Alpha Function and \(f_{\epsilon_0}(m)\)

This blog will compare the growth rate of the Alpha Function to the fast growing hierarchy function \(f_{\epsilon_0}(n)\). Starting with the following calculations at this link:

\(D(m,m,0,...,0) >> f_{\omega^2}(m)\) with approximately \(m\) zeros.


\(D(m,m,0,...,0,...,0,...,0) >> f_{\omega^2+1}(m)\) with approximately \(2^m\) zeros.


The following Alpha Functions are approximately equal to:

\(\alpha(2^{m+1}) >> D(m,m,0,...,0) >> f_{\omega^2}(m) = \alpha(2^m)\)

Without proof, the following is proposed:

\(\alpha(2^{2^{m+1}}) >> f_{\omega^2+1}(m)\)

Alternatively the following will be approximately correct:

\(\alpha(D(3,m)) = \alpha(2^{m+1}) >> D(m,m,0,...,0) >> f_{\omega^2}(m) = \alpha(2^m)\)


\(\alpha(D(4,m)) = \alpha(2^{2^{m+1}}) >> f_{\omega^2+1}(m)\)

then, without proof, the following is claimed:

\(\alpha(D(3+m,m)) >> f_{\omega^2+m}(m) >> f_{\omega^2+\omega}(m)\)

\(\alpha(D(1,m,0)) >> f_{\omega^2+\omega+1}(m)\)

\(\alpha(D(m,m,0)) >> f_{\omega^2+\omega.2}(m)\)

\(\alpha(D(1,m,0,0)) >> f_{\omega^2+\omega.2+1}(m)\)

\(\alpha(D(m,m,0,0)) >> f_{\omega^2+\omega.3}(m)\)

\(\alpha(\alpha(2^5.m)) = \alpha(D(1,m,0,0,0)) >> f_{\omega^2+\omega.3+1}(m)\)

\(\alpha(\alpha(2^{4+m})) >> f_{\omega^2+\omega.m}(m) = f_{\omega^2.2}(m)\)


Continuing without proof, it is claimed:

\(\alpha(\alpha(D(3,4+m))) = \alpha(\alpha(2^{4+m})) >> f_{\omega^2.2}(m)\)

\(\alpha(\alpha(D(3+m,m))) >> f_{\omega^2.2 + \omega}(m)\)

\(\alpha(\alpha(D(m,m,0))) >> f_{\omega^2.2 + \omega.2}(m)\)

\(\alpha(\alpha(D(m,m,0,0))) >> f_{\omega^2.2 + \omega.3}(m)\)

\(\alpha^3(2^m) = \alpha(\alpha(\alpha(2^m))) >> f_{\omega^2.3}(m)\)

\(\alpha^m(2^m) >> f_{\omega^3}(m)\)

\(\alpha^{m^{m-3}}(2^m) >> f_{\omega^{\omega}}(m)\)

\(\alpha^{m^{(m-3)^m}}(2^m) >> f_{\omega^{\omega^{\omega}}}(m)\)


Finally without proof, it is proposed:

\(\alpha^{m\uparrow\uparrow m}(2^m) >> f_{\epsilon_0}(m)\)

Questions and Comments

As always, any feedback on the assumptions made on this page, will be greatly appreciated.


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