British Journal of Mathematics & Computer Science 14(2): 1-11, 2016, Article no.BJMCS.23323 ISSN: 2231-0851

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Sequential Representation of Delta Operator in Finite Operator Calculus

A. Maheswaran1∗ and C. Elango1

1Department of Mathematical Sciences, Cardamom Planters’ Association College, Bodinayakanur, Tamilnadu, India.

Authors’ contributions

This work was carried out in collaboration between both authors. Author AM designed the study, performed the theoretical analysis, wrote the protocol, and wrote the first draft of the manuscript and managed literature searches. Author CE managed the analyses of the study and literature searches. Both authors read and approved the final manuscript.

Article Information DOI: 10.9734/BJMCS/2016/23323 Editor(s): (1) Andrej V. Plotnikov, Department of Applied and Calculus Mathematics and CAD, Odessa State Academy of Civil Engineering and Architecture, Ukraine. Reviewers: (1) Octav Olteanu, University Politehnica of Bucharest, Romania. (2) Anonymous, USA. (3) Ayman Shehata Mohammed Ahmed El-Shazly, Assiut University, Egypt. Complete Peer review History: http://sciencedomain.org/review-history/13141

Received: 25th November 2015 Accepted: 22nd January 2016 Original Research Article Published: 3rd February 2016

Abstract

In this article, the theme of Finite Operator Calculus of G.C.Rota is reintroduced by representing the delta operator as a numerical sequence.

Keywords: Shift invariant operator; delta operator; sequence of basic polynomials.

2010 Mathematics Subject Classification: 44A45, 05A10, 05A19.

1 Introduction

Operational Methods are used to reduce differential problems into algebraic problems. Kwasniewski, A. K. [1] proposed finite operator q-calculus by using q-delta operator and q-basic polynomial sequence. Emil C Popa [2] constructed Taylor formula in the Umbral calculus by using Q-integral

*Corresponding author: E-mail: [email protected]; Maheswaran and Elango; BJMCS, 14(2), 1-11, 2016; Article no.BJMCS.23323

operator. Octavian Agratini [3] derived an identity involving the generating function of a Binomial type sequence and the theoretical aspects are illustrated with many concrete examples. Several classical polynomials sets are discussed and analysed in Rainville, E.D. [4]. Heinrich Niederhausen [5] developed the finite operator calculus in several variables. Gian-Carlo Rota [6] contains a detailed study of delta operator and basic polynomials sequences and it serves as an introduction and a guide on combinatorics. After briefly summarizing and analysing Rota [6] critically, many new results evolved in this attempt are discussed.

This paper is made to introduce a new approach to delta operator as sequence of real numbers. This sequence is unique for any delta operator. Novel examples are constructed and explained to visualize the theory clearly. A table of delta operator and the corresponding sequences along with their properties is incorporated. Assigning different delta operators, corresponding set of basic polynomials sequence are obtained. Finally the effect of delta operator on exponential and circular functions are expressed in terms of infinite series.

2 Preliminaries

Let F be a Field of characteristic zero, preferably the real number field. Let p(x) be a polynomial in one variable defined over F . The set of such polynomials is denoted by P . A sequence of + polynomials is {pn(x)/n ∈ Z ∪ {0}}, where pn(x) is exactly of degree n. We study certain special type of operators in this section.

Definition 1. i An operator Ea is said to be a shift operator if Eap(x) = p(x + a), for all polynomials p(x) in one variable and for all real a in the field. ii A linear operator T which commutes with all shift operators is called a shift invariant. In symbols, TEa = EaT, ∀a ∈ F . iii A shift invariant operator Q satisfying that Qx is a non zero constant is called a delta operator. Thus every delta operator Q is a shift invariant. But a shift invariant operator need not be a delta operator.

The following results need some mention.

