Non-terminating q-Whippleの変換公式の両側類似
以下は, non-terminating $q$-Whippleの変換公式 の両側$q$超幾何級数への一般化である.
$w=a^2q/cde, a^3q^2=bcdefgh$とするとき,
\begin{align}
&\BQ44{a,f,g,h}{aq/b,aq/c,aq/d,aq/e}q\\
&=a\frac{(q,aq/f,aq/g,aq/h,b/a,c/a,d/a,e/a;q)_{\infty}}{(b,c,d,e,q/a,q/f,q/g,q/h;q)_{\infty}}\Q43{b,c,d,e}{aq/f,aq/g,aq/h}q\\
&\qquad+\frac ba\frac{(q,a,f,g,h,bq/c,bq/d,bq/e;q)_{\infty}}{(b,aq/b,aq/c,aq/d,aq/e,bf/a,bg/a,bh/a;q)_{\infty}}\Q43{b,bf/a,bg/a,bh/a}{bq/c,bq/d,bq/e}q\\
&\qquad+\frac{(a,b/a,wq/f,wq/g,wq/h,aq/wc,aq/wd,aq/we;q)_{\infty}}{(wq,q/w,c,d,e,bf/a,bg/a,bh/a;q)_{\infty}}\\
&\qquad\cdot\BQ88{\sqrt wq,-\sqrt wq,wc/a,wd/a,we/a,f,g,h}{\sqrt w,-\sqrt w,aq/c,aq/d,aq/e,wq/f,wq/g,wq/h}b
\end{align}
が成り立つ.
Baileyの4項変換公式
\begin{align}
&\Q{10}9{a,\sqrt aq,-\sqrt aq,b,c,d,e,f,g,h}{\sqrt a,-\sqrt a,aq/b,aq/c,aq/d,aq/e,aq/f,aq/g,aq/h}q\\
&\quad+\frac{(aq,b/a,c,d,e,f,g,h,bq/c,bq/d,bq/e,bq/f,bq/g,bq/h;q)_{\infty}}{(b^2q/a,a/b,aq/c,aq/d,aq/e,aq/f,aq/g,aq/h,bc/a,bd/a,be/a,bf/a,bg/a,bh/a;q)_{\infty}}\\
&\quad\qquad\cdot\, \Q{10}9{b^2/a,\sqrt{\frac{b^2}a}q,-\sqrt{\frac{b^2}a}q,b,bc/a,bd/a,be/a,bf/a,bg/a,bh/a}{\sqrt{\frac{b^2}a},-\sqrt{\frac{b^2}a},bq/a,bq/c,bq/d,bq/e,bq/f,bq/g,bq/h}q\\
&=\frac{(aq,b/a,wq/f,wq/g,w q/h,bf/w,bg/w,bh/w;q)_{\infty}}{(wq,b/w,aq/f,aq/g,aq/h,bf/a,bg/a,bh/a;q)_{\infty}}\Q{10}9{w,\sqrt wq,-\sqrt wq,b,wc/a,wd/a,we/a,f,g,h}{\sqrt w,-\sqrt w,wq/b,aq/c,aq/d,aq/e,wq/f,wq/g,wq/h}q\\
&\qquad +\frac{(aq,b/a,f,g,h,bq/f,bq/g,bq/h,wc/a,wd/a,we/a,abq/wc,abq/wd,abq/we;q)_{\infty}}{(b^2q/w,w/b,aq/c,aq/d,aq/e,aq/f,aq/g,aq/h,bc/a,bd/a,be/a,bf/a,bg/a,bh/a;q)_{\infty}}\\
&\qquad\qquad\cdot\,\Q{10}9{b^2/w,\sqrt{\frac{b^2}w}q,-\sqrt{\frac{b^2}w}q,b,bc/a,bd/a,be/a,bf/w,bg/w,bh/w}{\sqrt{\frac{b^2}w},-\sqrt{\frac{b^2}w},bq/w,abq/wc,abq/wd,abq/we,bq/f,bq/g,bq/h}q\qquad w=a^2q/cde, a^3q^2=bcdefgh
\end{align}
において, $a\mapsto aq^{-2n},f\mapsto fq^{-2n},g\mapsto gq^{-2n},h\mapsto hq^{-2n},w\mapsto q^{-4n}$とすると, 条件は$w=a^2q/cde, a^3q^2=bcdefgh$のままで
\begin{align}
&\Q{10}9{aq^{-2n},\sqrt aq^{1-n},-\sqrt aq^{1-n},b,c,d,e,fq^{-2n},gq^{-2n},hq^{-2n}}{\sqrt a q^{-n},-\sqrt aq^{-n},aq^{1-2n}/b,aq^{1-2n}/c,aq^{1-2n}/d,aq^{1-2n}/e,aq/f,aq/g,aq/h}q\\
