Surprise--I'm unhijacking this thread back to actual math content! On Sun, Oct 21, 2012 at 5:03 AM, Bill Gosper <billgosper@gmail.com> wrote:

http://dlmf.nist.gov/17.5#17.5.5 (or BHS (II.5)) gives

[image: \, _1\phi _1\left(a;c;q,\frac{c}{a}\right)=\frac{\left(\frac{c}{a};q\right)_{\infty }}{(c;q)_{\infty }}]

for |q|<1. This has the peculiar special case

[image: \, _1\phi _1(q;q z;q,z)=1-z]

independent of q. But this can be generalized to |q|≠1:

[image: \, _1\phi _1(q;q z;q,z)=\frac{z}{\left(\frac{1}{q z};\frac{1}{q}\right)_{\infty }}-z+1]

Can we similarly generalize BHS (II.5)? Maybe not. E.g.,

[image: \, _1\phi _1\left(q^2;c q^2;q,c\right)=(c-1) (c q-1) \left(\frac{c+q-c q}{c (q-1) \left(\frac{1}{c};\frac{1}{q}\right)_{\infty }}+1\right)]

We seem to be off to a rough start.

Sort of. [image: \, _1\phi _1\left(q^n;c q^n;q,c\right)=(c;q)_n \left(1-\frac{\sum _{k=0}^{n-1} \frac{\left(\frac{q}{c}\right)^k}{(q;q)_k}}{\left(\frac{1}{c};\frac{1}{q}\right)_{\infty }}\right)] Anyway, it beats ASCII art. As InputForm QHypergeometricPFQ[{q^n}, {c*q^n}, q, c] == QPochhammer[c, q, n]* (1 - Sum[(q/c)^k/QPochhammer[q, q, k], {k, 0, n - 1}]/ QPochhammer[1/c, 1/q])

I just implemented the base reciprocation formula for general rφs, which I was surprised BHS and DLMF 17.5 omit. Now I'm less surprised--it's a typesetting challenge.

Not really, so now I'm more surprised again. [image: _r\phi_s\left[\begin{array}{c}0,\text{...},0,a_1,\text{...},a_A\\0,\text{...},0,b_1,\text{...},b_B\end{array};q,z\right]=_{r+B-A+1}\phi_{s-B+A-1}\left[\begin{array}{c}0,\text{...},0,\frac{1}{a_1},\text{...},\frac{1}{a\!_A}\\0,\text{...},0 ,\frac{1}{b_1},\text{...},\frac{1}{b_B}\end{array};\ \frac{1}{q},\ \frac{a_1 \text{...} a\!_A z}{b_1 \text{...} b_B q}\right]] (Note the sensitivity to 0 parameters due to the newfangled BHS/Mma convention, which is really the right thing. But there's no way to translate this to Mma due to Mma's irredundant QHypergeometricPFQ notation leaving no way to specify the number of 0s.) --rwg

If instead of typesetting you see the alternate text, e.g.,

[image: \, _1\phi _1(q;q z;q,z)=\frac{z}{\left(\frac{1}{q z};\frac{1}{q}\right)_{\infty }}-z+1]

Then paste \, _1\phi _1(q;q z;q,z)=\frac{z}{\left(\frac{1}{q

z};\frac{1}{q}\right)_{\infty }}-z+1

into the window at www.texify.com . (Thanks, Neil!)

Neil has since found the infinitely superior http://www.codecogs.com/latex/eqneditor.php --rwg