{H}^{2}(t) + {c}_{3}{H}^{3}(t) + \cdots \,\right ){\mathrm{e}}^{xH(t)} & \\ & = J(H(t)){\mathrm{e}}^{xH(t)} = t{\mathrm{e}}^{xH(t)} =\sum\limits_{n=0}^{\infty }{C}_{n}(x){t}^{n+1} =\sum\limits_{n=0}^{\infty }{C}_{n-1}(x){t}^{n}& \\ \end{array}$$

(with ${C}_{-1}(x) := 0)$, which follows since H(t) and J(t) are formal inverses of one another. By comparing coefficients of t ⁿ in the relation directly above, we achieve $J[{C}_{n}(x)] = {C}_{n-1}(x)$ for n = 1, 2, 3, … and hence, ${C}_{n}(x) \equiv {B}_{n}(x)$ and Eq. (1.21) is established. Lastly, we now multiply Eq. (1.21) by A(t) as in Eq. (1.20), which yields

$$\begin{array}{rcl} A(t){\mathrm{e}}^{xH(t)} =\sum\limits_{n=0}^{\infty }{a}_{ n}{t}^{n}\sum\limits_{n=0}^{\infty }{B}_{ n}(x){t}^{n} =\sum\limits_{n=0}^{\infty }\sum\limits_{k=0}^{n}{a}_{ k}{B}_{n-k}(x){t}^{n} =\sum\limits_{n=0}^{\infty }{P}_{ n}(x){t}^{n},& &\end{array}$$

(1.23)

as a result of Theorem 1.3.

We also have the following.

Theorem 1.6.

The sequence of a _k -terms in Eq. (1.19)is exactly the same as those in Eq. (1.15).

Proof.

This statement follows immediately from the proof of Theorem 1.5, i.e., Eq. (1.23).

The next result plays an integral part in determining all of the Sheffer Type 0 orthogonal polynomials in the next section. This characterization is also interesting unto itself, as it is expressed entirely in terms of elements of the generating function (1.19).

Corollary 1.2.

${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ is of Type 0 if and only if sequences $\left \{{q}_{k,0}\right \}$ and $\left \{{q}_{k,1}\right \}$ exist such that

$$\begin{array}{rcl} \sum\limits_{k=1}^{\infty }({q}_{ k,0} + {q}_{k,1}x){P}_{n-k}(x) = n{P}_{n}(x), n = 1,2,\ldots,& &\end{array}$$

(1.24)

where

$$\left.\begin{array}{l} \sum\limits_{k=0}^{\infty }{q}_{k+1,0}{t}^{k} = A^{\prime}(t)/A(t) \\ \sum\limits_{k=0}^{\infty }{q}_{k+1,1}{t}^{k} = H^{\prime}(t) \end{array} \right \}$$

(1.25)

with A(t) and H(t) as defined in Eq. (1.20).

Proof.

We assume that ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ is a Type 0 set. We first consider the right-hand side of Eq. (1.24) and multiply it by t ⁿ and sum for $n = 0,1,2,\ldots$. This yields

$$\begin{array}{rcl} \sum\limits_{n=0}^{\infty }n{P}_{ n}(x){t}^{n} = t\sum\limits_{n=0}^{\infty }n{P}_{ n}(x){t}^{n-1} = t \frac{\mathrm{d}} {\mathrm{d}t}\left (A(t){\mathrm{e}}^{xH(t)}\right ) = t{\mathrm{e}}^{xH(t)}\left (xA(t)H^{\prime}(t) + A^{\prime}(t)\right ).& &\end{array}$$

(1.26)

Since ${J}^{k}[{P}_{n}(x)] = {P}_{n-k}(x)$, we see that

$$\sum\limits_{n=0}^{\infty }{J}^{k}[{P}_{ n}(x)]{t}^{n} = {t}^{k}\sum\limits_{n=k}^{\infty }{P}_{ n-k}(x){t}^{n-k} = {t}^{k}\sum\limits_{m=0}^{\infty }{P}_{ m}(x){t}^{m} = {t}^{k}A(t){\mathrm{e}}^{xH(t)}.$$

Therefore, we now multiply the left-hand side of Eq. (1.24) by t ⁿ, sum for $n = 0,1,2,\ldots$, and utilize (1.20) to obtain

$$\begin{array}{rcl} & \sum\limits_{n=0}^{\infty }\sum\limits_{k=1}^{\infty }({q}_{k,0} + {q}_{k,1}x){P}_{n-k}(x){t}^{n} =\sum\limits_{n=0}^{\infty }\sum\limits_{k=1}^{\infty }({q}_{k,0} + {q}_{k,1}x){J}^{k}[{P}_{n}(x)]{t}^{n}& \\ & \quad =\sum\limits_{k=1}^{\infty }({q}_{k,0} + {q}_{k,1}x)\sum\limits_{n=0}^{\infty }{J}^{k}[{P}_{n}(x)]{t}^{n} =\sum\limits_{k=1}^{\infty }({q}_{k,0} + {q}_{k,1}x){t}^{k}A(t){\mathrm{e}}^{xH(t)} & \\ & \quad = \left (t\sum\limits_{k=0}^{\infty }{q}_{k+1,0}{t}^{k} + tx\sum\limits_{k=0}^{\infty }{q}_{k+1,1}{t}^{k}\right )A(t){\mathrm{e}}^{xH(t)} = t{\mathrm{e}}^{xH(t)}\left (xA(t)H^{\prime}(t) + A^{\prime}(t)\right ). &\end{array}$$

(1.27)

Hence, we see that both Eqs. (1.26) and (1.27) are formally equal and from comparing coefficients of t ⁿ we obtain Eq. (1.24).

1.3 The Type 0 Orthogonal Polynomials

We now determine which Sheffer Type 0 sets are also orthogonal. We first assume that ${\left \{{Q}_{n}(x)\right \}}_{n=0}^{\infty }$ is an orthogonal set that satisfies the monic three-term relation of the form Eq. (1.4), which we write using the notation in [9]:

$$\begin{array}{rcl}{ Q}_{n}(x) = (x + {\lambda }_{n}){Q}_{n-1}(x) + {\mu }_{n}{Q}_{n-2}(x), {\mu }_{n}\neq 0, n = 1,2,\ldots.& &\end{array}$$

(1.28)

In our first theorem of this section, we determine a necessary and sufficient form that the recursion coefficients ${\lambda }_{n}$ and ${\mu }_{n}$ in Eq. (1.28) must have in order to characterize a Type 0 orthogonal set. First, we note that since ${\left \{{Q}_{n}(x)\right \}}_{n=0}^{\infty }$ is an orthogonal set, then so is ${\left \{{c}_{n}{Q}_{n}(x)\right \}}_{n=0}^{\infty }$. Now, we want to determine for which sets ${\left \{{Q}_{n}(x)\right \}}_{n=0}^{\infty }$ satisfying (1.28) do there exist a sequence of constants ${\left \{{c}_{n}\right \}}_{n=0}^{\infty }$ such that

$$\begin{array}{rcl}{ P}_{n}(x) = {c}_{n}{Q}_{n}(x), n = 0,1,2,\ldots & &\end{array}$$

(1.29)

is a Type 0 set. To accomplish this, we simultaneously use Eqs. (1.24) and (1.28), i.e., a characterization of a Type 0 set and a characterization of an orthogonal set.

Theorem 1.7.

A necessary and sufficient condition that an orthogonal polynomial set ${\left \{{Q}_{n}(x)\right \}}_{n=0}^{\infty }$ satisfying (1.28)be such that ${P}_{n}(x) = {c}_{n}{Q}_{n}(x)$ is of Type 0 for some ${c}_{n}\neq 0$ is that ${\lambda }_{n}$ and ${\mu }_{n}$ have the form

$$\begin{array}{rcl}{ \lambda }_{n} = \alpha + bn \mathrm{and} {\mu }_{n} = (n - 1)(c + dn)& &\end{array}$$

(1.30)

with $c + dn\neq 0$ for $n > 1.$

Proof.