&\quad+\frac{(aq^{1-2n},bq^{2n}/a,c,d,e,fq^{-2n},gq^{-2n},hq^{-2n},bq/c,bq/d,bq/e,bq^{2n+1}/f,bq^{2n+1}/g,bq^{2n+1}/h;q)_{\infty}}{(b^2q^{2n+1}/a,aq^{-2n}/b,aq^{1-2n}/c,aq^{1-2n}/d,aq^{1-2n}/e,aq/f,aq/g,aq/h,bcq^{2n}/a,bdq^{2n}/a,beq^{2n}/a,bf/a,bg/a,bh/a;q)_{\infty}}\\
&\quad\qquad\cdot\, \Q{10}9{b^2q^{2n}/a,\sqrt{\frac{b^2}a}q^{n+1},-\sqrt{\frac{b^2}a}q^{n+1},b,bcq^{2n}/a,bdq^{2n}/a,beq^{2n}/a,bf/a,bg/a,bh/a}{\sqrt{\frac{b^2}a}q^n,-\sqrt{\frac{b^2}a}q^n,bq^{2n+1}/a,bq/c,bq/d,bq/e,bq^{2n+1}/f,bq^{2n+1}/g,bq^{2n+1}/h}q\\
&=\frac{(aq^{1-2n},bq^{2n}/a,wq^{1-2n}/f,wq^{1-2n}/g,wq^{1-2n}q/h,bfq^{2n}/w,bgq^{2n}/w,bhq^{2n}/w;q)_{\infty}}{(wq^{1-4n},bq^{4n}/w,aq/f,aq/g,aq/h,bf/a,bg/a,bh/a;q)_{\infty}}\\
&\qquad\cdot\Q{10}9{wq^{-4n},\sqrt wq^{1-2n},-\sqrt wq^{1-2n},b,wcq^{-2n}/a,wdq^{-2n}/a,weq^{-2n}/a,fq^{-2n},gq^{-2n},hq^{-2n}}{\sqrt wq^{-2n},-\sqrt wq^{-2n},wq^{1-4n}/b,aq^{1-2n}/c,aq^{1-2n}/d,aq^{1-2n}/e,wq^{1-2n}/f,wq^{1-2n}/g,wq^{1-2n}/h}q\\
&\qquad +\frac{(aq^{1-2n},bq^{2n}/a,fq^{-2n},gq^{-2n},hq^{-2n},bq^{2n+1}/f,bq^{2n+1}/g,bq^{2n+1}/h,wcq^{-2n}/a,wdq^{-2n}/a,weq^{-2n}/a,abq^{2n+1}/wc,abq^{2n+1}/wd,abq^{2n+1}/we;q)_{\infty}}{(b^2q^{4n+1}/w,wq^{-4n}/b,aq^{1-2n}/c,aq^{1-2n}/d,aq^{1-2n}/e,aq/f,aq/g,aq/h,bcq^{2n}/a,bdq^{2n}/a,beq^{2n}/a,bf/a,bg/a,bh/a;q)_{\infty}}\\
&\qquad\qquad\cdot\,\Q{10}9{b^2q^{4n}/w,\sqrt{\frac{b^2}w}q^{2n+1},-\sqrt{\frac{b^2}w}q^{2n+1},b,bcq^{2n}/a,bdq^{2n}/a,beq^{2n}/a,bfq^{2n}/w,bgq^{2n}/w,bhq^{2n}/w}{\sqrt{\frac{b^2}w}q^{2n},-\sqrt{\frac{b^2}w}q^{2n},bq^{4n+1}/w,abq^{2n+1}/wc,abq^{2n+1}/wd,abq^{2n+1}/we,bq^{2n+1}/f,bq^{2n+1}/g,bq^{2n+1}/h}q
\end{align}
となる. 1つ目の項は
\begin{align}
&\Q{10}9{aq^{-2n},\sqrt aq^{1-n},-\sqrt aq^{1-n},b,c,d,e,fq^{-2n},gq^{-2n},hq^{-2n}}{\sqrt a q^{-n},-\sqrt aq^{-n},aq^{1-2n}/b,aq^{1-2n}/c,aq^{1-2n}/d,aq^{1-2n}/e,aq/f,aq/g,aq/h}q\\
&=\sum_{k=0}^{n-1}\frac{(aq^{-2n},\sqrt aq^{1-n},-\sqrt aq^{1-n},b,c,d,e,fq^{-2n},gq^{-2n},hq^{-2n};q)_k}{(q,\sqrt a q^{-n},-\sqrt aq^{-n},aq^{1-2n}/b,aq^{1-2n}/c,aq^{1-2n}/d,aq^{1-2n}/e,aq/f,aq/g,aq/h;q)_k}q^k\\
