Now, this was simply just an idea based on Kapiluka 's. This may have contradiction with the normal FGH, but I will try.
Suppose \(f_0(x)\) means \(x+1\) for all ordinals not just if \(x\in\mathbb{N}\). If so, we can say that
\(f_0(\omega)\) = \(\omega+1\).
Then, if we can continue this to transfinite ordinals with \(f_1(x)\) which would be \(f^{n}_0(x)\) times or \(2x\). You are most likely asking how this works for transfinite numbers. Here is what I propose:
For \(f^{\beta}\)
Iff \(\beta\in\mathbb{N}\): nest \(f\beta\) times.
Iff \(\beta\) is a limit ordinal: find the growth rate when nesting the function \(f\).
Iff \(\beta\) is a successor ordinal and \(\beta\notin\mathbb{N}\): nest \(f\) until you reach a limit ordinal.
If we follow the patterns that we see in normal FGH or using the ones we just described: \(f_2(x)\) = \(x*2^x\). We can get to \(f_3(\omega)\) like this:
\(f_2(f_2(\omega))\)=\(f_2(\omega*2^{(\omega)})\) = \(f_2(\omega*\omega)\) = \(2^{(\omega^2)}*\omega^2\)
= \(\omega^4\).
We can repeat this forever, and we get \(\omega^\omega\).
\(\therefore f_3(\omega) = \omega^\omega\). Let us say we generalize this to all countably transfinite ordinals.