On σ-type zero of Sheffer polynomials

Abstract

The main object of this paper is to investigate some properties of σ-type polynomials in one and two variables.

MSC:33C65, 33E99, 44A45, 46G25.

1 Introduction

In 1945, Thorne [1] obtained an interesting characterization of Appell polynomials by means of the Stieltjes integral. Srivastava and Manocha [2] discussed the Appell sets and polynomials. Dattoli et al. [3] studied the properties of the Sheffer polynomials. Recently Pintér and Srivastava [4] gave addition theorems for the Appell polynomials and the associated classes of polynomial expansions and some cases have also been discussed by Srivastava and Choi [5] in their book.

Appell sets may be defined by the following equivalent condition: $\{P_n(x)\}$, $n=0,1,2,\dots$ is an Appell set [6–8] ($P_n$ being of degree exactly n) if either

• (i) $P_n^{\prime }(x)=P_{n-1}(x)$, $n=0,1,2,\dots$ , or

• (ii) there exists a formal power series $A(t)={\sum }_{n=0}^{\mathrm{\infty }}{a}_n{t}^n$ ($a_0\ne 0$) such that

  $$A(t)exp(xt)=\sum _{n=0}^{\mathrm{\infty }}{P}_n(x)t^n.$$

Sheffer’s A-type classification

Let ${\varphi }_n(x)$ be a simple set of polynomials and let ${\varphi }_n(x)$ belong to the operator

$$J(x,D)=\sum _{k=0}^{\mathrm{\infty }}{T}_k(x)D^{k+1},$$

with $T_k(x)$ of degree ≤k. If the maximum degree of the coefficients $T_k(x)$ is m, then the set ${\varphi }_n(x)$ is of Sheffer A-type m. If the degree of $T_k(x)$ is unbounded as $k\to \mathrm{\infty }$, we say that ${\varphi }_n(x)$ is of Sheffer A-type ∞.

Polynomials of Sheffer A-type zero

Let ${\varphi }_n(x)$ be of Sheffer A-type zero. Then ${\varphi }_n(x)$ belong to the operator

$$J(D)=\sum _{k=0}^{\mathrm{\infty }}{c}_k{D}^{k+1},$$

in which $c_k$ are constants. Here $c_0\ne 0$ and $J{\varphi }_n={\varphi }_{n-1}$. Furthermore, since $c_k$ are independent of x for every k, a function $J(t)$ exists with the formal power series expansion

$$J(t)=\sum _{k=0}^{\mathrm{\infty }}{c}_k{t}^{k+1},c_0\ne 0.$$

Let $H(t)$ be the formal inverse of $J(t)$; that is,

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

Theorem (Rainville [9])

A necessary and sufficient condition that ${\varphi }_n(x)$ be of Sheffer A-type zero is that ${\varphi }_n(x)$ possess the generating function indicated in

$$A(t)exp(xH(t))=\sum _{n=0}^{\mathrm{\infty }}{\varphi }_n(x)t^n,$$

in which $H(t)$ and $A(t)$ have (formal) expansions

$$H(t)=\sum _{n=0}^{\mathrm{\infty }}{h}_n{t}^{n+1},h_0\ne 0,A(t)=\sum _{n=0}^{\mathrm{\infty }}{a}_n{t}^n,a_0\ne 0.$$

Theorem (Al-Salam and Verma [10])

Let $\{P_n(x)\}$ be a polynomial set. In order for $\{P_n(x)\}$ to be a Sheffer A-type zero, it is necessary and sufficient that there exist (formal) power series

$$H(t)=\sum _{j=1}^{\mathrm{\infty }}{h}_j{t}^j,h_1\ne 0,A_s(t)=\sum _{j=0}^{\mathrm{\infty }}{a}_j^{(s)}{t}^j\left(\mathit{\text{not all}}{a}_0^{(s)}\mathit{\text{are zero}}\right)$$

and

$$\sum _{j=1}^r{A}_j(t)exp(xH({\epsilon }_{j}t))=\sum _{n=0}^{\mathrm{\infty }}{P}_n(x)t^n,$$

where

$$J(D)P_n(x)=P_{n-r}(x)(n=r,r+1,\dots )\mathit{\text{where}}J(D)=\sum _{k=0}^{\mathrm{\infty }}{a}_k{D}^{k+r},a_0\ne 0$$

and r is a fixed positive integer. The function $A(t)$ may be called the determining function for the set $\{P_n(x)\}$.

