International Journal of Analysis and Applications Volume 16, Number 6 (2018), 856-867 URL: https://doi.org/10.28924/2291-8639 DOI: 10.28924/2291-8639-16-2018-856 IMPLICIT SUMMATION FORMULA FOR 2-VARIABLE LAGUERRE-BASED POLY-GENOCCHI POLYNOMIALS WASEEM A. KHAN∗, IDREES A. KHAN AND MOIN AHMAD Department of Mathematics, Faculty of Science, Integral University, Lucknow-226026, India ∗Corresponding author: waseem08 khan@rediffmail.com Abstract. The main object of this paper is to introduce a new class of Laguerre-based poly-Genocchi polynomials and investigate some properties for these polynomials and related to the Stirling numbers of the second kind. We derive summation formulae and general symmetry identities by using different analytical means and applying generating functions. 1. Introduction The generalized Bernoulli, Euler and Genocchi polynomials of (real or complex) order α are usually defined by means of the following generating functions (see [1-16]):( t et − 1 )α ext = ∞∑ n=0 B(α)n (x) tn n! , (| t |< 2π; 1α = 1), (1.1) ( 2 et + 1 )α ext = ∞∑ n=0 E(α)n (x) tn n! , (| t |< π; 1α = 1) (1.2) and ( 2t et + 1 )α ext = ∞∑ n=0 G(α)n (x) tn n! , (| t |< π; 1α = 1). (1.3) Received 2017-09-19; accepted 2017-12-07; published 2018-11-02. 2010 Mathematics Subject Classification. 33C45, 33C99, 05A10, 05A15. Key words and phrases. Laguerre polynomials, poly-Genocchi polynomials, Laguerre-based poly-Genocchi polynomials, summation formulae, symmetric identities. c©2018 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License. 856 https://doi.org/10.28924/2291-8639 https://doi.org/10.28924/2291-8639-16-2018-856 Int. J. Anal. Appl. 16 (6) (2018) 857 So that obviously Bn(x) = B 1 n(x),En(x) = E 1 n(x) and Gn(x) = G 1 n(x), (n ∈ N), where N0 = N∪{0}(N = 1, 2, 3, · · ·). The classical polylogarithmic function Lik(z) is defined by (see [2], [10]): Lik(z) = ∞∑ m=1 zm mk , (k ∈ Z). (1.4) The poly-Bernoulli numbers and polynomials are defined by following generating functions (see [7], [8], [9]): Lik(1 −e−t) et − 1 = ∞∑ n=0 B(k)n tn n! , (1.5) Lik(1 −e−t) et − 1 ext = ∞∑ n=0 B(k)n (x) tn n! . (1.6) In the case k = 1 in (1.5) and (1.6), we have B(1)n = Bn,B (1) n (x) = Bn(x). The poly-Genocchi numbers and polynomials are defined by following generating functions (see [14]): 2Lik(1 −e−t) et + 1 = ∞∑ n=0 G(k)n tn n! , (1.7) 2Lik(1 −e−t) et + 1 ext = ∞∑ n=0 G(k)n (x) tn n! . (1.8) In the case k = 1 in (1.7) and (1.8), we have G(1)n = Gn,G (1) n (x) = Gn(x). The 2-variable Laguerre polynomials (2-VLP) Ln(x,y), which is defined by (see [5]): 1 (1 −yt) exp ( −xt 1 −yt ) = ∞∑ n=0 Ln(x,y)t n, (| yt |< 1) (1.9) It is equivalently given by (see [6]). exp(yt)C0(xt) = ∞∑ n=0 Ln(x,y) tn n! , (1.10) where C0(x) denotes the 0 th order Tricomi function. The nth order Tricomi functions Cn(x) are defined as: Cn(x) = ∞∑ r=0 (−1)rxr r!