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Results & Lemmas (7)
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Lemma 1.
Lemma 1. Each of the following operational formulas holds true for the homogeneous q-difference operator defined by (2.1): rΦs …
Lemma 1. Each of the following operational formulas holds true for the homogeneous q-difference operator defined by (2.1): rΦs a1,a2,··· ,ar; b1,b2,··· ,bs; q;−zΘxy n (−1)nq−(n 2)Pn(y,x) o = Ψ(a,b)
Theorem 1
Theorem 1 (Extended generating function). For k ∈N and max |yt|,|zt| < 1, it is asserted that ∞ ∑ n=0 Ψ(a,b) n+k (x,y,z|q)(−1)n+kq(n+k 2 )…
Theorem 1 (Extended generating function). For k ∈N and max{|yt|,|zt|} < 1, it is asserted that ∞ ∑ n=0 Ψ(a,b) n+k (x,y,z|q)(−1)n+kq(n+k 2 ) tn (q;q)n = (xt;q)∞ tk(yt;q)∞ k ∑ j=0 (q−k,yt;q)j qj
Theorem 2
Theorem 2 (The Rogers formula for Ψ(a,b) n (x,y,z|q)). For max t ω,|yω|
Theorem 2 (The Rogers formula for Ψ(a,b) n (x,y,z|q)). For max t ω ,|yω|
Lemma 2.
Lemma 2. (see [21, Eq. (3.20)] and [4, Eq. (5.4)]) Each of the following generating relations holds true: ∞ ∑ n=0 φ(α) n (x|q)(λ;q)n tn…
Lemma 2. (see [21, Eq. (3.20)] and [4, Eq. (5.4)]) Each of the following generating relations holds true: ∞ ∑ n=0 φ(α) n (x|q)(λ;q)n tn (q;q)n = (λt;q)∞ (t;q)∞ 2Φ1 λ,α;
Theorem 3.
Theorem 3. Suppose that max |αq|,|vxtq|,|xztq| < 1. Then ∞ ∑ n=0 ψ(α) n (x|q)Ψ(a,b) n (u,v,z|q)(−1)nq(n+1 2 )tn (q;q)n = (q/x,uxtq;q)∞…
Theorem 3. Suppose that max{|αq|,|vxtq|,|xztq|} < 1. Then ∞ ∑ n=0 ψ(α) n (x|q)Ψ(a,b) n (u,v,z|q)(−1)nq(n+1 2 )tn (q;q)n = (q/x,uxtq;q)∞ (αq,vxtq;q)∞ ∞ ∑
Corollary 1.
Corollary 1. Let max |αq|,|axytq| < 1. Then ∞ ∑ n=0 ψ(α) n (x|q)ψ(a) n (y|q)(−1)nq(n+1 2 )tn (q;q)n = (q/x,xytq,xtq;q)∞ (αq,axytq;q)∞ 3Φ2 …
Corollary 1. Let max{|αq|,|axytq|} < 1. Then ∞ ∑ n=0 ψ(α) n (x|q)ψ(a) n (y|q)(−1)nq(n+1 2 )tn (q;q)n = (q/x,xytq,xtq;q)∞ (αq,axytq;q)∞ 3Φ2
Theorem 4.
Theorem 4. Let A(n) and B(n) satisfy the following relationship: ∞ ∑ n=0 A(n)Pn(v,u) = ∞ ∑ n=0 B(n)(xutq1−n;q)∞ (xvtq1−n;q)∞. (5.1) Then ∞ ∑
Theorem 4. Let A(n) and B(n) satisfy the following relationship: ∞ ∑ n=0 A(n)Pn(v,u) = ∞ ∑ n=0 B(n)(xutq1−n;q)∞ (xvtq1−n;q)∞ . (5.1) Then ∞ ∑
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