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The Story of The Unit Simulacrum and Whole PI

© 2004-2010

The names “The Bottom Up Solution”, “The Unit Simulacrum”, “Whole PI” and all equations, figures and schemes are copyrighted in 2004-2010 by William Craig Byrdwell.  No portion of the equations, figures or schemes may be used without acknowledging the copyright holder, William Craig Byrdwell. or Byrdwell.com

Chapter Four  —  The First Increment — Page Two

 

Now, the First Decrement: When we incremented the denominator, or decremented the ratio, we added 1 to the one in the denominator of the ratio, making (1/(1+1)) = 1/2 times PI, for p/2, which has a value of 1.5707963. This is also greater than One. Therefore p/2 ³ 1 would be the same Case as multiples of p. Thus, p, its First Increment and Its First Decrement would all be ³ 1. p/2 , p , and 2p would all be solved Case 2 in the Unit Simulacrum. These three variations on PI behave similarly to the Fibonacci Simulacrum, which was Case 2.2.2… in the Aligned notation.

We could increment the denominator again, and this would be the Second Decrement of the One ratio times PI. The Second Decrement is (1/(1+1+1))· p = p/3. This ratio, the Second Decrement of One Times PI is close to One (= 1.047197551). This is within 5.0% of One, 1.00, which in many cases can be considered approximately equal to one (100% » 1.00 + 5.0%). You can say that the Second Decrement of p·(1/1) is p·(1/(1+1+1)) and this is = p/3, which is approximately equal to One. This would mean that p·(1/(1+1+1)) » (1/1). So, it depends on your definition of ‘approximately’, and it always comes down to definitions. A question is: How certain can we be that our definition is correct? It can be related to percentage probability. What is the probability that the definition is correct? Or, in other words, in what percentage of circumstances will our definition be correct? Uncertainty is built into the observed system that we are trying to figure out. The ratio p/3 is still above One, and so would be a Case 2 solution, but it is very close the Critical Limit of the (1/1) ratio, which is 1.00. So, if p·(1/(1+1+1)) » (1/1), it can be operated like the (1/1) ratio in the Unit Simulacrum. The Critical Limit for the (1/1) ratio can be used for the p·(1/(1+1+1)) ratio since they’re approximately equal. The (1/1) ratio operated in the Unit Simulacrum gives a sum of 2. The (p·(1/(1+1+1))) ratio operated in the Unit Simulacrum gives a sum of approximately 2 (now only a 2.305471% approximation error). One is an inherent Critical Limit for ratios operated in the Unit Simulacrum. When the A/B ratio is 1.00 (1.05 for p·(1/(1+1+1)). At the Critical Limit, all solutions are equal, and so the solution can be solved using either a Case 1 or Case 2 solution. At the Critical Limit, all Cases are equal, and this constitutes the Simulacrum Equality. These are The Eight Possibilities for Observation. Thus, since p/3 » (1/1), the approximation opens up the possibility of using Case 1, which is a second Case having four more possibilities for observation. Since p/3 » (1/1), when the ratio is operated in the Unit Simulacrum, all Eight Solutions, can be used, which are equal at the Critical Limit of when the Ratio equals the One. Of course, we have to remember that we are in the Second Decrement of the One. The moral is, “Don’t forget the One!”.

Since the first decrement, the unit and the first increment of PI are: p/2 , p , 2p, and p/2 is closest to one, this is the ratio of PI that I first inserted into the Unit Simulacrum. Figure 31 shows every possible combination of 1 interacting with p/2 in the Unit Simulacrum. Figure 32 shows the simplified simulacrum sum of 1 and p/2 that results if we simplify ((p/2)/1) to just (p/2) and assume that only (p/2) is observed, not (1/(p/2)). Only Case 2, in which (p/2) ³ 1, is actually observed. The Eight Possibilities for Observation can be seen in Figure 31. The shows the sum all possible possibilities of A plus B in the Unit Simulacrum, without any limits placed on A, when B=1. In our observed reality, (p/2) ³ 1, so it is a Case 2 solution. Nevertheless, Figure 31 shows that there are Eight Possibilities for Observation of the sum of One plus p/2. Just because they are not actually observed does not keep them from being possibilities for observation, because each possibility gives the correct sum.

Continuing with the deconstruction of PI·(1/1), if the denominator is incremented a third time, the third increment in the denominator of PI·(1/1), could be called the Third Decrement. This would be: p·(1/(1+(1+1+1))), = (p/4), which means the Third Decrement is the first Decrement that is always less than (1/1). (p/4) = 0.7853981633974483. This is the first Decrement that would be solved as a Case 1 solution, (p/4) £ 1. The denominator is written as (1+(1+1+1)) to emphasize the fact that it is a solution in the Unit Simulacrum, which is (1+ ratio). The ratio in the case of the Third Decrement is (3/1). This is the ratio of the Decrement Number to the Unit Decrement. The Unit Decrement is the First Decrement, p·(1/(1+1)). We defined the terms First Decrement, the One variable, and the First Increment to describe the relationship between the numbers 1/2, 1 and 2 and also between p·(1/2), p·(1/1) and p·(2/1). The First Decrement constitutes the Unit Decrement.

From that point on, decrements will be measured in units of the first. Similarly, the First Increment constitutes the Unit Increment. Everything after the first will be measured in terms of the one that defined the category of First Increment. As I mentioned above, I noticed the equivalence between The Unit Simulacrum and the First Increment of a ratio. Thus, The Third Decrement = (1+(1/(1+(1+1+1)) = (1+(1/(1+(ThirdDec/UnitDec))). It is the Unit Simulacrum operated with the ratio (#Dec/UnitDec), where #Dec is the Decrement number, which is (3/1), and UnitDec is the First Decrement. #Dec is a variable that goes from 0 to 3, so that it can be divided into three Decrement times, tD. Three Decrement times would give td1, td2, and td3, where td is the ‘Decrement Time’, or how many Times UnitDecrement. We’ll assume the times are the same size, so the third Decrement Time would be equal to Three Times the Unit Decrement, which is td3 = 3· td1