(n + r)! , (n ∈ N0) (1.11) Int. J. Anal. Appl. 16 (6) (2018) 858 with the following generating function: exp ( t− x t ) = ∞∑ n=0 Cn(x)t n, (1.12) for t 6= 0 and for all finite x. From (1.9) and (1.10), we get Ln(x,y) = n! n∑ s=0 (−1)sxsyn−s (s!)2(n−s)! = ynLn(x/y). (1.13) Thus, we have Ln(x, 0) = (−1)nxn n! , Ln(0,y) = y n, Ln(x, 1) = Ln(x), (1.14) where Ln(x) are the classical Laguerre polynomials (see [1]). Now, we recall here the following definition as follows: The Stirling number of the first kind is given by (x)n = x(x− 1) · · ·(x−n + 1) = n∑ l=0 S1(n,l)x l, (n ≥ 0) (1.15) and the Stirling number of the second kind is defined by generating function: (et − 1)n = n! ∞∑ l=n S2(l,n) tl l! . (1.16) 2. 2-Variable Laguerre-based poly-Genocchi polynomials Let k ∈ Z, we inroduce 2-variable Laguerre-based poly-Genocchi polynomials by the following generating function: 2Lik(1 −e−t) et + 1 exp(yt)C0(xt) = ∞∑ n=0 LG (k) n (x,y) tn n! , (2.1) so that LG (k) n (x,y) = n∑ m=0   n m  G(k)n−mLm(x,y). (2.2) When x = y = 0, LG (k) n = G (k) n (0, 0) are called the poly-Genocchi numbers. For k = 1 in (2.1), we have 2Li1(1 −e−t) et + 1 exp(yt)C0(xt) = ∞∑ n=0 LGn(x,y) tn n! , (2.3) where LGn(x,y) is Laguerre-based Genocchi polynomials (see [13]). Thus, we have LG (k) n (x,y) = LGn(x,y), (n ≥ 0). Int. J. Anal. Appl. 16 (6) (2018) 859 On setting x = 0, (2.1) reduces to the known result of Kim et al. [14.,p.Eq.(4)4776]: 2Lik(1 −e−t) et + 1 exp(yt) = ∞∑ n=0 G(k)n (y) tn n! , (k ∈ Z). (2.4) Theorem 2.1. The following explicit summation formulae for Laguerre-based poly-Genocchi polynomials holds true: LG (2) n (x,y) = n∑ m=0   n m   Bmm! m + 1 LGn−m(x,y). (2.5) Proof. Using generating function for Laguerre-based poly-Genocchi polynomials (2.1), we have ∞∑ n=0 LG (k) n (x,y) tn n! = 2Lik(1 −e−t) et + 1 exp(yt)C0(xt) = 2 et + 1 exp(yt)C0(xt) ∫ t 0 1 ez − 1 ∫ t 0 1 ez − 1 · · · 1 ez − 1 ∫ t 0 z ez − 1 dz · · ·dz. In particular k = 2, we have ∞∑ n=0 LG (2) n (x,y) tn n! = 2 et + 1 exp(yt)C0(xt) ∫ t 0 z ez − 1 dz = ( ∞∑ m=0 tmBm m + 1 ) 2t et + 1 exp(yt)C0(xt) = ( ∞∑ m=0 tmBmm! m + 1m! )( ∞∑ n=0 LGn(x,y) tn n! ) . Replacing n by n−m in the r.h.s of above equation, we have ∞∑ n=0 LG (2) n (x,y) tn n! = ∞∑ n=0   n∑ m=0   n m   Bmm! m + 