SBA teaching ordinals

Mr. Xcoder

Nov 7, 2017 14:13

@SimplyBeautifulArt Here I am

I have a very basic understanding of what ordinals are from a vsauce / Numberphile (I don't remember) video.

Simply Beautiful Art

So our first thing is ∅, the empty set.

∅ = {}.

@Mr.Xcoder That's cool.

Mr. Xcoder

Ok, Ik about that

Simply Beautiful Art

Next, we have 0 = {∅}

This is how we define 0.

Mr. Xcoder

Wait. So we define 0 as {∅}

Simply Beautiful Art

Yup

Mr. Xcoder

Nov 7, 2017 14:14

go on.

Simply Beautiful Art

And from here on out, we define x = {y : y<x} (set of all numbers less than x)

For example, 1 = {0}

2 = {0,1}

etc.

Mr. Xcoder

set of all numbers less than x - Maybe non-negative integers.

Simply Beautiful Art

"numbers" refers to "ordinals", which are always greater than or equal to 0.

user202729

Non-negative.

Mr. Xcoder

@SimplyBeautifulArt ok. got it :D

Simply Beautiful Art

Nov 7, 2017 14:16

Now, we can simplify this a bit

A set X is equal to x if x is the smallest number greater than all numbers in X.

Mr. Xcoder

So $\{1,2,3\}=4$

Simply Beautiful Art

x = min{y : y > z, z ∈ X}

@Mr.Xcoder Yup

user202729

@Mr.Xcoder Should include 0.

Simply Beautiful Art

But you can also simplify this down to 4 = {3}

@user202729 Unnecessary.

Mr. Xcoder

cool

Simply Beautiful Art

Nov 7, 2017 14:18

So we can do stuff like 100 = {99}

user202729

@SimplyBeautifulArt Ok, I was thinking about the previous version of your message.

Simply Beautiful Art

And then we have infinite ordinals.

ω = {0, 1, 2, 3, ...}

Mr. Xcoder

... and we need extensions to ordinals? :P

Simply Beautiful Art

ω = smallest ordinal greater than any natural number.

Mr. Xcoder

Yes, I knew that from that video (excuse the pun)

Simply Beautiful Art

Nov 7, 2017 14:20

From there, we can go higher.

ω+1 = {ω}

ω+2 = {ω+1}

Mr. Xcoder

And we have $ω, ω+1, ω+2, ω+2, ω+3,...$

Simply Beautiful Art

ω2 = {ω, ω+1, ω+2, ω+3, ...}

user202729

$ω^2$?

Mr. Xcoder

No $ω_2$ I think.

Simply Beautiful Art

ω^2 = {0, ω, ω2, ω3, ω4, ω5, ...}

@Mr.Xcoder ω_1 and greater are not allowed for our purposes...

Mr. Xcoder

Nov 7, 2017 14:21

@SimplyBeautifulArt Then what is $ω2$?

Simply Beautiful Art

@Mr.Xcoder ω2 = ω+ω

user202729

$⍵ \times 2$.

Simply Beautiful Art

Smallest ordinal greater than ω + any natural number

Mr. Xcoder

Oh

Simply Beautiful Art

@user202729 Dot products please.

:P Small nitpicks

user202729

Nov 7, 2017 14:22

@SimplyBeautifulArt Why?

Simply Beautiful Art

@user202729 Eh, it's just how ordinals are done.

Anyways, we shall now translate these ordinals into finite numbers.

Mr. Xcoder

Oh I got it. I wonder what $w_n$ is though (now that I (mis)mentioned that)

user202729

So ⍵·2? (Funnily Enlist keyboard has both ⍵ and ·)

Simply Beautiful Art

@Mr.Xcoder ω_n is the nth ordinal with no bijection to lower ordinals... I can explain that later, but its not important here.

Mr. Xcoder

Ok. move on

user202729

Nov 7, 2017 14:24

The α-th infinite initial ordinal (from Wikipedia)

Mr. Xcoder

I remember that too

Simply Beautiful Art

H(0,n) = n
H({x},n) = H(x,n+1)
H({x_0, x_1, x_2, x_3, ...},n) = H(x_n,n)

Mr. Xcoder

But loosely

Simply Beautiful Art

These are the rules to the Hardy Hierarchy.

Mr. Xcoder

@SimplyBeautifulArt Wow let me process that :D

Simply Beautiful Art

Nov 7, 2017 14:25

It's not so terrible.

user202729

The x1 is x·1?

Simply Beautiful Art

No

Its a sequence

I clarified it.

user202729

So $x_i$ is a sequence where $i$ is ... natural number? Or ordinal number?

Simply Beautiful Art

i is a natural number.

Mr. Xcoder

@SimplyBeautifulArt $\{x\}$ was equal to $x+1$ wasn't it?

