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Problem of Goodstein function

What's the growth rate of Goodstein function? \(\varepsilon_0\)?

No. Unlike Hydra function and Worm function, which are comparable to \(f_{\varepsilon_0}(n)\), Goodstein function has some "cost" - \(G(2\uparrow\uparrow n)\approx f_{\varepsilon_0}(n)\). By the definition of "functional approximation":

\(f\ge^*g\) if there exists k such that for all n > 0, \(f(n+k)\ge g(n)\)

and

\(f\approx g\) if \(f\ge^*g\) and \(g\ge^*f\)

Goodstein function is not comparable to \(f_{\varepsilon_0}(n)\). Currently we don't have a compact notation to express the growth rate of it. However, in this blog post I won't consider this thing. Instead, I'll consider what it "costs" and extend it. Goodstein has a 3-level cost under \(\varepsilon_0\), since \(G(f_3(n))\approx f_{\varepsilon_0}(n)\); then what about a function with a \(\varepsilon_0\)-level cost under \(\varepsilon_0\)?

An \(\varepsilon_0\)-level cost function

The answer depends on the fundamental sequences.

Here I'll introduce a function with \(\varepsilon_0\)-level cost under \(\varepsilon_0\): \(FH(n)=f_{\sup\{\alpha<\varepsilon_0|H_\alpha(2)\le n+2\}}(2)\), where f and H are fast-growing hierarchy and Hardy hierarchy respectively, and the fundamental sequences are shown here.

Its behavior is quite similar to Goodstein function. We can write it in a similar form: \(G(n)=H_{\sup\{\alpha<\varepsilon_0|g_\alpha(2)\le n\}}(3)-3\), then the \(g_\alpha(2)\le n\) means the "cost". The \(\alpha\) is limited below \(\varepsilon_0\), so the cost is \(g_{\varepsilon_0}(n)\approx f_3(n)\). In FH function, the cost is \(H_{\varepsilon_0}(n)\), which is comparable to \(f_{\varepsilon_0}(n)\). And, \(FH(H_{\varepsilon_0}(n))\approx f_{\varepsilon_0}(n)\).

FH function is actually a fast-growing function. Consider the difference between the "\(\varepsilon_0\)" in \(f_{\varepsilon_0}(n)\) and in \(H_{\varepsilon_0}(n)\). From \(f_\alpha(n)=H_{\omega^\alpha}(n)\), we have \(H_{\varepsilon_0[k+1]}(n)=f_{\varepsilon_0[k]}(n)\). As a result, \(FH(H_{\varepsilon_0[k]}(n))\approx H_{\varepsilon_0[k+1]}(n)\), so \(FH^n(0)\approx H_{\varepsilon_0}(n)\approx f_{\varepsilon_0}(n)\).

This result suggests the FH function to have growth rate \(\varepsilon_0-1\), because \(f_{\varepsilon_0}(n)=f_{\varepsilon_0-1+1}(n)=f_{\varepsilon_0-1}^n(n)\) (though the \(f_{\varepsilon_0-1}\) doesn't really exist) and \(FH^n(n)\approx f_{\varepsilon_0}(n)\) show similar way.

Injection

This section is off-topic.

The first several values of FH function make me uncomfortable. Here are some small values:

\begin{eqnarray*} FH(0) &=& 3 \\ FH(1) &=& 4 \\ FH(2) &=& 8 \\ FH(3) &=& 8 \\ FH(4) &=& f_8(8) \\ FH(5) &=& f_8(8) \\ FH(6) &=& f_{\omega+1}(f_8(8)) \\ FH(7) &=& f_{\omega+1}(f_8(8)) \\ FH(8) &=& f_{\omega+1}(f_8(8)) \\ \vdots & & \vdots \\ FH(3\times2^{402653211}-3) &=& f_{\omega+1}(f_8(8)) \\ FH(3\times2^{402653211}-2) &=& f_{\omega^\omega}(f_{\omega+1}(f_8(8))) \end{eqnarray*}

It's due to the fast growing of HH while the ordinal increases.

This definition, shown below, may make me more comfortable. \(\sigma\) is a function from natural numbers to ordinals below \(\varepsilon_0\): \(\sigma(n,k)=\sup\{\alpha<\varepsilon_0|H_\alpha(k)\le n+2\}\). Then the a-sequence is defined as follows: Set \(a(n)_1=2\); if \(H_{\sigma(n,a(n)_k)}(a(n)_k)<n+2\), let \(a(n)_{k+1}=H_{\sigma(n,a(n)_k)}(a(n)_k)\); if \(H_{\sigma(n,a(n)_k)}(a(n)_k)=n+2\), then the a-sequence have only k terms, and define \(fh(n)=f_{\sigma(n,a(n)_k)}(f_{\sigma(n,a(n)_{k-1})}(\cdots f_{\sigma(n,a(n)_2)}(f_{\sigma(n,a(n)_1)}(2))))\).

fh function still has similar growth rate to FH function. Here are some small values of fh function:

\begin{eqnarray*} fh(0) &=& 3 \\ fh(1) &=& 4 \\ fh(2) &=& 8 \\ fh(3) &=& 16 \\ fh(4) &=& f_8(8) \\ fh(5) &=& 2f_8(8) \\ fh(6) &=& f_{\omega+1}(f_8(8)) \\ fh(7) &=& 2f_{\omega+1}(f_8(8)) \\ fh(8) &=& f_2(f_{\omega+1}(f_8(8))) \\ fh(9) &=& f_3(f_{\omega+1}(f_8(8))) \\ \vdots & & \vdots \\ fh(13) &=& f_7(f_{\omega+1}(f_8(8))) \\ fh(14) &=& f_\omega(f_{\omega+1}(f_8(8))) \\ fh(15) &=& 2f_\omega(f_{\omega+1}(f_8(8))) \\ fh(16) &=& f_{\omega+1}^2(f_8(8)) \\ fh(17) &=& 2f_{\omega+1}^2(f_8(8)) \\ fh(18) &=& f_{\omega+2}(f_{\omega+1}(f_8(8))) \\ fh(19) &=& 2f_{\omega+2}(f_{\omega+1}(f_8(8))) \\ fh(20) &=& f_{\omega+3}(f_{\omega+1}(f_8(8))) \\ \vdots & & \vdots \\ fh(29) &=& 2f_{\omega+7}(f_{\omega+1}(f_8(8))) \\ fh(30) &=& f_{\omega2}(f_{\omega+1}(f_8(8))) \\ fh(62) &=& f_{\omega3}(f_{\omega+1}(f_8(8))) \\ fh(1022) &=& f_{\omega7}(f_{\omega+1}(f_8(8))) \\ fh(2046) &=& f_{\omega^2}(f_{\omega+1}(f_8(8))) \\ fh(3\times2^{402653211}-2) &=& f_{\omega^\omega}(f_{\omega+1}(f_8(8))) \end{eqnarray*}

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