"Lebesgue identity"의 두 판 사이의 차이
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<h5 style="line-height: 2em; margin: 0px;">specializations</h5> | <h5 style="line-height: 2em; margin: 0px;">specializations</h5> | ||
− | * | + | * 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> | * [[Slater list|Slater's list]]<br> | ||
48번째 줄: | 48번째 줄: | ||
− | + | W(q)=\frac{\eta{2\tau}} | |
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+ | any relation to <math>\frac{\eta(\tau)^2}{\eta(2\tau)}</math> in the [[weight 1/2 eta quotients]] ???? | ||
64번째 줄: | 66번째 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%;">related items</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%;">related items</h5> | ||
− | * [[ | + | * [[1 Nahm's conjecture|Nahm's conjecture]]<br> |
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2010년 12월 3일 (금) 06:19 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 (rank 2 case)
\( \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}\)
specializations
- From the above, we can derive
\(\sum_{i,j\geq 0}\frac{q^{(i^2+2ij+2j^2)/2+i/2}}{(q)_{i}(q)_{j}}=(-q;q^2)_{\infty}(-q)_{\infty}\)
comparison with Rogers-Selberg identities
- Rogers-Selberg identities
\(AG_{3,3}(q)=\sum_{n_1,n_{2}\geq0}\frac{q^{n_{1}^2+2n_1n_2+2n_{2}^{2}}}{(q)_{n_1}(q)_{n_{2}}}=\prod_{r\neq 0,\pm 3 \pmod {7}}\frac{1}{1-q^r}=\frac{(q^3;q^7)_\infty (q^4; q^7)_\infty(q^7;q^7)_\infty}{(q)_\infty}\)
\(A(q)W(q)=AG_{3,3}(q)\)
where
\(W(q)=(-q)_{\infty}=\frac{(q^{2};q^{2})_{\infty}}{(q;q)_{\infty}}\) - Lebesgue's identity
\(\frac{W(q)^2}{W(q^2)}=\sum_{i,j\geq 0}\frac{q^{(i^2+2ij+2j^2)/2+i/2}}{(q)_{i}(q)_{j}}\)
(proof)
Note that from useful techniques in q-series
\((-q;q^{2})_{\infty}=\frac{(-q;q)_{\infty}}{(-q^{2};q^{2})_{\infty}}=\frac{(q^{2};q^{2})_{\infty}(q^{2};q^{2})_{\infty}}{(q^{4};q^{4})_{\infty}(q;q)_{\infty}}=\frac{W(q)}{W(q^2)}\)
Therefore
\((-q;q^2)_{\infty}(-q)_{\infty}=\frac{W(q)^2}{W(q^2)}\). ■
W(q)=\frac{\eta{2\tau}}
any relation to \(\frac{\eta(\tau)^2}{\eta(2\tau)}\) in the weight 1/2 eta quotients ????
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
- 구글 블로그 검색
- http://ncatlab.org/nlab/show/HomePage
experts on the field