# TheK Extended Laguerre Polynomials Involving Aαr,n,kxFrr,r>2

Authors: Adnan Khan; M. Haris Mateen; Ali Akgül; Md. Shajib Ali
Journal: Advances in Mathematical Physics
 (2022)

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## Abstract

In this manuscript, we present the generalized hypergeometric function of the typeFrr,r>2 and extension of the K Laguerre polynomial for the K extended Laguerre polynomials Ar,n,kαx. Additionally, we describe the K generating function, K recurrence relations, and KS Rodrigues formula.


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## Body

## 1. Introduction

Laguerre polynomials are utilized to investigate non-central Chi-square distribution. Many works are existed in the literature with implementation to classical orthogonal polynomials. There many extensions of Laguerre polynomials.Table 1
.
The extended Laguerre polynomialsAq,nαx Khan et al. [23]TheK extended Laguerre polynomials Ar,n,kαxAq,nαx=exq+αnn!Fqq−nq,−n+1q,⋯,−n+q−1q;q+αq,q+1+αq,⋯,2q+α−1q;xqAr,n,kαx=exrk+αn,kn;k!Frr,k−nr,k,−n+kr,k,⋯,−n+rk−1r,k;α+krr,k,α+rk+1r,k,⋯,α+2rk−1r,k;xr.If we putk=1 in our paper, then we get the result of Khan et al. [23].A large number of properties of Laguerre polynomials have been described in classical works, e.g., Erdélyi et al. [1] and Bell [2]; also we can refer to Wang and Guo [3] and Mathai [4].Chak [5] has given a representation for the Laguerre polynomials. Carlitz [6] proved the recurrence relations involving Laguerre polynomials. Al-Salam [7] proved several results involving Laguerre polynomials. Prabhakar [8] introduced that generating functions, integrals, and recurrence relations are obtained for the polynomials Znαx;k in xk..Andrews et al. [9], Chen and Srivastava [10], Trickovic and Stankovic [11], Radulescu [12], and Doha et al. [13] have done a lot of work for properties of Laguerre polynomials. Akbary et al. [14] can be referred for other application of Laguerre polynomials. Li [15], Aksoy et al. [16], Wang [17], and Krasikov and Zarkh [18] studied problems of permutation of polynomials; bijection that can induce polynomials with integer coefficients is modulo m.In this manuscript, we present the properties of the extending Laguerre polynomial includingFrr,r>2; we consider
(1)Lnαx=1+αnn!F11−n;1+α;x.Shively [19] extended the Laguerre polynomials as
(2)Rna,x=a2nn!anF11−n;a+n;x.Habibullah [20] demonstrated the Rodrigues formula as
(3)Rna+1,x=exx−α−nn!Dnxα+2ne−x,Lnαx=exx−αn!Dnxα+ne−x.Erdélyi et al. [1] introduced
(4)Dmxα+mLnα+mx=Γα+m+n+1Γα+n+1xαLnαx,D=ddx.Khan and Habibullah [21] introduced A2,nx=F22−n/2,−n+1/2;1/2,1;x2.Khan and Kalim [22] introduced
(5)A3,mαy=1+αmm!F33−m3,−m+13,−m+23;1+α3,2+α3,3+α3;y3.Khan et al. [23] proposed extended Laguerre polynomials Aαq,nx.Parashar [24] presented a new set of Laguerre polynomials Lnα,hx related to the Laguerre polynomials Lnαx. Sharma and Chongdar [25] proved an extension of bilateral generating functions of the modified Laguerre polynomials.Researchers [26–28] found additional properties of k gamma and k beta functions. Then, Mubeen and Habibullah [29] introduced k fractional integrals and discussed its application. Mubeen and Habibullah [30] introduced an integral representation of some k hypergeometric functions. Krasniqi [31] derived some properties of the k gamma and k beta function. Mubeen [32] proved the properties of confluent k integrals by using k fractional integrals. There is a tremendous scope to study k polynomials using k gamma, k beta, and k hypergeometric functions. Kokologiannaki and Krasniqi [33] introduced k analogue of the Riemann Zeta function and also proved some inequities relating to Riemann Zeta function and k gamma functions.Din et al. [34] understand the dynamical behavior such diseases; they fitted a susceptible-infectious quarantined model for human cases with constant proportions. Din et al. [35] investigated a newly constructed system of equation for hepatitis B disease in sense of Atanganaa–Baleanu Caputo (ABC) fractional order derivative. Din et al. [36] developed the analysis of a non-integer-order model for hepatitis B (HBV) under singular type Caputo fractional order derivative. They investigated proposed system for an approximate or semi-analytical solution using Laplace transform along with decomposition techniques by Adomian polynomial of nonlinear terms and some perturbation techniques of homotopy (HPM). Din [37] investigated the spread of such contagion by using a delayed stochastic epidemic model with general incidence rate, time-delay transmission, and the concept of cross immunity.Ain et al. [38] impression of activated charcoal is shaped by the fractional dynamics of the problem, which leads to speedy and low-cost first aid. Ain et al. [39] presented an impulsive differential equation system, which is useful in examining the effectiveness of activated charcoal in detoxifying the body with methanol poisoning. Din and Ain [40] developed a model based on a stochastic process that could be utilized to portray the effect of arbitrary-order derivatives. A nonlinear perturbation is used to study the proposed stochastic model with the help of white noises.Rehman et al.’s [41] unsaturated porous media were analyzed by solving Burger’s equation using the variational iterative modeling and homotopy perturbation method. Wang and Wang [42] described two different types of plasma models with variable coefficients by using the fractal derivative. Wang [43] investigated the fractal nonlinear dispersive Boussineq-like equation by variational perspective for the first time. The fractal variational principle of the fractal Boussineq-like equation was established via fractal semi-inverse method (FSM).


