"슬레이터 목록 (Slater's list)"의 두 판 사이의 차이

2011년 11월 12일 (토) 11:41 판

[Slater52-1] (4.1)
\(\sum_{r=0}^{n}\frac{(1-aq^{2r})(-1)^{r}q^{\frac{1}{2}(r^2+r)}(a)_{r}(c)_{r}(d)_{r}a^{r}}{(a)_{n+r+1}(q)_{n-r}(q)_{r}(aq/c)_{r}(aq/d)_{r}c^{r}d^{r}}=\frac{(aq/cd)_{n}}{(q)_{n}(aq/c)_{n}(aq/d)_{n}}\)

@@ 15번째 줄: / 15번째 줄: @@
 <h5 style="background-position: 0px 100%; font-size: 1.16em; margin: 0px; color: rgb(34, 61, 103); line-height: 3.42em; font-family: 'malgun gothic',dotum,gulim,sans-serif;">Group H</h5>
- '''[Slater52-1] '''(4.1)
+*   '''[Slater52-1] '''(4.1)<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>
-<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>
 * [[슬레이터 1]]
@@ 91번째 줄: / 89번째 줄: @@
 <h5>관련논문</h5>
+* '''[Slater52]'''Slater, L. J.[http://dx.doi.org/10.1112%2Fplms%2Fs2-54.2.147 Further identities of the Rogers-Ramanujan type] Proc. London Math. Soc.<br>1952s2-54: 147–167<br>
+* '''[Slater51]'''Slater, L. J. [http://dx.doi.org/10.1112/plms/s2-53.6.460 A New Proof of Rogers's Transformations of Infinite Series]Proc. London Math. Soc. 1951 s2-53: 460-475<br>
 * http://www.jstor.org/action/doBasicSearch?Query=