Durfee 사각형 항등식(Durfee rectangle identity)

개요

• (Durfee rectangle identity)

\(l \in \mathbb{N}\)에 대하여, 다음이 성립한다 \[\sum_{n,m\geq 0, n-m=l}\frac{q^{nm}}{(q)_n(q)_m}=\frac{1}{(q)_{\infty}}\] 또는 \[\sum_{n\geq 0}\frac{q^{n(n+l)}}{(q)_n(q)_{n+l}}=\frac{1}{(q)_{\infty}}\]

 

(증명)

q-가우스 합 을 이용하자. \[\sum_{n=0}^{\infty}\frac{(a,q)_{n}(b,q)_{n}}{(c ,q)_{n}(q ,q)_{n}}(\frac{c}{ab})^{n}=\frac{(c/a;q)_{\infty}(c/b;q)_{\infty}}{(c;q)_{\infty}(c/(ab);q)_{\infty}}\]

여기서 \(a\to\infty, b\to\infty,c=q^l\) 로 두면, 원하는 항등식을 얻는다. ■

 

 

(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/ ■

 

 

 

 

응용

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

(증명) \[ \begin{aligned} \frac{\sum_{l\geq 0}q^{\frac{a}{2}l^2+bl+c}}{(q)_{\infty}}&=\sum_{l\geq 0}\frac{q^{\frac{a}{2}l^2+bl+c}}{(q)_{\infty}} \end{aligned} \]

\(l=n-m\) 로 두면,  \[ \begin{aligned} {}&=\sum_{l\geq 0}\sum_{n,m\geq 0, n-m=l}\frac{q^{\frac{a}{2}l^2+bl+c}q^{nm}}{(q)_n(q)_m}\\ {}&=\sum_{n,m\geq 0}\frac{q^{nm+\frac{a}{2}(n-m)^2+b(n-m)+c}}{(q)_n(q)_m}=\sum_{n,m\geq 0}\frac{q^{\frac{1}{2}(an^2+(2-2a)mn+am^2)+b(n-m)+c}}{(q)_n(q)_m} \end{aligned} \]

■

 

 

 

 

역사

 

• http://www.google.com/search?hl=en&tbs=tl:1&q=
• 수학사 연표

 

 

메모

• http://www.springerlink.com/content/l842207736576587/
• http://siba-ese.unisalento.it/index.php/quadmat/article/download/6953/6317
• Math Overflow http://mathoverflow.net/search?q=