Verma-Jainによるnon-terminating nearly-poisedの変換公式
\begin{align}
W(a;b_1,\dots,b_r;z):=\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}{z}
\end{align}
とする. 今回は以下を示す.
\begin{align} &W\left(a;b,x,-x,y,-y,z,-z;-\frac{a^3q^3}{bx^2y^2z^2}\right)\\ &=\frac{(a^2q^2,a^2q^2/x^2y^2,a^2q^2/x^2z^2,a^2q^2/y^2z^2;q^2)_{\infty}}{(a^2q^2/x^2,a^2q^2/y^2,a^2q^2/z^2,a^2q^2/x^2y^2z^2;q^2)_{\infty}}\Q54{x^2,y^2,z^2,-aq/b,-aq^2/b}{x^2y^2z^2/a^2,a^2q^2/b^2,-aq,-aq^2}{q^2;q^2}\\ &\qquad+\frac{(x^2,y^2,z^2,a^4q^4/b^2x^2y^2z^2;q^2)_{\infty}}{(a^2q^2/x^2,a^2q^2/y^2,a^2q^2/z^2,x^2y^2z^2/a^2q^2;q^2)_{\infty}}\frac{(aq,-a^3q^{3}/x^2y^2z^2;q)_{\infty}}{(aq/b,-a^3q^{3}/bx^2y^2z^2;q)_{\infty}}\\ &\qquad\cdot\Q54{a^2q^2/x^2y^2,a^2q^2/x^2z^2,a^2q^2/y^2z^2,-a^3q^3/bx^2y^2z^2,-a^3q^4/bx^2y^2z^2}{a^2q^4/x^2y^2z^2,a^4q^4/b^2x^2y^2z^2,-a^3q^3/x^2y^2z^2,-a^3q^4/x^2y^2z^2}{q^2;q^2} \end{align}
Non-terminating $q$-Saalschützの和公式
\begin{align}
&\frac{(aq,aq/fg,aq/fh,aq/gh;q)_{\infty}}{(f,g,h,aq/fgh;q)_{\infty}}\Q32{f,g,h}{aq,fgh/a}q\\
&\qquad+\frac{(a^2q^2/fgh;q)_{\infty}}{(fgh/aq;q)_{\infty}}\Q32{aq/fg,aq/fh,aq/gh}{aq^2/fgh,a^2q^2/fgh}q=\frac{(aq/f,aq/g,aq/h;q)_{\infty}}{(f,g,h;q)_{\infty}}
\end{align}
において, $q\mapsto q^2,a\mapsto a^2q^{4r},f\mapsto x^2q^{2r},g\mapsto y^2q^{2r},h\mapsto z^2q^{2r}$とすると,
\begin{align}
&\frac{(a^2q^{4r+2},a^2q^2/x^2y^2,a^2q^2/x^2z^2,a^2q^2/y^2z^2;q^2)_{\infty}}{(x^2q^{2r},y^2q^{2r},z^2q^{2r},a^2q^{2-2r}/x^2y^2z^2;q^2)_{\infty}}\Q32{x^2q^{2r},y^2q^{2r},z^2q^{2r}}{a^2q^{4r+2},x^2y^2z^2q^{2r}/a^2}{q^2;q^2}\\
&\qquad+\frac{(a^4q^{2r+4}/x^2y^2z^2;q^2)_{\infty}}{(x^2y^2z^2q^{2r-2}/a^2;q^2)_{\infty}}\Q32{a^2q^2/x^2y^2,a^2q^2/x^2z^2,a^2q^2/y^2z^2}{a^2q^{4-2r}/x^2y^2z^2,a^4q^{4+2r}/x^2y^2z^2}{q^2;q^2}=\frac{(a^2q^{2r+2}/x^2,a^2q^{2r+2}/y^2,a^2q^{2r+2}/z^2;q^2)_{\infty}}{(x^2q^{2r},y^2q^{2r},z^2q^{2r};q^2)_{\infty}}
