# Which is larger, Graham's number or

1. Sep 15, 2013

### MathJakob

This isn't a homework problem or anything just simply curious. Which is a larger number, Graham's number or the estimated number of atoms in the universe factorial?

So Graham's number or $(10^{80})!$

2. Sep 15, 2013

### Boorglar

Well, as an upper bound, $\left(10^{80}\right)! < \left(10^{80}\right)^{\left(10^{80}\right)} = 10^{80 * 10^{80}} = 10^{8*10^{81}}$

which, if I am not mistaken, is far less than $g_1$.

3. Sep 15, 2013

### MathJakob

So $(10^{80})!$ isn't even as large as g1... and Graham's number is g64... WOW

4. Sep 15, 2013

### arildno

The Arildno number is even greater:

Arildno number =Graham's number+1.

Can I also get a WOW! from you?

5. Sep 15, 2013

### MathJakob

No. Your number is meaningless, it's never been used in any meaningful way.

6. Sep 15, 2013

### arildno

7. Sep 17, 2013

### fedaykin

Sorry to link to wikipedia, but you may be interested in TREE(3), and I can't find too much info on it.

A is the Ackermann function.

http://en.wikipedia.org/wiki/Kruskal's_tree_theorem

8. Sep 17, 2013

### johnqwertyful

Tip: if you can write out the number in any "normal" way, it's waaaaaaaaaaaaaaaaay smaller than Graham's number.

9. Sep 22, 2013

### Deedlit

What's amusing is that the above lower bound for TREE(3) is SO much smaller than TREE(3) that it gives people the wrong idea about it's size. Although the lower bound is larger than Graham's number, it's still at about the level ω+1 in the fast-growing hierarchy. TREE(3), on the other hand, is higher than the Small Veblen Ordinal in the fast-growing hierarchy, which is MUCH larger!

10. Sep 22, 2013

### arildno

It seems the Arildno number is a dwarf, after all..

11. Sep 22, 2013

### Mentallic

Poor thing, the Arildno number is coming under lots of fire in here!

I haven't really been able to find any sources that explain some of the larger operators in the fast growing hierarchy. For example,

http://googology.wikia.com/wiki/Fast-growing_hierarchy

gives a list of growth rates, but there's a lot of details being left out. Sadly, I only understand how to calculate up to and including $f_{\epsilon_0}(n)$.

And to do so, I use a calculator of course

12. Sep 23, 2013

### Deedlit

Good to see you found the Googology Wiki!

Beyond $\varepsilon_0$ we have:

$\varepsilon_1 = \varepsilon_0 ^ {\varepsilon_0 ^ {\varepsilon_0 ^\ddots}}$

$\varepsilon_2 = \varepsilon_1 ^ {\varepsilon_1 ^ {\varepsilon_1 ^\ddots}}$

$\varepsilon_\omega = \sup \lbrace \varepsilon_0, \varepsilon_1, \varepsilon_2, \ldots \rbrace$

and so on. Eventually we get to

$\zeta_0 = \varepsilon_{\varepsilon_{\varepsilon_\cdots}}$

We can set $\varphi(0, \alpha) = \omega^\alpha, \varphi(1, \alpha) = \varepsilon_\alpha, \varphi(2, \alpha) = \zeta_\alpha$. More generally,

$\varphi(\alpha+1, \beta) =$ the $\beta$th fixed point of $f(\gamma) = \varphi(\alpha, \gamma)$.

When $\alpha$ is a limit ordinal, $\varphi(\alpha, \beta)$ is the $\beta$th ordinal in the intersection in the ranges of $f(\delta) = \varphi(\gamma, \delta)$ for all $\gamma < \alpha$.

So, for example,

$\varphi(3, 0) = \varphi(2, \varphi(2, \varphi(2, \ldots)))$

$\varphi(\omega, 0) = \sup (\varphi(1, 0), \varphi (2, 0), \varphi (3, 0), \ldots)$

and so on. This takes us up to

$\Gamma_0 = \varphi(\varphi(\varphi(\ldots, 0),0),0)$

or alternatively, $\Gamma_0$ is the smallest ordinal $\alpha$ such that $\alpha = \varphi(\alpha, 0)$.

We can continue the notation with $\varphi(1, 0, 0) = \Gamma_0$, and $\varphi(1, 0, \alpha)$ is the $\alpha$th fixed point of $f(\beta) = \varphi(\beta, 0)$.

The ordinals $\varphi (1, \alpha, \beta)$ are defined analogously to $\varphi(\alpha, \beta)$, i.e. each function $f(\beta) = \varphi (1, \alpha+1, \beta)$ enumerates the fixed points of $g(\beta) = \varphi(1, \alpha, \beta)$, and at limit ordinals you enumerate the intersection of the ranges of previous ordinals. Then $\varphi(2, 0, \alpha)$ is the $\alpha$th fixed point of $f(\beta) = \varphi(1, \beta, 0)$, and you can construct the hierarchy $\varphi(2, \alpha, \beta)$ similarly. At limit ordinals you take the intersection of the ranges again, and so we define $\varphi (\alpha, \beta, \gamma)$ for all ordinals $\alpha, \beta, \gamma$. Then we can define

$\varphi (1, 0, 0, \alpha)$ to be the $\alpha$th fixed point of $f(\beta) = \varphi(\beta, 0, 0)$.