$(\Rightarrow )$ We assume that Eq. (1.29) holds and show that ${\lambda }_{n}$ and ${\mu }_{n}$ are as in Eq. (1.30). To begin, replace n with n − k + 1 in Eq. (1.28) and rewrite it in the following way:

$$\begin{array}{rcl} x{Q}_{n-k}(x) = {Q}_{n-k+1}(x) - {\lambda }_{n-k+1}{Q}_{n-k}(x) - {\mu }_{n-k+1}{Q}_{n-k-1}(x).& &\end{array}$$

(1.31)

Then, substituting the right-hand side of Eq. (1.29) into Eq. (1.24) and using Eq. (1.31), we see that

$$\begin{array}{rcl} n{c}_{n}{Q}_{n}(x)& =\sum\limits_{k=1}^{\infty }{q}_{k,0}{c}_{n-k}{Q}_{n-k}(x) +\sum\limits_{k=1}^{\infty }{q}_{k,1}{c}_{n-k}[x{Q}_{n-k}(x)] & \\ & =\sum\limits_{k=1}^{\infty }{q}_{k,0}{c}_{n-k}{Q}_{n-k}(x) +\sum\limits_{k=1}^{\infty }{q}_{k,1}{c}_{n-k}{Q}_{n-k+1}(x) & \\ & \quad -\sum\limits_{k=1}^{\infty }{q}_{k,1}{c}_{n-k}{\lambda }_{n-k+1}{Q}_{n-k}(x) -\sum\limits_{k=1}^{\infty }{q}_{k,1}{c}_{n-k}{\mu }_{n-k+1}{Q}_{n-k-1}(x).&\end{array}$$

(1.32)

We next compare coefficients of Q _j (x) for j = n, n − 1, n − 2 using Eq. (1.32) to obtain relationships involving the q’s and in turn develop expressions for the recursion coefficients ${\lambda }_{n}$ and ${\mu }_{n}$. First, by comparing the coefficients of Q _n (x) we see that $n{c}_{n} = {c}_{n-1}{q}_{1,1}$ and iterating this relationship, we obtain

$$\begin{array}{rcl}{ c}_{n} = {c}_{0}{q}_{1,1}^{n}/n!.& &\end{array}$$

(1.33)

After comparing the coefficients of Q _n − 1(x), we realize that

$$0 = {q}_{1,0}{c}_{n-1} + {q}_{2,1}{c}_{n-2} - {q}_{1,1}{c}_{n-1}{\lambda }_{n}.$$

Dividing both sides of this equation by c _n − 2 and using Eq. (1.33) yield

$$\begin{array}{rcl}{ \lambda }_{n} = \left ({q}_{1,0}{q}_{1,1} + {q}_{2,1}(n - 1)\right )/{q}_{1,1}^{2}.& &\end{array}$$

(1.34)

Lastly, upon comparing the coefficients of Q _n − 2(x), we achieve

$$0 = {q}_{2,0}{c}_{n-2} + {q}_{3,1}{c}_{n-3} - {q}_{2,1}{c}_{n-2}{\lambda }_{n-1} - {q}_{1,1}{c}_{n-1}{\mu }_{n}.$$

We also divide both sides of this equation by c _n − 2, use Eq. (1.33), and also call upon Eq. (1.34) to obtain

$$\begin{array}{rcl}{ \mu }_{n} = (n - 1)\left ({q}_{2,0}{q}_{1,1}^{2} - {q}_{ 1,0}{q}_{1,1}{q}_{2,1} - ({q}_{1,1}^{2} - {q}_{ 1,1}{q}_{3,1})(n - 2)\right )/{q}_{1,1}^{4}.& &\end{array}$$

(1.35)

Thus, we see that λ _n is at most linear in n and that μ _n is at most quadratic in n with a factor of (n − 1), i.e., ${\lambda }_{n}$ and ${\mu }_{n}$ satisfy (1.30).

$(\Leftarrow )$ We assume that ${\lambda }_{n}$ and ${\mu }_{n}$ agree with Eq. (1.30). We show that Eq. (1.29) is of Type 0. Now, if in Eq. (1.28) we replace Q _n (x) with c _n d _n Q _n (x) = d _n P _n (x) where ${c}_{n} := {\alpha }^{n}/n!$ ($\alpha \neq 0$) and ${d}_{n} := {c}_{n}^{-1}$ then our three-term recurrence relation (1.28), with ${\lambda }_{n}$ and ${\mu }_{n}$ as in Eq. (1.30), remains unaltered and we can therefore write

$${d}_{n}{P}_{n}(x) = (x + \alpha + bn){d}_{n-1}{P}_{n-1}(x) + (n - 1)(c + dn){d}_{n-2}{P}_{n-2}(x).$$

Dividing both sides of this relation by d _n defined above gives a relation of the form

$$\begin{array}{rcl} n{P}_{n}(x) = (\alpha x + \beta + \gamma n){P}_{n-1}(x) + (\delta + \epsilon n){P}_{n-2}(x),& & \\ \end{array}$$

where $\alpha \neq 0$, $\delta + \epsilon n\neq 0$ for n > 1 and with $\beta = {\alpha }^{2}$, $\gamma = b\alpha$, $\delta = c{\alpha }^{2}$, and $\epsilon = d{\alpha }^{2}$. This, of course, can be written in the form

$$x{P}_{n-1}(x) = {\alpha }^{-1}n{P}_{ n}(x) - {\alpha }^{-1}(\beta + \gamma n){P}_{ n-1}(x) - {\alpha }^{-1}(\delta + \epsilon n){P}_{ n-2}(x).$$

That is, we have expressed $x{P}_{n-1}(x)$ as a linear combination of P _n (x), P _n − 1(x), and P _n − 2(x). Therefore, sequences $\left \{{q}_{k,0}\right \}$ and $\left \{{q}_{k,1}\right \}$ exist such that

$${T}_{n} := ({q}_{1,0} + {q}_{1,1}x){P}_{n-1}(x) + ({q}_{2,0} + {q}_{2,1}x){P}_{n-2}(x) + \cdots = n{P}_{n}(x).$$

Furthermore, it can be shown that these q’s are related in the following way:

$$\begin{array}{rcl} & {q}_{k+2,1} = \gamma {q}_{k+1,1} + \epsilon {q}_{k,1}; {q}_{1,1} = \alpha, {q}_{2,1} = \alpha \gamma,&\end{array}$$

(1.36)

$$\begin{array}{rcl} & {q}_{k+1,0} = \frac{1} {\alpha }\left ({q}_{k,1}(\delta - (k - 1)\epsilon ) + {q}_{k+1,1}(\beta - \gamma k) + {q}_{k+2,1}(k + 1)\right ).&\end{array}$$

(1.37)

Thus, we have shown that ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ satisfies Corollary 1.2 and is therefore a Type 0 set.

As it turns out, we need the relations (1.36) and (1.37) to determine all of the Type 0 orthogonal sets. Now, if Eq. (1.36) is known, then Eq. (1.37) can be determined. Therefore, we solve (1.36) by considering the characteristic equation

$$\begin{array}{rcl}{ u}^{2} - \gamma u - \epsilon = 0,& &\end{array}$$

(1.38)

which has the solution

$${u}_{1,2} = \left (\gamma \pm \sqrt{{\gamma }^{2 } + 4\epsilon }\right )/2.$$

We analyze the different cases of the discriminant ${\gamma }^{2} + 4\epsilon$ and determine the value of q _k, 1, and therefore q _{k + 1, 0}, in each case. Thus, in each case, this leads to expressions for A(t) and H(t) in Eq. (1.20) and therefore the generating function (1.19), which yields the corresponding orthogonal set. For the repeated root cases $({\gamma }^{2} + 4\epsilon = 0)$ we use the general solution structure q _k, 1 = Au ₁ ^k + Bku ₁ ^k and for the distinct root cases $({\gamma }^{2} + 4\epsilon \neq 0)$, we use the general solution structure q _k, 1 = Au ₁ ^k + Bu ₂ ^k, where u ₁ and u ₂ are the roots listed above. There are a total of four cases, which we label as Ia, Ib, IIa and IIb. We work out the rigorous details of Case Ia and summarize the remaining cases.

