Slaterの変換公式3
前の記事 で以下の定理を示した.
\begin{align}
&\frac{(a_1q/a_{M+2},\dots,a_1q/a_{2M},q/a_{M+2},\dots,q/a_{2M},\sqrt{a_1},-\sqrt{a_1},q/\sqrt{a_1},-q/\sqrt{a_1};q)_{\infty}}{(a_1,\dots,a_{M+1},a_2/a_1,\dots,a_{M+1}/a_1;q)_{\infty}}\\
&\qquad\cdot\Q{2M}{2M-1}{a_1,\dots,a_{2M}}{a_1q/a_2,\dots,a_1q/a_{2M}}{-\frac{(a_1q)^M}{a_1\cdots a_{2M}}}\\
&=\frac{a_2(a_2q/a_{M+2},\dots,a_2q/a_{2M},a_1q/a_2a_{M+2},\dots,a_1q/a_2a_{2M},\sqrt{a_1}/a_2,-\sqrt{a_1}/a_2,a_2q/\sqrt{a_1},-a_2q/\sqrt{a_1};q)_{\infty}}{(a_2,a_2^2/a_1,a_2a_3/a_1,\dots,a_2a_{M+1}/a_1,a_1/a_2,a_3/a_2,\dots,a_{M+1}/a_2;q)_{\infty}}\\
&\qquad\cdot\Q{2M}{2M-1}{a_2^2/a_1,a_2,a_2a_3/a_1,\dots,a_2a_{2M}/a_1}{a_2q/a_1,a_2q/a_3,\dots,a_2q/a_{2M}}{-\frac{(a_1q)^{M}}{a_1\cdots a_{2M}}}\\
&\qquad+\mathrm{idem}(a_2;a_3,\dots,a_{M+1})
\end{align}
が成り立つ.
今回はこれを用いてwell-poised両側$q$超幾何級数に関するSlaterの変換公式を示す.
\begin{align} &\frac{(aq/b_1,\dots,aq/b_{2r},q/b_1,\dots,q/b_{2r},\sqrt{a},-\sqrt{a},q/\sqrt{a},-q/\sqrt{a};q)_{\infty}}{(a,q/a,a_1\dots,a_r,q/a_1,\dots,q/a_r,aq/a_1,\dots,aq/a_r,a_1/a,\dots,a_r/a;q)_{\infty}}\\ &\qquad\cdot\BQ{2r}{2r}{b_1,\dots,b_{2r}}{aq/b_1,\dots,aq/b_{2r}}{-\frac{(aq)^{r}}{b_1\cdots b_{2r}}}\\ &=\frac{a_1(a_1q/b_1,\dots,a_1q/b_{2r},aq/a_1b_1,\dots,aq/a_1b_{2r};q)_{\infty}}{(a_1,q/a_1,a/a_1,a_1q/a,a_1^2/a,aq/a_1^2,a_2/a_1,\dots,a_r/a_1,a_1q/a_2,\dots,a_1q/a_r;q)_{\infty}}\\ &\qquad\cdot\frac{(\sqrt{a}/a_1,-\sqrt{a}/a_1,a_1q/\sqrt{a},-a_1q/\sqrt{a};q)_{\infty}}{(a_1a_2/a,\dots,a_1a_{r}/a,aq/a_1a_2,\dots,aq/a_1a_r;q)_{\infty}}\\ &\qquad\cdot\BQ{2r}{2r}{a_1b_1/a,\dots,a_1b_{2r}/a}{a_1q/b_1,\dots,a_1q/b_{2r}}{-\frac{(aq)^{r}}{b_1\cdots b_{2r}}}\\ &\qquad+\mathrm{idem}(a_1;a_2,\dots,a_r)\\ \end{align}
定理1において, $M=2r+1$として, $a_{2r+3}=b_1,\dots, a_{4r+2}=b_{2r}, a_1\mapsto a, a_2\mapsto a_1,\dots,a_{2r+2}\mapsto a_{2r+1}$とすると,
