"Lebesgue identity"의 두 판 사이의 차이

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* '''[Alladi&Gordon1993] 278&279p'''<br>
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*  Lebesgue's identity<br><math>\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})</math><br>
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<h5 style="line-height: 2em; margin: 0px;">a 2x2 matrix</h5>
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*  Use q-binomial identity<br>  <math>(-z;q)_{n}= \sum_{r=0}^{n} \begin{bmatrix} n\\ r\end{bmatrix}_{q}q^{r(r-1)/2}z^r</math> and <math>(-zq;q)_{k}= \sum_{r=0}^{k} \begin{bmatrix} k\\ r\end{bmatrix}_{q}q^{r(r+1)/2}z^r</math><br>
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*  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>
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*  here we get a 2x2 matrix<br><math> \begin{bmatrix} 2 & 1  \\ 1 & 1 \end{bmatrix}</math><br>
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<h5 style="line-height: 2em; margin: 0px;">specializations</h5>
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*  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>
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* http://en.wikipedia.org/wiki/
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* http://www.proofwiki.org/wiki/
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* Princeton companion to mathematics([[2910610/attachments/2250873|Companion_to_Mathematics.pdf]])
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[[4909919|]]
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* '''[Alladi&Gordon1993]'''[http://dx.doi.org/10.1016/0097-3165%2893%2990061-C Partition identities and a continued fraction of Ramanujan]<br>
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** Krishnaswami Alladi and Basil Gordon, 1993   
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* http://www.ams.org/mathscinet
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* [http://www.zentralblatt-math.org/zmath/en/ ]http://www.zentralblatt-math.org/zmath/en/
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* http://arxiv.org/
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* http://www.pdf-search.org/
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* http://pythagoras0.springnote.com/
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* [http://math.berkeley.edu/%7Ereb/papers/index.html http://math.berkeley.edu/~reb/papers/index.html]
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* http://dx.doi.org/
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* http://arxiv.org/
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* [http://detexify.kirelabs.org/classify.html Detexify2 - LaTeX symbol classifier]
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* [http://pythagoras0.springnote.com/pages/1947378 수식표현 안내]
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* [http://www.research.att.com/%7Enjas/sequences/index.html The On-Line Encyclopedia of Integer Sequences]
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* http://functions.wolfram.com/
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*

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

 

 

related items

 

 

encyclopedia

 

 

books

 

[[4909919|]]

 

 

articles

 

 

 

question and answers(Math Overflow)

 

 

blogs

 

 

experts on the field

 

 

links