"Slater 92"의 두 판 사이의 차이

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8번째 줄: 8번째 줄:
  
 
<h5 style="line-height: 3.428em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic', dotum, gulim, sans-serif; font-size: 1.166em; background-image: ; background-color: initial; background-position: 0px 100%;">q-series identity</h5>
 
<h5 style="line-height: 3.428em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic', dotum, gulim, sans-serif; font-size: 1.166em; background-image: ; background-color: initial; background-position: 0px 100%;">q-series identity</h5>
 
 
 
  
 
<math>\sum_{n=0}^{\infty}\frac{(q^3;q^3)_{n}q^{n(n+1)}}{ (q)_{n}(q;q^{2})_n(q^2;q^2)_{n}}=\frac{(q^{9};q^{27})_{\infty}(q^{18};q^{27})_{\infty}(q^{27};q^{27})_{\infty}}{(q)_{\infty}}</math>
 
<math>\sum_{n=0}^{\infty}\frac{(q^3;q^3)_{n}q^{n(n+1)}}{ (q)_{n}(q;q^{2})_n(q^2;q^2)_{n}}=\frac{(q^{9};q^{27})_{\infty}(q^{18};q^{27})_{\infty}(q^{27};q^{27})_{\infty}}{(q)_{\infty}}</math>
30번째 줄: 28번째 줄:
 
 
 
 
  
<math>\frac{(1-x)(1-x^2)^2}{(1-x^3)}=x^2</math> 
+
<math>\frac{(1-x)(1-x^2)^2}{(1-x^3)}=x^2</math>
  
 
<math>x^3+3x^2-1=0</math>
 
<math>x^3+3x^2-1=0</math>

2010년 6월 23일 (수) 21:33 판

type of identity and specialization

 

 

q-series identity

\(\sum_{n=0}^{\infty}\frac{(q^3;q^3)_{n}q^{n(n+1)}}{ (q)_{n}(q;q^{2})_n(q^2;q^2)_{n}}=\frac{(q^{9};q^{27})_{\infty}(q^{18};q^{27})_{\infty}(q^{27};q^{27})_{\infty}}{(q)_{\infty}}\)

 

 

 

 

 

Bethe type equation (cyclotomic equation)

 

\(\frac{(1-x)(1-x^2)^2}{(1-x^3)}=x^2\)

\(x^3+3x^2-1=0\)

 

\(x, -y, -z^{-1}\)가 방정식 의 해

 

 

dilogarithm identity

\(L(x^3)-3L(x^2)-3L(x)=-\frac{7}{3}L(1)\)

 

 

related items

 

 

 

books

 

[[4909919|]]

 

 

articles