# Remarks on Homogeneous Al-Salam and Carlitz Polynomials

Authors: Jian-Ping Fang
Journal: Journal of Mathematics
 (2014)

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## Abstract

Several multilinear generating functions of the homogeneous Al-Salam and Carlitz polynomials are derived fromq-operator. In addition, two interesting relationships of product of this kind of polynomials are obtained.


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## Body

## 1. Introduction

The Al-Salam and Carlitz polynomials(1)ϕn(a)(x)=∑k=0n‍[nk](a;q)kxk
have been studied by many researchers for a long time. The history of these polynomials may go back to Al-Salam and Carlitz in 1965. Since then, these polynomials have been studied by many mathematicians [1–14].Recently, Cao [7] used Carlitz’s q-operators to study the following homogeneous Al-Salam and Carlitz polynomials:
(2)ϕn(a)(x,y)=∑k=0n‍[nk](a;q)kxkyn-k,y≠0
and he gave some linear generating functions of them. In this paper, we will research these polynomials by some construction of q-operator. With this method, some new multilinear generating functions can be easily derived. Firstly, we give the following three results which originated from the results about ϕn(a)(x) which appeared in [1, 2, 5–7, 9, 10].Theorem 1.
Ifm,n∈N, max⁡⁡{|xXz|,|xYz|,|yXz|,|yYz|}<1, then
(3)∑n=0∞‍ϕm+n(a)(x,y)ϕn(b)(X,Y)zn(q;q)n=ym(q;q)m(byXz,axXz;q)∞(yYz,yXz,xXz;q)∞×∑k=0∞‍(x/y)k(a,yYz,yXz;q)k(q,byXz,axXz;q)k(q;q)m-k×Φ23(aqk,yXzqk,bXYbyXzqk,axXzqk;q,xYz).Theorem 2.
Ifm,n∈N, max⁡⁡{|xt|,|yt|,|Xz|,|Yz|,|yYγ|,|yXγ|,|xYγ|}<1, then
(4)∑m,n,k=0∞‍ϕm+k(a)(x,y)ϕn+k(b)(X,Y)tmznγk(q;q)m(q;q)n(q;q)k=(bXz,axYγ;q)∞(Yz,Xz,yt,yYγ,xYγ;q)∞×∑m=0∞‍(yXγ)m(b,Yz;q)m(bXz;q)m×∑k=0∞‍(x/y)k(a,yt,yYγ;q)k(q,axYγ;q)k(q;q)m-k×Φ12(aqk,yYγqkaxYγqk;q,xt).Theorem 3 (cf. [7, Equation (1.9)]).
Ifk,m,n∈N, max⁡⁡{|xYz|,|xXz|,|Xyz|,|Yyz|}<1, then
(5)∑k=0∞‍ϕn+k(a)(x,y)ϕm+k(b)(X,Y)zk(q;q)k=ynYm(q;q)n(q;q)m(ybXz,xaXz;q)∞(yYz,yXz,xXz;q)∞×∑k=0∞‍(X/Y)k(b,yYz;q)k(q,ybXz;q)k(q;q)m-k×∑j=0∞‍(x/y)j(a,yYzqk,yXz;q)j(q,ybXzqk,xaXz;q)j(q;q)n-j×Φ23(aqj,yXzqj,bXYbyXzqk+j,axXzqj;q,xYzqk).Now further using our method, we can deduce more results of multilinear generating functions.Theorem 4.
IfG∈Z+, a1=q-G, |a1x1/y1|<1, then
(6)∑m1,m2,m3=0∞‍ϕm1+m2+m3(a1)(x1,y1)ϕm1+m3(a2)(x2,y2)×ϕm1+m2(a3)(x3,y3)t1m1t2m2t3m3(q;q)m1(q;q)m2(q;q)m3=(a1x1/y1,a3x3t2y1,a2x2t1y1y3;q)∞(x1/y1,t2y1y3,t3y1y2,t1y1y2y3,x3t2y1,x2t1y1y3;q)∞×∑j1=0∞‍(a3;q)j1(x3t1y1y2)j1(q;q)j1(q;q)j1×∑j2=0∞‍(x2/y2)j2(a2;q)j2(q;q)j2(q;q)j1-j2∑j3=0∞‍(x2t3y1)j3(a2qj2;q)j3(q;q)j3×∑j4=0∞‍(a1,x3t2y1,x2t1y1y3;q)j4(-1)j4q-(j4-1)j4/2(q,x1/(y1qj4);q)j4×(x1qj1+j3y1)j4(t3y1y2;q)j2+j4(t2y1y3;q)j1+j4(a3x3t2y1;q)j1+j4×(t1y1y2y3;q)j2+j3+j4(a2x2t1y1y3;q)j2+j3+j4,
provided that max⁡⁡{|tixjyl|,|t1xiyjyl|,|t1y1y2y3|}<1, where 1≤i≠j≠l≤3.Theorem 5.
