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## Fast Growing Hierarchy Function with $$\omega$$

I am exploring how the FGH functions will behave with an transfinite ordinal e.g. $$\omega$$ as the input parameter.

The results seem to be very interesting. The notation grows at the same rate as some Veblen Functions and it may be possible to extend it to grow at a comparable rate to Ordinal Collapsing Functions.

Also refer to my other Fundamental Sequences blogs for more information on my work.

There are many blogs out there that cover this some material. Here are the ones I am aware of:

## Summary Results

We start by recognising that:

$$f_0(\omega) = \omega + 1$$

Then

$$f_1(\omega) = \omega.2$$

$$f_2(\omega) >> \omega^2$$ because $$f_1^m(\omega) = \omega.2^m$$

$$f_2^2(\omega) >> \omega\uparrow\uparrow 2$$ because $$f_1^{m-2}(f_2(\omega)) >> \omega^m$$

$$f_3^2(\omega) >> \omega\uparrow\uparrow\omega = \varphi(1,0) = \epsilon_0$$ because $$f_2^{m.2 + 2}(\omega) >> \omega\uparrow\uparrow m$$

$$f_3^4(\omega) >> \varphi(1,1)$$

$$f_3^{f_0^2(f_1(m))}(\omega) >> \varphi(1,m)$$

$$f_4^2(\omega) >> f_3^{f_0^2(f_1(f_3^2(\omega))) - \omega}f_4(\omega) = f_3^{f_0^2(f_1(f_3^2(\omega)))}(\omega) >> \varphi(1,\varphi(1,0)) = \varphi^2(1,0_*)$$

$$f_4^3(\omega) >> f_3^{f_0^2(f_1(f_3^{f_1(f_3^2(\omega)))}(\omega))}(\omega) >> \varphi^3(1,0_*)$$

$$f_5(\omega) = f_4^{\omega}(\omega) >> \varphi^{\omega}(1,0_*) = \varphi(2,0) = \zeta_0$$

Without any proof, this notation seems to reach:

$$f_{m.2+1}(\omega) >> \varphi(m,0)$$

## Limitations of this Notation

There seems to be no limitation to using $$\omega$$ into the FGH functions and many other transfinite input ordinals can be used as seen above when the functions are nested.

The notation has a growth rate comparable to $$\varphi(\omega,0)$$

However it does not seem possible to extend these FGH functions to $$f_{\omega}(\gamma)$$ or beyond for any transfinite ordinal $$\gamma$$. This is the limit at which inconsistent results can not be avoided. Here is one example:

Let

$$\gamma = \omega + 1$$

Then

$$f_{\omega}(\gamma) = f_{\omega}(\omega + 1) = f_{\omega + 1}(\omega + 1) = f_{\omega}^{\omega + 1}(\omega + 1)$$

The problem here seems to be a lack of definition on how to diagonalise this step:

$$f_{\omega}(\omega + 1) = f_{\omega + 1}(\omega + 1)$$

It may be necessary to force the subscript of the FGH function to be finite if the input is transfinite. Of course, if the input if finite then the subscript can be transfinite as is the normal case for FGH functions.