Rahman-Vermaの二次変換公式

Andrewsによるterminating $q$-Watsonの和公式
\begin{align} \Q43{q^{-n},a^2q^n,b,-b}{a\sqrt q,-a\sqrt q,b^2}{q}&=\begin{cases} \displaystyle\frac{(q,a^2q/b^2;q^2)_{\frac n2}}{(a^2q,b^2q;q^2)_{\frac n2}}b^n&n:\mathrm{even}\\ 0&n:\mathrm{odd} \end{cases} \end{align}
より, 数列$A_n$に対して
\begin{align} \sum_{0\leq n}A_n\Q43{q^{-n},a^2q^n,b,-b}{a\sqrt q,-a\sqrt q,b^2}{q}&=\sum_{0\leq n}A_{2n}\frac{(q,a^2q/b^2;q^2)_{n}}{(a^2q,b^2q;q^2)_{n}}b^{2n} \end{align}
となる. ここで,
\begin{align} A_n=\frac{1-a^2q^{2n}}{1-a^2}\frac{(a^2,b^2,c,d,e,f;q)_n}{(q,a^2q/b^2,a^2q/c,a^2q/d,a^2q/e,a^2q/f;q)_n}\left(\frac{a^4q^2}{b^2cdef}\right)^n \end{align}
とすると,
\begin{align} &W\left(a^2;b^2,c,cq,d,dq,e,eq,f,fq;q^2;\left(\frac{a^4q^2}{b^2cdef}\right)^2\right)\\ &=\sum_{0\leq n}\frac{1-a^2q^{2n}}{1-a^2}\frac{(a^2,b^2,c,d,e,f;q)_n}{(q,a^2q/b^2,a^2q/c,a^2q/d,a^2q/e,a^2q/f;q)_n}\left(\frac{a^4q^2}{b^2cdef}\right)^n\\ &\qquad\cdot\Q43{q^{-n},a^2q^n,b,-b}{a\sqrt q,-a\sqrt q,b^2}{q} \end{align}
ここで, Watsonの変換公式 より
\begin{align} &\Q43{q^{-n},a^2q^n,b,-b}{a\sqrt q,-a\sqrt q,b^2}{q}\\ &=\frac{(a^2q/b^2,b;q)_n}{(a^2q/b,b^2;q)_n}b^n\Q87{a^2/b,q\sqrt{a^2/b},-q\sqrt{a^2/b},a\sqrt q/b,-a\sqrt q/b,b,a^2q^n,q^{-n}}{\sqrt{a^2/b},-\sqrt{a^2/b},a\sqrt q,-a\sqrt q,a^2q/b^2,q^{1-n}/b,a^2q^{n+1}/b}{-\frac qb} \end{align}
であるから, これを代入して
\begin{align} &W\left(a^2;b^2,c,cq,d,dq,e,eq,f,fq;q^2;\left(\frac{a^4q^2}{b^2cdef}\right)^2\right)\\ &=\sum_{0\leq n}\frac{1-a^2q^{2n}}{1-a^2}\frac{(a^2,b^2,c,d,e,f;q)_n}{(q,a^2q/b^2,a^2q/c,a^2q/d,a^2q/e,a^2q/f;q)_n}\left(\frac{a^4q^2}{b^2cdef}\right)^n\\ &\qquad\cdot \frac{(a^2q/b^2,b;q)_n}{(a^2q/b,b^2;q)_n}b^n\Q87{a^2/b,q\sqrt{a^2/b},-q\sqrt{a^2/b},a\sqrt q/b,-a\sqrt q/b,b,a^2q^n,q^{-n}}{\sqrt{a^2/b},-\sqrt{a^2/b},a\sqrt q,-a\sqrt q,a^2q/b^2,q^{1-n}/b,a^2q^{n+1}/b}{-\frac qb}\\ &=\sum_{0\leq n}\frac{1-a^2q^{2n}}{1-a^2}\frac{(a^2,b,c,d,e,f;q)_n}{(q,a^2q/b,a^2q/c,a^2q/d,a^2q/e,a^2q/f;q)_n}\left(\frac{a^4q^2}{bcdef}\right)^n\\ &\qquad\cdot \sum_{0\leq