IfG∈Z+, a1=q-G, then
(7)∑m1,m2,m3=0∞‍ϕm1+m2+m3(a1)(x1,y1)ϕm1+m2(a2)(x2,y2)×ϕm3(a3)(x3,y3)t1m1t2m2t3m3(q;q)m1(q;q)m2(q;q)m3=(a1x1/y1,a3x3t3y1,a2x2t2y1;q)∞(x1/y1,t3y1y3,t3y1x3,t1y1y2,t2y1y2,t2x2y1;q)∞×∑j1,j2=0∞‍(a2;q)j1(t1x2y1)j1(-1)j2q-(j2-1)j2/2(q;q)j1×(a1,t3y3y1,t3x3y1,t2x2y1,t1y2y1;q)j2(q,qx1/(y1qj2),a3t2x3y1;q)j2×(t2y2y1;q)j1+j2(a2t2x2y1;q)j1+j2(qj1x1y1)j2,
provided that max⁡{|y1y2t1|,|y1y2t2|, |y1y3t3|, |x3y1t3|, |x2y1t1|, |x1y1t2|,|a1x1/y1|}<1.Theorem 6.
Ifmi∈N, max⁡⁡{|x1|,|y1ti|,|x1ti|}<1, i=1,2,3, then
(8)∑m1,m2,m3=0∞‍ϕm1+m2+m3(a)(x1,y1)×t1m1t2m2(-t3)m3qm3(m3-1)/2(q;q)m1(q;q)m2(q;q)m3=(y1t3,ax1t2;q)∞(y1t1,y1t2,x1t2;q)∞×Φ23(a,y1t2,t3t1y1t3,ax1t2;q,x1t1).Theorem 7.
Ifmi∈N,n,G∈Z+, a=q-G, max⁡⁡{|yti|,|ax/y|}<1, i=1,2,…,n, then
(9)∑m1,…,mn=0∞‍ϕm1+⋯+mn(a)(x,y)t1m1⋯tnmn(q;q)m1⋯(q;q)mn=(ax/y;q)∞(x/y,yt1,yt2,…,ytn;q)∞×∑j=0∞‍(a,yt1,yt2,…,ytn;q)j(-1)jq-(j-1)j/2(q,x/(yqj);q)j(xy)j.Theorem 8.
Ifmi∈N, ti≠0, max⁡{|x|,|yti|,|xti|}<1, i=1,2,3,4, then
(10)∑m1,m2,m3,m4=0∞‍ϕm1+m2+m3+m4(a)(x,y)×(-t1)m1t2m2t3m3(-t4)m4qm1(m1-1)/2qm4(m4-1)/2(q;q)m1(q;q)m2(q;q)m3(q;q)m4=(yt1,yt4;q)∞(yt2,yt3;q)∞∑j=0∞‍(a,t1/t2,yt3;q)j(xt2)j(q,yt1,yt4;q)jaaaaaaaaaaaaaaaaa×Φ12(aqj,t4t3yt4qj;q,xt3).Theorem 9.
Ifmi∈N, max⁡⁡{|yiyti|,|xiyti|,|a0/x0|}<1, i=0,1,2,3, then
(11)∑m,m1,m2,m3=0∞‍ϕm1+m2+m3+s(a)(x,y)ϕm+r(a)×(x0,y0)ϕm1(a1)(x1,y1)ϕm2(a2)(x2,y2)×ϕm3(a3)(x3,y3)×tmt1m1t2m2t3m3(q;q)m(q;q)m1(q;q)m2(q;q)m3=(ysy0r(q;q)r(axy,a1yx1t1,a2yx2t2,a3yx3t3;q)∞)xxx×(xy)∞(xy,yy1t1,yx1t1,yy2t2,yx2t2,xxxxxjjcxxyy3t3,yx3t3,yy0t;qxy)∞)-1×∑j1,j2=0∞‍(x0/y0)j1(yx0t)j2(a0;q)j1+j2(q;q)j1(q;q)j2(q;q)r-j1×∑j3=0∞‍((a,yy1t1,yx1t1,yy2t2,yx2t2,yy3t3,yx3t3;q)j3×(yy0t;q)j1+j3)×((qyx,a1yx1t1,a2yx2t2,a3yx3t3;q)j3)-1×qj3(j2+s+1),
where t0=t, m0=m, and F∈Z+,a0=q-F.Theorem 10.
IfMi∈Z+, ai=q-Mi, xi≠0, and yi≠0, then
(12)∑m1,m2,m3,m4=0∞‍ϕm2+m3+m4(a1)(x1,y1)×ϕ1+m3+m4(a2)(x2,y2)ϕm1+m2+m4(a3)×(x3,y3)ϕm1+m2+m3(a4)(x4,y4)×(-t1)m1t2m2t3m3(-t4)m4qm1(m1-1)/2qm4(m4-1)/2(q;q)m1(q;q)m2(q;q)m3(q;q)m4=(a4x4y4,a3x3y3,a2x2y2,a1x1y1,t1y2y3y4,t4y1y2y3;q)∞×((x4y4,x3y3,x2y2,x1y1,t2y1y3y4,t3y1y2y4;q)∞)-1×∑k1,k2,k3,k4=0∞‍((a4;q)k4(a3;q)k3(a2;q)k2zzzzzzzzzzzzjz×(a1;q)k1qk1+k2+k3+k4)aaaaaaaaaaaaa×((q,qy4x4;q)k4(q,qy3x3;q)k3aaaaaaaaaaaaaaaa×(q,qy2x2;q)k2(q,qy1x1;q)k1)-1×(t1y2y3y4;q)k2+k3+k4(t4y1y2y3;q)k1+k2+k3(t2y1y3y4;q)k1+k3+k4(t3y1y2y4;q)k1+k2+k4,
provided that max⁡⁡{|aixi/yi|,|tiyjykyl|}<1, where 1≤i≠j≠k≠l≤4.Polynomials (2) evidently reduce to the Rogers-Szegö polynomials (cf. [9])
(13)ϕn(0)(x)=Hn(x)=∑k=0n‍[nk]xk
when a=0 and y=1. And when y=1 they reduce to the common Al-Salam and Carlitz polynomials (1) (cf. [9]).So now we take some special cases for checking.Leta=b=0, x=Y=1 in (3); we have the following.Corollary 11 (cf. [6, Theorem 1.1] or [9, Equation (4.1)]).