Theorem 1. i) If Q is a delta operator, then Qa = 0 for every constant 0a0. ii) If p(x) is a polynomial of degree n, then Qp(x) is a polynomial of degree n − 1. Proof. Refer G-C Rota [6]

n n−1 + The usual derivative D is a delta operator. But the operator defined by Q(x ) = x , n ∈ Z ∪{0}, will not be a delta operator , since it is not shift invariant.

n n−1 + Taking Q(x ) = an x where an is a real constant, for n ∈ Z and assuming Q to be a delta a n a n operator, E Q(x ) = QE (x ) implies an = na1. In this case, the delta operator becomes a constant multiple of the usual derivative operator D. This leads to the theorem :

n n−1 + Theorem 2. ”If Q is a delta operator and Q(x ) = an x , where an is a real constant, n ∈ Z , then Q = kD where k is a real constant and D is the usual derivative”.

Definition 2. Let Q be the delta operator, A polynomial sequence pn(x) is called the sequence of

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basic polynomials for Q if:

1. p0(x) = 1

2. pn(0) = 0, whenever n > 0

3. Qpn(x) = np(n−1)(x)

A trivial example for basic polynomials sequence is {xn}. We found that certain delta operator will be a constant multiple of the usual derivative D. We take Q = kD. Consequently the corresponding xn + basic polynomials sequence with respect to this delta operator is { kn /n ∈ Z ∪ {0}}. Theorem 3. Every delta operator has a unique sequence of basic polynomials. Proof. Refer to G-C Rota [6]

3 The Main Results

Now we attempt to formulate the delta operator represented in terms of a sequence of real numbers.

n + Theorem 4. For the monomial {x : n ∈ Z ∪ {0}}, and for each αr an arbitary real value,

n ! n X n n−r Q(x ) = α x . (3.1) r r r=1 Proof. If n = 1, then from the definition of delta operator, Q(x) is a non zero constant.

Let it be α1. Therefore, Q(x) = α1 6= 0 and hence the result is true for n = 1 2 Let n = 2. Construct Q(x ) = c0 x + c1, by (ii) Theorem 1. Since Q is shift invariant, EaQ(x2) = QEa(x2). a 2 a 2 E Q(x ) = c0x + c1 + 2aα1 and QE (x ) = c0x + c0a + c1 implies c0 = 2α1 c1 is a new independent constant which may be taken as α2. Hence 2 Q(x ) = 2α1x + α2

Therefore, the result is true for n = 2. Let us assume that the result is true for all n = k. Therefore ,

k ! ! ! ! k X k k−r k k−1 k k−2 k k−r Q(x ) = α x = α x + α x + ··· + α x + ··· + α (3.2) r r 1 1 2 2 r r k r=1

n Since {x } is a basic polynomial sequence, it satisfies Qpn(x) = npn−1(x) and hence we have, Q(xk) = k xk−1 (3.3)

From (3), we see that the delta operator Q is a usual derivative D. From (2) and (3) , ! ! ! k k k α xk−1 + α xk−2 + ··· + α xk−r + ··· + α = k xk−1 (3.4) 1 1 2 2 r r k

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By comparing the corresponding terms, we have α1 = 1 and αj = 0, j = 2, 3, ··· k Therefore, the result is true for n = k means that

α1 = 1 and αj = 0 (j = 2, 3, ··· k) (3.5) Now we have to show that this result is true for n = k + 1

Q(xk+1) = Q(xk x)

= Q(xk) x + Q(x) xk (for the basic polynomial sequence {xn},Q = D) ( ! ! ! ) k k k = α xk−1 + α xk−2 + ··· + α xk−r + ··· + α x + α xk 1 1 2 2 r r k 1 ! ! k k = α (k xk + xk) + α xk−1 + α xk−2 + ··· + α x 1 2 2 3 3 k = (k + 1) xk by (5)

Thus we have Q(xk+1) = (k + 1) xk (3.6)

On other hand, using the property that Qpn(x) = n pn−1(x) , we have k+1 k Q(x ) = (k + 1) pk(x) = (k + 1) x (3.7) From (3.6) and (3.7) , we conclude that the result is true for all n = k + 1 Thus we proved the Theorem 1.

