For his ntuplication merit badge, a young Scout asked me how to get the usual factorial ntuplication formula n^(n*z+1/2)*prod((z-i/n)!,i,0,n-1)/(2*%pi)^((n-1)/2) = (n*z)! n - 1 /===\ n z + 1/2 | | i n | | (z - -)! | | n i = 0 ------------------------- = (n z)! n - 1 ----- 2 (2 pi) as the q->1 limit of the q-factorial ntuplication formula (1 - q^n)^(n * z - ((n - 1)/2)) * qpoch(q, q, inf) * product(faq(z - (k/n), q^n),k,0,n - 1)/((1 - q)^(n * z) * qpoch(q^n, q^n, inf)^n) = faq(n * z, q) n - 1 n - 1 n z - ----- /===\ n 2 | | k n (q; q) (1 - q ) | | faq(z - -, q ) inf | | n k = 0 -------------------------------------------------- = faq(n z, q), n n n n z (q ; q ) (1 - q) inf with faq(z,q):=(1-q)^-z*prod((1-q^n)/(1-q^(n+z)),n,1,inf) inf /===\ n | | 1 - q | | ---------- | | z + n n = 1 1 - q faq(z, q) := ---------------- z (1 - q) (The kid's a little slow, and needs to take his shoes off to do transfinite induction.) Using the obvious definition of pi_q, %pi[q] := faq(-1/2,q^2) = sqrt(1-q^2)*qpoch(q^2,q^2,inf)/qpoch(q,q^2,inf) 2 2 2 (q ; q ) sqrt(1 - q ) 1 2 inf pi := faq(- -, q ) = ------------------------ q 2 2 (q; q ) inf I get, for n a power of 2, prod(((q^(2^k) + 1)^(n * z - 2^(k - 1) + 1/2)/(%pi[q^(2^k)])^(2^(k - 1))),k,0,lg(n) - 1)* product(faq(z - (k/n), q^n),k,0,n - 1) = faq(n * z, q) lg(n)-1 k k - 1 n - 1 /===\ 2 n z - 2 + 1/2 /===\ | | (q + 1) | | k n ( | | ---------------------------) | | faq(z - -, q ) = faq(n z, q) | | k - 1 | | n k = 0 2 k = 0 pi k 2 q For general n, |q|<1, the product on the left converges when extended to inifinity, so you can pull the usual hack of dividing two infinite products to get the finite one, but this doesn't say how to actually do the n=3 case: 'limit((1-q^3)^(3*z-1)*qpoch(q,q,inf)/((1-q)^(3*z)*qpoch(q^3,q^3,inf)^3),q,1,minus) = 3^(3*z+1/2)/(2*%pi) 3 3 z - 1 3 z + 1/2 (1 - q ) qpoch(q, q, inf) 3 limit -------------------------------- = ----------. q -> 1- 3 z 3 3 3 2 pi (1 - q) qpoch (q , q , inf) Asking Mma 7.0, In:=Assuming[Abs[q] < 1, Limit[(((1 - q^3)^(3*z - 1)* QPochhammer[q, q])/((1 - q)^(3*z)*(QPochhammer[q^3, q^3])^3)), q -> 1] -> ((3^(3*z + 1/2))/(2*Pi))] "\!\(\* StyleBox[\"\\\"Warning: Contradictory assumption(s) \\\"\", \"MT\"]\)\ \!\(\* StyleBox[ RowBox[{ RowBox[{ RowBox[{\"Abs\", \"[\", \ RowBox[{\"1\", \"+\", \"q\"}], \"]\"}], \"<\", \"1\"}], \"&&\", \ RowBox[{\"0\", \"<\", \"q\", \"<\", FractionBox[\"1\", \ \"1073741824\"]}]}], \"MT\"]\)\!\(\* StyleBox[\"\"\", \"MT\"]\) \ encountered In:=Assuming[Abs[q] < 1, Series[(((1 - q^3)^(3*z - 1)* QPochhammer[q, q])/((1 - q)^(3*z)*(QPochhammer[q^3, q^3])^3)), {q, 1, 2}] -> ((3^(3*z + 1/2))/(2*Pi))] [A very large output was generated. Here is a sample of it:] [. . .] The solution is to write the q-pochhammers as etas, (1-q^n)^(n*z-(n-1)/2)*(q^n)^(n/24)*eta(q)*(prod(faq(z-k/n,q^n),k,0,n-1))/((1-q)^(n*z)*q^(1/24)*eta(q^n)^n) = faq(n*z,q) n - 1 n - 1 n z - ----- /===\ n 2 n n/24 | | k n (1 - q ) (q ) eta(q) | | faq(z - -, q ) | | n k = 0 -------------------------------------------------------- = faq(n z, q) n z 1/24 n n (1 - q) q eta (q ) and use the Jacobi imaginary transformation block([radexpand:false],eta(q),%%=theta_imtrans(%%)) eta(q) = sqrt(-2*%pi/log(q))*eta(%e^(4*%pi^2/log(q))) 2 4 pi ------ 2 pi log(q) eta(q) = sqrt(- ------) eta(%e ) log(q) to get 'limit((1-q^n)^(n*z-(n-1)/2)*(q^n)^(n/24)*eta(q)/((1-q)^(n*z)*q^(1/24)*eta(q^n)^n),q,1,minus) = n^(n*z+1/2)/(2*%pi)^((n-1)/2) n - 1 n z - ----- n 2 n n/24 n z + 1/2 (1 - q ) (q ) eta(q) n limit ----------------------------------- = ----------- q -> 1- n z 1/24 n n n - 1 (1 - q) q eta (q ) ----- 2 (2 pi) This approach is less evident in Mma, whose DedekindEta takes argument tau = log(q)/(2i pi) instead of q. --rwg SELF-EVIDENT FIELD EVENTS