* Bill Gosper <billgosper@gmail.com> [Feb 18. 2012 07:42]:
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Linearly combining other results gives a thetalike gfcn for the reciprocal pochhammers 1/(x;q)_k in terms of the gfcn for the unreciprocated pochhammers (t;q)_k !
Sum[x^k*QPochhammer[t, q, k], {k, 0, Infinity}] == Sum[(q^(-(k/2) + k^2/2)*t^k*x^k)/((-1)^k*QPochhammer[x, q, 1 + k]), {k, 0, Infinity}] --rwg
I'll check this one next.
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Sum[t^n/(1 - q^n*x), {n, 0, Infinity}] == Sum[((-1)^k*q^(k^2/2 - k/2)*QPochhammer[q, q, k]*t^k*x^k)/ (QPochhammer[t, q, k + 1]*QPochhammer[x, q, k + 1]), {k, 0, Infinity}]
Like I said: nuts. I'll bet yours|Fine's is in there somewhere. --rwg
Yes, this one as well: %%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Alternate identity for fast computation} Fine's second transformation is \cite[relation (12.2), p.13]{Fine} %(also (17.6.10) in \cite{DLMF}) \[ % (1-t)*sum(n=0,N, t^n * qbin(a*q,q,n)/qbin(b*q,q,n) ); %(1-t)\,F(a,b;t) = (1-t)\,\sum_{n\geq{}0}{ \frac{(a\,q;q)_n}{(b\,q;q)_n} \, t^n } = % % sum(n=0,N, (qbin(b/a,q,n)) / (qbin(b*q,q,n) * qbin(t*q,q,n)) * (-a*t)^n * q^((n^2+n)/2) ); \sum_{n\geq{}0}{ \frac{(b/q;q)_n}{(b\,q;q)_n\,(t\,q;q)_n} \, (-a\,t)^{n}\,q^{(n^2+n)/2} } \] % Setting $b=a\,q$, then replacing $a$ by $a/q$ and dividing by $(1-a)\,(1-t)$, and finally replacing $a$ by $x$ gives \begin{equation} % sum(n=0,N, t^n/(1-x*q^n) ); \sum_{n\geq{}0}{ \frac{t^n}{1-x\,q^{n}} } = % % sum(n=0,N, (qbin(q,q,n)) / (qbin(x,q,n+1)*qbin(t,q,n+1)) * (-x*t)^n * q^((n^2-n)/2) ); \sum_{n\geq{}0}{ \frac{ (q;q)_n }{ (x;q)_{n+1} \, (t;q)_{n+1} } \, (x\,t)^n\, q^{(n^2-n)/2} } \end{equation} % Again the symmetry between $x$ and $t$ is evident from the right side. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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