qザールシュッツの和公式

qザールシュッツの和公式はハイネの変換式から導かれる。ハイネの変換式を反復すると

${\displaystyle {\begin{aligned}{_{2}\phi _{1}}\left[{\begin{matrix}a,b\\c\end{matrix}};q,z\right]&={\frac {(b;q)_{\infty }(az;q)_{\infty }}{(c;q)_{\infty }(z;q)_{\infty }}}{_{2}\phi _{1}}\left[{\begin{matrix}z,{\frac {c}{b}}\\az\end{matrix}};q,b\right]\\&={\frac {(b;q)_{\infty }(az;q)_{\infty }}{(c;q)_{\infty }(z;q)_{\infty }}}\cdot {\frac {({\frac {b}{c}};q)_{\infty }(bz;q)_{\infty }}{(az;q)_{\infty }(b;q)_{\infty }}}{_{2}\phi _{1}}\left[{\begin{matrix}b,{\frac {abz}{c}}\\bz\end{matrix}};q,{\frac {c}{b}}\right]\\&={\frac {(b;q)_{\infty }(az;q)_{\infty }}{(c;q)_{\infty }(z;q)_{\infty }}}\cdot {\frac {({\frac {b}{c}};q)_{\infty }(bz;q)_{\infty }}{(az;q)_{\infty }(b;q)_{\infty }}}\cdot {\frac {({\frac {abz}{c}};q)_{\infty }(c;q)_{\infty }}{(bz;q)_{\infty }({\frac {c}{b}};q)_{\infty }}}{_{2}\phi _{1}}\left[{\begin{matrix}{\frac {c}{b}},{\frac {c}{a}}\\c\end{matrix}};q,{\frac {abz}{c}}\right]\\&={_{2}\phi _{1}}\left[{\begin{matrix}{\frac {c}{a}},{\frac {c}{b}}\\c\end{matrix}};q,{\frac {abz}{c}}\right]{\frac {({\frac {abz}{c}};q)_{\infty }}{(z;q)_{\infty }}}\\\end{aligned}}}$

となり、q二項定理を用いて

${\displaystyle {\begin{aligned}{_{2}\phi _{1}}\sum _{n=0}^{\infty }{\frac {(a;q)_{n}(b;q)_{n}}{(q;q)_{n}(c;q)_{n}}}z^{n}&=\sum _{m=0}^{\infty }{\frac {({\frac {c}{a}};q)_{m}({\frac {c}{b}};q)_{m}}{(q;q)_{m}(c;q)_{m}}}\left({\frac {abz}{c}}\right)^{m}\sum _{k=0}^{\infty }{\frac {({\frac {ab}{c}};q)_{k}}{(q;q)_{k}}}z^{k}\\\end{aligned}}}$

となる。${\displaystyle z^{n}}$の係数を比べると

${\displaystyle {\begin{aligned}{\frac {(a;q)_{n}(b;q)_{n}}{(q;q)_{n}(c;q)_{n}}}&=\sum _{m=0}^{\infty }{\frac {({\frac {c}{a}};q)_{m}({\frac {c}{b}};q)_{m}}{(q;q)_{m}(c;q)_{m}}}\left({\frac {ab}{c}}\right)^{m}{\frac {({\frac {ab}{c}};q)_{n-m}}{(q;q)_{n-m}}}\\&=\sum _{m=0}^{\infty }{\frac {({\frac {c}{a}};q)_{m}({\frac {c}{b}};q)_{m}}{(q;q)_{m}(c;q)_{m}}}\left({\frac {ab}{c}}\right)^{m}{\frac {({\frac {ab}{c}};q)_{n}(q^{1+n-m};q)_{m}}{(q;q)_{n}({\frac {ab}{c}}q^{n-m};q)_{m}}}\end{aligned}}}$

であるが、qポッホハマー記号の変換式${\displaystyle (aq^{-m+1};q)_{m}=(-a)^{m}q^{-m(m-1)/2}\left(a^{-1};q\right)_{m}}$を用いて、

${\displaystyle {\begin{aligned}&{\frac {(a;q)_{n}(b;q)_{n}}{(q;q)_{n}(c;q)_{n}}}=\sum _{m=0}^{\infty }{\frac {({\frac {c}{a}};q)_{m}({\frac {c}{b}};q)_{m}}{(q;q)_{m}(c;q)_{m}}}{\frac {({\frac {ab}{c}};q)_{n}(q^{-n};q)_{m}}{(q;q)_{n}({\frac {c}{ab}}q^{-n+1};q)_{m}}}q^{m}\\&{\frac {(a;q)_{n}(b;q)_{n}}{({\frac {ab}{c}};q)_{n}(c;q)_{n}}}=\sum _{m=0}^{\infty }{\frac {({\frac {c}{a}};q)_{m}({\frac {c}{b}};q)_{m}}{(q;q)_{m}(c;q)_{m}}}{\frac {(q^{-n};q)_{m}}{({\frac {c}{ab}}q^{-n+1};q)_{m}}}q^{m}={_{3}\phi _{2}}\left[{\begin{matrix}{\frac {c}{a}},{\frac {c}{b}},q^{-n}\\c,{\frac {c}{ab}}q^{-n+1}\end{matrix}};q,q\right]\end{aligned}}}$

を得る。${\displaystyle a,b}$を${\displaystyle {\tfrac {c}{a}},{\tfrac {c}{b}}}$に置き換えて

${\displaystyle {_{3}\phi _{2}}\left[{\begin{matrix}a,b,q^{-n}\\c,{\frac {ab}{c}}q^{-n+1}\end{matrix}};q,q\right]={\frac {({\frac {c}{a}};q)_{n}({\frac {c}{b}};q)_{n}}{({\frac {c}{ab}};q)_{n}(c;q)_{n}}}}$

を得る。

1. Wolfram Mathworld: q-Saalschütz Sum