Polynomial of σ-type zero [9, 11]

Let $\{p_n(x)\}$ be a simple set of polynomials that belongs to the operator

$$\begin{array}{c}J(x,\sigma )=\sum _{k=0}^{\mathrm{\infty }}{T}_k(x){\sigma }^{k+1},\hfill \\ \sigma =D\prod _{i=1}^q(xD+b_i-1),D=\frac{d}{dx},(J(x,\sigma )p_n(x)=p_{n-1}(x)),\hfill \end{array}$$

where $b_i$ are constants, not equal to zero or a negative integer, and $T_k(x)$ are polynomials of degree ≤k. We can say that this set is of σ-type m if the maximum degree of $T_k(x)$ is m, $m=0,1,2,\dots$ .

A necessary and sufficient condition that ${\varphi }_n(x)$ be of σ-type zero, with

$$\sigma =D\prod _{i=1}^q(xD+b_i-1),$$

is that ${\varphi }_n(x)$ possess the generating function

$$A{(t)}_0{F}_q(-;b_1,b_2,\dots ,b_q;xH(t))=\sum _{n=0}^{\mathrm{\infty }}{\varphi }_n(x)t^n,$$

in which $H(t)$ and $A(t)$ have (formal) expansions

$$H(t)=\sum _{n=0}^{\mathrm{\infty }}{h}_n{t}^{n+1},h_0\ne 0,$$

and

$$A(t)=\sum _{n=0}^{\mathrm{\infty }}{a}_n{t}^n,a_0\ne 0.$$

Since ${\varphi }_n(x)$ belongs to the operator $J(\sigma )={\sum }_{k=0}^{\mathrm{\infty }}{c}_k{\sigma }^{k+1}$, where $c_k$ are constant and $c_0\ne 0$.

2 Main results

Theorem 1 If $p_n(x)$ is a polynomial set, then $p_n(x)$ is of σ-type zero with $\sigma =D{\prod }_{m=1}^q(xD+b_m-1)$. It is necessary and sufficient condition that there exist formal power series

$$H(t)=\sum _{n=0}^{\mathrm{\infty }}{h}_n{t}^{n+1},h_0\ne 0,$$

and

$$A_i(t)=\sum _{n=0}^{\mathrm{\infty }}{a}_n^{(i)}{t}^n\left(\mathit{\text{not all}}{a}_0^{(i)}\mathit{\text{are zero}}\right)$$

such that

$$\sum _{i=1}^r{A}_i{(t)}_0{F}_q(-;b_1,b_2,\dots ,b_q;xH({\epsilon }_{i}t))=\sum _{n=0}^{\mathrm{\infty }}{p}_n(x)t^n,$$

(1)

where $\theta =xD$.

Proof Let $y_i{=}_0{F}_q(-;b_1,b_2,\dots ,b_q;z_i)$, where $i=1,2,\dots ,r$, be a solution of the following differential equation:

$$[\theta \prod _{m=1}^q(xD+b_m-1)-z_i]y_i=0,\theta =xD,D=\frac{d}{dx}.$$

On substituting $z_i=xH({\epsilon }_{i}t)$ and keeping t as a constant, where

$$\sigma =D\prod _{m=1}^q(xD+b_m-1),\theta =xD,$$

we get

$$[xD\prod _{m=1}^q(\theta +b_m-1)-xH(z_i)]y_i=0.$$

This can also be written as

$$\sigma y_i=H({\epsilon }_{i}t)y_i$$

or

$${\sigma }_0{F}_q(-;b_1,b_2,\dots ,b_q;xH({\epsilon }_{i}t))=H{({\epsilon }_{i}t)}_0{F}_q(-;b_1,b_2,\dots ,b_q;xH({\epsilon }_{i}t)).$$