k}\frac{(1-a^2q^{2k}/b)(a^2/b,a\sqrt q/b,-a\sqrt q/b,b,a^2q^n,q^{-n};q)_k}{(1-a^2/b)(q,a\sqrt q,-a\sqrt q,a^2q/b^2,q^{1-n}/b,a^2q^{n+1}/b;q)_k}\left(-\frac qb\right)^k\\ &=\sum_{0\leq k}\frac{(1-a^2q^{2k}/b)(a^2/b,a\sqrt q/b,-a\sqrt q/b,b;q)_k}{(1-a^2/b)(q,a\sqrt q,-a\sqrt q,a^2q/b^2;q)_k}\left(-1\right)^k\\ &\qquad\cdot\sum_{0\leq n}\frac{1-a^2q^{2n}}{1-a^2}\frac{(a^2;q)_{n+k}(b;q)_{n-k}(c,d,e,f;q)_n}{(a^2q/b;q)_{n+k}(q;q)_{n-k}(a^2q/c,a^2q/d,a^2q/e,a^2q/f;q)_n}\left(\frac{a^4q^2}{bcdef}\right)^n\\ &=\sum_{0\leq k}\frac{(1-a^2q^{2k}/b)(a^2/b,a\sqrt q/b,-a\sqrt q/b,b;q)_k}{(1-a^2/b)(q,a\sqrt q,-a\sqrt q,a^2q/b^2;q)_k}\left(-\frac{a^4q^2}{bcdef}\right)^k\\ &\qquad\cdot\frac{(a^2q;q)_{2k}(c,d,e,f;q)_k}{(a^2q/b;q)_{2k}(a^2q/c,a^2q/d,a^2q/e,a^2q/f;q)_k}\\\ &\qquad\cdot W\left(a^2q^{2k};b,cq^k,dq^k,eq^k,fq^k;\frac{a^4q^2}{bcdef}\right) \end{align}
を得る. ここで, $b=q^{-n}, f=a^4q^{n+1}/cde $とすると, Jacksonの和公式 から

\begin{align} &W\left(a^2q^{2k};q^{-n},cq^k,dq^k,eq^k,a^4q^{n+k+1}/cde;q\right)\\ &=\frac{(a^2q^{2k+1},a^2q/cd,a^2q/ce,a^2q/de;q)_n}{(a^2q^{k+1}/c,a^2q^{k+1}/d,a^2q^{k+1}/e,a^2q^{1-k}/cde;q)_n} \end{align}
であるから,

\begin{align} &W\left(a^2;q^{-2n},c,cq,d,dq,e,eq,\frac{a^4q^{n+1}}{cde},\frac{a^4q^{n+2}}{cde};q^2;q^2\right)\\ &=\sum_{0\leq k}\frac{(1-a^2q^{n+2k})(a^2q^n,aq^n\sqrt q,-aq^n\sqrt q,q^{-n};q)_k}{(1-a^2/b)(q,a\sqrt q,-a\sqrt q,a^2q^{2n+1};q)_k}\left(-q\right)^k\\ &\qquad\cdot\frac{(a^2q;q)_{2k}(c,d,e,a^4q^{n+1}/cde;q)_k}{(a^2q^{n+1};q)_{2k}(a^2q/c,a^2q/d,a^2q/e,cdeq^{-n}/a^2;q)_k}\\\ &\qquad\cdot \frac{(a^2q^{2k+1},a^2q/cd,a^2q/ce,a^2q/de;q)_n}{(a^2q^{k+1}/c,a^2q^{k+1}/d,a^2q^{k+1}/e,a^2q^{1-k}/cde;q)_n}\\ &=\frac {(a^2q,a^2q/cd,a^2q/ce,a^2q/de;q)_n}{(a^2q/c,a^2q/d,a^2q/e,a^2q/cde;q)_n}\sum_{0\leq k}\frac{(1-a^2q^{n+2k})(a^2q^n,aq^n\sqrt q,-aq^n\sqrt q,q^{-n};q)_k}{(1-a^2/b)(q,a\sqrt q,-a\sqrt q,a^2q^{2n+1};q)_k}\left(-q^{n+1}\right)^k\\ &\qquad\cdot\frac{(c,d,e,a^4q^{n+1}/cde;q)_k}{(a^2q^{n+1}/c,a^2q^{n+1}/d,a^2q^{n+1}/e,cde/a^2;q)_k}\\ &=\frac {(a^2q,a^2q/cd,a^2q/ce,a^2q/de;q)_n}{(a^2q/c,a^2q/d,a^2q/e,a^2q/cde;q)_n}\\ &\qquad\cdot W(a^2q^n;c,d,e,aq^{n+\frac 12},-aq^{n+\frac 12},a^4q^{n+1}/cde,q^{-n};-q^{n+1}) \end{align}
