Re: [math-fun] How I miss Mizan Rahman, or, what the heck is this?

7 Jul 2016

      George Andrews got it from Bailey's ₂Ψ₂ transformation (rather than finding
it scrawled on the wall of an ancient tomb.)  Merely with elementary
manipulations, starting with partial fractions, I reduced it to
            n
          q
Sum[---------------, {n, -∞, ∞}] ==
          1 + 2 n 2
    (1 - q       )

               2   2 4
  QPochhammer[q , q ]
  --------------------
                  2 4
  QPochhammer[q, q ]

which is presumably known, since it's a simple q-extension of

In[1653]:= Sum[(2 n - 1)^-2, {n, -∞, ∞}]

Out[1653]= π^2/4.

Exercises:  Evaluate

         3 n + 1
        q
Sum[---------------, {n, -∞, ∞}]
          1 + 2 n 2
    (1 - q       )

and
          n
         q
Sum[------------, {n, -∞, ∞}]
         1 + 2 n
    1 - q

Spoilers:
               2   2 4
  QPochhammer[q , q ]
  --------------------
                  2 4
  QPochhammer[q, q ]

and 0 (off the unit circle).

Problem:  n integer, t real, L=Limit[sin nt,n->∞].
Then L=0 or it doesn't exist?  --rwg

On Sat, Jul 2, 2016 at 11:19 AM, Bill Gosper <billgosper@gmail.com> wrote:
...
Out[1131]=
                   k
                  q
Sum -------------------------------,
          1 - 2 k 2       1 + 2 k 2
    (1 - q       )  (1 - q       )
k,-∞,∞
=
                       1    2 4
  (1 + q) QFactorial[-(-), q ]
                       2
  -----------------------------
                  2 4
            (1 - q )
q-Pochhammers in the sum but not the summand?  "q-Mittag Leffler"?
It q-retracts to
                  16
          Sum[-----------, {k,-∞,∞}] ==
                      2 2
              (1 - 4 k )
      2
  2 π
Mathematica is dumbstruck.  For minutes on end.
Are these actually common?
Test for q=1/2:
In[1132]:= N[Activate[%1131/.q->.5]]
Out[1132]= True
Term ratio:
In[1180]:= DiscreteRatio[%1131[[1,1]],k]
Out[1180]= (q (1-q^(1-2 k))^2 (1-q^(1+2 k))^2)/((1-q^(-1-2 k))^2 (1-q^(3+2
k))^2)
Nine terms of the sum give 27 term agreement with the product, despite the
lack
of q^k²:
In[1189]:=
Series[%1131/.\[Infinity]->9/.-\[Infinity]->-9//Activate,{q,0,29}]
Out[1189]= 1+5 q+12 q^2+24 q^3+44 q^4+68 q^5+102 q^6+150 q^7+202 q^8+270
q^9+354 q^10+446 q^11+561 q^12+693 q^13+838 q^14+1002 q^15+1195 q^16+1407
q^17+1638 q^18+1906 q^19+2179 q^20+2491 q^21+2842 q^22+3202 q^23+3610
q^24+4042 q^25+4504 q^26+5008 q^27+5549 q^28+6114 q^29+O[q]^30==1+5 q+12
q^2+24 q^3+44 q^4+68 q^5+102 q^6+150 q^7+202 q^8+270 q^9+354 q^10+446
q^11+561 q^12+693 q^13+838 q^14+1002 q^15+1195 q^16+1407 q^17+1638
q^18+1906 q^19+2179 q^20+2491 q^21+2842 q^22+3202 q^23+3610 q^24+4042
q^25+4504 q^26+5008 q^27+5550 q^28+6114 q^29+O[q]^30
In[1190]:= Subtract@@%
Out[1190]= -q^28+O[q]^30
I got it by mildly abusing the (bilateral) "q-Dixon's theorem" in
http://math.sun.ac.za/~hproding/pdffiles/Triples-long.pdf
which is not the (unilateral) q-Dixon's I get from path invariant
matrices.  --rwg