&\qquad+\sum_{-n\leq k}\frac{(aq^{-2n},\sqrt aq^{1-n},-\sqrt aq^{1-n},b,c,d,e,fq^{-2n},gq^{-2n},hq^{-2n};q)_{k+2n}}{(q,\sqrt a q^{-n},-\sqrt aq^{-n},aq^{1-2n}/b,aq^{1-2n}/c,aq^{1-2n}/d,aq^{1-2n}/e,aq/f,aq/g,aq/h;q)_{k+2n}}q^{k+2n}\\
&=\sum_{k=0}^{n-1}\frac{(aq^{-2n},\sqrt aq^{1-n},-\sqrt aq^{1-n},b,c,d,e,fq^{-2n},gq^{-2n},hq^{-2n};q)_k}{(q,\sqrt a q^{-n},-\sqrt aq^{-n},aq^{1-2n}/b,aq^{1-2n}/c,aq^{1-2n}/d,aq^{1-2n}/e,aq/f,aq/g,aq/h;q)_k}q^k\\
&\qquad+\frac{1-aq^{2n}}{q^{2n}-a}\frac{(b,c,d,e,q/a,q/f,q/g,q/h;q)_{2n}}{(q,aq/f,aq/g,aq/h,b/a,c/a,d/a,e/a;q)_{2n}}\\
&\qquad\cdot\sum_{-n\leq k}\frac{(a,\sqrt aq^{1+n},-\sqrt aq^{1+n},bq^{2n},cq^{2n},dq^{2n},eq^{2n},f,g,h;q)_{k}}{(q^{2n+1},\sqrt a q^{n},-\sqrt aq^{n},aq/b,aq/c,aq/d,aq/e,aq^{2n+1}/f,aq^{2n+1}/g,aq^{2n+1}/h;q)_{k}}q^k\\
&\to \Q43{b,c,d,e}{aq/f,aq/g,aq/h}q-\frac 1a\frac{(b,c,d,e,q/a,q/f,q/g,q/h;q)_{\infty}}{(q,aq/f,aq/g,aq/h,b/a,c/a,d/a,e/a;;q)_{\infty}}\BQ44{a,f,g,h}{aq/b,aq/c,aq/d,aq/e}q\qquad n\to\infty
\end{align}
となる. 2つ目の項は
\begin{align}
&\frac{(aq^{1-2n},bq^{2n}/a,c,d,e,fq^{-2n},gq^{-2n},hq^{-2n},bq/c,bq/d,bq/e,bq^{2n+1}/f,bq^{2n+1}/g,bq^{2n+1}/h;q)_{\infty}}{(b^2q^{2n+1}/a,aq^{-2n}/b,aq^{1-2n}/c,aq^{1-2n}/d,aq^{1-2n}/e,aq/f,aq/g,aq/h,bcq^{2n}/a,bdq^{2n}/a,beq^{2n}/a,bf/a,bg/a,bh/a;q)_{\infty}}\\
&\quad\qquad\cdot\, \Q{10}9{b^2q^{2n}/a,\sqrt{\frac{b^2}a}q^{n+1},-\sqrt{\frac{b^2}a}q^{n+1},b,bcq^{2n}/a,bdq^{2n}/a,beq^{2n}/a,bf/a,bg/a,bh/a}{\sqrt{\frac{b^2}a}q^n,-\sqrt{\frac{b^2}a}q^n,bq^{2n+1}/a,bq/c,bq/d,bq/e,bq^{2n+1}/f,bq^{2n+1}/g,bq^{2n+1}/h}q\\
&=\frac{(1/a,q/f,q/g,q/h;q)_{2n}}{(bq/a,c/a,d/a,e/a;q)_{2n}}\\
&\qquad\cdot\frac{(aq,bq^{2n}/a,c,d,e,f,g,h,bq/c,bq/d,bq/e,bq^{2n+1}/f,bq^{2n+1}/g,bq^{2n+1}/h;q)_{\infty}}{(b^2q^{2n+1}/a,a/b,aq/c,aq/d,aq/e,aq/f,aq/g,aq/h,bcq^{2n}/a,bdq^{2n}/a,beq^{2n}/a,bf/a,bg/a,bh/a;q)_{\infty}}\\
&\quad\qquad\cdot\, \Q{10}9{b^2q^{2n}/a,\sqrt{\frac{b^2}a}q^{n+1},-\sqrt{\frac{b^2}a}q^{n+1},b,bcq^{2n}/a,bdq^{2n}/a,beq^{2n}/a,bf/a,bg/a,bh/a}{\sqrt{\frac{b^2}a}q^n,-\sqrt{\frac{b^2}a}q^n,bq^{2n+1}/a,bq/c,bq/d,bq/e,bq^{2n+1}/f,bq^{2n+1}/g,bq^{2n+1}/h}q\\
