Watsonの8φ7変換公式

$q$-Saalschützの和公式 より,
\begin{align} \frac{(b,c;q)_n}{(aq/b,aq/c;q)_n}&=\left(\frac{bc}{aq}\right)^n\Q32{aq/bc,aq^n,q^{-n}}{aq/b,aq/c}{q} \end{align}
であるから,
\begin{align} &\Q{r+4}{r+3}{a,b,c,a_1,\dots,a_r,q^{-n}}{aq/b,aq/c,b_1,\dots,b_{r+1}}{x}\\ &=\sum_{k=0}^n\frac{(a,a_1,\dots,a_r,q^{-n};q)_k}{(b_1,\dots,b_{r+1},q;q)_k}\left(\frac{bcx}{aq}\right)^k\sum_{j=0}^k\frac{(aq/bc,aq^k,q^{-k};q)_j}{(aq/b,aq/c,q;q)_j}q^j\\ &=\sum_{j=0}^n\frac{(aq/bc;q)_j}{(aq/b,aq/c,q;q)_j}q^j\sum_{k=j}^n\frac{(a,a_1,\dots,a_r,q^{-n};q)_k}{(b_1,\dots,b_{r+1},q;q)_k}(aq^k,q^{-k};q)_j\left(\frac{bcx}{aq}\right)^k\\ &=\sum_{j=0}^n\frac{(aq/bc;q)_j}{(aq/b,aq/c,q;q)_j}q^j\sum_{k=j}^n\frac{(a_1,\dots,a_r,q^{-n};q)_k}{(b_1,\dots,b_{r+1};q)_k}\frac{(a;q)_{k+j}}{(q;q)_{k-j}}(-1)^jq^{\binom j2-kj}\left(\frac{bcx}{aq}\right)^k\\ &=\sum_{j=0}^n\frac{(aq/bc;q)_j}{(aq/b,aq/c,q;q)_j}q^{-\binom j2}\left(-\frac{bcx}{aq}\right)^j\sum_{k=0}^{n-j}\frac{(a_1,\dots,a_r,q^{-n};q)_{k+j}}{(b_1,\dots,b_{r+1};q)_{k+j}}\frac{(a;q)_{k+2j}}{(q;q)_{k}}\left(\frac{bcx}{aq^{j+1}}\right)^k\\ &=\sum_{j=0}^n\frac{(aq/bc,a_1,\dots,a_r,q^{-n};q)_j(a;q)_{2j}}{(aq/b,aq/c,b_1,\dots,b_{r+1},q;q)_j}q^{-\binom j2}\left(-\frac{bcx}{aq}\right)^j\sum_{k=0}^{n-j}\frac{(aq^{2j},a_1q^j,\dots,a_rq^j,q^{j-n};q)_{k}}{(b_1q^j,\dots,b_{r+1}q^j,q;q)_{k}}\left(\frac{bcx}{aq^{j+1}}\right)^k \end{align}
となって示される.

定理1より,
\begin{align} \Q43{a,\sqrt aq,-\sqrt aq,q^{-n}}{\sqrt a,-\sqrt a,aq^{n+1}}{q^n}&=\sum_{k=0}^n\frac{(-q^{-1},q^{-n};q)_k(a;q)_{2k}}{(\sqrt a,-\sqrt a,aq^{n+1},q;q)_k}q^{-\binom k2}q^{(n+1)k}\Q21{aq^{2k},q^{k-n}}{aq^{n+k+1}}{-q^{n+1-k}} \end{align}
ここで, Bailey-Daumの和公式
\begin{align} \Q21{a,b}{aq/b}{-\frac qb}&=\frac{(-q;q)_{\infty}(aq,aq^2/b^2;q^2)_{\infty}}{(aq/b,-q/b;q)_{\infty}} \end{align}
を用いると,
\begin{align} \Q21{aq^{2k},q^{k-n}}{aq^{n+k+1}}{-q^{n+1-k}}&=\frac{(-q;q)_{\infty}(aq^{2k+1},aq^{2n+2};q^2)_{\infty}}{(aq^{n+k+1},-q^{n-k+1};q)_{\infty}} \end{align}
であるから,
\begin{align} \Q43{a,\sqrt aq,-\sqrt aq,q^{-n}}{\sqrt a,-\sqrt a,aq^{n+1}}{q^n}&=\sum_{k=0}^n\frac{(-q^{-1},q^{-n};q)_k(a;q)_{2k}}{(\sqrt a,-\sqrt a,aq^{n+1},q;q)_k}q^{-\binom k2}q^{(n+1)k}\frac{(-q;q)_{\infty}(aq^{2k+1},aq^{2n+2};q^2)_{\infty}}{(aq^{n+k+1},-q^{n-k+1};q)_{\infty}}\\ &=\frac{(aq^{2n+2};q)_{\infty}}{(aq;q)_{\infty}}\sum_{k=0}^n\frac{(-q^{-1},-q^{-n};q)_k(aq;q^2)_k(aq^{2k+1};q^2)_{\infty}(aq;q)_{n+k}(-q;q)_{n-k}}{(aq^{n+1},q;q)_k}q^{(n+1)k-\binom k2}\\ &=\frac{(aq^{2n+2},aq;q^2)_{\infty}(aq,-q;q)_n}{(aq;q)_{\infty}}\sum_{k=0}^n\frac{(-q^{-1},q^{-n};q)_k}{(-q^{-n};q)_k}q^k \end{align}
ここで, $q$-Vandermondeの恒等式 , $0< n$のとき,
\begin{align} \sum_{k=0}^n\frac{(-q^{-1},q^{-n};q)_k}{(-q^{-n};q)_k}q^k&=\frac{(q^{1-n};q)_n}{(-q^{-n};q)_n}(-q^{-1})^n=0 \end{align}
であり, $n=0$のとき,
\begin{align} \Q43{a,\sqrt aq,-\sqrt aq,q^{-n}}{\sqrt a,-\sqrt a,aq^{n+1}}{q^n}=1 \end{align}
であるから補題を得る.

