
#1
Sep1513, 12:11 PM

P: 153

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
Sep1513, 01:42 PM

PF Gold
P: 161

Well, as an upper bound, [itex]\left(10^{80}\right)! < \left(10^{80}\right)^{\left(10^{80}\right)} = 10^{80 * 10^{80}} = 10^{8*10^{81}} [/itex]
which, if I am not mistaken, is far less than [itex]g_1[/itex]. 



#3
Sep1513, 01:46 PM

P: 153

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




#4
Sep1513, 01:48 PM

Sci Advisor
HW Helper
PF Gold
P: 12,016

Which is larger, Graham's number or
The Arildno number is even greater:
Arildno number =Graham's number+1. Can I also get a WOW! from you? 



#5
Sep1513, 02:23 PM

P: 153





#6
Sep1513, 02:27 PM

Sci Advisor
HW Helper
PF Gold
P: 12,016





#7
Sep1713, 12:21 PM

P: 124

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%27s_tree_theorem 



#8
Sep1713, 09:48 PM

P: 308





#9
Sep2213, 01:43 PM

P: 13

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 fastgrowing hierarchy. TREE(3), on the other hand, is higher than the Small Veblen Ordinal in the fastgrowing hierarchy, which is MUCH larger!




#10
Sep2213, 01:57 PM

Sci Advisor
HW Helper
PF Gold
P: 12,016





#11
Sep2213, 07:48 PM

HW Helper
P: 3,436

http://googology.wikia.com/wiki/Fastgrowing_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 [itex]f_{\epsilon_0}(n)[/itex]. And to do so, I use a calculator of course 



