슬레이터 3

q-가우스 합 에서 얻어진 다음 결과를 이용
\(\delta_n=\frac{(y)_n(z)_n x^n}{y^n z^n}\), \(\gamma_n=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}\)
\(\gamma_{n}=\sum_{r=0}^{\infty}\frac{\delta_{n+r}}{(x)_{r+2n}(q)_{r}}\)
다음의 특수한 경우
\(x=q^2, y\to\infty, z\to\infty\).
얻어진 켤레 베일리 쌍
\(\delta_n=q^{n^2+n}\)
\(\gamma_n=\frac{(1-q)}{(q)_{\infty}}q^{n^2-n}\)

다음을 이용 [Slater51] (4.2)
\(\sum_{r=-[n/2]}^{r=[n/2]}\frac{(1-aq^{4r})(q^{-n})_{2r}a^{2r}q^{2nr+r}(d)_{q^2,r}(e)_{q^2,r}}{(1-a)(aq^{n+1})_{2r}d^re^r(aq^2/d)_{q^2,r}(aq^2/e)_{q^2,r}}=\frac{(q^2/a,aq/d,aq/e,aq^2/de;q^2)_{\infty}}{(q,q^2/d,q^2/e,a^2q/de;q^2)_{\infty}}\frac{(q)_{n}(aq)_{n}(a^2/de)_{q^2,n}}{(aq)_{q^2,n}(aq/d)_{n}(aq/e)_{n}}\)
다음의 특수한 경우
\(a=q,d\to 0,e\to 0 \)
얻어진 베일리 쌍
\(\alpha_{2r}=2\), \(\alpha_{2r+1}=-2\)
\(\beta_n=\sum_{r=0}^{n}\frac{\alpha_r}{(x)_{n-r}(q)_{n+r}}=\sum_{r=0}^{n}\frac{\alpha_r}{(q^{2})_{n-r}(q)_{n+r}}=\frac{(-1)^n}{(q^2;q^2)_{n}}\)

항등식
\(\prod_{n=1}^{\infty}(1-q^{2n-1})=\sum_{n=1}^{\infty}\frac{(-1)^nq^{n^2}}{(q^2;q^2)_n}\)

베일리 쌍(Bailey pair)과 베일리 보조정리
\(\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\sum_{n=0}^{\infty}\beta_n\delta_{n}\)
\(\sum_{n=0}^{\infty}\beta_n\delta_{n}=\sum_{n=0}^{\infty}=\frac{(-1)^nq^{n^2+n}}{(q^2;q^2)_{n}}\)
\(\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\frac{(-q)_{\infty}}{(q)_{\infty}}\sum_{n=0}^{\infty}(-1)^{n}(q^{\frac{3n^2+n}{2}}-q^{\frac{3n^2+5n+2}{2}})=(-q)_{\infty}\)

목차