Conversation started Mar 10, 2017 at 20:26.
Xam
Xam
Mar 10, 2017 20:26
@SimplyBeautifulArt what does "what is the largest number you could write on a sheet of paper?" mean?
I mean try to write the largest number you can on a sheet of paper. For example, 100000000!!!!!!!!
(Factorial = !)
You may also use any made up function as long as you properly define it
Xam
Xam
Well, I suppose I can take googol!
But that definitely doesn't use the full paper
(Hint: use superultramegahyperoperations)
Lmao, great description @FineMan
Mar 10, 2017 20:30
Or just hyperoperations. Your choice. :)
Xam
Xam
Then I don't know how to make such a large numbers xD
What are "hyperoperations"?
I didn't know until Simply told me yesterday.
Here comes the description... drumroll for Simply!!!
The first hyperoperation is addition
The second is repeated addition, or multiplication
The third is ... Which is exponentiation
The fourth is... Repeated exponentiation
Etc.
Mildly large numbers come out of this
@Xam Do you know ordinals and fundamental sequences?
Xam
Xam
So that is an iteration of addition, multiplication and exponentiation?
Mar 10, 2017 20:34
Yes
it is repeated operations of the previous operation
hence the name hyper-operation
Xam
Xam
Nope, I don't know :(
Well, we can actually produce numbers larger than anything you could feasibly represent in hyperoperators
Xam
Xam
Using ordinals?
Do you have MathJax running @Xam ?
Xam
Xam
Yes
Mar 10, 2017 20:35
Yes, using ordinals. =D
Xam
Xam
Ordinals like $\omega$?
Yes.

Well, we have a function defined as follows:

$f_0(n)=n+1$

$f_{\alpha+1}(n)=\underbrace{f_\alpha(f_\alpha(\dots f_\alpha(n)\dots))}_{n\ f_\alpha's}$

$f_\alpha(n)=f_{\alpha[n]}(n)$
Basically, we define $\omega$ as follows:

$\omega=\sup\{1,2,3,\dots\}$
Then $\omega[n]$ is the nth term of it's defined sequence
For example, $\omega[3]=3$
Now, the fun part. First of all, for whole numbers $k$, we have
$$f_k(n)\approx H_{k+1}(n,n)$$
where $H_k$ is the $k$th hyperoperator with $k=0$ being $a+b$.
We then have, using our $\omega$,
$$f_\omega(n)=f_{\omega[n]}(n)=f_n(n)$$
Still not too bad. You can still approximate this in terms of hyperoperators. But then we have
Xam
Xam
And that function produces large numbers?
$$f_{\omega+1}(n)=\underbrace{f_\omega(f_\omega(\dots f_\omega(n)\dots))}_n$$
@Xam Yes, would you like to see an explicit example?
Hate to interrupt all these dollar signs, but how do you activate MathJax in chat?
Xam
Xam
Mar 10, 2017 20:41
That's something interesting :D
I use that script
@Wojowu Hi, explaining the FGH to some pepes
Xam
Xam
I supposed to be better than Robjohn's script
It's supposed*
A small demonstration:
$$f_{\omega+1}(3)=f_\omega(f_\omega(f_\omega(3)))=f_\omega(f_\omega(f_3(3)))\approx f_\omega(f_\omega(H_4(3)))$$
What browser do you use?
Xam
Xam
Mar 10, 2017 20:43
@SimplyBeautifulArt what is FGH? I can't figure out.
Opera
@Xam Fast growing hierarchy
Xam
Xam
Ah I see xd
Note that $$H_4(3)=3^{3^{3^3}}$$
Xam
Xam
Yeah?
$$\approx10^{3\text{ trillion}}$$
Mar 10, 2017 20:44
OMG! No more dollar signs in chat!! :D
@FineMan I KNOW!!! BEST THING EVER
Xam
Xam
FGH is related to "computability theory, computational complexity theory and proof theory"
Yes, it is very useful
Xam
Xam
Wow, that's something I would like to learn
@SimplyBeautifulArt how did you learn all that?
Now notice that $$f_\omega(f_\omega(3))\approx f_\omega(10^{3\text{ trillion}})$$
YouTube plus Wikipedia. And I didn't learn ALL of that
Xam
Xam
Mar 10, 2017 20:46
@FineMan it's great, right?
$$=f_{10^{3\text{ trillion}}}(10^{3\text{ trillion}})$$
So we saw that $f_3(3)$ was humongous, right?
(@Wojowu Actually, I think I may have done a small calculation error lol, but they get the point)
Xam
Xam
Yes, that's right :D
It's epic. And does this work on other sites?
So This is humongous
Not just SE?
Mar 10, 2017 20:47
@FineMan Yes, it works on any page. Very funny when you go to some economy page with all the dollar signs hehe
LOL.
And we're not even done! This is only $f_\omega(f_\omega(3))$. We want to take the $f_\omega$ of that!
And so this is fairly large, but it gets worse
Xam
Xam
So you can get something like HUMONGOUS, right?
@Xam Do you know Graham's number?
Xam
Xam
No, tell me about it
Mar 10, 2017 20:49
$$f_{\omega+1}(64)>\text{Graham's number}$$
Xam
Xam
Wow xd
I'll leave that story for another day, since we're already way past Graham's number
Consider the following:
$$f_{\omega+2}(3)=f_{\omega+1}(f_{\omega+1}(f_{\omega+1}(3)))$$
Xam
Xam
Ok, no problem
Lol, no problem indeed, because we may as well just go straight to $\omega+k$, right?
Xam
Xam
what happens if we use $2\omega$? Can we?
Mar 10, 2017 20:50
Well, we have $\omega2=\sup\{\omega+1,\omega+2,\omega+3,\dots\}$
Thus,
$$f_{\omega2}(3)=f_{\omega+3}(3)$$
Likewise,
$$f_{\omega3}(3)=f_{\omega2+3}(3)$$
Xam
Xam
What about $\omega^{\omega}$?
etc. you get the main idea
... these $\omega$s are ringing a bell from a vid I watched long ago. When do you get to $\aleph$?
Whoa whoa, hold up, we need to define our numbers!
@FineMan That's a completely different type of number, so it can't compare
I saw a VSauce video on very big numbers. I recall none of it but I do remember there were a lot of greek and hebrew letters. :)
Mar 10, 2017 20:53
First,
$$\omega^2=\sup\{\omega,\omega2,\omega3,\omega4,\dots\}\\2\omega^2=\sup\{\omega^2+\omega,\omega^2+\omega2,\omega^2+\omega3,\dots\}$$
@FineMan Ah, yes... very large numbers. How to count past infinity.
Xam
Xam
Oh, I got it.
In general, if we have, say, $3\omega^5$, we split it up into $2\omega^5+1\omega^5$, then "diagonolize"
I think I'll watch it again one day, but @Xam might want to see it soon: youtube.com/watch?v=SrU9YDoXE88
Diagonolizing is just the process where we write it out and take the nth term for our function's purposes.
