Type III TAG Results: Online Resources

Analogous to Type II TAG is the fact that CR3 is nested into the denominator of CR2 and CR2 is nested into the denominator of CR1for Type III TAG, producing a three-level deep nested construct.Specifically, the sum of the numerator and denominator of CR3 make up the denominator of CR2, and the sum of the numerator and denominator of CR2 make up the [DAG]+ that is in the denominator of CR1. Furthermore, the sum of the numerator and the denominator of CR1 makes up the sum of all ions.

The nested equations can readily be observed by examining the Case 1.1.1 and Case 1.1.2 solutions. When [AB]+ is 1 (=100%), indicated by Case 1.1.1, the [BC]+/[AB]+ ratio, CR3, becomes simply [BC]+/1, and the sum of [AB]++[BC]+ is simply the Unit Simulacrum solution 1+([BC]+/[AB]+), or 1+CR3. And of course, [AB]++[BC]+ is the denominator of CR2, the [AC]+/([AB]++[BC]+) ratio. So, when [AC]+/([AB]++[BC]+) is multiplied by 1+([BC]+/[AB]+)=[AB]++[BC]+, the denominator of CR2 cancels out giving simply [AC]+, as seen in Scheme IIIA. In this way, CR3 is nested into the denominator of CR2, so that [AC]+=CR2x(1+CR3).

For Case 1.1.2, the situation is similar, except now [BC]+ =1, and the sum of [AB]++[BC]+ is the Unit Simulacrum inverse solution 1+1/([BC]+/[AB]+), or 1+1/CR3. When this solution for [AB]++[BC]+ is multiplied times CR2, [AC]+/([AB]++[BC]+), the denominator again cancels out, giving [AC]+=CR2x(1+1/CR3). Thus, the sum of [AB]++[BC]+ is either 1+ratio or 1+1/ratio, and the Unit Simulacrum sum of [AB]++[BC]+ is 1+CR3 or 1+1/CR3, depending on Case 1.1.1 or Case 1.1.2. Either way, [AC]+ is arrived at by multiplying CR2, the [AC]+/([AB]++[BC]+) ratio, times the sum in the denominator, in the form of the Unit Simulacrum sum solution, to cancel out the denominator and give [AC]+. Thus, it can be seen that the Unit Simulacrum sum of CR3 (1+CR3 or 1+1/CR3) is the sum that is nested into the denominator of CR2.

Similarly, the Unit Simulacrum sum of CR2 is nested into the denominator of CR1. It is easiest and most direct to observe this relationship in the Case 1.2.n examples. In both Case 1.2.1 and Case 1.2.2, [AC]+=1. In those Cases, the [DAG]+ is simply 1+1/CR2, or 1+1/([AC]+/([AB]++[BC]+)), which becomes 1+1/(1/([AB]++[BC]+)), which simplifies to 1+([AB]++[BC]+). When [AC]+=1, the sum of all [DAG]+ fragments, [AC]++[AB]++[BC]+ is 1+([AB]++[BC]+). Then, the value of [MH]+ is obtained by multiplying CR1x[DAG]+, in the form of [MH]+/[DAG]+x(1+([AB]++[BC]+), where (1+([AB]++[BC]+) is the Unit Simulacrum inverse solution 1+1/([AC]+/([AB]++[BC]+)), or 1+1/CR2. To summarize Case 1.2.n, [MH]+=CR1x(1+1/CR2), where 1+1/CR2 is equal to the sum in the denominator of CR1.