定理1より,
\begin{align} \Q65{a,\sqrt aq,-\sqrt aq,b,c,q^{-n}}{\sqrt a,-\sqrt a,aq/b,aq/c,aq^{n+1}}{\frac{aq^{n+1}}{bc}}&=\sum_{k=0}^n\frac{(aq/bc,\sqrt aq,-\sqrt aq,q^{-n};q)_k(a;q)_{2k}}{(aq/b,aq/c,\sqrt a,-\sqrt a,aq^{n+1},q;q)_k}q^{-\binom k2}(-q^n)^k\\ &\qquad\qquad\cdot \Q43{aq^{2k},\sqrt aq^{k+1},-\sqrt aq^{k+1},q^{k-n}}{\sqrt aq^k,-\sqrt aq^k,aq^{n+k+1}}{q^{n-k}} \end{align}
補題2より,
\begin{align} \Q43{aq^{2k},\sqrt aq^{k+1},-\sqrt aq^{k+1},q^{k-n}}{\sqrt aq^k,-\sqrt aq^k,aq^{n+k+1}}{q^{n-k}}&=\delta_{n,k} \end{align}
だから,
\begin{align} \Q65{a,\sqrt aq,-\sqrt aq,b,c,q^{-n}}{\sqrt a,-\sqrt a,aq/b,aq/c,aq^{n+1}}{\frac{aq^{n+1}}{bc}}&=\frac{(aq/bc,\sqrt aq,-\sqrt aq,q^{-n};q)_n(a;q)_{2n}}{(aq/b,aq/c,\sqrt a,-\sqrt a,aq^{n+1},q;q)_n}q^{-\binom n2}(-q^n)^n\\ &=\frac{(aq,aq/bc;q)_n}{(aq/b,aq/c;q)_n} \end{align}
となって補題を得る.

定理1より,
\begin{align} &\Q87{a,\sqrt aq,-\sqrt aq,b,c,d,e,q^{-n}}{\sqrt a,-\sqrt a,aq/b,aq/c,aq/d,aq/e,aq^{n+1}}{\frac{a^2q^{2+n}}{bcde}}\\ &=\sum_{k=0}^n\frac{(aq/bc,\sqrt aq,-\sqrt aq,d,e,q^{-n};q)_k(a;q)_{2k}}{(aq/b,aq/c,\sqrt a,-\sqrt a,aq/d,aq/e,aq^{n+1},q;q)_k}q^{-\binom k2}\left(\frac{aq^{n+1}}{de}\right)^k\\ &\qquad\qquad \cdot\, \Q65{aq^{2k},\sqrt aq^{k+1},-\sqrt aq^{k+1},dq^k,eq^k,q^{k-n}}{\sqrt aq^k,-\sqrt aq^k,aq^{k+1}/d,aq^{k+1}/e,aq^{n+k+1}}{\frac{aq^{n-k+1}}{de}} \end{align}
ここで, 補題3より,
\begin{align} &\Q65{aq^{2k},\sqrt aq^{k+1},-\sqrt aq^{k+1},dq^k,eq^k,q^{k-n}}{\sqrt aq^k,-\sqrt aq^k,aq^{k+1}/d,aq^{k+1}/e,aq^{n+k+1}}{\frac{aq^{n-k+1}}{de}}\\ &=\frac{(aq^{2k+1},aq/de;q)_{n-k}}{(aq^{k+1}/d,aq^{k+1}/e;q)_{n-k}}\\ &=\frac{(aq;q)_{n+k}}{(aq;q)_{2k}}\frac{(aq/d,aq/e;q)_k}{(aq/d,aq/e;q)_n}\frac{(aq/de;q)_n}{(deq^{-n}/a;q)_k}\left(-\frac{de}a\right )^kq^{\binom k2-nk}\\ &=\frac{(aq,aq/de;q)_n}{(aq/d,aq/e;q)_n}\frac{(aq/d,aq/e,aq^{n+1};q)_k}{(aq;q)_{2k}(deq^{-n}/a;q)_k}\left(-\frac{de}{aq^n}\right)^kq^{\binom k2} \end{align}
より,
\begin{align} &\Q87{a,\sqrt aq,-\sqrt aq,b,c,d,e,q^{-n}}{\sqrt a,-\sqrt a,aq/b,aq/c,aq/d,aq/e,aq^{n+1}}{\frac{a^2q^{2+n}}{bcde}}\\ &=\frac{(aq,aq/de;q)_n}{(aq/d,aq/e;q)_n}\sum_{k=0}^n\frac{(aq/bc,d,e,q^{-n};q)_k}{(aq/b,aq/c,deq^{-n}/a,q;q)_k}q^k \end{align}
となって定理が示される.