&\to\frac{(1/a,q/f,q/g,q/h,aq,c,d,e,f,g,h,bq/c,bq/d,bq/e;q)_{\infty}}{(bq/a,c/a,d/a,e/a,a/b,aq/c,aq/d,aq/e,aq/f,aq/g,aq/h,bf/a,bg/a,bh/a;q)_{\infty}}\\
&\quad\qquad\cdot\, \Q43{b,bf/a,bg/a,bh/a}{bq/c,bq/d,bq/e}q\qquad n\to\infty\\
\end{align}
となる. 3つ目の項は$|b|<1$として,
\begin{align}
&\frac{(aq^{1-2n},bq^{2n}/a,wq^{1-2n}/f,wq^{1-2n}/g,wq^{1-2n}q/h,bfq^{2n}/w,bgq^{2n}/w,bhq^{2n}/w;q)_{\infty}}{(wq^{1-4n},bq^{4n}/w,aq/f,aq/g,aq/h,bf/a,bg/a,bh/a;q)_{\infty}}\\
&\qquad\cdot\Q{10}9{wq^{-4n},\sqrt wq^{1-2n},-\sqrt wq^{1-2n},b,wcq^{-2n}/a,wdq^{-2n}/a,weq^{-2n}/a,fq^{-2n},gq^{-2n},hq^{-2n}}{\sqrt wq^{-2n},-\sqrt wq^{-2n},wq^{1-4n}/b,aq^{1-2n}/c,aq^{1-2n}/d,aq^{1-2n}/e,wq^{1-2n}/f,wq^{1-2n}/g,wq^{1-2n}/h}q\\
&=\frac{(aq^{1-2n},wq^{1-2n}/f,wq^{1-2n}/g,wq^{1-2n}/h;q)_{2n}}{(wq^{1-4n};q)_{4n}}\frac{(aq,bq^{2n}/a,wq/f,wq/g,wq/h,bfq^{2n}/w,bgq^{2n}/w,bhq^{2n}/w;q)_{\infty}}{(wq,bq^{4n}/w,aq/f,aq/g,aq/h,bf/a,bg/a,bh/a;q)_{\infty}}\\
&\qquad\cdot\frac{(wq^{-4n},\sqrt wq^{1-2n},-\sqrt wq^{1-2n},b,wcq^{-2n}/a,wdq^{-2n}/a,weq^{-2n}/a,fq^{-2n},gq^{-2n},hq^{-2n};q)_{2n}}{(q,\sqrt wq^{-2n},-\sqrt wq^{-2n},wq^{1-4n}/b,aq^{1-2n}/c,aq^{1-2n}/d,aq^{1-2n}/e,wq^{1-2n}/f,wq^{1-2n}/g,wq^{1-2n}/h;q)_{2n}}q^{2n}\\
&\qquad\cdot\sum_{-2n\leq k}\frac{(wq^{-2n},\sqrt wq,-\sqrt wq,bq^{2n},wc/a,wd/a,we/a,f,g,h;q)_k}{(q^{2n+1},\sqrt w,-\sqrt w,wq^{1-2n}/b,aq/c,aq/d,aq/e,wq/f,wq/g,wq/h;q)_k}q^k\\
&=\frac{(aq^{1-2n},wcq^{-2n}/a,wdq^{-2n}/a,weq^{-2n}/a,fq^{-2n},gq^{-2n},hq^{-2n},b;q)_{2n}}{(wq^{-2n},wq^{1-4n}/b,aq^{1-2n}/c,aq^{1-2n}/d,aq^{1-2n}/e,q;q)_{2n}}q^{2n}\\
&\qquad\cdot\frac{(aq,bq^{2n}/a,wq/f,wq/g,wq/h,bfq^{2n}/w,bgq^{2n}/w,bhq^{2n}/w;q)_{\infty}}{(wq,bq^{4n}/w,aq/f,aq/g,aq/h,bf/a,bg/a,bh/a;q)_{\infty}}\\
&\qquad\cdot\sum_{-2n\leq k}\frac{(wq^{-2n},\sqrt wq,-\sqrt wq,bq^{2n},wc/a,wd/a,we/a,f,g,h;q)_k}{(q^{2n+1},\sqrt w,-\sqrt w,wq^{1-2n}/b,aq/c,aq/d,aq/e,wq/f,wq/g,wq/h;q)_k}q^k\\
&=\frac{(1/a,aq/wc,aq/wd,aq/we,q/f,q/g,q/h,b;q)_{2n}}{(q/w,bq^{2n}/w,c/a,d/a,e/a,q;q)_{2n}}\\
&\qquad\cdot\frac{(aq,bq^{2n}/a,wq/f,wq/g,wq/h,bfq^{2n}/w,bgq^{2n}/w,bhq^{2n}/w;q)_{\infty}}{(wq,bq^{4n}/w,aq/f,aq/g,aq/h,bf/a,bg/a,bh/a;q)_{\infty}}\\
