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

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imported>Pythagoras0님의 2013년 5월 12일 (일) 10:46 판
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introduction



q-hypergeometric series

identities

\[\sum_{n=0}^{\infty}\frac{q^{n^2+2n}(-q;q^{2})_{n}}{ (q^{2};q^{2})_{n}}=1/(q^{3};q^{8})_{\infty}(q^{4};q^{8})_{\infty}(q^{5};q^{8})_{\infty}\]

\[\sum_{n=0}^{\infty}\frac{q^{n^2}(-q;q^{2})_{n}}{ (q^{2};q^{2})_{n}}=1/(q^{1};q^{8})_{\infty}(q^{4};q^{8})_{\infty}(q^{7};q^{8})_{\infty}\]

  • G. E. Andrews, The Theory of Partitions, 1976, Corollary 2.7., page 21,

\[\sum_{n=0}^{\infty}\frac{q^{n^2+n}(-q;q^{2})_{n}}{ (q^{2};q^{2})_{n}}=1/(q^{1};q^{8})_{\infty}(q^{5};q^{8})_{\infty}(q^{6};q^{8})_{\infty}\label{GA}\]

  • is \ref{GA} modular?


lifting to rank 2 case

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

Note that we have used the identity $$ (-q)_{n}=\frac{(q^2;q^2)_{n}}{(q;q)_{n}} $$

  • one can obtain the following matrix from the above hypergeometric series

\[ \begin{bmatrix} 4 & 2 \\ 2 & 2 \end{bmatrix}\]





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  • G. E. Andrews, The Theory of Partitions, Addison-Wesley, 1976
    • Sec 7.4


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