\end{align}
を得る. この両辺に
\begin{align}
\frac{1-aq^{2r}}{1-a}\frac{(a,b;q)_r}{(q,aq/b;q)_r}\left(-\frac{a^3q^3}{bx^2y^2z^2}\right)^r
\end{align}
を掛けて足し合わせると, 右辺は
\begin{align}
\frac{(a^2q^2/x^2,a^2q^2/y^2,a^2q^2/z^2;q^2)_{\infty}}{(x^2,y^2,z^2;q^2)_{\infty}}W\left(a;b,x,-x,y,-y,z,-z;-\frac{a^3q^3}{bx^2y^2z^2}\right)
\end{align}
となる. 左辺は
\begin{align}
&\sum_{0\leq r}\frac{1-aq^{2r}}{1-a}\frac{(a,b;q)_r}{(q,aq/b;q)_r}\left(-\frac{a^3q^3}{bx^2y^2z^2}\right)^r\frac{(a^2q^{4r+2},a^2q^2/x^2y^2,a^2q^2/x^2z^2,a^2q^2/y^2z^2;q^2)_{\infty}}{(x^2q^{2r},y^2q^{2r},z^2q^{2r},a^2q^{2-2r}/x^2y^2z^2;q^2)_{\infty}}\Q32{x^2q^{2r},y^2q^{2r},z^2q^{2r}}{a^2q^{4r+2},x^2y^2z^2q^{2r}/a^2}{q^2;q^2}\\
&\qquad+\sum_{0\leq r}\frac{1-aq^{2r}}{1-a}\frac{(a,b;q)_r}{(q,aq/b;q)_r}\left(-\frac{a^3q^3}{bx^2y^2z^2}\right)^r\frac{(a^4q^{2r+4}/x^2y^2z^2;q^2)_{\infty}}{(x^2y^2z^2q^{2r-2}/a^2;q^2)_{\infty}}\Q32{a^2q^2/x^2y^2,a^2q^2/x^2z^2,a^2q^2/y^2z^2}{a^2q^{4-2r}/x^2y^2z^2,a^4q^{4+2r}/x^2y^2z^2}{q^2;q^2}
\end{align}
この第1項は
\begin{align}
&\sum_{0\leq r}\frac{1-aq^{2r}}{1-a}\frac{(a,b;q)_r}{(q,aq/b;q)_r}\left(-\frac{a^3q^3}{bx^2y^2z^2}\right)^r\frac{(a^2q^{4r+2},a^2q^2/x^2y^2,a^2q^2/x^2z^2,a^2q^2/y^2z^2;q^2)_{\infty}}{(x^2q^{2r},y^2q^{2r},z^2q^{2r},a^2q^{2-2r}/x^2y^2z^2;q^2)_{\infty}}\Q32{x^2q^{2r},y^2q^{2r},z^2q^{2r}}{a^2q^{4r+2},x^2y^2z^2q^{2r}/a^2}{q^2;q^2}\\
&=\frac{(a^2q^2,a^2q^2/x^2y^2,a^2q^2/x^2z^2,a^2q^2/y^2z^2;q^2)_{\infty}}{(x^2,y^2,z^2,a^2q^2/x^2y^2z^2;q^2)_{\infty}}\\
&\qquad\cdot\sum_{0\leq r}\frac{1-aq^{2r}}{1-a}\frac{(a,b;q)_r}{(q,aq/b;q)_r}\left(\frac{aq^3}{b}\right)^r\frac {q^{r(r-1)}(x^2,y^2,z^2;q^2)_r}{(x^2y^2z^2/a^2;q^2)_r(a^2q^2;q^2)_{2r}}\sum_{0\leq k}\frac{(x^2q^{2r},y^2q^{2r},z^2q^{2r};q^2)_k}{(q^2,a^2q^{4r+2},x^2y^2z^2q^{2r}/a^2;q^2)_k}q^{2k}\\