1 LGn−m(x,y)   tn n! . On equating the coefficients of the like powers of t in both sides, we get (2.5). Remark 2.1. On setting x = 0, Theorem (2.1) reduces to the known result of Kim et al. [14.,p. 4777, Theorem (2.1)]. Corollary 2.1. For n ≥ 0, we have G(2)n (y) = n∑ m=0   n m   Bmm! m + 1 Gn−m(y). (2.6) Theorem 2.2. For n ≥ 1, the degree of LG (k) n (x,y) is n-1. We have LG (k) n (x,y) n = n−1∑ m=0   n− 1 m   G(k)m+1 m + 1 Ln−m−1(x,y). (2.7) Proof. From (2.1), we have ∞∑ n=0 LG (k) n (x,y) tn n! = 2Lik(1 −e−t) 1 −e−t exp(yt)C0(xt) Int. J. Anal. Appl. 16 (6) (2018) 860 = ( ∞∑ m=0 G(k)m tm m! )( ∞∑ n=0 Ln(x,y) tn n! ) Replacing n by n−m in above equation and comparing the coefficients of tn, we get LG (k) n (x,y) = n∑ m=0   n m  G(k)m Ln−m(x,y), (n ≥ 0). (2.8) From (2.8), we have LG (k) n (x,y) n = n−1∑ m=0   n− 1 m   G(k)m+1 m + 1 Ln−m−1(x,y), (n ≥ 1) (2.9) Therefore by (2.9), we obtain the result (2.7). Remark 2.2. For x = 0, Theorem (2.2) reduces to the known result of Kim et al. [14.,p. 4778, Theorem (2.2)]. Corollary 2.2. For n ≥ 1, the degree of G(k)n (x) is n-1. We have G (k) n (y) n = n−1∑ m=0   n− 1 m   G(k)m+1 m + 1 yn−m−1. (2.10) Theorem 2.3. For n ≥ 0, we have LG (k) n (x,y) = n∑ p=0 p+1∑ l=1 (−1)l+p+1l!S2(p + 1, l) lk(p + 1)   n p   LGn−p(x,y). (2.11) Proof. By using (2.1), we can be written as ∞∑ n=0 LG (k) n (x,y) tn n! = ( Lik(1 −e−t) t )( 2t et + 1 exp(yt)C0(xt) ) . (2.12) Now 1 t Lik(1 −e−t) = 1 t ∞∑ l=1 (1 −e−t)l lk = 1 t ∞∑ l=1 (−1)l lk (1 −e−t)l = 1 t ∞∑ l=1 (−1)l lk l! ∞∑ p=l (−1)pS2(p,l) tp p! = 1 t ∞∑ p=1 p∑ l=1 (−1)l+p lk l!S2(p,l) tp p! = ∞∑ p=0 ( p+1∑ l=1 (−1)l+p+1 lk l! S2(p + 1, l) p + 1 ) tp p! . (2.13) From equations (2.12) and (2.13), we get ∞∑ n=0 LG (k) n (x,y) tn n! = ∞∑ p=0 ( p+1∑ l=1 (−1)l+p+1 lk l! S2(p + 1, l) p + 1 tp p! )( ∞∑ n=0 LGn(x,y) tn n! ) . Int. J. Anal. Appl. 16 (6) (2018) 861 Replacing n by n−p in the r.h.s. of above equation and comparing the coefficients of tn in both sides, we arrive at the desired result (2.11). Remark 2.3. For x = 0, Theorem (2.3) reduces to the known result of Kim et al. [14.,p. 4779, Theorem (2.3)]. Corollary 2.3. For n ≥ 0, we have G(k)n (y) = n∑ p=0 p+1∑ l=1 (−1)l+p+1l!S2(p + 1, l) lk(p + 1)   n p  Gn−p(y). (2.14) Theorem 2.4. For n ≥ 1, we have LG (k) n (x,y + 1) + LG (k) n (x,y) = 2 n∑ p=1 p∑ l=1 (−1)l+p lk l!S2(p,l)   n p  Ln−p(x,y). (2.15) Proof. By using definition (2.1), we have ∞∑ n=0 LG (k) n (x,y + 1) tn n! + ∞∑ n=0 LG (k) n (x,y) tn n! = 2Lik(1 −e−t) et + 1 exp((y + 1)t)C0(xt) + 2Lik(1 −e−t) et + 1 exp(yt)C0(xt) = 