FANDOM


Definition

Base case:

\(f_0(\alpha)=\alpha+1\)


Successor ordinal:

\(f_{\beta+1}(\alpha)=f_{\beta}^\alpha(\alpha)\) for any \(\alpha\) and \(\beta\).


\(f_{\beta}^{\gamma+1}(\alpha)=f_{\beta}(f_{\beta}^{\gamma}(\alpha))\)

\(f_{\beta}^{\gamma+\omega}(\alpha)=\text{lim}\{f_{\beta}^{\gamma}(\alpha),f_{\beta}(f_{\beta}^{\gamma}(\alpha)),f_{\beta}(f_{\beta}(f_{\beta}^{\gamma}(\alpha))),...\}\)

\(f_{\beta}^{\gamma}(\alpha)=\text{lim}_{n\mapsto\omega}f_{\beta}^{\gamma[n]}(\alpha)\)


Limit ordinal:

\(f_{\beta}(\alpha)=f_{\beta[\alpha]}(\alpha)\) for \(\alpha<\omega\)

\(f_{\beta}(\alpha)=\text{lim}_{n\mapsto\omega}f_{\beta[n]}(\alpha)\) for \(\alpha≥\omega\)

Examples&Analysis

\(f_{2}(\omega2) = f_1^{\omega2}(\omega2) = f_1^{\omega}(f_1^{\omega}(\omega2)) = f_1^{\omega}(\omega^2) = \omega^3\)

\(f_{2}(\omega^2) = \omega^\omega\)

\(f_{2}(\omega^3) = \omega^{\omega^2}\)

...

\(f_{3}(\omega) = \varepsilon_0\)

\(f_1^{\omega}f_{3}(\omega) = \varepsilon_0\omega\)

\(f_{2}(f_{3}(\omega)) = \varepsilon_0^2\)

\(f_{3}(\omega2) = f_{2}^\omega(f_{3}(\omega)) = \varepsilon_0^\omega\)

...

\(f_{4}(\omega) = \zeta_0\)

\(f_{2}(f_{4}(\omega)) = \zeta_0^\omega\)

\(f_{2}^\omega(f_{4}(\omega)) = \zeta_0^{\omega^\omega}\)

\(f_{2}^{\omega^2}(f_{4}(\omega)) = \zeta_0^{\omega^{\omega^2}}\)

\(f_{2}^{f_{3}(\omega)}(f_{4}(\omega)) = \zeta_0^{\varepsilon_0}\)

\(f_{3}(f_{4}(\omega)) = \zeta_0^{\zeta_0}\)

\(f_{4}(\omega2) = \varepsilon_{\zeta_0+1}\)

\(f_{5}(\omega2) = \zeta_{\eta_0+1}\)

\(f_{\alpha}(\omega2) = \varphi(\alpha-3,\varphi(\alpha-2,0))\) for \(\alpha<\omega\).


\(f_{\omega}(\omega2)\) =

\(\text{lim}_{n\mapsto\omega}f_{\omega[n]}(\omega2)\) =

\(\text{lim}_{n\mapsto\omega}\{f_{0}(\omega2),f_{1}(\omega2),f_{2}(\omega2),f_{3}(\omega2),...\}\) =

\(\varphi(\omega,\varphi(\omega,0))\)

It turns out that \(f_{\omega}(\omega^2)\) = \(\varphi(\omega+1,0)\), and

\(f_{\omega}(\omega^3)\) = \(\varphi(\omega+2,0)\)

\(f_{\omega}(\omega^\omega)\) = \(\varphi(\omega2,0)\)

\(f_{\omega}(\varepsilon_0)\) = \(\varphi(\varepsilon_0,0)\)

\(f_{\omega+1}(\omega)\) = \(\varphi(1,0,0)\)

\(f_{\omega+1}(\omega2)\) = \(\varphi(1,0,1)\)

\(f_{\omega+1}(\omega^2)\) = \(\varphi(1,0,\omega)\)

\(f_{\omega+2}(\omega)\) = \(\varphi(1,1,0)\)

\(f_{\omega2}(\omega)\) = \(\varphi(1,\omega,0)\)

\(f_{\omega2+1}(\omega)\) = \(\varphi(2,0,0)\)

\(f_{\omega3+1}(\omega)\) = \(\varphi(3,0,0)\)

\(f_{\omega^2}(\omega)\) = \(\varphi(\omega,0,0)\)

\(f_{\omega^2+1}(\omega)\) = \(\varphi(1,0,0,0)\)

\(f_{\omega^\omega}(\omega)\) = \(\vartheta(\Omega^\omega)\)

\(f_{\omega^\omega+1}(\omega)\) = \(\vartheta(\Omega^\Omega)\)

\(f_{\varepsilon_0}(\omega)\) = \(\vartheta(\varepsilon_{\Omega+1})\)

\(f_{\varepsilon_0}(\omega2)\) = \(\vartheta(\varepsilon_{\Omega+2})\)

\(f_{f_{\omega}(\omega)}(\omega)\) = \(\vartheta(\varphi(\omega,\Omega+1))\)

\(f_{f_{\omega}(\omega)+1}(\omega)\) = \(\vartheta(\varphi(\Omega,1))\)

\(f_{f_{\omega}(\omega)+2}(\omega)\) = \(\vartheta(\varphi(\Omega,1)+1)\)