Case Ia.

${\gamma }^{2} + 4\epsilon = 0$ and $\gamma \neq 0$In this case, we see that Eq. (1.38) yields ${u}_{1} = {u}_{2} = \gamma /2$ as a solution, which implies

$${q}_{k,1} = A{(\gamma /2)}^{k} + Bk{(\gamma /2)}^{k}.$$

Using our initial conditions in Eq. (1.36), we have the following system:

$$\begin{array}{rcl} & {q}_{1,1} = \frac{1} {2}A\gamma + \frac{1} {2}B\gamma = \alpha, & \\ & {q}_{2,1} = \frac{1} {4}A{\gamma }^{2} + \frac{1} {4}B{\gamma }^{2} = \alpha \gamma,& \\ \end{array}$$

which has the solutions A = 0 and B = 2α ∕ γ and thus,

$$\begin{array}{rcl}{ q}_{k,1} = k\alpha {(\gamma /2)}^{k-1}.& &\end{array}$$

(1.39)

By substituting the right-hand side of Eq. (1.36) into Eq. (1.37) and then using Eq. (1.39) accordingly, after some algebraic manipulations, we obtain the following:

$${q}_{k+1,0} ={ \left (\frac{\gamma } {2}\right )}^{k-1}\left (\frac{1} {2}(\beta \gamma + 2\delta )k + \frac{1} {2}(\beta + \gamma )\gamma + \frac{1} {2}({\gamma }^{2} + 4\epsilon )k\right ).$$

Then, recalling that the discriminant is zero in this case, we obtain

$$\begin{array}{rcl}{ q}_{k+1,0} ={ \left (\frac{\gamma } {2}\right )}^{k-1}\left (\frac{1} {2}(\beta \gamma + 2\delta )k + \frac{1} {2}(\beta + \gamma )\gamma \right ).& &\end{array}$$

(1.40)

For this case, we can now determine the series H(t) and A(t) as in Eq. (1.20) using Eq. (1.25). We first substitute (1.39) into our H′(t) expression of Eq. (1.25) and then integrate, which leads to the geometric series

$$H(t) = \alpha t\sum\limits_{k=0}^{\infty }{\left (\frac{\gamma t} {2} \right )}^{k} = \frac{2\alpha t} {2 - \gamma t}.$$

It is also worth mentioning that the formal inverse of H(t) above is therefore readily determined to be

$$J(t) = \frac{2t} {2\alpha + \gamma t}.$$

For A(t), we first notice that we have the following first-order differential equation:

$$A^{\prime}(t) -\sum\limits_{k=0}^{\infty }{q}_{ k+1,0}{t}^{k}A(t) = 0,$$

which, via integrating factor, has the solution

$$A(t) = \mu \exp \left [\int \nolimits \nolimits \sum\limits_{k=0}^{\infty }{q}_{ k+1,0}{t}^{k}\mathrm{d}t\right ].$$

We then substitute (1.40) into the result directly above, evaluate the sum and integrate, which eventually yields

$$A(t) = \mu {\left (\frac{2 - \gamma t} {2} \right )}^{-2\delta (\gamma -2)/{\gamma }^{2} }\exp \left (\frac{2t(\beta \gamma + 2\delta )} {\gamma (2 - \gamma t)} \right ),$$

with $(-2\delta (\gamma - 2)/{\gamma }^{2})$ not equal to a nonnegative integer, so that ${\mu }_{n}\neq 0$ is satisfied.

Case Ib.

${\gamma }^{2} + 4\epsilon = 0$ and $\gamma = 0$ In this case, we also have $\epsilon = 0$. This gives

$${q}_{1,1} = \alpha, {q}_{k,1} = 0 (k > 1), {q}_{1,0} = \beta, {q}_{2,0} = \delta \mathrm{and} {q}_{k,0} = 0 (k > 2)$$

and thus,

$$H(t) = \alpha t, J(t) = t/\alpha \mathrm{and} A(t) = \mu \exp \left (\beta t + \frac{1} {2}\delta {t}^{2}\right ),$$

with $\delta \neq 0$, so that ${\mu }_{n}\neq 0$ holds. For Cases IIa and IIb that follow $({\gamma }^{2} + 4\epsilon \neq 0)$, we first derive the general q _k, 1 and q _{k + 1, 0} terms. We have

$$\begin{array}{rcl}{ q}_{k,1} = \frac{\alpha ({u}_{1}^{k} - {u}_{2}^{k})} {\sqrt{{\gamma }^{2 } + 4\epsilon }} \mathrm{and} {q}_{k+1,0} = \frac{\lambda {u}_{1}^{k} - \sigma {u}_{2}^{k}} {\sqrt{{\gamma }^{2 } + 4\epsilon }} & &\end{array}$$

(1.41)

with

$$\begin{array}{rcl} \lambda := \delta + 2\epsilon + (\beta + \gamma ){u}_{1} \mathrm{and} \sigma := \delta + 2\epsilon + (\beta + \gamma ){u}_{2}.& & \\ \end{array}$$

Case IIa.

${\gamma }^{2} + 4\epsilon \neq 0$ and $\epsilon = 0$ In this case, we consequently have $\gamma \neq 0$. Via Eq. (1.41), we observe that

$$\begin{array}{rcl} H(t)& =& \alpha {\int \nolimits \nolimits }_{0}^{t} \frac{\mathrm{d}\tau } {1 - \gamma \tau } = -\frac{\alpha } {\gamma }\ln (1 - \gamma t), J(t) = \frac{1} {\gamma }\left (1 -\exp \left (-\frac{\gamma t} {\alpha } \right )\right ), \\ A(t)& =& \mu {(1 - \gamma t)}^{-(\beta \gamma +{\gamma }^{2}+\delta )/{\gamma }^{2} }\exp \left (-\frac{\delta t} {\gamma } \right ), \\ \end{array}$$

with $\delta \neq 0$ so that ${\mu }_{n}\neq 0$.

Case IIb.

${\gamma }^{2} + 4\epsilon \neq 0$ and $\epsilon \neq 0$ Here, from Eq. (1.41), one can achieve

$$\begin{array}{rcl} H(t)& = \alpha {\int \nolimits \nolimits }_{0}^{t} \frac{\mathrm{d}\tau } {1-\gamma \tau -\epsilon {\tau }^{2}} = \alpha \frac{\ln (({u}_{2}t-1)/({u}_{1}t-1))} {\sqrt{{\gamma }^{2 } +4\epsilon }}, & \\ J(t)& = \frac{\exp \left ( \frac{t} {\alpha }\sqrt{{\gamma }^{2 } +4\epsilon }\right )-1} {{u}_{1}\exp \left ( \frac{t} {\alpha }\sqrt{{\gamma }^{2 } +4\epsilon }\right )-{u}_{2}}, A(t) = \mu \frac{{(1-{u}_{2}t)}^{{h}_{2}}} {{(1-{u}_{1}t)}^{{h}_{1}}},& \\ \end{array}$$

with

$${h}_{i} = \frac{{u}_{i}(\beta + \gamma ) + (\delta + 2\epsilon )} {{u}_{i}\sqrt{{\gamma }^{2 } + 4\epsilon }}, i = 1,2.$$

We next state the main result of this section and for simplicity, we redefine the parameters involved in each of our H(t) and A(t) expressions in the same manner as Sheffer. This result displays the general forms of each of the generating functions for the orthogonal Type 0 sets.

Theorem 1.8.

