Gasper-RahmanによるSears-Carlitzの公式の拡張

BaileyのNearly-Poised${}_5{\varphi }_4$の変換公式
$$\begin{array}{rl}& {}_5{\varphi }_4[\begin{array}{c}a,b,c,d,q^{-n}\\ aq/b,aq/c,aq/d,a^2{q}^{-n}/w^2\end{array};q]\\ & =\frac{(wq/a,w^{2}q/a;q{)}_n}{(wq,w^{2}q/a^2;q{)}_n}{}_{12}{\varphi }_{11}[\begin{array}{c}w,\sqrt{w}q,-\sqrt{w}q,wb/a,wc/a,wd/a,\sqrt{a},-\sqrt{a},\sqrt{aq},-\sqrt{aq},w^2{q}^{n+1}/a,q^{-n}\\ \sqrt{w},-\sqrt{w},aq/b,aq/c,aq/d,wq/\sqrt{a},-wq/\sqrt{a},w\sqrt{q/a},-w\sqrt{q/a},a{q}^{-n}/w,w{q}^{n+1}\end{array};q](w=a^{2}q/bcd)\end{array}$$
において, $d\mapsto d{q}^n$とすると, $w=a^2{q}^{1-n}/bcd$であり,
$$\begin{array}{rl}& {}_5{\varphi }_4[\begin{array}{c}a,b,c,d{q}^n,q^{-n}\\ aq/b,aq/c,a{q}^{1-n}/d,b^2{c}^2{d}^2{q}^{n-2}/a^2\end{array};q]\\ & =\frac{(a{q}^{2-n}/bcd,a^3{q}^{3-2n}/b^2{c}^2{d}^2;q{)}_n}{(a^2{q}^{2-n}/bcd,a^2{q}^{3-2n}/b^2{c}^2{d}^2;q{)}_n}{}_{12}{\varphi }_{11}[\begin{array}{c}w,\sqrt{w}q,-\sqrt{w}q,a{q}^{1-n}/cd,a{q}^{1-n}/bd,aq/bc,\sqrt{a},-\sqrt{a},\sqrt{aq},-\sqrt{aq},a^3{q}^{3-n}/b^2{c}^2{d}^2,q^{-n}\\ \sqrt{w},-\sqrt{w},aq/b,aq/c,a{q}^{1-n}/d,wq/\sqrt{a},-wq/\sqrt{a},w\sqrt{q/a},-w\sqrt{q/a},bcd/aq,a^2{q}^2/bcd\end{array};q]\end{array}$$
よって, $n\to \mathrm{\infty }$とすると,
$$\begin{array}{rl}& {}_3{\varphi }_2[\begin{array}{c}a,b,c\\ aq/b,aq/c\end{array};\frac{d}{a}]\\ & =\underset{n\to \mathrm{\infty }}{lim}\frac{(a{q}^{2-n}/bcd,a^3{q}^{3-2n}/b^2{c}^2{d}^2;q{)}_n}{(a^2{q}^{2-n}/bcd,a^2{q}^{3-2n}/b^2{c}^2{d}^2;q{)}_n}{}_{12}{\varphi }_{11}[\begin{array}{c}w,\sqrt{w}q,-\sqrt{w}q,a{q}^{1-n}/cd,a{q}^{1-n}/bd,aq/bc,\sqrt{a},-\sqrt{a},\sqrt{aq},-\sqrt{aq},a^3{q}^{3-n}/b^2{c}^2{d}^2,q^{-n}\\ \sqrt{w},-\sqrt{w},aq/b,aq/c,aq/d,wq/\sqrt{a},-wq/\sqrt{a},w\sqrt{q/a},-w\sqrt{q/a},bcd/aq,a^2{q}^2/bcd\end{array};q]\\ & =\frac{(bcd/aq;q{)}_{\mathrm{\infty }}}{(bcd/a^{2}q;q{)}_{\mathrm{\infty }}}\underset{n\to \mathrm{\infty }}{lim}{}_{12}{\varphi }_{11}[\begin{array}{c}w,\sqrt{w}q,-\sqrt{w}q,a{q}^{1-n}/cd,a{q}^{1-n}/bd,aq/bc,\sqrt{a},-\sqrt{a},\sqrt{aq},-\sqrt{aq},a^3{q}^{3-n}/b^2{c}^2{d}^2,q^{-n}\\ \sqrt{w},-\sqrt{w},aq/b,aq/c,a{q}^{1-n}/d,wq/\sqrt{a},-wq/\sqrt{a},w\sqrt{q/a},-w\sqrt{q/a},bcd/aq,a^2{q}^2/bcd\end{array};q]\end{array}$$