@FineMan Xam already knows his ordinals I think
Oh. OK. Whatever ordinals are. :P
Mar 10, 2017 20:55
Anyways, we may finally reach @Xam 's proposal.
$$\omega^\omega=\sup\{\omega,\omega^2,\omega^3,\omega^4,\dots\}$$
Xam
Xam
I read something about ordinals, but I haven't studied it properly.
Poof. I think the universe exploded.
Xam
Xam
@SimplyBeautifulArt lol
But we can go further!
First, I need to define some rules.
Xam
Xam
Yes, do it!
Mar 10, 2017 20:56
$$\omega^{\omega+1}=\omega^\omega\times\omega$$
$$=\sup\{\omega^\omega \times1, \omega^\omega \times2, \omega^\omega \times3,\dots\}$$
Sorry, the MathJax starts to break if I don't put spaces in the coding XD
The key: Always reduce! Then diagonolize!
Now, we can safely do things like this:
$$\omega^{\omega^\omega}$$
Xam
Xam
And so on right?
When we have things in the exponents that don't reduce down to 'ground level', we must diagonolize the exponents first
Xam
Xam
I see
$$\omega^{\omega^\omega}=\omega^{ \sup\{\omega, \omega^2, \omega^3,\dots\}}$$
Xam
Xam
@FineMan youtube.com/watch?v=uWwUpEY4c8o you might like this
Mar 10, 2017 21:00
Clearly, we are going to have a crazy time trying to diagonolize this. Anyways, we move on to the next 'level'
This level when put into the fast growing hierarchy produces functions that grow so fast, you can't do basic PA with it:
$$\epsilon_0=\sup\{\omega,\omega^\omega, \omega^{\omega^\omega},\dots\}$$
And then we have our first fixed-point.
$$\omega^{\epsilon_0}=\epsilon_0$$
Funny point in the storyline, because it's not too hard to get past this.
$$\omega^{\epsilon_0+1}=\omega^{\epsilon_0}\omega=\epsilon_0\omega$$
Xam
Xam
Epsilon zero? nice
We then diagonolize the omega out, which gives us a sum of $\epsilon_0$, which we then diagonolize one by one (and remember folks! Whenever you can, reduce the fast growing hierarchy down by turning it into repeated iterations of itself (see rule 2))
Anyways, we can then produce a higher ordinal!
$$\epsilon_1=\sup\{\epsilon_0+1, \omega^{\epsilon_0+ 1}, \omega^{\omega^{\epsilon_0+ 1}}, \dots\}$$
Xam
Xam
That's amusing!
Anyways, just absorb how absurdly large this number is after you put it into the fast growing hierarchy
Of course, we move on.
$$\epsilon_2=\sup\{\epsilon_1+1, \omega^{\epsilon_1+ 1}, \omega^{\omega^{\epsilon_1+ 1}}, \dots\}$$
We keep doing this until we reach $\epsilon_\omega$,
$$\epsilon_\omega=\sup\{ \epsilon_0, \epsilon_1, \epsilon_2,\dots\}$$
Xam
Xam
Is HUMONGOUS?
Mar 10, 2017 21:07
moral of the story at this step. You diagonolize the inside part first. Then work your way out, which is how we diagonolize something like this:
$$\epsilon_{\epsilon_0}=\epsilon_{\sup\{ \omega, \omega^\omega,\dots \}}$$
Xam
Xam
Interesting.
Diagonolize and diagonolize until you can't, then turn, say, the $\epsilon_{\omega+1}$ into $\omega^{\omega^{\epsilon_\omega+1}}$
@FineMan You good there man? XD
@projectilemotion Hey, glad you could stop by lol
Xam
Xam
And so on, right?
@SimplyBeautifulArt Hey, how are you?
Anyways, we reach a totally higher level now:
$$\zeta_0=\sup\{\epsilon_0, \epsilon_{\epsilon_0},\epsilon_{\epsilon_{\epsilon_0}}, \dots\}$$
@projectilemotion I'm doing good, you?
Xam
Xam
Mar 10, 2017 21:09
Wow
@Xam And so, we end up with a thing known as Veblen Hierarchy
@SimplyBeautifulArt Not bad, thanks
Which keeps this going forever
Xam
Xam
I see
Well, not forever. You kind of stop at $\omega$, because that is like forever, but not, if you get what I mean
And this is where I stop and say Veblen Hierarchy is not enough
Xam
Xam
Mar 10, 2017 21:11
Yeah, I got it
Because you can actually produce ordinals so large, they don't fit inside Veblen hierarchy
I believe, in terms of Mor's ordinal collapsing function, the supremum of all (small) Veblen hierarchy is $\psi(\Omega^\omega)$
Which, as you can see, is some pretty compact notation that conveys an unsettling powerful message
Xam
Xam
That's amazing.
Well, there is an upper bound to $\psi(\alpha)$ though. That is where we introduce $\psi_1(\alpha)$
And you can see where the story goes
 
Conversation ended Mar 10, 2017 at 21:13.