[math-fun] $500 pi approximation contest
From: http://www.kaidy.com/PiReward.htm Alas, someone already won the contest with a remarkably elegant submission. rwg, you'll just have to calculate faster next time. Magic of 355 / 113 355 / 113 is a 99.9999% accurate approximation of pi. 355 / 113 = 3.1415929203 pi = 3.1415926535 The difference = 0.0000002668 Can you find a BETTER fraction to more closely approximate the value of pi? The first one to do so will earn a reward of $500. (NOTE: The fraction you submit must be simplified to lowest terms, so not exactly equivalent to 355 / 113, and the numerator and denominator must each be whole numbers.) Click here to see if someone has come up with a BETTER solution. And from the "winner" page: We have a winner! John McKiernan was the earliest to submit 6,283,185,307/2,000,000,000. Thank you to all of our other participants.
Quoting Jason <jason@lunkwill.org>:
John McKiernan was the earliest to submit 6,283,185,307/2,000,000,000.
Do I see a pattern here? with all the zillion digits of pi, there are probably some prime numbers lurking as subsequences here and there. From then on, it's off to win a prize! - hvm ------------------------------------------------- www.correo.unam.mx UNAMonos Comunicándonos
This *must* be a come-on--it's too stupid to be real: (c2) cfexpand([3,7,15,1,292/2]) [ 52163 355 ] (d2) [ ] [ 16604 113 ] (c3) dfloat(%[1]/%[2]-%pi) (d3) [- 2.66213257216208d-7, 2.66764189404967d-7]
Quoting Jason <jason@lunkwill.org>:
John McKiernan was the earliest to submit 6,283,185,307/2,000,000,000.
Do I see a pattern here? with all the zillion digits of pi, there are probably some prime numbers lurking as subsequences here and there. From then on, it's off to win a prize!
- hvm Who needs primes? Just reduce the fractions.
participants (3)
-
Jason -
mcintosh@servidor.unam.mx -
rwg@sdf.lonestar.org