Gasper-Rahmanの四次変換公式

前の記事 で, Gasper-Rahmanの二次変換公式を示した. 今回は同様の手法でGasper-Rahmanによる四次変換公式
\begin{align} &W(ac^2/b;a^2b^2/q^2,ac/b,c,cq/b,cq^2/b,cq^3/b,cq^4/b;q^4;q^4)\\ &\qquad+\frac {(1-c)(1-ac/b)(cq/b,a^2b^3/cq^2;q)_{\infty}(ac^2q^4/b,ab^3/c^2q^2;q^4)_{\infty}}{(1-a^2b^2/cq^2)(1-ab^3/cq^2)(ac,ab^2/cq;q)_{\infty}(c^2q^2/ab^3,a^3b^5/c^2;q^4)_{\infty}}\\ &\qquad\cdot W(a^3b^5/c^2q^4;a^2b^2/q^2,a^2b^2/cq^2,ab^3/cq^2,ab^2/cq,ab^2/c,ab^2q/c,ab^2q^2/c;q^4;q^4)\\ &\qquad+\frac {ab^2(a,b,cq/b;q)_{\infty}(ac^2q^4/b,ab^3/c^2q^2,a^2b^2/q^2;q^4)_{\infty}}{cq^2(1-a^2b^2/cq^2)(ac,ab^2/cq^2,cq^3/ab^2;q)_{\infty}(cq^4,acq^4/b,ab^3/cq^2;q^4)_{\infty}}\\ &\qquad\qquad\cdot \Q11{a^2b^2/cq^2}{a^2b^2q^2/c}{q^4;\frac{ab^3q^2}c}\\ &=\frac{(ab;q)_{\infty}(ab/q;q^2)_{\infty}(ac^2q^4/b,ab^3/c^2q^2;q^4)_{\infty}}{(acq,ab^2/cq^2;q)_{\infty}}\\ &\qquad\cdot\sum_{0\leq k}\frac{1-acq^{5k}}{1-ac}\frac{(a,b;q)_k(a^2b^2/q^2;q^4)_k(cq/b;q)_{3k}}{(cq^3/ab^2;q)_k(ab;q)_{2k}(ab/q;q^2)_k(cq^4,acq^4/b;q^4)_k}q^k \end{align}
を示したいと思う. 前の記事と同様に Gasper-Rahmanのbibasic超幾何級数の和公式 において$m=0, c=q^{-n}$とした式
\begin{align} &\sum_{k=0}^n\frac{(1-adp^kq^k)(1-bp^kq^{-k}/d)}{(1-ad)(1-b/d)}\frac{(a,b;p)_k(q^{-n},ad^2q^n/b;q)_k}{(dq,adq/b;q)_k(adpq^n,bpq^{-n}/d;p)_k}q^k\\ &=\frac{(1-d)(1-ad/b)(1-adq^n)(1-dq^n/b)}{(1-ad)(1-d/b)(1-dq^n)(1-adq^n/b)} \end{align}
から始める. $q=p^4$としてから, $d\mapsto c,p\mapsto q$とすると
\begin{align} &\sum_{k=0}^n\frac{(1-acq^{5k})(1-bq^{-3k}/c)}{(1-ac)(1-b/c)}\frac{(a,b;q)_k(q^{-4n},ac^2q^{4n}/b;q^4)_k}{(cq^4,acq^4/b;q^4)_k(acq^{4n+1},bq^{1-4n}/c;q)_k}q^{4k}\\ &=\frac{(1-c)(1-ac/b)(1-acq^{4n})(1-cq^{4n}/b)}{(1-ac)(1-c/b)(1-cq^{4n})(1-acq^{4n}/b)} \end{align}
となる. 両辺に
\begin{align} \frac{(1-ac^2q^{8n}/b)(ac^2/b,a^2b^2/q^2;q^4)_n(c/b;q)_{4n}}{(1-ac^2/b)(q^4,c^2q^6/ab^3;q^4)_n(acq;q)_{4n}}q^{4n} \end{align}
を掛けて足し合わせると
\begin{align} W(a;b_1,\dots,b_r;q;x):=\Q{r+3}{r+2}{a,\sqrt aq,-\sqrt aq,b_1,\dots,b_r}{\sqrt a,-\sqrt a,aq/b_1,\dots,aq/b_r}{q;x} \end{align}
として