The following Table 1. contains first few polynomials Q(xn), for each degree n.

n Q(xn)

1 α1

2 2 α1 x + α2 2 3 3 α1 x + 3 α2 x + α3 3 2 4 4 α1 x + 6 α2 x + 4 α3 x + α3 4 3 2 5 5 α1 x + 10 α2 x + 10 α3 x + 5 α4 x + α5 5 4 3 2 6 6 α1 x + 15 α2 x + 20 α3 x + 15 α4 x + 6 α5 x + α6 6 5 4 3 2 5 7 7 α1 x + 21 α2 x + 35 α3 x + 35 α4 x + 21 α5 x + 7 α6 x + α7 7 6 5 4 3 2 8 8 α1 x + 28 α2 x + 56 α3 x + 70 α4 x + 56 α5 x + 28 α6 x + 8 α7 x + α8 8 7 6 5 4 3 2 9 9 α1 x + 36 α2 x + 84 α3 x + 126 α4 x + 126 α5 x + 84 α6 x + 36 α7 x + 9 α8 x + α9

n Here, Q(x ) has n independent parameters, αi, (i = 1, 2, 3 . . . n). These parameters are unique. Allowing n being large, we get an infinite sequence of real numbers. Hence to represent a delta operator, we need to consider only corresponding infinite sequence of real numbers. The values of αi ’s uniquely fix the delta operator and hence the delta operator is characterized uniquely. We conclude that any delta operator may be fixed uniquely by an infinite sequence of real numbers. Different delta operators are analysed and characterised in Table 2.

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Table 2. Characterization of Delta Operator as Infinite Sequence

S.No. Q(xn) Characterization of delta operator n−1 1 nx α1 = 1 and α2 = α3 = ··· = αn = 0. n−1 2 knx α1 = k and α2 = α3 = ··· = αn = 0 . n
n−1 n
n−2 1 x + 2 x + ··· 3 n
n−r αn = 1, for all n ≥ 1 + r x + ··· + 1. n
n−1 n
n−2 1 x + 2 2 x + ··· 4 n
n−r αn = n, for all n ≥ 1 . +r r x + ··· + n n
n−1 n
n−2 − 1 x + 2 x + ··· n 5 rn
n−r r αn = (−1) , n ≥ 1 . (−1) r x + ··· + (−1) n
n−1 n
n−2 1 x + 2! 2 x + ··· 6 n
n−r αn = n!, for all n ≥ 1 . +r! r x + ··· + n! n
xn−1 + 1 n
xn−2 + ··· 7 1 2 2 1 1 n
n−r 1 αn = n , for all n ≥ 1. + r r x + ··· + n n
xn−1 + 1 n
xn−2 + ··· 8. 1 2! 2 1 1 n
n−r 1 αn = n! , for all n ≥ 1 . + r! r x + ··· + n! Remarks: Table 1 i) In row 1, Delta operator is the usual derivative. ii) In row 2, Delta operator is the constant multiple of usual derivative. iii) In row 3, Q(xn) = (x + 1)n − xn. n n P n
n−r iv) In row 4, Q(x ) = r r x . r=1 v) In row 5, Q(xn) = (−1)n[(x + 1)n − xn]. n n P n
n−r vi) In row 6, Q(x ) = r! r x . r=1 0 vii) The values of αn s in row 7 are reciprocal of row 4. 0 viii) The values of αn s in row 8 are reciprocal of row 6.

4 Basic Polynomials Sequences for Different Delta Operators

For each delta operator assigned, there exists a unique basic polynomials sequence. This basic polynomial sequence may be found out using the above values of Q(xn) in Table 1 which is strictly representing the delta operator.

From (3.1), we obtain the following proposition.

