"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>
  
 <br> 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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*  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>
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This corresponds to <math>\frac{\eta(\tau)^2}{\eta(2\tau)}</math> in the [[weight 1/2 eta quotients]]
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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]] ????
  
 
 
 
 
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* [[#]]<br>
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* [[1 Nahm's conjecture|Nahm's conjecture]]<br>
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2010년 12월 3일 (금) 07:19 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 ????

 

 

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