BaileyのNearly-poised 5φ4, 7φ6の変換公式

Jacksonの${}_8\phi_7$和公式 より, $w=a^2q/bcd$としたとき,
\begin{align} \Q87{w,\sqrt wq,-\sqrt wq,wb/a,wc/a,wd/a,aq^n,q^{-n}}{\sqrt w,-\sqrt w,aq/b,aq/c,aq/d,wq^{1-n}/a,wq^{n+1}}{q}&=\frac{(b,c,d,wq;q)_n}{(aq/b,aq/c,aq/d,a/w;q)_n} \end{align}
より,
\begin{align} \frac{(a,b,c,d;q)_n}{(aq/b,aq/c,aq/d,q;q)_n}&=\frac{(a,a/w;q)_n}{(wq,q;q)_n}\sum_{k=0}^n\frac{(w,\sqrt wq,-\sqrt wq,wb/a,wc/a,wd/a,aq^n,q^{-n};q)_k}{(\sqrt w,-\sqrt w,aq/b,aq/c,aq/d,wq^{1-n}/a,wq^{n+1},q;q)_k}q^k\\ &=\sum_{k=0}^n\frac{(w,\sqrt wq,-\sqrt wq,wb/a,wc/a,wd/a;q)_k(a;q)_{n+k}(a/w;q)_{n-k}}{(\sqrt w,-\sqrt w,aq/b,aq/c,aq/d,q;q)_k(wq;q)_{n+k}(q;q)_{n-k}}\left(\frac aw\right)^k \end{align}
であるから,

\begin{align} &\Q54{a,b,c,d,q^{-n}}{aq/b,aq/c,aq/d,a^2q^{-n}/w^2}q\\ &=\sum_{0\leq k}\frac{(q^{-n};q)_k}{(a^2q^{-n}/w^2;q)_k}q^k\sum_{j=0}^k\frac{(w,\sqrt wq,-\sqrt wq,wb/a,wc/a,wd/a;q)_j(a;q)_{k+j}(a/w;q)_{k-j}}{(\sqrt w,-\sqrt w,aq/b,aq/c,aq/d,q;q)_j(wq;q)_{k+j}(q;q)_{k-j}}\left(\frac aw\right)^j\\ &=\sum_{0\leq j}\frac{(w,\sqrt wq,-\sqrt wq,wb/a,wc/a,wd/a;q)_j}{(\sqrt w,-\sqrt w,aq/b,aq/c,aq/d,q;q)_j}\left(\frac aw\right)^j\sum_{0\leq k}\frac{(q^{-n};q)_k(wq;q)_{k+j}(a/w;q)_{k-j}}{(a^2q^{-n}/w^2;q)_k(a;q)_{k+j}(q;q)_{k-j}}q^k\\ &=\sum_{0\leq j}\frac{(w,\sqrt wq,-\sqrt wq,wb/a,wc/a,wd/a,q^{-n};q)_j(a;q)_{2j}}{(\sqrt w,-\sqrt w,aq/b,aq/c,aq/d,a^2q^{-n}/w^2,q;q)_j(wq;q)_{2j}}\left(\frac {aq}w\right)^j\sum_{0\leq k}\frac{(q^{j-n},aq^{2j},a/w;q)_k}{(wq^{2j+1},a^2q^{j-n}/w^2,q;q)_k}q^k \end{align}
ここで, $q$-Saalschützの和公式より,
\begin{align} \sum_{0\leq k}\frac{(q^{j-n},aq^{2j},a/w;q)_k}{(wq^{2j+1},a^2q^{j-n}/w^2,q;q)_k}q^k&=\frac{(wq/a,w^2q^{2j+1}/a;q)_{n-j}}{(wq^{2j+1},w^2q/a^2;q)_{n-j}}\\ &=\frac{(wq/a,w^2q/a;q)_n}{(wq,w^2q/a^2;q)_n}\frac{(a^2q^{-n}/w^2,w^2q^{n+1}/a;q)_j(wq;q)_{2j}}{(aq^{-n}/w,wq^{n+1};q)_j(w^2q/a^2;q)_{2j}}\left(\frac wa\right)^j \end{align}
より,
\begin{align} &\Q54{a,b,c,d,q^{-n}}{aq/b,aq/c,aq/d,a^2q^{-n}/w^2}q\\ &=\sum_{0\leq j}\frac{(w,\sqrt wq,-\sqrt wq,wb/a,wc/a,wd/a,q^{-n};q)_j(a;q)_{2j}}{(\sqrt w,-\sqrt w,aq/b,aq/c,aq/d,a^2q^{-n}/w^2,q;q)_j(wq;q)_{2j}}\left(\frac {aq}w\right)^j\\ &\qquad\cdot\frac{(wq/a,w^2q/a;q)_n}{(wq,w^2q/a^2;q)_n}\frac{(a^2q^{-n}/w^2,w^2q^{n+1}/a;q)_j(wq;q)_{2j}}{(aq^{-n}/w,wq^{n+1};q)_j(w^2q/a^2;q)_{2j}}\left(\frac wa\right)^j\\ &=\frac{(wq/a,w^2q/a;q)_n}{(wq,w^2q/a^2;q)_n}\sum_{0\leq j}\frac{(w,\sqrt wq,-\sqrt wq,wb/a,wc/a,wd/a,w^2q^{n+1}/a,q^{-n};q)_j(a;q)_{2j}}{(\sqrt w,-\sqrt w,aq/b,aq/c,aq/d,aq^{-n}/w,wq^{n+1},q;q)_j(w^2q/a^2;q)_{2j}}q^j \end{align}
ここで, $(a;q)_{2j}=(\sqrt a,-\sqrt a,\sqrt{aq},-\sqrt{aq};q)_j$などを用いて, 定理を得る.