&=\frac{(a^2q^2,a^2q^2/x^2y^2,a^2q^2/x^2z^2,a^2q^2/y^2z^2;q^2)_{\infty}}{(x^2,y^2,z^2,a^2q^2/x^2y^2z^2;q^2)_{\infty}}\\
&\qquad\cdot\sum_{0\leq r}\frac{1-aq^{2r}}{1-a}\frac{(a,b;q)_r}{(q,aq/b;q)_r}\left(\frac{aq^{r}}{b}\right)^r\sum_{r\leq k}\frac{(x^2,y^2,z^2;q^2)_{k}}{(q^2;q^2)_{k-r}(a^2q^2;q^2)_{k+r}(x^2y^2z^2/a^2;q^2)_{k}}q^{2k}\\
&=\frac{(a^2q^2,a^2q^2/x^2y^2,a^2q^2/x^2z^2,a^2q^2/y^2z^2;q^2)_{\infty}}{(x^2,y^2,z^2,a^2q^2/x^2y^2z^2;q^2)_{\infty}}\\
&\qquad\cdot\sum_{0\leq k}\frac{(x^2,y^2,z^2;q^2)_{k}}{(q^2,a^2q^2,x^2y^2z^2/a^2;q^2)_{k}}q^{2k}\sum_{0\leq r}\frac{1-aq^{2r}}{1-a}\frac{(a,b,-q^{-k},q^{-k};q)_r}{(q,aq/b,aq^{k+1},-aq^{k+1};q)_r}\left(-\frac{aq^{2k+1}}{b}\right)^r
\end{align}
ここで,
Rogersの${}_6\phi_5$和公式
より
\begin{align}
\sum_{0\leq r}\frac{1-aq^{2r}}{1-a}\frac{(a,b,-q^{-k},q^{-k};q)_r}{(q,aq/b,-aq^{k+1},aq^{k+1};q)_r}\left(-\frac{aq^{2k+1}}{b}\right)^r&=\frac{(aq,-aq^{k+1}/b;q)_k}{(aq/b,-aq^{k+1};q)_k}\\
&=\frac{(a^2q^2,-aq/b,-aq^2/b;q^2)_k}{(a^2q^2/b^2,-aq,-aq^2;q^2)_k}
\end{align}
となるからこれを代入すると
\begin{align}
&\frac{(a^2q^2,a^2q^2/x^2y^2,a^2q^2/x^2z^2,a^2q^2/y^2z^2;q^2)_{\infty}}{(x^2,y^2,z^2,a^2q^2/x^2y^2z^2;q^2)_{\infty}}\\
&\qquad\cdot\sum_{0\leq k}\frac{(x^2,y^2,z^2;q^2)_{k}}{(q^2,a^2q^2,x^2y^2z^2/a^2;q^2)_{k}}q^{2k}\sum_{0\leq r}\frac{1-aq^{2r}}{1-a}\frac{(a,b,-q^{-k},q^{-k};q)_r}{(q,aq/b,aq^{k+1},-aq^{k+1};q)_r}\left(-\frac{aq^{2k+1}}{b}\right)^r\\
&=\frac{(a^2q^2,a^2q^2/x^2y^2,a^2q^2/x^2z^2,a^2q^2/y^2z^2;q)_{\infty}}{(x^2,y^2,z^2,a^2q^2/x^2y^2z^2;q^2)_{\infty}}\Q54{x^2,y^2,z^2,-aq/b,-aq^2/b}{x^2y^2z^2/a^2,a^2q^2/b^2,-aq,-aq^2}{q^2;q^2}
\end{align}
を得る. 次に先ほどの第2項は
\begin{align}