Proposition 1. For the characterization of delta operator being α1 = 1 and n + α2 = α3 = ··· = αn = 0, the basic polynomial sequence is {x }, where n ∈ Z ∪ {0}

n n−1 Proof. If we take α1 = 1 and α2 = α3 = ··· = αn = 0 , then from (3.1) , Q(x ) = n x .

Taking p0(x) = 1, let p1(x) = a1x, [the constant term c0 is already omitted by Rota [6] in the proof of the above theorem 2 ].

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Using Qpn(x) = npn−1(x) , we get a1 = 1 , giving p1(x) = x. 2 2 Let p2(x) = b1x + b2 x. Using Qpn(x) = npn−1(x) , we get b1 = 1 and b2 = 0 ⇒ p2(x) = x . 3 2 Let p3(x) = c1x + c2x + c3x and Qpn(x) = npn−1(x) , we get c1 = 1, c2 = 0 and c3 = 0 ⇒ 3 p3(x) = x .

4 5 n Proceeding like this , we get p4(x) = x , p5(x) = x , ··· pn(x) = x Thus the basic polynomial sequence corresponding to the delta operator Q is :

n + {x }, where n ∈ Z ∪ {0}. Remark 1. In Proposition 1, the delta operator Q is usual derivative D.

Therefore, by Proposition 1, we obtain the following corollary.

Corollary 1. For α1 = k and α2 = α3 = ··· = αn = 0, the basic polynomial sequence corresponding to the delta operator Q is: xn { /n ∈ + ∪ {0}}. kn Z

Remark 2. In Corollary 1, the delta operator Q is a constant multiple of usual derivative D.

From (3.1)in the Theorem 4, we obtain the following proposition.

Proposition 2. For αn = n, for all n > 1, the basic polynomial sequence for the delta operator Q is : n−1 + {x(x − n) }, n ∈ Z ∪ {0}.

Proof. If we take αn = n, for all n > 1, then by equation (3.1), n n
n−1 n
n−2 n
n−r Q(x ) = 1 x + 2 2 x + ··· + r r x + ··· + n. For n = 1, 2, 3, ··· , we have Q(x) = 1,Q(x2) = 2x + 2,Q(x3) = 3x2 + 6x + 3, Q(x4) = 4x3 + 12x2 + 12x + 4 and so on. Let p0(x) = 1 and construct p1(x) = a1x For n = 1, Qpn(x) = npn−1(x) ⇒ Qp1(x) = 1p0(x). Qp1(x) = a1α1 and 1p0(x) = 1 ⇒ p1(x) = x 2 Let p2(x) = b1x + b2x . For n = 2, Qpn(x) = npn−1(x) implies Qp2(x) = 2p1(x). Qp2(x) = 2b1x + 2b1 + b2 and 2p1(x) = 2x ⇒ b1 = 1 and b2 = −2 2 Hence p2(x) = x − 2x = x(x − 2) 3 2 Let p3(x) = c1x + c2x + c3x For n = 3, Qpn(x) = npn−1(x) implies Qp3(x) = 3p2(x). 2 Qp3(x) = x (3α1c1) + x(3α2c1 + 2α1c2) + c1α3 + c2α2 + c3α1 2 and 3p2(x) = 3x − 6x ⇒ c1 = 1, c2 = −6 and c3 = 9 3 2 2 Hence p3(x) = x − 6x + 9x = x(x − 3) 3 By similar procedure, p4(x) = x(x − 4) n−1 Proceeding in this way, we get pn(x) = x(x − n) Thus the basic polynomial sequence for the delta operator Q is :

n−1 + {x(x − n) }, n ∈ Z ∪ {0}.

From (3.1), in the main result, we obtain the following proposition.

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Proposition 3. For αn = 1 for all n > 1, the basic polynomials sequence corresponding to the delta operator Q is :

+ {x(x − 1)(x − 2) ··· (x − n + 1)}, n ∈ Z ∪ {0}.