定理1において, $c=\sqrt aq,d=-\sqrt aq$とすると, $w=-a/bq$であり,
\begin{align} &\Q54{a,\sqrt aq,-\sqrt aq,b,q^{-n}}{\sqrt a,-\sqrt a,aq/b,b^2q^{2-n}}{q}\\ &=\frac{(-1/b,a/b^2q;q)_n}{(-a/b,1/b^2q;q)_n}\Q87{w,\sqrt wq,-\sqrt wq,wb/a,\sqrt{aq},-\sqrt{aq},w^2q^{n+1}/a,q^{-n}}{\sqrt w,-\sqrt w,aq/b,w\sqrt{q/a},-w\sqrt{q/a},aq^{-n}/w,wq^{n+1}}q \end{align}
Jacksonの${}_8\phi_7$和公式 より,
\begin{align} &\Q87{w,\sqrt wq,-\sqrt wq,wb/a,\sqrt{aq},-\sqrt{aq},w^2q^{n+1}/a,q^{-n}}{\sqrt w,-\sqrt w,aq/b,w\sqrt{q/a},-w\sqrt{q/a},aq^{-n}/w,wq^{n+1}}q\\ &=\frac{(-a/b,\sqrt{aq}/b,-\sqrt{aq}/b,1/bq;q)_n}{(aq/b,\sqrt{a/q}/b,-\sqrt{a/q}/b,-1/b;q)_n} \end{align}
だから,
\begin{align} &\frac{(-1/b,a/b^2q;q)_n}{(-a/b,1/b^2q;q)_n}\frac{(-a/b,\sqrt{aq}/b,-\sqrt{aq}/b,1/bq;q)_n}{(aq/b,\sqrt{a/q}/b,-\sqrt{a/q}/b,-1/b;q)_n}\\ &=\frac{(a/b^2q,1/bq;q)_n(aq/b^2;q^2)_n}{(aq/b,1/b^2q;q)_n(a/b^2q;q^2)_n}\\ &=\frac{(a/b^2q,1/bq;q)_n}{(aq/b,1/b^2q;q)_n}\frac{1-aq^{2n-1}/b^2}{1-a/b^2q} \end{align}

