"Lebesgue identity"의 두 판 사이의 차이

수학노트
둘러보기로 가기 검색하러 가기
imported>Pythagoras0
 
(사용자 2명의 중간 판 10개는 보이지 않습니다)
1번째 줄: 1번째 줄:
 
==introduction==
 
==introduction==
 
* {{수학노트|url=르벡_항등식}}
 
* {{수학노트|url=르벡_항등식}}
* '''[Alladi&Gordon1993] 278&279p'''
 
:<math>f(a,c)=\sum_{k\geq 0}\frac{a^{k}q^{k(k-1)/2}(-cq)_{k}}{(q)_{k}}</math><br>
 
  
 
 
  
==fermionic form expression==
+
==fermionic formula==
 +
* '''[Alladi&Gordon1993] 278&279p'''
 
:<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+2j^2-i}{2}}}{(q)_{i}(q)_{j}}\label{faz}</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+2j^2-i}{2}}}{(q)_{i}(q)_{j}}\label{faz}</math>
  
12번째 줄: 10번째 줄:
  
 
We use the q-binomial identity (see [[useful techniques in q-series]])
 
We use the q-binomial identity (see [[useful techniques in q-series]])
:<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>(-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>
 
:<math>(-zq;q)_{k}= \sum_{r=0}^{k} \begin{bmatrix} k\\ r\end{bmatrix}_{q}q^{r(r+1)/2}z^r.</math>
  
22번째 줄: 20번째 줄:
 
\end{aligned}
 
\end{aligned}
 
</math>
 
</math>
By putting <math>j=r</math> and <math>k=i+j</math>,  
+
By putting <math>r=j</math> and <math>k=i+j</math>,  
 
:<math>
 
:<math>
 
\begin{aligned}
 
\begin{aligned}
28번째 줄: 26번째 줄:
 
{}&=\sum_{i,j\geq 0}\frac{a^{i+j}z^{j}q^{\frac{i^2+2ij+2j^2-i}{2}}}{(q)_{i}(q)_{j}}
 
{}&=\sum_{i,j\geq 0}\frac{a^{i+j}z^{j}q^{\frac{i^2+2ij+2j^2-i}{2}}}{(q)_{i}(q)_{j}}
 
\end{aligned}
 
\end{aligned}
</math> 
+
</math>
  
here we get a 2x2 matrix ([[rank 2 case]])
+
here we get a 2x2 matrix ([[rank 2 case]])
 
:<math> \begin{bmatrix} 1 & 1  \\ 1 & 2 \end{bmatrix}</math>
 
:<math> \begin{bmatrix} 1 & 1  \\ 1 & 2 \end{bmatrix}</math>
  
 
+
==specilizations : Lebesgue's identity==
 
 
 
 
 
 
==Lebesgue's identity==
 
 
 
 
*  Put a=q, c=z. we get Lebesgue's identity.
 
*  Put a=q, c=z. we get Lebesgue's identity.
:<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>
+
:<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>
 
*  special case : we get a rank 2 form of Lebesgue's identity
 
*  special case : we get a rank 2 form of Lebesgue's identity
:<math>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}</math><br>
+
:<math>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}</math>
  
 
+
  
 
+
  
 
==specializations==
 
==specializations==
 
* we expect to find five vectors for linear terms
 
* we expect to find five vectors for linear terms
$$\vec{b}=(1/2,-1),(0,0),(1/2,0),(1/2,1),(1,1)$$
+
:<math>\vec{b}=(1/2,-1),(0,0),(1/2,0),(1/2,1),(1,1)</math>
 +
* for a complete list, see [[Ramanujan-Göllnitz-Gordon continued fraction]]
 +
 
  
 
===Theorem===
 
===Theorem===
For $\vec{b}=(1/2,0)$,
+
For <math>\vec{b}=(1/2,0)</math>,
 
:<math>f(q,q^{-1})=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+2j^2}{2}+\frac{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},</math>
 
:<math>f(q,q^{-1})=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+2j^2}{2}+\frac{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},</math>
For $\vec{b}=(1/2,1)$,
+
For <math>\vec{b}=(1/2,1)</math>,
 
:<math>f(q,1)=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+2j^2}{2}+\frac{i}{2}+j}}{(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}}</math>
 
:<math>f(q,1)=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+2j^2}{2}+\frac{i}{2}+j}}{(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}}</math>
 +
  
 
===proof===
 
===proof===
70번째 줄: 66번째 줄:
  
 
* [[Slater list]]
 
