"Lebesgue identity"의 두 판 사이의 차이

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<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%;">introduction==
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==introduction==
  
 
* [http://pythagoras0.springnote.com/pages/7958020 르벡 항등식]
 
* [http://pythagoras0.springnote.com/pages/7958020 르벡 항등식]
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<h5 style="line-height: 2em; margin: 0px;">fermionic form expression==
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==fermionic form expression==
  
 
<math>f(a,z)=\sum_{k\geq 0}\frac{a^{k}q^{k(k-1)/2}(-zq)_{k}}{(q)_{k}}=\sum_{i,j\geq 0}\frac{a^{i+j}z^{j}q^{\frac{i^2+2ij+j^2-i}{2}}}{(q)_{i}(q)_{j}}</math>
 
<math>f(a,z)=\sum_{k\geq 0}\frac{a^{k}q^{k(k-1)/2}(-zq)_{k}}{(q)_{k}}=\sum_{i,j\geq 0}\frac{a^{i+j}z^{j}q^{\frac{i^2+2ij+j^2-i}{2}}}{(q)_{i}(q)_{j}}</math>
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<h5 style="line-height: 2em; margin: 0px;">Lebesgue's identity==
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==Lebesgue's identity==
  
 
*  Put a=q, c=z. we get the Lebesgue's identity.<br><math>f(q,z)=\sum_{k\geq 0}\frac{q^{k}q^{k(k-1)/2}(-zq)_{k}}{(q)_{k}}=\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>
 
*  Put a=q, c=z. we get the Lebesgue's identity.<br><math>f(q,z)=\sum_{k\geq 0}\frac{q^{k}q^{k(k-1)/2}(-zq)_{k}}{(q)_{k}}=\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;">specializations==
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==specializations==
  
 
(Theorem)
 
(Theorem)
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<h5 style="line-height: 2em; margin: 0px;">continued fraction expression==
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==continued fraction expression==
  
 
* [[rank 2 continued fraction]]<br>
 
* [[rank 2 continued fraction]]<br>
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<h5 style="line-height: 2em; margin: 0px;">related dilogarithm identity==
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==related dilogarithm identity==
  
 
 
 
 
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<h5 style="line-height: 2em; margin: 0px;">comparison with Rogers-Selberg identities==
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==comparison with Rogers-Selberg identities==
  
 
* [[Rogers-Selberg identities]]<br><math>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}</math><br><math>A(q)W(q)=AG_{3,3}(q)</math><br> where<br><math>W(q)=(-q)_{\infty}=\frac{(q^{2};q^{2})_{\infty}}{(q;q)_{\infty}}</math><br>
 
* [[Rogers-Selberg identities]]<br><math>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}</math><br><math>A(q)W(q)=AG_{3,3}(q)</math><br> where<br><math>W(q)=(-q)_{\infty}=\frac{(q^{2};q^{2})_{\infty}}{(q;q)_{\infty}}</math><br>
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<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%;">history==
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==history==
  
 
* http://www.google.com/search?hl=en&tbs=tl:1&q=
 
* http://www.google.com/search?hl=en&tbs=tl:1&q=
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==related items==
  
 
* [[1 Nahm's conjecture|Nahm's conjecture]]<br>
 
* [[1 Nahm's conjecture|Nahm's conjecture]]<br>
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==encyclopedia==
  
 
* http://en.wikipedia.org/wiki/
 
* http://en.wikipedia.org/wiki/
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==books==
  
 
 
 
 
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==articles==
  
 
*  New Proofs of Identities of Lebesgue and Göllnitz via Tilings<br>
 
*  New Proofs of Identities of Lebesgue and Göllnitz via Tilings<br>

2012년 10월 28일 (일) 16:37 판

introduction

  • [Alladi&Gordon1993] 278&279p
    \(f(a,c)=\sum_{k\geq 0}\frac{a^{k}q^{k(k-1)/2}(-cq)_{k}}{(q)_{k}}\)

 

 

 

fermionic form expression

\(f(a,z)=\sum_{k\geq 0}\frac{a^{k}q^{k(k-1)/2}(-zq)_{k}}{(q)_{k}}=\sum_{i,j\geq 0}\frac{a^{i+j}z^{j}q^{\frac{i^2+2ij+j^2-i}{2}}}{(q)_{i}(q)_{j}}\)

(proof)

We use the q-binomial identity useful techniques in q-series

 \((-z;q)_{n}= \sum_{r=0}^{n} \begin{bmatrix} n\\ r\end{bmatrix}_{q}q^{r(r-1)/2}z^r\) and \((-zq;q)_{k}= \sum_{r=0}^{k} \begin{bmatrix} k\\ r\end{bmatrix}_{q}q^{r(r+1)/2}z^r\)

\(f(a,z)=\sum_{k\geq 0}\frac{a^{k}q^{k(k-1)/2}(-zq)_{k}}{(q)_{k}}\)

\(=\sum_{k\geq 0}\frac{a^kq^{k(k-1)/2}}{(q)_{k}}\sum_{r=0}^{k} \begin{bmatrix} k\\ r\end{bmatrix}_{q}q^{r(r+1)/2}z^r\). Put \(j=r\) and \(i=k-j\).