We can then define $\varphi(\alpha, \beta, \gamma, \delta), \varphi(\alpha, \beta, \gamma, \delta, \epsilon)$, and so on. The general definition for the n-ary Veblen funciton is:

$\varphi (\alpha) = \omega^{\alpha}$

$\varphi (\alpha_1, \alpha_2, \ldots, \alpha_n + 1, 0, \ldots, 0, \beta)$ is the $\beta$th fixed point of the function $f(\gamma) = \varphi(\alpha_1, \alpha_2, \ldots, \alpha_n, \gamma, 0, \ldots, 0)$

When $\alpha_n$ is a limit ordinal, $\varphi (\alpha_1, \alpha_2, \ldots, \alpha_n, 0, \ldots, 0, \beta)$ is the $\beta$th ordinal in the intersection of the ranges of $f(\delta) = \varphi(\alpha_1, \alpha_2, \ldots, \alpha_{n-1}, \gamma, \delta, 0, \ldots, 0)$ for all $\gamma < \alpha_n$.

We define the Small Veblen Ordinal as

$\sup (\varphi (1, 0), \varphi(1, 0, 0), \varphi(1, 0, 0), \ldots)$.

TREE(n) is larger than the fast-growing hierarchy at the level of the Small Veblen Ordinal. (how much further is not known.).

I'll go one step further: we can extend the n-ary Veblen function to transfinitely many places. Obviously, we can't write out transfinitely many variables, so we need to modify our notation: instead of

$\varphi(\alpha, \beta, \gamma, \delta, \epsilon)$

we write

$\varphi(\alpha @ 4, \beta @ 3, \gamma @ 2, \delta @ 1, \epsilon @ 0)$.

So we append "@ n" to every variable, where n represents the index of the variable. This allows us to skip variables that are 0, and so we can notate things like $\varphi (1 @ \omega, \alpha @ 0)$. $\varphi(1 @ \omega, \alpha @ 0)$ is defined as the $\alpha$th ordinal that is a fixed point of $f(\beta) = \varphi (\beta @ n)$ for all $n < \omega$.

More generally, we define

$\varphi(\alpha_1 @ \beta_1, \alpha_2 @ \beta_2, \ldots, \alpha_n + 1 @ \beta_n + 1, \gamma @ 0)$ is the $\gamma$th fixed point of $f(\delta) = \varphi(\alpha_1 @ \beta_1, \alpha_2 @ \beta_2, \ldots, \alpha_n @ \beta_n + 1, \delta @ \beta_n)$

When $\alpha_n$ is a limit ordinal, $\varphi(\alpha_1 @ \beta_1, \alpha_2 @ \beta_2, \ldots, \alpha_n @ \beta_n + 1, \gamma @ 0)$ is the $\gamma$th ordinal in the intersection of the ranges of $f(\epsilon) = \varphi(\alpha_1 @ \beta_1, \alpha_2 @ \beta_2, \ldots, \delta @ \beta_n + 1, \epsilon @ \beta_n)$ for all $\delta < \alpha_n$

When $\beta_n$ is a limit ordinal, $\varphi(\alpha_1 @ \beta_1, \alpha_2 @ \beta_2, \ldots, \alpha_n + 1 @ \beta_n, \gamma @ 0)$ is the $\gamma$th ordinal in the intersection of the ranges of $f(\epsilon) = \varphi(\alpha_1 @ \beta_1, \alpha_2 @ \beta_2, \ldots, \alpha_n @ \beta_n, \epsilon @ \delta)$ for all $\delta < \beta_n$

When $\alpha_n$ and $\beta_n$ are limit ordinals, $\varphi(\alpha_1 @ \beta_1, \alpha_2 @ \beta_2, \ldots, \alpha_n @ \beta_n, \gamma @ 0$ is the $\gamma$th ordinal in the intersection of the ranges of $f(\zeta) = \varphi(\alpha_1 @ \beta_1, \alpha_2 @ \beta_2, \ldots, \delta @ \beta_n, \zeta @ \epsilon)$ for all $\delta < \alpha_n, \epsilon < \beta_n$

This defines $\varphi(\alpha_1 @ \beta_1, \ldots, \alpha_n @ \beta_n)$ for all $\alpha_i$ and $\beta_i$. This notation is known as Schutte's Klammersymbolen.

The smallest ordinal $\alpha$ such that $\alpha = \varphi (1 @ \alpha)$ is known as the Large Veblen Ordinal. I would think that TREE(n) would not reach the Large Veblen Ordinal in the fast-growing hierarchy, but I don't think this is known.

Phew! I hope this was at least somewhat comprehensible.