\begin{align}
&\frac{(aq/b_1,\dots,aq/b_{2r},q/b_1,\dots,q/b_{2r},\sqrt{a},-\sqrt{a},q/\sqrt{a},-q/\sqrt{a};q)_{\infty}}{(a,a_1\dots,a_{2r+1},a_1/a,\dots,a_{2r+1}/a;q)_{\infty}}\\
&\qquad\cdot\Q{4r+2}{4r+1}{a,a_1,\dots,a_{2r+1},b_1,\dots,b_{2r}}{aq/a_1,\dots,aq/a_{2r+1},aq/b_1,\dots,aq/b_{2r}}{-\frac{(aq)^{2r+1}}{aa_1\cdots a_{2r+1}b_1\cdots b_{2r}}}\\
&=\frac{a_1(a_1q/b_1,\dots,a_1q/b_{2r},aq/a_1b_1,\dots,aq/a_1b_{2r},\sqrt{a}/a_1,-\sqrt{a}/a_1,a_1q/\sqrt{a},-a_1q/\sqrt{a};q)_{\infty}}{(a_1,a_1^2/a,a_1a_2/a,\dots,a_1a_{2r+1}/a,a/a_1,a_2/a_1,\dots,a_{2r+1}/a_1;q)_{\infty}}\\
&\qquad\cdot\Q{4r+2}{4r+1}{a_1^2/a,a_1,a_1a_2/a,\dots,a_1a_{2r+1}/a,a_1b_1/a,\dots,a_1b_{2r}/a}{a_1q/a,a_1q/a_2,\dots,a_1q/a_{2r+1},a_1q/b_1,\dots,a_1q/b_{2r}}{-\frac{(aq)^{2r+1}}{a a_1\cdots a_{2r+1}b_1\cdots b_{2r}}}\\
&\qquad+\mathrm{idem}(a_1;a_2,\dots,a_{2r+1})
\end{align}
となる. ここで, $a_{r+1}=aq/a_1,\dots,a_{2r}=aq/a_r, a_{2r+1}=q$とすると,
\begin{align}
&\frac{(aq/b_1,\dots,aq/b_{2r},q/b_1,\dots,q/b_{2r},\sqrt{a},-\sqrt{a},q/\sqrt{a},-q/\sqrt{a};q)_{\infty}}{(q,a,q/a,a_1\dots,a_r,aq/a_1,\dots,aq/a_r,a_1/a,\dots,a_r/a,q/a_1,\dots,q/a_r;q)_{\infty}}\\
&\qquad\cdot\Q{2r+1}{2r}{q,b_1,\dots,b_{2r}}{aq/b_1,\dots,aq/b_{2r}}{-\frac{(aq)^{r}}{b_1\cdots b_{2r}}}\\
&=\frac{a_1(a_1q/b_1,\dots,a_1q/b_{2r},aq/a_1b_1,\dots,aq/a_1b_{2r},\sqrt{a}/a_1,-\sqrt{a}/a_1,a_1q/\sqrt{a},-a_1q/\sqrt{a};q)_{\infty}}{(a_1,a_1^2/a,a_1a_2/a,\dots,a_1a_{r}/a,q,a_1q/a_2,\dots,a_1q/a_r,a_1q/a,a/a_1,a_2/a_1,\dots,a_r/a_1,aq/a_1^2,aq/a_1a_2,\dots,aq/a_1a_r,q/a_1;q)_{\infty}}\\
&\qquad\cdot\Q{2r+1}{2r}{q,a_1b_1/a,\dots,a_1b_{2r}/a}{a_1q/b_1,\dots,a_1q/b_{2r}}{-\frac{(aq)^{r}}{b_1\cdots b_{2r}}}\\
&\qquad+\mathrm{idem}(a_1;a_2,\dots,a_r)\\
