[seqfan] Re: OT: q-binomial-oids (ascii art included)

[Joerg Arndt]

Jacobi!  (this was hard to find)


> The relations are respectively the special cases (a,b)=(-1,0) and
> (a,b)=(0,1) of an identity due to Jacobi:
>
> ___M-1         n          ___n-1     M-k     ___n-1     k
> | |    (1-a x q )         | |    (1-q   )    | |    (b q -a)
> | |n=0            \~~ M   | |k=0             | |k=0             n   n (n-1)/2
> -----------------= >      -----------------------------------  x   q
> ___M-1         n  /__ n=0  ___n-1     k     ___n-1       k
> | |    (1-b x q )          | |    (1-q )    | |    (1-b q )
> | |n=0                     | |k=0           | |k=0
>

To be found on p.795 of
  W. P. Johnson:
  {How Cauchy Missed Ramanujan's ${}_1\psi_1$ Summation},
  American Mathematical Monthly,
  vol.111, no.9, pp.791-800, November-2004




* Joerg Arndt <arndt at jjj.de> [Mar 18. 2010 08:02]:
> [crossposted to seqfan because we have several
>  people here who might be able to help]
>
>
> I can find the following in the literature:
>
> >                               ___n-1    M-k
> >                               | |   (1-q   )
> >    ___M-1      n      \~~ M   | |k=0            n    n (n-1)/2
> >    | |   (1+x q )  =   >      --------------   x    q
> >    | |n=0             /__ n=0  ___n      k
> >                                | |   (1-q )
> >                                | |k=1
>
>
> Now I came up with
>
> >                                       ___n-1    M-k
> >                                       | |   (1-q   )
> >          1            \~~ M           | |k=0                     n    n (n-1)
> >    --------------  =   >      -------------------------------   x    q
> >    ___M-1      n      /__ n=0 ___n-1      k    ___n-1      k
> >    | |   (1-x q )             | |   (1-q q )   | |   (1-x q )
> >    | |n=0                     | |k=0           | |k=0
>
> This is certainly known.
> Can anyone point out where this is given?
>
>
> LaTeX sources are
>
> \prod_{n=0}^{M-1}{(1+x\,q^n)}  & = &
>  \sum_{n=0}^{M}{
>    \frac{\prod_{k=0}^{n-1}{(1-q^{M-k})}}
>         {\prod_{k=1}^{n}{(1-q^k)}}
>    \, x^n \, q^{n\,(n-1)/2} }
>
> and
>
> \frac{1}{\prod_{n=0}^{M-1}{(1-x\,q^n)}}  & = &
>  \sum_{n=0}^{M}{
>    \frac{\prod_{k=0}^{n-1}{(1-q^{M-k})}}
>         {\prod_{k=0}^{n-1}{(1-q\,q^k)} \, \prod_{k=0}^{n-1}{(1-x\,q^k)}}
>    \, x^n \, q^{n\,(n-1)} }
>
>
> [off to the beers, I'll be back tomorrow]
>
>
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