[Apologies, Funsters, for the repetition. The formulas in my PS, while true, were tautological and belied the text. "Luckily", I also neglected to bcc the eavesdroppers, so this stone in the head kills two in the bush. --Louis B. Mayer. (Not really.)] Dr. q himself, Mizan Rahman, gets the bilateral version as a limiting case of Bailey's 6psi6, saying
This bilateral sum looked too familiar to overlook.It's q-analogue is a special case of the well-known 6psi6 summation formula that you will find,for instance in our book,(5.3.1).Just let a go to q^(2),then set e=a^(1/2),let b,c,d,go to 1/a,1/b,1/c,respectively,then simplify.Note that a q-analogue of your sum is not a 3psi3,rather a 4psi4..In fact the relevant formula is (5.3.3),which is a q-analogue of Dixon's formula.In the simplification process,you have to remember that the limit of (q/a;q)_inf/(q/a^(1/2);q)_inf,as a goes to q^(2),exists,and equals 2(1-q^(-2)).
Equation numbers refer to his (& Gasper's) book, Basic Hypergeometric Series. Without even knowing this, I should have gone after the more general sum with the index shifted by an arbitrary phase parameter. Mizan then prophetically admonished
I hope this will help realize that many of the summation formulas that may appear new are really special or limiting cases of some long-known formulas in different notations.Or in some other disguise.
Sure enough, retracing my Macsyma steps reveals that the unilateral version, inf ==== \ (i - a)! (i - b)! (i - c)!
-------------------------------------- = / (i + a + 1)! (i + b + 1)! (i + c + 1)! ==== i = 0
(- a - 1)! (- b - 1)! (- c - 1)! (c + b + a)! 1 -------------------------------- (-------------------------- - --------), 2 (b + a)! (c + a)! (c + b)! a! b! c! is just the limit a->0 of Dougall's very-well-poised 5F4 (2.1.7). The same trick applied to Jackson's q-extension (2.7.1) (which erroneously claims to extend (2.1.6)) gives stringpoch(intosum(multthru(q, sum(q^n*(q^(n+1)+1)*qpoch(b*q,c*q,a*q,q,n)/(a^n*b^n*c^n*qpoch(q^2/b,q^2/c,q^2/a,q,n)),n,0,inf) =(a-q)*(b-q)*(c-q)/((a-1)*(b-1)*(c-1)*q) -(1-q/a)*(1-q/b)*(1-q/c)*qpoch(q,q/(b*c),q/(a*b),q/(a*c)),q,inf) /(qpoch(1/b,1/c,1/a,q/(a*b*c),q,inf)*q))) oo n + 1 n + 1 ==== (b q, c q, a q; q) q (q + 1) \ n
--------------------------------------- = / 2 2 2 ==== n n n q q q n = 0 a b c (--, --, --; q) b c a n
(a - q) (b - q) (c - q) ----------------------- (a - 1) (b - 1) (c - 1) q q q q q q (q, ---, ---, ---; q) (1 - -) (1 - -) (1 - -) b c a b a c oo a b c - ------------------------------------------------. 1 1 1 q (-, -, -, -----; q) b c a a b c oo What makes these peculiar is their failure to "terminate on the left". I.e. the summand's denominator lacks the "obligatory" i! or (q;q)_n present in nearly all sums that come out in Gamma (or q-Gamma) functions. Also mysterious is the difficulty of recurrence-relating them to their contiguous neighbors. OK, so they weren't news. If you want not so well known, try www.tweedledum.com/rwg/idents.htm . I'll bet with enough tawa salmon and masala chai, I could "q" that 4F3[125/128]. --rwg PS: Why didn't I try this earlier?: (c1) makelist(sum((i-a)!*(i-b)!*(i-c)!/((i+a+1)!*(i+b+1)!*(i+d)!),i,0,inf),d,d,d+2) inf ==== \ (i - a)! (i - b)! (i - c)! (d1) [ > ----------------------------------, / (i + a + 1)! (i + b + 1)! (i + d)! ==== i = 0 inf ==== \ (i - a)! (i - b)! (i - c)!
--------------------------------------, / (i + a + 1)! (i + b + 1)! (i + d + 1)! ==== i = 0
inf ==== \ (i - a)! (i - b)! (i - c)!
--------------------------------------] / (i + a + 1)! (i + b + 1)! (i + d + 2)! ==== i = 0
(c2) (sign_uses_hashing:false,tlimswitch:true,substpart(factor(piece),contiguate(%),1,1)) (d2) (d + a + 1) (d + b + 1) (d + c + 1) inf ==== \ (i - a)! (i - b)! (i - c)! ( > --------------------------------------) / (i + a + 1)! (i + b + 1)! (i + d + 2)! ==== i = 0 2 /(d + c + 2 b + 2 a + 1) - (2 d + 2 c d + 3 b d + 3 a d + 3 d + b c inf ==== \ (i - a)! (i - b)! (i - c)! + a c + c + b + a + 1) ( > --------------------------------------) / (i + a + 1)! (i + b + 1)! (i + d + 1)! ==== i = 0 inf ==== \ (i - a)! (i - b)! (i - c)! /(d + c + 2 b + 2 a + 1) + > ---------------------------------- = / (i + a + 1)! (i + b + 1)! (i + d)! ==== i = 0 (a + 1) (- a)! (b + 1) (- b)! (- c)! (d + 2) -------------------------------------------------- (a + 1)! (b + 1)! (d + c + 2 b + 2 a + 1) (d + 2)! Answer: Because this isn't very useful--it doesn't tell us any values of the contiguous series because [with d = c+1, e.g.] we only know the last of the three.