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It is possible to accelerate growth of hierarchy if FGH will be nested in each step of this hierarchy. Let's define

s_0(n)=n+1,

s_{\alpha+1}(n)=FGH_{\beta}(s_{\alpha}(n)),

s_{\alpha}(n)=s_{\alpha[n]}(n) iff \alpha is a limit ordinal.

Let \beta = \omega that means FGH_{\omega}(s_{\alpha}(n)) is equal to f_{\omega}(n) for usual FGH with f_0 (n)=s_{\alpha}(n)

(in other words f_0 (n)=s_{\alpha}(n), f_{\alpha+1} (n)=f_{\alpha}^n (n), f_{\alpha} (n)=f_{\alpha[n]} (n) iff \alpha is a limit ordinal).

In this case s_{\alpha+1}(n)=FGH_{\omega}(s_{\alpha}(n))\approx s_{\alpha}(n)\uparrow^n n

where s_{\alpha}(n)\uparrow b=s_{\alpha}^b(n)=\underbrace{s_{\alpha}(s_{\alpha}(\cdots(s_{\alpha}(s_{\alpha}(n)))\cdots))}_{b \quad s_{\alpha}'s}

and s_{\alpha}(n)\uparrow^{k+1} b=\underbrace{s_{\alpha}(n)\uparrow^{k}(s_{\alpha}(n)\uparrow^{k}(\cdots(s_{\alpha}(n)\uparrow^{k}  n)\cdots))}_{b \quad s_{\alpha}'s}.

Example n=3: then s_0(3)=3+1=4,

s_1(n)=f_{\omega}(n) \approx2\uparrow^{n-1}n and s_1(3)\approx 2 \uparrow \uparrow 3.

For the calculation s_2(3) we must start new FGH (let it be f'_{\alpha} (n) to show difference) with f'_0 (n)=s_{1}(n)\approx2\uparrow^{n-1}n

then f'_{1} (3) \approx
  \left. 
 \begin{matrix} 
    &&\underbrace{2 \uparrow \uparrow \cdots \uparrow \uparrow}3\\
    & &\underbrace{2 \uparrow \uparrow \cdots \uparrow \uparrow} 3\\ 
    & & 2 \uparrow \uparrow 3\\
 \end{matrix} 
\right \} \text {3 layers}\approx\lbrace3,3,1,2\rbrace \approx f_{\omega+1} (3)

f'_{2} (3) \approx \lbrace3,3,2,2\rbrace \approx f_{\omega+2} (3)

and f'_{\omega} (3)=f'_{3} (3) \approx \lbrace3,3,3,2\rbrace \approx f_{\omega+3} (3)

This way s_2(n)=f'_{\omega} (n)=f'_{n} (n) \approx \lbrace n,n,n,2\rbrace \approx f_{\omega.2} (n)

and s_m (n) \approx \lbrace n,n,n,m\rbrace \approx f_{\omega.m} (n)

After this we must overcome the limit ordinal \omega (Indeed FGH_{\omega}(f(x)) can be expressed as huge amount of iterations of some function f(x) )

s_{\omega+1}(3)=FGH_{\omega}(s_{\omega}(3)) \approx s_{\omega}(3) \uparrow^3 3 \approx f_{\omega^2+\omega} (n)

and so on.

So if \beta=\omega then s_{\omega^{\alpha}}(n) \approx f_{\omega^{\alpha+1}}(n)=f_{\omega^{\alpha}.\omega}(n) (comparing with usual FGH with f_0(n)=n+1) and that is why in this case s_{\omega^{\omega}+1}(n) \approx f_{\omega^{\omega}+1}(n). As I remember, it is named "catching ordinal" (\omega^\omega for this case)

To go further we can imagine for example FGH_{\beta}=FGH_F

s_{\alpha+1}(n)=FGH_F(s_{\alpha}(n)) where FGH_F(s_{\alpha}(n))=f_{F}(n) for usual FGH with f_0(n)=s_{\alpha}(n), where F=\theta_{\varepsilon_{\Omega_\omega + 1}}(0) is Takeuti-Feferman-Buchholz ordinal - The supremum of the range of the Feferman theta function,

or even FGH_{\beta}=FGH_R

s_{\alpha+1}(n)=FGH_R(s_{\alpha}(n)) where FGH_R(s_{\alpha}(n))=f_{R}(n) for usual FGH with f_0(n)=s_{\alpha}(n), where R= \psi_{\omega_1}(\chi_{\varepsilon_{\text {M+1}}}(0)) is Rathjen's ordinal - The supremum of the range of the Rathjen's psi function (And, as I know, largest ordinal which is well-defined in professional math).

This way we can reach s_{R}(n). Let s_{R}(10^{100}) is Roogol.

Extensions

Let's define

s'_0(n)=n+1,

s'_{\alpha+1}(n)=SFGH_{R}(s'_{\alpha}(n)),

s'_{\alpha}(n)=s'_{\alpha[n]}(n) iff \alpha is a limit ordinal.

where SFGH_{R}(s'_{\alpha}(n)) is equal to s_{R}(n) for SFGH with s_0 (n)=s'_{\alpha}(n)

Let's name this hierachy as SFGH^1 and notate s'_{\alpha}(n)=s_{\alpha}^{(1)}(n). After this same way we can define s_{\alpha}^{(2)}(n) and s_{\alpha}^{(3)}(n) and so on.

Let m(n)=s_{R}^{(n)}(n) and Roogolplex is equal to m(10^{100})

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