&\sum_{0\leq r}\frac{1-aq^{2r}}{1-a}\frac{(a,b;q)_r}{(q,aq/b;q)_r}\left(-\frac{a^3q^3}{bx^2y^2z^2}\right)^r\frac{(a^4q^{2r+4}/x^2y^2z^2;q^2)_{\infty}}{(x^2y^2z^2q^{2r-2}/a^2;q^2)_{\infty}}\Q32{a^2q^2/x^2y^2,a^2q^2/x^2z^2,a^2q^2/y^2z^2}{a^2q^{4-2r}/x^2y^2z^2,a^4q^{4+2r}/x^2y^2z^2}{q^2;q^2}\\
&=\frac{(a^4q^4/x^2y^2z^2;q^2)_{\infty}}{(x^2y^2z^2/a^2q^2;q^2)_{\infty}}\sum_{0\leq r}\frac{1-aq^{2r}}{1-a}\frac{(a,b;q)_r}{(q,aq/b;q)_r}\left(-\frac{a^3q^3}{bx^2y^2z^2}\right)^r\frac{(x^2y^2z^2/a^2q^2;q^2)_r}{(a^4q^4/x^2y^2z^2;q^2)_{2r}}\\
&\qquad\cdot\sum_{0\leq k}\frac{(a^2q^2/x^2y^2,a^2q^2/x^2z^2,a^2q^2/y^2z^2;q^2)_k}{(q^2,a^2q^{4-2r}/x^2y^2z^2,a^4q^{4+2r}/x^2y^2z^2;q^2)_k}q^{2k}\\
&=\frac{(a^4q^4/x^2y^2z^2;q^2)_{\infty}}{(x^2y^2z^2/a^2q^2;q^2)_{\infty}}\sum_{0\leq k}\frac{(a^2q^2/x^2y^2,a^2q^2/x^2z^2,a^2q^2/y^2z^2;q^2)_k}{(q^2;q^2)_k}q^{2k}\\
&\qquad\cdot\sum_{0\leq r}\frac{1-aq^{2r}}{1-a}\frac{(a,b;q)_r}{(q,aq/b;q)_r}\left(\frac{aq^r}b\right)^r\frac{1}{(a^2q^4/x^2y^2z^2;q^2)_{k-r}(a^4q^4/x^2y^2z^2;q^2)_{k+r}}\\
&=\frac{(a^4q^4/x^2y^2z^2;q^2)_{\infty}}{(x^2y^2z^2/a^2q^2;q^2)_{\infty}}\sum_{0\leq k}\frac{(a^2q^2/x^2y^2,a^2q^2/x^2z^2,a^2q^2/y^2z^2;q^2)_k}{(q^2,a^2q^4/x^2y^2z^2,a^4q^4/x^2y^2z^2;q^2)_k}q^{2k}\\
&\qquad\cdot\sum_{0\leq r}\frac{1-aq^{2r}}{1-a}\frac{(a,b,xyzq^{-k-1}/a,-xyzq^{-k-1}/a;q)_r}{(q,aq/b,a^2q^{k+2}/xyz,-a^2q^{k+2}/xyz;q)_r}\left(-\frac{a^3q^{2k+3}}{bx^2y^2z^2}\right)^r
\end{align}
ここで,
Rogersの${}_6\phi_5$和公式
より
\begin{align}
&\sum_{0\leq r}\frac{1-aq^{2r}}{1-a}\frac{(a,b,xyzq^{-k-1}/a,-xyzq^{-k-1}/a;q)_r}{(q,aq/b,a^2q^{k+2}/xyz,-a^2q^{k+2}/xyz;q)_r}\left(-\frac{a^3q^{2k+3}}{bx^2y^2z^2}\right)^r\\
&=\frac{(aq,a^2q^{k+2}/bxyz,a^2q^{k+2}/bxyz,-a^3q^{2k+3}/x^2y^2z^2;q)_{\infty}}{(aq/b,a^2q^{k+2}/xyz,-a^2q^{k+2}/xyz,-a^3q^{2k+3}/bx^2y^2z^2;q)_{\infty}}\\
