(Thought funsters might enjoy eavesdropping.) Jason, I'm still occupied by your pi question. I believe we've already rejected montecarloing or integrating Buffon's needle. How about integrating the area under sqrt(1-x^2)? Besides the numerous numerical methods, one can expand via the binomial thm and termwise integrate. Expanding at x=0 gives the slow series i 1/2 inf (- 1) ( ) ==== i %pi \ --- = > ------------, 4 / 2 i + 1 ==== i = 0 but expanding at x=1 gives one bit/term: i 1/2 inf (- 1) ( ) ==== i %pi \ --- = > ------------------. 4 / i - 3/2 ==== (2 i + 3) 2 i = 0 Better yet, only integrate from sqrt(3)/2 to 1: inf ==== i + 3/2 \ (2 - sqrt(3)) 1/2 3 sqrt(3) %pi = 12 > -------------------- ( ) + --------- / i i 2 === (2 i + 3) (- 4) i = 0 gives > 1 decimal digit/term (ratio = .06699). Expansion at points other than 0 and 1 produces a powerseries whose coefficients are themselves sums. But these obey a three term recurrence, so the double sum reduces to a 4x4 matrix product. Strangely, integrate(expansion at 1/sqrt(2),x,0,1) produces a divergent series. How about inf ==== 2 \ 1 %pi > -- = ---- ? / 2 6 ==== n n = 1 (I have an Euleroid derivation of this in the form of an annotated Macsyma notebook, starting with the atan x series.) Here's a neat one: the average value of gcd([k sqrt(2)],[k sqrt(3)])^-2 should approach pi^2/15, for large intervals of k. ([]:=floor) Too bad you don't like logs. I have *really* magic ways to get pi/ln 2 and (pi/ln 2)^2, the latter using a random bit stream. Heck, pi is basically log(-1). Ah, how about (d5) t(n) := if n = 0 then .2d1 else (t(n - 1), 2 %% ---------------------------------) %% (--, sqrt(%% + 1) + sqrt(1 - %%)) n 2 (c6) t(24) (d6) 3.14159265358979d0 or (d7) s(n) := if n = 0 then .4d1 else (s(n - 1), 2 %% --------------------) 2 %% 1 + sqrt(1 + ------) n + 1 4 (c8) s(25) (d8) 3.14159265358979d0 ? --rwg POINT SOURCE PROSECUTION