Operating $J(\sigma )$ on both sides of Equation (1) yields

$$\begin{array}{rl}J(\sigma )\sum _{n=0}^{\mathrm{\infty }}{p}_n(x)t^n& =J(\sigma )\sum _{i=1}^r{A}_i{(t)}_0{F}_q(-;b_1,b_2,\dots ,b_q;xH({\epsilon }_{i}t))\\ =\sum _{i=1}^r{A}_i(t)J{(H({\epsilon }_{i}t))}_0{F}_q(-;b_1,b_2,\dots ,b_q;xH({\epsilon }_{i}t))\\ =t\sum _{n=0}^{\mathrm{\infty }}{p}_n(x)t^n\\ =\sum _{n=1}^{\mathrm{\infty }}{p}_{n-1}(x)t^n.\end{array}$$

Therefore, $J(\sigma )p_0(x)=0$ and $J(\sigma )p_n(x)=p_{n-1}(x)$, $n\ge 1$.

Since $J(\sigma )$ is independent of x, using the definition of σ-type [9, 11], we arrive at the conclusion that $p_n(x)$ is σ-type zero.

Conversely, suppose $p_n(x)$ is of σ-type zero and belongs to the operator $J(\sigma )$. Now $q_n(x)$ is a simple set of polynomials, we can write

$${\sum _{i=1}^r}_0{F}_q(-;b_1,b_2,\dots ,b_q;xH({\epsilon }_{i}t))=\sum _{n=0}^{\mathrm{\infty }}{p}_n(x)t^n,$$

(2)

where ${\epsilon }_1,{\epsilon }_2,\dots ,{\epsilon }_r$ are the roots of unity.

Since $q_n(x)$ is a simple set, there exists a sequence $c_k$ [10], independent of n, such that

$$p_n(x)=\sum _{k=0}^n{c}_{n-k}{q}_k(x)$$

and

$$\sum _{n=0}^{\mathrm{\infty }}{p}_n(x)t^n=\sum _{n=0}^{\mathrm{\infty }}\sum _{k=0}^n{c}_{n-k}{q}_k(x)t^n.$$

On replacing n by $n+k$, this becomes

$$\begin{array}{rl}\sum _{n=0}^{\mathrm{\infty }}{p}_n(x)t^n& =\sum _{n=0}^{\mathrm{\infty }}\sum _{k=0}^{\mathrm{\infty }}{c}_n{q}_k(x)t^{n+k}\\ =\sum _{k=0}^{\mathrm{\infty }}{q}_k(x)t^k\sum _{n=0}^{\mathrm{\infty }}{c}_n{t}^n.\end{array}$$

Setting $c_n=a_n^{(i)}$ (i is independent of n, where $i=1,2,\dots ,r$), this becomes

$$\begin{array}{rl}\sum _{n=0}^{\mathrm{\infty }}{p}_n(x)t^n& =\sum _{k=0}^{\mathrm{\infty }}{q}_k(x)t^k\sum _{n=0}^{\mathrm{\infty }}{a}_n^{(i)}{t}^n,\text{by using Equation (2), we get}\\ =\sum _{i=1}^r{A}_i{(t)}_0{F}_q(-;b_1,b_2,\dots ,b_q;xH({\epsilon }_{i}t)).\end{array}$$

This completes the proof. □

Theorem 2 A necessary and sufficient condition that $p_n(x)$ be of σ-type zero and there exist a sequence $h_k$, independent of x and n, such that

$$\sum _{i=1}^r{\epsilon }_i^n{h}_{n-1}\psi ({\epsilon }_{i}t)=\sigma p_n(x),$$

(3)

where $\psi ({\epsilon }_{i}t)=A_i{(t)}_0{F}_q(-;b_1,b_2,\dots ,b_q;xH({\epsilon }_{i}t))$.

