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On σ-type zero of Sheffer polynomials

An Erratum to this article was published on 06 March 2015

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, is an Appell set [68] ( P n being of degree exactly n) if either

  1. (i)

    P n (x)= P n 1 (x), n=0,1,2, , or

  2. (ii)

    there exists a formal power series A(t)= n = 0 a n t n ( a 0 0) such that

    A(t)exp(xt)= n = 0 P n (x) t n .

Sheffer’s A-type classification

Let ϕ n (x) be a simple set of polynomials and let ϕ n (x) belong to the operator

J(x,D)= k = 0 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 ϕ n (x) is of Sheffer A-type m. If the degree of T k (x) is unbounded as k, we say that ϕ n (x) is of Sheffer A-type ∞.

Polynomials of Sheffer A-type zero

Let ϕ n (x) be of Sheffer A-type zero. Then ϕ n (x) belong to the operator

J(D)= k = 0 c k D k + 1 ,

in which c k are constants. Here c 0 0 and J ϕ n = ϕ 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)= k = 0 c k t k + 1 , c 0 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 ϕ n (x) be of Sheffer A-type zero is that ϕ n (x) possess the generating function indicated in

A(t)exp ( x H ( t ) ) = n = 0 ϕ n (x) t n ,

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

H(t)= n = 0 h n t n + 1 , h 0 0,A(t)= n = 0 a n t n , a 0 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)= j = 1 h j t j , h 1 0, A s (t)= j = 0 a j ( s ) t j ( not all a 0 ( s ) are zero )

and

j = 1 r A j (t)exp ( x H ( ε j t ) ) = n = 0 P n (x) t n ,

where

J(D) P n (x)= P n r (x)(n=r,r+1,) where J(D)= k = 0 a k D k + r , a 0 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

J ( x , σ ) = k = 0 T k ( x ) σ k + 1 , σ = D i = 1 q ( x D + b i 1 ) , D = d d x , ( J ( x , σ ) p n ( x ) = p n 1 ( x ) ) ,

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, .

A necessary and sufficient condition that ϕ n (x) be of σ-type zero, with

σ=D i = 1 q (xD+ b i 1),

is that ϕ n (x) possess the generating function

A ( t ) 0 F q ( ; b 1 , b 2 , , b q ; x H ( t ) ) = n = 0 ϕ n (x) t n ,

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

H(t)= n = 0 h n t n + 1 , h 0 0,

and

A(t)= n = 0 a n t n , a 0 0.

Since ϕ n (x) belongs to the operator J(σ)= k = 0 c k σ k + 1 , where c k are constant and c 0 0.

2 Main results

Theorem 1 If p n (x) is a polynomial set, then p n (x) is of σ-type zero with σ=D m = 1 q (xD+ b m 1). It is necessary and sufficient condition that there exist formal power series

H(t)= n = 0 h n t n + 1 , h 0 0,

and

A i (t)= n = 0 a n ( i ) t n ( not all a 0 ( i ) are zero )

such that

i = 1 r A i ( t ) 0 F q ( ; b 1 , b 2 , , b q ; x H ( ε i t ) ) = n = 0 p n (x) t n ,
(1)

where θ=xD.

Proof Let y i = 0 F q (; b 1 , b 2 ,, b q ; z i ), where i=1,2,,r, be a solution of the following differential equation:

[ θ m = 1 q ( x D + b m 1 ) z i ] y i =0,θ=xD,D= d d x .

On substituting z i =xH( ε i t) and keeping t as a constant, where

σ=D m = 1 q (xD+ b m 1),θ=xD,

we get

[ x D m = 1 q ( θ + b m 1 ) x H ( z i ) ] y i =0.

This can also be written as

σ y i =H( ε i t) y i

or

σ 0 F q ( ; b 1 , b 2 , , b q ; x H ( ε i t ) ) =H ( ε i t ) 0 F q ( ; b 1 , b 2 , , b q ; x H ( ε i t ) ) .

Operating J(σ) on both sides of Equation (1) yields

J ( σ ) n = 0 p n ( x ) t n = J ( σ ) i = 1 r A i ( t ) 0 F q ( ; b 1 , b 2 , , b q ; x H ( ε i t ) ) = i = 1 r A i ( t ) J ( H ( ε i t ) ) 0 F q ( ; b 1 , b 2 , , b q ; x H ( ε i t ) ) = t n = 0 p n ( x ) t n = n = 1 p n 1 ( x ) t n .

Therefore, J(σ) p 0 (x)=0 and J(σ) p n (x)= p n 1 (x), n1.

