"Durfee 사각형 항등식(Durfee rectangle identity)"의 두 판 사이의 차이

2011년 7월 29일 (금) 06:48 판

(Durfee rectangle identity)
For any \(l \in \mathbb{Z}\),
\(\sum_{n,m\geq 0, n-m=l}\frac{q^{nm}}{(q)_n(q)_m}=\frac{1}{(q)_{\infty}}\) or
For any \(l \in \mathbb{N}\),
\(\sum_{n\geq 0}\frac{q^{n(n+l)}}{(q)_n(q)_{n+l}}=\frac{1}{(q)_{\infty}}\)

\(\sum_{n=0}^\infty p(n)q^n = 1+\sum_{n=1}\frac{q^{n^2}}{(1-q)^2(1-q^2)^2\cdots(1-q^n)^2}\)

(증명)

@@ 1번째 줄: / 1번째 줄: @@
+*  (Durfee rectangle identity)<br> For any <math>l \in \mathbb{Z}</math>,<br><math>\sum_{n,m\geq 0, n-m=l}\frac{q^{nm}}{(q)_n(q)_m}=\frac{1}{(q)_{\infty}}</math> or <br> For any <math>l \in \mathbb{N}</math>,<br><math>\sum_{n\geq 0}\frac{q^{n(n+l)}}{(q)_n(q)_{n+l}}=\frac{1}{(q)_{\infty}}</math><br>
+<math>\sum_{n=0}^\infty p(n)q^n = 1+\sum_{n=1}\frac{q^{n^2}}{(1-q)^2(1-q^2)^2\cdots(1-q^n)^2}</math>
+(증명)
+http://cfranc.wordpress.com/2009/11/24/an-identity-of-ramanujan/ ■
+http://www.springerlink.com/content/l842207736576587/
+http://siba-ese.unisalento.it/index.php/quadmat/article/download/6953/6317