となって1つ目の等式が得られる. 次に,
\begin{align} &W\left(a^2;b^2,c,cq,d,dq,e,eq,f,fq;q^2;\left(\frac{a^4q^2}{b^2cdef}\right)^2\right)\\ &=\sum_{0\leq k}\frac{(1-a^2q^{2k}/b)(a^2/b,a\sqrt q/b,-a\sqrt q/b,b;q)_k}{(1-a^2/b)(q,a\sqrt q,-a\sqrt q,a^2q/b^2;q)_k}\left(-\frac{a^4q^2}{bcdef}\right)^k\\ &\qquad\cdot\frac{(a^2q;q)_{2k}(c,d,e,f;q)_k}{(a^2q/b;q)_{2k}(a^2q/c,a^2q/d,a^2q/e,a^2q/f;q)_k}\\\ &\qquad\cdot W\left(a^2q^{2k};b,cq^k,dq^k,eq^k,fq^k;\frac{a^4q^2}{bcdef}\right) \end{align}
において, $f=q^{-n},e=a^4q^{n+1}/bcd$とすると, Jacksonの和公式 より
\begin{align} &W\left(a^2q^{2k};b,cq^k,dq^k,a^4q^{n+k+1}/bcd,q^{k-n};q\right)\\ &=\frac{(a^2q^{2k+1},a^2q^{k+1}/bc,a^2q^{k+1}/bd,a^2q/cd;q)_{n-k}}{(a^2q^{2k+1}/b,a^2q^{k+1}/c,a^2q^{k+1}/d,a^2q/bcd;q)_{n-k}} \end{align}
であるから,
\begin{align} &W\left(a^2;b^2,c,cq,d,dq,a^4q^{n+1}/bcd,a^4q^{n+2}/bcd,q^{-n},q^{1-n};q^2;q^2\right)\\ &=\sum_{0\leq k}\frac{(1-a^2q^{2k}/b)(a^2/b,a\sqrt q/b,-a\sqrt q/b,b;q)_k}{(1-a^2/b)(q,a\sqrt q,-a\sqrt q,a^2q/b^2;q)_k}\left(-q\right)^k\\ &\qquad\cdot\frac{(a^2q;q)_{2k}(c,d,a^4q^{n+1}/bcd,q^{-n};q)_k}{(a^2q/b;q)_{2k}(a^2q/c,a^2q/d,bcdq^{-n}/a^2,a^2q^{n+1};q)_k}\\\ &\qquad\cdot\frac{(a^2q^{2k+1},a^2q^{k+1}/bc,a^2q^{k+1}/bd,a^2q/cd;q)_{n-k}}{(a^2q^{2k+1}/b,a^2q^{k+1}/c,a^2q^{k+1}/d,a^2q/bcd;q)_{n-k}}\\ &=\frac {(a^2q,a^2q/bc,a^2q/bd,a^2q/cd;q)_n}{(a^2q/b,a^2q/c,a^2q/d,a^2q/bcd;q)_n}\sum_{0\leq k}\frac{(1-a^2q^{2k}/b)(a^2/b,a\sqrt q/b,-a\sqrt q/b,b;q)_k}{(1-a^2/b)(q,a\sqrt q,-a\sqrt q,a^2q/b^2;q)_k}\left(-q/b\right)^k\\ &\qquad\cdot\frac{(c,d,a^4q^{n+1}/bcd,q^{-n};q)_k}{(a^2q^{n+1}/b,a^2q/bc,a^2q/bd,cdq^{-n}/a^2;q)_k}\\ &=\frac {(a^2q,a^2q/bc,a^2q/bd,a^2q/cd;q)_n}{(a^2q/b,a^2q/c,a^2q/d,a^2q/bcd;q)_n}W(a^2/b;a\sqrt q/b,-a\sqrt q/b,b,c,d,a^2q^{n+1}/bcd,q^{-n};q;-q/b) \end{align}
ここで, $b\mapsto e$として2つ目の等式を得る.