Ifm,n∈N, then
(14)∑n=0∞‍Hm+n(y)Hn(X)zn(q;q)n=ym(yXz2;q)∞(z,Xz,yz,yX;q)∞×3Φ1(q-m,yz,yXzyXz2;q,qmy),
provided that max⁡⁡{|z|,|yz|,|Xz|,|yXz|}<1.Remark 12 (from [6, Equation (3.1)]).
We know that Corollary11 is equivalent to Theorem 1.1 given in [6]. And if we take m=0 and y=Y=1, (3) turns to [9, Equation (1.2)].Takinga=b=0,x=Y=1 in (4), we have the following.Corollary 13 (cf. [6, Theorem 1.2] or [9, Equation (1.3)]).
Ifm,n∈N, then
(15)∑m,n,k=0∞‍Hm+k(y)Hn+k(X)tmznγk(q;q)m(q;q)n(q;q)k=(ytγ,yXzγ;q)∞(t,z,Xz,yt,γ,yγ,yXγ;q)∞×Φ23(z,yt,yγytγ,yXzγ;q,Xγ),
provided that max⁡⁡{|t|,|z|,|γ|,|yt|,|yγ|,|Xz|,|Xγ|,|yXγ|}<1.Remark 14.
Using Hall’s transformation [15, Equation (3.2.7)]
(16)Φ23(a,b,cd,e;q,de(abc))=(e/a,de/bc;q)∞(e,de/abc;q)∞Φ23(a,db,dcd,de(bc);q,ea),
we find that Corollary 13 is equivalent to Theorem 1.2 given in [6]. And if we take y=Y=1, then, with simplifying, (4) turns to [9, Equation (3.5)].Takinga=b=0 and x=X=1 in (5), we have the following.Corollary 15 (cf. [6, Theorem 1.3] or [9, Equation (1.4)]).
Ifm,n,k∈N, then
(17)∑k=0∞‍Hn+k(y)Hm+k(Y)zk(q;q)k=(yYz2;q)∞(z,yz,Yz,yYz;q)∞×∑k=0m‍∑j=0n‍[mk][mj]Ym-kyn-j(yYz;q)k+j(yYz2;q)k+j×(Yz;q)k(yz;q)j,
provided that max⁡⁡{|z|,|yz|,|Yz|,|yYz|}<1.The rest of this paper is organized as follows. In Section2, we will give some notations and lemmas. In Section 3, we give the proofs of theorems. Section 4 describes the relationship between Ψn(a,b;x1,y1;z) (cf. [4, Equation (2.2)]) and ϕn(a)(x,y). In addition, an interesting relationship between the polynomial multiplications is given.


## 2. Notations and Some Lemmas

In this paper, we apply the standard notations that follow from [15] and assume that 0<|q|<1; the q-shifted factorial and its compact factorials are defined, respectively, by
(18)(a;q)0=1;(a;q)n=∏k=0n-1‍(1-aqk),(a;q)∞=∏k=0∞‍(1-aqk)
and (a1,…,am;q)n=(a1;q)n,…,(am;q)n, where n=0,1,…,∞. And we use N to denote the set of nonnegative integers and Z+ to denote the set of positive integers.For any complex numberα, we have
(19)(a;q)α=(a;q)∞(aqα;q)∞.Theq-binomial coefficient and the q-binomial theorem are given by
(20)[nk]=(q;q)n(q;q)k(q;q)n-k,∑m=0∞‍(a;q)m(q;q)mym=(ay;q)∞(y;q)∞,
respectively.The basic hypergeometric seriesΦsr is defined by
(21)Φsr(a1,…,arb1,…,bs;q,x)=∑n=0∞‍(a1,a2,…,ar;q)n(q,b1,…,bs;q)n×[(-1)nqn(n-1)/2]1+s-rxn.Theq-derivative operator Dq and q-shifted operator η (cf. [1, 6, 7, 9, 10, 16–23]), acting on the variable x, are defined by
(22)Dq{f(x)}=f(x)-f(xq)x,η{f(x)}=f(xq).Theq-Leibnitz rule for the product of two functions (cf. [17–25]) is given by
(23)Dqn{f(x)g(x)}=∑k=0n‍[nk]qk(k-n)Dqk{f(x)}Dqn-k{g(xqk)}.From definition, we can easily obtainDq=x-1(1-η); by mathematical induction, we have the following proposition.Proposition 16 (cf. [20, Equation (6)]).