A polynomial set ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ is Type 0 and orthogonal if and only if $A(t){\mathrm{e}}^{xH(t)}$ is of one of the following forms:

$$\begin{array}{rcl} & A(t){\mathrm{e}}^{xH(t)} = \mu {(1 - bt)}^{c}\exp \left (\frac{dt+atx} {1-bt} \right ), abc\mu \neq 0,&\end{array}$$

(1.42)

$$\begin{array}{rcl} & A(t){\mathrm{e}}^{xH(t)} = \mu \exp \left (t(b + ax) + c{t}^{2}\right ), ac\mu \neq 0,&\end{array}$$

(1.43)

$$\begin{array}{rcl} & A(t){\mathrm{e}}^{xH(t)} = \mu {\mathrm{e}}^{ct}{(1 - bt)}^{d+ax}, abc\mu \neq 0,&\end{array}$$

(1.44)

$$\begin{array}{rcl} & A(t){\mathrm{e}}^{xH(t)} = \mu {(1 - t/c)}^{{d}_{1}+x/a}{(1 - t/b)}^{{d}_{2}-x/a}, abc\mu \neq 0, b\neq c.&\end{array}$$

(1.45)

Proof.

This statement follows from the above analysis.

By judiciously choosing each of the parameters in Eqs. (1.42)–(1.45) we can achieve all of the Sheffer Type 0 orthogonal polynomials previously discussed. For emphasis, we write each of these parameter selections below and then display the corresponding generating function as it appears in contemporary literature.

In Eq. (1.42), we select the parameters as μ = 1, $a = -1$, b = 1, c =  − (α + 1), and d = 0 to obtain

$$\begin{array}{rcl} \sum\limits_{n=0}^{\infty }{L}_{ n}^{(\alpha )}(x){t}^{n} = {(1 - t)}^{-(\alpha +1)}\exp \left ( \frac{xt} {t - 1}\right ).& & \\ \end{array}$$

With the assignments μ = 1, a = 2, b = 0, and c =  − 1 in Eq. (1.43), we have

$$\begin{array}{rcl} \sum\limits_{n=0}^{\infty }\frac{{H}_{n}(x){t}^{n}} {n!} =\exp (2xt - {t}^{2}).& & \\ \end{array}$$

If in Eq. (1.44) we choose $\mu = 1$, a = 1, b = 1 ∕ a, c = 1, and d = 0, then we achieve

$$\begin{array}{rcl} \sum\limits_{n=0}^{\infty }\frac{{C}_{n}(x;a){t}^{n}} {n!} ={ \mathrm{e}}^{t}{\left (1 - \frac{t} {a}\right )}^{x}.& & \\ \end{array}$$

In Eq. (1.45), we select $\mu = 1$, a = 1, b = 1, $c =\mathrm{ arbitrary} \mathrm{constant}$, d ₁ = 0, and d ₂ =  − β leading to

$$\begin{array}{rcl} \sum\limits_{n=0}^{\infty }\frac{{(\beta )}_{n}} {n!} M(x;\beta,c){t}^{n} ={ \left (1 -\frac{t} {c}\right )}^{x}{(1 - t)}^{-(x+\beta )}.& & \\ \end{array}$$

Taking $\mu = 1$, a =  − i, b = e^iϕ, c = e^− iϕ, and ${d}_{1} = {d}_{2} = -\lambda$ in Eq. (1.45) leads to

$$\begin{array}{rcl} \sum\limits_{n=0}^{\infty }{P}_{ n}^{(\lambda )}(x;\phi ){t}^{n} = {(1 -{\mathrm{e}}^{i\phi }t)}^{-\lambda +ix}{(1 -{\mathrm{e}}^{-i\phi }t)}^{-\lambda -ix}.& & \\ \end{array}$$

Lastly, selecting μ = 1, a = 1, b =  − 1, c = p ∕ (1 − p), d ₁ = 0, and d ₂ = N in Eq. (1.45) yields

$$\begin{array}{rcl} \sum\limits_{n=0}^{N}\mathrm{C}(N,n){K}_{ n}(x;p,N){t}^{n} ={ \left (1 -\frac{1 - p} {p} t\right )}^{x}{(1 + t)}^{N-x},& & \\ \end{array}$$

for $x = 0,1,2,\ldots,N,$ where $\mathrm{C}(N,n)$ denotes the binomial coefficient.

Refer to [6] and the references therein for definitions and characterizations of each of these A-Type 0 sets. For additional analyses regarding these orthogonal sets, also consider [3,  4,  5,  8,  10].

1.4 An Overview of the Classification of Type

We now discuss the essential details of the three kinds of characterizations that Sheffer developed in [9], which are entitled A-Type, B-Type and C-Type. The definition of Type is dependent on which characterization of Type 0 that is to be generalized. That is, each of the Sheffer Types generalizes a certain Type 0 characterizing structure.

1.4.1 The Sheffer A-Type Classification

We deem all of the characterizations up to this juncture as A-Type k and therefore restate Definition 1.6 accordingly.

Definition 1.9.

Let the set $\mathcal{S} :={ \left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ correspond to the unique operator J. Then, $\mathcal{S}$ is of A-Type k if the coefficients ${\left \{{L}_{j}(x)\right \}}_{j=0}^{\infty }$, as defined in Eq. (1.12), are such that $\deg ({L}_{j}(x)) \leq k$ for all i and there exists at least one ${L}_{i}(x) \in {\left \{{L}_{j}(x)\right \}}_{j=0}^{\infty }$ such that $\deg ({L}_{i}(x)) = k$. If ${\left \{{L}_{j}(x)\right \}}_{j=0}^{\infty }$ is unbounded, then $\mathcal{S}$ is of A-Type $\infty$.

We also will make use of the following result.

Theorem 1.9.

The set ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ is of A-Type 0 if and only if a sequence of constants $\left \{{s}_{n}\right \}$ exist such that

$$\begin{array}{rcl}{ P^{\prime}}_{n}(x) = {s}_{1}{P}_{n-1}(x) + {s}_{2}{P}_{n-2}(x) + \cdots + {s}_{n}{P}_{0}(x), n = 1,2,3,\ldots,& &\end{array}$$

(1.46)

where the elements of the sequence $\left \{{s}_{n}\right \}$ are the same as those in Eq. (1.18).

Proof.

We differentiate (1.19) with respect to x and see that

$$\sum\limits_{n=0}^{\infty }{P^{\prime}}_{ n}(x){t}^{n} = H(t)A(t){\mathrm{e}}^{xH(t)} =\sum\limits_{n=0}^{\infty }{P}_{ n}(x)H(t){t}^{n}.$$

Then, from comparing coefficients of t ⁿ in

$$\sum\limits_{n=0}^{\infty }{P^{\prime}}_{ n}(x){t}^{n} =\sum\limits_{n=0}^{\infty }{P}_{ n}(x)[{s}_{1}t + {s}_{2}{t}^{2} + {s}_{ 3}{t}^{3} + \cdots \,]{t}^{n},$$

we obtain our result. It is readily seen that this argument is both necessary and sufficient.

1.4.2 The Sheffer B-Type Classification

To begin, we state that Theorem 1.9 can be extended in a natural way. Consider the structure

$${T}_{0}(x){P}_{n-1}(x) + {T}_{1}(x){P}_{n-2}(x) + \cdots + {T}_{n-1}(x){P}_{0}(x), n = 1,2,3,\ldots.$$

Then, by successively setting $n = 1,2,\ldots$, we observe that given a set ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$, a unique sequence of polynomials $\left \{{T}_{n}(x)\right \}$ exists such that

$$\begin{array}{rcl}{ P^{\prime}}_{n}(x) = {T}_{0}(x){P}_{n-1}(x) + {T}_{1}(x){P}_{n-2}(x) + \cdots + {T}_{n-1}(x){P}_{0}(x), n = 1,2,\ldots & &\end{array}$$

(1.47)

with T ₀(x)≠0. However, $\deg ({T}_{n}(x)) \leq n$ since, e.g., Eq. (1.47) readily reduces to Eq. (1.46) if ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ is of A-Type 0. The following statement immediately follows.