Proof. If we take αn = 1 for all n > 1, then by equation (3.1), n n
n−1 n
n−2 n
n−r Q(x ) = 1 x + 2 x + ··· + r x + ··· + 1. For n = 1, 2, 3, ··· ., we have Q(x) = 1,Q(x2) = 2x + 1,Q(x3) = 3x2 + 3x + 1, Q(x4) = 4x3 + 6x2 + 4x + 1 and so on. Take p0(x) = 1 and construct p1(x) = a1x Using Qpn(x) = npn−1(x), we get Qp1(x) = 1p0(x). Qp1(x) = a1 and 1p0(x) = 1 implies p1(x) = x 2 Let p2(x) = b1x +b2x . Using Qp2(x) = 2p1(x), we get b1 = 1 and b2 = −1 implies p2(x) = x(x−1) 3 2 Let p3(x) = c1x + c2x + c3x . Using Qp3(x) = 3p2(x), we get c1 = 1 c2 = −3 and c3 = 2 implies p3(x) = x(x − 1)(x − 2) By similar procedure p4(x) = x(x − 1)(x − 2)(x − 3) Proceeding in this way, we get pn(x) = x(x − 1)(x − 2) ··· (x − n + 1) Hence the basic polynomials sequence corresponding to the delta operator Q is:

+ {x(x − 1)(x − 2) ··· (x − n + 1)}, n ∈ Z ∪ {0}.

By applying the same procedure as above, we obtain several basic polynomials sequences that are listed below.

n Corollary 2. If αn = (−1) , n > 1 , then the basic polynomial sequence corresponding to the delta operator Q is:

n + (−1) {x(x − 1)(x − 2) ··· (x − n + 1)}, n ∈ Z ∪ {0}.

Corollary 3. If αn = 2, n > 1 , then the basic polynomial sequence corresponding to the delta operator Q is :

1 x(x − 1)(x − 2) ··· (x − n + 1)}, n ∈ + ∪ {0}. 2n Z

Corollary 4. If αn = −2, n > 1 , then the basic polynomial sequence corresponding to the delta operator Q is :

1 {(−1)n x(x − 1)(x − 2) ··· (x − n + 1)}, n ∈ + ∪ {0} 2n Z . Corollary 5. If αn = k, n > 1 , then the basic polynomial sequence corresponding to the delta operator Q is :

1 { x(x − 1)(x − 2) ··· (x − n + 1)}, n ∈ + ∪ {0} kn Z . Corollary 6.If αn = −k, n > 1 , then the basic polynomial sequence corresponding to the delta operator Q is :

(−1n) { x(x − 1)(x − 2) ··· (x − n + 1)}, n ∈ + ∪ {0} kn Z

7 Maheswaran and Elango; BJMCS, 14(2), 1-11, 2016; Article no.BJMCS.23323

n n It is interesting to note that if αn = (k) and (−k) , both cases, leads to a tedious process to get pn(x) in a standard form. The above results are shown vividly in Table 3.(appendix) Remarks: Table 3 i) In row 1, the basic polynomial sequence to the usual derivative is {xn}. ii) In row 2, The basic polynomial sequence corresponding to constant multiple k of the usual xn derivative is { kn }. n−1 iii) In row 3, The basic polynomial sequence corresponding to αn = n, for all n > 1, is {x (x−n) }. iv) The basic polynomial sequence given in row 4 is related to ’Newtonian’ polynomials [4]. 1 v) It is very interesting to study the polynomial for αn = n , for all n > 1 . vi) Let us note the the difference in the row 4 and row 5. The sequential characterization of both are the same except sign changes alternatively. But there is wide difference between the corresponding polynomial sequence . 1 vii) The basic polynomial sequence in row 6 is 2n times of polynomials in row 4. viii)The polynomials in row 6 and row 7 are merely same except slight changes of sign. x
n! ix) The basic polynomial sequence in row 8 can be simplified in a standard form as pn(x) = n kn . x) The polynomials in row 8 and row 9 are merely same except slight changes of sign and they are related to the classical ’Newtonian’ polynomial [4]. n n xi) If αn = k and αn = (−k) , then there is a difficulty to get pn(x) in a standard form.