&=\frac{(a^4q^4/b^2x^2y^2z^2;q^2)_{\infty}(aq,-a^3q^{3}/x^2y^2z^2;q)_{\infty}}{(a^4q^4/x^2y^2z^2;q^2)_{\infty}(aq/b,-a^3q^{3}/bx^2y^2z^2;q)_{\infty}}\frac{(a^4q^4/x^2y^2z^2;q^2)_k(-a^3q^3/bx^2y^2z^2;q)_{2k}}{(a^4q^4/b^2x^2y^2z^2;q^2)_k(-a^3q^3/x^2y^2z^2;q)_{2k}}
\end{align}
であるから, これを代入して,
\begin{align}
&\frac{(a^4q^4/x^2y^2z^2;q^2)_{\infty}}{(x^2y^2z^2/a^2q^2;q^2)_{\infty}}\sum_{0\leq k}\frac{(a^2q^2/x^2y^2,a^2q^2/x^2z^2,a^2q^2/y^2z^2;q^2)_k}{(q^2,a^2q^4/x^2y^2z^2,a^4q^4/x^2y^2z^2;q^2)_k}q^{2k}\\
&\qquad\cdot\sum_{0\leq r}\frac{1-aq^{2r}}{1-a}\frac{(a,b,xyzq^{-k-1}/a,-xyzq^{-k-1}/a;q)_r}{(q,aq/b,a^2q^{k+2}/xyz,-a^2q^{k+2}/xyz;q)_r}\left(-\frac{a^3q^{2k+3}}{bx^2y^2z^2}\right)^r\\
&=\frac{(a^4q^4/b^2x^2y^2z^2;q^2)_{\infty}}{(x^2y^2z^2/a^2q^2;q^2)_{\infty}}\frac{(aq,-a^3q^{3}/x^2y^2z^2;q)_{\infty}}{(aq/b,-a^3q^{3}/bx^2y^2z^2;q)_{\infty}}\\
&\qquad\cdot\Q54{a^2q^2/x^2y^2,a^2q^2/x^2z^2,a^2q^2/y^2z^2,-a^3q^3/bx^2y^2z^2,-a^3q^4/bx^2y^2z^2}{a^2q^4/x^2y^2z^2,a^4q^4/bx^2y^2z^2,-a^3q^3/x^2y^2z^2,-a^3q^4/x^2y^2z^2}{q^2;q^2}
\end{align}
を得る.
まとめると
\begin{align}
&\frac{(a^2q^2/x^2,a^2q^2/y^2,a^2q^2/z^2;q^2)_{\infty}}{(x^2,y^2,z^2;q^2)_{\infty}}W\left(a;b,x,-x,y,-y,z,-z;-\frac{a^3q^3}{bx^2y^2z^2}\right)\\
&=\frac{(a^2q^2,a^2q^2/x^2y^2,a^2q^2/x^2z^2,a^2q^2/y^2z^2;q^2)_{\infty}}{(x^2,y^2,z^2,a^2q^2/x^2y^2z^2;q^2)_{\infty}}\Q54{x^2,y^2,z^2,-aq/b,-aq^2/b}{x^2y^2z^2/a^2,a^2q^2/b^2,-aq,-aq^2}{q^2;q^2}\\
&\qquad+\frac{(a^4q^4/b^2x^2y^2z^2;q^2)_{\infty}}{(x^2y^2z^2/a^2q^2;q^2)_{\infty}}\frac{(aq,-a^3q^{3}/x^2y^2z^2;q)_{\infty}}{(aq/b,-a^3q^{3}/bx^2y^2z^2;q)_{\infty}}\\
&\qquad\cdot\Q54{a^2q^2/x^2y^2,a^2q^2/x^2z^2,a^2q^2/y^2z^2,-a^3q^3/bx^2y^2z^2,-a^3q^4/bx^2y^2z^2}{a^2q^4/x^2y^2z^2,a^4q^4/bx^2y^2z^2,-a^3q^3/x^2y^2z^2,-a^3q^4/x^2y^2z^2}{q^2;q^2}
\end{align}
となって定理を得る.