Theorem 1.10.

For every set ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ there exist a unique sequence of polynomials $\left \{{T}_{n}(x)\right \}$ , with $\deg ({T}_{n}(x)) \leq n$ , such that Eq. (1.47)holds.

The converse of this theorem does not hold, as we have the following result.

Theorem 1.11.

Given a sequence of polynomials $\left \{{T}_{n}(x)\right \}$ , with $\deg ({T}_{n}(x)) \leq n$ , there exist infinitely many sets ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ such that Eq. (1.47)is satisfied.

Proof.

This result is immediate, since the constant term of P _n (x) in Eq. (1.47) can be made arbitrary.

Based on the above results, we have the following definition.

Definition 1.10.

A set ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ is of B-Type k if the maximum degree of the polynomials $\left \{{T}_{n}(x)\right \}$ in Eq. (1.47) is k. Otherwise, ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ is classified as B-Type $\infty$.

We can now establish the following statement.

Theorem 1.12.

A set ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ is of B-Type 0 if and only if it is of A-Type 0.

Proof.

$(\Rightarrow )$ If ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ is of B-Type 0, then $\left \{{T}_{n}(x)\right \}$ in Eq. (1.47) must be a sequence of constants. Moreover, Eq. (1.47) is reduced to Eq. (1.46) and thus ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ is an A-Type 0 set.

$(\Leftarrow )$ If ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ is of A-Type 0, then Eq. (1.46) holds, which is Eq. (1.47) with constant coefficients. Thus, ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ is of B-Type 0.

We also have the classification of B-Type k sets below.

Theorem 1.13.

The characterization(1.47)is equivalent to the relation

$$\begin{array}{rcl} H(x,t) = A(t)\exp \left [t{\int \nolimits \nolimits }_{0}^{x}T(\xi,t)\mathrm{d}\xi \right ],& &\end{array}$$

(1.48)

with A(t) as in Eq. (1.20).

Proof.

$(\Rightarrow )$ First, we define

$$\begin{array}{rcl} H := H(x,t) =\sum\limits_{k=0}^{\infty }{P}_{ k}(x){t}^{k},& &\end{array}$$

(1.49)

$$\begin{array}{rcl} T := T(x,t) =\sum\limits_{k=0}^{\infty }{T}_{ k}(x){t}^{k}.& &\end{array}$$

(1.50)

We now consider

$$\begin{array}{rcl} tHT = t\sum\limits_{k=0}^{\infty }{P}_{ k}(x){t}^{k}\sum\limits_{k=0}^{\infty }{T}_{ k}(x){t}^{k}.& &\end{array}$$

(1.51)

We can find the coefficient of t ⁿ in the right-hand side of Eq. (1.51) by considering the following structure, which is t multiplied by the general term in each of the sums of Eq. (1.51):

$$t{P}_{{k}_{0}}{t}^{{k}_{0} }{T}_{{k}_{1}}{t}^{{k}_{1} }$$

and finding all nonnegative integer solutions to the equation $1 + {k}_{0} + {k}_{1} = n$, i.e., the sum of the t-exponents. For the solution k ₀ = n − 1 and k ₁ = 0, we obtain T ₀(x)P _n − 1(x) and for the solution k ₀ = n − 2 and k ₁ = 1, we achieve T ₁(x)P _n − 2(x). Continuing in this fashion, we see that the coefficient of t ⁿ turns out to be

$${T}_{0}(x){P}_{n-1}(x) + {T}_{1}(x){P}_{n-2}(x) + \cdots + {T}_{n-1}(x){P}_{0}(x) = {P^{\prime}}_{n}(x)$$

via Eq. (1.47). Thus, the right-hand side of Eq. (1.51) is $\sum\limits_{n=0}^{\infty }{P^{\prime}}_{n}(x){t}^{n} = \partial H(x,t)/\partial x$ and we have constructed the following first-order differential equation, as Eq. (1.51) becomes

$$\begin{array}{rcl} \frac{\partial } {\partial x}H - tTH = 0; H(0,t) =\sum\limits_{n=0}^{\infty }{P}_{ n}(0){t}^{n} = A(t),& &\end{array}$$

(1.52)

with A(t) as in Eq. (1.20). Furthermore, we have shown that this relation is equivalent to Eq. (1.47). The general solution to Eq. (1.52) can be determined via the integrating factor $\exp \left [-{\int \nolimits \nolimits }_{0}^{x}tT(\xi,t)\mathrm{d}\xi \right ]$ to be

$$H(x,t) = A(t)\exp \left [t{\int \nolimits \nolimits }_{0}^{x}T(\xi,t)\mathrm{d}\xi \right ],$$

where A(t) is an arbitrary power series with a nonzero constant term. However, in considering the initial condition of Eq. (1.52), we see that A(t) must be as in Eq. (1.20). This argument is certainly both necessary and sufficient.

Next, we note that if we write T(x, t) in Eq. (1.50) as a power series in x, as opposed to t, the coefficients of each of the x ⁿ-terms are power series in t. To facilitate this, momentarily let T _n (x) = c _n, n x ⁿ + c _{n, n − 1} x ^n − 1 + ⋯ + c _n, 0, and then we have

$$\begin{array}{rcl} T(x,t) = ({c}_{0,0} + {c}_{1,0}t + {c}_{2,0}{t}^{2} + \cdots \,)& +({c}_{ 1,1}t + {c}_{2,1}{t}^{2} + {c}_{ 3,1}{t}^{3} + \cdots \,)x & \\ & +({c}_{2,2}{t}^{2} + {c}_{3,2}{t}^{3} + {c}_{4,2}{t}^{4} + \cdots \,){x}^{2} + \cdots \,.& \\ \end{array}$$

Letting x = ξ above, integrating this result with respect to ξ from 0 to x and multiplying by t, we see that

$$t{\int \nolimits \nolimits }_{0}^{x}T(\xi,t)\mathrm{d}\xi = x{H}_{ 1}(t) + {x}^{2}{H}_{ 2}(t) + {x}^{3}{H}_{ 3}(t) + \cdots \,,$$

where each H _i (t) is a power series in t that starts with the t ⁱ-term (or higher if the coefficient of the t ⁱ-term is zero), i.e., H _i (t) = h _i, i t ⁱ + h _{i, i + 1} t ^i + 1 + h _{i, i + 2} t ^i + 2 + ⋯ . However, H ₁(t) must begin with a linear term in t, since by Eq. (1.47) T ₀(x)≠0 and h _1, 1 = T ₀(x) by construction. Thus, we can write (1.48) in the following way:

$$\begin{array}{rcl} H(x,t) = A(t)\exp \left [x{H}_{1}(t) + {x}^{2}{H}_{ 2}(t) + {x}^{3}{H}_{ 3}(t) + \cdots \,\right ], {h}_{1,1}\neq 0.& &\end{array}$$

(1.53)

Now, for ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ to be of B-Type k, it is necessary and sufficient that T(x, t) is a polynomial in x of degree k. This restriction is equivalent to terminating the sum $\left [x{H}_{1}(t) + {x}^{2}{H}_{2}(t) + {x}^{3}{H}_{3}(t) + \cdots \,\right ]$ in Eq. (1.53) at k + 1 via the presence of the integral in Eq. (1.48). Thus, we obtain (1.54) in Theorem 1.14 below, which is the culmination of the above analysis.

Theorem 1.14.