Proof If $p_n(x)$ is of σ-type zero, then it follows from Theorem 1 that

$$\sum _{n=0}^{\mathrm{\infty }}{p}_n(x)t^n=\sum _{i=1}^r{A}_i{(t)}_0{F}_q(-;b_1,b_2,\dots ,b_q;xH({\epsilon }_{i}t)).$$

This can be written as

$$\begin{array}{rl}\sum _{n=0}^{\mathrm{\infty }}\sigma p_n(x)t^n& =\sum _{i=1}^r{A}_i(t){\sigma }_0{F}_q(-;b_1,b_2,\dots ,b_q;xH({\epsilon }_{i}t))\\ =\sum _{i=1}^{r}H({\epsilon }_{i}t)A_i{(t)}_0{F}_q(-;b_1,b_2,\dots ,b_q;xH({\epsilon }_{i}t))\\ =\sum _{i=1}^r\left(\sum _{n=0}^{\mathrm{\infty }}{h}_n{\epsilon }_i^{n+1}{t}^{n+1}\right)A_i{(t)}_0{F}_q(-;b_1,b_2,\dots ,b_q;xH({\epsilon }_{i}t))\\ =\sum _{n=1}^{\mathrm{\infty }}\sum _{i=1}^r\left({\epsilon }_i^n{h}_{n-1}\right)\psi ({\epsilon }_{i}t)t^n.\end{array}$$

Thus

$$\sigma p_n(x)=\sum _{i=1}^r{\epsilon }_i^n{h}_{n-1}\psi ({\epsilon }_{i}t).$$

This completes the proof. □

3 Sheffer polynomials in two variables [12]

Let $p_n(x,y)$ be of σ-type zero. Then $p_n(x,y)$ belongs to an operator $J(\sigma )={\sum }_{k=0}^{\mathrm{\infty }}{c}_k{\sigma }^{k+1}$, in which $c_k$ are constants and $c_0\ne 0$.

Since

$$J(\sigma )p_n(x,y)=p_{n-1}(x,y),n\ge 1,$$

where

$$\begin{array}{c}{D}_x=\frac{\partial }{\partial x},D_y=\frac{\partial }{\partial y},\theta =x\frac{\partial }{\partial x},\varphi =y\frac{\partial }{\partial y},\hfill \\ {\sigma }_x=D_x\prod _{m=1}^p(\theta +b_m-1),{\sigma }_y=D_y\prod _{s=1}^q(\theta +b_s-1),\hfill \end{array}$$

and

$$J((G+H)(t))=((G+H)J(t))=t,\sigma ={\sigma }_x+{\sigma }_y.$$

Theorem 3 A necessary and sufficient condition that $p_n(x,y)$ be of σ-type zero, with

$${\sigma }_x=D_x\prod _{m=1}^p(\theta +b_m-1),{\sigma }_y=D_y\prod _{s=1}^q(\theta +b_s-1),\sigma ={\sigma }_x+{\sigma }_y,$$

is that $p_n(x,y)$ possess a generating function in

$$\sum _{i=1}^r{A}_i{(t)}_0{F}_p{(-;b_1,b_2,\dots ,b_p;xG({\epsilon }_{i}t))}_0{F}_q(-;c_1,c_2,\dots ,c_q;yH({\epsilon }_{i}t))=\sum _{n=0}^{\mathrm{\infty }}{p}_n(x,y)t^n,$$

(4)

in which

$$\begin{array}{c}G(t)=\sum _{n=0}^{\mathrm{\infty }}{g}_n{t}^{n+1},g_0\ne 0,\hfill \\ H(t)=\sum _{n=0}^{\mathrm{\infty }}{h}_n{t}^{n+1},h_0\ne 0,\hfill \\ A_i(t)=\sum _{n=0}^{\mathrm{\infty }}{a}_n^{(i)}{t}^n\left(\mathit{\text{not all}}{a}_0^{(i)}\mathit{\text{are zero}}\right)\hfill \end{array}$$

and i is independent of n.