## 2. Extended Polynomials

Lemma 1.
Ifk,j∈ℤ+ and n is any non-negative integer. Then, we will get
(6)−nrkj−n+1rkj⋯−n+r−1rkj=−1rkjn!rrkjn−rkj!.Proof.
(7)−nrkj−n+1rkj⋯−n+r−1rkj=−nr−nr+1−nr+2⋯−nr+kj−1−n+1r−n+1r+1−n+1r+2⋯−n+1r+kj−1−n+r−1r−n+r−1r+1−n+r−1r+2⋯−n+r−1r+kj−1=−nr−n+rr−n+2rr⋯−n+rkj−rr−n+1r−n+r+1r−n+2r+1r⋯−n+rkj−r+1r−n+r−1r−n+2r−1r−n+3r−1r⋯−n+rkj−1r.
By simplification we get our desired result.Lemma 2.
Ifk∈ℤ+ and n is any non-negative integer, thus
(8)αkn=kknαknα+1kn⋯α+k−1kn.
Rainville [44] (p 22)).Lemma 3.
Assume thatk∈ℤ+ and n is any non-negative integer. Then, we reach
(9)∑n=0∞∑k=0nBk,n=∑n=0∞∑k=0∞Bk,n+k.
Rainville [44] (p 57)).Lemma 4.
Assume thatk∈ℤ+ and n is any non-negative integer. Thus, we have
(10)∑n=0∞∑k=0∞Bk,n=∑n=0∞∑k=0nBk,n−k.
Rainville [44] (p 56)).