Since J(σ) 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(σ). Now q n (x) is a simple set of polynomials, we can write

i = 1 r 0 F q ( ; b 1 , b 2 , , b q ; x H ( ε i t ) ) = n = 0 p n (x) t n ,
(2)

where ε 1 , ε 2 ,, ε 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)= k = 0 n c n k q k (x)

and

n = 0 p n (x) t n = n = 0 k = 0 n c n k q k (x) t n .

On replacing n by n+k, this becomes

n = 0 p n ( x ) t n = n = 0 k = 0 c n q k ( x ) t n + k = k = 0 q k ( x ) t k n = 0 c n t n .

Setting c n = a n ( i ) (i is independent of n, where i=1,2,,r), this becomes

n = 0 p n ( x ) t n = k = 0 q k ( x ) t k n = 0 a n ( i ) t n , by using Equation (2), we get = i = 1 r A i ( t ) 0 F q ( ; b 1 , b 2 , , b q ; x H ( ε i t ) ) .

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

i = 1 r ε i n h n 1 ψ( ε i t)=σ p n (x),
(3)

where ψ( ε i t)= A i ( t ) 0 F q (; b 1 , b 2 ,, b q ;xH( ε i t)).

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

n = 0 p n (x) t n = i = 1 r A i ( t ) 0 F q ( ; b 1 , b 2 , , b q ; x H ( ε i t ) ) .

This can be written as

n = 0 σ p n ( x ) t n = i = 1 r A i ( t ) σ 0 F q ( ; b 1 , b 2 , , b q ; x H ( ε i t ) ) = i = 1 r H ( ε i t ) A i ( t ) 0 F q ( ; b 1 , b 2 , , b q ; x H ( ε i t ) ) = i = 1 r ( n = 0 h n ε i n + 1 t n + 1 ) A i ( t ) 0 F q ( ; b 1 , b 2 , , b q ; x H ( ε i t ) ) = n = 1 i = 1 r ( ε i n h n 1 ) ψ ( ε i t ) t n .

Thus

σ p n (x)= i = 1 r ε i n h n 1 ψ( ε 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(σ)= k = 0 c k σ k + 1 , in which c k are constants and c 0 0.

Since

J(σ) p n (x,y)= p n 1 (x,y),n1,

where

D x = x , D y = y , θ = x x , ϕ = y y , σ x = D x m = 1 p ( θ + b m 1 ) , σ y = D y s = 1 q ( θ + b s 1 ) ,

and

J ( ( G + H ) ( t ) ) = ( ( G + H ) J ( t ) ) =t,σ= σ x + σ y .

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

σ x = D x m = 1 p (θ+ b m 1), σ y = D y s = 1 q (θ+ b s 1),σ= σ x + σ y ,

is that p n (x,y) possess a generating function in

i = 1 r A i ( t ) 0 F p ( ; b 1 , b 2 , , b p ; x G ( ε i t ) ) 0 F q ( ; c 1 , c 2 , , c q ; y H ( ε i t ) ) = n = 0 p n (x,y) t n ,
(4)

in which

G ( t ) = n = 0 g n t n + 1 , g 0 0 , H ( t ) = n = 0 h n t n + 1 , h 0 0 , A i ( t ) = n = 0 a n ( i ) t n ( not all a 0 ( i ) are zero )

and i is independent of n.

Proof Let u i = 0 F p (; b 1 , b 2 ,, b p ; z i ) and v i = 0 F q (; c 1 , c 2 ,, c q ; w i ) be the solutions of the following differential equations:

[ θ z m = 1 p ( θ z + b m 1 ) z i ] u i =0, θ z =z z ,

and

[ ϕ w s = 1 q ( ϕ w + c s 1 ) z i ] w i =0, ϕ w =w w .

On substituting z i =xG( ε i t), w i =yH( ε i t) and keeping t as a constant, where θ=x x = θ z , ϕ=y y = ϕ w , we get

θ m = 1 p (θ+ b m 1) u i =xG( ε i t) u i

and

ϕ s = 1 q (ϕ+ c s 1) w i =yH( ε i t) w i .

This can also be written as

σ 0 F p ( ; b 1 , b 2 , , b p ; x G ( ε i t ) ) 0 F q ( ; c 1 , c 2 , , c q ; y H ( ε i t ) ) = { G ( ε i t ) + H ( ε i t ) } 0 F p ( ; b 1 , b 2 , , b p ; x G ( ε i t ) ) 0 F q ( ; c 1 , c 2 , , c q ; y H ( ε i t ) ) .

