| Topic: |
Science > Physics |
| User: |
"dedanoe" |
| Date: |
28 Sep 2006 04:44:04 AM |
| Object: |
the k-maker made the healer k (checkmate dude) |
taken from: http://dedanoe.tripod.com/solving_primes.html
Reducing the Prime vs. non-Prime problem down to 1 against 0
Primes are whole numbers that consist whole number of times only ONE
(divided with ONE they return their self) and their self (divided with
their self they return ONE). And ONE is GOD product and geometric
middle between ALL and NOTHING:
1 * 1 = 2 * 0.5 = 4 * 0.25 = 8 * 0.125 = ... = Infinity * Zero
I knew that one day I would reduce the Prime versus non-Prime problem
down to 1 against 0. I knew because from the first time I met them
Primes I had a hunch that they are the numbers of the Imbalance - the
numbers with disturbed reciprocity while the non-Primes i.e. Composites
are the numbers of the Balance - the reciprocal once. Here is the
way:
Function(k) = Round(k * Sqrt(N)) * Round(Sqrt(N) / k) - N = 0
The real number k here is the so-called Healer k. The Healer k turns k
* Sqrt(N) and Sqrt(N) / k into whole numbers. If N is Prime then this
equation has no real solution for k in range between 1 and
Trunc(Sqrt(N)). Otherwise, if N is non-Prime i.e. Composite then there
is at least one such k that solves it! You can solve k with methods of
approximations. However, since most probably no one knows how to
differentiate Round(x), the methods of approximations that implement
differential calculus like Newton's Tangent Method are useless. In
this case, most useful are the methods of approximations like my
Universal Leverian Method of Approximations which is general case of
the method Regula Falsi. You simply guess two values for k: k1 and k2
and do the following:
repeat
if |function(k1)| < |function(k2)| then begin
k := k1 - function(k1) * (k1 - k2) / (function(k1) -
function(k2) );
k2 := k1; k1 := k;
end else if |function(k2)| < |function(k1)| then begin
k := k2 - function(k2) * (k2 - k1) / (function(k2) -
function(k1) );
k1 := k2; k2 := k;
end else Randomize(k1, k2);
until |k1 - k2| < Acceptable_Error;
Well, how do you like me now? We are going to have a problem here...
Listen to me you mother fucker: You have to like me because if you turn
me down my need for your love will become miraculous and I will kick
your ***** and put you 2 meters underground and I'll make you a tree
with brunches and roots! You got that?
Dobri Karagorgov - DedaNoe
schizo-paranoid-lever-sex-pert
http://dedanoe.tripod.com
.
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