Proof Let $u_i{=}_0{F}_p(-;b_1,b_2,\dots ,b_p;z_i)$ and $v_i{=}_0{F}_q(-;c_1,c_2,\dots ,c_q;w_i)$ be the solutions of the following differential equations:

$$[{\theta }_z\prod _{m=1}^p({\theta }_z+b_m-1)-z_i]u_i=0,{\theta }_z=z\frac{\partial }{\partial z},$$

and

$$[{\varphi }_w\prod _{s=1}^q({\varphi }_w+c_s-1)-z_i]w_i=0,{\varphi }_w=w\frac{\partial }{\partial w}.$$

On substituting $z_i=xG({\epsilon }_{i}t)$, $w_i=yH({\epsilon }_{i}t)$ and keeping t as a constant, where $\theta =x\frac{\partial }{\partial x}={\theta }_z$, $\varphi =y\frac{\partial }{\partial y}={\varphi }_w$, we get

$$\theta \prod _{m=1}^p(\theta +b_m-1)u_i=xG({\epsilon }_{i}t)u_i$$

and

$$\varphi \prod _{s=1}^q(\varphi +c_s-1)w_i=yH({\epsilon }_{i}t)w_i.$$

This can also be written as

$$\begin{array}{c}{\sigma }_0{F}_p{(-;b_1,b_2,\dots ,b_p;xG({\epsilon }_{i}t))}_0{F}_q(-;c_1,c_2,\dots ,c_q;yH({\epsilon }_{i}t))\hfill \\ ={\{G({\epsilon }_{i}t)+H({\epsilon }_{i}t)\}}_0{F}_p{(-;b_1,b_2,\dots ,b_p;xG({\epsilon }_{i}t))}_0{F}_q(-;c_1,c_2,\dots ,c_q;yH({\epsilon }_{i}t)).\hfill \end{array}$$

Operating $J(\sigma )$ on both sides of Equation (4) yields

$$\begin{array}{r}J(\sigma )\sum _{n=0}^{\mathrm{\infty }}{p}_n(x,y)t^n\\ =J(\sigma )\sum _{i=1}^r{A}_i{(t)}_0{F}_p{(-;b_1,b_2,\dots ,b_p;xG({\epsilon }_{i}t))}_0{F}_q(-;c_1,c_2,\dots ,c_q;yH({\epsilon }_{i}t))\\ =\sum _{i=1}^r{A}_i(t)J{((G+H)({\epsilon }_{i}t))}_0{F}_p{[-;b_1,b_2,\dots ,b_p;xG({\epsilon }_{i}t)]}_0{F}_q[-;c_1,c_2,\dots ,c_q;yH({\epsilon }_{i}t)]\\ =t\sum _{n=0}^{\mathrm{\infty }}{p}_n(x,y)t^n\\ =\sum _{n=1}^{\mathrm{\infty }}{p}_{n-1}(x,y)t^n.\end{array}$$

Therefore, $J(\sigma )p_0(x,y)=0$ and $J(\sigma )p_n(x,y)=p_{n-1}(x,y)$, $n\ge 1$.

Since $J(\sigma )$ is independent of x and y, thus we arrive at the conclusion that $p_n(x,y)$ is of σ-type zero.

Conversely, suppose $p_n(x,y)$ is of σ-type zero and belongs to the operator $J(\sigma )$. Now $q_n(x,y)$ is a simple set of polynomials. We can write

$${\sum _{i=1}^r}_0{F}_p{(-;b_1,b_2,\dots ,b_p;xG({\epsilon }_{i}t))}_0{F}_q(-;c_1,c_2,\dots ,c_q;yH({\epsilon }_{i}t))=\sum _{n=0}^{\mathrm{\infty }}{p}_n(x,y)t^n.$$

(5)

Since $q_n(x,y)$ is a simple set, there exists a sequence $c_k$, independent of n, such that

$$p_n(x,y)=\sum _{k=0}^n{c}_{n-k}{q}_k(x,y)$$

and

$$\sum _{n=0}^{\mathrm{\infty }}{p}_n(x,y)t^n=\sum _{n=0}^{\mathrm{\infty }}\sum _{k=0}^n{c}_{n-k}{q}_k(x,y)t^n.$$

