Non-terminating q-Saalschützの和公式の両側類似

Non-terminating Jacksonの和公式
\begin{align} &\Q87{a,\sqrt aq,-\sqrt aq,b,c,d,e,f}{\sqrt a,-\sqrt a,aq/b,aq/c,aq/d,aq/e,aq/f}q\\ &=\frac ba\frac{(aq,bq/a,bq/c,bq/d,bq/e,bq/f,c,d,e,f;q)_{\infty}}{(aq/b,aq/c,aq/d,aq/e,aq/f,bc/a,bd/a,be/a,bf/a,b^2q/a;q)_{\infty}}\\ &\qquad\cdot\Q87{b^2/a,\sqrt{\frac{b^2}a}q,-\sqrt{\frac{b^2}a}q,b,bc/a,bd/a,be/a,bf/a}{\sqrt{\frac{b^2}a},-\sqrt{\frac{b^2}a},bq/a,bq/c,bq/d,bq/e,bq/f}q\\ &\qquad+\frac{(aq,aq/cd,aq/ce,aq/cf,aq/de,aq/df,aq/ef,b/a;q)_{\infty}}{(aq/c,aq/d,aq/e,aq/f,bc/a,bd/a,be/a,bf/a;q)_{\infty}}\qquad a^2q=bcdef \end{align}
において, $a\mapsto aq^{-2n},e\mapsto eq^{-2n},f\mapsto fq^{-2n}$とすると, 条件は$a^2q=bcdef$のままで
\begin{align} &\Q87{aq^{-2n},\sqrt aq^{1-n},-\sqrt aq^{1-n},b,c,d,eq^{-2n},fq^{-2n}}{\sqrt aq^{-n},-\sqrt aq^{-n},aq^{1-2n}/b,aq^{1-2n}/c,aq^{1-2n}/d,aq/e,aq/f}q\\ &=\frac baq^{2n}\frac{(aq^{1-2n},bq^{2n+1}/a,bq/c,bq/d,bq^{2n+1}/e,bq^{2n+1}/f,c,d,eq^{-2n},fq^{-2n};q)_{\infty}}{(aq^{1-2n}/b,aq^{1-2n}/c,aq^{1-2n}/d,aq/e,aq/f,bcq^{2n}/a,bdq^{2n}/a,be/a,bf/a,b^2q^{2n+1}/a;q)_{\infty}}\\ &\qquad\cdot\Q87{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,be/a,bf/a}{\sqrt{\frac{b^2}a}q^n,-\sqrt{\frac{b^2}a}q^n,bq^{2n+1}/a,bq/c,bq/d,bq^{2n}/e,bq^{2n}/f}q\\ &\qquad+\frac{(aq^{1-2n},aq^{1-2n}/cd,aq/ce,aq/cf,aq/de,aq/df,aq^{2n+1}/ef,bq^{2n}/a;q)_{\infty}}{(aq^{1-2n}/c,aq^{1-2n}/d,aq/e,aq/f,bcq^{2n}/a,bdq^{2n}/a,be/a,bf/a;q)_{\infty}} \end{align}
が成り立つ. 左辺は
\begin{align} &\Q87{aq^{-2n},\sqrt aq^{1-n},-\sqrt aq^{1-n},b,c,d,eq^{-2n},fq^{-2n}}{\sqrt aq^{-n},-\sqrt aq^{-n},aq^{1-2n}/b,aq^{1-2n}/c,aq^{1-2n}/d,aq/e,aq/f}q\\ &=\sum_{k=0}^{n-1}\frac{(aq^{-2n},\sqrt aq^{1-n},-\sqrt aq^{1-n},b,c,d,eq^{-2n},fq^{-2n};q)_k}{(\sqrt aq^{-n},-\sqrt aq^{-n},aq^{1-2n}/b,aq^{1-2n}/c,aq^{1-2n}/d,aq/e,aq/f,q;q)_k}q^k\\ &\qquad+\sum_{k=-n}^{\infty}\frac{(aq^{-2n},\sqrt aq^{1-n},-\sqrt aq^{1-n},b,c,d,eq^{-2n},fq^{-2n};q)_{k+2n}}{(\sqrt aq^{-n},-\sqrt aq^{-n},aq^{1-2n}/b,aq^{1-2n}/c,aq^{1-2n}/d,aq/e,aq/f,q;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,eq^{-2n},fq^{-2n};q)_k}{(\sqrt aq^{-n},-\sqrt aq^{-n},aq^{1-2n}/b,aq^{1-2n}/c,aq^{1-2n}/d,aq/e,aq/f,q;q)_k}q^k\\ &\qquad+\frac{1-aq^{2n}}{q^{2n}-a}\frac{(q/a,b,c,d,q/e,q/f;q)_{2n}}{(b/a,c/a,d/a,aq/e,aq/f,q;q)_{2n}}\\ &\qquad\cdot\sum_{k=-n}^{\infty}\frac{(a,\sqrt aq^{1+n},-\sqrt aq^{1+n},bq^{2n},cq^{2n},dq^{2n},e,f;q)_{k}}{(\sqrt aq^{n},-\sqrt aq^{n},aq/b,aq/c,aq/d,aq^{2n+1}/e,aq^{2n+1}/f,q^{2n+1};q)_{k}}q^{k}\\ &\to \Q32{b,c,d}{aq/e,aq/f}q\\ &\qquad-\frac{1}{a}\frac{(q/a,b,c,d,q/e,q/f;q)_{\infty}}{(b/a,c/a,d/a,aq/e,aq/f,q;q)_{\infty}}\BQ33{a,e,f}{aq/b,aq/c,aq/d}q\qquad n\to\infty \end{align}
となる. 右辺第1項は
\begin{align} &\frac baq^{2n}\frac{(aq^{1-2n},bq^{2n+1}/a,bq/c,bq/d,bq^{2n+1}/e,bq^{2n+1}/f,c,d,eq^{-2n},fq^{-2n};q)_{\infty}}{(aq^{1-2n}/b,aq^{1-2n}/c,aq^{1-2n}/d,aq/e,aq/f,bcq^{2n}/a,bdq^{2n}/a,be/a,bf/a,b^2q^{2n+1}/a;q)_{\infty}}\\ &\qquad\cdot\Q87{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,be/a,bf/a}{\sqrt{\frac{b^2}a}q^n,-\sqrt{\frac{b^2}a}q^n,bq^{2n+1}/a,bq/c,bq/d,bq^{2n}/e,bq^{2n}/f}q\\ &=\frac baq^{2n}\frac{(aq^{1-2n},eq^{-2n},fq^{-2n};q)_{2n}}{(aq^{1-2n}/b,aq^{1-2n}/c,aq^{1-2n}/d;q)_{2n}}\frac{(aq,bq^{2n+1}/a,bq/c,bq/d,bq^{2n+1}/e,bq^{2n+1}/f,c,d,e,f;q)_{\infty}}{(aq/b,aq/c,aq/d,aq/e,aq/f,bcq^{2n}/a,bdq^{2n}/a,be/a,bf/a,b^2q^{2n+1}/a;q)_{\infty}}\\ &\qquad\cdot\Q87{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,be/a,bf/a}{\sqrt{\frac{b^2}a}q^n,-\sqrt{\frac{b^2}a}q^n,bq^{2n+1}/a,bq/c,bq/d,bq^{2n}/e,bq^{2n}/f}q\\ &=\frac ba\frac{(1/a,q/e,q/f;q)_{2n}}{(b/a,c/a,d/a;q)_{2n}}\frac{(aq,bq^{2n+1}/a,bq/c,bq/d,bq^{2n+1}/e,bq^{2n+1}/f,c,d,e,f;q)_{\infty}}{(aq/b,aq/c,aq/d,aq/e,aq/f,bcq^{2n}/a,bdq^{2n}/a,be/a,bf/a,b^2q^{2n+1}/a;q)_{\infty}}\\ &\qquad\cdot\Q87{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,be/a,bf/a}{\sqrt{\frac{b^2}a}q^n,-\sqrt{\frac{b^2}a}q^n,bq^{2n+1}/a,bq/c,bq/d,bq^{2n}/e,bq^{2n}/f}q\\ &\to \frac ba\frac{(1/a,q/e,q/f;q)_{\infty}}{(b/a,c/a,d/a;q)_{\infty}}\frac{(aq,bq/c,bq/d,c,d,e,f;q)_{\infty}}{(aq/b,aq/c,aq/d,aq/e,aq/f,be/a,bf/a;q)_{\infty}}\Q32{b,be/a,bf/a}{bq/c,bq/d}q\qquad n\to\infty \end{align}
右辺第2項は