#12
Sep2313, 08:58 PM

P: 13

Good to see you found the Googology Wiki!
Beyond [itex]\varepsilon_0[/itex] we have: [itex]\varepsilon_1 = \varepsilon_0 ^ {\varepsilon_0 ^ {\varepsilon_0 ^\ddots}}[/itex] [itex]\varepsilon_2 = \varepsilon_1 ^ {\varepsilon_1 ^ {\varepsilon_1 ^\ddots}}[/itex] [itex]\varepsilon_\omega = \sup \lbrace \varepsilon_0, \varepsilon_1, \varepsilon_2, \ldots \rbrace [/itex] and so on. Eventually we get to [itex]\zeta_0 = \varepsilon_{\varepsilon_{\varepsilon_\cdots}}[/itex] We can set [itex]\varphi(0, \alpha) = \omega^\alpha, \varphi(1, \alpha) = \varepsilon_\alpha, \varphi(2, \alpha) = \zeta_\alpha[/itex]. More generally, [itex]\varphi(\alpha+1, \beta) = [/itex] the [itex]\beta[/itex]th fixed point of [itex]f(\gamma) = \varphi(\alpha, \gamma)[/itex]. When [itex]\alpha[/itex] is a limit ordinal, [itex]\varphi(\alpha, \beta)[/itex] is the [itex]\beta[/itex]th ordinal in the intersection in the ranges of [itex]f(\delta) = \varphi(\gamma, \delta)[/itex] for all [itex]\gamma < \alpha[/itex]. So, for example, [itex]\varphi(3, 0) = \varphi(2, \varphi(2, \varphi(2, \ldots)))[/itex] [itex]\varphi(\omega, 0) = \sup (\varphi(1, 0), \varphi (2, 0), \varphi (3, 0), \ldots)[/itex] and so on. This takes us up to [itex]\Gamma_0 = \varphi(\varphi(\varphi(\ldots, 0),0),0)[/itex] or alternatively, [itex]\Gamma_0[/itex] is the smallest ordinal [itex]\alpha[/itex] such that [itex]\alpha = \varphi(\alpha, 0)[/itex]. We can continue the notation with [itex]\varphi(1, 0, 0) = \Gamma_0[/itex], and [itex]\varphi(1, 0, \alpha)[/itex] is the [itex]\alpha[/itex]th fixed point of [itex]f(\beta) = \varphi(\beta, 0)[/itex]. The ordinals [itex]\varphi (1, \alpha, \beta)[/itex] are defined analogously to [itex]\varphi(\alpha, \beta)[/itex], i.e. each function [itex]f(\beta) = \varphi (1, \alpha+1, \beta)[/itex] enumerates the fixed points of [itex]g(\beta) = \varphi(1, \alpha, \beta)[/itex], and at limit ordinals you enumerate the intersection of the ranges of previous ordinals. Then [itex]\varphi(2, 0, \alpha)[/itex] is the [itex]\alpha[/itex]th fixed point of [itex]f(\beta) = \varphi(1, \beta, 0)[/itex], and you can construct the hierarchy [itex]\varphi(2, \alpha, \beta)[/itex] similarly. At limit ordinals you take the intersection of the ranges again, and so we define [itex]\varphi (\alpha, \beta, \gamma)[/itex] for all ordinals [itex]\alpha, \beta, \gamma[/itex]. Then we can define [itex]\varphi (1, 0, 0, \alpha)[/itex] to be the [itex]\alpha[/itex]th fixed point of [itex]f(\beta) = \varphi(\beta, 0, 0)[/itex]. We can then define [itex]\varphi(\alpha, \beta, \gamma, \delta), \varphi(\alpha, \beta, \gamma, \delta, \epsilon)[/itex], and so on. The general definition for the nary Veblen funciton is: [itex]\varphi (\alpha) = \omega^{\alpha}[/itex] [itex]\varphi (\alpha_1, \alpha_2, \ldots, \alpha_n + 1, 0, \ldots, 0, \beta)[/itex] is the [itex]\beta[/itex]th fixed point of the function [itex]f(\gamma) = \varphi(\alpha_1, \alpha_2, \ldots, \alpha_n, \gamma, 0, \ldots, 0)[/itex] When [itex]\alpha_n[/itex] is a limit ordinal, [itex]\varphi (\alpha_1, \alpha_2, \ldots, \alpha_n, 0, \ldots, 0, \beta)[/itex] is the [itex]\beta[/itex]th ordinal in the intersection of the ranges of [itex]f(\delta) = \varphi(\alpha_1, \alpha_2, \ldots, \alpha_{n1}, \gamma, \delta, 0, \ldots, 0)[/itex] for all [itex]\gamma < \alpha_n[/itex]. We define the Small Veblen Ordinal as [itex]\sup (\varphi (1, 0), \varphi(1, 0, 0), \varphi(1, 0, 0), \ldots)[/itex]. TREE(n) is larger than the fastgrowing 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 nary Veblen function to transfinitely many places. Obviously, we can't write out transfinitely many variables, so we need to modify our notation: instead of [itex]\varphi(\alpha, \beta, \gamma, \delta, \epsilon)[/itex] we write [itex]\varphi(\alpha @ 4, \beta @ 3, \gamma @ 2, \delta @ 1, \epsilon @ 0)[/itex]. 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 [itex]\varphi (1 @ \omega, \alpha @ 0)[/itex]. [itex]\varphi(1 @ \omega, \alpha @ 0)[/itex] is defined as the [itex]\alpha[/itex]th ordinal that is a fixed point of [itex]f(\beta) = \varphi (\beta @ n)[/itex] for all [itex]n < \omega[/itex]. More generally, we define [itex]\varphi(\alpha_1 @ \beta_1, \alpha_2 @ \beta_2, \ldots, \alpha_n + 1 @ \beta_n + 1, \gamma @ 0)[/itex] is the [itex]\gamma[/itex]th fixed point of [itex]f(\delta) = \varphi(\alpha_1 @ \beta_1, \alpha_2 @ \beta_2, \ldots, \alpha_n @ \beta_n + 1, \delta @ \beta_n)[/itex] When [itex]\alpha_n[/itex] is a limit ordinal, [itex]\varphi(\alpha_1 @ \beta_1, \alpha_2 @ \beta_2, \ldots, \alpha_n @ \beta_n + 1, \gamma @ 0)[/itex] is the [itex]\gamma[/itex]th ordinal in the intersection of the ranges of [itex]f(\epsilon) = \varphi(\alpha_1 @ \beta_1, \alpha_2 @ \beta_2, \ldots, \delta @ \beta_n + 1, \epsilon @ \beta_n)[/itex] for all [itex]\delta < \alpha_n[/itex] When [itex]\beta_n[/itex] is a limit ordinal, [itex]\varphi(\alpha_1 @ \beta_1, \alpha_2 @ \beta_2, \ldots, \alpha_n + 1 @ \beta_n, \gamma @ 0)[/itex] is the [itex]\gamma[/itex]th ordinal in the intersection of the ranges of [itex]f(\epsilon) = \varphi(\alpha_1 @ \beta_1, \alpha_2 @ \beta_2, \ldots, \alpha_n @ \beta_n, \epsilon @ \delta)[/itex] for all [itex]\delta < \beta_n[/itex] When [itex]\alpha_n[/itex] and [itex]\beta_n[/itex] are limit ordinals, [itex]\varphi(\alpha_1 @ \beta_1, \alpha_2 @ \beta_2, \ldots, \alpha_n @ \beta_n, \gamma @ 0[/itex] is the [itex]\gamma[/itex]th ordinal in the intersection of the ranges of [itex]f(\zeta) = \varphi(\alpha_1 @ \beta_1, \alpha_2 @ \beta_2, \ldots, \delta @ \beta_n, \zeta @ \epsilon)[/itex] for all [itex]\delta < \alpha_n, \epsilon < \beta_n[/itex] This defines [itex]\varphi(\alpha_1 @ \beta_1, \ldots, \alpha_n @ \beta_n)[/itex] for all [itex]\alpha_i[/itex] and [itex]\beta_i[/itex]. This notation is known as Schutte's Klammersymbolen. The smallest ordinal [itex]\alpha[/itex] such that [itex]\alpha = \varphi (1 @ \alpha)[/itex] is known as the Large Veblen Ordinal. I would think that TREE(n) would not reach the Large Veblen Ordinal in the fastgrowing hierarchy, but I don't think this is known. Phew! I hope this was at least somewhat comprehensible. 


Register to reply 
Related Discussions  
determine if the next number will be larger or smaller  General Math  2  
How to represent a small change in a larger number  General Math  1  
Graham's number  Linear & Abstract Algebra  5  
Why is it when I divide by a number greater than 1, I get a larger number?  General Math  2  
Graham's Number compared to... anything?  General Math  16 