[math-fun] Limit puzzle
(-1/2 + Sum[(a^n*QPochhammer[x/a, q, n])/QPochhammer[a*x, q, n], {n, 0, Infinity}])/ (1 + a) as a -> -1+ . (In terms of q-hypergeometric(s) in x and q). This gave me quite a tussle. If you crack it, please say how. After all else failed, I StringReverse["]}ytinifnI ,1 ,k{ ,)]k ,q ,x-[remmahhcoPQ*]k + 1 ,q ,1-[remmahhcoPQ( /)]k + 1- ,q ,q[remmahhcoPQ *k^x*)k*)k + 1-(*)2/1((^q*k^)1-(([muS*2 + 4/1 si ,1-=a rof ,hcihw ]}ytinifnI ,1 ,k{ ,)]k ,q ,x*a[remmahhcoPQ*]k + 1 ,q ,a[remmahhcoPQ( \ /)]k + 1- ,q ,q*2^a[remmahhcoPQ*k^x*)k*)k + 1-(*)2/1((^q*k^)1-(([muS *)a + 1-(- )))a + 1-(*2(/1(- :1+a eht delecnac hcihw morf eno gnidnif yllanif ,dnatimil eht demrofsnart hcihw sruotnoc xirtam dehcraes ylmodnar tsuj"] --rwg The spellchecker likes demrofsnart? Oh, it's on the same line as (").
On Sun, Feb 26, 2012 at 12:32 AM, Bill Gosper <billgosper@gmail.com> wrote:
(-1/2 + Sum[(a^n*QPochhammer[x/a, q, n])/QPochhammer[a*x, q, n], {n, 0, Infinity}])/ (1 + a)
as a -> -1+ . (In terms of q-hypergeometric(s) in x and q).
Here's a much better answer as two Lambert series: Limit[(-1/2 + Sum[(a^n*QPochhammer[x/a, q, n])/QPochhammer[a*x, q, n], {n, 0, Infinity}])/(1 + a), a -> -1, Direction -> -1] == 1/4 + 2*Sum[(-x)^n/(1 - q^(2*n)), {n, 1, Infinity}] + x*Sum[q^n/(1 + q^n*x), {n, 0, Infinity}] This came from combining Heine's first and second transformations http://dlmf.nist.gov/17.6 (17.6.6, 17.6.7) to get QHypergeometricPFQ[{a, b}, {c}, q, z]== QHypergeometricPFQ[{c/b, z}, {a z}, q, b] * QPochhammer[b, q] QPochhammer[a z, q])/ (QPochhammer[c, q] QPochhammer[z, q]) This transforms the sum in the limitand into one with numerator parameter a^2, which means all its terms after the 0th will contain a removable factor of 1-a^2 = (1-a)(1+a), and the 0th minus 1/2 and then divided by a+1 contributes one of the Lambert sums as a -> -1. 17.6.6 and 17.6.7 (call them hein1 and hein2) along with hein0:= a<->b, generate a group of order 12. Heine's 3rd (17.6.8) is just hein2 hein0 hein1 hein0. So why did dlmf list these three and not two nor twelve? Presumably, hein3 is not obvious, but the rest should be, as each one becomes four when pre- (switch the slots) and post- (switch a and b) "multiplied" by hein0. But I didn't spot my a^2 until generating the whole group and then plugging in. Likewise for a contour result from the matrix system I've been thrashing with Joerg. I thought it was actually new. --rwg rwg> This gave me quite a
tussle. If you crack it, please say how. After all else failed, I
StringReverse["]}ytinifnI ,1 ,k{ ,)]k ,q ,x-[remmahhcoPQ*]k + 1 ,q ,1-[remmahhcoPQ( /)]k + 1- ,q ,q[remmahhcoPQ *k^x*)k*)k + 1-(*)2/1((^q*k^)1-(([muS*2 + 4/1 si ,1-=a rof ,hcihw
]}ytinifnI ,1 ,k{ ,)]k ,q ,x*a[remmahhcoPQ*]k + 1 ,q ,a[remmahhcoPQ( \
/)]k + 1- ,q ,q*2^a[remmahhcoPQ*k^x*)k*)k + 1-(*)2/1((^q*k^)1-(([muS *)a + 1-(- )))a + 1-(*2(/1(-
:1+a eht delecnac hcihw morf eno gnidnif yllanif ,dnatimil eht demrofsnart hcihw sruotnoc xirtam dehcraes ylmodnar tsuj"]
--rwg
The spellchecker likes demrofsnart? Oh, it's on the same line as (").
* Bill Gosper <billgosper@gmail.com> [Feb 27. 2012 14:27]:
On Sun, Feb 26, 2012 at 12:32 AM, Bill Gosper <billgosper@gmail.com> wrote:
(-1/2 + Sum[(a^n*QPochhammer[x/a, q, n])/QPochhammer[a*x, q, n], {n, 0, Infinity}])/ (1 + a)
as a -> -1+ . (In terms of q-hypergeometric(s) in x and q).
Here's a much better answer as two Lambert series:
Limit[(-1/2 + Sum[(a^n*QPochhammer[x/a, q, n])/QPochhammer[a*x, q, n], {n, 0, Infinity}])/(1 + a), a -> -1, Direction -> -1] == 1/4 + 2*Sum[(-x)^n/(1 - q^(2*n)), {n, 1, Infinity}] + x*Sum[q^n/(1 + q^n*x), {n, 0, Infinity}]
This came from combining Heine's first and second transformations http://dlmf.nist.gov/17.6 (17.6.6, 17.6.7) to get QHypergeometricPFQ[{a, b}, {c}, q, z]== QHypergeometricPFQ[{c/b, z}, {a z}, q, b] * QPochhammer[b, q] QPochhammer[a z, q])/ (QPochhammer[c, q] QPochhammer[z, q])
This transforms the sum in the limitand into one with numerator parameter a^2, which means all its terms after the 0th will contain a removable factor of 1-a^2 = (1-a)(1+a), and the 0th minus 1/2 and then divided by a+1 contributes one of the Lambert sums as a -> -1.
[...]
Not sure this is pertinent here, but see relation (3.1) on p.604 of %\bibitem{Chan}{Song Heng Chan: % {Generalized Lambert Series Identities}, % Proceedings of the London Mathematical Society, vol.91, no.3, pp.598-622, (2005). %}%% rel.(3.1): expression for bilateral Lambert series Can send the pdf in case the paper isn't open access. IIRC there is something similar in Fine, but can only point to his relation (18.4) on p.20 right now.
participants (2)
-
Bill Gosper -
Joerg Arndt