Rahmanによるwell-poised 2φ1の積公式

Heineの変換公式 より,
\begin{align} \Q21{a,b}{aq/b}{\frac{x^2q}b}&=\frac{(a,x^2q;q)_{\infty}}{(aq/b,x^2q/b;q)_{\infty}}\Q21{q/b,x^2q/b}{x^2q}a\\ \Q21{a,b}{aq/b}{\frac{y^2q}b}&=\frac{(aq/b^2,y^2q;q)_{\infty}}{(aq/b,y^2q/b;q)_{\infty}}\Q21{by^2,b}{y^2q}{\frac{aq}{b^2}} \end{align}
であるから,
\begin{align} &\Q21{a,b}{aq/b}{\frac{x^2q}b}\Q21{a,b}{aq/b}{\frac{y^2q}b}\\ &=B\Q21{q/b,x^2q/b}{x^2q}a\Q21{b,by^2}{y^2q}{\frac{aq}{b^2}}\\ &=B\sum_{0\leq k}\sum_{j=0}^k\frac{(q/b,x^2q/b;q)_{k-j}}{(x^2q,q;q)_{k-j}}a^{k-j}\frac{(b,by^2;q)_j}{(y^2q,q;q)_j}\left(\frac{aq}{b^2}\right)^j\\ &=B\sum_{0\leq k}\frac{(q/b,x^2q/b;q)_k}{(x^2q,q;q)_k}a^k\Q43{b,by^2,q^{-k}/x^2,q^{-k}}{y^2q,bq^{-k},bq^{-k}/x^2}{q} \end{align}
ここで, Watsonの変換公式 より,
\begin{align} &\Q43{b,by^2,q^{-k}/x^2,q^{-k}}{y^2q,bq^{-k},bq^{-k}/x^2}{q}\\ &=\frac{(x^2q,x^2y^2q;q)_k}{(bx^2y^2q,x^2q/b;q)_k}W(bx^2y^2;b,bx^2,by^2,x^2y^2q^{k+1},q^{-k};q/b) \end{align}
であるから,
\begin{align} &\Q21{a,b}{aq/b}{\frac{x^2q}b}\Q21{a,b}{aq/b}{\frac{y^2q}b}\\ &=B\sum_{0\leq k}\frac{(q/b,x^2y^2q;q)_k}{(bx^2y^2q,q;q)_k}a^kW(bx^2y^2;b,bx^2,by^2,x^2y^2q^{k+1},q^{-k};q/b)\\ &=B\sum_{0\leq k}\frac{(q/b,x^2y^2q;q)_k}{(bx^2y^2q,q;q)_k}a^k\sum_{0\leq j}\frac{(1-bx^2y^2q^{2j})(bx^2y^2,b,bx^2,by^2,x^2y^2q^{k+1},q^{-k};q)_j}{(1-bx^2y^2)(q,x^2y^2q,y^2q,x^2q,bq^{-k},bxyq^{k+1};q)_j}\left(\frac qb\right)^j\\ &=B\sum_{0\leq j}\frac{(1-bx^2y^2q^{2j})(bx^2y^2,b,bx^2,by^2;q)_j}{(1-bx^2y^2)(q,x^2y^2q,y^2q,x^2q;q)_j}\left(\frac{q}{b^2}\right)^j\sum_{0\leq k}\frac{(q/b;q)_{k-j}(x^2y^2q;q)_{j+k}}{(q;q)_{k-j}(bx^2y^2q;q)_{j+k}}a^k\\ &=B\sum_{0\leq j}\frac{(1-bx^2y^2q^{2j})(bx^2y^2,b,bx^2,by^2;q)_j(x^2y^2q;q)_{2j}}{(1-bx^2y^2)(q,x^2y^2q,y^2q,x^2q;q)_j(bx^2y^2q;q)_{2j}}\left(\frac{aq}{b^2}\right)^j\Q21{q/b,x^2y^2q^{2j+1}}{bx^2y^2q^{2j+1}}{a} \end{align}
ここで, Gasper-Rahmanの二次変換公式 と, non-terminating Watsonの変換公式 より