&\qquad+\frac{aq(aq^2/a_1b_1,\dots,aq^2/a_1b_{2r},a_1/b_1,\dots,a_1/b_{2r},a_1/\sqrt aq,-a_1/\sqrt{a}q,\sqrt aq^2/a_1,-\sqrt aq^2/a_1;q)_{\infty}}{a_1(aq/a_1,aq^2/a_1^2,aq^2/a_1a_2,\dots,aq^2/a_1a_r,q,a_2q/a_1,\dots,a_rq/a_1,q^2/a_1,a_1/q,a_1/a_2,\dots,a_1/a_r,a_1^2/aq,a_1a_2/aq,\dots,a_1a_r/aq,a_1/a;q)_{\infty}}\\
&\qquad\cdot\Q{2r+1}{2r}{q,b_1q/a_1,\dots,b_{2r}q/a_1}{aq^2/a_1b_1,\dots,aq^2/a_1b_{2r}}{-\frac{(aq)^{r}}{b_1\cdots b_{2r}}}\\
&\qquad+\mathrm{idem}(a_1;a_2,\dots,a_r)\\
&\qquad+\frac{q(q^2/b_1,\dots,q^2/b_{2r},a/b_1,\dots,a/b_{2r},\sqrt{a}/q,-\sqrt{a}/q,q^2/\sqrt{a},-q^2/\sqrt{a};q)_{\infty}}{(q,q^2/a,a_1q/a,\dots,a_rq/a,q^2/a_1,\dots,q^2/a_r,a/q,a_1/q,\dots,a_r/q,a/a_1,\dots,a/a_r;q)_{\infty}}\\
&\qquad\cdot\Q{2r+1}{2r}{q,b_1q/a,\dots,b_{2r}q/a}{q^2/b_1,\dots,q^2/b_{2r}}{-\frac{(aq)^{r}}{b_1\cdots b_{2r}}}\\
&=\frac{a_1(a_1q/b_1,\dots,a_1q/b_{2r},aq/a_1b_1,\dots,aq/a_1b_{2r},\sqrt{a}/a_1,-\sqrt{a}/a_1,a_1q/\sqrt{a},-a_1q/\sqrt{a};q)_{\infty}}{(q,a_1,q/a_1,a/a_1,a_1q/a,a_1^2/a,aq/a_1^2,a_1a_2/a,\dots,a_1a_{r}/a,a_1q/a_2,\dots,a_1q/a_r,a_2/a_1,\dots,a_r/a_1,aq/a_1a_2,\dots,aq/a_1a_r;q)_{\infty}}\\
&\qquad\cdot\Q{2r+1}{2r}{q,a_1b_1/a,\dots,a_1b_{2r}/a}{a_1q/b_1,\dots,a_1q/b_{2r}}{-\frac{(aq)^{r}}{b_1\cdots b_{2r}}}\\
&\qquad+\mathrm{idem}(a_1;a_2,\dots,a_r)\\
&\qquad-\frac{(aq)^{r}(aq^2/a_1b_1,\dots,aq^2/a_1b_{2r},a_1/b_1,\dots,a_1/b_{2r},a_1q/\sqrt a,-a_1q/\sqrt{a},\sqrt a/a_1,-\sqrt a/a_1;q)_{\infty}}{a_1^{2r-1}(q,a_1,q/a_1,a/a_1,a_1q/a,a_1^2/a,aq/a_1^2,aq/a_1a_2,\dots,aq/a_1a_r,a_2/a_1,\dots,a_r/a_1,a_1q/a_2,\dots,a_1q/a_r,a_1a_2/a,\dots,a_1a_r/a;q)_{\infty}}\\
&\qquad\cdot\Q{2r+1}{2r}{q,b_1q/a_1,\dots,b_{2r}q/a_1}{aq^2/a_1b_1,\dots,aq^2/a_1b_{2r}}{-\frac{(aq)^{r}}{b_1\cdots b_{2r}}}\\
&\qquad+\mathrm{idem}(a_1;a_2,\dots,a_r)\\
