슬레이터 3

항등식
\(\prod_{n=1}^{\infty}(1-q^{2n-1})=\sum_{n=1}^{\infty}\frac{(-1)^nq^{n^2}}{(q^2;q^2)_n}\)
q-초기하급수에 대한 오일러공식의 특별한 경우
\(\prod_{n=0}^{\infty}(1+zq^n)=1+\sum_{n\geq 1}\frac{q^{n(n-1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)} z^n\) 에서 \(z=-q^{1/2}\) 인 경우

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
\(x=q, y\to\infty, z\to\infty\).
얻어진 켤레 베일리 쌍 (relative to 1)
\(\delta_n=q^{n^2}\)
\(\gamma_n=\frac{q^{n^2}}{(q)_{\infty}}\)

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 (not confirmed)
\(a=q,d\to 0,e\to 0 \)
\(\frac{(q^2/a,aq/d,aq/e,aq^2/de;q^2)_{\infty}}{(q,q^2/d,q^2/e,a^2q/de;q^2)_{\infty}}\to 1\)
\(\frac{(q)_{n}(aq)_{n}(a^2/de)_{q^2,n}}{(aq)_{q^2,n}(aq/d)_{n}(aq/e)_{n}}\to \frac{(-1)^n (q)_n (a q)_n}{(a q,q^2)_n q^{2n} }\)
얻어진 베일리 쌍 (relative to 1)
\(\alpha_{0}=1\),\(\alpha_{2r}=2\), \(\alpha_{2r+1}=-2\)
\(\beta_n=\sum_{r=0}^{n}\frac{\alpha_r}{(x)_{n-r}(q)_{n+r}}=\sum_{r=0}^{n}\frac{\alpha_r}{(q)_{n-r}(q)_{n+r}}=\frac{(-1)^n}{(q^2;q^2)_{n}}\)

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
\(\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\sum_{n=0}^{\infty}\beta_n\delta_{n}\)
\(\sum_{n=0}^{\infty}\beta_n\delta_{n}=\sum_{n=0}^{\infty}\frac{(-1)^nq^{n^2}}{(q^2;q^2)_{n}}\)
\(\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\frac{1+2\sum_{n=1}^{\infty}(-1)^{n}q^{n^2}}{(q)_{\infty}}=\prod_{n=1}^{\infty}(1-q^{2n-1})\)

목차