## FANDOM

10,828 Pages

After all of the different opinions on what my notation has as its order in comparison to the fast growing hierarchy (most of the incorrect opinions were mine), I would like to check that the fast growing hierarchy works in the way I now think it does, so I am going to give an example evaluation of $f_{\epsilon_0}(3)$ to check I have the right idea about how it works. If I don't, can you please point out what is wrong so I can finally find out for certain how this thing actually works. Due to how huge some of the expressions will get, I will just focus on the center expression and ignore all of the others generated by iteration, because if I did not, the expression would become huge and unmanageable, although even with this it still does a pretty good job at huge and unmanageable. So, this is how I believe it works:

$f_{\epsilon_0}(3)$

= $f_{\omega^{\omega^{\omega}}}(3)$

= $f_{\omega^{\omega^3}}(3)$

= $f_{\omega^{(\omega^2\cdot3)}}(3)$

= $f_{\omega^{(\omega^2\cdot2)+(\omega^2)}}(3)$

= $f_{\omega^{(\omega^2\cdot2)+(\omega\cdot3)}}(3)$

= $f_{\omega^{(\omega^2\cdot2)+(\omega\cdot2)+(\omega)}}(3)$

= $f_{\omega^{(\omega^2\cdot2)+(\omega\cdot2)+3}}(3)$

= $f_{\omega^{(\omega^2\cdot2+(\omega\cdot2)+2)}\cdot3}(3)$

= $f_{(\omega^{(\omega^2\cdot2+(\omega\cdot2)+2)}\cdot2)+(\omega^{(\omega^2\cdot2+(\omega\cdot2)+2)})}(3)$

= $f_{(\omega^{(\omega^2\cdot2+(\omega\cdot2)+2)}\cdot2)+(\omega^{(\omega^2\cdot2+(\omega\cdot2)+1)}\cdot3)}(3)$

= $f_{(\omega^{(\omega^2\cdot2+(\omega\cdot2)+2)}\cdot2)+(\omega^{(\omega^2\cdot2+(\omega\cdot2)+1)}\cdot2)+(\omega^{(\omega^2\cdot2+(\omega\cdot2))}\cdot3)}(3)$

= $f_{(\omega^{(\omega^2\cdot2+(\omega\cdot2)+2)}\cdot2)+(\omega^{(\omega^2\cdot2+(\omega\cdot2)+1)}\cdot2)+(\omega^{(\omega^2\cdot2+(\omega\cdot2))}\cdot2)+(\omega^{(\omega^2+\omega+3)})}(3)$

= $f_{(\omega^{(\omega^2\cdot2+(\omega\cdot2)+2)}\cdot2)+(\omega^{(\omega^2\cdot2+(\omega\cdot2)+1)}\cdot2)+(\omega^{(\omega^2\cdot2+(\omega\cdot2))}\cdot2)+(\omega^{(\omega^2+\omega+2)}\cdot2)+(\omega^{(\omega^2+\omega+1)}\cdot2)+(\omega^{(\omega^2+3)}\cdot3)}(3)$

This expression is getting extremely complex, so from now on $x_1 = (\omega^{(\omega^2\cdot2+(\omega\cdot2)+2)}\cdot2)+$

$(\omega^{(\omega^2\cdot2+(\omega\cdot2)+1)}\cdot2)+(\omega^{(\omega^2\cdot2+(\omega\cdot2))}\cdot2)+(\omega^{(\omega^2+\omega+2)}\cdot2)+(\omega^{(\omega^2+\omega+1)}\cdot2)$.

Now, continuing to evaluate the expressions:

= $f_{x_1+(\omega^{(\omega^2+3)}\cdot3)}(3)$

= $f_{x_1+(\omega^{(\omega^2+3)}\cdot2)+(\omega^{(\omega^2+2)}\cdot2)+(\omega^{(\omega^2+1)}\cdot2)+(\omega^{(\omega^2)}\cdot3)}(3)$

= $f_{x_1+(\omega^{(\omega^2+3)}\cdot2)+(\omega^{(\omega^2+2)}\cdot2)+(\omega^{(\omega^2+1)}\cdot2)+(\omega^{(\omega\cdot2+3)}\cdot3)}(3)$

= $f_{x_1+(\omega^{(\omega^2+3)}\cdot2)+(\omega^{(\omega^2+2)}\cdot2)+(\omega^{(\omega^2+1)}\cdot2)+(\omega^{(\omega\cdot2+3)}\cdot2)+(\omega^{(\omega\cdot2+2)}\cdot2)+(\omega^{(\omega\cdot2+1)}\cdot2)+(\omega^{(\omega\cdot2)}\cdot3)}(3)$

Again this is getting very complex, so from now on, $x_2 = x_1+(\omega^{(\omega^2+3)}\cdot2)+(\omega^{(\omega^2+2)}\cdot2)+$

$(\omega^{(\omega^2+1)}\cdot2)+(\omega^{(\omega\cdot2+3)}\cdot2)+(\omega^{(\omega\cdot2+2)}\cdot2)+(\omega^{(\omega\cdot2+1)}\cdot2)$.

So continuing with the expressions:

= $f_{x_2+(\omega^{(\omega\cdot2)}\cdot3)}(3)$

= $f_{x_2+(\omega^{(\omega\cdot2)}\cdot2)+(\omega^{(\omega+3)})}(3)$

= $f_{x_2+(\omega^{(\omega\cdot2)}\cdot2)+(\omega^{(\omega+2)}\cdot2)+(\omega^{(\omega+1)}\cdot2)+(\omega^{\omega}\cdot3)}(3)$

= $f_{x_2+(\omega^{(\omega\cdot2)}\cdot2)+(\omega^{(\omega+2)}\cdot2)+(\omega^{(\omega+1)}\cdot2)+(\omega^{\omega}\cdot2)+(\omega^2\cdot2)+(\omega\cdot2)+3}(3)$

This singular 3 on the end will cause the whole function to iterate three times with the 3 decreased to 2. Then the center one of these will iterate again, and the center again. Then very slowly, the expressions with omega in can be evaluated to give (a lot) more iteration, before eventually evaluating down into a huge number of functions with the center one being $f_3(3)$, which can actually be evaluated to a number:

Using the Wainer hierarchy, $f_0(n) = n + 1$, and so $f_1(n) = 2n$, and so $f_2(n) = 2\times2\times2\cdots\times n$ with n 2's, and therefore $f_2(n) = n\cdot2^n$. Using this, we find that $f_2(f_2(f_2(3)))$ = $f_2(f_2(24))$ = $f_2(402653184)$, and this could be calculated. Of course, there would be a huge amount of other very complex functions surrounding this, and it would be incredibly difficult to get any kind of solution not involving ordinals or the FGH. Please leave a comment saying whether or not you think this is correct, and if not, why.