&\qquad\cdot\sum_{-2n\leq k}\frac{(wq^{-2n},\sqrt wq,-\sqrt wq,bq^{2n},wc/a,wd/a,we/a,f,g,h;q)_k}{(q^{2n+1},\sqrt w,-\sqrt w,wq^{1-2n}/b,aq/c,aq/d,aq/e,wq/f,wq/g,wq/h;q)_k}q^k\\
&\to \frac{(1/a,aq/wc,aq/wd,aq/we,q/f,q/g,q/h,b;q)_{\infty}}{(q/w,c/a,d/a,e/a,q;q)_{\infty}}\frac{(aq,wq/f,wq/g,wq/h;q)_{\infty}}{(wq,aq/f,aq/g,aq/h,bf/a,bg/a,bh/a;q)_{\infty}}\\
&\qquad\cdot\BQ88{\sqrt wq,-\sqrt wq,wc/a,wd/a,we/a,f,g,h}{\sqrt w,-\sqrt w,aq/c,aq/d,aq/e,wq/f,wq/g,wq/h}q\qquad n\to\infty
\end{align}
4つ目の項は
\begin{align}
&\frac{(aq^{1-2n},bq^{2n}/a,fq^{-2n},gq^{-2n},hq^{-2n},bq^{2n+1}/f,bq^{2n+1}/g,bq^{2n+1}/h,wcq^{-2n}/a,wdq^{-2n}/a,weq^{-2n}/a,abq^{2n+1}/wc,abq^{2n+1}/wd,abq^{2n+1}/we;q)_{\infty}}{(b^2q^{4n+1}/w,wq^{-4n}/b,aq^{1-2n}/c,aq^{1-2n}/d,aq^{1-2n}/e,aq/f,aq/g,aq/h,bcq^{2n}/a,bdq^{2n}/a,beq^{2n}/a,bf/a,bg/a,bh/a;q)_{\infty}}\\
&\qquad\qquad\cdot\,\Q{10}9{b^2q^{4n}/w,\sqrt{\frac{b^2}w}q^{2n+1},-\sqrt{\frac{b^2}w}q^{2n+1},b,bcq^{2n}/a,bdq^{2n}/a,beq^{2n}/a,bfq^{2n}/w,bgq^{2n}/w,bhq^{2n}/w}{\sqrt{\frac{b^2}w}q^{2n},-\sqrt{\frac{b^2}w}q^{2n},bq^{4n+1}/w,abq^{2n+1}/wc,abq^{2n+1}/wd,abq^{2n+1}/we,bq^{2n+1}/f,bq^{2n+1}/g,bq^{2n+1}/h}q\\
&=\frac{(aq^{1-2n},fq^{-2n},gq^{-2n},hq^{-2n},wcq^{-2n}/a,wdq^{-2n}/a,weq^{-2n}/a;q)_{2n}}{(wq^{-4n}/b;q)_{4n}(aq^{1-2n}/c,aq^{1-2n}/d,aq^{1-2n}/e;q)_{2n}}\\
&\qquad\cdot\frac{(a,bq^{2n}/a,f,g,h,bq^{2n+1}/f,bq^{2n+1}/g,bq^{2n+1}/h,wc/a,wd/a,we/a,abq^{2n+1}/wc,abq^{2n+1}/wd,abq^{2n+1}/we;q)_{\infty}}{(b^2q^{4n+1}/w,w/b,aq/c,aq/d,aq/e,aq/f,aq/g,aq/h,bcq^{2n}/a,bdq^{2n}/a,beq^{2n}/a,bf/a,bg/a,bh/a;q)_{\infty}}\\
&\qquad\qquad\cdot\,\Q{10}9{b^2q^{4n}/w,\sqrt{\frac{b^2}w}q^{2n+1},-\sqrt{\frac{b^2}w}q^{2n+1},b,bcq^{2n}/a,bdq^{2n}/a,beq^{2n}/a,bfq^{2n}/w,bgq^{2n}/w,bhq^{2n}/w}{\sqrt{\frac{b^2}w}q^{2n},-\sqrt{\frac{b^2}w}q^{2n},bq^{4n+1}/w,abq^{2n+1}/wc,abq^{2n+1}/wd,abq^{2n+1}/we,bq^{2n+1}/f,bq^{2n+1}/g,bq^{2n+1}/h}q\\
&=\frac{(1/a,q/f,q/g,q/h,aq/wc,aq/wd,aq/we;q)_{2n}}{(bq/w;q)_{4n}(c/a,d/a,e/a;q)_{2n}}b^{2n}\\