\(f_{f_{\omega}(\omega)2+1}(\omega)\) = \(\vartheta(\varphi(\Omega,1)2)\)

\(f_{f_{\omega}(\omega+1)}(\omega)\) = \(\vartheta(\varphi(\Omega+1,1))\)

\(f_{f_{\omega}(\omega2)}(\omega)\) = \(\vartheta(\varphi(\Omega+\omega,1))\)

\(f_{f_{\omega}(\omega2+1)}(\omega)\) = \(\vartheta(\varphi(\Omega2,1))\)

\(f_{f_{\omega+1}(\omega)}(\omega)\) = \(\vartheta(\Omega_2)\)

\(f_{f_{f_{\omega+1}(\omega)}(\omega)}(\omega)\) = \(\vartheta(\Omega_3)\)

\(f_{f_{f_{f_{\omega+1}(\omega)}(\omega)}(\omega)}(\omega)\) = \(\vartheta(\Omega_4)\)

\(f_{\Omega}(\omega)\) = \(\vartheta(\Omega_\omega)\)

Proof \(f_2^\omega(\omega2) = f_3(\omega)\)

\(f_2^\omega(\omega2) = f_3(\omega)\)

This can be seen as following:

Define the following sequences:

A1 = \(\omega\)

B1 = \(\omega2\)

An = \(f_{2}(\text{A}(n-1))\)

Bn = \(f_{2}(\text{B}(n-1))\)

As An < Bn < A(n+1), their limits are equal.

1st Extension

\(\Omega\) has a higher level than anything expressible using \(0\), \(1\), \(\omega\) and \(f\).

\(\Omega_2\) has a higher level than anything expressible using \(0\), \(1\), \(\omega\), \(\Omega\) and \(f\).

In general, \(\Omega_k\) has a higher level than anything expressible using \(0\), \(1\), \(\omega\), \(\Omega_n\) for any \(n<k\) and \(f\).

Now the definition, if \(\Omega_k\) has the lowest level:

\(f_{\Omega_k+\alpha}(n)[1] = f_{\alpha}(n)\)

\(f_{\Omega_k+\alpha}(n)[s] = f_{\alpha+f_{\Omega_k+\alpha}(\Omega_{k-1})[s-1]}(n)\), where \(\omega\) is the same as \(\Omega_0\).

The limit is \(f_{\alpha \mapsto \Omega_\alpha}(\omega)\), which is equal to \(C(C_I(\omega))\)

2nd Extension

We define \(f_I(\alpha)=\Omega_{\alpha}\).

Fast growing I hierarchy Normal \(\psi_I\) function
\(f_{I+1}(\omega)\) \(\psi_I(0)\)
\(f_{I+1}(\omega2)\) \(\psi_I(1)\)
\(f_{I+1}(\omega^2)\) \(\psi_I(\omega)\)
\(f_{I+1}(\omega^\omega)\) \(\psi_I(\omega^\omega)\)
\(f_{I+2}(\omega)\) \(\psi_I(\Omega)\)
\(f_I^\omega(f_{I+2}(\omega))\) \(\psi_I(\Omega+1)\)
\(f_{I+2}(\omega2)\) \(\psi_I(\Omega2)\)
\(f_{I+2}(\omega^2)\) \(\psi_I(\Omega\omega)\)
\(f_{I+3}(\omega)\) \(\psi_I(\Omega^2)\)
\(f_{I+\omega}(\omega)\) \(\psi_I(\Omega^{\omega})\)
\(f_{I+\omega+1}(\omega)\) \(\psi_I(\Omega^{\Omega})\)
\(f_{I+\Omega}(\omega)\) \(\psi_I(\Omega_\omega)\)
\(f_{I+\Omega2}(\omega)\) \(\psi_I(\Omega_{\Omega_\omega})\)

\(f_{I+\Omega\omega}(\omega)\)

\(\psi_I(\psi_I(0))\)

After that, it gets very quick bigger, until \(f_{I2}(n)\), you can expand it by the following rules:

If I has the lowest level (there are only other I's or powers of I), then:

\(f_{I+\alpha}(n)[1] = f_{\alpha}(n)\)

\(f_{I+\alpha}(n)[k] = f_{\alpha+f_{I+\alpha}(n)[k-1]}(n)\)


\(f_{I2}(n)\) can be written as \(f_{f_1(I)}(n)\),

use your imagination to find the limit of this extension!

Analysis part II

\(f_{f_{I+1}(\omega)}(\omega) = C(C_I(\omega))\)

\(f_{f_{I+\Omega}(\omega)}(\omega) = C(C_I(\omega+1))\)

\(f_{f_{I+\Omega\omega}(\omega)}(\omega) = C(C_I(\omega2))\)

\(f_{f_{I+\Omega\omega^2}(\omega)}(\omega) = C(C_I(\omega^2))\)

\(f_{f_{I2}(\omega)}(\omega) = C(C_I(I))\)

\(f_{f_{f_3(I)}(\omega)}(\omega) = f_{f_{\varepsilon_{I+1}}(\omega)}(\omega) = C(C_I(\varepsilon_{I+1}))\)

...

Ad blocker interference detected!


Wikia is a free-to-use site that makes money from advertising. We have a modified experience for viewers using ad blockers

Wikia is not accessible if you’ve made further modifications. Remove the custom ad blocker rule(s) and the page will load as expected.