```Date: Jul 30, 2013 11:11 AM
Author: Peter Percival
Subject: Re: Calculating Pi using polygon sides sum / radius

jonas.thornvall@gmail.com wrote:> Let say there was a rounding error due to some flaw in the used> arithmetic and it turn out that Pi actually was rational at some> fairly early digit.>> But i guess that is ruled out,It is, you'll find a proof of pi's irrationality in books on realanalysis.  Here's a sketch:Let I_n(alpha) = integral_{-1}^1 (1 - x^2)^n cos alpha x dx.  Show that,if n > 1, alpha^2 = 2n(2n - 1)I_{n-1} - 4n(n - 1)I_{n-2}.  Show that,if n > 0, alpha^{2n + 1}I_n(alpha) = n!(P sin alpha + Q cos alpha),where P and Q are polynomials in alpha with integral coefficients ofdegree < 2n + 1.  Show that, if pi/2 = b/a, a and b integers, thenb^{2n + 1}I_n(pi/2)/n! would be an integer.  Now consider large n...> i will try to calculate Pi with my hexagonal javascript, sum the> vertices as i double them up and calculate the new hypotenuse.>> It will not be 22/7 but who knows maybe the dividend turn out to be> less then a million before making it rational?>-- Nam Nguyen in sci.logic in the thread 'Q on incompleteness proof'on 16/07/2013 at 02:16: "there can be such a group where informallyit's impossible to know the truth value of the abelian expressionAxy[x + y = y + x]".
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