この左辺は
\begin{align}
&W\left(a;b,x,-x,y,-y,z,-z;-\frac{a^3q^3}{bx^2y^2z^2}\right)\\
&=\sum_{0\leq n}\frac{(1-aq^{2n})(a,b;q)_n(x^2,y^2,z^2;q^2)_n}{(1-a)(q,aq/b;q)_n(a^2q^2/x^2,a^2q^2/y^2,a^2q^2/z^2;q^2)_n}\left(-\frac{a^3q^3}{bx^2y^2z^2}\right)^n\\
&=\sum_{0\leq n}\frac{(a,b;q)_n(aq^2,x^2,y^2,z^2;q^2)_n}{(q,aq/b;q)_n(a,a^2q^2/x^2,a^2q^2/y^2,a^2q^2/z^2;q^2)_n}\left(-\frac{a^3q^3}{bx^2y^2z^2}\right)^n\\
\end{align}
と表すこともできる. $N$を非負整数として, 定理1において$z=q^{-N}$とすると, 第二項が消えて以下の系を得る.
$N$を非負整数とするとき,
\begin{align}
&W\left(a;b,x,-x,y,-y,q^{-N},-q^{-N};-\frac{a^3q^{2N+3}}{bx^2y^2}\right)\\
&=\frac{(a^2q^2,a^2q^2/x^2y^2;q^2)_N}{(a^2q^2/x^2,a^2q^2/y^2;q^2)_N}\Q54{x^2,y^2,q^{-2N},-aq/b,-aq^2/b}{x^2y^2q^{-2N}/a^2,a^2q^2/b^2,-aq,-aq^2}{q^2;q^2}
\end{align}
が成り立つ.
これは
\begin{align}
&\sum_{n=0}^N\frac{(a,b;q)_n}{(q,aq/b;q)_n}\frac{(aq^2,x^2,y^2,q^{-2N};q^2)_n}{(a,a^2q^2/x^2,a^2q^2/y^2,a^2q^{2N+2};q^2)_n}\left(-\frac{a^3q^{2N+3}}{bx^2y^2}\right)^n\\
&=\frac{(a^2q^2,a^2q^2/x^2y^2;q^2)_N}{(a^2q^2/x^2,a^2q^2/y^2;q^2)_N}\Q54{x^2,y^2,q^{-2N},-aq/b,-aq^2/b}{x^2y^2q^{-2N}/a^2,a^2q^2/b^2,-aq,-aq^2}{q^2;q^2}
\end{align}
と書き換えられる. 両辺の和を$n\to N-n$として逆順にしてから変数を付け替えることによって, これは
terminatingな場合のnearly-poised変換公式
に一致することが分かる. また系1の右辺は
Carlitzの和公式
\begin{align}
\Q43{q^{-2n},e^{2}q^{2n-2},c,cq}{c^2,e,eq}{q^2;q^2}&=\frac{(-q,e/c;q)_n}{(-c,e;q)_n}c^n\frac{1-eq^{n-1}}{1-eq^{2n-1}}
\end{align}
に現れる${}_4\phi_3$の一般化になっているようである.
定理1において$z=a/x$とすると
\begin{align}
&W\left(a;b,x,-x,y,-y,a/x,-a/x;-\frac{aq^3}{by^2}\right)\\
&=\frac{(a^2q^2,a^2q^2/x^2y^2,q^2,x^2q^2/y^2;q^2)_{\infty}}{(a^2q^2/x^2,a^2q^2/y^2,x^2q^2,q^2/y^2;q^2)_{\infty}}\Q43{x^2,a^2/x^2,-aq/b,-aq^2/b}{a^2q^2/b^2,-aq,-aq^2}{q^2;q^2}\\
&\qquad+\frac{(x^2,y^2,a^2/x^2,a^2q^4/b^2y^2;q^2)_{\infty}}{(a^2q^2/x^2,a^2q^2/y^2,x^2q^2,y^2/q^2;q^2)_{\infty}}\frac{(aq,-aq^{3}/y^2;q)_{\infty}}{(aq/b,-aq^{3}/by^2;q)_{\infty}}\\
&\qquad\cdot\Q54{a^2q^2/x^2y^2,q^2,x^2q^2/y^2,-aq^3/by^2,-aq^4/by^2}{q^4/y^2,a^2q^4/b^2y^2,-aq^3/y^2,-aq^4/y^2}{q^2;q^2}
\end{align}
となる. 第一項の${}_4\phi_3$が$y$に依存していないのが面白いところである.