\begin{align} &\frac{(aq^{1-2n},aq^{1-2n}/cd,aq/ce,aq/cf,aq/de,aq/df,aq^{2n+1}/ef,bq^{2n}/a;q)_{\infty}}{(aq^{1-2n}/c,aq^{1-2n}/d,aq/e,aq/f,bcq^{2n}/a,bdq^{2n}/a,be/a,bf/a;q)_{\infty}}\\ &=\frac{(aq^{1-2n},aq^{1-2n}/cd;q)_{2n}}{(aq^{1-2n}/c,aq^{1-2n}/d;q)_{2n}}\frac{(aq,aq/cd,aq/ce,aq/cf,aq/de,aq/df,aq^{2n+1}/ef,bq^{2n}/a;q)_{\infty}}{(aq/c,aq/d,aq/e,aq/f,bcq^{2n}/a,bdq^{2n}/a,be/a,bf/a;q)_{\infty}}\\ &=\frac{(1/a,cd/a;q)_{2n}}{(c/a,d/a;q)_{2n}}\frac{(aq,aq/cd,aq/ce,aq/cf,aq/de,aq/df,aq^{2n+1}/ef,bq^{2n}/a;q)_{\infty}}{(aq/c,aq/d,aq/e,aq/f,bcq^{2n}/a,bdq^{2n}/a,be/a,bf/a;q)_{\infty}}\\ &\to \frac{(1/a,cd/a;q)_{\infty}}{(c/a,d/a;q)_{\infty}}\frac{(aq,aq/cd,aq/ce,aq/cf,aq/de,aq/df;q)_{\infty}}{(aq/c,aq/d,aq/e,aq/f,be/a,bf/a;q)_{\infty}}\qquad n\to\infty \end{align}
となる. よって, これらより,
\begin{align} &\Q32{b,c,d}{aq/e,aq/f}q\\ &\qquad-\frac{1}{a}\frac{(q/a,b,c,d,q/e,q/f;q)_{\infty}}{(b/a,c/a,d/a,aq/e,aq/f,q;q)_{\infty}}\BQ33{a,e,f}{aq/b,aq/c,aq/d}q\\ &=\frac ba\frac{(1/a,q/e,q/f;q)_{\infty}}{(b/a,c/a,d/a;q)_{\infty}}\frac{(aq,bq/c,bq/d,c,d,e,f;q)_{\infty}}{(aq/b,aq/c,aq/d,aq/e,aq/f,be/a,bf/a;q)_{\infty}}\Q32{b,be/a,bf/a}{bq/c,bq/d}q\\ &\qquad+\frac{(1/a,cd/a;q)_{\infty}}{(c/a,d/a;q)_{\infty}}\frac{(aq,aq/cd,aq/ce,aq/cf,aq/de,aq/df;q)_{\infty}}{(aq/c,aq/d,aq/e,aq/f,be/a,bf/a;q)_{\infty}} \end{align}
を得る. ${}_3\psi_3$について表すと,
\begin{align} &\BQ33{a,e,f}{aq/b,aq/c,aq/d}q\\ &=a\frac{(b/a,c/a,d/a,aq/e,aq/f,q;q)_{\infty}}{(q/a,b,c,d,q/e,q/f;q)_{\infty}}\Q32{b,c,d}{aq/e,aq/f}q\\ &\qquad-a\frac{(b/a,c/a,d/a,aq/e,aq/f,q;q)_{\infty}}{(q/a,b,c,d,q/e,q/f;q)_{\infty}}\frac ba\frac{(1/a,q/e,q/f;q)_{\infty}}{(b/a,c/a,d/a;q)_{\infty}}\frac{(aq,bq/c,bq/d,c,d,e,f;q)_{\infty}}{(aq/b,aq/c,aq/d,aq/e,aq/f,be/a,bf/a;q)_{\infty}}\Q32{b,be/a,bf/a}{bq/c,bq/d}q\\ &\qquad-a\frac{(b/a,c/a,d/a,aq/e,aq/f,q;q)_{\infty}}{(q/a,b,c,d,q/e,q/f;q)_{\infty}}\frac{(1/a,cd/a;q)_{\infty}}{(c/a,d/a;q)_{\infty}}\frac{(aq,aq/cd,aq/ce,aq/cf,aq/de,aq/df;q)_{\infty}}{(aq/c,aq/d,aq/e,aq/f,be/a,bf/a;q)_{\infty}}\\ &=a\frac{(b/a,c/a,d/a,aq/e,aq/f,q;q)_{\infty}}{(q/a,b,c,d,q/e,q/f;q)_{\infty}}\Q32{b,c,d}{aq/e,aq/f}q\\ &\qquad+\frac ba\frac{(q,a,bq/c,bq/d,e,f;q)_{\infty}}{(b,aq/b,aq/c,aq/d,be/a,bf/a;q)_{\infty}}\Q32{b,be/a,bf/a}{bq/c,bq/d}q\\ &\qquad+\frac{(q,b/a,cd/a,a,aq/cd,aq/ce,aq/cf,aq/de,aq/df;q)_{\infty}}{(b,c,d,q/e,q/f,aq/c,aq/d,be/a,bf/a;q)_{\infty}} \end{align}
となって示すべき等式が得られる.