슬레이터 1

개요

얻어지는 항등식은 오일러의 오각수정리(pentagonal number theorem) 의 경우와 같다\[\prod_{n=1}^{\infty}(1-q^n)=1+\sum_{n=1}^{\infty}(-1)^{n}(q^{\frac{3 n^2-n}{2}}+q^{\frac{3 n^2+n}{2}})=\sum_{n=-\infty}^\infty(-1)^nq^{n(3n-1)/2}\]

q-가우스 합 에서 얻어진 다음 결과를 이용\[\delta_n=\frac{(y)_n(z)_n x^n}{y^n z^n}\], \(\gamma_n=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}\)\[\gamma_{n}=\sum_{r=0}^{\infty}\frac{\delta_{n+r}}{(x)_{r+2n}(q)_{r}}\]

Specialize\[x=q, y\to\infty, z\to\infty\].
Bailey pair\[\delta_n=q^{n^2}\]\[\gamma_n=\frac{1}{(q)_{\infty}}q^{n^2}\]\[\gamma_L=\sum_{r=L}^{\infty}\frac{\delta_r}{(q)_{r-L}(aq)_{r+L}}=\sum_{r=0}^{\infty}\frac{\delta_{r+L}}{(q)_{r}(aq)_{r+2L}}\]

Use the following [Slater51] (4.1)\[\sum_{r=0}^{n}\frac{(1-aq^{2r})(-1)^{r}q^{\frac{1}{2}(r^2+r)}(a)_{r}(c)_{r}(d)_{r}a^{r}}{(a)_{n+r+1}(q)_{n-r}(q)_{r}(aq/c)_{r}(aq/d)_{r}c^{r}d^{r}}=\frac{(aq/cd)_{n}}{(q)_{n}(aq/c)_{n}(aq/d)_{n}}\]
Specialize\[a=1,c=0,d=\infty\]
Bailey pair\[\alpha_{0}=1\], \(\alpha_{r}=(-1)^{r}(1+q^r)q^{\frac{1}{2}r(r-1)}\)\[\beta_{0}=1\], \(\beta_{r}=0\)\[\beta_n=\sum_{r=0}^{n}\frac{\alpha_r}{(x)_{n-r}(q)_{n+r}}=\sum_{r=0}^{n}\frac{(-1)^{r}(1+q^r)q^{\frac{1}{2}r(r-1)}}{(q)_{n-r}(q)_{n+r}}=0\]

Bailey pairs\[\delta_n=q^{n^2}\]\[\gamma_n=\frac{1}{(q)_{\infty}}q^{n^2}\]\[\alpha_{0}=1\], \(\alpha_{r}=(-1)^{r}(1+q^r)q^{\frac{1}{2}r(r-1)}\)\[\beta_{0}=1\], \(\beta_{r}=0\)

\(\prod_{n=1}^{\infty}(1-q^n)=1+\sum_{n=1}^{\infty}(-1)^{n}(q^{\frac{3 n^2-n}{2}}+q^{\frac{3 n^2+n}{2}})=\sum_{n=-\infty}^\infty(-1)^nq^{n(3n-1)/2}\)

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