ここで, $d=a^{2}qx/bc$とすると, $w=q^{-n}/x$であり,
$$\begin{array}{rl}& {}_3{\varphi }_2[\begin{array}{c}a,b,c\\ aq/b,aq/c\end{array};\frac{aqx}{bc}]\\ & =\frac{(ax;q{)}_{\mathrm{\infty }}}{(x;q{)}_{\mathrm{\infty }}}\underset{n\to \mathrm{\infty }}{lim}{}_{12}{\varphi }_{11}[\begin{array}{c}w,\sqrt{w}q,-\sqrt{w}q,b{q}^{-n}/ax,c{q}^{-n}/ax,aq/bc,\sqrt{a},-\sqrt{a},\sqrt{aq},-\sqrt{aq},q^{1-n}/a{x}^2,q^{-n}\\ \sqrt{w},-\sqrt{w},aq/b,aq/c,bc{q}^{-n}/ax,wq/\sqrt{a},-wq/\sqrt{a},w\sqrt{q/a},-w\sqrt{q/a},ax,q/x\end{array};q]\end{array}$$
$n$を奇数として, $n=2m+1$とすると,
$$\begin{array}{rl}& \underset{n\to \mathrm{\infty }}{lim}{}_{12}{\varphi }_{11}[\begin{array}{c}w,\sqrt{w}q,-\sqrt{w}q,b{q}^{-n}/ax,c{q}^{-n}/ax,aq/bc,\sqrt{a},-\sqrt{a},\sqrt{aq},-\sqrt{aq},q^{1-n}/a{x}^2,q^{-n}\\ \sqrt{w},-\sqrt{w},aq/b,aq/c,bc{q}^{-n}/ax,wq/\sqrt{a},-wq/\sqrt{a},w\sqrt{q/a},-w\sqrt{q/a},ax,q/x\end{array};q]\\ & =\underset{n\to \mathrm{\infty }}{lim}\sum _{k=0}^m\frac{(w,\sqrt{w}q,-\sqrt{w}q,b{q}^{-n}/ax,c{q}^{-n}/ax,aq/bc,\sqrt{a},-\sqrt{a},\sqrt{aq},-\sqrt{aq},q^{1-n}/a{x}^2,q^{-n};q{)}_k}{(\sqrt{w},-\sqrt{w},aq/b,aq/c,bc{q}^{-n}/ax,wq/\sqrt{a},-wq/\sqrt{a},w\sqrt{q/a},-w\sqrt{q/a},ax,q/x,q;q{)}_k}{q}^k\\ & +\underset{n\to \mathrm{\infty }}{lim}\sum _{k=0}^m\frac{(w,\sqrt{w}q,-\sqrt{w}q,b{q}^{-n}/ax,c{q}^{-n}/ax,aq/bc,\sqrt{a},-\sqrt{a},\sqrt{aq},-\sqrt{aq},q^{1-n}/a{x}^2,q^{-n};q{)}_{n-k}}{(\sqrt{w},-\sqrt{w},aq/b,aq/c,bc{q}^{-n}/ax,wq/\sqrt{a},-wq/\sqrt{a},w\sqrt{q/a},-w\sqrt{q/a},ax,q/x,q;q{)}_{n-k}}{q}^{n-k}\\ & =\sum _{k=0}^{\mathrm{\infty }}\frac{(aq/bc,\sqrt{a},-\sqrt{a},\sqrt{aq},-\sqrt{aq};q{)}_k}{(aq/b,aq/c,ax,q/x,q;q{)}_k}{q}^k\\ & +\underset{n\to \mathrm{\infty }}{lim}\frac{(w,\sqrt{w}q,-\sqrt{w}q,b{q}^{-n}/ax,c{q}^{-n}/ax,aq/bc,\sqrt{a},-\sqrt{a},\sqrt{aq},-\sqrt{aq},q^{1-n}/a{x}^2,q^{-n};q{)}_n}{(\sqrt{w},-\sqrt{w},aq/b,aq/c,bc{q}^{-n}/ax,wq/\sqrt{a},-wq/\sqrt{a},w\sqrt{q/a},-w\sqrt{q/a},ax,q/x,q;q{)}_n}{q}^n\\ & \cdot \sum _{k=0}^m\frac{(q^{1-n}/\sqrt{w},-q^{1-n}/\sqrt{w},b{q}^{-n}/a,c{q}^{-n}/a,axq/bc,x\sqrt{a},-x\sqrt{a},x\sqrt{aq},-x\sqrt{aq},q^{1-n}/ax,x{q}^{-n},q^{-n};q{)}_k}{(xq,q^{-n}/\sqrt{w},-q^{-n}/\sqrt{w},axq/b,axq/c,bc{q}^{-n}/a,q^{1-n}/\sqrt{a},-q^{1-n}/\sqrt{a},q^{1-n}/\sqrt{aq},-q^{1-n}/\sqrt{aq},a{x}^2,q;q{)}_k}{q}^k\\ & =\sum _{k=0}^{\mathrm{\infty }}\frac{(aq/bc,\sqrt{a},-\sqrt{a},\sqrt{aq},-\sqrt{aq};q{)}_k}{(aq/b,aq/c,ax,q/x,q;q{)}_k}{q}^k\\ & -x\underset{n\to \mathrm{\infty }}{lim}\frac{(xq,axq/b,axq/c,aq/bc,\sqrt{a},-\sqrt{a},\sqrt{aq},-\sqrt{aq},a{x}^2;q{)}_n}{(aq/b,aq/c,axq/bc,x\sqrt{a},-x\sqrt{a},x\sqrt{aq},-x\sqrt{aq},ax,q/x,q;q{)}_n}\\ & \cdot \sum _{k=0}^{\mathrm{\infty }}\frac{(axq/bc,x\sqrt{a},-x\sqrt{a},x\sqrt{aq},-x\sqrt{aq};q{)}_k}{(xq,axq/b,axq/c,a{x}^2,q;q{)}_k}{q}^k\\ & =\sum _{k=0}^{\mathrm{\infty }}\frac{(aq/bc,\sqrt{a},-\sqrt{a},\sqrt{aq},-\sqrt{aq};q{)}_k}{(aq/b,aq/c,ax,q/x,q;q{)}_k}{q}^k\\ & -x\frac{(xq,axq/b,axq/c,aq/bc,a;q{)}_{\mathrm{\infty }}}{(aq/b,aq/c,axq/bc,ax,q/x,q;q{)}_{\mathrm{\infty }}}\sum _{k=0}^{\mathrm{\infty }}\frac{(axq/bc,x\sqrt{a},-x\sqrt{a},x\sqrt{aq},-x\sqrt{aq};q{)}_k}{(xq,axq/b,axq/c,a{x}^2,q;q{)}_k}{q}^k\\ & =\sum _{k=0}^{\mathrm{\infty }}\frac{(aq/bc,\sqrt{a},-\sqrt{a},\sqrt{aq},-\sqrt{aq};q{)}_k}{(aq/b,aq/c,ax,q/x,q;q{)}_k}{q}^k\\ & +\frac{(x,axq/b,axq/c,aq/bc,a;q{)}_{\mathrm{\infty }}}{(aq/b,aq/c,axq/bc,ax,1/x,q;q{)}_{\mathrm{\infty }}}\sum _{k=0}^{\mathrm{\infty }}\frac{(axq/bc,x\sqrt{a},-x\sqrt{a},x\sqrt{aq},-x\sqrt{aq};q{)}_k}{(xq,axq/b,axq/c,a{x}^2,q;q{)}_k}{q}^k\end{array}$$
となるから,
$$\begin{array}{rl}& {}_3{\varphi }_2[\begin{array}{c}a,b,c\\ aq/b,aq/c\end{array};\frac{axq}{bc}]\\ & =\frac{(ax;q{)}_{\mathrm{\infty }}}{(x;q{)}_{\mathrm{\infty }}}{}_5{\varphi }_4[\begin{array}{c}aq/bc,\sqrt{a},-\sqrt{a},\sqrt{aq},-\sqrt{aq}\\ aq/b,aq/c,ax,q/x\end{array};q]\\ & +\frac{(a,aq/bc,axq/b,axq/c;q{)}_{\mathrm{\infty }}}{(aq/b,aq/c,axq/bc,1/x;q{)}_{\mathrm{\infty }}}{}_5{\varphi }_4[\begin{array}{c}axq/bc,x\sqrt{a},-x\sqrt{a},x\sqrt{aq},-x\sqrt{aq}\\ axq/b,axq/c,xq,a{x}^2\end{array};q]\end{array}$$
を得る.