User:Alemagno12/Sandbox/fffffff

Brief refresher on the fast growing hierarchy for finite subscripts:

$ f_k(n)=\begin{cases}n+1,&k=0\\f_{k-1}^n(n),&k>0\end{cases} $

where we use function iteration notation. It may easily be derived that

$ f_1(n)=2n $

$ f_2(n)=n2^n $

which are left to the readers as induction exercises.

I then define tetration with a modified top exponent as follows:

$ \operatorname{Tet}(a,b,c)=\underbrace{a\widehat{~}a\widehat{~}\dots\widehat{~}a\widehat{~}}_cb=\begin{cases}b,&c=0\\a\widehat{~}\operatorname{Tet}(a,b,c-1),&c>0\end{cases} $

I shall then prove the following bounds for all $ n>0 $:

$ n\operatorname{Tet}(2^n,1,n)\le f_3(n)\le\operatorname{Tet}(2\sqrt[n]n,n,n) $

We do this by seeing it holds for $ n=1 $ and then applying induction over $ j $ with $ n>1 $ in the following:

$ n\operatorname{Tet}(2^n,1,j)\le f_2^j(n)\le\operatorname{Tet}(2\sqrt[n]n,n,j) $

Trivially it is true for $ j=1 $. Assume it holds for some arbitrary $ k>0 $, and we shall prove it holds for $ k+1 $.

$ \begin{align}n\operatorname{Tet}(2^n,1,k+1)&=n(2\widehat{~}\operatorname{Tet}(2^n,1,k))\\&\le\operatorname{Tet}(2^n,1,j)(2\widehat{~}\operatorname{Tet}(2^n,1,k))\\&=f_2(\operatorname{Tet}(2^n,1,k))\\&\le f_2(f_2^k(n))\\&=f_2^{k+1}(n)\\&=f_2(f_2^k(n))\\&=f_2^k(n)(2\widehat{~} f_2^k(n))\\&=2\widehat{~}(f_2^k(n)+\log_2(f_2^k(n)))\\&=2\widehat{~}\left(f_2^k(n)\left(1+\frac{\log_2(f_2^k(n))}{f_2^k(n)}\right)\right)\\&\le2\widehat{~}\left(f_2^k(n)\left(1+\frac{\log_2(n)}n\right)\right)\\&=(2\sqrt[n]n)\widehat{~}f_2^k(n)\\&\le(2\sqrt[n]n)\widehat{~}\operatorname{Tet}(2\sqrt[n]n,n,k)\\&=\operatorname{Tet}(2\sqrt[n]n,n,k+1)\end{align} $

Q.E.D.