User blog:B1mb0w/Strong D Function Calculations

The Strong D Function will be calculated in this blog hopefully up to a growth rate of \(f_{\epsilon_0}\). This blog will also refer to from omega to epsilon nought where required to map \(D\) to the ordinal hierarchy index of f.

Strong D Function Overview
From the Strong D Function blog, we can start with these formulas:

Formula and Calculations
Next we use the number of iterations required to get to \(\epsilon_0\) :

\(n\uparrow\uparrow n\) from \(1\) to \(\epsilon_0\) in the f index

Multiplying by another n-1 will calculate the number of iterations when we start from adding 1 to the f exponent for the number of nested functions to 1 and then to \(\epsilon_0\)

We then need to apply \((n-1).n\uparrow\uparrow n\) iterations to the \(D(2,0,x)\) to get this formula.

\(D(2,0,(n-1).n\uparrow\uparrow n) >> f_{\epsilon_0}(n)\) for any reasonable value of n

The uparrow calculations can be replaced by \(f_4(n)\) as follows:

\(f_3(n) >> 2\uparrow\uparrow n\) but if we are conservative then:

\(f_4(n) >> (n-1).n\uparrow\uparrow n\)

\(D(2,0,(n-1).n\uparrow\uparrow n) = D(2,0,f_4(n)) >> f_{\epsilon_0}(n)\) for any reasonable value of n

Refer to mapped example for n=3. A higher lower bound results and in the case n=3 we have:

\(D(2,0,3\uparrow\uparrow 3) >> f_{\epsilon_0}(3)\)

General Rule
The general rule then to compare growth rate of the Strong D Function to \(f_{\epsilon_0}(n)\) is:

\(D(2,0,D(5,n-2)) >> D(2,0,f_4(n)) >> f_{\epsilon_0}(n)\)

Using these formulas for the Strong D Function

\(D(3,0,0) = D(2,D(2,3,3),D(2,3,3)) = D(2,0,D(2,3,3)-1+D(2,3,3)^2/2+D(2,3,3).3/2 + 2/2)\)

\( >> D(2,0,D(2,3,3)^2/2) >> D(2,0,D(2,3,3).100)\)  the constant 100 is arbitrary, but true nonetheless

\(D(2,m,n) = D(2,0,n-1+(m+2).(m+1)/2)\)

\(D(2,3,3) = D(2,0,2-1+2^2/2+2.3/2 + 2/2) = D(2,0,1+2+3+1) = D(2,0,7)\)

then

\(D(3,0,0) >> D(2,0,D(2,0,7).100) >> D(2,0,D(5,n-2)) >> f_{\epsilon_0}(n)\) for any reasonable value of n

because

\(D(2,0,7) >> f_{\omega.2+2}(n) >> D(5,n-2) >> f_4(n)\) for any reasonable value of n

and finally

\(D(3,0,n) >> f_{\epsilon_0}(n)\)