\(=\sum_{i,j\geq 0}\frac{a^{i+j}z^{j}q^{(i+j)(i+j-1)/2+j(j+1)/2}}{(q)_{i}(q)_{j}}\)

\(=\sum_{i,j\geq 0}\frac{a^{i+j}z^{j}q^{\frac{i^2+2ij+j^2-i}{2}}}{(q)_{i}(q)_{j}}\)  ■

  • here we get a 2x2 matrix (rank 2 case)
    \( \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}\)

 

 

Lebesgue's identity

  • Put a=q, c=z. we get the Lebesgue's identity.
    \(f(q,z)=\sum_{k\geq 0}\frac{q^{k}q^{k(k-1)/2}(-zq)_{k}}{(q)_{k}}=\sum_{k\geq 0}\frac{q^{k(k+1)/2}(-zq)_{k}}{(q)_{k}}=(-zq^2;q^2)_{\infty}(-q)_{\infty}=\prod_{m=1}^{\infty} (1+zq^{2m})(1+q^{m})\)
  • special case : we get a rank 2 form of the Lebesgue's identity
    \(f(q,z)=\sum_{k\geq 0}\frac{q^{k}q^{k(k-1)/2}(-zq)_{k}}{(q)_{k}}=\sum_{i,j\geq 0}\frac{z^{j}q^{\frac{i^2+2ij+j^2+i+2j}{2}}}{(q)_{i}(q)_{j}}=(-zq^2;q^2)_{\infty}(-q)_{\infty}\)

 

 

specializations

(Theorem)

\(f(q,q^{-1})=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+j^2+i}{2}}}{(q)_{i}(q)_{j}}=(-q;q^2)_{\infty}(-q)_{\infty}=\frac{(q^{2};q^{2})_{\infty}^3}{(q;q)_{\infty}^2(q^{4};q^{4})_{\infty}}=\frac{(q^2;q^4)_{\infty}}{(q;q^4)_{\infty}^2(q^3;q^4)_{\infty}^2}\)

\(f(q,1)=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+2j^2}{2}+\frac{i+2j}{2}}}{(q)_{i}(q)_{j}}=(-q^2;q^2)_{\infty}(-q)_{\infty}=\frac{(q^4;q^4)_{\infty}}{(q;q)_{\infty}}=\frac{1}{(q^1;q^4)_{\infty}(q^2;q^4)_{\infty}(q^3;q^4)_{\infty}}\)

(proof)

useful techniques in q-series

\((-q)_{n}=\frac{(q^2;q^2)_{n}}{(q;q)_{n}}\)

\((-q;q^{2})_{n}=\frac{(-q;q)_{n}}{(-q^{2};q^{2})_{n}}=\frac{(q^{2};q^{2})_{n}(q^{2};q^{2})_{n}}{(q^{4};q^{4})_{n}(q;q)_{n}}=\frac{(q^{2};q^{4})_{n}}{(q^{1};q^{4})_{n}(q^{3};q^{4})_{n}}\) .

\((-q^2;q^{2})_{n}=\frac{(q^4;q^4)_{n}}{(q^2;q^2)_{n}}=\frac{1}{(q^2;q^4)_{n}}\)

Therefore

\(f(q,q^{-1})=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+j^2+i}{2}}}{(q)_{i}(q)_{j}}=(-q;q^2)_{\infty}(-q)_{\infty}=\frac{(q^{2};q^{2})_{\infty}^3}{(q;q)_{\infty}^2(q^{4};q^{4})_{\infty}}=\frac{(q^2;q^4)_{\infty}}{(q;q^4)_{\infty}^2(q^3;q^4)_{\infty}^2}\)

\(f(q,1)=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+2j^2}{2}+\frac{i+2j}{2}}}{(q)_{i}(q)_{j}}=(-q^2;q^2)_{\infty}(-q)_{\infty}=\frac{(q^4;q^4)_{\infty}}{(q;q)_{\infty}}=\frac{1}{(q^1;q^4)_{\infty}(q^2;q^4)_{\infty}(q^3;q^4)_{\infty}}\). ■

 