Iff(x) is any analytical function, then
(24)Dqn{f(x)}=x-n∑k=0n‍(q-n;q)k(q;q)k(qη)k{f(x)}.In [19, 20], we had established the following q-operator structure:
(25)Φ01(b-;q,cDq)=∑n=0∞‍(b;q)n(cDq)n(q;q)n.
For convenient, we use F(a;xDq)y to denote the operator Φ01(a-;q,xDq) acting on variable y. From this definition we can easily get the following Lemma.Lemma 17.
One has(26)ϕn(a)(x,y)=F(a;xDq)y{yn}.Proof.
For(27)Dq(k){yn}=(q;q)n(q;q)n-kyn-k,
we have
(28)F(a;xDq)y{yn}=∑k=0∞‍(a;q)kxk(q;q)k(q;q)n(q;q)n-kyn-k=∑k=0n‍[nk](a;q)kxkyn-k,
which completes the proof.It is the reason why we employ this operator to study the properties of homogeneous Al-Salam and Carlitz polynomials.From Proposition16, we may easily obtain the following identity.Lemma 18.
IfG∈Z+, a=q-G, |ax/y|<1 and f(x) is any analytical function, then
(29)F(a;xDq)y{f(y)}=(ax/y;q)∞(x/y;q)∞∑k=0∞‍(-1)k(a;q)k(x/y)k(q,x/(yqk);q)kaaaaaaaaaaaaaaa×q-k(k-1)/2f(yqk).Proof.
Applying Proposition16, LHS of Lemma 18 comes to
(30)∑n=0∞‍(a;q)n(x/y)n(q;q)n×∑k=0n‍(q;q)nqk(q;q)k(q;q)n-k(-1)kqk(k-1)/2-nkf(yqk)=∑k=0∞‍(-1)k(a;q)k(x/y)k(q;q)kq-k(k-1)/2f(yqk)×∑n=0∞‍(aqk;q)n(q;q)n(xyqk)n=∑k=0∞‍(-1)k(a;q)k(x/y)k(q;q)kq-k(k-1)/2f(yqk)×(ax/y;q)∞(x/(yqk);q)∞=(ax/y;q)∞(x/y;q)∞∑k=0∞‍(-1)k(a;q)k(x/y)k(q,x/(yqk);q)kdccddddddddddddd×q-k(k-1)/2f(yqk).
We complete the proof.Remark 19.
To calculate the inner sum of the above, under the conditiona=q-G, q-binomial theorem is usable.Lemma 20 (cf. [20, Lemma 1]).
If|sc|<1, |tc|<1, then
(31)F(b;cDq)a{(aω;q)∞(as,at;q)∞}=(aω,bct;q)∞(as,at,ct;q)∞Φ23(b,at,ωsaω,bct;q,sc).Proof.
LHS of Lemma20 equates to
(32)∑n=0∞‍(b;q)ncn(q;q)n∑k=0n‍[nk]qk(k-n)Dqk{(aω;q)∞(as;q)∞}×Dqn-k{1(atqk;q)∞}=∑n=0∞‍(b;q)ncn(q;q)n×∑k=0n‍[nk]qk(k-n)sk(ω/s;q)k(aωqk;q)∞(as;q)∞(tqk)n-k(atqk;q)∞=(aω;q)∞(as,at;q)∞∑k=0∞‍(at,ω/s;q)k(q,aω;q)ksk×∑n=k∞‍(b;q)n(q;q)n-kcntn-k=(aω;q)∞(as,at;q)∞∑k=0∞‍(at,ω/s;q)k(q,aω;q)k(sc)k×∑m=0∞‍(b;q)m+k(q;q)m(tc)m=(aω;q)∞(as,at;q)∞∑k=0∞‍(at,ω/s,b;q)k(q,aω;q)k(sc)k×∑m=0∞‍(bqk;q)m(q;q)m(tc)m=(aω;q)∞(as,at;q)∞∑k=0∞‍(at,ω/s,b;q)k(q,aω;q)k(sc)k(bctqk;q)∞(ct;q)∞=(aω,bct;q)∞(as,at,ct;q)∞Φ23(b,at,ωsaω,bct;q,sc).
We complete the proof.The special cases of Lemma20 as taking t=0, ω=t=0, and ω=0, respectively, will be frequently used.Lemma 21.
For|sc|<1,
(33)F(b;cDq)a{(aω;q)∞(as;q)∞}=(aω;q)∞(as;q)∞Φ12(b,ωsaω;q,sc).Lemma 22.
If|sc|<1, then
(34)F(b;cDq)a{1(as;q)∞}=(bcs;q)∞(as,cs;q)∞.Lemma 23.
For|sc|<1, |tc|<1,
(35)F(b;cDq)a{1(as,at;q)∞}=(bct;q)∞(as,at,ct;q)∞Φ12(b,atbct;q,sc).Throughout this paper, we also often use the following property.Lemma 24.
