"슬레이터 1"의 두 판 사이의 차이

2011년 11월 12일 (토) 12:17 판

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

@@ 26번째 줄: / 26번째 줄: @@
 *  Use the following <br><math>\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}}</math><br>
 *  Specialize<br><math>a=1,c=0,d=\infty</math><br>
-*  Bailey pair<br><math>\alpha_{0}=1</math>, <math>\alpha_{n}=(-1)^{n}q^{n^2}(1-q^{2n+1})/(1-q)</math><br><math>\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}{(q)_{n}(-q)_{n}}</math><br>
+*   <br> Bailey pair<br><math>\alpha_{0}=1</math>, <math>\alpha_{r}=(-1)^{n}(1+q^r)q^{\frac{1}{2}r(r-1)}</math><br><math>\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}{(q)_{n}(-q)_{n}}</math><br>