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Another Ramanujan Trick
09-29-2019, 01:28 PM
Post: #5
ttw Offline
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RE: Another Ramanujan Trick
(Sorry about the typo.)

The method works by developing the continued fraction for Pi^4. The continued fraction for Pi has a 292 real early and 355/113 comes from stopping just before the 292.

The continued fraction for Pi^4 is (97; 2, 2, 3, 1, 16539...) so one stops before the 16539.

Continued fractions (to anything) give a sequence of p(i))/q(i) of numerators and denominators. The error in the ith term is less than 1 part in q(i)*(q(i)+q(i-1)). (Assuming I'm not off 1 in counting.)
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Messages In This Thread
Another Ramanujan Trick - ttw - 09-29-2019, 06:45 AM
RE: Another Ramanujan Trick - Thomas Okken - 09-29-2019, 09:17 AM
RE: Another Ramanujan Trick - ijabbott - 10-03-2019, 07:21 AM
RE: Another Ramanujan Trick - ijabbott - 10-03-2019, 08:23 PM
RE: Another Ramanujan Trick - ttw - 09-29-2019 01:28 PM
RE: Another Ramanujan Trick - Albert Chan - 09-29-2019, 03:23 PM
RE: Another Ramanujan Trick - ttw - 09-29-2019, 05:09 PM
RE: Another Ramanujan Trick - Helix - 10-19-2019, 11:59 PM
RE: Another Ramanujan Trick - grsbanks - 10-19-2019, 05:46 PM
RE: Another Ramanujan Trick - rprosperi - 10-19-2019, 11:43 PM



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