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

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http://bomber0.myid.net/ (토론)님의 2010년 10월 8일 (금) 09:21 판
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introduction
  • Göllnitz
    \(1+q+{q^{2} \over 1+q^{3} + } {q^{4} \over 1+q^{5}+} {q^{6} \over \cdots}=\frac{(q^{3};q^{8})_{\infty}(q^{4};q^{8})_{\infty}(q^{5};q^{8})_{\infty}}{(q^{1};q^{8})_{\infty}(q^{4};q^{8})_{\infty}(q^{7};q^{8})_{\infty}}=\frac{(q^{3};q^{8})_{\infty}(q^{5};q^{8})_{\infty}}{(q^{1};q^{8})_{\infty}(q^{7};q^{8})_{\infty}}\)
  • [Gordon1965]
    \(1+{q \over 1+q^2 + } {q^3 \over 1+q^4+} {q^5 \over 1+q^6} } \cdots=\frac{(q^{2};q^{8})_{\infty}(q^{3};q^{8})_{\infty}(q^{7};q^{8})_{\infty}}{(q^{1};q^{8})_{\infty}(q^{5};q^{8})_{\infty}(q^{6};q^{8})_{\infty}}\)

 

 

 

Ramanujan's result
  • Berndt, notebook V entry 22 p. 50
    \({1 \over 1+} {q+q^2 \over 1+} {q^4 \over 1+} {q^3+q^6 \over 1+}{q^8 \over 1+\cdots} =\frac{(q^{1};q^{8})_{\infty}(q^{7};q^{8})_{\infty}}{(q^{3};q^{8})_{\infty}(q^{5};q^{8})_{\infty}}\)

 

 

modular function
  • fractional power
    \(q^{1/2}({1 \over {1+q}} {q^2 \over 1+q^3 + } {q^4 \over 1+q^5 + {}} {q^6 \over 1+q^7} } \cdots)=q^{1/2}\frac{(q^{1};q^{8})_{\infty}(q^{7};q^{8})_{\infty}}{(q^{3};q^{8})_{\infty}(q^{5};q^{8})_{\infty}}\)
  • [Duke2005] (9.4)

 

 

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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