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## 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

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

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

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

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

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

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

...