Göllnitz-Gordon identities and Ramanujan-Göllnitz-Gordon continued fractions

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http://bomber0.myid.net/ (토론)님의 2012년 8월 26일 (일) 14:08 판
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introduction

 

 

q-hypergeometric series
  • Slater 34
    \(\sum_{n=0}^{\infty}\frac{q^{n^2+2n}(-q;q^{2})_{n}}{ (q^{2};q^{2})_{n}}=1/(q^{3};q^{8})_{\infty}(q^{4};q^{8})_{\infty}(q^{5};q^{8})_{\infty}}\)
  • Slater 36
    \(\sum_{n=0}^{\infty}\frac{q^{n^2}(-q;q^{2})_{n}}{ (q^{2};q^{2})_{n}}=1/(q^{1};q^{8})_{\infty}(q^{4};q^{8})_{\infty}(q^{7};q^{8})_{\infty}}\)
  • lifting to rank 2 case

Using the Guass formula (in useful techniques in q-series)

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

we can rewrite \((-q;q^{2})_{n}\) as

\((-q;q^{2})_{n}=\sum_{r=0}^{n} \begin{bmatrix} n\\ r\end{bmatrix}_{q}q^{r^{2}}\)

Therefore,

\(\sum_{n=0}^{\infty}\frac{q^{n^2}(-q;q^{2})_{n}}{ (q^{2};q^{2})_{n}}=\sum_{n=0}^{\infty}\frac{q^{n^2}\sum_{r=0}^{n} \begin{bmatrix} n\\ r\end{bmatrix}_{q}q^{r^{2}}}{ (q^{2};q^{2})_{n}}\)

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

 

and

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

 

one can obtain the following matrix from the above hypergeometric series\[ \begin{bmatrix} 4 & 2 \\ 2 & 2 \end{bmatrix}\]

 

 

 

 

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