A set is of B-Type k if and only if

$$\begin{array}{rcl} & H(x,t) = A(t)\mathrm{exp}\left [x{H}_{1}(t) + \cdots + {x}^{k+1}{H}_{k+1}(t)\right ], & \\ & \mathrm{with} {H}_{i}(t) = {h}_{i,i}{t}^{i} + {h}_{i,i+1}{t}^{i+1} + \cdots \,, {h}_{1,1}\neq 0, i = 1,2,\ldots,k + 1.&\end{array}$$

(1.54)

1.4.3 The Sheffer C-Type Classification

The last classification that Sheffer developed is entitled C-Type. As was done in the previous section, we establish all of the theorems necessary for understanding this classification, beginning with the following.

Theorem 1.15.

For each set ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ , there exist a unique sequence of polynomials ${\left \{{U}_{n}(x)\right \}}_{n=0}^{\infty }$ , with $\deg ({U}_{n}(x)) \leq n$ , such that

$$\begin{array}{rcl} n{P}_{n}(x) = {U}_{1}(x){P}_{n-1}(x) + {U}_{2}(x){P}_{n-2}(x) + \cdots + {U}_{n}(x){P}_{0}(x), n = 1,2,\ldots.& &\end{array}$$

(1.55)

Proof.

Analogous to establishing (1.47) in Theorem 1.10, we set $n = 1,2,\ldots$ in Eq. (1.55) and then successively and uniquely determine the set ${\left \{{U}_{n}(x)\right \}}_{n=0}^{\infty }$. Through this process, it becomes clear that no polynomial U _n (x) can exceed the degree of its subscript. The result therefore follows.

This leads us to the following definition.

Definition 1.11.

A set ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ is classified as C-Type k if the maximum degree of the polynomials $\left \{{U}_{n}(x)\right \}$ in Eq. (1.55) is k + 1. Else, ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ is classified as C-Type ∞.

We note that we use k + 1 in the above definition, as opposed to k as in our previous Type definitions, since otherwise, Eq. (1.55) would not be satisfied. As an example, consider the case when each U _i (x) is constant.

Now, we define

$$U := U(x,t) =\sum\limits_{n=0}^{\infty }{U}_{ n+1}(x){t}^{n}.$$

Then, calling upon Eq. (1.55) and using the same methodology that was used to construct (1.52) in the previous section, we derive the following first-order differential equation:

$$\frac{\partial } {\partial t}H - UH = 0; H(x,0) = {P}_{0}(x) = {a}_{0}$$

with H : = H(x, t) as in Eq. (1.49). This can easily be solved using the integrating factor $\exp \left [-{\int \nolimits \nolimits }_{0}^{t}U(x,\tau )\mathrm{d}\tau \right ]$, yielding the solution

$$\begin{array}{rcl} H(x,t) = {a}_{0}\exp \left [{\int \nolimits \nolimits }_{0}^{t}U(x,\tau )\mathrm{d}\tau \right ],& &\end{array}$$

(1.56)

after incorporating our initial condition. Then, from comparing (1.56) with Eq. (1.48), we have

$${a}_{0}\exp \left [{\int \nolimits \nolimits }_{0}^{t}U(x,\tau )\mathrm{d}\tau \right ] = A(t)\mathrm{exp}\left [t{\int \nolimits \nolimits }_{0}^{x}T(\xi,t)\mathrm{d}\xi \right ]$$

and taking the natural logarithm of both sides of this relation leads to

$$\begin{array}{rcl} \ln {a}_{0} +{ \int \nolimits \nolimits }_{0}^{t}U(x,\tau )\mathrm{d}\tau =\ln A(t) + t{\int \nolimits \nolimits }_{0}^{x}T(\xi,t)\mathrm{d}\xi.& &\end{array}$$

(1.57)

Next, differentiate (1.57) with respect to t in order to obtain

$$\begin{array}{rcl} U(x,t) = \frac{A^{\prime}(t)} {A(t)} +{ \int \nolimits \nolimits }_{0}^{x} \frac{\partial } {\partial t}\left (tT(\xi,t)\right )\mathrm{d}\xi & &\end{array}$$

(1.58)

and also differentiate (1.57) with respect to x, which leads to

$$\begin{array}{rcl} tT(x,t) ={ \int \nolimits \nolimits }_{0}^{t} \frac{\partial } {\partial x}U(x,\tau )\mathrm{d}\tau.& &\end{array}$$

(1.59)

We now substitute (1.59) into Eq. (1.58) and then put this new expression for U(x, t) into Eq. (1.56), which leads to

$$\begin{array}{rcl} H(x,t) = A(t)\exp \left [{\int \nolimits \nolimits }_{0}^{t}{ \int \nolimits \nolimits }_{0}^{x} \frac{\partial } {\partial \xi }U(\xi,\tau )\xi \mathrm{d}\tau \right ].& &\end{array}$$

(1.60)

We see that the exponent in Eq. (1.60) is a polynomial in x of degree k + 1 if and only if the polynomial of maximal degree in ${\left \{{U}_{n}(x)\right \}}_{n=1}^{\infty }$ is of degree k + 1, i.e., if and only if the set ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ for which the sequence ${\left \{{U}_{n}(x)\right \}}_{n=1}^{\infty }$ corresponds to is of C-Type k. Under this assumption, the exponent in Eq. (1.60) has the following structure:

$$\tilde{{U}}_{1}(x)t +\tilde{ {U}}_{2}(x){t}^{2} + \cdots +\tilde{ {U}}_{ k+1}(x){t}^{k+1},$$

which is the same form as the exponent in Eq. (1.48) when it corresponds to a B-Type 0 set. Thus, Eq. (1.60) can therefore be reduced to Eq. (1.54) using the same type of manipulation that was used in the previous section. Hence, we have proven the following theorem.

Theorem 1.16.

A set ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ is of C-Type k if and only if it is of B-Type k.

1.4.4 A Summary of the Rainvilleσ-Type Classification

To complete our summary of Type, we conclude with a natural extension of Sheffer’s classification, entitled $\sigma$ -Type, which was constructed by E.D. Rainville and originally appeared in [8]. Now, we have seen that various results were ascertained from the generating function (1.19), which we now know characterizes A-Type 0 sets. Moreover, we know that for $y =\exp \left [xH(t)\right ]$ with $D := \mathrm{d}/\mathrm{d}x$ we have

$$Dy = H(t)y.$$

We intend to then modify the operator D so that it behaves in a similar fashion and also obtain a generating relation for the most basic type of sets classified by this modified operator. That is, if we replace D above by another differential operator, say σ, and the exponential function $\exp (z)$ by another function, e.g., F(z), then we want to find $\sigma$ such that

$$\sigma F(z) = F(z).$$

From there, we can construct an analogue of Eq. (1.19) in Theorem 1.5. This leads to the following definition of our differential operator $\sigma$ (using the same notation as in [8]).

Definition 1.12.

We define the differential operator $\sigma$ as follows:

$$\sigma := D{\prod \nolimits }_{i=1}^{q}(\theta + {b}_{ i} - 1), D := \frac{\mathrm{d}} {\mathrm{d}x} \mathrm{and} \theta := xD,$$

with ${b}_{i}\neq 0$ and b _i not equal to a negative integer.

In fact, this operator is a degree-lowering operator. To facilitate its application on polynomials and its degree-lowering nature, consider acting on a monomial, like x ², with q = 1 and b ₁ = 1. The function of the q and b _i -terms will become more transparent in Theorem 1.17 at the end of this section.

We now define $\sigma$ -Type.

Definition 1.13.

Let ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ be a set such that

$${J}_{\sigma }\left [{P}_{n}(x)\right ] =\sum\limits_{k=0}^{\infty }{T}_{ k}(x){\sigma }^{k+1}{P}_{ n}(x) = {P}_{n-1}(x),$$

with $\deg ({T}_{j}(x)) \leq k$. If the polynomial of maximal degree in the set of coefficients T _j (x) _j = 0 ^∞6 is of degree k, then we classify ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ as $\sigma$ -Type k. If the coefficients ${\left \{{T}_{j}(x)\right \}}_{j=0}^{\infty }$ are unbounded, then ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ is $\sigma$ -Type $\infty$.