Operating J(σ) on both sides of Equation (4) yields

J ( σ ) n = 0 p n ( x , y ) t n = J ( σ ) i = 1 r A i ( t ) 0 F p ( ; b 1 , b 2 , , b p ; x G ( ε i t ) ) 0 F q ( ; c 1 , c 2 , , c q ; y H ( ε i t ) ) = i = 1 r A i ( t ) J ( ( G + H ) ( ε i t ) ) 0 F p [ ; b 1 , b 2 , , b p ; x G ( ε i t ) ] 0 F q [ ; c 1 , c 2 , , c q ; y H ( ε i t ) ] = t n = 0 p n ( x , y ) t n = n = 1 p n 1 ( x , y ) t n .

Therefore, J(σ) p 0 (x,y)=0 and J(σ) p n (x,y)= p n 1 (x,y), n1.

Since J(σ) 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(σ). Now q n (x,y) is a simple set of polynomials. We can write

i = 1 r 0 F p ( ; b 1 , b 2 , , b p ; x G ( ε i t ) ) 0 F q ( ; c 1 , c 2 , , c q ; y H ( ε i t ) ) = n = 0 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)= k = 0 n c n k q k (x,y)

and

n = 0 p n (x,y) t n = n = 0 k = 0 n c n k q k (x,y) t n .

On replacing n by n+k, this becomes

= n = 0 k = 0 c n q k ( x , y ) t n + k = k = 0 q k ( x , y ) t k n = 0 c n t n .

Setting c n = a n ( i ) (i is independent of n, where i=1,2,,r), this becomes

= k = 0 q k ( x , y ) t k n = 0 a n ( i ) t n = i = 1 r A i ( t ) 0 F p ( ; b 1 , b 2 , , b p ; x G ( ε i t ) ) 0 F q ( ; c 1 , c 2 , , c q ; y H ( ε i t ) ) .

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

i = 1 r ε i n ( g n 1 + h n 1 )υ( ε i t)=σ p n (x,y),
(6)

where υ( ε i t)= A i ( t ) 0 F p ( ; b 1 , b 2 , , b p ; x G ( ε i t ) ) 0 F q (; c 1 , c 2 ,, c q ;yH( ε i t)).

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

n = 0 p n (x,y) t n = i = 1 r A i ( t ) 0 F p ( ; b 1 , b 2 , , b p ; x G ( ε i t ) ) 0 F q ( ; c 1 , c 2 , , c q ; y H ( ε i t ) ) .

This can be written as

n = 0 σ p n ( x , y ) t n = i = 1 r A i ( t ) σ 0 F p ( ; b 1 , b 2 , , b p ; x G ( ε i t ) ) 0 F q ( ; c 1 , c 2 , , c q ; y H ( ε i t ) ) = i = 1 r ( G + H ) ( ε i t ) A i ( t ) 0 F p ( ; b 1 , b 2 , , b p ; x G ( ε i t ) ) × 0 F q ( ; c 1 , c 2 , , c q ; y H ( ε i t ) ) = i = 1 r ( n = 0 ( g n + h n ) ε i n + 1 t n + 1 ) A i ( t ) 0 F p ( ; b 1 , b 2 , , b p ; x G ( ε i t ) ) × 0 F q ( ; c 1 , c 2 , , c q ; y H ( ε i t ) ) = n = 1 i = 1 r ( ε i n ( g n 1 + h n 1 ) ) υ ( ε i t ) t n .

Thus

σ p n (x,y)= i = 1 r ε i n ( g n 1 + h n 1 )υ( ε i t),

where υ( ε i t)= A i ( t ) 0 F p ( ; b 1 , b 2 , , b p ; x G ( ε i t ) ) 0 F q (; c 1 , c 2 ,, c q ;yH( ε i t)). This completes the proof. □

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Acknowledgements

Dedicated to Professor Hari M Srivastava.

The authors are indebted to the referees for their valuable suggestions which led to a better presentation of paper.

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Correspondence to Ajay K Shukla.

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The authors contributed equally and significantly in writing this article. The authors read and approved the final manuscript.

An erratum to this article is available at http://dx.doi.org/10.1186/s13660-015-0605-8.

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Shukla, A.K., Rapeli, S.J. & Shah, P.V. On σ-type zero of Sheffer polynomials. J Inequal Appl 2013, 241 (2013). https://doi.org/10.1186/1029-242X-2013-241

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Keywords

  • Appell sets
  • differential operator
  • Sheffer polynomials
  • generalized Sheffer polynomials