* [[Slater list]]
===search for other b's===
+
 
* $\vec{b}=(1/2,0)$, http://oeis.org/A080054
+
 
:<math>f_{A,B,0}=\frac{(q^2;q^4)_{\infty}}{(q;q^4)_{\infty}^2(q^3;q^4)_{\infty}^2}</math>
 
* $\vec{b}=(1/2,1)$, http://oeis.org/A001935
 
:<math>f_{A,B,0}=\frac{1}{(q^1;q^4)_{\infty}(q^2;q^4)_{\infty}(q^3;q^4)_{\infty}}</math>
 
* $\vec{b}=(1/2,-1)$, http://oeis.org/A001935
 
$$
 
f_{A,B,0}=\frac{2}{(q^1;q^4)_{\infty}(q^2;q^4)_{\infty}(q^3;q^4)_{\infty}}
 
$$
 
* $\vec{b}=(0,0)$, http://oeis.org/A036016
 
$$
 
\frac{\left(Q^3;Q^8\right){}_{\infty } \left(Q^5;Q^8\right){}_{\infty } \left(Q^8;Q^8\right){}_{\infty }}{\left(Q;Q^4\right){}_{\infty } \left(Q^3;Q^4\right){}_{\infty } \left(Q^4;Q^4\right){}_{\infty }} =\frac{1}{\left(Q;Q^8\right){}_{\infty } \left(Q^7;Q^8\right){}_{\infty } \left(Q^4;Q^8\right){}_{\infty }}=\frac{1}{\left(q^{1/2};q^4\right){}_{\infty } \left(q^{7/2};q^4\right){}_{\infty } \left(q^{2};q^4\right){}_{\infty }}
 
$$
 
where $Q^2=q$
 
* $\vec{b}=(1,1)$, http://oeis.org/A036015
 
$$
 
\frac{1}{\left(Q^3;Q^8\right){}_{\infty } \left(Q^4;Q^8\right){}_{\infty } \left(Q^5;Q^8\right){}_{\infty }}=\frac{1}{\left(q^{3/2};q^4\right){}_{\infty } \left(q^{2};q^4\right){}_{\infty } \left(q^{5/2};q^4\right){}_{\infty }}
 
$$
 
where $Q^2=q$
 
* $\frac{\eta(4\tau)}{\eta(\tau)}$
 
  
 
==continued fraction expression==
 
==continued fraction expression==
  
* [[rank 2 continued fraction]]<br>
+
* [[rank 2 continued fraction]]
 
* '''[Alladi&Gordon1993] 277-278p'''
 
* '''[Alladi&Gordon1993] 277-278p'''
 
* Let <math>f(a,c)=\sum_{k\geq 0}\frac{a^{k}q^{k(k-1)/2}(-cq)_{k}}{(q)_{k}}</math> as above
 
* Let <math>f(a,c)=\sum_{k\geq 0}\frac{a^{k}q^{k(k-1)/2}(-cq)_{k}}{(q)_{k}}</math> as above
 
* consider the following continued fractions
 
* consider the following continued fractions
 
:<math>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}</math>
 
:<math>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}</math>
:<math>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}</math><br> where
+
:<math>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}</math> where
 
:<math>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{q^{\frac{i^2+2ij+2j^2}{2}+\frac{i}{2}}}{(q)_{i}(q)_{j}}</math>
 
:<math>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{q^{\frac{i^2+2ij+2j^2}{2}+\frac{i}{2}}}{(q)_{i}(q)_{j}}</math>
 
and  
 
and  
:<math>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{q^{\frac{i^2+2ij+2j^2}{2}+\frac{i}{2}+j}}{(q)_{i}(q)_{j}}</math><br>
+
:<math>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{q^{\frac{i^2+2ij+2j^2}{2}+\frac{i}{2}+j}}{(q)_{i}(q)_{j}}</math>
 
*  applications
 
*  applications
 
:<math>R^N(1,1)=f(q,q^{-1})=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+2j^2}{2}+\frac{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}</math>
 
:<math>R^N(1,1)=f(q,q^{-1})=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+2j^2}{2}+\frac{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}</math>
:<math>R^{D}(1,1)=f(q,1)=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+2j^2}{2}+\frac{i}{2}+j}}{(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}}</math><br>
+
:<math>R^{D}(1,1)=f(q,1)=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+2j^2}{2}+\frac{i}{2}+j}}{(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}}</math>
 
*  continued fraction
 
*  continued fraction
 
:<math>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}}</math>
 
:<math>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}}</math>
  
 
 