Simply Beautiful Art

Nov 7, 2017 14:26

In any case, we always take x_n, where n is a natural number.

So it wouldn't really matter much.

@Mr.Xcoder Indeed :-)

Mr. Xcoder

Then how is $H(\{x\},n) = H(x,n+1)$. Do we redefine the above?

user202729

The former one is $H(\{x\},n)$.

Simply Beautiful Art

@Mr.Xcoder What do you mean? It is the definition of H.

user202729

Well, probably because of LaTeX strip off {}'s.

Simply Beautiful Art

For example,
H(ω+1,2)
= H({ω},2)
= H(ω,3)
= H({0, 1, 2, 3, ...},3)
= H(3,3)
= H({2},3)
= H(2,4)
= H({1},4)
= H(1,5)
= H({0},5)
= H(0,6)
= 6

Mr. Xcoder

Nov 7, 2017 14:28

I don't understand the transition between $H(\{0, 1, 2, 3, ...\},3) $ and $H(3, 3)$

user202729

Index-of.

By the third rule.

Mr. Xcoder

Oh got it

user202729

Wikipedia has the same definition too. (some people somehow think that Wikipedia is hard to understand)

Mr. Xcoder

@SimplyBeautifulArt Ok.

Simply Beautiful Art

Simple enough?

Mr. Xcoder

Nov 7, 2017 14:29

Yes.

Simply Beautiful Art

Now we can get into some syntax.

{x} = x+1

Let X = {x_0, x_1, x_2, ...}

Let X[n] = x_n.

ω[n] = n

(a+b)[n] = a+(b[n])

(a∙b)[n] = a∙(b[n])

(a^b)[n] = a^(b[n])

user202729

How do you define + and ·? The same as in normal number?

And the ^ is exponentiation?

Simply Beautiful Art

a∙(b+1) = a∙b+a
a^(b+1) = (a^b)∙a

Yes, they are the ordinal forms of addition, multiplication, and exponentiation.

So, now we can simplify our notation a little bit.

H(0,n) = n
H(x+1,n) = H(x,n+1)
H(x,n) = H(x[n],n)

user202729

If $x$ has finite element you use the second definition, and if $x$ has infinite element you use the third definition?

Simply Beautiful Art

@user202729 yes.

Mr. Xcoder

Nov 7, 2017 14:34

I got this too

Simply Beautiful Art

Also, ω[n] = n

Mr. Xcoder

3 mins ago, by Simply Beautiful Art

ω[n] = n

Yes

Simply Beautiful Art

Same example as before using new syntax:
H(ω+1,2)
= H(ω,3)
= H(ω[3],3)
= H(3,3)
= H(2,4)
= H(1,5)
= H(0,6)
= 6

@Mr.Xcoder Oh lol

user202729

Are you using 1-based or 0-based?

Simply Beautiful Art

@user202729 Hm?

user202729

Nov 7, 2017 14:35

Because $⍵$ is defined as $\{0,1,2,3,...\}$

Mr. Xcoder

@user202729 Doesn't matter. $x=x$ regardless of indexing

user202729

Apparently 0-based.

Simply Beautiful Art

Oh, then yes, 0-based.

Now attempt to calculate H(ω^ω,2)

user202729

Wait a minute

You define $x+1$ as $\{x\}$

Simply Beautiful Art

*::60 second counter starts::*

Mr. Xcoder

Nov 7, 2017 14:37

Give us some time. I'll post in a tio link for spoilers.

Simply Beautiful Art

@user202729 yes

Mr. Xcoder

@SimplyBeautifulArt Please make that 300 seconds :D

user202729

so infinite values that is not a successor of any value have set size = infinite, while infinite value that is a successor of other value has set size = 1?

(although that finite set contains other infinte set)

Simply Beautiful Art

@Mr.Xcoder *::adds on 270 seconds::*

@user202729 yeah.

@user202729 and yeah

Mr. Xcoder

Wait I forgot how $ω^ω$ is defined >_>

Simply Beautiful Art

Nov 7, 2017 14:39

(a^b)[n] = a^(b[n]) if b is infinite set.

user202729

4?

Mr. Xcoder

Ok thanks.

Simply Beautiful Art

@user202729 Nope, definitely bigger than H(ω+1,2) = 6.

Mr. Xcoder

Also please spoiler first.

user202729

>! doesn't works here?

Simply Beautiful Art

Nov 7, 2017 14:40

No idea how I should do a spoiler. You want TIO link?

user202729

I don't like TIO either.

Simply Beautiful Art

>! Spoiler test

Mr. Xcoder

@SimplyBeautifulArt Yes.

user202729

So basically the result is base64-encoded and deflated.