If|sc|<1, |tc|<1, then
(36)F(b;cDq)a{aM(aω;q)∞(as,at;q)∞}=aM(q;q)M(aω,bct;q)∞(as,at,ct;q)∞×∑k=0∞‍(c/a)k(b,as,at;q)k(q,aω,bct;q)k(q;q)M-k×Φ23(bqk,atqk,ωsaωqk,bctqk;q,sc).Proof.
We find that LHS of Lemma24 equates to
(37)∑n=0∞‍(b;q)ncn(q;q)n∑k=0n‍[nk]qk(k-n)Dqk{aM}×Dqn-k{(aωqk;q)∞(asqk,atqk;q)∞}=∑k=0∞‍(q;q)MaM-k(q;q)k(q;q)M-k×∑n=k∞‍(b;q)ncnqk(k-n)(q;q)n-kDqn-k{(aωqk;q)∞(asqk,atqk;q)∞}=aM(q;q)M∑k=0∞‍(c/a)k(b;q)k(q;q)k(q;q)M-kF(bqk;(cqk)Dq)aaaaaaaaaaaaaa×{(aωqk;q)∞(asqk,atqk;q)∞}=aM(q;q)M(aω,bct;q)∞(as,at,ct;q)∞×∑k=0∞‍(c/a)k(b,as,at;q)k(q,aω,bct;q)k(q;q)M-kaaaaaaa×Φ23(bqk,atqk,ωsaωqk,bctqk;q,sc).
This completes the proof.


## 3. Proof of Theorems

Proof of Theorem1.
LHS of (3) is equal to
(38)∑n=0∞‍F(a;xDq)y{y(m+n)}F(b;XDq)Y{Y(n)}zn(q;q)n=F(a;xDq)yF(b;XDq)Y{ym(Yyz;q)∞}=F(a;xDq)y{ym(ybXz;q)∞(yYz,yXz;q)∞}=ym(q;q)m(byXz,axXz;q)∞(yYz,yXz,xXz;q)∞×∑k=0∞‍(x/y)k(a,yYz,yXz;q)k(q,byXz,axXz;q)k(q;q)m-k×Φ23(aqk,yXzqk,bXYbyXzqk,axXzqk;q,xYz).
We complete the proof.Proof of Theorem2.
LHS of (4) is equal to
(39)F(a;xDq)y{1(yt:q)∞F(b;XDq)Y{1(Yyγ,Yz;q)∞}}=(bXz;q∞)(Yz,Xz;q)∞∑m=0∞‍(b,Yz;q)m(Xγ)m(q;bXz)maaaaaaaaaaaaaaaaa×F(a;xDq)y{ym(yt,yYγ:q)∞};
then using Lemma 24 and simplifying, we can get the desired result.Proof of Theorem3.
LHS of (5) is equal to
(40)F(a;xDq)y{ynF(b;XDq)Y{Ym(Yyz;q)∞}}=Ym(q;q)m∑k=0∞‍(X/Y)k(b;q)k(q;q)k(q;q)m-kF(a;xDq)yaaaaaaaaaaaaaa×{yn(ybXzqk;q)∞(yYzqk,yXz;q)∞}=ynYm(q;q)n(q;q)m(ybXz,xaXz;q)∞(yYz,yXz,xXz;q)∞×∑k=0∞‍(X/Y)k(b,yYz;q)k(q,ybXz;q)k(q;q)m-k×∑j=0∞‍(x/y)j(a,yYzqk,yXz;q)j(q,ybXzqk,xaXz;q)j(q;q)n-jaaaaaaaa×Φ23(aqj,yXzqj,bXYbyXzqk+j,axXzqj;q,xYzqk).
This completes the proof.Proof of Theorem4.
LHS of (6) is equal to
(41)∑m1,m2,m3=0∞‍F(a1;x1Dq)y1{y1m1+m2+m3}F(a2;x2Dq)y2×{y2m1+m3}F(a3;x3Dq)y3{y3m1+m2}×t1m1t2m2t3m3(q;q)m1(q;q)m2(q;q)m3=F(a1;x1Dq)y1F(a2;x2Dq)y2F(a3;x3Dq)y3×{1(t1y1y2y3,t2y1y3,t3y1y2;q)∞}=F(a1;x1Dq)y1F(a2;x2Dq)y2×{{1(t1y1y2y3,t2y1y3;q)∞}1(t3y1y2;q)∞F(a3;x3Dq)y3×{1(t1y1y2y3,t2y1y3;q)∞}}=F(a1;x1Dq)y1F(a2;x2Dq)y2×{(a3x3t2y1;q)∞(t2y1y3,x3t2y1;q)∞∑j1=0∞‍(a3,t2y1y3;q)j1(q,a3x3t2y1;q)j1×(x3t1y1y2)j1(t3y1y2,t1y1y2y3;q)∞∑j1=0∞‍}=F(a1;x1Dq)y1×{(a3x3t2y1;q)∞(t2y1y3,x3t2y1;q)∞∑j1=0∞‍(a3,t2y1y3;q)j1(q,a3x3t2y1;q)j1×F(a2;x2Dq)y2{(x3t1y1y2)j1(t3y1y2,t1y1y2y3;q)∞}∑j1=0∞‍}=∑j1=0∞‍(a3;q)j1(x3t1y2)j1(q;q)j1(q;q)j1×∑j2=0∞‍(x2/y2)j2(a2;q)j2(q;q)j2(q;q)j1-j2×∑j3=0∞‍(x2t3)j3(a2qj2;q)j3(q;q)j3F(a1;x1Dq)y1×{(qj1)1y1j1+j3(a3x3t2y1qj1,a2x2t1y1y3qj2+j3;q)∞)×((t2y1y3qj1,t3y1y2qj2,t1y1y2y3qj2+j3,x3t2y1,x2t1y1y3;qqj1)∞)-1}.
