SECRIST%OAK.SAINET.MFENET@lll-mfe.arpa (12/18/85)
Date: Sat, 14-DEC-1985 10:58 EST To: Info-Micro@BRL-VGR.Arpa Message-ID: <[OAK.SAINET.MFENET].8F8D6F60.008E77DA.SECRIST> Organization: Science Applications Int'l. Corp., Oak Ridge, Tenn. Geographic-Location: 36 01' 42" N, 84 14' 14" W CompuServe-ID: [71636,52] X-VMS-Mail-To: ARPA%"Info-Micro@BRL-VGR.Arpa" Does anybody out there know the algorithm for computing Pi to n-digits ? I don't mean 22/7 or a quickie approximate-it-in-a-loop-in-basic... I mean computing it to like 100,000 places. Not that this has much practical value, of course, but just as a nifty computing curiosity. Thanks. Richard SECRIST%OAK.SAInet.MFEnet@LLL-MFE.Arpa
doug@terak.UUCP (Doug Pardee) (12/20/85)
> Does anybody out there know the algorithm for computing Pi to n-digits ?
I don't know a fast one. One that works, but will take forever, is
4/1 - 4/3 + 4/5 - 4/7 + 4/9 - 4/11 + 4/13 - 4/15 + 4/17 - 4/19 + ...
--
Doug Pardee -- CalComp -- {hardy,savax,seismo,decvax,ihnp4}!terak!dougLINDSAY@tl-20b.arpa (12/22/85)
Pi = 16 arctan 1/5 - 4 arctan 1/239 arctan x = x - x**3/3 + x**5/5 ... -------
LINDSAY@tl-20b.arpa (12/24/85)
I've been asked for the provenance of the formula I gave. I suggest looking up the "Gregory-Leibniz equation", which is a close relative of arctan x = x - x**3/3 + x**5/5 ... and dates to the 1600's. Various people have fiddled with argument values and trig transforms: Gauss's version Pi = 48 arctan 1/18 + 32 arctan 1/57 - 20 arctan 1/239 Stormer's version Pi = 24 arctan 1/8 + 8 arctan 1/57 + 4 arctan 1/239 the one I gave Pi = 16 arctan 1/5 - 4 arctan 1/239 In 1961, someone calculated Pi to 100,000 places, using both Gauss's version and Stormer's version. Both gave the same answer (except for roundoff). However, Gauss's version ran in half the time. I suspect my version is the slowest of the three. -------
ags@pucc-h (Dave Seaman) (12/24/85)
In article <842@brl-tgr.ARPA> SECRIST%OAK.SAINET.MFENET@lll-mfe.arpa writes: >Does anybody out there know the algorithm for computing Pi to n-digits ? A reasonably fast series for pi is given by pi/4 = 4 * arctan(1/5) - arctan(1/239) where arctan x = x - x^3/3 + x^5/5 - x^7/7 + ... so that pi/4 = 4 * [1/5 - 1/(3*5^3) + 1/(5*5^5) - 1/(7*5^7) + ...] - [1/239 - 1/(3*239^3) + 1/(5*239^5) - 1/(7*239^7) + ...] -- Dave Seaman pur-ee!pucc-h!ags