&\qquad\cdot\frac{(a,bq^{2n}/a,f,g,h,bq^{2n+1}/f,bq^{2n+1}/g,bq^{2n+1}/h,wc/a,wd/a,we/a,abq^{2n+1}/wc,abq^{2n+1}/wd,abq^{2n+1}/we;q)_{\infty}}{(b^2q^{4n+1}/w,w/b,aq/c,aq/d,aq/e,aq/f,aq/g,aq/h,bcq^{2n}/a,bdq^{2n}/a,beq^{2n}/a,bf/a,bg/a,bh/a;q)_{\infty}}\\
&\qquad\qquad\cdot\,\Q{10}9{b^2q^{4n}/w,\sqrt{\frac{b^2}w}q^{2n+1},-\sqrt{\frac{b^2}w}q^{2n+1},b,bcq^{2n}/a,bdq^{2n}/a,beq^{2n}/a,bfq^{2n}/w,bgq^{2n}/w,bhq^{2n}/w}{\sqrt{\frac{b^2}w}q^{2n},-\sqrt{\frac{b^2}w}q^{2n},bq^{4n+1}/w,abq^{2n+1}/wc,abq^{2n+1}/wd,abq^{2n+1}/we,bq^{2n+1}/f,bq^{2n+1}/g,bq^{2n+1}/h}q\\
&\to 0\qquad n\to\infty
\end{align}
となる. よってこれらを合わせると
\begin{align}
&\Q43{b,c,d,e}{aq/f,aq/g,aq/h}q-\frac 1a\frac{(b,c,d,e,q/a,q/f,q/g,q/h;q)_{\infty}}{(q,aq/f,aq/g,aq/h,b/a,c/a,d/a,e/a;;q)_{\infty}}\BQ44{a,f,g,h}{aq/b,aq/c,aq/d,aq/e}q\\
&\qquad+\frac{(1/a,q/f,q/g,q/h,aq,c,d,e,f,g,h,bq/c,bq/d,bq/e;q)_{\infty}}{(bq/a,c/a,d/a,e/a,a/b,aq/c,aq/d,aq/e,aq/f,aq/g,aq/h,bf/a,bg/a,bh/a;q)_{\infty}}\\
&\quad\qquad\cdot\, \Q43{b,bf/a,bg/a,bh/a}{bq/c,bq/d,bq/e}q\\
&=\frac{(1/a,aq/wc,aq/wd,aq/we,q/f,q/g,q/h,b;q)_{\infty}}{(q/w,c/a,d/a,e/a,q;q)_{\infty}}\frac{(aq,wq/f,wq/g,wq/h;q)_{\infty}}{(wq,aq/f,aq/g,aq/h,bf/a,bg/a,bh/a;q)_{\infty}}\\
&\qquad\cdot\BQ88{\sqrt wq,-\sqrt wq,wc/a,wd/a,we/a,f,g,h}{\sqrt w,-\sqrt w,aq/c,aq/d,aq/e,wq/f,wq/g,wq/h}q
\end{align}
を得る. これより,
\begin{align}
&\BQ44{a,f,g,h}{aq/b,aq/c,aq/d,aq/e}q\\
&=a\frac{(q,aq/f,aq/g,aq/h,b/a,c/a,d/a,e/a;;q)_{\infty}}{(b,c,d,e,q/a,q/f,q/g,q/h;q)_{\infty}}\Q43{b,c,d,e}{aq/f,aq/g,aq/h}q\\
&\qquad+a\frac{(q,aq/f,aq/g,aq/h,b/a,c/a,d/a,e/a;;q)_{\infty}}{(b,c,d,e,q/a,q/f,q/g,q/h;q)_{\infty}}\frac{(1/a,q/f,q/g,q/h,aq,c,d,e,f,g,h,bq/c,bq/d,bq/e;q)_{\infty}}{(bq/a,c/a,d/a,e/a,a/b,aq/c,aq/d,aq/e,aq/f,aq/g,aq/h,bf/a,bg/a,bh/a;q)_{\infty}}\\
&\quad\qquad\cdot\, \Q43{b,bf/a,bg/a,bh/a}{bq/c,bq/d,bq/e}q\\
&\qquad-a\frac{(q,aq/f,aq/g,aq/h,b/a,c/a,d/a,e/a;;q)_{\infty}}{(b,c,d,e,q/a,q/f,q/g,q/h;q)_{\infty}}\frac{(1/a,aq/wc,aq/wd,aq/we,q/f,q/g,q/h,b;q)_{\infty}}{(q/w,c/a,d/a,e/a,q;q)_{\infty}}\\
&\qquad\cdot\frac{(aq,wq/f,wq/g,wq/h;q)_{\infty}}{(wq,aq/f,aq/g,aq/h,bf/a,bg/a,bh/a;q)_{\infty}}\\
&\qquad\cdot\BQ88{\sqrt wq,-\sqrt wq,wc/a,wd/a,we/a,f,g,h}{\sqrt w,-\sqrt w,aq/c,aq/d,aq/e,wq/f,wq/g,wq/h}q\\
