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

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imported>Pythagoras0님의 2013년 5월 12일 (일) 13:45 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

First, we make change $q^2$ into $q$ to get the right form of the identity

Using the Gauss formula (in useful techniques in q-series) \[\prod_{r=0}^{n-1}(1+zq^r)=(1+z)(1+zq)\cdots(1+zq^{n-1})= \sum_{r=0}^{n} \begin{bmatrix} n\\ r\end{bmatrix}_{q}q^{r(r-1)/2}z^r\]

Therefore, \[ \begin{align} \sum_{n=0}^{\infty}\frac{q^{n^2/2+\beta n/2}(-z;q)_{n}}{ (q;q)_{n}}&=\sum_{n=0}^{\infty}\frac{q^{n^2/2+\beta n/2}\sum_{r=0}^{n} \begin{bmatrix} n\\ r\end{bmatrix}_{q}q^{r(r-1)/2}z^j}{ (q;q)_{n}} \\ &=\sum_{i,j\geq 0}\frac{q^{\frac{(i+j)^2+j^2+\beta(i+j)-j}{2}}z^j}{(q)_{i}(q)_{j}}=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+2j^2+\beta i+(\beta-1)j}{2}}z^j}{(q)_{i}(q)_{j}} \end{align} \]

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

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

specializations

  • $z=q^{1/2}$ and $\beta=2$

$$ \sum_{n=0}^{\infty}\frac{q^{n^2/2+n}(-q^{1/2};q)_{n}}{ (q;q)_{n}}=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+2j^2}{2}+(i+j)}}{(q)_{i}(q)_{j}}=\frac{1}{\left(q^{3/2};q^4\right){}_{\infty } \left(q^{2};q^4\right){}_{\infty } \left(q^{5/2};q^4\right){}_{\infty }} $$

  • $z=q^{1/2}$ and $\beta=0$

$$ \sum_{n=0}^{\infty}\frac{q^{n^2/2+n}(-q^{1/2};q)_{n}}{ (q;q)_{n}}=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+2j^2}{2}}}{(q)_{i}(q)_{j}}=\frac{1}{\left(q^{1/2};q^4\right){}_{\infty } \left(q^{7/2};q^4\right){}_{\infty } \left(q^{2};q^4\right){}_{\infty }} $$

  • $z=q^{1/2}$ and $\beta=1$

$$ \sum_{n=0}^{\infty}\frac{q^{n^2/2+n}(-q^{1/2};q)_{n}}{ (q;q)_{n}}=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+2j^2}{2}+\frac{i+j}{2}}}{(q)_{i}(q)_{j}}=\frac{1}{\left(q^{1/2};q^4\right){}_{\infty } \left(q^{5/2};q^4\right){}_{\infty } \left(q^{3};q^4\right){}_{\infty }} $$


history


related items


computational resource



encyclopedia


books

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


expositions


articles

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