(I didn't look at the paper -- this stuff might be in it.) The FOR loop iteration [u=sqrt(2u+2)/2] maps cos x -> cos(x/2). FA computes 2^n sin(x/2^n) * (2 + sec(x/2^n)). If we write y for x/2^n, the FA computation produces 2^n * (2 sin y + tan y). 2 sin y + tan y is a cute expression to cancel the y^3 term in the power series. This amounts to Archimedes' calculation of pi using a polygon of 3*2^n sides, With the 2sin+tan trick making better use of the side length, getting better Convergence: the error is reduced by a factor of 16 for each iteration. This matches JJs numeric results. I'm a little puzzled by Plouffe's results. The numeric results seem to indicate an error reduction of 2.5 for each additional Iteration: (3979-1591)/(10000-4000) = .398 digits/iteration = log10(.4). The ES loop is the same cos x -> cos(x/2) step; but the ES power series would be Expected to cancel out terms up through y^11 in the sine power series. This Should give a 2^12 = 4096 reduction in the error, for each step. Even if the Coefficients in the power series are suboptimal, the reduction should be an Integral power of 4, which 2.5 is not. A puzzle. Numerical coincidence: 597/256 ~= 7/3. Rich -----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Joerg Arndt Sent: Wednesday, March 29, 2017 1:57 AM To: math-fun <math-fun@mailman.xmission.com> Subject: [EXTERNAL] Re: [math-fun] about a certain François Mendzina Essomba and surprising pi formulas Linear convergence also for the one on the first page of the pdf (and to 2*Pi, not Pi), using Pari/gp: u=0.5; fa(n, u)=2^n*sqrt(1-u*u)*(2+1/u); { for (n=1, 20, a = fa(n, u); print([n, a]); u = sqrt(2*u + 2)/2; ); } [ 1, 6.928203230275509174109785366023489467771221015241522512] [ 2, 6.309401076758503058036595122007829822590407005080507504] [ 3, 6.284698261089313849362810669937794318042660390635843197] [ 4, 6.283278112439984585093422611211121983684728189547233237] [ 5, 6.283191080816779590245594957769397689710707177378823561] [ 6, 6.283185667617591716261043297794296056715862349224554543] [ 7, 6.283185329700498681764470474408453804919470421723461451] [ 8, 6.283185308587042544838278336102423311351107299137065124] [ 9, 6.283185307267550904067606378734738531926108069644277105] [10, 6.283185307185084228980090604450773526344097135976414163] [11, 6.283185307179930086043689611234615188055755351473937852] [12, 6.283185307179607952489170979060409544946653401185434692] [13, 6.283185307179587819147935530420307409300710608427640595] [14, 6.283185307179586560814200845559546060792694428128468339] [15, 6.283185307179586482168343873547465250587906278534678652] [16, 6.283185307179586477252977835387205202554516100676935709] [17, 6.283185307179586476945767458355165431733785104475147206] [18, 6.283185307179586476926566809796178203582845220233949936] [19, 6.283185307179586476925366769261327677722211068633393702] [20, 6.283185307179586476925291766727900866354340044649732273] Next! Best regards, jj * Simon Plouffe <simon.plouffe@gmail.com> [Mar 27. 2017 18:51]:
Hello,
I programmed the first one,
is is not like the AGM, the number of valid digits is proportional to n, something like approx. (5/2)^n,
here is the maple routine :
# F.M. Essomba formula
es:=proc(n) option remember; if n = 0 then 1/2 elif n = 1 then sqrt(3.0)/2 else sqrt(2.0*es(n-1) + 2)/2.0 end if: end;
a:=proc(n) 3*2.0^n/315*sqrt(1.0-es(n)^2)*(256 + 597*es(n) - 1024*es(n)^2 + 542*es(n)^3-56*es(n)^5);end;
a(10000) differs from Pi at the 3979'th position. a (4000) differs from Pi at the 1591'th position,
confirming that it converges rather slowly.
It is not a record calculation, but certainly original (IMHO).
I have not tried the other ones yet.
Have a nice day and bonne journée.
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