Thus Table-3 contains a study of sequence of basic polynomials corresponding to different delta operators. Now we get a way opened to study the classical special polynomials by a new approach namely the finite operator calculus method.

5 Effect of Q on ex

The theory of finite operator calculus may be extended to algebraic and transcendental functions. The effect of delta operator on the exponential functions is interesting.

x 1 2 1 3 From Q(e ) = Q(1) + Q(x) + 2 ! Q(x ) + 3 ! Q(x ) + ... and equation(3.1) , we get 1 1 1 Q(ex) = α + (2 α x + α ) + (3 α x2 + 3 α x + α ) + (4 α x3 + 6 α x2 + 4 α x + α ) + 1 2 ! 1 2 3 ! 1 2 3 4 ! 1 2 3 4 1 (5 α x4 + 10 α x3 + 10 α x2 + 5 α x + α ) + ... 5 ! 1 2 3 4 5 x 1 1 = α (1 + + x2 + x3 + ... ) 1 1! 2 ! 3 ! 1 x 1 1 + α (1 + + x2 + x3 + ... ) 2! 2 1! 2 ! 3 ! 1 x 1 1 + α (1 + + x2 + x3 + ... ) 3! 3 1! 2 ! 3 ! 1 x 1 1 + α (1 + + x2 + x3 + ... ) + ... 4! 4 1! 2 ! 3 ! 1 1 1 = α ex + α ex + α ex + α ex + ··· 1 2! 2 3! 3 4! 4

Therefore, we obtained the following Proposition.

Proposition 4. For real constant αr (r = 1, 2, 3, ··· ),

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∞ x x x X αr Q(e ) = α e + e . 1 r! r=2 The effect of delta operator on (sin x) function is also interesting.

1 3 1 5 1 7 Q(sin x) = Q(x) − 3! Q(x ) + 5! Q(x ) − 7 Q(x ) + ··· 1 Q(sin x) = α − (3 α x2 + 3 α x + α ) + 1 3 ! 1 2 3 1 (5 α x4 + 10 α x3 + 10 α x2 + 5 α x + α ) + 5 ! 1 2 3 4 5 1 (7α x6 + 21α x5 + 35α x4 + 35α x3 + 21α x2 + 7α x + α ) + ... 7 1 2 3 4 5 6 7

x2 x4 x6 = α (1 − + − + ... ) 1 2! 4 ! 6 ! 1 x3 x5 − α (x − + − ... ) 2! 2 3! 5 ! 1 x2 x4 x6 − α (1 − + − + ... ) 3! 3 2! 4 ! 6 ! 1 x3 x5 + α (x − + + ... ) 4! 4 3! 5 !

Therefore, we obtained the following Proposition.

Proposition 5. For real constant αr (r = 1, 2, 3, ··· ), 1 1 1 Q(sin x) = α (cos x) − α (sin x) − α (cos x) + α (sin x) + ··· 1 2! 2 3! 3 4! 4

By similar procedure, we can easy to obtain the following Proposition.

Proposition 6. For real constant αr (r = 1, 2, 3, ··· ), 1 1 1 Q(cos x) = α (− sin x) − α (cos x) + α (sin x) + α (cos x) − · · · 1 2! 2 3! 3 4! 4

Thus the effect of delta operator on the exponential and circular functions are expressed in terms of infinite series. The first term is a constant multiple of usual derivative of the function. This constant become the first parameter α1 in the sequential characterization of the delta operator. If we take α1 = 1, α2, = α3 = ··· = 0, then the delta operator becomes the usual derivative and thus the above relations reduce to the classical counterparts in terms of differential operator.