\begin{align} &W(ac^2/b;a^2b^2/q^2,ac/b,c,cq/b,cq^2/b,cq^3/b,cq^4/b;q^4;q^4)\\ &=\sum_{0\leq n}\frac{(1-ac^2q^{8n}/b)(ac^2/b,a^2b^2/q^2;q^4)_n(c/b;q)_{4n}}{(1-ac^2/b)(q^4,c^2q^6/ab^3;q^4)_n(acq;q)_{4n}}q^{4n}\\ &\qquad\cdot\sum_{k=0}^n\frac{(1-acq^{5k})(1-bq^{-3k}/c)}{(1-ac)(1-b/c)}\frac{(a,b;q)_k(q^{-4n},ac^2q^{4n}/b;q^4)_k}{(cq^4,acq^4/b;q^4)_k(acq^{4n+1},bq^{1-4n}/c;q)_k}q^{4k}\\ &=\sum_{0\leq k}\frac{(1-acq^{5k})(1-bq^{-3k}/c)}{(1-ac)(1-b/c)}\frac{(a,b;q)_k}{(cq^4,acq^4/b;q^4)_k}\left(\frac{c}{b}\right)^kq^{3\binom k2+3k}\\ &\qquad\cdot\sum_{0\leq n}\frac{(1-ac^2q^{8n}/b)(ac^2/b;q^4)_{n+k}(a^2b^2/q^2;q^4)_n(c/b;q)_{4n-k}}{(1-ac^2/b)(q^4;q^4)_{n-k}(c^2q^6/ab^3;q^4)_n(acq;q)_{4n+k}}q^{4n}\\ &=\sum_{0\leq k}\frac{(1-acq^{5k})(1-bq^{-3k}/c)}{(1-ac)(1-b/c)}\frac{(a,b;q)_k}{(cq^4,acq^4/b;q^4)_k}\left(\frac{c}{b}\right)^kq^{3\binom k2+7k}\\ &\qquad\cdot\sum_{0\leq n}\frac{(1-ac^2q^{8n+8k}/b)(ac^2/b;q^4)_{n+2k}(a^2b^2/q^2;q^4)_{n+k}(c/b;q)_{4n+3k}}{(1-ac^2/b)(q^4;q^4)_{n}(c^2q^6/ab^3;q^4)_{n+k}(acq;q)_{4n+5k}}q^{4n}\qquad n\mapsto n+k\\ &=\sum_{0\leq k}\frac{1-acq^{5k}}{1-ac}\frac{(a,b;q)_k(a^2b^2/q^2;q^4)_k(ac^2/b;q^4)_{2k}(cq/b;q)_{3k}}{(cq^4,acq^4/b,c^2q^6/ab^3;q^4)_k(acq;q)_{5k}}\left(\frac{c}{b}\right)^kq^{3\binom k2+4k}\\ &\qquad W(ac^2q^{8k}/b;a^2b^2q^{4k-2},cq^{3k}/b,cq^{3k+1}/b,cq^{3k+2}/b,cq^{3k+3}/b;q^4;q^4) \end{align}
を得る. ここで, non-terminating Jacksonの和公式 より
\begin{align} &W(ac^2q^{8k}/b;a^2b^2q^{4k-2},cq^{3k}/b,cq^{3k+1}/b,cq^{3k+2}/b,cq^{3k+3}/b;q^4;q^4)\\ &=\frac{(ac^2q^{8k+4}/b,abq^{2k+3},abq^{2k+2},abq^{2k+1},abq^{2k+1},abq^{2k},abq^{2k-1},ab^3q^{-4k-2}/c^2;q^4)_{\infty}}{(acq^{5k+4},acq^{5k+3},acq^{5k+2},acq^{5k+1},ab^2q^{-k-2}/c,ab^2q^{-k-1}/c,ab^2q^{-k}/c,ab^2q^{1-k}/c;q^4)_{\infty}}\\ &\qquad+\frac{ab^3q^{-4k-2}}{c^2}\frac{(ac^2q^{8k+4}/b,ab^3q^{2-4k}/c^2,a^2b^3q^{k+2}/c,a^2b^3q^{k+1}/c,a^2b^3q^{k}/c,a^2b^3q^{k-1}/c,cq^{3k}/b,cq^{3k+1}/b,cq^{3k+2}/b,cq^{3k+3}/b;q^4)_{\infty}}{(c^2q^{4k+6}/ab^3,acq^{5k+4},acq^{5k+3},acq^{5k+2},acq^{5k+1},ab^2q^{-k-2}/c,ab^2q^{-k-1}/c,ab^2q^{-k}/c,ab^2q^{1-k}/c,a^3b^5/c^2;q^4)_{\infty}}\\ &\qquad\cdot W(a^3b^5/c^2q^4;a^2b^2q^{4k-2},ab^2q^{-k-2}/c,ab^2q^{-k-1},ab^2q^{-k}/c,ab^2q^{1-k};q^4;q^4)\\ &=\frac{(abq^{2k};q)_{\infty}(abq^{2k-1};q^2)_{\infty}(ac^2q^{8k+4}/b,ab^3q^{-4k-2}/c^2;q^4)_{\infty}}{(acq^{5k+1},ab^2q^{-k-2}/c;q)_{\infty}}\\ &\qquad+\frac{ab^3q^{-4k-2}}{c^2}\frac{(a^2b^3q^{k-1}/c,cq^{3k}/b;q)_{\infty}(ac^2q^{8k+4}/b,ab^3q^{2-4k}/c^2;q^4)_{\infty}}{(acq^{5k+1},ab^2q^{-k-2}/c;q)_{\infty}(c^2q^{4k+6}/ab^3,a^3b^5/c^2;q^4)_{\infty}}\\ &\qquad\cdot W(a^3b^5/c^2q^4;a^2b^2q^{4k-2},ab^2q^{-k-2}/c,ab^2q^{-k-1}/c,ab^2q^{-k}/c,ab^2q^{1-k}/c;q^4;q^4) \end{align}
であるから, これを代入すると,
\begin{align} &W(ac^2/b;a^2b^2/q^2,ac/b,c,cq/b,cq^2/b,cq^3/b,cq^4/b;q^4;q^4)\\ &=\sum_{0\leq k}\frac{1-acq^{5k}}{1-ac}\frac{(a,b;q)_k(a^2b^2/q^2;q^4)_k(ac^2q^4/b;q^4)_{2k}(cq/b;q)_{3k}}{(cq^4,acq^4/b,c^2q^6/ab^3;q^4)_k(acq;q)_{5k}}\left(\frac{c}{b}\right)^kq^{3\binom k2+4k}\\ &\qquad \frac{(abq^{2k};q)_{\infty}(abq^{2k-1};q^2)_{\infty}(ac^2q^{8k+4}/b,ab^3q^{-4k-2}/c^2;q^4)_{\infty}}{(acq^{5k+1},ab^2q^{-k-2}/c;q)_{\infty}}\\ &\qquad+\sum_{0\leq k}\frac{1-acq^{5k}}{1-ac}\frac{(a,b;q)_k(a^2b^2/q^2;q^4)_k(ac^2q^4/b;q^4)_{2k}(cq/b;q)_{3k}}{(cq^4,acq^4/b,c^2q^6/ab^3;q^4)_k(acq;q)_{5k}}\left(\frac{c}{b}\right)^kq^{3\binom k2+4k}\\ &\qquad\cdot\frac{ab^3q^{-4k-2}}{c^2}\frac{(a^2b^3q^{k-1}/c,cq^{3k}/b;q)_{\infty}(ac^2q^{8k+4}/b,ab^3q^{2-4k}/c^2;q^4)_{\infty}}{(acq^{5k+1},ab^2q^{-k-2}/c;q)_{\infty}(c^2q^{4k+6}/ab^3,a^3b^5/c^2;q^4)_{\infty}}\\ &\qquad\cdot W(a^3b^5/c^2q^4;a^2b^2q^{4k-2},ab^2q^{-k-2}/c,ab^2q^{-k-1}/c,ab^2q^{-k}/c,ab^2q^{1-k}/c;q^4;q^4)\\ &=\frac{(ab;q)_{\infty}(ab/q;q^2)_{\infty}(ac^2q^4/b,ab^3/c^2q^2;q^4)_{\infty}}{(acq,ab^2/cq^2;q)_{\infty}}\\ &\qquad\cdot\sum_{0\leq k}\frac{1-acq^{5k}}{1-ac}\frac{(a,b;q)_k(a^2b^2/q^2;q^4)_k(cq/b;q)_{3k}}{(ab;q)_{2k}(ab/q;q^2)_k(cq^4,acq^4/b,c^2q^6/ab^3;q^4)_k}\left(\frac{c}{b}\right)^kq^{3\binom k2+4k}\frac{(ab^2/cq^2;q)_{-k}}{(ab^3/c^2q^2;q^4)_{-k}}\\ &\qquad+\frac{ab^3}{c^2q^2}\frac {(c/b,a^2b^3/cq;q)_{\infty}(ac^2q^4/b,ab^3q^2/c^2;q^4)_{\infty}}{(acq,ab^2/cq^2;q)_{\infty}(c^2q^6/ab^3,a^3b^5/c^2;q^4)_{\infty}}\\ &\qquad\cdot\sum_{0\leq k}\frac{(1-acq^{5k})(1-bq^{-3k}/c)}{(1-ac)(1-b/c)}\frac{(a,b;q)_k(a^2b^2/q^2;q^4)_k}{(a^2b^3/cq;q)_k(cq^4,acq^4/b;q^4)_k}\left(\frac{c}{b}\right)^kq^{3\binom