Durfee 사각형 항등식(Durfee rectangle identity)

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

 

 

\(\frac{\theta}{\eta}=\frac{\sum_{l\geq 0}q^{\frac{a}{2}l^2+bl+c}}{(q)_{\infty}}=\sum_{n,m\geq 0}\frac{q^{\frac{1}{2}(an^2+(2-2a)mn+am^2)+b(n-m)+c}}{(q)_n(q)_m}\)

(pf)

\(\frac{\sum_{l\geq 0}q^{\frac{a}{2}l^2+bl+c}}{(q)_{\infty}}=\sum_{l\geq 0}\frac{q^{\frac{a}{2}l^2+bl+c}}{(q)_{\infty}}=\sum_{l\geq 0}\sum_{n,m\geq 0, n-m=l}\frac{q^{\frac{a}{2}l^2+bl+c}q^{nm}}{(q)_n(q)_m}\sum_{n,m\geq 0}\frac{q^{nm+\frac{a}{2}(n-m)^2+b(n-m)+c}}{(q)_n(q)_m}=\sum_{n,m\geq 0}\frac{q^{\frac{1}{2}(an^2+(2-2a)mn+am^2)+b(n-m)+c}}{(q)_n(q)_m}\)

where we used \(l=n-m\). ■

 

(따름정리)

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

 

(증명)

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