2Lik(1 −e−t) exp(y)tC0(xt) = ∞∑ p=1 ( 2 p∑ l=1 (−1)l+p lk l!S2(p,l) ) tp p! exp(yt)C0(xt) = ( ∞∑ p=1 ( 2 p∑ l=1 (−1)l+p lk l!S2(p,l) ) tp p! )( ∞∑ n=0 Ln(x,y) tn n! ) . Replacing n by n−p in the above equation and comparing the coefficients of tn in both sides, we obtain the result (2.15). Remark 2.4. Taking x = 0, Theorem 2.4 reduces to the known result of Kim et al. [14.,p. 4780, Theorem (2.4)]. Corollary 2.4. For n ≥ 1, we have G(k)n (y + 1) + G (k) n (y) = 2 n∑ p=1 p∑ l=1 (−1)l+p lk l!S2(p,l)   n p  yn−p. (2.16) Theorem 2.5. For d ∈ N with d ≡ 1(mod2), we have LG (k) n (x,y) = n∑ p=0   n p  dn−p−1 p+1∑ l=0 d−1∑ a=0 (−1)l+p+1l!S2(p + 1, l) lk (−1)aLGn−p( a + y d ,x). (2.17) Int. J. Anal. Appl. 16 (6) (2018) 862 Proof. From equation (2.1), we can be written as ∞∑ n=0 LG (k) n (x,y) tn n! = 2Lik(1 −e−t) et + 1 exp(yt)C0(xt) = ( 2Lik(1 −e−t) t )( 2t edt + 1 d−1∑ a=0 (−1)a exp((a + y)t)C0(xt) ) = ( ∞∑ p=0 ( p+1∑ l=1 (−1)l+p+1 lk l! S2(p + 1, l) p + 1 ) tp p! )( ∞∑ n=0 dn−1 d−1∑ a=0 (−1)aLGn( a + y d ,x) tn n! ) . Replacing n by n−p in above equation and comparing the coefficient of tn in both sides, we get (2.17). Remark 2.5. For x = 0, Theorem 2.5 reduces to the known result of Kim et al. [14.,p. 4780]. Corollary 5. For d ∈ N with d ≡ 1(mod2), we have G(k)n (y) = n∑ p=0   n p  dn−p−1 p+1∑ l=0 d−1∑ a=0 (−1)l+p+1l!S2(p + 1, l) lk (−1)aGn−p( a + y d ). (2.18) 3. Summation formulae for Laguerre-based poly-Genocchi polynomials In this section, we establish summation formula for Laguerre-based poly-Genocchi polynomials by using series techniques method. Theorem 3.1. The following implicit summation formulae for Laguerre-based poly-Genocchi polynomials LG (k) n (x,y) holds true: LG (k) l+p(x,z) = l,p∑ m,n=0   l m     p n   (z −y)m+nLG(k)l+p−m−n(x,y). (3.1) Proof. Replacing t by t + u and rewrite the generating function (2.1) as 2Lik(1 − (e)−(t+u)) et+u + 1 C0(x(t + u)) = e −y(t+u) ∞∑ l,p=0 LG (k) l+p(x,y) tl l! up p! . (3.2) Replacing y by z in the above equation and equating the resulting equation to the above equation, we get e(z−y)(t+u) ∞∑ m,l=0 LG (k) l+p(x,y) tl l! up p! = ∞∑ l,p=0 LG (k) l+p(x,z) tl l! up p! . (3.3) On expanding exponential function (3.3) gives ∞∑ N=0 [(z −y)(t + u)]N N! ∞∑ l,p=0 LG (k) l+p(x,y) tl l! up p! = ∞∑ l,p=0 LG (k) l+p(x,z) tl l! up p! (3.4) which on using formula [16, p.52(2)] ∞∑ N=0 f(N) (x + y)N N! = ∞∑ n,m=0 f(n + m) xn n! ym m! , (3.5) Int. J. Anal. Appl. 16 (6) (2018) 863 in the left hand side becomes ∞∑ m,n=0 (z −x)m+ntmun m!n! ∞∑ l,p=0 HG (k) l+p(x,y) tl l! up p! = ∞∑ l,p=0 