\begin{align} &\Q21{x^2y^2q^{2j+1},q/b}{bx^2y^2q^{2j+1}}{a}\\ &=\frac{(aq/b,a^2x^2y^2q^{2j+2}/b^2;q)_{\infty}}{(ax^2y^2q^{2j+2}/b,a^2q/b^2;q)_{\infty}}\\ &\qquad\cdot W(ax^2y^2q^{2j+1}/b;aq/b^2,xyq^{j+\frac 12},-xyq^{j+\frac 12},xyq^{j+1},-xyq^{j+1};a)\\ &=\frac{(aq/b,a^2x^2y^2q^{2j+2}/b^2;q)_{\infty}}{(ax^2y^2q^{2j+2}/b,a^2q/b^2;q)_{\infty}}\\ &\qquad\cdot\bigg(\frac{(ax^2y^2q^{2j+2}/b,-aq/b,a\sqrt q/b,-a\sqrt q/b;q)_{\infty}}{(axyq^{j+\frac 32}/b,-axyq^{j+\frac32}/b,-axyq^{j+1}/b,aq^{-j}/bxy;q)_{\infty}}\Q43{bxyq^{j},xyq^{j+\frac 12},-xyq^{j+\frac 12},-xyq^{j+1}}{bx^2y^2q^{2j+1},axyq^{j+1}/b,bxyq^{j+1}/a}{q}\\ &\qquad+\frac{(ax^2y^2q^{2j+2}/b,bxyq^j,xyq^{j+\frac 12},-xyq^{j+\frac 12},-xyq^{j+1},a^2q/b^2,axyq^{j+1};q)_{\infty}}{(bx^2y^2q^{2j+1},axyq^{j+1}/b,axyq^{j+\frac 32}/b,-axyq^{j+\frac 32}/b,-axyq^{j+1}/b,a,bxyq^j/a;q)_{\infty}}\Q43{a,a\sqrt q/b,-a\sqrt q/b,-aq/b}{axyq^{j+1},a^2q/b^2,aq^{1-j}/bxy}q\bigg)\\ &=\frac{(axyq^{j+1}/b;q)_{\infty}}{(aq^{-j}/bxy;q)_{\infty}}\Q43{bxyq^{j},xyq^{j+\frac 12},-xyq^{j+\frac 12},-xyq^{j+1}}{bx^2y^2q^{2j+1},axyq^{j+1}/b,bxyq^{j+1}/a}{q}\\ &\qquad+\frac{(aq/b,bxyq^j,x^2y^2q^{2j+1},axyq^{j+1};q)_{\infty}}{(xyq^{j+1},bx^2y^2q^{2j+1},a,bxyq^j/a;q)_{\infty}}\Q43{a,a\sqrt q/b,-a\sqrt q/b,-aq/b}{axyq^{j+1},a^2q/b^2,aq^{1-j}/bxy}q\\ &=\frac{(axyq/b;q)_{\infty}}{(a/bxy;q)_{\infty}}\frac{(-bxyq/a)^jq^{\binom j2}}{(bxyq/a,axyq/b;q)_{j}}\Q43{bxyq^{j},xyq^{j+\frac 12},-xyq^{j+\frac 12},-xyq^{j+1}}{bx^2y^2q^{2j+1},axyq^{j+1}/b,bxyq^{j+1}/a}{q}\\ &\qquad+\frac{(aq/b,bxy,x^2y^2q,axyq;q)_{\infty}}{(xyq,bx^2y^2q,a,bxy/a;q)_{\infty}}\frac{(xyq,bxy/a;q)_j(bx^2y^2q;q)_{2j}}{(bxy,axyq;q)_j(x^2y^2q;q)_{2j}}\Q43{a,a\sqrt q/b,-a\sqrt q/b,-aq/b}{axyq^{j+1},a^2q/b^2,aq^{1-j}/bxy}q\\ \end{align}
であるから, これを代入して,
\begin{align} &\Q21{a,b}{aq/b}{\frac{x^2q}b}\Q21{a,b}{aq/b}{\frac{y^2q}b}\\ &=B\frac{(axyq/b;q)_{\infty}}{(a/bxy;q)_{\infty}}\sum_{0\leq j}\frac{(1-bx^2y^2q^{2j})(bx^2y^2,b,bx^2,by^2;q)_j(x^2y^2q;q)_{2j}}{(1-bx^2y^2)(q,x^2y^2q,y^2q,x^2q;q)_j(bx^2y^2q;q)_{2j}}\left(\frac{aq}{b^2}\right)^j\\ &\qquad\cdot\frac{(-bxyq/a)^jq^{\binom j2}}{(bxyq/a,axyq/b;q)_{j}}\Q43{bxyq^{j},xyq^{j+\frac 12},-xyq^{j+\frac 12},-xyq^{j+1}}{bx^2y^2q^{2j+1},axyq^{j+1}/b,bxyq^{j+1}/a}{q}\\ &\qquad+B\frac{(aq/b,x^2y^2q,bxy,axyq;q)_{\infty}}{(a,bxyq,xyq,bxy/a;q)_{\infty}}\sum_{0\leq