&\qquad+\frac{q(q^2/b_1,\dots,q^2/b_{2r},a/b_1,\dots,a/b_{2r},\sqrt{a}/q,-\sqrt{a}/q,q^2/\sqrt{a},-q^2/\sqrt{a};q)_{\infty}}{(q,q^2/a,a_1q/a,\dots,a_rq/a,q^2/a_1,\dots,q^2/a_r,a/q,a_1/q,\dots,a_r/q,a/a_1,\dots,a/a_r;q)_{\infty}}\\
&\qquad\cdot\Q{2r+1}{2r}{q,b_1q/a,\dots,b_{2r}q/a}{q^2/b_1,\dots,q^2/b_{2r}}{-\frac{(aq)^{r}}{b_1\cdots b_{2r}}}\\
&=\frac{a_1(a_1q/b_1,\dots,a_1q/b_{2r},aq/a_1b_1,\dots,aq/a_1b_{2r},\sqrt{a}/a_1,-\sqrt{a}/a_1,a_1q/\sqrt{a},-a_1q/\sqrt{a};q)_{\infty}}{(q,a_1,q/a_1,a/a_1,a_1q/a,a_1^2/a,aq/a_1^2,a_1a_2/a,\dots,a_1a_{r}/a,a_1q/a_2,\dots,a_1q/a_r,a_2/a_1,\dots,a_r/a_1,aq/a_1a_2,\dots,aq/a_1a_r;q)_{\infty}}\\
&\qquad\cdot\left(\Q{2r+1}{2r}{q,a_1b_1/a,\dots,a_1b_{2r}/a}{a_1q/b_1,\dots,a_1q/b_{2r}}{-\frac{(aq)^{r}}{b_1\cdots b_{2r}}}-\frac{\left(1-\frac{b_1}{a_1}\right)\cdots \left(1-\frac{b_{2r}}{a_1}\right)}{\left(1-\frac{aq}{a_1b_1}\right)\cdots\left(1-\frac{aq}{a_1b_{2r}}\right)}\frac{(aq)^r}{b_1\cdots b_{2r}}\Q{2r+1}{2r}{q,b_1q/a_1,\dots,b_{2r}q/a_1}{aq^2/a_1b_1,\dots,aq^2/a_1b_{2r}}{-\frac{(aq)^{r}}{b_1\cdots b_{2r}}}\right)\\
&\qquad+\mathrm{idem}(a_1;a_2,\dots,a_r)\\
&\qquad+\frac{q(q^2/b_1,\dots,q^2/b_{2r},a/b_1,\dots,a/b_{2r},\sqrt{a}/q,-\sqrt{a}/q,q^2/\sqrt{a},-q^2/\sqrt{a};q)_{\infty}}{(q,q^2/a,a_1q/a,\dots,a_rq/a,q^2/a_1,\dots,q^2/a_r,a/q,a_1/q,\dots,a_r/q,a/a_1,\dots,a/a_r;q)_{\infty}}\\
&\qquad\cdot\Q{2r+1}{2r}{q,b_1q/a,\dots,b_{2r}q/a}{q^2/b_1,\dots,q^2/b_{2r}}{-\frac{(aq)^{r}}{b_1\cdots b_{2r}}}\\
&=\frac{a_1(a_1q/b_1,\dots,a_1q/b_{2r},aq/a_1b_1,\dots,aq/a_1b_{2r},\sqrt{a}/a_1,-\sqrt{a}/a_1,a_1q/\sqrt{a},-a_1q/\sqrt{a};q)_{\infty}}{(q,a_1,q/a_1,a/a_1,a_1q/a,a_1^2/a,aq/a_1^2,a_1a_2/a,\dots,a_1a_{r}/a,a_1q/a_2,\dots,a_1q/a_r,a_2/a_1,\dots,a_r/a_1,aq/a_1a_2,\dots,aq/a_1a_r;q)_{\infty}}\\
&\qquad\cdot\BQ{2r}{2r}{a_1b_1/a,\dots,a_1b_{2r}/a}{a_1q/b_1,\dots,a_1q/b_{2r}}{-\frac{(aq)^{r}}{b_1\cdots b_{2r}}}\\
