Oops, I've missed a bunch of opportunities to specialize parameters to get nonugliness from that system. E.g., note the quadrinomial in the UR k (2nd) matrix numerator: {{{(-q^(i + k) + q^(i + j + n))/(-1 + q^(k + n)), 1}, {0, 1}}, {{(q^(i + n)*(q^j - q^(2*k))*(q^j - q^(1 + 2*k)))/((q^(i + j) - q^k)*(-1 + q^(i + k))* (-q^(1 + k) + q^(j + n))*(-1 + q^(k + n))), (q^k*(q^(1 + i + j) - q^(1 + k) + q^(j + n) - q^(i + j + k + n)))/ ((q^(i + j) - q^k)*(-1 + q^(i + k))*(-q^(1 + k) + q^(j + n)))}, {0, 1}}} It will factor for j=-2i and j=2-2n (among other cases). The former leads to 1 == (QPochhammer[-x, q]*Sum[(-1)^n*q^(((-1 + n)*n)/2)*x^n, {n, 0, Infinity}])/ (QPochhammer[q, q]*QPochhammer[x, q]*QPochhammer[x^2, q^2]) - Sum[(q^(1 + n)*QPochhammer[-(x/q), q, 1 + n]*QPochhammer[q*x^2, q^2, n])/ (QPochhammer[q, q, 1 + n]*QPochhammer[x, q, 1 + n]*QPochhammer[x^2, q, n]), {n, 0, Infinity}] and the Lambert sum Sum[q^n/(1 - q^(2*n)*x), {n, 0, Infinity}] == (QPochhammer[-q, q^2]*QPochhammer[q^4, q^4]*Sum[(-1)^n*q^n^2*x^n, {n, 0, Infinity}])/ (QPochhammer[q, q^2]*QPochhammer[q^2/x, q^2]*QPochhammer[x, q^2, Infinity]) - ((q + x)*Sum[(q^(1 + 2*n)*QPochhammer[-q, q^2, n]*QPochhammer[q^4, q^4, n])/ (QPochhammer[q, q^2, 1 + n]*QPochhammer[q^2/x, q^2, 1 + n]* QPochhammer[x, q^2, 1 + n]), {n, 0, Infinity}])/x and their common generalization Sum[(a^n*QPochhammer[x/a, q, n])/QPochhammer[a*x, q, n], {n, 0, Infinity}] == (QPochhammer[-a, q]*QPochhammer[a^2*q, q^2]*Sum[(-1)^n*q^(((-1 + n)*n)/2)*x^n, {n, 0, Infinity}])/(QPochhammer[a, q]*QPochhammer[(a*q)/x, q]* QPochhammer[a*x, q]) - (a*(q + x)*Sum[(q^n*QPochhammer[-a, q, n]*QPochhammer[a^2*q, q^2, n])/ (QPochhammer[a, q, 1 + n]*QPochhammer[(a*q)/x, q, 1 + n]* QPochhammer[a*x, q, n]), {n, 0, Infinity}])/x Note the "half theta series". --rwg On Thu, Feb 23, 2012 at 1:52 PM, Bill Gosper <billgosper@gmail.com> wrote:
On Thu, Feb 16, 2012 at 12:09 PM, Bill Gosper <billgosper@gmail.com>wrote:
Despite the pygalgia of wrapping all my summands in HoldForms,
pygodynia! The insidious bugs this causes are hard to believe.
I'm methodically constructing all the non-ugly identities generated by the contours producing Joerg's Lambert series, where ugly:= trinomials or worse in a summand.
