Puzzle: Assuming[0 < q < 1, Limit[QPochhammer[a*c, q]/QPochhammer[b*c, q], c -> \[Infinity]]] Or maybe Assuming[0 < q < 1, Limit[QPochhammer[a*x, q]* QPochhammer[b*x, q]/QPochhammer[c*x, q]/QPochhammer[a*b*x/c, q], x -> \[Infinity]]] Back to Joerg's sum. This time I got Sum[t^n/(1 - q^n*x), {n, 0, Infinity}] == (QPochhammer[q, q]*Sum[t^n*QPochhammer[x, q, n], {n, 0, Infinity}])/ (QPochhammer[q/t, q]*QPochhammer[x, q]) - Sum[(q^k*QPochhammer[q, q, k - 1])/(QPochhammer[q/t, q, k]* QPochhammer[x, q, k]), {k, 1, Infinity}]/t Note the middle sum is the g.f of the finite qpochhammers of two fixed arguments. Shouldn't that sum be easy? Is it in Fine? I left my BHS at Neil's. (But I can never find stuff in there anyway.) Trying the g.f. sum, I hit this weirdy: Sum[(c^n*QPochhammer[b, q, n])/(b^n*QPochhammer[c*q, q, n]), {n, 0, Infinity}] == (b*(-1 + c))/(-b + c) independent of q. Plotting for b=2/3,c=1/3, it holds somewhat past q=1, and then develops a very interesting collection of poles. --rwg On Mon, Feb 6, 2012 at 5:46 PM, Bill Gosper <billgosper@gmail.com> wrote:
Joerg>From the department of formulas that could possibly be found in some paper from 200 years ago.
Semi-recently I made the observation (and posted it here) that sum(n>=0, x*q^n/(1-x*q^n) can be computed as sum(n>=0, q^(n^2+n)*x^(n+1)*(1 - x*q^(2*n+1)) / ((1 - x*q^n)*(1 - q*q^n))) and this allows the fast computation of sums of inverse Fibs (and infinitely more sums of reciprocal order-two linrecs) without splitting into even and odd part (yes, RWG, AIM304 exists).
YOW, I'd forgotten that stunt!
After much effort I found that sum(n>=0, t^n / (1-x*q^n) )
Random aside: Per my recent (generalized) contiguous Heine mail, with c=a and d=b q = x q and z=t, sum(((t^n)/(1 - q^n * x)),n,0,inf) = ((a * (1 - q * t) * x * sum(((q^(2 * n) * t^n)/(1 - q^n * x)),n,0,inf) - ( - q * t * x + q * x - a * q * t + a) * sum(((q^n * t^n)/(1 - q^n * x)),n,0,inf) - q + a)/(q * (t - 1)))
independent of a!
Joerg>can be computed as sum(n=0, S, (1-x*t*q^(2*n))*(x*t)^n*q^(n^2) / ( (1-x*q^n)*(1-t*q^n)) ) (which btw. is symmetric in x and t).
This is driving me nuts. The 3x3 system mentioned in that Heine mail refuses to triangularize for this Lambert case, which is degenerate enough to permit discarding two parameters from the Heine N matrix. One of these, call it k, can be via a reversible (unit determinant) "recoord" (our new name for "transformal"). But k then persists in the other matrices, and K, the k-bumpng matrix, does not become the 3x3 identity! Thus, wandering around in the k-n plane is a no-op, because, by path invariance, N(k,n).K(k,n+1) = K(k,n).N(k+1,n), but N(k,n)=N(k+1,n)! There must be some use for this.
Rolling back to a 1998 Heine 2x2 system (that was born triangular), then replacing n+1 by n+h by shifting i,j,k,n by 1-h,1-h,1-h,h-1, then specializing h to i*t (and discarding the I matrix), then q^t->t, gives the J,K,N matrices
{{{-(((1 - q^(n + j))*(1 - 1/(q^(k - j - 2)*t))*t)/(q^(n + j + 1)*(1 - 1/q^(k - j - 1)))), -((1 - q^(n + k - 1))/ (q^(n + k - 1)*(1 - 1/q^(k - j - 1))))}, {0, 1}}, {{-(((q^(n + k - 1)*(1 - q^(k - j))*t)/(1 - q^(n + k)))*(1 - q^(k - j - 1)*t)), 1/(1 - q^(k - j - 1)*t)}, {0, 1}}, {{(q^(k - j - 1)*(1 - q^(n + j))*t)/(1 - q^(n + k)), 1}, {0, 1}}}
Then pegging j at x and closing a rectangle based at k=x+1,n=0, then q^x->x gives a *different* theta-convergent series symmetric in t and x:
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}]
(Dialog) In[649]:= Simplify[ FunctionExpand[Series[%[[2]] /. \[Infinity] -> 6, {t, 0, 6}]]]
(Dialog) Out[649]= SeriesData[t, 0, {(1 - x)^(-1), (1 - q x)^(-1), ( 1 - q^2 x)^(-1), (1 - q^3 x)^(-1), (1 - q^4 x)^(-1), ( 1 - q^5 x)^(-1), (1 - q^6 x)^(-1)}, 0, 7, 1]
Like I said: nuts. I'll bet yours|Fine's is in there somewhere. --rwg
Now this one turns out to be easily obtained by setting a := -b in Fine's "versatile" relation (14-1) (p.15 in "Basic Hypergeometric Series and Applications"). Anybody with one bit of interest in q-things:
Buy this book, it's a marvel.