Setting(42)f(y1)=(y1j1+j3(a3x3t2y1qj1,a2x2t1y1y3qj2+j3;q)∞)×((t2y1y3qj1,t3y1y2qj2,t1y1y2y3qj2+j3,x3t2y1,x2t1y1y3;qqj2+j3)∞)-1,
then by applying Lemma 18 and simplifying, we get
(43)F(a1;x1Dq)y1{f(y1)}=(y1j1+j3(a1x1y1,a3x3t2y1,a2x2t1y1y3;q)∞)×(x1y1)∞(x1y1,t2y1y3,t3y1y2,t1y1y2y3,x3t2y1,x2t1y1y3;qx1y1)∞)-1×∑j4=0∞‍(a1,x3t2y1,x2t1y1y3;q)j4(-1)j4q-(j4-1)j4/2(q,x1/(y1qj4);q)j4×(x1qj1+j3y1)j4(t3y1y2;q)j2+j4(t2y1y3;q)j1+j4(a3x3t2y1;q)j1+j4×(t1y1y2y3;q)j2+j3+j4(a2x2t1y1y3;q)j2+j3+j4;
substituting it into (41), we complete the theorem.Proof of Theorem5.
LHS of (7) is equal to
(44)F(a1;x1Dq)y1F(a2;x2Dq)y2F(a3;x3Dq)y3×{1(y1y2t1,y1y2t2,y1y3t3;q)∞}=F(a1;x1Dq)y1F(a2;x2Dq)y2×{(a3x3y1t3;q)∞(y1y3t3,y1x3t3;q)∞1(y1y2t1,y1y2t2;q)∞}=∑j1=0∞‍(a2;q)j1(t1x2)j1(q;q)j1F(a1;x1Dq)y1×{×y1y2t2qj1,y1x2t2;q)∞)-1(y1j1(a3y1x3t3,a2y1x2t2qj1;q)∞)×(×y1y2t2qj1,y1x2t2;q)∞(qj1y1y3t3,y1x3t3,y1y2t1,222222222222y1y2t2qj1,y1x2t2;q)∞)-1}.
Applying Lemma18 and simplifying, we get the theorem. This completes the proof.Proof of Theorems6 and 7.
Using Lemma17, then Theorem 6 can be obtained by Lemma 20 directly.
Using Lemma17 and then applying Lemma 18, we can get Theorem 7.Proof of Theorem8.
LHS of (10) is equal to
(45)F(a;xDq)y{(yt1,yt4;q)∞(yt2,yt3;q)∞}=∑n=0∞‍(a;q)nxn(q;q)nDqn{(yt1,yt4;q)∞(yt2,yt3;q)∞}=∑n=0∞‍(a;q)nxn(q;q)n×∑k=0n‍[nk]qk(k-n)Dqk{(yt1;q)∞(yt2;q)∞}×Dqn-k{(yt4qk;q)∞(yt3qk;q)∞}=(yt1,yt4;q)∞(yt2,yt3;q)∞∑n=0∞‍(a;a)n(xt3)n(q;q)n×∑k=0n‍(q;q)n(t2/t3)k(q;q)k(q;q)n-k(t1/t2,yt3;q)k(t4/t3;q)n-k(ytq;q)k=(yt1,yt4;q)∞(yt2,yt3;q)∞∑j=0∞‍(a,t1/t2,yt3;q)j(xt2)j(q,yt1,yt4;q)j×Φ12(aqj,t4t3yt4qj;q,xt3);
we complete the proof.Proof of Theorem9.
LHS of (11) equates to
(46)F(a;xDq)yF(a0;x0Dq)y0F(a1;x1Dq)y1×F(a2;x2Dq)y2F(a3;x3Dq)y3×{ysy0r(yy0t,yy1t1,yy2t2,yy3t3;q)∞}=F(a;xDq)yF(a0;x0Dq)y0F(a1;x1Dq)y1×F(a2;x2Dq)y2{ysy0r(yy0t,yy1t1,yy2t2;q)∞dddddddddddddddd×(a3yx3t3;q)∞((yy3t3,yx3t3;q)∞)}=F(a;xDq)yF(a0;x0Dq)y0F(a1;x1Dq)y1×{ysy0r(yy0t,yy1t1;q)∞(a2yx2t2;q)∞((yy2t2,yx2t2;q)∞)×(a3yx3t3;q)∞((yy3t3,yx3t3;q)∞)}=F(a;xDq)yF(a0;x0Dq)y0{ysy0r(yy0t;q)∞}×{(a1yx1t1;q)∞((yy1t1,yx1t1;q)∞)(a2yx2t2;q)∞((yy2t2,yx2t2;q)∞)×(a3yx3t3;q)∞((yy3t3,yx3t3;q)∞)}.