Here's a not-too-ugly Lambert: = trinomial:
Sum[q^n/(1 - a*q^n), {n, 0, Infinity}] == Sum[(a^(2*k)*q^((k*(1 + 3*k))/2)*(1 + q^k*(q - a*q^(1 + k)))*QPochhammer[q, q, k]^2)/ ((-1)^k*(1 + q^(1 + k))*QPochhammer[a, q, 1 + k]*QPochhammer[q, q, 1 + 2*k]), {k, 0, Infinity}]
More generally,
Sum[(j^(2*n)*q^n*QPochhammer[a/(j*q), q, n])/QPochhammer[a, q, n], {n, 0, Infinity}] == Sum[(a^(2*k)*j^k*q^((3*(-1 + k)*k)/2)*(1 + j*q^k*(q - a*q^k))*QPochhammer[j*q, q, k]^2)/ ((-1)^k*(1 + j*q^(1 + k))*QPochhammer[a, q, k]*QPochhammer[j^2*q, q, 1 + 2*k]), {k, 0, Infinity}]
This will make a nice test of the pPhiq notator I'm trying to write. pPhiq notation has even more drawbacks than pFq. --rwg
One of these non-uglies does something novel: The base leg producing the Lambert sum gets multiplied by zero, leaving the somewhat peculiar identity
Sum[t^n/QPochhammer[x, q, n], {n, 0, Infinity}] == Sum[(q^(-1 + j)*(q - x)*QPochhammer[q^(1 + j), q])/ (QPochhammer[q^j*t, q]*QPochhammer[q^(-1 + j)*x, q]), {j, 0, Infinity}],
whose lhs is the generating fcn of the reciprocals of (x;q)_n for nonnegative integer n. I quote it here on the off chance it lacks a more conventional derivation.
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
Note that the HoldForm workaround is fundamentally wrong because HoldForm[f[n]] is not a function of n !:
In[903]:= Sum[HoldForm[f[n]], {n, 69}]
Out[903]=69 f[n]
I have a Sum[(-1)^(2n)*f[n],{n,Infinity}]. Morbidly, I wondered what it would take to punt the (-1)^(2n) automatically. Sneaking Simplify etc. into the summand under the HoldForm does nothing. Completely releasing the HoldForm "only" took a minute or so, but still wouldn't punt the (-1)^(2n). FullSimplify on the unwholesome whole Sum(s) has now been running for three days. It is not a large expression:
Sum[i^n/(1 - b*q^n), {n, 0, Infinity}] == (QPochhammer[q, q]*Sum[(-1)^(2*n)*i^(-1 + n)*QPochhammer[b, q, -1 + n], {n, 1, Infinity}])/(QPochhammer[q/i, q]*QPochhammer[b, q, Infinity]) - Sum[(q^n*QPochhammer[q, q, -1 + n])/(i*QPochhammer[b, q, n]* QPochhammer[q/i, q, n]), {n, 1, Infinity}]
On Fri, Feb 10, 2012 at 4:47 AM, Bill Gosper <billgosper@gmail.com> wrote:
The very simple 4D matrix system for Joerg's Lambert series is I(i, j, k, n) := [(q^(n + k - 1) * (1 - (1/(q^(k - j - i - 1))))/(1 - q^i)), (1 - q^(n + k - 1)/(1 - q^i)); 0, 1], J(i, j, k, n) := [(1 - (1/(q^(k - j - i - 1)))) * (1 - q^(n + j))/(1 - (1/(q^(k - j - 1)))), - (1 - q^(n + k - 1))/(q^(k - j - 1) * (1 - (1/(q^(k - j - 1))))); 0, 1], K(i, j, k, n) := [(1 - q^(k - j))/((1 - q^(k - j - i)) * (1 - q^(n + k))), - (q^(k - j - i)/(1 - q^(k - j - i))); 0, 1], N(i, j, k, n) := [q^i * (1 - q^(n + j))/(1 - q^(n + k)), 1; 0, 1]
Prod(N(t,x,x+1,n),n,0,oo) computes the Lambert series. The contour in the n-i plane, (t,x,x+1,0)...(t,x,x+1,oo) ...(oo,x,x+1,oo) = (t,x,x+1,0)...(oo,x x+1,0) ...(oo,x,x+1,oo) directly computes Joerg's symmetry observation:
sum(t^n/(1 - q^n * x),n,0,inf) = sum(x^i/(1 - q^i * t),i,0,inf) --rwg Aggrieved at not finding Joerg's Theta-convergent, Pochhammer-free identity, I exhaustively searched all computationally feasible coordinate changes of this 4D system. The winner was i=t-1, j=-1, leaving the extremely simple 2D system [ n + 2 k + i - 1 ] [ n + 2 k + i 1 - q ] [ q ----------------------------- ] [km(k, n) := [ k + i n + k - 1 ], [ (1 - q ) (1 - q ) ] [ ] [ 0 1 ]
[ k + i 1 ] [ q -------------- ] nm(k, n) := [ n + k - 1 ]] [ 1 - q ] [ ] [ 0 1 ] (the specialization of j to -1 permits "sidestepping" to almost pure sum notation, ruling out Pochhammers from the contour.) Running these through MProd, Julian's new matrix product to sum converter,
{MProd[{{q^(i + 1), 1/(1 - q^n)}, {0, 1}}, {n, x, Infinity}] , MProd[{{0, -(1/(q^(k + i) - 1))}, {0, 1}}, {k, 1, Infinity}]} -> {MProd[{{q^(x + 2*k + i), (1 - q^(x + 2*k + i - 1))/((1 - q^(k + i))*(1 - q^(x + k - 1)))}, {0, 1}}, {k, 1, Infinity}] , {{0, 1}, {0, 1}}}
elicited from Mma a bunch of bogus nonconvergence complaints, then a bunch of infectious and gratuitous "Indeterminate"s, and ultimately an utterly useless and incorrect "False". Changing Equal to Rule and Dot to List, and then changing them back: In[101]:= Dot @@ # & /@ (% /. Indeterminate -> 0) /. Rule -> Equal
Out[101]= {{0, Sum[(q^(1 + i))^(k39 - x)/(1 - q^k39), {k39, x, Infinity}]}, {0, 1}} == {{0, Sum[(q^((-1 + k43)*(i + k43 + x))*(1 - q^(-1 + i + 2*k43 + x)))/((1 - q^(i + k43))*(1 - q^(-1 + k43 + x))), {k43, 1, Infinity}]}, {0, 1}}
At last! --rwg So now we know that it's possible to exclude Pochhammers from a system before choosing the contour. Further recoordinatizations of that 2D system are guaranteed to remain Pochhammer free. Other integer j should also work, but probably won't yield anything we couldn't get from Joerg's identity plus partial fractions. Path invariance is preserved by differentiating and integrating wrt to variables not appearing in the upper left [1,1] element, but this is unlikely to yield anything we couldn't get from differentiating and integrating the actual sums.
(This is weird: Firefox will *only* work in virtual XP when I'm home, and *only* work in the Lion OS at the Tastebuds restaurant. What will happen in Zieglerville?)
While Mma 8.04 knows the q-binomial and q-exponential sums, it appears not to know the q-Gauss (Heine) nor q-Dixon, e.g. QHypergeometricPFQ[{a, (-Sqrt[a])*q, b, c}, {-Sqrt[a], (a*q)/b, (a*q)/c}, q, (Sqrt[a]*q)/(b*c)] == (QPochhammer[a*q, q]* QPochhammer[(Sqrt[a]*q)/b, q]* QPochhammer[(Sqrt[a]*q)/c, q]*QPochhammer[(a*q)/(b*c), q])/ (QPochhammer[Sqrt[a]*q, q]*QPochhammer[(a*q)/b, q]* QPochhammer[(a*q)/c, q]* QPochhammer[(Sqrt[a]*q)/(b*c), q])
It can, however, test this symbolically by expanding at q=0.
On Wed, Feb 8, 2012 at 12:01 AM, Bill Gosper <billgosper@gmail.com> wrote:
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
On Sat, Feb 11, 2012 at 2:52 AM, Bill Gosper <billgosper@gmail.com> wrote: -
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.