User blog comment:Mh314159/Alpha numbers (and beyond)/@comment-35470197-20191007022858/@comment-35470197-20191009102233

Good definitions. They look much clearer. I guess that \(\alpha(x)\) in your comment means \(\alpha[x]\).

For a function \(f(x)\), I denote by \(o(f)\) the ordinal approximately corresponding to \(f\) with respect to FGH. We have \begin{eqnarray*} o(\alpha[0_0,x]) & = & 0 \ o(\alpha[(a+1)_0,x]) & = & o(\alpha[a,x+a+1]) + 2 \\ o(\alpha[a_{b+1},x]) & = & o(\alpha[a_b,x+b+1]) + 2, \end{eqnarray*} and hence \begin{eqnarray*} o(\alpha[0_b,x]) & = & 2b \\ o(\alpha[(a+1)_b,x]) & = & o(\alpha[a,x+a+1]) + 2(1+b). \end{eqnarray*} It implies \begin{eqnarray*} o(\alpha[0,x]) & = & \left( \sup_{b \in \omega} o(\alpha[0_{1+b},x]) \right) + 1 = 2(1 + \omega) + 1 \\ o(\alpha[a+1,x]) & = & \left( \sup_{b \in \omega} o((a+1)_b,x) \right) + 1 & = & o(\alpha[a,x+a+1]) + 2(1+\omega) + 1, \end{eqnarray*} and hence \begin{eqnarray*} o(\alpha[a,x]) & = & 2(1 + \omega)(1+a) + 1. \end{eqnarray*} We obtain \begin{eqnarray*} o(\alpha_0[x]) & = & \left( \sup_{a \in \omega} o(\alpha[a,x]) \right) + 1 = 2(1 + \omega)^2 + 1. \\ o(\alpha_{b+1}[x]) & = & o(\alpha_b[x]) + 2, \end{eqnarray*} and hence \begin{eqnarray*} o(\alpha_b[x]) & = & 2(1 + \omega)^2 + 1 + 2b, \end{eqnarray*} It implies \begin{eqnarray*} o(\alpha[x]) & = & \left( \sup_{b \in \omega} 2(1 + \omega)^2 + 1 + 2b \right) + 1 = 2(1 + \omega)^2 + 1 + 2 \omega + 1 \\ o(\alpha^x[x]) & = & o(\alpha[x]) + 1 = 2(1 + \omega)^2 + 1 + 2 \omega + 2, \end{eqnarray*} and hence the limit of this notation corresponds to \(\omega^2 + \omega + 2\), which is of level 8 with respect to my googological ruler if I am correct.

The point is that the strengthening of the form \(f(x) = g^{f(x-1)}(x)\) contributes to the ordinal as "+2", and the diagonalisation of the form \(g(x) = h_{g(x-1)}(x)\) contributes to the ordinal as "sup +1". In order to get stronger, you need to iterate the latter pattern. For example, even if you repeat the former pattern 100 times, it does not contributes to the ordinal as "+ω".