슬레이터 14

로저스-라마누잔 항등식 의 하나
\(\sum_{n=0}^{\infty}\frac{q^{n(n+1)}}{ (q)_{n}}=\frac{(q^{1};q^{5})_{\infty}(q^{4};q^{5})_{\infty}(q^{5};q^{5})_{\infty}}{(q)_{\infty}}=\frac{1}{(q^{2};q^{5})_{\infty}(q^{3};q^{5})_{\infty}}\)

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
\(\sum_{r=0}^{n}\frac{(1-aq^{2r})(-1)^{r}q^{\frac{1}{2}(r^2+r)}(a)_{r}(c)_{r}(d)_{r}a^{r}}{(a)_{n+r+1}(q)_{n-r}(q)_{r}(aq/c)_{r}(aq/d)_{r}c^{r}d^{r}}=\frac{(aq/cd)_{n}}{(q)_{n}(aq/c)_{n}(aq/d)_{n}}\)
다음의 특수한 경우
\(a=q,c\to\infty,d\to\infty\)
얻어진 베일리 쌍 (relative to 1)
\(\alpha_{0}=1\), \(\alpha_{n}=(-1)^{n}q^{\frac{3}{2}n^2}(q^{\frac{3}{2}n}+q^{-\frac{3}{2}n})\)
\(\beta_n=\frac{1}{(q)_{n}}\)
\(\beta_n=\sum_{r=0}^{n}\frac{\alpha_r}{(x)_{n-r}(q)_{n+r}}=\sum_{r=0}^{n}\frac{\alpha_r}{(q)_{n-r}(q)_{n+r}}=\frac{q^n}{(q)_{n}}\)

베일리 쌍과 켤레 베일리 쌍
\(\delta_n=q^{n^2}\)
\(\gamma_n=\frac{q^{n^2}}{(q)_{\infty}}\)
\(\alpha_{0}=1\), \(\alpha_{n}=(-1)^{n}q^{\frac{3}{2}n^2}(q^{\frac{3}{2}n}+q^{-\frac{3}{2}n})\)
\(\beta_n=\frac{q^n}{(q)_{n}}\)

항등식
\(\prod_{n=1}^{\infty}(1+q^n)=\sum_{n=1}^{\infty}\frac{q^{n(n+1)/2}}{(q)_n}\sim \frac{1}{\sqrt{2}}\exp(\frac{\pi^2}{12t}+\frac{t}{24})\)

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

목차