"Durfee 사각형 항등식(Durfee rectangle identity)"의 두 판 사이의 차이
(피타고라스님이 이 페이지의 이름을 Durfee 사각형 항등식(Durfee rectangle identity)로 바꾸었습니다.) |
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| − | * (Durfee rectangle identity)<br><math>l \in \mathbb{ | + | * (Durfee rectangle identity)<br><math>l \in \mathbb{N}</math>,<br><math>\sum_{n,m\geq 0, n-m=l}\frac{q^{nm}}{(q)_n(q)_m}=\frac{1}{(q)_{\infty}}</math> 또는<br><math>\sum_{n\geq 0}\frac{q^{n(n+l)}}{(q)_n(q)_{n+l}}=\frac{1}{(q)_{\infty}}</math><br> |
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(따름정리) | (따름정리) | ||
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<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> | <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> | ||
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(증명) | (증명) | ||
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http://cfranc.wordpress.com/2009/11/24/an-identity-of-ramanujan/ ■ | http://cfranc.wordpress.com/2009/11/24/an-identity-of-ramanujan/ ■ | ||
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| 27번째 줄: | 43번째 줄: | ||
(pf) | (pf) | ||
| − | <math>\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}}= | + | <math>\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}}</math> |
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| + | <math>l=n-m</math> 로 두면, | ||
| − | <math>l | + | <math>=\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}</math> |
<math>=\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}</math> | <math>=\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}</math> | ||
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2011년 11월 15일 (화) 04:38 판
- (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)
(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,
\(=\sum_{l\geq 0}\sum_{n,m\geq 0, n-m=l}\frac{q^{\frac{a}{2}l^2+bl+c}q^{nm}}{(q)_n(q)_m}\)
\(=\sum_{n,m\geq 0}\frac{q^{nm+\frac{a}{2}(n-m)^2+b(n-m)+c}}{(q)_n(q)_m}=\sum_{n,m\geq 0}\frac{q^{\frac{1}{2}(an^2+(2-2a)mn+am^2)+b(n-m)+c}}{(q)_n(q)_m}\)
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http://www.springerlink.com/content/l842207736576587/
http://siba-ese.unisalento.it/index.php/quadmat/article/download/6953/6317