k2+3k}\frac{(ab^2/cq^2;q)_{-k}}{(ab^3q^2/c^2;q^4)_{-k}}\\ &\qquad\cdot W(a^3b^5/c^2q^4;a^2b^2q^{4k-2},ab^2q^{-k-2}/c,ab^2q^{-k-1}/c,ab^2q^{-k}/c,ab^2q^{1-k}/c;q^4;q^4)\\ &=\frac{(ab;q)_{\infty}(ab/q;q^2)_{\infty}(ac^2q^4/b,ab^3/c^2q^2;q^4)_{\infty}}{(acq,ab^2/cq^2;q)_{\infty}}\\ &\qquad\cdot\sum_{0\leq k}\frac{1-acq^{5k}}{1-ac}\frac{(a,b;q)_k(a^2b^2/q^2;q^4)_k(cq/b;q)_{3k}}{(cq^3/ab^2;q)_k(ab;q)_{2k}(ab/q;q^2)_k(cq^4,acq^4/b;q^4)_k}q^k\\ &\qquad-\frac {(c/b,a^2b^3/cq;q)_{\infty}(ac^2q^4/b,ab^3/c^2q^2;q^4)_{\infty}}{(acq,ab^2/cq^2;q)_{\infty}(c^2q^2/ab^3,a^3b^5/c^2;q^4)_{\infty}}\\ &\qquad\cdot\sum_{0\leq k}\frac{(1-acq^{5k})(1-bq^{-3k}/c)}{(1-ac)(1-b/c)}\frac{(a,b;q)_k(a^2b^2/q^2,cq^2/ab^3;q^4)_k}{(cq^3/ab^2,a^2b^3/cq;q)_k(cq^4,acq^4/b;q^4)_k}q^{4k}\\ &\qquad\cdot W(a^3b^5/c^2q^4;a^2b^2q^{4k-2},ab^2q^{-k-2}/c,ab^2q^{-k-1}/c,ab^2q^{-k}/c,ab^2q^{1-k}/c;q^4;q^4) \end{align}
が得られる. ここで, 第2項の和は足し合わせる順番を入れ替えることによって,
\begin{align} &\sum_{0\leq k}\frac{(1-acq^{5k})(1-bq^{-3k}/c)}{(1-ac)(1-b/c)}\frac{(a,b;q)_k(a^2b^2/q^2,cq^2/ab^3;q^4)_k}{(cq^3/ab^2,a^2b^3/cq;q)_k(cq^4,acq^4/b;q^4)_k}q^{4k}\\ &\qquad\cdot W(a^3b^5/c^2q^4;a^2b^2q^{4k-2},ab^2q^{-k-2}/c,ab^2q^{-k-1}/c,ab^2q^{-k}/c,ab^2q^{1-k}/c;q^4;q^4)\\ &=\sum_{0\leq k}\frac{(1-acq^{5k})(1-bq^{-3k}/c)}{(1-ac)(1-b/c)}\frac{(a,b;q)_k(a^2b^2/q^2,cq^2/ab^3;q^4)_k}{(cq^3/ab^2,a^2b^3/cq;q)_k(cq^4,acq^4/b;q^4)_k}q^{4k}\\ &\qquad\cdot \sum_{0\leq j}\frac{(1-a^3b^5q^{8j-4}/c^2)(a^3b^5/c^2q^4,a^2b^2q^{4k-2};q^4)_j(ab^2q^{-k-2}/c;q)_{4j}}{(1-a^3b^5/cq^4)(q^4,ab^3q^{2-4k}/c^2;q^4)_j(a^2b^3q^{k-1}/c;q)_{4j}}q^{4j}\\ &=\sum_{0\leq j}\frac{(1-a^3b^5q^{8j-4}/c^2)(a^3b^5/c^2q^4;q^4)_j}{(1-a^3b^5/cq^4)(q^4;q^4)_j}q^{4j}\\ &\qquad\cdot\sum_{0\leq k}\frac{(1-acq^{5k})(1-bq^{-3k}/c)}{(1-ac)(1-b/c)}\frac{(a,b;q)_k(a^2b^2/q^2;q^4)_{j+k}(cq^2/ab^3;q^4)_k}{(cq^3/ab^2;q)_k(cq^4,acq^4/b;q^4)_k(a^2b^3/cq;q)_{4j+k}}q^{4k}\frac{(ab^2q^{-k-2}/c;q)_{4j}}{(ab^3q^{2-4k}/c^2;q^4)_j}\\ &=\sum_{0\leq j}\frac{(1-a^3b^5q^{8j-4}/c^2)(a^2b^2/q^2,a^3b^5/c^2q^4;q^4)_j(ab^2/cq^2;q)_{4j}}{(1-a^3b^5/cq^4)(q^4,ab^3q^2/c^2;q^4)_j(a^2b^3/cq;q)_{4j}}q^{4j}\\ &\qquad\cdot\sum_{0\leq k}\frac{(1-acq^{5k})(1-bq^{-3k}/c)}{(1-ac)(1-b/c)}\frac{(a,b;q)_k(a^2b^2q^{4j-2},cq^{2-4j}/ab^3;q^4)_k}{(cq^4,acq^4/b;q^4)_k(c^2q^{3-4j}/ab^2,a^2b^3q^{4j-1}/c;q)_k}q^{4k} \end{align}
