"Lebesgue identity"의 두 판 사이의 차이

수학노트
둘러보기로 가기 검색하러 가기
69번째 줄: 69번째 줄:
  
 
<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">articles</h5>
 
<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">articles</h5>
 
 
 
  
 
* '''[Alladi&Gordon1993]'''[http://dx.doi.org/10.1016/0097-3165%2893%2990061-C Partition identities and a continued fraction of Ramanujan]<br>
 
* '''[Alladi&Gordon1993]'''[http://dx.doi.org/10.1016/0097-3165%2893%2990061-C Partition identities and a continued fraction of Ramanujan]<br>

2010년 12월 1일 (수) 12:07 판

introduction
  • [Alladi&Gordon1993] 278&279p
  • Lebesgue's identity
    \(\sum_{k\geq 0}\frac{q^{k(k+1)/2}(-zq)_{k}}{(q)_{k}}=(-zq^2;q^2)_{\infty}(-q)_{\infty}=\prod_{m=1}^{\infty} (1+zq^{2m})(1+q^{m})\)

 

 

a 2x2 matrix
  • Use q-binomial identity
     \((-z;q)_{n}= \sum_{r=0}^{n} \begin{bmatrix} n\\ r\end{bmatrix}_{q}q^{r(r-1)/2}z^r\) and \((-zq;q)_{k}= \sum_{r=0}^{k} \begin{bmatrix} k\\ r\end{bmatrix}_{q}q^{r(r+1)/2}z^r\)
  • we get a rank 2 form of the Lebesgue's identity
    \(\sum_{k\geq 0}\frac{q^{k(k+1)/2}(-zq)_{k}}{(q)_{k}}=\sum_{i,j\geq 0}\frac{z^{j}q^{(i+j)(i+j+1)/2+j(j+1)/2}}{(q)_{i}(q)_{j}}=(-zq^2;q^2)_{\infty}(-q)_{\infty}\) where \(i=k-j\).
  • here we get a 2x2 matrix
    \( \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}\)

 

 

specializations
  • From the above, we can derive
    \(\sum_{i,j\geq 0}\frac{q^{(i^2+2ij+2j^2)/2+i/2}}{(q)_{i}(q)_{j}}=(-q;q^2)_{\infty}(-q)_{\infty}\)
  • Slater's list

 

 

history

 

 

related items

 

 

encyclopedia

 

 

books

 

[[4909919|]]

 

 

articles

 

 

question and answers(Math Overflow)

 

 

blogs

 

 

experts on the field

 

 

links