"Lebesgue identity"의 두 판 사이의 차이

2010년 12월 2일 (목) 16:45 판

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

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 (rank 2 case)
\( \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}\)

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

[[4909919|]]

[Alladi&Gordon1993]Partition identities and a continued fraction of Ramanujan
- Krishnaswami Alladi and Basil Gordon, 1993

@@ 13번째 줄: / 13번째 줄: @@
 *  we get a rank 2 form of the Lebesgue's identity<br><math>\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}</math> where <math>i=k-j</math>.<br>
-*  here we get a 2x2 matrix<br><math> \begin{bmatrix} 2 & 1  \\ 1 & 1 \end{bmatrix}</math><br>
+*  here we get a 2x2 matrix ([[rank 2 case]])<br><math> \begin{bmatrix} 2 & 1  \\ 1 & 1 \end{bmatrix}</math><br>
@@ 23번째 줄: / 23번째 줄: @@
 *  From the above, we can derive<br><math>\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}</math><br>
 * [[Slater list|Slater's list]]<br>