## 3. TheK Extended Laguerre Polynomials Ar,n,kαx

We describe theK extended Laguerre polynomial set Ar,n,kαx as
(11)Ar,n,kαx=exrk+αn,kn;k!Frr,k−nr,k,−n+kr,k,⋯,−n+rk+1r,k;α+krr,k,α+rk+1r,k,⋯,α+2rk−1r,k;xr,where α∈ℝ,n,r,k∈ℤ+.Theorem 5.
IfAr,n,kαx, are the K extended Laguerre polynomials. Then
(12)Ar,n,kαx=exrk+αn,k∑j=0n/rk−1rkjn−rkj;k!rk+αrkjxrkjrkj;k!,(13)α∈ℝ,n,r,k∈ℤ+.Proof.
Consider(14)Ar,n,kαx=exrk+αn,kn;k!Fqq,k−nr,k,−n+kr,k,⋯,−n+rk+1r,k;;xrα+rkr,k,α+rk+1r,k,⋯,α+2rk−1r,k.=exrk+αn,kn;k!×∑j=0n/rk−n/r,kj−n+k/r,kj⋯−n+rk+1/r,kjα+kq/r,kjα+qk+1/r,kj⋯α+2rk−1/r,kjxrkjrkj;k!.
By using Lemma (1)
(15)Ar,n,kαx=exrk+αn,kn;k!×∑j=0n/rk−1rkjn;k!rrjn−rkj;k!α+rk/r,kjα+rk+1/r,kj⋯α+2rk−1/r,kjxrkjrkj;k!.
Now, by applying Lemma (2), we get our desired result.


## 4.K Generating Functions

Theorem 6.
Suppose thatn,j,k∈ℤ+. Thus, we reach
(16)∑n=0∞∑j=0n/rk−1rkjextnn−rkj;k!rk+αrkjxrkjrkj;k!=exMktF0r,k−−;rk+αr;k,rk+1+αr;k,⋯,2rk+α−1r;k;−xtrr.Proof.
We have(17)∑n=0∞∑j=0n/rk−1rkjextnn−rkj;k!rk+αrkjxrkjrkj;k!=∑n=0∞∑j=0∞−1rkjextn+rkjn;k!rk+αrkjxrkjrkj;k!=ex∑n=0∞tnn;k!∑j=0∞−1rkjtrkjrk+αrkjxrkjrkj;k!=exMkt∑j=0∞−xtrkjrk+αrkjrkj;k!
By applying Lemma (2), we get
(18)∑n=0∞∑j=0n/rk−1rkjextnn−rkj;k!rk+αrkjxrkjrkj;k!=exMkt×∑j=0∞−xtrkjrrkjrk+α/r;kjrk+1+α/r;kj⋯2rk+α−1/r;kjrkj;k!.
After simplification, we get our result.Corollary 7.
Suppose thatα∈ℝand n,r,j,k∈ℤ+. Thus, we reach
(19)∑n=0∞Aαr,n,kxtnrk+αn,k=exMktF0r,k−−;−xtrrrk+αr;k,rk+1+αr;k,⋯,2rk+α−1r;k;.Proof.
From Equation (12), we acquire
(20)∑n=0∞Aαr,n,kxrk+αn,ktn=∑n=0∞∑j=0n/rk−1rkjn−rkj;k!rk+αrkjxrkjrkj;k!tn.
Then, we have our result.Theorem 8.
Ifc∈ℤ+, then
(21)∑n=0∞cn,kAαr,n,kxtnα+rkn,k=ex1−ktkc/k×Frr,kcr,k,c+kr,k,⋯,c+rk+1r,k;−xt1−ktk1/krα+rkr,k,α+rk+1r,k,⋯,α+2rk−1r,k;.Proof.
From Equation (20), we note that
(22)∑n=0∞cnAαr,n,kxrk+αn,ktn=∑n=0∞cnex∑j=0n/rk−1rkjn−rkj;k!rk+αrkjxrkjrkj;k!tn.
We get(23)∑n=0∞cn,kAαr,n,kxtnrk+αn,k=∑n=0∞∑j=0∞cn+rkj,kextn+rkjn;k!−1rjxrkjrk+αrkj,krkj!=∑j=0∞∑n=0∞c+rkjn,ktnn;k!crj,kα+rkrj,kex−xtrkjrkj;k!,
Sincecn+rkj,k=c+rkjn,kcrj,k, and 1−ktk−m/k=∑n=0∞mn,ktn/n;k! it thus implies that
(24)∑n=0∞cn,kAαr,n,kxtnα+rkn,k=∑j=0∞crkj,k1−tc+rkjα+rkrj,kex−xtrkjrkj;k!=ex1−ktkc/k∑j=0∞crkj,krk+αrj,k−xt/1−ktk1/krjkrjk;k!Corollary 9.
Assume thatα∈ℝ and n,r,j,k∈ℤ+. Thus, we reach
(25)∑n=0∞Aαr,n,kxtn=11−ktkα+qk/kexpx−2xt1−t.Proof.
We choosec=r+α in Equation (21). We can reach the desired results.