 
 
 
 
==related dilogarithm identity==
 
  
 
 
  
 
+
 
 
 
 
  
 
==comparison with Rogers-Selberg identities==
 
==comparison with Rogers-Selberg identities==
124번째 줄: 94번째 줄:
 
* [[Rogers-Selberg identities]]
 
* [[Rogers-Selberg identities]]
 
:<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>
 
:<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>
:<math>A(q)W(q)=AG_{3,3}(q)</math><br> where
+
:<math>A(q)W(q)=AG_{3,3}(q)</math> where
:<math>W(q)=(-q)_{\infty}=\frac{(q^{2};q^{2})_{\infty}}{(q;q)_{\infty}}</math><br>
+
:<math>W(q)=(-q)_{\infty}=\frac{(q^{2};q^{2})_{\infty}}{(q;q)_{\infty}}</math>
 
*  Lebesgue's identity
 
*  Lebesgue's identity
:<math>\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}}</math><br>
+
:<math>\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}}</math>
  
 
+
  
 
(proof)
 
(proof)
139번째 줄: 109번째 줄:
 
:<math>(-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)}</math>. ■
 
:<math>(-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)}</math>. ■
  
 
+
  
 
+
  
 
:<math>W(q)=\frac{\eta(2\tau)}{\eta(\tau)}</math>
 
:<math>W(q)=\frac{\eta(2\tau)}{\eta(\tau)}</math>
150번째 줄: 120번째 줄:
 
* see [[eta product and eta quotient]] also
 
* see [[eta product and eta quotient]] also
  
 
+
  
 
+
  
 
==history==
 
==history==
158번째 줄: 128번째 줄:
 
* http://www.google.com/search?hl=en&tbs=tl:1&q=
 
* http://www.google.com/search?hl=en&tbs=tl:1&q=
  
 
+
  
 
+
  
 
==related items==
 
==related items==
  
* [[1 Nahm's conjecture|Nahm's conjecture]]<br>
+
* [[1 Nahm's conjecture|Nahm's conjecture]]
* [[3-states Potts model]]<br>
+
* [[3-states Potts model]]
  
 
+
 +
==computational resource==
 +
* https://docs.google.com/file/d/0B8XXo8Tve1cxZjNHc0hNUmxfREk/edit
  
 
+
 
 
==encyclopedia==
 
 
 
* http://en.wikipedia.org/wiki/
 
* http://www.scholarpedia.org/
 
* http://www.proofwiki.org/wiki/
 
 
 
  
 
==articles==
 
==articles==
  
* New Proofs of Identities of Lebesgue and Göllnitz via Tilings<br>
+
* Little, David P., and James A. Sellers. 2009. “New Proofs of Identities of Lebesgue and Göllnitz via Tilings.” Journal of Combinatorial Theory, Series A 116 (1) (January): 223–231. doi:10.1016/j.jcta.2008.05.004. http://www.sciencedirect.com/science/article/pii/S0097316508000782
** DP Little, 2007
+
* Chen, Sin-Da, and Sen-Shan Huang. 2005. “On the Series Expansion of the Göllnitz-Gordon Continued Fraction.” International Journal of Number Theory 1 (1): 53–63. doi:10.1142/S1793042105000030. http://www.worldscientific.com/doi/abs/10.1142/S1793042105000030
 
 
 
* '''[Alladi&Gordon1993]''' Alladi, Krishnaswami, and Basil Gordon. 1993. [http://dx.doi.org/10.1016/0097-3165%2893%2990061-C Partition identities and a continued fraction of Ramanujan] <em>Journal of Combinatorial Theory, Series A</em> 63 (2) (July): 275-300. doi:10.1016/0097-3165(93)90061-C.
 
* '''[Alladi&Gordon1993]''' Alladi, Krishnaswami, and Basil Gordon. 1993. [http://dx.doi.org/10.1016/0097-3165%2893%2990061-C Partition identities and a continued fraction of Ramanujan] <em>Journal of Combinatorial Theory, Series A</em> 63 (2) (July): 275-300. doi:10.1016/0097-3165(93)90061-C.
  