HG (k) l+p(z,y) tl l! up p! (3.6) Now replacing l by l−m, p by p−n and using the lemma [16, p.100(1)] in the left hand side of (3.6), we get ∞∑ m,n=0 ∞∑ l,p=0 (z −x)m+n m!n! LG (k) l+p−m−n(x,y) tl (l−m)! up (p−n)! = ∞∑ l,p=0 LG (k) l+p(x,z) tl l! up p! . (3.7) Finally on equating the coefficients of the like powers of t and u in the above equation, we get the required result. Remark 3.1. Taking l = 0 in assertion (3.1) of Theorem 3.1, we deduce the following consequence of Theorem 3.1. Corollary 3.1. The following summation formula for Laguerre-based poly-Genocchi polynomials HG (k) n (z,y) holds true: LG (k) p (x,z) = p∑ n=0   p n   (z −y)nLG(k)p−n(x,y). (3.8) Remark 3.2. Replacing z by z + y in (3.8), we obtain LG (k) p (x,z + y) = p∑ n=0   p n  znLG(k)p−n(x,y). (3.9) Theorem 3.2. The following summation formula for Laguerre-based poly-Genocchi polynomials HG (k) n (z,y) holds true: LG (k) n (x,y + u) = n∑ j=0   n j  ujLG(k)n−j(x,y). (3.10) Proof. Using (2.1), we can be written as ∞∑ n=0 LG (k) n (x,y + u) tn n! = 2Lik(1 −e−t) et + 1 exp((y + u)t)C0(xt) = ( ∞∑ n=0 LG (k) n (x,y) tn n! ) ∞∑ j=0 uj tj j!   Now replacing n by n− j and comparing the coefficients of tn in both sides, we obtain (3.10). Int. J. Anal. Appl. 16 (6) (2018) 864 Theorem 3.3. The following summation formula for Laguerre-based poly-Genocchi polynomials HG (k) n (z,y) holds true: LG (k) n (x + w,y + u) = n∑ m=0   n m   LG (k) n−m(x,y)Lm(u,w). (3.11) Proof. From (2.1) and (1.10), we have 2Lik(1 − (e)−t) et + 1 exp((y + u)t)C0((x + w)t) = ( ∞∑ n=0 LG (k) n (x,y) tn n! )( ∞∑ m=0 Lm(u,w) tm m! ) . Now replacing n by n−m and comparing the coefficients of tn in both sides, we get (3.11). Theorem 3.4. The following summation formula for Laguerre-based poly-Genocchi polynomials LG (k) n (x,y) holds true: LG (k) n (x,y + 1) = n∑ m=0   n m   LG (k) n−m(x,y). (3.12) Proof. Using definition (2.1), we have ∞∑ n=0 LG (k) n (x,y + 1) tn n! − ∞∑ n=0 LG (k) n (x,y) tn n! = 2Lik(1 −e−t) et + 1 exp(yt)C0(xt)(e t − 1) = ( ∞∑ n=0 LG (k) n (x,y) tn n! )( ∞∑ m=0 tm m! ) − ∞∑ n=0 LG (k) n (x,y) tn n! = ∞∑ n=0 n∑ m=0 LG (k) n−m(x,y) tn (n−m)!m! − ∞∑ n=0 LG (k) n (x,y) tn n! . Finally, equating the coefficients of the like powers of tn, we get (3.12). 