## 5.K Recurrence Relations

Theorem 10.
Assume thatα∈ℝ and n,j,k∈ℤ+. Thus, we reach
(26)xDAαr,n,kx=n+xAαr,n,kx−rk+α+n−1Aαr,n−1,kx,D=ddx.Proof.
From Equation (16)
(27)∑n=0∞Aαr,n,kxtnrk+αn,k=exMktF0r,k−−;−xtrrrk+αr;k,rk+1+αr;k,⋯,2rk+α−1r;k;.
Let(28)σr,n,kx=Aαr,n,kxα+rkn,k.
Suppose that(29)F0r,k−−;−xtrrrk+αr;k,rk+1+αr;k,⋯,2rk+α−1r;k;=ψxrtrr.
Then(30)F=exMktψxrtrr=∑n=0∞σr,n,kxtn.
By taking partial derivatives,(31)∂F∂x=exMktψ+xr−1trexMktψ′,(32)∂F∂t=exMktψ+xrtr−1exMktψ′,(33)x∂F∂x−t∂F∂t=xF−tF.
Since(34)F=∑n=0∞σr,n,kxtn,
therefore∂F/∂x=∑n=0∞σr,n,k′xtn, and t∂F/∂t=∑n=0∞nσr,n,kxtn.
Equation (33), then yields
(35)x∑n=0∞σr,n,k′xtn−∑n=0∞nσr,n,kxtn=x∑n=0∞σr,n,kxtn−∑n=0∞σr,n,kxtn+1=x∑n=0∞σr,n,kxtn−∑n=1∞σr,n−1,kxtn.
We getσr,0′x=0, and for n>1, we get our result.Theorem 11.
Ifα∈ℝ and n≥2, then
(36)DAαr,n,kx=DAαr,n−1,kx+Aαr,n,kx−2Aαr,n−1,kx.Proof.
By (25), we reach
(37)1−t−rk−αexpx1−2t1−t=∑n=0∞Aαr,n,kxtn.
Let(38)F=Atexpx1−2t1−t=∑n=0∞yr,n,kxtn,(39)∂F∂x=1−2t1−tAtexpx1−2t1−t,(40)1−t∂F∂x=1−2tAtexpx1−2t1−t.
By using Equation (38), we obtain
(41)1−t∂F∂x=1−2tF.(42)SinceF=∑n=0∞yr,n,kxtn,wereach∂F∂x=∑n=0∞yr,n,k′xtn.
Equation (41) can be expressed as
(43)∑n=0∞yr,n,k′xtn−∑n=0∞yr,n,k′xtn+1=∑n=0∞yr,n,kxtn−2∑n=0∞yr,n,kxtn+1.
We getyr,0,k′x=0,yr,1,k′x=0, and for n>2, we get our result.Theorem 12.
Ifα∈ℝ and n≥r, then
(44)DAαr,n,kx=Aαr,n,kx−∑j=0n−1Aαr,j,kx.Proof.
We have(45)∂F∂x=1−t1−tF.
Applying Equation (38) yields
(46)∂F∂x=1−t1−t∑n=0∞yr,n,kxtn.
By using Equation (42), we obtain