When [AC]+ is not equal to 1, for instance Case 1.1.1 and 1.1.2, the sum of all [DAG]+ fragments is not simply 1+1/CR2, but is instead 1+1/CR2 times the Unit Simulacrum solution (1+ratio or 1+1/ratio) of CR3, as follows: (1+CR2)x((1+CR3) or (1+1/CR3)). Specifically, for Case 1.1.1, the [MH]+ is found by multiplying [MH]+/[DAG]+ times [DAG]+, where [DAG]+ is given by (1+([AC]+/([AB]++[BC]+)))x(1+([BC]+/[AB]+)). As discussed above, for Case 1.1.1 [AB]+=1, and the denominator of CR2 is given as the Unit Simulacrum solution 1+CR3. Therefore, the denominator ([AB]++[BC]+) cancels out when multiplied by (1+([BC]+/1), since ([AB]++[BC]+)=(1+([BC]+/1), leaving 1x(1+([BC]+/1))+[AC]+, which equals [AB]++[BC]++[AC]+, or [DAG]+. To summarize Case 1.1.1, [MH]+ is given by CR1x(1+CR2)x(1+CR3). Of course, when [BC]+=1 instead of [AB]+, the solution contains the inverse of CR3, or [MH]+=CR1x(1+CR2)x(1+1/CR3). In both these Cases, the denominator of CR2 is cancelled out by the Unit Simulacrum solution of CR3, either 1+CR3 or 1+1/CR3, and the denominator of CR1 is cancelled out by the Unit Simulacrum solution for CR2 times the Unit Simulacrum solution for CR3, [DAG]+=(1+CR2)x((1+CR3) or (1+1/CR3)).

In the solutions discussed above, the answers depended on which fragment was “carrying the unit”, or which fragment was =1 (100%). When [AC]+ “carried the unit”, Case 1.2.n, the [DAG]+ was found using CR2 by itself, and the [MH]+ was found by multiplying [MH]+/[DAG]+ x [DAG]+, or CR1x(1+1/CR2). When [AB]+ or [BC]+ carried the unit, as in Case 1.1.1 and 1.1.2, the [DAG]+ was found by multiplying (1+CR2)x((1+CR3) or (1+1/CR3)). In all Case 2.n.n solutions, the [MH]+ carries the unit and has a value of 1.


Online Resource1The Unit Simulacrum (US), or First Specified Form of a Simulacrum (FSFS) where one value is specified to be 1, and the other value, B, can represent any value. The default Case 1 has 1  B because this simulacrum came from the FGFS (Fig. 1) having “A” set to 1. A simulacrum is composed of the Simulacrum Sum, Possibilities to Observe, two Cases, and eight solutions


Online Resource 2 The Unit Unit Simulacrum (UUS), where both values are specified to be 1 and the default Case 1 has 1  1. A simulacrum is composed of the Simulacrum Sum, Possibilities to Observe, two Cases, and eight solutions


Online Resource 3 The Unit Identity Simulacrum (UIS), where one value is specified to be 1 and the other value is specified to be 0. The default Case 1 has 0  1. A simulacrum is composed of the Simulacrum Sum, Possibilities to Observe, two Cases, and eight solutions


Online Resource 4 The General Form of the Anti Simulacrum (GFAS), where A and B can represent any values, and the default Case 1 has A  B. A simulacrum is composed of the Simulacrum Sum, Possibilities to Observe, two Cases, and eight solutions


Online Resource 5 The Infinity Unit Simulacrum (IUS), where one value is specified to be 1 and the other value is specified to be . This is Case 2 by default, since . A simulacrum is composed of the Simulacrum Sum, Possibilities to Observe, two Cases, and eight solutions



Online Resource 6a The infinitely nested First Decrement or First Deconstruction, which is a First Increment in the denominator, 1/(1+1). This produces a continued fraction. b 1+ the infinitely nested First Decrement, or the First Increment of the infinitely nested First Decrement, 1+1/(1+1…), which produces another continued fraction. The first gives lesser Fibonacci ratios (<1), the second gives greater Fibonacci ratios (>1). The lesser Fibonacci ratios converge to 1/, while the greater Fibonacci ratios converge to .






Online Resource 7 The First Increment of (1/), which gives the Phi Unit Simulacrum, and the once, twice, and infinitely nested Phi continued fractions. Phi is the Golden Ratio, or Golden Mean,  = 1.618033989… (=(Sqrt(5)+1)/2)