&=a\frac{(q,aq/f,aq/g,aq/h,b/a,c/a,d/a,e/a;q)_{\infty}}{(b,c,d,e,q/a,q/f,q/g,q/h;q)_{\infty}}\Q43{b,c,d,e}{aq/f,aq/g,aq/h}q\\
&\qquad+\frac ba\frac{(q,a,f,g,h,bq/c,bq/d,bq/e;q)_{\infty}}{(b,aq/b,aq/c,aq/d,aq/e,bf/a,bg/a,bh/a;q)_{\infty}}\Q43{b,bf/a,bg/a,bh/a}{bq/c,bq/d,bq/e}q\\
&\qquad+\frac{(a,b/a,wq/f,wq/g,wq/h,aq/wc,aq/wd,aq/we;q)_{\infty}}{(wq,q/w,c,d,e,bf/a,bg/a,bh/a;q)_{\infty}}\\
&\qquad\cdot\BQ88{\sqrt wq,-\sqrt wq,wc/a,wd/a,we/a,f,g,h}{\sqrt w,-\sqrt w,aq/c,aq/d,aq/e,wq/f,wq/g,wq/h}q\\
\end{align}
定理1を${}_8\psi_8$について表すと
\begin{align}
&\frac{(wq/f,wq/g,wq/h,aq/wc,aq/wd,aq/we;q)_{\infty}}{(wq,q/w,c,d,e,bf/a,bg/a,bh/a;q)_{\infty}}\\
&\qquad\cdot\BQ88{\sqrt wq,-\sqrt wq,wc/a,wd/a,we/a,f,g,h}{\sqrt w,-\sqrt w,aq/c,aq/d,aq/e,wq/f,wq/g,wq/h}b\\
&=\frac 1{(a,b/a;q)_{\infty}}\BQ44{a,f,g,h}{aq/b,aq/c,aq/d,aq/e}q\\
&\qquad+\frac{(q,aq/f,aq/g,aq/h,c/a,d/a,e/a;q)_{\infty}}{(aq,1/a,b,c,d,e,q/f,q/g,q/h;q)_{\infty}}\Q43{b,c,d,e}{aq/f,aq/g,aq/h}q\\
&\qquad+\frac{(q,f,g,h,bq/c,bq/d,bq/e;q)_{\infty}}{(bq/a,a/b,b,aq/c,aq/d,aq/e,bf/a,bg/a,bh/a;q)_{\infty}}\Q43{b,bf/a,bg/a,bh/a}{bq/c,bq/d,bq/e}q
\end{align}
となる. $c\mapsto ac/w,d\mapsto ad/w,e\mapsto ae/w$とすれば, $a=w^2q/cde, b=w^3q^2/cdefgh$となり,
\begin{align}
&\frac{(wq/f,wq/g,wq/h,q/c,q/d,q/e;q)_{\infty}}{(wq,q/w,wq/cd,wq/ce,wq/de,wq/fg,wq/fh,wq/gh;q)_{\infty}}\\
&\qquad\cdot\BQ88{\sqrt wq,-\sqrt wq,c,d,e,f,g,h}{\sqrt w,-\sqrt w,wq/c,wq/d,wq/e,wq/f,wq/g,wq/h}{\frac{w^3q^2}{cdefgh}}\\
&=\frac 1{(w^2q/cde,wq/fgh;q)_{\infty}}\BQ44{w^2q/cde,f,g,h}{fgh/w,wq/c,wq/d,wq/e}q\\
&\qquad+\frac{(q,w^2q^2/cdef,w^2q^2/cdeg,w^2q^2/cdeh,c/w,d/w,e/w;q)_{\infty}}{(w^2q^2/cde,cde/w^2q,w^3q/cdefgh,wq/cd,wq/ce,wq/de,q/f,q/g,q/h;q)_{\infty}}\Q43{w^3q^2/cdefgh,wq/cd,wq/ce,wq/de}{w^2q^2/cdef,w^2q^2/cdeg,w^2q^2/cdeh}q\\
&\qquad+\frac{(q,f,g,h,w^2q^2/cfgh,w^2q^2/dfgh,w^2q^2/efgh;q)_{\infty}}{(wq^2/fgh,fgh/wq,w^3q^2/cdefgh,wq/c,wq/d,wq/e,wq/fg,wq/fh,wq/gh;q)_{\infty}}\Q43{w^3q^2/cdefgh,wq/fg,wq/fh,wq/gh}{w^2q^2/cfgh,w^2q^2/dfgh,w^2q^2/efgh}q
\end{align}
を得る. よって変数を置き換えると, 以下のようになる.