(47)∑n=0∞yr,n,k′xtn=∑n=0∞yr,n,kxtn−∑n=0∞tn+1∑n=0∞yr,n,kxtn=∑n=0∞yr,n,kxtn−∑n=0∞∑j=0∞yr,j,kxtjtn+1
By using Lemma (4), we get
(48)∑n=0∞yr,n,k′xtn=∑n=0∞yr,n,kxtn−∑n=0∞∑j=0nyr,j,kxtn+1=∑n=0∞yr,n,kxtn−∑n=1∞∑j=0n−1yr,j,kxtnThen, we haveyr,0,k′x=0,yr,1,k′x=0, and for n>r,(49)yr,n,k′x=yr,n,kx−∑j=0n−1yr,j,kx.We get our desired result.Theorem 13.
Suppose thatα∈ℝand n≥r+1. Thus, we get
(50)nAαr,n,kx=3x−rk−αAαr,n−1,kx−rk+α+n−2Aαr,n−2,kx.Proof.
We have(51)0=nAαr,n,kx−xDAαr,n−1,kx+2x−rk−α−n+1Aαr,n−1,kx,nAαr,n,kx=xDAαr,n−1,kx−2x−rk−α−n+1Aαr,n−1,kx.
Then, after simplification, we get our result.Theorem 14.
Assume thatα∈ℝand n,r,j,k∈ℤ+. Thus, we obtain
(52)A1+αr,n−1,kx+Aαr,n,kx=A1+αr,n,kx.Proof.
From Equation (12), we obtain
(53)A1+αr,n−1,kx=exrk+1+αn−1,k∑j=0n−1/rk−1rkjn−1−rkj;k!r+1+αrkjxrkjrkj;k!,so thatAαr,n,kx=exrk+αn,k∑j=0n/rk−1rkj/n−rkj;k!rk+αrkjxrkj/rkj;k!.
Then, we acquire(54)A1+αr,n−1,kx+Aαr,n,kx=exrk+1+αn−1,k∑j=0n−1rk−1rkjn−1−rkj;k!r+1+αrkjxrkjrkj;k!exrk+αn,k∑j=0nrk−1rkjn−rkj;k!rk+αrkjxrkjrkj;k!=ex∑j=0n−1rkrk+α+n−1!−1rkjn−1−rkj;k!r+α+rkj!xrkjrkj;k!+∑k=0nrkrk+α+n−1!−1rkjn−rkj;k!r+α+rkj−1!xrkjrkj;k!=ex∑j=0n−1rkrk+α+n−1!−1rkjn−1−rkj;k!r+α+rkj!xrkjrkj;k!+∑k=0n−1rkrk+α+n−1!−1rkjn−rkj;k!r+α+rkj−1!xrkjrkj;k!+xrknrkn;k!=ex∑j=0n−1rkr+α+n−1!xrkj−1rkjrkj;k!1n−1−rkj;k!r+α+rkj!+1n−rkj;k!r+α+rkj−1!+xrknrkn;k!=ex∑j=0n−1rkr+α+n−1!−1rkjn−rkj;k!r+α+rkj!r+α+nxrkjrkj;k!+xrknrkn;k!=ex∑j=0n−1rkr+α+n!−1rkjn−rkj;k!r+α+rkj!xrkjrkj;k!+xrknrkn;k!=exr+1+αn,k∑j=0n−1rk−1rkjn−rkj;k!r+1+αrj,kxrkjrkj;k!+xrknrkn;k!=exr+1+αn,k∑j=0nrk−1rkjn−rkj;k!r+1+αrj,kxrkjrkj;k!=A1+αr,n,kx.


