[math-fun] (further) generalized Lambert series with Theta-convergence
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.
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.
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 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 - 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.
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 - 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.
Despite the pygalgia of wrapping all my summands in HoldForms, 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. 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 Sat, Feb 11, 2012 at 2:52 AM, Bill Gosper <billgosper@gmail.com> wrote:
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 - 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.
[Last update about 75% in.] 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, 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. 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}],
which is somewhat less peculiar when properly simplified: Sum[t^n/QPochhammer[x, q, 1 + n], {n, 0, Infinity}] == Sum[(q^j*QPochhammer[q^(1 + j), q])/(QPochhammer[q^j*t, q]* QPochhammer[q^j*x, q]), {j, 0, Infinity}] equivalently QHypergeometricPFQ[{q, 0}, {x}, q, t] == (QHypergeometricPFQ[{t, x/q}, {0}, q, q]*QPochhammer[q, q])/ (QPochhammer[t, q]*QPochhammer[x, q])
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.
We really need a repository of such results. (It's probably in BHS, but I can never find anything in there w/o help from George or Mizan.) Presumably, q-holonomy is at or near the point of automatically proving these, but what if you only have one side of the eqn, and wonder what other forms are known? Even if Mma's Sum is disciplined by FunctionExpand, the latter is already overloaded and underspecific. There's a surprising wealth of stuff in DLMF chap 17 (including four identities by Fine). Could this somehow be sufficiently aggrandized? It has a fairly lame assortment of contiguity relations, but the right way to do these is with matrices. But matrices are impractical w/o an underlying math engine. I.e., we need a "live DLMF". Maybe someday in Mma.
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 just finished after 2.48 CPU 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 Sat, Feb 11, 2012 at 2:52 AM, Bill Gosper <billgosper@gmail.com> wrote:
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".
(Actually, some of the Indeterminates were our fault.)
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.
And fake poles? ( http://gosper.org/poles.png)
--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!
You might have wondered how simply bisecting a sum qualifies as a stunt. We're actually talking about the unlikely-looking transformation 2*Sum[1/(q^(4*n) - 1), {n, 1, Infinity}] == Sum[1/((-1)^n*q^n - (-1)^n), {n, 1, Infinity}] + Sum[1/(q^n - (-1)^n), {n, 1, Infinity}] --rwg
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.
* Bill Gosper <billgosper@gmail.com> [Feb 18. 2012 07:42]:
[...]
which is somewhat less peculiar when properly simplified:
Sum[t^n/QPochhammer[x, q, 1 + n], {n, 0, Infinity}] == Sum[(q^j*QPochhammer[q^(1 + j), q])/(QPochhammer[q^j*t, q]* QPochhammer[q^j*x, q]), {j, 0, Infinity}]
From the equation I get ( N := \infty )
qbin(t,q,N) * qbin(x,q,N) / qbin(q,q,N) * sum(n=0,N, t^n / qbin(x,q,n+1) ) == sum(j=0,N, ( q^j * qbin(t,q,j) * qbin(x,q,j) ) / ( qbin(q,q,j) ) ) Now the second one is invariant against exchange of x and t, and so is the factor of the first one preceding the sum. Hence we have: sum(n=0,N, t^n / qbin(x,q,n+1) ) == sum(n=0,N, x^n / qbin(t,q,n+1) ) Now this a nice counterpart of (my/Osler's) sum(n=0,N, t^n / (1 - x*q^n) ) == sum(n=0,N, x^n / (1 - t*q^n) )
equivalently QHypergeometricPFQ[{q, 0}, {x}, q, t] == (QHypergeometricPFQ[{t, x/q}, {0}, q, q]*QPochhammer[q, q])/ (QPochhammer[t, q]*QPochhammer[x, q])
Which may well be in http://dlmf.nist.gov/ If so, can you point to it?
[...]
So now there's a \section{News from planet Gosper}
On 2/17/2012 3:25 PM, Bill Gosper wrote:
We really need a repository of such results. (It's probably in BHS, but I can never find anything in there w/o help from George or Mizan.) Presumably, q-holonomy is at or near the point of automatically proving these, but what if you only have one side of the eqn, and wonder what other forms are known?
Even if Mma's Sum is disciplined by FunctionExpand, the latter is already overloaded and underspecific.
There's a surprising wealth of stuff in DLMF chap 17 (including four identities by Fine). Could this somehow be sufficiently aggrandized? It has a fairly lame assortment of contiguity relations, but the right way to do these is with matrices. But matrices are impractical w/o an underlying math engine. I.e., we need a "live DLMF". Maybe someday in Mma.
For a while now, I've been thinking about building a live database of formulas, with some backend with a math engine. In fact, probably a year ago, I started building it. But then I realized that I, alone, probably wouldn't have the patience to digitize most of the A&S or DLMF or etc. formulas in S-exp syntax. Maybe if there's interest, I'll resume work on it again! -Robert
Could you facilitate volunteer contributions, a la OEIS? --rwg On Tue, Feb 21, 2012 at 12:34 PM, Robert Smith <quadricode@gmail.com> wrote:
On 2/17/2012 3:25 PM, Bill Gosper wrote:
We really need a repository of such results. (It's probably in BHS, but I can never find anything in there w/o help from George or Mizan.) Presumably, q-holonomy is at or near the point of automatically proving these, but what if you only have one side of the eqn, and wonder what other forms are known?
