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 real

analysis. 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 of

degree < 2n + 1. Show that, if pi/2 = b/a, a and b integers, then

b^{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 informally

it's impossible to know the truth value of the abelian expression

Axy[x + y = y + x]".