&\qquad+\mathrm{idem}(a_1;a_2,\dots,a_r)\\
&\qquad+\frac{q(q^2/b_1,\dots,q^2/b_{2r},a/b_1,\dots,a/b_{2r},\sqrt{a}/q,-\sqrt{a}/q,q^2/\sqrt{a},-q^2/\sqrt{a};q)_{\infty}}{(q,q^2/a,a_1q/a,\dots,a_rq/a,q^2/a_1,\dots,q^2/a_r,a/q,a_1/q,\dots,a_r/q,a/a_1,\dots,a/a_r;q)_{\infty}}\\
&\qquad\cdot\Q{2r+1}{2r}{q,b_1q/a,\dots,b_{2r}q/a}{q^2/b_1,\dots,q^2/b_{2r}}{-\frac{(aq)^{r}}{b_1\cdots b_{2r}}}
\end{align}
よって,
\begin{align}
&\frac{a_1(a_1q/b_1,\dots,a_1q/b_{2r},aq/a_1b_1,\dots,aq/a_1b_{2r};q)_{\infty}}{(q,a_1,q/a_1,a/a_1,a_1q/a,a_1^2/a,aq/a_1^2,a_1a_2/a,\dots,a_1a_{r}/a,a_1q/a_2,\dots,a_1q/a_r;q)_{\infty}}\\
&\qquad\cdot\frac{(\sqrt{a}/a_1,-\sqrt{a}/a_1,a_1q/\sqrt{a},-a_1q/\sqrt{a};q)_{\infty}}{(a_2/a_1,\dots,a_r/a_1,aq/a_1a_2,\dots,aq/a_1a_r;q)_{\infty}}\\
&\qquad\cdot\BQ{2r}{2r}{a_1b_1/a,\dots,a_1b_{2r}/a}{a_1q/b_1,\dots,a_1q/b_{2r}}{-\frac{(aq)^{r}}{b_1\cdots b_{2r}}}\\
&\qquad+\mathrm{idem}(a_1;a_2,\dots,a_r)\\
&=\frac{(aq/b_1,\dots,aq/b_{2r},q/b_1,\dots,q/b_{2r},\sqrt{a},-\sqrt{a},q/\sqrt{a},-q/\sqrt{a};q)_{\infty}}{(q,a,q/a,a_1\dots,a_r,aq/a_1,\dots,aq/a_r,a_1/a,\dots,a_r/a,q/a_1,\dots,q/a_r;q)_{\infty}}\\
&\qquad\cdot\Q{2r+1}{2r}{q,b_1,\dots,b_{2r}}{a_1q/b_1,\dots,a_1q/b_{2r}}{-\frac{(aq)^{r}}{b_1\cdots b_{2r}}}\\
&\qquad-\frac{q(q^2/b_1,\dots,q^2/b_{2r},a/b_1,\dots,a/b_{2r},\sqrt{a}/q,-\sqrt{a}/q,q^2/\sqrt{a},-q^2/\sqrt{a};q)_{\infty}}{(q,q^2/a,a_1q/a,\dots,a_rq/a,q^2/a_1,\dots,q^2/a_r,a/q,a_1/q,\dots,a_r/q,a/a_1,\dots,a/a_r;q)_{\infty}}\\
&\qquad\cdot\Q{2r+1}{2r}{q,b_1q/a,\dots,b_{2r}q/a}{q^2/b_1,\dots,q^2/b_{2r}}{-\frac{(aq)^{r}}{b_1\cdots b_{2r}}}\\
&=\frac{(aq/b_1,\dots,aq/b_{2r},q/b_1,\dots,q/b_{2r},\sqrt{a},-\sqrt{a},q/\sqrt{a},-q/\sqrt{a};q)_{\infty}}{(q,a,q/a,a_1\dots,a_r,aq/a_1,\dots,aq/a_r,a_1/a,\dots,a_r/a,q/a_1,\dots,q/a_r;q)_{\infty}}\\