\begin{align} &\frac{(aq/e,aq/f,aq/g,q/b,q/c,q/d;q)_{\infty}}{(aq,q/a,aq/bc,aq/bd,aq/cd,aq/ef,aq/eg,aq/fg;q)_{\infty}}\\ &\qquad\cdot\BQ88{\sqrt aq,-\sqrt aq,b,c,d,e,f,g}{\sqrt a,-\sqrt a,aq/b,aq/c,aq/d,aq/e,aq/f,aq/g}{\frac{a^3q^2}{bcdefg}}\\ &=\frac 1{(a^2q/bcd,aq/efg;q)_{\infty}}\BQ44{a^2q/bcd,e,f,g}{efg/a,aq/b,aq/c,aq/d}q\\ &\qquad+\frac{(q,a^2q^2/bcde,a^2q^2/bcdf,a^2q^2/bcdg,b/a,c/a,d/a;q)_{\infty}}{(a^2q^2/bcd,bcd/a^2q,a^3q/bcdefg,aq/bc,aq/bd,aq/cd,q/e,q/f,q/g;q)_{\infty}}\\ &\qquad\cdot\Q43{a^3q^2/bcdefg,aq/bc,aq/bd,aq/cd}{a^2q^2/bcde,a^2q^2/bcdf,a^2q^2/bcdg}q\\ &\qquad+\frac{(q,e,f,g,a^2q^2/befg,a^2q^2/cefg,a^2q^2/defg;q)_{\infty}}{(aq^2/efg,efg/aq,a^3q^2/bcdefg,aq/b,aq/c,aq/d,aq/ef,aq/eg,aq/fg;q)_{\infty}}\\ &\qquad\cdot\Q43{a^3q^2/bcdefg,aq/ef,aq/eg,aq/fg}{a^2q^2/befg,a^2q^2/cefg,a^2q^2/defg}q \end{align}
古典極限
以下ガンマ関数の積を$\Gamma(a_1,\dots,a_r):=\Gamma(a_1)\cdots \Gamma(a_r)$のように略記する. 定理1の$q\to 1$の極限を考えると以下を得る.
$w=1+2a-c-d-e, 2+3a=b+c+d+e+f+g+h$とするとき,
\begin{align}
&\H44{a,f,g,h}{1+a-b,1+a-c,1+a-d,1+a-e}1\\
&=\frac{\Gamma(b,c,d,e,1-a,1-f,1-g,1-h)}{\Gamma(1+a-f,1+a-g,1+a-h,b-a,c-a,d-a,e-a)}\F43{b,c,d,e}{1+a-f,1+a-g,1+a-h}1\\
&\qquad+\frac{\Gamma(b,1+a-b,1+a-c,1+a-d,1+a-e,b+f-a,b+g-a,b+h-a)}{\Gamma(a,f,g,h,1+b-c,1+b-d,1+b-e)}\\
&\qquad\cdot\F43{b,b+f-a,b+g-a,b+h-a}{1+b-c,1+b-d,1+b-e}1\\
&\qquad+\frac{\Gamma(1+w,1-w,c,d,e,b+f-a,b+g-a,b+h-a)}{\Gamma(a,b-a,1+w-f,1+w-g,1+w-h,1+a-w-c,1+a-w-d,1+a-w-e)}\\
&\qquad\cdot\H77{1+\frac w2,w+c-a,w+d-a,w+e-a,f,g,h}{\frac w2,1+a-c,1+a-d,1+a-e,1+w-f,1+w-g,1+w-h}1
\end{align}
が成り立つ.
定理2の$q\to 1$の極限を考えると以下を得る.
\begin{align} &\frac{\Gamma(1+a,1-a,1+a-b-c,1+a-b-d,1+a-c-d,1+a-e-f,1+a-e-g,1+a-f-g)}{\Gamma(1+a-e,1+a-f,1+a-g,1-b,1-c,1-d)}\\ &\qquad\cdot\H77{1+\frac a2,b,c,d,e,f,g}{\frac a2,1+a-b,1+a-c,1+a-d,1+a-e,1+a-f,1+a-g}{1}\\ &=\Gamma(1+2a-b-c-d,1+a-e-f-g)\H44{1+2a-b-c-d,e,f,g}{e+f+g-a,1+a-b,1+a-c,1+a-d}1\\ &\qquad+\frac{\Gamma(2+2a-b-c-d,b+c+d-2a-1,1+3a-b-c-d-e-f-g,1+a-b-c,1+a-b-d,1+a-c-d,1-e,1-f,1-g)}{\Gamma(2+2a-b-c-d-e,2+2a-b-c-d-f,2+2a-b-c-d-g,b-a,c-a,d-a)}\\ &\qquad\cdot\F43{2+3a-b-c-d-e-f-g,1+a-b-c,1+a-b-d,1+a-c-d}{2+2a-b-c-d-e,2+2a-b-c-d-f,2+2a-b-c-d-g}1\\ &\qquad+\frac{\Gamma(2+a-e-f-g,e+f+g-a-1,2+3a-b-c-d-e-f-g,1+a-b,1+a-c,1+a-d,1+a-e-f,1+a-e-g,1+a-f-g)}{\Gamma(e,f,g,2+2a-b-e-f-g,2+2a-c-e-f-g,2+2a-d-e-f-g)}\\ &\qquad\cdot\F43{2+3a-b-c-d-e-f-g,1+a-e-f,1+a-e-g,1+a-f-g}{2+2a-b-e-f-g,2+2a-c-e-f-g,2+2a-d-e-f-g}1 \end{align}
定理4において$g\to\infty$を考えると, Baileyの${}_6H_6$変換公式 が得られる.