## 6.K Differential Equation

Theorem 15.
Assume thatα∈ℝ and n≥q. Thus, we reach
(55)xD2Aαr,n,kx+rk+α−3xDAr,n,kαx+2x+n−rk−αAr,n,kαx=0.Proof.
We have(56)xD2Ar,n,kαx+DAr,n,kαx=n+xDAr,n,kαx+Ar,n,kαx−rk+α+n−1DAr,n−1,kαx.
By using Equation (36), we get
(57)xD2Ar,n,kαx+DAr,n,kαx=n+xDAr,n,kαx+Ar,n,kαx−rk+α+n−1DAr,n,kαx−Ar,n,kαx+2Ar,n−1,kαx,or
(58)xD2Ar,n,kαx+rk+α−xDAr,n,kαx=rk+α+nAαr,n,kx−2rk+α+n−1Ar,n−1,kαx.
By using Equation (26), we have
(59)xD2Ar,n,kαx+rk+α−xDAr,n,kαx=rk+α+nAr,n,kαx+2xDAr,n,kαx−2n+xDAr,n,kαx,
or(60)xD2Ar,n,kαx+rk+α−3xDAr,n,kαx+2x+n−rk−αAr,n,kαx=0.


## 7.K Rodrigues Formula

Theorem 16.
Assume thatα∈ℝ and n,j,k∈ℤ+. Thus, we reach
(61)Ar,n,kαx=x−rk−1−αe2xn;k!Dknxrk−1+α+nke−x.Proof.
We take into consideration theK extended Laguerre polynomials involving
r>22F2,r>2,(62)Ar,n,kαx=exrk+αn,kn;k!Frr,k−nr,k,−n+kr,k,⋯,−n+rk+1r,k;α+rkr,k,α+rk+1r,k,⋯,α+2rk−1r,k;xr.
By Theorem (12), we have
(63)Ar,n,kαx=exn;k!∑j=0n/rkn;k!n−rkj;k!rkj;k!rk+αn,kxrkjrk+αrj,k=exx−rk−1−αn;k!∑j=0n/r−1rkjn;k!n−rkj!rkj;k!α+rkn,kxrkj+α+rk−1α+rkrj,k.
SinceDnk−rjkxn+α+r−1=α+rkn,kxrj+α+r−1/α+rkrj,k, we get
(64)Ar,n,kαx=x−r−1−αe2xn;k!∑j=0n/rkn;k!n−rjk;k!rj!−1rkje−xDnk−rkjxn+α+rk−1=x−r−1−αe2xn;k!∑j=0n/rknCrjk;kDnk−rkjxn+α+rk−1Drkje−x
Then, we get our desired result.


## 8. Special Properties

Theorem 17.
Suppose thatα,β∈ℝ and n,j,r,k∈ℤ+. Thus, we acquire
(65)Ar,n,kαx=∑j=0n/rkα−βrkjAr,n−rkjβxrkj;k!.Proof.
From Equation (25),
(66)∑n=0∞Ar,n,kαxtn=11−trk+αexpx−2xt1−t.
Also, consider(67)11−trk+αexpx1−2t1−t=1−t−α−β1−t−rk−βexpx1−2t1−t,∑n=0∞Ar,n,kαxtn=1−t−α−β∑n=0∞Ar,n,kβxtn=∑n=0∞α−βrntrnrn;k!∑n=0∞Ar,n,kβxtn=∑n=0∞∑j=0∞α−βrkjtrkjAr,n,kβxtnrkj;k!.
By using Lemma (4) yields
(68)∑n=0∞Ar,n,kαxtn=∑n=0∞∑j=0n/rkα−βrkjtrkjAr,n−rkj,kβxtn−rkjrkj;k!=∑n=0∞∑j=0n/rkα−βrkjAr,n−rkj,kβxtnrkj;k!.
Then, we get our result.Theorem 18.
Ifα∈ℝ and n,j,k∈ℤ+, then
(69)Ar,n,kα+β+qkx+y=∑j=0n/rkAr,n−rkj,kβyAr,rkj,kαx.Proof.
Consider(70)1−t−rk−αexpx1−2t1−t1−t−rk−βexpy1−2t1−t=1−t−rk−α+β+rkexpx+y1−2t1−t.
Then, we get(71)∑n=0∞Aαr,n,kxtn∑n=0∞Aβr,n,kytn=∑n=0∞Aα+β+qr,n,kx+ytn.
By using Lemma (4), we acquire
(72)∑n=0∞Aα+β+qkr,n,kx+ytn=∑n=0∞∑j=0n/rkAβr,n−rkj,kyAαr,rkj,kxtn.
On comparing the coefficients oftn, we acquire our result.


## 9. Conclusion

We constructed theK extended Laguerre polynomials Ar,n,kαx relied on the Frr,r>2. We acquired K generating functions, K recurrence relations and K Rodrigues formula for these K extended Laguerre polynomials. We will use the integral transformations on the results of K extended Laguerre polynomials in our future works (Table 1). We can also apply Laplace transformation on our results.



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