"Lebesgue identity"의 두 판 사이의 차이
Pythagoras0 (토론 | 기여) |
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| (사용자 3명의 중간 판 48개는 보이지 않습니다) | |||
| 1번째 줄: | 1번째 줄: | ||
| − | + | ==introduction== | |
| + | * {{수학노트|url=르벡_항등식}} | ||
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| − | + | ==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> | |
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| − | <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+ | ||
(proof) | (proof) | ||
| − | We use the q-binomial identity [[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>(-zq;q)_{k}= \sum_{r=0}^{k} \begin{bmatrix} k\\ r\end{bmatrix}_{q}q^{r(r+1)/2}z^r.</math> | ||
| − | + | Then the LHS of \ref{faz} can be written as | |
| + | :<math> | ||
| + | \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} | ||
| + | </math> | ||
| + | By putting <math>r=j</math> and <math>k=i+j</math>, | ||
| + | :<math> | ||
| + | \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} | ||
| + | </math> ■ | ||
| − | <math> | + | * here we get a 2x2 matrix ([[rank 2 case]]) |
| + | :<math> \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}</math> | ||
| − | <math>=\sum_{k\geq 0}\frac{ | + | ==specilizations : 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> | ||
| + | * 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> | ||
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| − | * | + | ==specializations== |
| + | * we expect to find five vectors for linear terms | ||
| + | :<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]] | ||
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| − | + | ===Theorem=== | |
| + | 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> | ||
| + | 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> | ||
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| − | + | ===proof=== | |
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| − | + | Let us use the following identities from [[useful techniques in q-series]] | |
| − | + | :<math>(-q)_{n}=\frac{(q^2;q^2)_{n}}{(q;q)_{n}}</math> | |
| − | + | :<math>(-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}}</math> . | |
| − | + | :<math>(-q^2;q^{2})_{n}=\frac{(q^4;q^4)_{n}}{(q^2;q^2)_{n}}=\frac{1}{(q^2;q^4)_{n}}</math> | |
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| − | [[useful techniques in q-series]] | ||
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| − | <math>(-q)_{n}=\frac{(q^2;q^2)_{n}}{(q;q)_{n}}</math> | ||
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| − | <math>(-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}}</math> . | ||
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| − | <math>(-q^2;q^{2})_{n}=\frac{(q^4;q^4)_{n}}{(q^2;q^2)_{n}}=\frac{1}{(q^2;q^4)_{n}}</math> | ||
Therefore | Therefore | ||
| + | :<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,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>■ | ||
| − | + | * [[Slater list]] | |
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| − | * [[Slater | ||
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| − | + | ==continued fraction expression== | |
| − | < | + | * [[rank 2 continued fraction]] |
| + | * '''[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 | ||
| + | * 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>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> | ||
| + | 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> | ||
| + | * 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^{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 | ||
| + | :<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> | ||
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| − | + | ==comparison with Rogers-Selberg identities== | |
| − | * [[Rogers-Selberg identities]] | + | * [[Rogers-Selberg identities]] |
| − | * Lebesgue's identity | + | :<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> where | ||
| + | :<math>W(q)=(-q)_{\infty}=\frac{(q^{2};q^{2})_{\infty}}{(q;q)_{\infty}}</math> | ||
| + | * 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> | ||
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(proof) | (proof) | ||
Note that from [[useful techniques in q-series]] | Note that from [[useful techniques in q-series]] | ||
| − | + | :<math>(-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)}</math> | |
| − | <math>(-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)}</math> | ||
Therefore | Therefore | ||
| + | :<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>. ■ | ||
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| − | <math>\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}}</math> | + | :<math>W(q)=\frac{\eta(2\tau)}{\eta(\tau)}</math> |
| + | :<math>W(q^2)=\frac{\eta(4\tau)}{\eta(2\tau)}</math> | ||
| + | :<math>\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}}</math> | ||
| + | :<math>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}}</math> | ||
| − | + | * see [[eta product and eta quotient]] also | |
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| − | + | ==history== | |
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* 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]] | |
| + | * [[3-states Potts model]] | ||
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| + | ==computational resource== | ||
| + | * https://docs.google.com/file/d/0B8XXo8Tve1cxZjNHc0hNUmxfREk/edit | ||
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| − | + | ==articles== | |
| − | + | * 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 | |
| + | * 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. | ||
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| − | + | [[분류:개인노트]] | |
| − | + | [[분류:q-series]] | |
| − | + | [[분류:migrate]] | |
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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 \(k=i+j\), \[ \begin{aligned} {}&=\sum_{i,j\geq 0}\frac{a^{i+j}z^{j}q^{(i+j)(i+j-1)/2+j(j+1)/2}}{(q)_{i}(q)_{j}}\\ {}&=\sum_{i,j\geq 0}\frac{a^{i+j}z^{j}q^{\frac{i^2+2ij+2j^2-i}{2}}}{(q)_{i}(q)_{j}} \end{aligned} \] ■
- here we get a 2x2 matrix (rank 2 case)
\[ \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}\]
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
\[\vec{b}=(1/2,-1),(0,0),(1/2,0),(1/2,1),(1,1)\]
- for a complete list, see Ramanujan-Göllnitz-Gordon continued fraction
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 \[(-q;q^2)_{\infty}(-q)_{\infty}=\frac{(q^{2};q^{2})_{\infty}^3}{(q;q)_{\infty}^2(q^{4};q^{4})_{\infty}}=\frac{W(q)^2}{W(q^2)}\]. ■
\[W(q)=\frac{\eta(2\tau)}{\eta(\tau)}\] \[W(q^2)=\frac{\eta(4\tau)}{\eta(2\tau)}\] \[\frac{W(q)^2}{W(q^2)}=\frac{(q^{2};q^{2})_{\infty}^3}{(q;q)_{\infty}^2(q^{4};q^{4})_{\infty}}=\frac{\eta(2\tau)^3}{\eta(\tau)^2\eta(4\tau)}=\sum_{i,j\geq 0}\frac{q^{(i^2+2ij+2j^2)/2+i/2}}{(q)_{i}(q)_{j}}\] \[W(q^2)W(q)=\frac{\eta(4\tau)}{\eta(\tau)}=q^{1/8}\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+2j^2}{2}+\frac{i+2j}{2}}}{(q)_{i}(q)_{j}}=\frac{q^{1/8}(q^4;q^4)_{\infty}}{(q;q)_{\infty}}=\frac{q^{1/8}}{(q^1;q^4)_{\infty}(q^2;q^4)_{\infty}(q^3;q^4)_{\infty}}\]
- see eta product and eta quotient also
history
computational resource
articles
- 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
- 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. 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.