j}\frac{(1-bx^2y^2q^{2j})(bx^2y^2,b,bx^2,by^2;q)_j(x^2y^2q;q)_{2j}}{(1-bx^2y^2)(q,x^2y^2q,y^2q,x^2q;q)_j(bx^2y^2q;q)_{2j}}\left(\frac{aq}{b^2}\right)^j\\ &\qquad\cdot \frac{(xyq,bxy/a;q)_j(bx^2y^2q;q)_{2j}}{(bxy,axyq;q)_j(x^2y^2q;q)_{2j}}\Q43{a,a\sqrt q/b,-a\sqrt q/b,-aq/b}{axyq^{j+1},a^2q/b^2,aq^{1-j}/bxy}q\\ &=B\frac{(axyq/b;q)_{\infty}}{(a/bxy;q)_{\infty}}\sum_{0\leq j}\frac{(1-bx^2y^2q^{2j})(bx^2y^2,b,bx^2,by^2;q)_j}{(1-bx^2y^2)(q,x^2y^2q,y^2q,x^2q,xyq,bxy;q)_j}\left(-\frac{xyq^2}{b}\right)^jq^{\binom j2}\\ &\qquad\cdot \sum_{0\leq k}\frac{(bxy,xy\sqrt q,-xy\sqrt q,-xyq;q)_{j+k}}{(q;q)_k(axyq/b,bxyq/a;q)_{j+k}(bx^2y^2q;q)_{2j+k}}q^k\\ &\qquad+B\frac{(aq/b,x^2y^2q,bxy,axyq;q)_{\infty}}{(a,bxyq,xyq,bxy/a;q)_{\infty}}\sum_{0\leq j}\frac{(1-bx^2y^2q^{2j})(bx^2y^2,b,bx^2,by^2,xyq;q)_j}{(1-bx^2y^2)(q,x^2y^2q,y^2q,x^2q,bxy;q)_j}\left(\frac{aq}{b^2}\right)^j\\ &\qquad\cdot \sum_{0\leq k}\frac{(a,a\sqrt q/b,-a\sqrt q/b,-aq/b;q)_k(bxyq^{-k}/a;q)_j}{(q,a^2q/b^2,aq/bxy;q)_k(axyq;q)_{j+k}}q^{k+jk}\\ &=B\frac{(axyq/b;q)_{\infty}}{(a/bxy;q)_{\infty}}\sum_{0\leq k}\frac{(bxy,xy\sqrt q,-xy\sqrt q,-xyq;q)_{k}}{(q,axyq/b,bxyq/a,bx^2y^2q;q)_{k}}q^k\\ &\qquad\cdot \sum_{0\leq j}\frac{(1-bx^2y^2q^{2j})(bx^2y^2,b,bx^2,by^2,xyq,q^{-k};q)_j}{(1-bx^2y^2)(q,x^2y^2q,y^2q,x^2q,bxy,bx^2y^2q^{k+1};q)_j}\left(\frac{xyq^{k+1}}{b}\right)^j\\ &\qquad+B\frac{(aq/b,x^2y^2q,bxy,axyq;q)_{\infty}}{(a,bxyq,xyq,bxy/a;q)_{\infty}}\sum_{0\leq k}\frac{(a,a\sqrt q/b,-a\sqrt q/b,-aq/b;q)_k}{(q,a^2q/b^2,aq/bxy,axyq;q)_k}q^{k}\\ &\qquad\cdot \sum_{0\leq j}\frac{(1-bx^2y^2q^{2j})(bx^2y^2,b,bx^2,by^2,xyq,bxyq^{-k}/a;q)_j}{(1-bx^2y^2)(q,x^2y^2q,y^2q,x^2q,bxy,axyq^{k+1};q)_j}\left(\frac{aq^{k+1}}{b^2}\right)^j\\ &=B\frac{(axyq/b;q)_{\infty}}{(a/bxy;q)_{\infty}}\sum_{0\leq k}\frac{(bxy,xy\sqrt q,-xy\sqrt q,-xyq;q)_k}{(q,axyq/b,bxyq/a,bx^2y^2q;q)_k}q^k\\ &\qquad\cdot W(bx^2y^2;b,bx^2,by^2,xyq,q^{-k};xyq^{k+1}/b)\\ &\qquad+B\frac{(aq/b,x^2y^2q,bxy,axyq;q)_{\infty}}{(a,bxyq,xyq,bxy/a;q)_{\infty}}\sum_{0\leq k}\frac{(a,a\sqrt q/b,-a\sqrt q/b,-aq/b;q)_k}{(q,a^2q/b^2,axyq,aq/bxy;q)_k}q^k\\ &\qquad\cdot W(bx^2y^2;b,bx^2,by^2,xyq,bxyq^{-k}/a;aq^{k+1}/b^2) \end{align}
となって示すべき等式が得られた.