I said
[...]which will obey nonlinear, three term relations among pi(q), pi(q^a), and pi(q^b), for rational a and b. (Actually, we want Pi(q):=pi(q)/(1-q^2), which changes q^(6k^2) to q^(6(k^2-1/24)) in our series.) E.g., 2 2 2 Pi (q) Pi (q ) 4 = ------------- - -------, 2 4 2 4 Pi(q ) Pi(q ) Pi (q )
9 2 3 9 9 3 Pi(q) Pi(q ) + Pi (q ) = sqrt(Pi(q) Pi(q )) (3 Pi(q ) + Pi(q)), . . .
which Gene pointed out (in the 70s) give quadratic and cubic methods for pi computation. I.e., you solve the first one, say, for Pi(q), and substitute sqrt(q) for q. Then choose an initial q (say 2^-4) for which Pi(q) and Pi(q^2) are very easy to evaluate, and use the recurrence to get Pi(sqrt q), Pi(sqrt(sqrt q)), Pi(q^(1/8)),..., Pi(q^2^-n). But this leads to a nice
PUZZLE (or at least trick question): q^2^-n converges to 1 linearly with n: (2.d0^-4)^2^-9 = 0.99459942348363d0, and pi(q^2^-n) converges linearly to pi: 2 theta (0, q) theta (0, q) (1 - q ) 2 3 pi(q) = ---------------------------------- 2 subst(%th(2),q,%) pi(0.99459942348363d0) = 3.12464151346649d0 So how can we claim this is a quadratic method? --rwg Answer: Under the Jacobi imaginary transformation, 2 2 %pi %pi ------ ------ log(q) log(q) 2 %pi theta (0, %e ) theta (0, %e ) (1 - q ) 3 4 pi(q) = ----------------------------------------------------, 1 2 log(-) q so it is the quantity -n 2 n 2 n 1 2 %pi 2 %pi 2 2 log(-) pi(q ) ------- ------- q log(q) log(q) ---------------- = %pi theta (0, %e ) theta (0, %e ) 2 3 4 -- n n 2 2 (1 - q ) that converges quadratically, not pi(q^2^-n). (And thus requires a valuation of log q to get pi.) --------------------------------- Catch up on fall's hot new shows on Yahoo! TV. Watch previews, get listings, and more!