User blog:D57799/Ranging FGH before omega

Here I will range FGH before $$f_\omega(n)$$.

The basic definitions:

1. f_0(n)=n+1

2. f_{\alpha+1}(n)={f_\alpha}^n(n)

FGH here is smaller than \(f_\omega(n)\), so the situation of when \(\alpha\) is a ordinal is ignored.

Lower bound
Obviously the lower bound of \(f_m(n)\) is \(2\uparrow^{m-1}n\), so let's prove it.

1. Take \(f_2(n)\) as our starting point,  \(f_2(n)=n2^n> 2\uparrown\). \( n2^n \) grows faster than \(2\uparrown\), the inequality holds.

2. Assume that \(f_m(n)>2\uparrow^{m-1}n\),

\(f_{m+1}(n)=\underbrace{f_m(f_m(...(f_m}_{n}(n))...))\)

\(2\uparrow^mn= \underbrace{ 2\uparrow^{m-1}2\uparrow^{m-1}2...\uparrow^{m-1}2 }{n}\) Because that  \( f_m(n) \) overgrows  \(2\), and that every  \( f_m(n) \) overgrows  \(2 \uparrow^{m-1}n\), we know that  \( f_{m+1}(n) \) overgrows  \(2 \uparrow^mn\).

Thus, for every m > 1 ,    \( f_m(n)> 2 \uparrow^{m-1}n\)

Upper bound
Accurate result proved. Update later.