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Durfee 사각형 항등식(Durfee rectangle identity)

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  • (Durfee rectangle identity)

    l \in \mathbb{N},

    \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}

(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}}

l=n-m 로 두면, 

=\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}

■

 

 

 

http://www.springerlink.com/content/l842207736576587/

http://siba-ese.unisalento.it/index.php/quadmat/article/download/6953/6317

 

 

 

 

History

Last edited on 11/15/2011 03:41 by 피타고라스

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