On replacing n by $n+k$, this becomes

$$\begin{array}{c}=\sum _{n=0}^{\mathrm{\infty }}\sum _{k=0}^{\mathrm{\infty }}{c}_n{q}_k(x,y)t^{n+k}\hfill \\ =\sum _{k=0}^{\mathrm{\infty }}{q}_k(x,y)t^k\sum _{n=0}^{\mathrm{\infty }}{c}_n{t}^n.\hfill \end{array}$$

Setting $c_n=a_n^{(i)}$ (i is independent of n, where $i=1,2,\dots ,r$), this becomes

$$\begin{array}{c}=\sum _{k=0}^{\mathrm{\infty }}{q}_k(x,y)t^k\sum _{n=0}^{\mathrm{\infty }}{a}_n^{(i)}{t}^n\hfill \\ =\sum _{i=1}^r{A}_i{(t)}_0{F}_p{(-;b_1,b_2,\dots ,b_p;xG({\epsilon }_{i}t))}_0{F}_q(-;c_1,c_2,\dots ,c_q;yH({\epsilon }_{i}t)).\hfill \end{array}$$

This completes the proof. □

Theorem 4 A necessary and sufficient condition that $p_n(x,y)$ be of σ-type zero and there exist sequences $g_k$ and $h_k$ , independent of x, y and n, such that

$$\sum _{i=1}^r{\epsilon }_i^n(g_{n-1}+h_{n-1})\upsilon ({\epsilon }_{i}t)=\sigma p_n(x,y),$$

(6)

where $\upsilon ({\epsilon }_{i}t)=A_i{(t)}_0{F}_p{(-;b_1,b_2,\dots ,b_p;xG({\epsilon }_{i}t))}_0{F}_q(-;c_1,c_2,\dots ,c_q;yH({\epsilon }_{i}t))$.

Proof If $p_n(x,y)$ is of σ-type zero, then it follows from Theorem 3 that

$$\sum _{n=0}^{\mathrm{\infty }}{p}_n(x,y)t^n=\sum _{i=1}^r{A}_i{(t)}_0{F}_p{(-;b_1,b_2,\dots ,b_p;xG({\epsilon }_{i}t))}_0{F}_q(-;c_1,c_2,\dots ,c_q;yH({\epsilon }_{i}t)).$$

This can be written as

$$\begin{array}{rl}\sum _{n=0}^{\mathrm{\infty }}\sigma p_n(x,y)t^n=& \sum _{i=1}^r{A}_i(t){\sigma }_0{F}_p{(-;b_1,b_2,\dots ,b_p;xG({\epsilon }_{i}t))}_0{F}_q(-;c_1,c_2,\dots ,c_q;yH({\epsilon }_{i}t))\\ =& \sum _{i=1}^r(G+H)({\epsilon }_{i}t)A_i{(t)}_0{F}_p(-;b_1,b_2,\dots ,b_p;xG({\epsilon }_{i}t))\\ {\times }_0{F}_q(-;c_1,c_2,\dots ,c_q;yH({\epsilon }_{i}t))\\ =& \sum _{i=1}^r(\sum _{n=0}^{\mathrm{\infty }}(g_n+h_n){\epsilon }_i^{n+1}{t}^{n+1})A_i{(t)}_0{F}_p(-;b_1,b_2,\dots ,b_p;xG({\epsilon }_{i}t))\\ {\times }_0{F}_q(-;c_1,c_2,\dots ,c_q;yH({\epsilon }_{i}t))\\ =& \sum _{n=1}^{\mathrm{\infty }}\sum _{i=1}^r({\epsilon }_i^n(g_{n-1}+h_{n-1}))\upsilon ({\epsilon }_{i}t)t^n.\end{array}$$

Thus

$$\sigma p_n(x,y)=\sum _{i=1}^r{\epsilon }_i^n(g_{n-1}+h_{n-1})\upsilon ({\epsilon }_{i}t),$$

where $\upsilon ({\epsilon }_{i}t)=A_i{(t)}_0{F}_p{(-;b_1,b_2,\dots ,b_p;xG({\epsilon }_{i}t))}_0{F}_q(-;c_1,c_2,\dots ,c_q;yH({\epsilon }_{i}t))$. This completes the proof. □