6 Status and Further Directions

In this short extension of the theory, we limited our study to the effect of delta operator on the simplest transcendental functions. This method may be extended to higher transcendental functions. The basic polynomials sequence in the q-monodiffric sense with its delta operator is discussed in Maheswaran, A. [7]. It is also desirable that the theory of finite operator calculus may be extended to the study of delta operator equations. The simplest of such equation is Qf(x) = 0 and this problem become interesting in the development of finite operator calculus.

9 Maheswaran and Elango; BJMCS, 14(2), 1-11, 2016; Article no.BJMCS.23323

Competing Interests

The authors declare that no competing interests exist.

References

[1] Kwasniewski AK. On deformations of finite operator calculus. Circolo Matematico di Palermo. 2000; Serie II(Suplemento no 63):141-148. [2] Emil C. Popa. On an expansion theorem in the finite operator calculus of G. C. Rota. General Mathematics. 2008;16(4):149-154. [3] Octavian Agratini. Binomial polynomials and their application in approximation theory. Conferenza Tenuta il, Babes-Bolyai University, Romania; 2000. [4] Rainville ED.Special functions. Chalsea Publishing Company; 1960. [5] Heinrich Niederhausen. Finite operator calculus with application to linear recursions. Florida Atlantic University, Boca Roton. [6] Rota GC.Finite operator calculus. Academic Press, London; 1975. [7] Maheswaran A. Extension of finite operator calculus to q-monodiffric theory. M. Phil Dissertation, Bharathiyar University, Coimbatore, Tamil Nadu, India; 1988.

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Appendix Table 3. Basic Polynomials Sequences For Different Delta Operators.

0 n S.No Values Of αis Q(x ) pn(x) α1 = 1 and n−1 n 1 α2 = α3 = ··· nx x = αn = 0 α1 = k and n−1 xn 2 α2 = α3 = ··· knx kn . = αn = 0 n
n−1 n
n−2 1 x + 2 2 x + ··· n−1 3 αn = n n
n−r x (x − n) +r r x + ··· + n n
n−1 n
n−2 1 x + 2 x + ··· {x(x − 1)(x − 2) 4 αn = 1 n
n−r + r x + ··· + 1 ··· (x − n + 1)} n
n−1 n
n−2 n n − 1 x + 2 x + ··· + (−1) {x(x + 1)(x + 2) ··· 5 αn = (−1) rn
n−r n (−1) r x + ··· + (−1) +(x + n − 1)} n
n−1 n
n−2 1 2 1 x + 2 2 x + ··· { 2n x(x − 1)(x − 2) 6 αn = 2 n
n−r +r + 2 r x + ··· + 2 ··· (x − n + 1)} n
n−1 n
n−2 n 1 −2 1 x − 2 2 x − · · · {(−1) 2n x(x − 1)(x − 2) 7 αn = −2 n
n−r +r − 2 r x + · · · − 2 (x-3)··· (x − n + 1)} n
n−1 n
n−2 1 k ( 1 x + 2 x + ··· { kn x(x − 1)(x − 2) 8 αn = k n
n−r + r x + ··· + 1) ··· (x − n + 1)} −k (n
xn−1 + n
xn−2 + ··· (−1n) 1 2 { kn x(x − 1)(x − 2) 9 αn = −k n
n−r + r x + ··· + 1) ··· (x − n + 1)} x p1(x) = k . 1 2 1 n
n−1 2 n
n−2 p2(x) = k2 x − k x n k 1 x + k 2 x + ··· 1 3 3 2 2 10 αn = k rn
n−r n p3(x) = k3 x − k2 x + k x. + k r x + ··· + k p4(x) = 1 4 8 3 14 7 k4 x − k3 x + k2 − k x. x p1(x) = − k . n
n−1 2 n
n−2 1 2 1 n −k 1 x + k 2 x + ··· p2(x) = k2 x + k x 11 αn = (−k) rn
n−r n (−k) r x + ··· + (−k) p3(x) = 1 3 3 2 2 −( k3 x + k2 x + k x).

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