ここで, Gasper-Rahmanのbibasic超幾何級数の和公式 より,
\begin{align} &\sum_{0\leq k}\frac{(1-acq^{5k})(1-bq^{-3k}/c)}{(1-ac)(1-b/c)}\frac{(a,b;q)_k(a^2b^2q^{4j-2},c^2q^{2-4j}/ab^3;q^4)_k}{(cq^4,acq^4/b;q^4)_k(cq^{3-4j}/ab^2,a^2b^3q^{4j-1}/c;q)_k}q^{4k}\\ &=\frac{(1-a)(1-b)(1-a^2b^2q^{4j-2})(1-c^2q^{2-4j}/ab^3)}{c(1-ac)(1-b/c)(1-a^2b^2q^{4j-2}/c)(1-cq^{2-4j}/ab^3)}\\ &\qquad\left(\frac{(aq,bq;q)_{\infty}(a^2b^2q^{4j+2},c^2q^{6-4j}/ab^3;q^4)_{\infty}}{(cq^4,acq^4/b;q^4)_{\infty}(cq^{3-4j}/ab^2,a^2b^3q^{4j-1}/c;q)_{\infty}}-\frac{(1-ab^2q^{4j-2}/c)(1-cq^{2-4j}/a^2b^3)(1-1/c)(1-b/ac)}{(1-q^{2-4j}/a^2b^2)(1-ab^3q^{4j-2}/c^2)(1-1/a)(1-1/b)}\right)\\ &=\frac{1}{c(1-ac)(1-b/c)(1-a^2b^2q^{4j-2}/c)(1-cq^{2-4j}/ab^3)}\frac{(a,b;q)_{\infty}(a^2b^2q^{4j-2},c^2q^{2-4j}/ab^3;q^4)_{\infty}}{(cq^4,acq^4/b;q^4)_{\infty}(cq^{3-4j}/ab^2,a^2b^3q^{4j-1}/c;q)_{\infty}}\\ &\qquad-\frac{ac(1-ab^2q^{4j-2}/c)(1-a^2b^3q^{4j-2}/c)(1-1/c)(1-b/ac)}{(1-ac)(1-b/c)(1-a^2b^2q^{4j-2}/c)(1-ab^3q^{4j-2}/c)}\\ &=-\frac{ab^3}{c^2q^2(1-ac)(1-b/c)(1-a^2b^2q^{4j-2}/c)(1-ab^3q^{4j-2}/c)}\frac{(a,b;q)_{\infty}(a^2b^2/q^2,c^2q^{2}/ab^3;q^4)_{\infty}}{(cq^4,acq^4/b;q^4)_{\infty}(cq^3/ab^2,a^2b^3/cq;q)_{\infty}}\frac{(cq^3/ab^2;q)_{-4j}(a^2b^3/cq;q)_{4j}}{(a^2b^2/q^2;q^4)_j(c^2q^2/ab^3;q^4)_{-j}}q^{4j}\\ &\qquad+\frac{(1-c)(1-ac/b)(1-ab^2q^{4j-2}/c)(1-a^2b^3q^{4j-2}/c)}{(1-ac)(1-c/b)(1-a^2b^2q^{4j-2}/c)(1-ab^3q^{4j-2}/c)}\\ &=\frac{(1-c)(1-ac/b)(1-ab^2q^{4j-2}/c)(1-a^2b^3q^{4j-2}/c)}{(1-ac)(1-c/b)(1-a^2b^2q^{4j-2}/c)(1-ab^3q^{4j-2}/c)}\\ &\qquad-\frac{ab^3}{c^2q^2(1-ac)(1-b/c)}\frac{(a,b;q)_{\infty}(a^2b^2/q^2,c^2q^{2}/ab^3;q^4)_{\infty}}{(cq^4,acq^4/b;q^4)_{\infty}(cq^3/ab^2,a^2b^3/cq;q)_{\infty}}\\ &\qquad\cdot\frac{(a^2b^3/cq;q)_{4j}(ab^3q^2/c^2;q^4)_j}{(1-a^2b^2q^{4j-2}/c)(1-ab^3q^{4j-2}/c)(ab^2/cq^2;q)_{4j}(a^2b^2/q^2;q^4)_j}\left(-\frac{a^3b^5}{c^2}\right)^jq^{12\binom j2} \end{align}
