Changing the denominators to the closest multiple of 6, and we have pi*sqrt(3)/6 = (5/6) (7/6) (11/12) (13/12) (17/18) (19/18) (23/24) ... Warut On Mon, Feb 14, 2011 at 4:09 PM, Christian Boyer <cboyer@club-internet.fr> wrote:
In www.mathpages.com/home/kmath477.htm, explanation of:
3 5 7 11 13 17 19 π/4 = - × - × - × -- × -- × -- × -- × ... 4 4 8 12 12 16 20
Here the numerators are the odd primes, and the denominators are the closest numbers of the form 4n.
Christian.
-----Message d'origine----- De : math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] De la part de Bill Gosper Envoyé : lundi 14 février 2011 08:24 À : math-fun@mailman.xmission.com Objet : Re: [math-fun] Euler's crazy pi product
interesting how if you start with 1 you get a line of slope 1. you can also include all integers.
π/4 = 1/2 × 2/2 × 3/2 × 4/4 × 5/6 × 6/6 × 7/6 ...
if I've got that right.
For the n-1/2 cases, round up and round down produce divergent products. Round to even
gives 4*(1/4)!^2/sqrt(2*pi) . At last, a mathematical justification for round to even! --rwg
On Sun, Feb 13, 2011 at 9:36 PM, Bill Gosper <billgosper@gmail.com <http://gosper.org/webmail/src/compose.php?send_to=billgosper%40gmail.com>> wrote:
http://math.ucr.edu/home/baez/week127.html finds in Lennart Berggren, Jonathan Borwein and Peter Borwein, π: A Source Book, Springer-Verlag, New York, 1997, "the following weirdly beautiful formula due to Euler, which unfortunately is not explained:" (Then how can they call it a sourcebook?--rwg)
3 5 7 11 13 17 19 π/2 = - × - × - × -- × -- × -- × -- × ... 2 6 6 10 14 18 18
"Here the numerators are the odd primes, and the denominators are the closest numbers of the form 4n+2." E.g., for 10, 100, ..., 10^6 terms, In[209]:=
Table[Block[{p=2},Nest[N[#*(p=NextPrime[p])/(2+4*Round[(p-2)/4]),9]&,2 ,10^n]],{n,6}]
Out[209]= {3.10152045, 3.13398462, 3.13772561, 3.14073685, 3.14143290, 3.14157195}
Why the heck?? Was Euler a preincarnation of Ramanujan? --rwg
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