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From: OsherD on 2 May 2010 20:47 From Osher Doctorow The Fibonacci numbers actually start counting from 0: 1) F0 = 0 2) F1 = 1 Then F2 = F1 + F0 = 1 + 0 = 1. So my referring to 1, 1 as initial values was incorrect - 0 and 1 are the first two values. Osher Doctorow
From: OsherD on 2 May 2010 21:03 From Osher Doctorow Some arguably quark-triple relevant aspects of sqrt(3) (see Wikipedia's "Square root of 3", Wolfram's "Equilateral Triangle", Wolfram's "Theodorus' Constant", etc.): 1) Each angle of an equilateral triangle is pi/3 (which is 60 degrees). 2) tan(60 degrees) = sqrt(3). 3) The space diagonal of a unit cube has length sqrt(3). 4) The area of an equilateral triangle is (1/4)sqrt(3) s^2 where s is the length of any side. 5) The altitude of an equilateral triangle is sqrt(3)/2. 6) The area of the circumcircle of an equilateral triangle is (1/3)pi s^2, where s is the length of any side. Osher Doctorow
From: OsherD on 2 May 2010 21:10 From Osher Doctorow The altitude of an equilateral triangle is [sqrt(3)/2)]s where s is the length of any side. I omitted the s (typo). Osher Doctorow
From: OsherD on 2 May 2010 21:25 From Osher Doctorow Here is a rather curious result resulting to (1 +/- sqrt(3)/2: 1) [(1/2)(1 + sqrt(3))]^2 + [(1/2)(1 - sqrt(3))]^2 = i^2 = -1 (so the roots are sides of an imaginary right triange of hypoteneuse i). because the first square on the left is (1/2)(sqrt(3)i - 1), while the second square on the left is (-1/2)(1 + sqrt(3)i). So the values of x from Gn = Gn-1 - Gn-2 obtained as (1 +/- sqrt(3))/2 (partly analogous to the Golden ratio from the Fibonacci numbers) are sides of an imaginary right triangle with hypoteneuse i. Osher Doctorow
From: Igor on 3 May 2010 09:39
On May 2, 8:38 pm, OsherD <mdocto...(a)ca.rr.com> wrote: > From Osher Doctorow > > We have seen that, in analogy with the Fibonacci Numbers F_n or Fn: > > 1) Fn = Fn-1 + Fn-2, n > 1 > > the Probable Causation/Influence Differences y - x from P(A-->B) = 1 + > y - x in their sequence representation: > > 2) Gn = Gn-1 - Gn-2 > > are related to a "limiting type" result partly analogous to the > Fibonacci quotient Fn+1/Fn limit phi, where phi is the Golden Mean or > Golden Ratio = (1 + sqrt(5))/2. The partial analog for Gn is: > > 3) x = [1 +/- sqrt(3) i ]/2 > > The exactly same procedure that produces phi from Fn, namely dividing > both sides of (1) by Fn-1 and taking the limit as n --> infinity, > yields (3) from (2) except that the limit in a strict sense does not > exist (not surprising since it contains the imaginary i while Gn are > real-valued). > > Since (3) arguably represents Unification of the 4 Fundamental > Interactions as explained in the recent post here, it involves "new > mathematics" and "new physics". > > There are some interesting consequences of putting various initial > values G1, G2 into (2) and then calculating the resulting sequences of > Gn, although unlike Fn where F1 = 1 = F2, the Gn sequences are quite > different: > > 4) If G2 - G1 is different from both G1 and G2 and neither is 0 (and > G2 - G1 is not 0), then Gn takes on 6 different values in sequence > which keep repeating, namely G1, G2, G2 - G1, and their negatives. > > 5) If G2 - G1 equals one of G1 or G2 (usually G1 in that case) and > neither is 0 and they are unequal, then Gn takes on 4 different values > in sequence, namely G1, G2, and their negatives. If one of G1 or G2 > is 0 or they are equal, then Gn has fewer than 4 different values. > > The 6 in (4) is 3! = 3 times 2 times 1, which again arguably relates > back to quarks (the number of ways of selecting one quark from 3 or > one edge from a triangle). The 4 in (5) appears to relate to the > "degeneration" of the triangle from 3 to 2 sides. > > For example, if we select G1 = 2 and G2 = 10, then we get: > > 6) G1 = 2, G2 = 10, G3 = 8, G4 = -2, G5 = -10, G2 = -8, G7 = 2, etc. > (repeats). > > Osher Doctorow But how is this related to quantum gravity? |