4. Identities for 2-variable Laguerre-based poly-Genocchi polynomials In this section, we derive general symmetry identities for 2-variable Laguerre-based poly-Genocchi polyno- mials LG (k) n (x,y) by applying the generating function(2.1). Such type of identities have been introduced by several authors (see [11], [12], [13], [15]). Theorem 4.1. Let a,b > 0 and a 6= b, x,y ∈ R, n ≥ 0, then the following identity holds true: n∑ m=0   n m  an−mbmLG(k)n−m(bx,by)LG(k)m (au,aw) = n∑ m=0   n m  ambn−mLG(k)n−m(ax,ay)LG(k)m (bu,bw). (4.1) Int. J. Anal. Appl. 16 (6) (2018) 865 Proof. Let G(t) = ( (2Lik(1 −e−at)(2Lik(1 −e−bt)) (eat + 1)(Abt −B−bt) ) exp(ab(y + u)t)C0(abxt)C0(abwt). (4.2) Since G(t) is symmetric in a and b and G(t) can written as G(t) = ∞∑ n=0 LG (k) n (bx,by) (at)n n! ∞∑ m=0 LG (k) m (au,aw) (bt)m m! G(t) = ∞∑ n=0   n∑ m=0   n m  an−mbmLG(k)n−m(bx,by)LG(k)m (au,aw)   tn n! . (4.3) Similarly, we can show that G(t) = ∞∑ n=0 LG (k) n (ax,ay) (bt)n n! ∞∑ m=0 LG (k) m (bu,bw) (at)m m! G(t) = ∞∑ n=0   n∑ m=0   n m  ambn−mLG(k)n−m(ax,ay)LG(k)m (bu,bw)   tn n! . (4.4) Comparing the coefficients of t n n! in (4.3) and (4.4), we arrive at the desired result. Remark 4.1. On setting b = 1 in Theorem 4.1, we get n∑ m=0   n m  an−mLG(k)n−m(x,y)LG(k)m (au,aw) = n∑ m=0   n m  amLG(k)n−m(ax,ay)LG(k)m (u,w). Theorem 4.2. Let a,b > 0 and a 6= b, x,y ∈ R and n ≥ 0, then the following identity holds true: n∑ m=0   n m  a−1∑ i=0 b−1∑ j=0 LG (k) n−m ( by + b a i + j,bx ) LG (k) m (au,aw)b man−m = n∑ m=0   n m   b−1∑ i=0 a−1∑ j=0 LG (k) n−m ( ay + a b i + j,ax ) LG (k) m (bu,bw)a mbn−m. (4.5) Proof. Let G(t) = ( (2Lik(1 −e−at))(2Lik(1 −e−bt)) (eat + 1)2(ebt + 1)2 ) (eabt + 1)2 exp(ab(y + u)t)C0(abxt)C0(abwt) G(t) = ( 2Lik(1 −e−at) eat + 1 ) exp(abyt)C0(abxt) ( eabt + 1 ebt + 1 )( 2Lik(1 −e−bt) ebt + 1 ) ×exp(abut)C0(abwt) ( eabt + 1 eat + 1 ) Int. J. Anal. Appl. 16 (6) (2018) 866 = ( 2Lik(1 −e−at) (eat + 1 ) exp(abyt)C0(abxt) a−1∑ i=0 (−1)iebti ( 2Lik(1 −e−bt) ebt + 1 ) ×exp(abut)C0(abwt) b−1∑ j=0 (−1)jeatj. = ( 2Lik(1 −e−t) eat + 1 ) C0(abxt) a−1∑ i=0 b−1∑ j=0 (−1)i+je(by+ b a i+j)at ∞∑ m=0 LG (k) m (au,aw) (bt)m m! = ∞∑ n=0 a−1∑ i=0 b−1∑ j=0 LG (k) n ( by + b a i + j,bx ) (at)n n! ∞∑ m=0 LG (k) m (au,aw) (bt)m (m)! G(t) = ∞∑ n=0   n∑ m=0   n m  a−1∑ i=0 b−1∑ j=0 (−1)i+jLG (k) n−m ( by + b a i + j,bx ) LG (k) m (au,aw)b man−m   tn n! . (4.6) On the other hand G(t) = ∞∑ n=0   n∑ m=0   n m   b−1∑ i=0 a−1∑ j=0 (−1)i+jLG (k) n−m ( ay + a b i + j,ax ) LG (k) m (bu,bw)a mbn−m   tn n! . (4.7) On comparing the coefficients of t n n! in (4.6) and (4.7), we arrive at the desired result (4.5). Acknowledgement. All authors would like to thank Integral University, Lucknow, India, for providing the manuscript number IU/R&D/2017-MCN000240 for the present research work. References [1] L. C. 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