であるから, これを代入して
\begin{align} &\sum_{0\leq k}\frac{(1-acq^{5k})(1-bq^{-3k}/c)}{(1-ac)(1-b/c)}\frac{(a,b;q)_k(a^2b^2/q^2,cq^2/ab^3;q^4)_k}{(cq^3/ab^2,a^2b^3/cq;q)_k(cq^4,acq^4/b;q^4)_k}q^{4k}\\ &\qquad\cdot W(a^3b^5/c^2q^4;a^2b^2q^{4k-2},ab^2q^{-k-2}/c,ab^2q^{-k-1}/c,ab^2q^{-k}/c,ab^2q^{1-k}/c;q^4;q^4)\\ &=\sum_{0\leq j}\frac{(1-a^3b^5q^{8j-4}/c^2)(a^2b^2/q^2,a^3b^5/c^2q^4;q^4)_j(ab^2/cq^2;q)_{4j}}{(1-a^3b^5/cq^4)(q^4,ab^3q^2/c^2;q^4)_j(a^2b^3/cq;q)_{4j}}q^{4j}\\ &\qquad\cdot\frac{(1-c)(1-ac/b)(1-ab^2q^{4j-2}/c)(1-a^2b^3q^{4j-2}/c)}{(1-ac)(1-c/b)(1-a^2b^2q^{4j-2}/c)(1-ab^3q^{4j-2}/c)}\\ &\qquad-\frac{ab^3}{c^2q^2(1-ac)(1-b/c)}\frac{(a,b;q)_{\infty}(a^2b^2/q^2,c^2q^{2}/ab^3;q^4)_{\infty}}{(cq^4,acq^4/b;q^4)_{\infty}(cq^3/ab^2,a^2b^3/cq;q)_{\infty}}\\ &\qquad\cdot\sum_{0\leq j}\frac{(1-a^3b^5q^{8j-4}/c^2)(a^2b^2/q^2,a^3b^5/c^2q^4;q^4)_j(ab^2/cq^2;q)_{4j}}{(1-a^3b^5/cq^4)(q^4,ab^3q^2/c^2;q^4)_j(a^2b^3/cq;q)_{4j}}q^{4j}\\ &\qquad\cdot\frac{(a^2b^3/cq;q)_{4j}(ab^3q^2/c^2;q^4)_j}{(1-a^2b^2q^{4j-2}/c)(1-ab^3q^{4j-2}/c)(ab^2/cq^2;q)_{4j}(a^2b^2/q^2;q^4)_j}\left(-\frac{a^3b^5}{c^2}\right)^jq^{12\binom j2}\\ &=\frac{(1-c)(1-ac/b)(1-ab^2/cq^2)(1-a^2b^3/cq^2)}{(1-ac)(1-c/b)(1-a^2b^2/cq^2)(1-ab^3/cq^2)}\sum_{0\leq j}\frac{(1-a^3b^5q^{8j-4}/c^2)(a^2b^2/q^2,a^3b^5/c^2q^4;q^4)_j(ab^2/cq;q)_{4j}}{(1-a^3b^5/cq^4)(q^4,ab^3q^2/c^2;q^4)_j(a^2b^3/cq^2;q)_{4j}}q^{4j}\\ &\qquad\cdot\frac{(1-a^2b^2/cq^2)(1-ab^3/cq^2)}{(1-a^2b^2q^{4j-2}/c)(1-ab^3q^{4j-2}/c)}\\ &\qquad-\frac{ab^3}{c^2q^2(1-ac)(1-b/c)}\frac{(a,b;q)_{\infty}(a^2b^2/q^2,c^2q^{2}/ab^3;q^4)_{\infty}}{(cq^4,acq^4/b;q^4)_{\infty}(cq^3/ab^2,a^2b^3/cq;q)_{\infty}}\\ &\qquad\cdot\sum_{0\leq j}\frac{(1-a^3b^5q^{8j-4}/c^2)(a^3b^5/c^2q^4;q^4)_j}{(1-a^3b^5/cq^4)(q^4;q^4)_j}\frac{1}{(1-a^2b^2q^{4j-2}/c)(1-ab^3q^{4j-2}/c)}\left(-\frac{a^3b^5q^4}{c^2}\right)^jq^{12\binom j2}\\ &=\frac{(1-c)(1-ac/b)(1-ab^2/cq^2)(1-a^2b^3/cq^2)}{(1-ac)(1-c/b)(1-a^2b^2/cq^2)(1-ab^3/cq^2)}\\ &\qquad\cdot W(a^3b^5/c^2q^4;a^2b^2/q^2,a^2b^2/cq^2,ab^3/cq^2,ab^2/cq,ab^2/c,ab^2q/c,ab^2q^2/c;q^4;q^4)\\ &\qquad-\frac{ab^3}{c^2q^2(1-ac)(1-b/c)(1-a^2b^2/cq^2)(1-ab^3/cq^2)}\frac{(a,b;q)_{\infty}(a^2b^2/q^2,c^2q^{2}/ab^3;q^4)_{\infty}}{(cq^4,acq^4/b;q^4)_{\infty}(cq^3/ab^2,a^2b^3/cq;q)_{\infty}}\\ &\qquad\cdot \Q{5}{7}{a^3b^5/c^2q^4,\sqrt{a^3b^5q^4/c^2},-\sqrt{a^3b^5q^4/c^2},a^2b^2/cq^2,ab^3/cq^2}{\sqrt{a^3b^5/c^2q^4},-\sqrt{a^3b^5/c^2q^4},a^2b^2q^2/c,ab^3q^2/c,0,0,0}{\frac{a^3b^5q^4}{c^2}} \end{align}
となる. よって,
\begin{align} &W(ac^2/b;a^2b^2/q^2,ac/b,c,cq/b,cq^2/b,cq^3/b,cq^4/b;q^4;q^4)\\ &=\frac{(ab;q)_{\infty}(ab/q;q^2)_{\infty}(ac^2q^4/b,ab^3/c^2q^2;q^4)_{\infty}}{(acq,ab^2/cq^2;q)_{\infty}}\\ &\qquad\cdot\sum_{0\leq k}\frac{1-acq^{5k}}{1-ac}\frac{(a,b;q)_k(a^2b^2/q^2;q^4)_k(cq/b;q)_{3k}}{(cq^3/ab^2;q)_k(ab;q)_{2k}(ab/q;q^2)_k(cq^4,acq^4/b;q^4)_k}q^k\\ &\qquad-\frac {(c/b,a^2b^3/cq;q)_{\infty}(ac^2q^4/b,ab^3/c^2q^2;q^4)_{\infty}}{(acq,ab^2/cq^2;q)_{\infty}(c^2q^2/ab^3,a^3b^5/c^2;q^4)_{\infty}}\\ &\qquad\cdot\frac{(1-c)(1-ac/b)(1-ab^2/cq^2)(1-a^2b^3/cq^2)}{(1-ac)(1-c/b)(1-a^2b^2/cq^2)(1-ab^3/cq^2)}\\ &\qquad\cdot W(a^3b^5/c^2q^4;a^2b^2/q^2,a^2b^2/cq^2,ab^3/cq^2,ab^2/cq,ab^2/c,ab^2q/c,ab^2q^2/c;q^4;q^4)\\ &\qquad+\frac {(c/b,a^2b^3/cq;q)_{\infty}(ac^2q^4/b,ab^3/c^2q^2;q^4)_{\infty}}{(acq,ab^2/cq^2;q)_{\infty}(c^2q^2/ab^3,a^3b^5/c^2;q^4)_{\infty}}\\ &\qquad\cdot\frac{ab^3}{c^2q^2(1-ac)(1-b/c)(1-a^2b^2/cq^2)(1-ab^3/cq^2)}\frac{(a,b;q)_{\infty}(a^2b^2/q^2,c^2q^{2}/ab^3;q^4)_{\infty}}{(cq^4,acq^4/b;q^4)_{\infty}(cq^3/ab^2,a^2b^3/cq;q)_{\infty}}\\ &\qquad\cdot \Q{5}{7}{a^3b^5/c^2q^4,\sqrt{a^3b^5q^4/c^2},-\sqrt{a^3b^5q^4/c^2},a^2b^2/cq^2,ab^3/cq^2}{\sqrt{a^3b^5/c^2q^4},-\sqrt{a^3b^5/c^2q^4},a^2b^2q^2/c,ab^3q^2/c,0,0,0}{\frac{a^3b^5q^4}{c^2}}\\ &=\frac{(ab;q)_{\infty}(ab/q;q^2)_{\infty}(ac^2q^4/b,ab^3/c^2q^2;q^4)_{\infty}}{(acq,ab^2/cq^2;q)_{\infty}}\\ &\qquad\cdot\sum_{0\leq k}\frac{1-acq^{5k}}{1-ac}\frac{(a,b;q)_k(a^2b^2/q^2;q^4)_k(cq/b;q)_{3k}}{(cq^3/ab^2;q)_k(ab;q)_{2k}(ab/q;q^2)_k(cq^4,acq^4/b;q^4)_k}q^k\\ &\qquad-\frac {(1-c)(1-ac/b)(cq/b,a^2b^3/cq^2;q)_{\infty}(ac^2q^4/b,ab^3/c^2q^2;q^4)_{\infty}}{(1-a^2b^2/cq^2)(1-ab^3/cq^2)(ac,ab^2/cq;q)_{\infty}(c^2q^2/ab^3,a^3b^5/c^2;q^4)_{\infty}}\\ &\qquad\cdot W(a^3b^5/c^2q^4;a^2b^2/q^2,a^2b^2/cq^2,ab^3/cq^2,ab^2/cq,ab^2/c,ab^2q/c,ab^2q^2/c;q^4;q^4)\\ &\qquad-\frac {ab^2(a,b,cq/b;q)_{\infty}(ac^2q^4/b,ab^3/c^2q^2,a^2b^2/q^2;q^4)_{\infty}}{cq^2(1-a^2b^2/cq^2)(1-ab^3/cq^2)(ac,ab^2/cq^2,cq^3/ab^2;q)_{\infty}(a^3b^5/c^2,cq^4,acq^4/b;q^4)_{\infty}}\\ &\qquad\cdot \Q{5}{7}{a^3b^5/c^2q^4,\sqrt{a^3b^5q^4/c^2},-\sqrt{a^3b^5q^4/c^2},a^2b^2/cq^2,ab^3/cq^2}{\sqrt{a^3b^5/c^2q^4},-\sqrt{a^3b^5/c^2q^4},a^2b^2q^2/c,ab^3q^2/c,0,0,0}{\frac{a^3b^5q^4}{c^2}} \end{align}
を得る. ここで, Watsonの変換公式 において, $b,e,N\to\infty$として得られる式
\begin{align} \Q57{a,\sqrt aq,-\sqrt aq,c,d}{\sqrt a,-\sqrt a,aq/c,aq/d}{\frac{a^2q^2}{cd}}&=\frac{(aq;q)_{\infty}}{(aq/d;q)_{\infty}}\Q11{d}{aq/c}{\frac{aq}d} \end{align}
を用いると,
\begin{align} &\Q{5}{7}{a^3b^5/c^2q^4,\sqrt{a^3b^5q^4/c^2},-\sqrt{a^3b^5q^4/c^2},a^2b^2/cq^2,ab^3/cq^2}{\sqrt{a^3b^5/c^2q^4},-\sqrt{a^3b^5/c^2q^4},a^2b^2q^2/c,ab^3q^2/c,0,0,0}{\frac{a^3b^5q^4}{c^2}}\\ &=\frac{(a^3b^5/c^2;q^4)_{\infty}}{(ab^3q^2/c;q^4)_{\infty}}\Q11{a^2b^2/cq^2}{a^2b^2q^2/c}{q^4;\frac{ab^3q^2}c} \end{align}
であるからこれを代入して
\begin{align} &W(ac^2/b;a^2b^2/q^2,ac/b,c,cq/b,cq^2/b,cq^3/b,cq^4/b;q^4;q^4)\\ &=\frac{(ab;q)_{\infty}(ab/q;q^2)_{\infty}(ac^2q^4/b,ab^3/c^2q^2;q^4)_{\infty}}{(acq,ab^2/cq^2;q)_{\infty}}\\ &\qquad\cdot\sum_{0\leq k}\frac{1-acq^{5k}}{1-ac}\frac{(a,b;q)_k(a^2b^2/q^2;q^4)_k(cq/b;q)_{3k}}{(cq^3/ab^2;q)_k(ab;q)_{2k}(ab/q;q^2)_k(cq^4,acq^4/b;q^4)_k}q^k\\ &\qquad-\frac {(1-c)(1-ac/b)(cq/b,a^2b^3/cq^2;q)_{\infty}(ac^2q^4/b,ab^3/c^2q^2;q^4)_{\infty}}{(1-a^2b^2/cq^2)(1-ab^3/cq^2)(ac,ab^2/cq;q)_{\infty}(c^2q^2/ab^3,a^3b^5/c^2;q^4)_{\infty}}\\ &\qquad\cdot W(a^3b^5/c^2q^4;a^2b^2/q^2,a^2b^2/cq^2,ab^3/cq^2,ab^2/cq,ab^2/c,ab^2q/c,ab^2q^2/c;q^4;q^4)\\ &\qquad-\frac {ab^2(a,b,cq/b;q)_{\infty}(ac^2q^4/b,ab^3/c^2q^2,a^2b^2/q^2;q^4)_{\infty}}{cq^2(1-a^2b^2/cq^2)(ac,ab^2/cq^2,cq^3/ab^2;q)_{\infty}(cq^4,acq^4/b,ab^3/cq^2;q^4)_{\infty}}\\ &\qquad\qquad\cdot \Q11{a^2b^2/cq^2}{a^2b^2q^2/c}{q^4;\frac{ab^3q^2}c} \end{align}
つまり以下が得られた.