Compare the growth rate of $f(x)$ and $g(x)$ and show how much quicker the quicker of the two grow

Question

Which of the 2 grows quicker? $f(x)$, $g(x)$, or both grow equally as quick?

Definitions

Notation

$a [x] b = a ↑^{x-2} b$

Functions

$f(x) := \begin{cases} 1, & \text{if } x=0\\ f(x-1) [f(x-1)] f(x-1), & \text{if } x<0 \end{cases}$

$g(x) := \begin{cases} 3 [6] 3, & \text{if } x=0\\ 3 [g(x-1)] 3, & \text{if } x<0 \end{cases}$

Note

$g(64)$ is Graham’s number.

My conjecture

$f(x)$ is faster

My reasoning

Both the operands and operator change for every $x$.

Questions

Is my conjecture true or false?

If so, how much quicker?

How do you prove it?

I suppose the $x$'s and $n$'s should be the same in the definitions? — Simply Beautiful Art, Aug 30 '20 at 14:01
@SimplyBeautifulArt Ah yes, they should be. Thanks for catching the mistake. I’ve edited the question. — Aphrontos, Aug 31 '20 at 02:11

score 0 · Accepted Answer · answered Aug 30 '20 at 14:00

We may note the first values of $f$ are

$$f(0)=1$$ $$f(1)=2$$ $$f(2)=4$$ $$f(3)=4[4]4\gg6$$ $$f(4)\gg3[6]3=g(0)$$

so that we easily have by induction

$$f(n+4)\ge g(n)$$

but also $f(3)\ll g(0)$.

Now note that

\begin{align}g(n+1)&=3[g(n)]3\\&=3[g(n)-1](3[g(n)-1]3)\tag1\\&\ge3[g(n)-1](3+g(n))\tag2\\&>(3[g(n)-1]3)[g(n)-1]g(n)\tag3\\&>(g(n)-1)[g(n)-1](g(n)-1)\tag4\end{align}

so that by induction,

$$g(n)-1\ge f(n+3)$$

where the steps are given by

$(1):$ Definition of the hyperoperator.

$(2):$ Using the fact that $3[k-1]3\ge3+k$ for any natural $k$.

$(3):$ Using Knuth's arrow theorem to change the base.

$(4):$ Again reducing with $3[g(n)-1]3\ge3+g(n)-1>g(n)-1$.

Thanks! So, if I understand the proof correctly, f(n) grows more slowly than g(n), for the same n. — Aphrontos, Aug 31 '20 at 02:21