I said inf /===\ 2 | | 4 n %pi
(d93) | | ------------------- = --- | | (2 n - 1) (2 n + 1) 2 n = 1
which (Wallis) presumably got from Euler's infinite product for sin(x)/x.
Dick Askey chided that this formula preceded Euler by several decades, and pointed me at https://ceres.math.wisc.edu/horde/imp/attachment.php?u=askey&t=1153038348&f=... Sergey V (not N!) Khrushchev's reconstruction of Lord Brouncker's 1655 derivation of the first continued fraction for pi, which I would notate 4 2 1 -- + 1 = K(2, (2 k + 1) ) = 2 + -----------, pi 9 2 + ------- 25 2 + --- . . . whose reciprocal is termwise equivalent to Gregory's series, pi/4 = 1 - 1/3 + 1/5 - ... Using the matrix methods of gosper.org/stanfordn2.dvi, the corresponding CF for 1/atan(x) 2 2 2 2 K(x - 2 k (x - 1) + 1, (2 k + 1) x ) = --------------------------------------- - x. x The CF transformation in the next section, gosper.org/stanfordn3.dvi, makes this 2 2 K((2 k + 1) (1 - x) (x + 1), 4 (k + 1) x ) -------------------------------------------, x instead of the usual 2 2 K(2 k + 1, (k + 1) x ) -----------------------. x But note that x -> 1 in the previous gives Wallis's product, modulo regrouping. The transformation applied to the usual atan CF gives asec(x) 1 ------------ = ---------------------------------------. 2 2 2 sqrt(x - 1) (2 k + 1) (x - 1) 1 - x K(x + 2 k, -------------------) + ----- 4 2 A&S seems to give a CF for asin(x)/sqrt(1-x^2) but not just asin. The term-for-term formula, after a tweak, is 1 1 ------- = - asin(x) x x - ---------------------------------------------------------------, 1 2 2 3 3 2 2 K(2 (k + -) x + (k + 1) (2 k + 3), - 4 (k + 1) (k + -) x ) 2 2 which may have been too ugly for them. The transformation formula only applies to K(linear,quadratic), but I can probably find one that works. Also, Alec M. remarked of no discussion of the hypergeometric approach, which I believe most practical for enormous precision. My favorite is the Chudnovsky formula inf /===\ [ - (2 k - 1) (6 k - 5) (6 k - 1) 545140134 k - 531548725 ] | | [ ] | | [ 3 3 ] | | [ 0 72 53360 k ] k = 1 sqrt(10005) (homogeneously) = -------------- 25625606400 pi with which they extracted biwyuns of Saganic verses (MLB's term). Other series with higher "digits/term" have a quadratic surd in their term ratio, thus requiring > twice the computation/term. The AGM methods (incuding that pi_q method I mentioned) are asymptotically better, but the crossover point is astronomical. But I claim anyone computing digits instead of partial quotients is wasting computer time. --rwg