Even if Mma's Sum is disciplined by FunctionExpand, the latter is already overloaded and underspecific.
There's a surprising wealth of stuff in DLMF chap 17 (including four identities by Fine). Could this somehow be sufficiently aggrandized? It has a fairly lame assortment of contiguity relations, but the right way to do these is with matrices. But matrices are impractical w/o an underlying math engine. I.e., we need a "live DLMF". Maybe someday in Mma.
For a while now, I've been thinking about building a live database of formulas, with some backend with a math engine. In fact, probably a year ago, I started building it. But then I realized that I, alone, probably wouldn't have the patience to digitize most of the A&S or DLMF or etc. formulas in S-exp syntax.
Maybe if there's interest, I'll resume work on it again!
-Robert
* Bill Gosper <billgosper@gmail.com> [Feb 18. 2012 07:42]:
[...]
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.
[...]
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. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
[...]
* Joerg Arndt <arndt@jjj.de> [Feb 22. 2012 11:19]:
* Bill Gosper <billgosper@gmail.com> [Feb 18. 2012 07:42]:
[...]
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.
Done. \\ Your \\ default(echo,1); \\ > > 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}] \\ is, in pari/gp: t1 = sum(n=0,N, x^n * qbin(t,q,n) ) t2 = sum(n=0,N, ( (-x*t)^n * q^((n^2-n)/2) ) / qbin(x,q,n+1) ) t1-t2 \\ == zero \\ ... which is Fine's relation (6.1): t3 = sum(n=0,N, x^n * qbin(q*t,q,n) ) t4 = sum(n=0,N, ( (-x*t)^n * q^((n^2+n)/2) ) / qbin(x,q,n+1) ) t3-t4 \\ == zero
[...]
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 Sat, Feb 11, 2012 at 2:52 AM, Bill Gosper <billgosper@gmail.com> wrote:
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 - 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.
* Bill Gosper <billgosper@gmail.com> [Feb 11. 2012 15:10]:
[...] 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
OK, here is what I get: jj: k39, k43 |--> k { {0, Sum[(q^(1 + i))^(k - x)/(1 - q^k), {k, x,Infinity}]}, {0, 1} } == { {0, Sum[(q^((-1 + k)*(i + k + x))*(1 - q^(-1 + i + 2*k + x))) / ((1 - q^(i + k))*(1 - q^(-1 + k + x))), {k, 1, Infinity}]}, {0, 1} } jj: x |--> 1 { {0, Sum[(q^(1 + i))^(k - 1)/(1 - q^k), {k, 1, Infinity}]}, {0, 1} } == { {0, Sum[(q^((-1 + k)*(i + k + 1))*(1 - q^(i + 2*k))) / ((1 - q^(i + k))*(1 - q^(k))), {k, 1, Infinity}]}, {0, 1} } jj: i |--> 0 { {0, Sum[q^(k-1)/(1 - q^k), {k, 1, Infinity}]}, {0, 1} } == { {0, Sum[ ( q^((k-1)*(k+1)) * (1 - q^(2*k)) ) / ( (1 - q^(k)) * (1 - q^(k)) ), {k, 1, Infinity}]}, {0, 1} } And this is Clausen's \sum_{n\geq{}1}{ \frac{q^n}{1-q^{n}} } = \sum_{n\geq{}1}{ \frac{1+q^{n}}{1-q^{n}} \, q^{n^2} } (given on p.95 of \bibitem{Clausen}{Thomas Clausen: {Beitrag zur Theorie der Reihen}, Journal f\"{u}r die reine und angewandte Mathematik, vol.3, pp.92-95, (1828).} %% http://www.digizeitschriften.de/main/dms/toc/?PPN=PPN243919689_0003 ) Will mention this matrix magic in the newly launched section. ... but now it's Beer o'clock thanks, jj
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* Joerg Arndt <arndt@jjj.de> [Feb 21. 2012 18:09]:
* Bill Gosper <billgosper@gmail.com> [Feb 11. 2012 15:10]:
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OK, here is what I get: [...]
And this is Clausen's [...]
I tried to write that up, but then my brain exploded. (There were just to many changes in variables and notation in what was in this mail). Could define K(k,n) and N(k,n) and show which product gives Clausen's expressions? If possible, please omit any variables that are specialized out of existence later.
* Bill Gosper <billgosper@gmail.com> [Feb 07. 2012 11:40]:
[jj:] without splitting into even and odd part (yes, RWG, AIM304 exists).
YOW, I'd forgotten that stunt!
That one needed a small correction, see http://www.jjj.de/lambert-paper/ (Warning: very first iteration, just spend today over it; certainly contains typos (tell me if you spot one!)). What I referred to as "Fine's versatile relation"
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"). appears to be known as "Rogers-Fine identity", this term is used in my draft.
Later I'll process your other mails (big THANKS btw.!)
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participants (3)
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Bill Gosper -
Joerg Arndt -
Robert Smith