Based on this definition, we see that polynomials of $\sigma$ -Type 0 satisfy the following form:

$${J}_{\sigma }\left [{P}_{n}(x)\right ] =\sum\limits_{k=0}^{\infty }{c}_{ k}{\sigma }^{k+1}{P}_{ n}(x) = {P}_{n-1}(x),$$

where each c _k is a constant and ${c}_{0}\neq 0$. Since each of the c _k ’s is a constant, analogous to our A-Type 0 analysis, there exists a generating function for ${J}_{\sigma }$, which we call ${J}_{\sigma }(t)$, and a corresponding inverse ${H}_{\sigma }(t)$ such that ${J}_{\sigma }({H}_{\sigma }(t)) = {H}_{\sigma }({J}_{\sigma }(t)) = t$:

$$\begin{array}{rcl}{ J}_{\sigma }(t) =\sum\limits_{n=0}^{\infty }{c}_{ n}{t}^{n+1}, {c}_{ 0}\neq 0 \mathrm{and} {H}_{\sigma }(t) =\sum\limits_{n=0}^{\infty }{h}_{ n}{t}^{n+1}, {h}_{ 0}\neq 0.& & \\ \end{array}$$

This leads us to the main characterization theorem for $\sigma$ -Type 0 sets.

Theorem 1.17.

A set is of $\sigma$-Type 0 , with $\sigma$ as defined in Definition 1.12,if and only if ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ satisfies the generating function

$$A{(t)}_{0}{F}_{q}(-;{b}_{1},{b}_{2},\cdots \,,{b}_{q};xH(t)) =\sum\limits_{n=0}^{\infty }{P}_{ n}(x){t}^{n}$$

with H(t) as defined above and A(t) as in Eq. (1.20).

In Theorem 1.17, we wrote the generating function in the generalized hypergeometric form, which is defined as

$${ }_{r}{F}_{s}\left (\begin{array}{c|} {a}_{1},\ldots,{a}_{r} \\ {b}_{1},\ldots,{b}_{s} \end{array} x\right ) =\sum\limits_{k=0}^{\infty }\frac{{\left ({a}_{1},\ldots,{a}_{r}\right )}_{k}} {{\left ({b}_{1},\ldots,{b}_{s}\right )}_{k}} \frac{{x}^{k}} {k!},$$

(1.61)

where the Pochhammer symbol (a) _k is

$$\begin{array}{rcl}{ (a)}_{k} := a\left (a + 1\right )\left (a + 2\right )\cdots \left (a + k - 1\right ),{ \left (a\right )}_{0} := 1,& &\end{array}$$

(1.62)

and

$${({a}_{1},\ldots,{a}_{j})}_{k} := {({a}_{1})}_{k}\ldots {({a}_{j})}_{k}.$$

We now see that the selection of q is dependent on the number of denominator parameters (the b _i ’s) in the generating function of Theorem 1.17. For a proof of Theorem 1.17, the interested reader can refer to [8].

1.5 A Brief Discussion of Meixner’s Analysis

In 1934, J. Meixner published [7] (written in German) in which he considered the generating relation (1.19) (with the same assumptions on the A(t) and H(t) as in Sheffer’s work [9]) to be the definition of a certain class of polynomials. From there, he determined all sets that satisfy this relation that were also orthogonal and reached the same conclusions as Sheffer did in [9]. In other words, Meixner determined all orthogonal sets ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ that satisfy

$$\begin{array}{rcl} f(t){\mathrm{e}}^{xu(t)} =\sum\limits_{n=0}^{\infty }\frac{{P}_{n}(x)} {n!} {t}^{n}; f(0) = 1, u(0) = 0, \frac{\mathrm{d}} {\mathrm{d}t}u(0) = 1,& &\end{array}$$

(1.63)

which we have written using Meixner’s notation, which we essentially adhere to throughout this section.

In a similar manner as Sheffer, Meixner defined a general degree-lowering, linear, differential operator of infinite order, which we call J as opposed to his “t” to avoid confusion with t-parameters, which satisfies a certain commutation relation with the differential operator D. Moreover, J(t) is a formal power series without a constant term and with a unitary linear coefficient. The formal power series inverse of J(t) was defined to be u(t), i.e., u(J(t)) = J(u(t)) = t.

Throughout his work, Meixner considered ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ to be a set of monic polynomials. With this convention, he utilized the following monic three-term recurrence relation:

$$\begin{array}{rcl}{ P}_{n+1}(x) = (x + {l}_{n+1}){P}_{n}(x) + {k}_{n+1}{P}_{n-1}(x), n = 0,1,2,\ldots & &\end{array}$$

(1.64)

with ${l}_{n+1} \in \mathbb{R}$ and ${k}_{n+1} \in {\mathbb{R}}^{-}$ and demonstrated that

$$\begin{array}{rcl} J(D){P}_{n}(x) = n{P}_{n-1}(x).& &\end{array}$$

(1.65)

Now, we act on Eq. (1.64) with J(D) as in Eq. (1.65) and obtain

$$\begin{array}{rcl} (n + 1){P}_{n}(x) = (x + {l}_{n+1})n{P}_{n-1}(x) + J^{\prime}(D){P}_{n}(x) + {k}_{n+1}(n - 1){P}_{n-2}(x),& &\end{array}$$

(1.66)

where J′(D) is of course the derivative of the formal power series J(D). Then, we replace n with n − 1 in Eq. (1.64) and multiply both sides by n to obtain

$$\begin{array}{rcl} n{P}_{n}(x) = (x + {l}_{n})n{P}_{n-1}(x) + {k}_{n}n{P}_{n-2}(x).& &\end{array}$$

(1.67)

Next, we subtract (1.67) from Eq. (1.66) which yields

$$\left (1 - J^{\prime}(D)\right ){P}_{n}(x) = ({l}_{n+1} -{l}_{n})n{P}_{n-1}(x)+\left (\frac{{k}_{n+1}} {n} - \frac{{k}_{n}} {n - 1}\right )n(n-1){P}_{n-2}(x).$$

We then replace n with n + 1 in the recursion coefficients above leading to

$$\begin{array}{rcl} \left (1 - J^{\prime}(D)\right ){P}_{n}(x) = ({l}_{n+2} - {l}_{n+1})n{P}_{n-1}(x) + \left ( \frac{{k}_{n+2}} {n + 1} -\frac{{k}_{n+1}} {n} \right )n(n - 1){P}_{n-2}(x).& & \\ \end{array}$$

We assign

$$\begin{array}{rcl} & \lambda := {l}_{n+1} - {l}_{n}, n = 1,2,\ldots,&\end{array}$$

(1.68)

$$\begin{array}{rcl} & \kappa := \frac{{k}_{n+1}} {n} - \frac{{k}_{n}} {n-1}, n = 2,3,\ldots &\end{array}$$

(1.69)

giving

$$\begin{array}{rcl} \left (1 - J^{\prime}(D)\right ){P}_{n}(x) = \lambda n{P}_{n-1}(x) + \kappa n(n - 1){P}_{n-2}(x), n = 0,1,2,\ldots & & \\ \end{array}$$

and via Eq. (1.65) we see that this recurrence can be written as

$$\begin{array}{rcl} \left (1 - J^{\prime}(D)\right ){P}_{n}(x) = \lambda J(D){P}_{n}(x) + \kappa {J}^{2}(D){P}_{ n}(x), n = 0,1,2,\ldots & &\end{array}$$

(1.70)

From rewriting Eqs. (1.68) and (1.69) as

$$\begin{array}{rcl}{ l}_{n+1} = {l}_{n} + \lambda \mathrm{and} {k}_{n+1} = n\left ( \frac{{k}_{n}} {n - 1} + \kappa \right )& & \\ \end{array}$$

and iterating, we see that ${l}_{n+1} = {l}_{1} + n\lambda$ and ${k}_{n+1} = {k}_{2} + (n - 1)\kappa$. We then substitute these recursion coefficients into Eq. (1.64) to obtain the following three-term recurrence relation:

$$\begin{array}{rcl}{ P}_{n+1}(x) = (x + {l}_{1} + n\lambda ){P}_{n}(x) + n({k}_{2} + (n - 1)\kappa ){P}_{n-1}(x), n = 0,1,2,\ldots & &\end{array}$$

(1.71)

with k ₂ < 0 and $\kappa \leq 0$ from the original restrictions imposed upon Eq. (1.64).