&\qquad\cdot\left(\Q{2r+1}{2r}{q,b_1,\dots,b_{2r}}{aq/b_1,\dots,aq/b_{2r}}{-\frac{(aq)^{r}}{b_1\cdots b_{2r}}}-\frac{\left(1-\frac{b_1}a\right)\cdots \left(1-\frac{b_{2r}}a\right)}{\left(1-\frac{q}{b_1}\right)\cdots \left(1-\frac{q}{b_{2r}}\right)}\frac{(aq)^r}{b_1\cdots b_{2r}}\Q{2r+1}{2r}{q,b_1q/a,\dots,b_{2r}q/a}{q^2/b_1,\dots,q^2/b_{2r}}{-\frac{(aq)^{r}}{b_1\cdots b_{2r}}}\right)\\
&=\frac{(aq/b_1,\dots,aq/b_{2r},q/b_1,\dots,q/b_{2r},\sqrt{a},-\sqrt{a},q/\sqrt{a},-q/\sqrt{a};q)_{\infty}}{(q,a,q/a,a_1\dots,a_r,aq/a_1,\dots,aq/a_r,a_1/a,\dots,a_r/a,q/a_1,\dots,q/a_r;q)_{\infty}}\\
&\qquad\cdot\BQ{2r}{2r}{b_1,\dots,b_{2r}}{aq/b_1,\dots,aq/b_{2r}}{-\frac{(aq)^{r}}{b_1\cdots b_{2r}}}
\end{align}
となるので, 両辺に$(q;q)_{\infty}$を掛けて示すべき等式が得られる.
定理2において, $r\mapsto r+1$として, $a_r=b_{2r+1}=\sqrt aq, a_{r+1}=b_{2r+2}=-\sqrt{a}q$とすると, 最後の2つの項は消えるので, 以下の系を得る.
\begin{align} &\frac{(aq/b_1,\dots,aq/b_{2r},q/b_1,\dots,q/b_{2r};q)_{\infty}}{(aq,q/a,a_1,\dots,a_{r-1},q/a_1,\dots,q/a_{r-1},aq/a_1,\dots,aq/a_{r-1},a_1/a,\dots,a_{r-1}/a;q)_{\infty}}\\ &\qquad\cdot\BQ{2r+2}{2r+2}{\sqrt aq,-\sqrt aq,b_1,\dots,b_{2r}}{\sqrt a,-\sqrt a,aq/b_1,\dots,aq/b_{2r}}{\frac{a^rq^{r-1}}{b_1\cdots b_{2r}}}\\ &=\frac{(a_1q/b_1,\dots,a_1q/b_{2r},aq/a_1b_1,\dots,aq/a_1b_{2r};q)_{\infty}}{(a_1,q/a_1,aq/a_1,a_1/a,a_1^2q/a,aq/a_1^2,a_2/a_1,\dots,a_{r-1}/a_1,a_1q/a_2,\dots,a_1q/a_{r-1};q)_{\infty}}\\ &\qquad\cdot\frac{1}{(a_1a_2/a,\dots,a_1a_{r-1}/a,aq/a_1a_2,\dots,aq/a_1a_{r-1};q)_{\infty}}\\ &\qquad\cdot\BQ{2r+2}{2r+2}{a_1q/\sqrt a,-a_1q/\sqrt a,a_1b_1/a,\dots,a_1b_{2r}/a}{a_1/\sqrt a,-a_1/\sqrt a,a_1q/b_1,\dots,a_1q/b_{2r}}{\frac{a^rq^{r-1}}{b_1\cdots b_{2r}}}\\ &\qquad+\mathrm{idem}(a_1;a_2,\dots,a_{r-1}) \end{align}
少し長くなったので, この公式の具体例については次の記事に書こうと思う.