古典極限
そのまま定理1の古典極限を考えると右辺がwell-poised${}_6F_5$, 左辺が${}_3F_2$となるので, Whippleの変換公式を用いれば${}_3F_2$の三項変換公式になってしまうが, $a\mapsto -a$と置き換えた式
\begin{align}
&\Q{10}9{-a,\sqrt{-a}q,-\sqrt{-a}q,b,x,-x,y,-y,z,-z}{\sqrt{-a},-\sqrt{-a},-aq/b,-aq/x,aq/x,-aq/y,aq/y,-aq/z,aq/z}{\frac{a^3q^3}{bx^2y^2z^2}}\\
&=\frac{(a^2q^2,a^2q^2/x^2y^2,a^2q^2/x^2z^2,a^2q^2/y^2z^2;q^2)_{\infty}}{(a^2q^2/x^2,a^2q^2/y^2,a^2q^2/z^2,a^2q^2/x^2y^2z^2;q^2)_{\infty}}\Q54{x^2,y^2,z^2,aq/b,aq^2/b}{x^2y^2z^2/a^2,a^2q^2/b^2,aq,aq^2}{q^2;q^2}\\
&\qquad+\frac{(x^2,y^2,z^2,a^4q^4/b^2x^2y^2z^2;q^2)_{\infty}}{(a^2q^2/x^2,a^2q^2/y^2,a^2q^2/z^2,x^2y^2z^2/a^2q^2;q^2)_{\infty}}\frac{(aq,a^3q^{3}/x^2y^2z^2;q)_{\infty}}{(aq/b,a^3q^{3}/bx^2y^2z^2;q)_{\infty}}\\
&\qquad\cdot\Q54{a^2q^2/x^2y^2,a^2q^2/x^2z^2,a^2q^2/y^2z^2,a^3q^3/bx^2y^2z^2,a^3q^4/bx^2y^2z^2}{a^2q^4/x^2y^2z^2,a^4q^4/bx^2y^2z^2,a^3q^3/x^2y^2z^2,a^3q^4/x^2y^2z^2}{q^2;q^2}
\end{align}
の古典極限を考えると以下を得る.
\begin{align} &\F43{b,x,y,z}{1+a-x,1+a-y,1+a-z}{1}\\ &=\frac{\Gamma(1+a-x)\Gamma(1+a-y)\Gamma(1+a-z)\Gamma(1+a-x-y-z)}{\Gamma(1+a)\Gamma(1+a-x-y)\Gamma(1+a-x-z)\Gamma(1+a-y-z)}\F54{x,y,z,\frac{1+a-b}2,\frac{2+a-b}2}{x+y+z-a,1+a-b,\frac{1+a}2,\frac{2+a}2}1\\ &\qquad+\frac{\Gamma(1+a-x)\Gamma(1+a-y)\Gamma(1+a-z)\Gamma(x+y+z-a-1)\Gamma(1+a-b)\Gamma(3+3a-b-2x-2y-2z)}{\Gamma(x)\Gamma(y)\Gamma(z)\Gamma(2+2a-b-x-y-z)\Gamma(1+a)\Gamma(3+3a-2x-2y-2z)}\\ &\qquad\cdot\F54{1+a-x-y,1+a-x-z,1+a-y-z,\frac{3+3a-b}{2}-x-y-z,\frac{4+3a-b}{2}-x-y-z}{2+a-x-y-z,2+2a-b-x-y-z,\frac{3+3a}2-x-y-z,\frac{4+3a}2-x-y-z}1 \end{align}
特に$z=-N$とすれば2つ目の項が消えて, 古典的なnearly-poisedの変換公式を得る.