$w=a^2q/bcd$とする. 先ほどの証明の途中で用いた
\begin{align} \frac{(a,b,c,d;q)_n}{(aq/b,aq/c,aq/d,q;q)_n}&=\sum_{k=0}^n\frac{(w,\sqrt wq,-\sqrt wq,wb/a,wc/a,wd/a;q)_k(a;q)_{n+k}(a/w;q)_{n-k}}{(\sqrt w,-\sqrt w,aq/b,aq/c,aq/d,q;q)_k(wq;q)_{n+k}(q;q)_{n-k}}\left(\frac aw\right)^k \end{align}
より,
\begin{align} &\Q54{a,\sqrt aq,-\sqrt aq,b,c,d,q^{-n}}{\sqrt a,-\sqrt a,aq/b,aq/c,aq/d,a^2q^{2-n}/w^2}q\\ &=\sum_{0\leq k}\frac{(\sqrt aq,-\sqrt aq,q^{-n};q)_k}{(\sqrt a,-\sqrt a,a^2q^{2-n}/w^2;q)_k}q^k\sum_{j=0}^k\frac{(w,\sqrt wq,-\sqrt wq,wb/a,wc/a,wd/a;q)_j(a;q)_{k+j}(a/w;q)_{k-j}}{(\sqrt w,-\sqrt w,aq/b,aq/c,aq/d,q;q)_j(wq;q)_{k+j}(q;q)_{k-j}}\left(\frac aw\right)^j\\ &=\sum_{0\leq j}\frac{(w,\sqrt wq,-\sqrt wq,wb/a,wc/a,wd/a;q)_j}{(\sqrt w,-\sqrt w,aq/b,aq/c,aq/d,q;q)_j}\left(\frac aw\right)^j\sum_{0\leq k}\frac{(\sqrt aq,-\sqrt aq,q^{-n};q)_k(wq;q)_{k+j}(a/w;q)_{k-j}}{(\sqrt a,-\sqrt a,a^2q^{2-n}/w^2;q)_k(a;q)_{k+j}(q;q)_{k-j}}q^k\\ &=\sum_{0\leq j}\frac{(w,\sqrt wq,-\sqrt wq,wb/a,wc/a,wd/a,\sqrt aq,-\sqrt aq,q^{-n};q)_j(a;q)_{2j}}{(\sqrt w,-\sqrt w,aq/b,aq/c,aq/d,a^2q^{2-n}/w^2,\sqrt a,-\sqrt aq;q)_j(wq;q)_{2j}}\left(\frac {aq}w\right)^j\sum_{0\leq k}\frac{(\sqrt aq^{j+1},-\sqrt aq^{j+1},q^{j-n},aq^{2j},a/w;q)_k}{(\sqrt aq^j,-\sqrt aq^j,wq^{2j+1},a^2q^{2+j-n}/w^2,q;q)_k}q^k \end{align}
ここで, 補題1
\begin{align} &\sum_{0\leq k}\frac{(\sqrt aq^{j+1},-\sqrt aq^{j+1},q^{j-n},aq^{2j},a/w;q)_k}{(\sqrt aq^j,-\sqrt aq^j,wq^{2j+1},a^2q^{2+j-n}/w^2,q;q)_k}q^k\\ &=\frac{(w^2q^{2j-1}/a,w/aq;q)_{n-j}}{(wq^{2j+1},w^2/a^2q;q)_{n-j}}\frac{1-w^2q^{2n-1}/a}{1-w^2q^{2j-1}/a}\\ &=\frac{(w^2/aq,w/aq;q)_n}{(wq,w^2/a^2q;q)_n}{}\frac{(wq;q)_{2j}(w^2q^{n-1}/a,a^2q^{2-n}/w^2;q)_{j}}{(w^2/aq;q)_{2j}(wq^{n+1},aq^{2-n}/w;q)_j}\frac{1-w^2q^{2n-1}/a}{1-w^2q^{2j-1}/a}\\ &=\frac{(w^2/aq,w/aq;q)_n}{(wq,w^2/a^2q;q)_n}{}\frac{(wq;q)_{2j}(w^2q^{n-1}/a,a^2q^{2-n}/w^2;q)_{j}}{(w^2/a;q)_{2j}(wq^{n+1},aq^{2-n}/w;q)_j}\left(\frac{w}{a}\right)^j\frac{1-w^2q^{2n-1}/a}{1-w^2/aq} \end{align}

\begin{align} &\Q76{a,\sqrt aq,-\sqrt aq,b,c,d,q^{-n}}{\sqrt a,-\sqrt a,aq/b,aq/c,aq/d,a^2q^{2-n}/w^2}{q}\\ &\sum_{0\leq j}\frac{(w,\sqrt wq,-\sqrt wq,wb/a,wc/a,wd/a,\sqrt aq,-\sqrt aq,q^{-n};q)_j(a;q)_{2j}}{(\sqrt w,-\sqrt w,aq/b,aq/c,aq/d,a^2q^{2-n}/w^2,\sqrt a,-\sqrt a,q;q)_j(wq;q)_{2j}}\left(\frac {aq}w\right)^j\\ &\qquad\cdot\frac{(w^2/aq,w/aq;q)_n}{(wq,w^2/a^2q;q)_n}{}\frac{(wq;q)_{2j}(w^2q^{n+1}/a,a^2q^{2-n}/w^2;q)_{j}}{(w^2/a;q)_{2j}(wq^{n+1},aq^{2-n}/w;q)_j}\left(\frac{w}{a}\right)^j\frac{1-w^2q^{2n-1}/a}{1-w^2/aq}\\ &=\frac{(w^2/aq,w/aq;q)_n}{(wq,w^2/a^2q;q)_n}\frac{1-w^2q^{2n-1}/a}{1-w^2/aq}\\ &\qquad\cdot \sum_{0\leq j}\frac{(w,\sqrt wq,-\sqrt wq,wb/a,wc/a,wd/a,w^2q^{n-1}/a,q^{-n};q)_j(aq;q)_{2j}}{(\sqrt w,-\sqrt w,aq/b,aq/c,aq/d,aq^{2-n}/w,wq^{n+1},q;q)_j(w^2/a;q)_{2j}}q^j\\ \end{align}
ここで, $(aq;q)_{2j}=(\sqrt{aq},-\sqrt{aq},\sqrt aq,-\sqrt aq;q)_j$などを用いて定理を得る.