If we set
(47)f(y)=F(a0;x0Dq)y0{ysy0r(yy0t;q)∞}×{(a1yx1t1;q)∞((yy1t1,yx1t1;q)∞)(a2yx2t2;q)∞((yy2t2,yx2t2;q)∞)aaa×(a3yx3t3;q)∞((yy3t3,yx3t3;q)∞)},
using Lemma 24, we get f(y); then employing Lemma 18, we can get the desired result.Proof of Theorem10.
LHS of (12) equates to
(48)F(a1;x1Dq)y1F(a2;x2Dq)y2F(a3;x3Dq)y3×F(a4;x4Dq)y4{(t1y2y3y4,t4y1y2y3;q)∞(t2y1y3y4,t3y1y2y4;q)∞};
then using Lemma 18 four times we can get the result.


## 4. Some Other Cases and a Homogeneousq-Mehler’s Formula

In this section we obtain Mehler’s formula involving aΦ23 series and then give an interesting relationship between the polynomial multiplications.Theorem 25 (cf. [1, Equation (1.17)] or [7, Lemma 12]).
One has(49)∑n=0∞‍ϕn(a)(x,y)ϕn(b)(X,Y)tn(q;q)n=(axYt,byXt;q)∞(xYt,yXt,yYt;q)∞Φ23(a,b,yYtaxYt,byXt;q,xXt).Proof.
LHS of Theorem25 is equal to
(50)∑n=0∞‍ϕn(b)(X,Y)tn(q;q)nF(a;xDq)y{yn}=F(a;xDq)y{∑n=0∞‍∑k=0n‍(b;q)kXkYn-k(ty)n(q;q)k(q;q)n-k}=F(a;xDq)y{(btXy;q)∞(tXy;q)∞1(tYy;q)∞}=(axYt,byXt;q)∞(xYt,yXt,yYt;q)∞×Φ23(a,b,yYtaxYt,byXt;q,xXt).
This completes the proof.Remark 26.
Theorem25 can be reduced to the common q-Mehler’s formula (cf. [1, Equation (1.6)]) by setting a=b=0 and y=Y=1. Employing this formula, we can show that the polynomials Ψn(a,b;x1,y1;z) defined by Andrews [4] can be expressed by ϕn(a)(x,y). Using Sear’s nonterminating Φ23 transformation formula [15, page 241, Equation III.10]
(51)Φ23(A,B,CD,E;q,DEABC)=(B,DE/AB,DE/BC;q)∞(D,E,DE/ABC;q)∞×Φ23(DB,EB,DEABCDEAB,DEBC;q,B),
then the right-hand side of (49) reduces to
(52)(a,xyXYt2,bxXt;q)∞(xYt,yXt,yYt,xXt;q)∞×Φ23(byXta,xYt,xXtxyXYt2,bxXt;q,a).
Then setting xX=z, yX=y1, and xY=x1, we have (cf. [4, Theorem 4.1])
(53)∑n=0∞‍ϕn(a)(x,y)ϕn(b)(X,Y)tn(q;q)n=(a,bzt,x1y1t2;q)∞(t,y1t,zt,x1t;q)∞×Φ23(bty1a,zt,x1tbzt,x1y1t2;q,xXt).Comparison of the above identity with Theorem 4.1 that appeared in [4] gives the following.Corollary 27 (cf. [4, Corollary 3.1]).
IfxX=z, yX=y1, and xY=x1, then
(54)Ψn(a,b;x1,y1;z)=ϕn(a)(x,y)ϕn(b)(X,Y).
Combined with Theorem 6, if we replace (x,y,X,Y,a,b,t) by (x1,y1,1,t2/t1,a1,t3/t1,t1), respectively, and then equate the coefficient of t1N, we have the following.Corollary 28.
One has(55)ϕN(a)(x1,y1)ϕN(b)(1,Y)(q;q)N=∑m1+m2+m3=N‍ϕm1+m2+m3(a)(x1,y1)×Ym2bm3(-1)m3q(m3-1)m3/2(q;q)m1(q;q)m2(q;q)m3.Theorem 29 (cf. [1, Equation (1.17)]).
Ifm,n∈N, then
(56)∑m,n=0∞‍ϕm+n(a1)(x1,y1)ϕm(a2)(x2,y2)ϕn(a3)×(x3,y3)tmzn(q;q)m(q;q)n=(a1x1x3z,a2x2y1t,a3x3y1z;q)∞(x1x3z,x2y1t,x3y1z,y1y2t,y1y3z;q)∞×∑j1,j2,j3=0∞‍(x1y3z)j1(x1x2t)j2(x1y2t)j3(q;q)j1(q;q)j2(q;q)j3×(a2x2/y2,x2y1t;q)j3(a3x3z/(x2t);q)j2(a2x2y1t;q)j3×(y1y3z;q)j2+j3(a1,x3y1z;q)j1+j2+j3(a3x3y1z;q)j2+j3(a1x1x3z;q)j1+j2+j3,
provided that max⁡{|x1y3z|,|x3y1z|, |x1y2t|,|x2y1t|, |y1y2t|,|y1y3z|, |a1x1x3z|}<1.Proof.