  
 
[[분류:개인노트]]
 
[[분류:개인노트]]
[[분류:thesis]]
 
 
[[분류:q-series]]
 
[[분류:q-series]]
 +
[[분류:migrate]]

2020년 11월 16일 (월) 06:09 기준 최신판

introduction


fermionic formula

  • [Alladi&Gordon1993] 278&279p

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

(proof)

We use the q-binomial identity (see 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 \[(-zq;q)_{k}= \sum_{r=0}^{k} \begin{bmatrix} k\\ r\end{bmatrix}_{q}q^{r(r+1)/2}z^r.\]

Then the LHS of \ref{faz} can be written as \[ \begin{aligned} f(a,z)& =\sum_{k\geq 0}\frac{a^{k}q^{k(k-1)/2}(-zq)_{k}}{(q)_{k}}\\ {}&=\sum_{k\geq 0}\frac{a^kq^{k(k-1)/2}}{(q)_{k}}\sum_{r=0}^{k} \begin{bmatrix} k\\ r\end{bmatrix}_{q}q^{r(r+1)/2}z^r \end{aligned} \] By putting \(r=j\) and \(k=i+j\), \[ \begin{aligned} {}&=\sum_{i,j\geq 0}\frac{a^{i+j}z^{j}q^{(i+j)(i+j-1)/2+j(j+1)/2}}{(q)_{i}(q)_{j}}\\ {}&=\sum_{i,j\geq 0}\frac{a^{i+j}z^{j}q^{\frac{i^2+2ij+2j^2-i}{2}}}{(q)_{i}(q)_{j}} \end{aligned} \] ■

\[ \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}\]

specilizations : Lebesgue's identity

  • Put a=q, c=z. we get 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 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

  • we expect to find five vectors for linear terms

\[\vec{b}=(1/2,-1),(0,0),(1/2,0),(1/2,1),(1,1)\]


Theorem

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


proof

Let us use the following identities from 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 \[f(q,q^{-1})=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+2j^2}{2}+\frac{i}{2}}}{(q)_{i}(q)_{j}}=(-q;q^2)_{\infty}(-q)_{\infty}=\frac{(q^{2};q^{2})_{\infty}^3}{(q;q)_{\infty}^2(q^{4};q^{4})_{\infty}}=\frac{(q^2;q^4)_{\infty}}{(q;q^4)_{\infty}^2(q^3;q^4)_{\infty}^2}\] \[f(q,1)=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+2j^2}{2}+\frac{i}{2}+j}}{(q)_{i}(q)_{j}}=(-q^2;q^2)_{\infty}(-q)_{\infty}=\frac{(q^4;q^4)_{\infty}}{(q;q)_{\infty}}=\frac{1}{(q^1;q^4)_{\infty}(q^2;q^4)_{\infty}(q^3;q^4)_{\infty}}.\]■


continued fraction expression

  • rank 2 continued fraction
  • [Alladi&Gordon1993] 277-278p
  • Let \(f(a,c)=\sum_{k\geq 0}\frac{a^{k}q^{k(k-1)/2}(-cq)_{k}}{(q)_{k}}\) as above
  • consider the following continued fractions

\[F(a,c)=\frac{f(a,c)}{f(aq,c)}=1+a+\frac{acq}{1+aq} {\ \atop+} \frac{acq^2}{1+aq^2}{\ \atop+} \frac{acq^3}{1} {\ \atop+\dots}\] \[R(a,b)=\frac{f(a,a^{-1}b)}{f(aq,a^{-1}b)}-a=\frac{R^{N}(a,b)}{R^{D}(a,b)}=1+\frac{bq}{1+aq} {\ \atop+} \frac{bq^2}{1+aq^2}{\ \atop+} \frac{bq^3}{1} {\ \atop+\dots}\] where \[R^{N}(a,b)=f(q,a^{-1}b)-af(aq,a^{-1}b)=f(aq,a^{-1}bq^{-1})=\sum_{k\geq 0}\frac{a^{k}q^{k(k+1)/2}(-a^{-1}b)_{k}}{(q)_{k}}=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+2j^2}{2}+\frac{i}{2}}}{(q)_{i}(q)_{j}}\] and \[R^{D}(a,b)=f(aq,a^{-1}b)=\sum_{k\geq 0}\frac{a^{k}q^{k(k+1)/2}(-a^{-1}bq)_{k}}{(q)_{k}}=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+2j^2}{2}+\frac{i}{2}+j}}{(q)_{i}(q)_{j}}\]

  • applications

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

  • continued fraction

\[R(1,1)=\frac{R^{N}(1,1)}{R^{D}(1,1)}=1+{q \over 1+q + } {q^2 \over 1+q^2+} {q^3 \over 1+q^3} \cdots=\frac{(q^2;q^4)_{\infty}^2}{(q^1;q^4)_{\infty}(q^3;q^4)_{\infty}}\]



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



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



history



related items


computational resource


articles