Now, from Eq. (1.70), it follows that

$$J^{\prime}(u(t)) = 1 - \lambda t - \kappa {t}^{2}.$$

Differentiating both sides of the relation J(u(t)) = t tells us that J′(u(t)) = 1 ∕ u′(t). Thus, our relation directly above becomes

$$\begin{array}{rcl} J^{\prime}(u(t)) = 1 - \lambda t - \kappa {t}^{2} = \frac{1} {u^{\prime}(t)}.& &\end{array}$$

(1.72)

By setting x = 0, we note that the generating function (1.63) turns into

$$f(t) =\sum\limits_{n=0}^{\infty }\frac{{P}_{n}(0){t}^{n}} {n!}.$$

Thus, multiplying both sides of Eq. (1.71) by t ⁿ ∕ n! , setting x = 0 and summing for n = 0, 1, 2, … lead to the differential equation

$$\frac{f^{\prime}(t)} {f(t)} = \frac{{k}_{2}t} {1 - \lambda t - \kappa {t}^{2}}.$$

Factoring $1 - \lambda t - \kappa {t}^{2}$ as $(1 - \alpha t)(1 - \beta t)$ gives

$$\begin{array}{rcl} \frac{f^{\prime}(t)} {f(t)} = \frac{{k}_{2}t} {(1 - \alpha t)(1 - \beta t)}, \alpha,\beta \in \mathbb{C}.& &\end{array}$$

(1.73)

We can now exhaust every possible combination of α and $\beta$ (and incorporate $\lambda$ and $\kappa$ as well), substitute each of them into Eqs. (1.72) and (1.73), and solve the resulting differential equations. In each case, the solution to Eq. (1.72) will yield an expression for $u(t)$ and the solution to Eq. (1.73) will yield an expression for f(t). In substituting these into Eq. (1.63), we achieve a generating function for each orthogonal set.

Below, we write the results for each of these aforementioned cases. In each case, part (i) denotes the solutions to Eqs. (1.72) and (1.73), which respectively yield our expressions for u(t) and f(t). In part (ii), we write each of the resulting generating functions in their rescaled form, so they appear as they do in the contemporary literature.

Case I.

The Hermite Polynomials: α = β = 0 $(\lambda = \kappa = 0)$

$$\begin{array}{rcl} & (i) u(t) = t \mathrm{and} f(t) =\exp \left ({k}_{2}{t}^{2}/2\right )& \\ & (ii) \exp \left (2xt - {t}^{2}\right ) =\sum\limits_{n=0}^{\infty }\frac{H(x)} {n!} {t}^{n} & \\ \end{array}$$

Case II.

The Laguerre Polynomials: $\alpha = \beta \neq 0$

$$\begin{array}{rcl} & (i) u(t) = \frac{t} {1-\alpha t} \mathrm{and} f(t) = {(1 - \alpha t)}^{{k}_{2}/{\alpha }^{2} }\exp \left ( \frac{t} {1-\alpha t} \frac{{k}_{2}} {\alpha } \right )& \\ & (ii) {(1 - t)}^{-(\alpha +1)}\exp \left ( \frac{xt} {t-1}\right ) =\sum\limits_{n=0}^{\infty }{L}_{n}^{(\alpha )}(x){t}^{n} & \\ \end{array}$$

Case III.

The Charlier Polynomials: $\alpha \neq 0$ and $\beta = 0$ ($\kappa = 0$)

$$\begin{array}{rcl} & (i) u(t) = -\frac{1} {\alpha }\ln (1 - \alpha t) \mathrm{and} f(t) = {(1 - \alpha t)}^{-{k}_{2}/{\alpha }^{2} }{\mathrm{e}}^{-{k}_{2}t/\alpha }& \\ & (ii){ \mathrm{e}}^{t}{\left (1 - \frac{t} {\alpha }\right )}^{x} =\sum\limits_{ n=0}^{\infty }\frac{{C}_{n}(x;\alpha )} {n!} {t}^{n} & \\ \end{array}$$

Case IV.

Meixner determined $two$ orthogonal sets that stem from this case. The general u(t) and f(t) are as follows:

$$\begin{array}{rcl} u(t) = \frac{1} {\alpha - \beta }\ln \left (\frac{1 - \beta t} {1 - \alpha t}\right ) \mathrm{and} f(t) ={ \left (\frac{{(1 - \beta t)}^{1/\beta }} {{(1 - \alpha t)}^{1/\alpha }}\right )}^{{k}_{2}/(\alpha -\beta )}.& & \\ \end{array}$$

• (a) The Meixner Polynomials: $\alpha \neq \beta$ and $\alpha,\beta \in \mathbb{R}$ ($\kappa \neq 0$)

  $$\begin{array}{rcl}{ \left (1 -\frac{t} {c}\right )}^{x}{(1 - t)}^{-(x+\beta )} =\sum\limits_{n=0}^{\infty }\frac{{(\beta )}_{n}} {n!} M(x;\beta,c){t}^{n}.& & \\ \end{array}$$

• (b) The Meixner–Pollaczek Polynomials: $\alpha \neq \beta$, $\alpha$ and $\beta$ complex conjugates ($\kappa \neq 0$)

$$\begin{array}{rcl}{ (1 -{\mathrm{e}}^{i\phi }t)}^{-\lambda +ix}{(1 -{\mathrm{e}}^{-i\phi }t)}^{-\lambda -ix} =\sum\limits_{n=0}^{\infty }{P}_{ n}^{(\lambda )}(x;\phi ){t}^{n}.& & \\ \end{array}$$

Remark 1.1.

The Krawtchouk polynomials are the $third$ orthogonal set that comes from Case IV above. These polynomials were not included in either Meixner’s or Sheffer’s work.

Example 1.5.

As a simple example of the scaling process, we see that in Case I above, u(t) = t and $f(t) =\exp \left ({k}_{2}{t}^{2}/2\right )$, so that

$$f(t){\mathrm{e}}^{xu(t)} =\exp \left (xt + \frac{1} {2}{k}_{2}{t}^{2}\right ).$$

Thus, we can obtain the generating relation in Case I by simply choosing k ₂ =  − 1 ∕ 2 and rescaling t via $t \rightarrow 2t$.

1.5.1 Al-Salam’s Extension of Meixner’s Characterization

To briefly supplement our discussion of Meixner’s analysis, we state that W. A. Al-Salam extended the results of Meixner, and therefore Sheffer, in [1]. Namely, he showed that the left-hand side of Eq. (1.63) can be replaced with $\exp \left (Q(x,t)\right )$, where Q(x, t) is a polynomial in x with coefficients that are functions of t, as seen below:

$$\begin{array}{rcl} & \exp \left (Q(x,t)\right ) =\sum\limits_{n=0}^{\infty }{P}_{n}(x)\frac{{t}^{n}} {n!}, & \\ & Q(x,t) =\sum\limits_{j=0}^{k}{x}^{j}{a}^{(j)}(t), k \geq 1, {a}^{(j)}(t) =\sum\limits_{r=0}^{\infty }{a}_{r}^{(j)}{t}^{r}, j = 0,1,2,\ldots,k& \\ \end{array}$$

and that the resulting orthogonal polynomials ${\left \{{P}_{n}(x)\right \}}_{n=0}^{\infty }$ will be the same as those achieved by Meixner and Sheffer. This showed that the conditions on the generating function (1.63) can be weakened without yielding new orthogonal sets.