Applying Lemma17 and then using q-binomial theorem, we get that the LHS of (56) comes to
(57)F(a1,x1Dq)y1F(a2,x2Dq)y2F(a3,x3Dq)y3×{1(y1y2t,y1y3z;q)∞}.
Twice applying Lemma22, the above identity is equal to
(58)F(a1,x1Dq)y1{(a2x2y1t,a3x3y1z;q)∞(y1y2t,x2y1t,y1y3z,x3y1z;q)∞}=∑n=0∞‍(a1;q)nx1n(q;q)nDqn×{(a2x2y1t,a3x3y1z;q)∞(y1y2t,x2y1t,y1y3z,x3y1z;q)∞}.
For (58), by using q-Leibnitz rule three times and using q-binomial theorem, we can achieve the proof after simplification.Takinga1=a2=a3=0, y1=y2=y3=1, we have the following.Corollary 30 (cf. [9, Equation (4.4)]).
One has(59)∑m,n=0∞‍Hm+n(x1)Hm(x2)Hn(x3)tmzn(q;q)m(q;q)n=(x1x3z2;q)∞(x1z,x1x3z,x2t,x3z,t,z;q)∞×∑j1,j2=0∞‍(x2t;q)j2(z,x3z;q)j1+j2(x1t)j1+j2x2j1(q;q)j1(q;q)j2(x1x3z2;q)j1+j2.Remark 31.
Corollary30 is equivalent to [9, Equation (4.4)] after applying q-Heine transformation formula to simplify. Note that the process of proof is only using q-Leibnitz rule, so we can remove the restrictive condition a=q-F,b=q-G in [7, Theorem 3]. Employing the same method, we recover Theorem 7 as follows.Theorem 32.
Ifmi∈N, max⁡{|yti|,|xti|}<1, i=1,2,…,n, then
(60)∑m1,…,mn=0∞‍ϕm1+⋯+mn(a)(x,y)t1m1⋯tnmn(q;q)m1⋯(q;q)mn=(axt1;q)∞(xt1,yt1,yt2,…,ytn;q)∞×∑j1,j2,…,jn-1=0∞‍(a,yt1;q)j1+j2+⋯+jn-1xj1+j2+⋯+jn-1(axt1;q)j1+j2+⋯+jn-1(q;q)j1(q;q)j2×((yt2;q)j2+⋯+jn-1,…,(ytn-2;q)jn-2+jn-1×(ytn-1;q)jn-1t2j1t3j2,…,tnjn-1)×((q;q)j3,…,(q;q)jn-1)-1.Proof .
Using Lemma17, LHS of (60) is equal to
(61)F(a;xDq)y{1(ytn,ytn-1,…,yt1;q)∞}=∑k=0∞‍(a;q)kxk(q;q)kDqk{1(ytn,ytn-1,…,yt1;q)∞}=∑k=0∞‍(a;q)kxk(q;q)k×∑j1=0k‍[kj1]qj1(j1-k)Dqj1{1(ytn,ytn-1,…,yt2;q)∞}×Dqk-j1{1(yt1j1;q)∞}=∑k=0∞‍(a;q)kxk(q;q)k×∑j1=0k‍[kj1]Dqj1{1(ytn,ytn-1,…,yt2;q)∞}×t1k-j1(yt1;q)j1(yt1;q)∞.
Exchanging the order of summation and usingq-binomial, theorem (61) is equal to
(62)(axt1;q)∞(xt1,yt1;q)∞∑j1=0∞‍(a,yt1;q)j1xj1(q,axt1;q)j1×Dqj1{1(ytn,ytn-1,…,yt2;q)∞}=(axt1;q)∞(xt1,yt1;q)∞∑j1=0∞‍(a,yt1;q)j1xj1(q,axt1;q)j1×Dqj1{1(ytn,ytn-1,…,yt3;q)∞×1(yt2;q)∞}.
Applying q-Leibnitz and exchanging the order, (62) is equal to
(63)(axt1;q)∞(xt1,yt1,yt2;q)∞×∑j2=0∞‍(yt2;q)j2(q;q)j2Dqj2{1(ytn,ytn-1,…,yt3;q)∞}×∑j1=0∞‍(a,yt1;q)j1+j2xj1+j2t2j2(q;q)j2(axt1;q)j1+j2.
By induction, similar proof can be performed to get the desired result.Takinga=0, y=1 in (60), we have the following.Corollary 33.
Ifmi∈N, max⁡⁡{|ti|,|xti|}<1, i=1,2,…,n, then
(64)∑m1,…,mn=0∞‍Hm1+⋯+mn(x)t1m1⋯tnmn(q;q)m1⋯(q;q)mn=1(xt1,t1,t2,…,tn;q)∞×∑j1,j2,…,jn-1=0∞‍(t1;q)j1+j2+⋯+jn-1xj1+j2+⋯+jn-1(q;q)j1(q;q)j2×(tnjn-1(t2;q)j2+⋯+jn-1⋯(tn-2;q)jn-2+jn-1×(tn-1;q)jn-1t2j1t3j2⋯tnjn-1)×((q;q)j3⋯(q;q)jn-1)-1.



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*Source: 523013-2014-07-17.xml*