Lebesgue identity
http://bomber0.myid.net/ (토론)님의 2010년 9월 25일 (토) 23:42 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
\(\sum_{i,j\geq 0}\frac{q^{(i^2+2ij+2j^2)/2+i/2}}{(q)_{i}(q)_{j}}=(-q;q^2)_{\infty}(-q)_{\infty}\)
history
encyclopedia
- http://en.wikipedia.org/wiki/
- http://www.scholarpedia.org/
- http://www.proofwiki.org/wiki/
- Princeton companion to mathematics(Companion_to_Mathematics.pdf)
books
- 2010년 books and articles
- http://gigapedia.info/1/
- http://gigapedia.info/1/
- http://www.amazon.com/s/ref=nb_ss_gw?url=search-alias%3Dstripbooks&field-keywords=
[[4909919|]]
articles
- [Alladi&Gordon1993]Partition identities and a continued fraction of Ramanujan
- Krishnaswami Alladi and Basil Gordon, 1993
- http://www.ams.org/mathscinet
- [1]http://www.zentralblatt-math.org/zmath/en/
- http://arxiv.org/
- http://www.pdf-search.org/
- http://pythagoras0.springnote.com/
- http://math.berkeley.edu/~reb/papers/index.html
- http://dx.doi.org/
question and answers(Math Overflow)
blogs
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experts on the field