 

continued fraction expression

  • rank 2 continued fraction
  • [Alladi&Gordon1993] 277-278p
    Let \(f(a,c)=\sum_{k\geq 0}\frac{a^{k}q^{k(k-1)/2}(-cq)_{k}}{(q)_{k}}\).
    \(F(a,c)=\frac{f(a,c)}{f(aq,c)}=1+a+\frac{acq}{1+aq} {\ \atop+} \frac{acq^2}{1+aq^2}{\ \atop+} \frac{acq^3}{1} {\ \atop+\dots}\)
    \(R(a,b)=\frac{f(a,a^{-1}b)}{f(aq,a^{-1}b)}-a=\frac{R^{N}(a,b)}{R^{D}(a,b)}=1+\frac{bq}{1+aq} {\ \atop+} \frac{bq^2}{1+aq^2}{\ \atop+} \frac{bq^3}{1} {\ \atop+\dots}\)
    where
    \(R^{N}(a,b)=f(q,a^{-1}b)-af(aq,a^{-1}b)=f(aq,a^{-1}bq^{-1})=\sum_{k\geq 0}\frac{a^{k}q^{k(k+1)/2}(-a^{-1}b)_{k}}{(q)_{k}}=\sum_{i,j\geq 0}\frac{a^{i}b^{j}q^{\frac{i^2+2ij+j^2+i}{2}}}{(q)_{i}(q)_{j}}\)
    \(R^{D}(a,b)=f(aq,a^{-1}b)=\sum_{k\geq 0}\frac{a^{k}q^{k(k+1)/2}(-a^{-1}bq)_{k}}{(q)_{k}}=\sum_{i,j\geq 0}\frac{a^{i}b^{j}q^{\frac{i^2+2ij+j^2+i+2j}{2}}}{(q)_{i}(q)_{j}}\)
  • application
    \(R^N(1,1)=f(q,q^{-1})=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+j^2+i}{2}}}{(q)_{i}(q)_{j}}=(-q;q^2)_{\infty}(-q)_{\infty}=\frac{(q^{2};q^{2})_{\infty}^3}{(q;q)_{\infty}^2(q^{4};q^{4})_{\infty}}=\frac{(q^2;q^4)_{\infty}}{(q^1;q^4)_{\infty}^2(q^3;q^4)_{\infty}^2}\)
    \(R^{D}(1,1)=f(q,1)=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+2j^2+i+2j}{2}}}{(q)_{i}(q)_{j}}=(-q^2;q^2)_{\infty}(-q)_{\infty}=\frac{(q^4;q^4)_{\infty}}{(q;q)_{\infty}}=\frac{1}{(q^1;q^4)_{\infty}(q^2;q^4)_{\infty}(q^3;q^4)_{\infty}}\)
  • continued fraction
    \(R(1,1)=\frac{R^{N}(1,1)}{R^{D}(1,1)}=1+{q \over 1+q + } {q^2 \over 1+q^2+} {q^3 \over 1+q^3} } \cdots=\frac{(q^2;q^4)_{\infty}^2}{(q^1;q^4)_{\infty}(q^3;q^4)_{\infty}}\)

 

 

related dilogarithm identity

 

 

 

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

\((-q;q^2)_{\infty}(-q)_{\infty}=\frac{(q^{2};q^{2})_{\infty}^3}{(q;q)_{\infty}^2(q^{4};q^{4})_{\infty}}=\frac{W(q)^2}{W(q^2)}\). ■

 

 

\(W(q)=\frac{\eta(2\tau)}{\eta(\tau)}\)

\(W(q^2)=\frac{\eta(4\tau)}{\eta(2\tau)}\)

\(\frac{W(q)^2}{W(q^2)}=\frac{(q^{2};q^{2})_{\infty}^3}{(q;q)_{\infty}^2(q^{4};q^{4})_{\infty}}=\frac{\eta(2\tau)^3}{\eta(\tau)^2\eta(4\tau)}=\sum_{i,j\geq 0}\frac{q^{(i^2+2ij+2j^2)/2+i/2}}{(q)_{i}(q)_{j}}\)

\(W(q^2)W(q)=\frac{\eta(4\tau)}{\eta(\tau)}=q^{1/8}\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+2j^2}{2}+\frac{i+2j}{2}}}{(q)_{i}(q)_{j}}=\frac{q^{1/8}(q^4;q^4)_{\infty}}{(q;q)_{\infty}}=\frac{q^{1/8}}{(q^1;q^4)_{\infty}(q^2;q^4)_{\infty}(q^3;q^4)_{\infty}}\)

eta product and eta quotient

 

 

history

 

 

related items

 

 

encyclopedia

 

 

books

 

4909919

 

 

articles

  • New Proofs of Identities of Lebesgue and Göllnitz via Tilings
    • DP Little, 2007
  • Alladi, Krishnaswami, and Basil Gordon. 1993. “Partition identities and a continued fraction of Ramanujan.” Journal of Combinatorial Theory, Series A 63 (2) (July): 275-300. doi:10.1016/0097-3165(93)90061-C.
  • [Alladi&Gordon1993]Partition identities and a continued fraction of Ramanujan ,Krishnaswami Alladi and Basil Gordon, 1993