User blog:DrCeasium/fast growing hierarchy example

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. So, this is how I believe it works:

$$f_{\epsilon_0}(3)$$

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

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

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

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

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

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

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

-> (focusing on the center expression) $$f_{\omega^{(\omega^2\cdot2+2)}\cdot2}(3)$$

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

-> (expanding and focusing on the center again as before) $$f_{\omega^{(\omega^2\cdot2+2)}}(3)$$

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

-> (expanding the .2+3 as before and focusing) $$f_{\omega^{(\omega^2\cdot2+1)}}(3)$$

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

-> (again expanding the .2+3 and focusing) $$f_{\omega^{(\omega^2\cdot2)}}(3)$$

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

-> (repeatedly expanding the +3 in the power to give .2+3, and expanding and focusing that) $$f_{\omega^{(\omega^2)}}(3)$$

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

-> (as before, remove all of the non-ordinal numbers, expanding and focusing) $$f_{\omega^{\omega}}(3)$$

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

-> (again, remove all of the non-ordinal numbers by expanding and focusing) $$f_{\omega}(3)$$

= $$f_3(3)$$

= $$f_2(f_2(f_2(3)))$$.

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 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.