Simply Beautiful Art

tio.run/##KypNqvz/3@L/fwA

^ Don't click unless you wanna check ur answer

Mr. Xcoder

Nov 7, 2017 14:42

@SimplyBeautifulArt This is what I have so far.

user202729

Weird spoiler method:

Simply Beautiful Art

Seems good

user202729

thisisaspoilerandthelinkisnotexistent.com/…

(there is no actual spoiler)

Simply Beautiful Art

...

user202729

So result = thisisaspoilerandthelinkisnotexistent.com/8 ?

Simply Beautiful Art

Nov 7, 2017 14:43

Nah, I like TIO better x'D

user202729

But I can't base64 encode and deflate in my head. This one is definitely easier.

Mr. Xcoder

@SimplyBeautifulArt Could you please check again?

user202729

Yay I was correct.

Simply Beautiful Art

@Mr.Xcoder Use the syntactic approach plz.

12 mins ago, by Simply Beautiful Art

a∙(b+1) = a∙b+a
a^(b+1) = (a^b)∙a

Mr. Xcoder

@SimplyBeautifulArt What do you mean?

Simply Beautiful Art

Nov 7, 2017 14:45

Simplify using that.

Then apply the other rules.

user202729

@user202729 There is one drawback: the content must be sufficiently short.

Mr. Xcoder

@SimplyBeautifulArt Better?

Also I'll have to go in 5'.

user202729

Also put = twice is not a good idea.

Simply Beautiful Art

@Mr.Xcoder You should rewrite like this.

user202729

Big dot • and not small dot ·?

Mr. Xcoder

Nov 7, 2017 14:48

Aww but I'll have to go and don't have time to fix. Rushing for now

Simply Beautiful Art

@Mr.Xcoder Feel free to come back anytime.

@user202729 I don't have the small dot on my keyboard.

so...

user202729

Go make another keyboard (layout).

Simply Beautiful Art

@user202729 @Mr.Xcoder by the time I finish, you guys will be blown away by the size of TREE(3).

:-)

Btw, once you get this down, you can approximately encode things like the Ackermann function @Mr.Xcoder

Hello @EriktheOutgolfer

Mr. Xcoder

@SimplyBeautifulArt Here we go \o/

not sure if I used proper notation. gtg o/

Simply Beautiful Art

@Mr.Xcoder I'd prefer this

and cya

@user202729 shall we continue?

And do you happen to know the Ackermann function?

user202729

Nov 7, 2017 14:51

Yes.

Simply Beautiful Art

H(ω^k,n) ≈ Ack(k,n)

So these numbers get very big, very fast.

Graham's number ≈ H(ω^(ω+1),64)

Conway's chained arrow notation goes up to about H(ω^(ω^2),n) I think.

And the Goodstein sequence grows so fast, we can't express them using H(a,n) yet.

To do this, we need a new number.

ε_0 = {1, ω, ω^ω, ω^ω^ω, ω^ω^ω^ω, ...}
(Note that a^b^c = a^(b^c))

Under this, Goodstein(2^2^2^...n times) ≈ H(ε_0,n)

Feel free to try and calculate something like H(ε_0,3)

(too big for me to post final result)

All good so far? @user202729

user202729

Yes.

Simply Beautiful Art

And then we can do nightmarish things like H(ε_0^ε_0,n)

While we're at it, we may as well make
ε_1 = {1, ε_0, ε_0^ε_0, ε_0^ε_0^ε_0, ...}
ε_2 = {1, ε_1, ε_1^ε_1, ε_1^ε_1^ε_1, ...}
etc.

user202729

@SimplyBeautifulArt So I'm trying to calculate this, but there is an issue: How can I detect whether a set is finite or not?

(in other word, should I apply rule 2 or 3?)

Simply Beautiful Art

@user202729 As far as syntax goes, check if there's a '+1' at the end.

ε_ω = {ε_0, ε_1, ε_2, ...}

ε_(ω+1) = {1, ε_ω, ε_ω^ε_ω, ε_ω^ε_ω^ε_ω, ...}

@user202729 Would you like me to post my H(ε_0,3)? Spoiler/no-spoiler.

user202729

Nov 7, 2017 15:06

But you said that it is too large?

Simply Beautiful Art

Yeah. I mean my expansion/work

(skipping a lot of steps)

user202729

I manage to reduce it to like $H(⍵^{⍵^2·2+⍵·2+2}·3,3)$

but it becomes too tedious so I give up.

Simply Beautiful Art

H(ε_0,3)
= H(ω^ω^ω,3)
= H(ω^ω^3,3)
= H(ω^((ω^2)∙3),3)
= H(ω^((ω^2)∙2+ω∙3),3)
= H(ω^((ω^2)∙2+ω∙2+3),3)
= H((ω^((ω^2)∙2+ω∙2+2))∙3,3)
= H((ω^((ω^2)∙2+ω∙2+2))∙2+(ω^((ω^2)∙2+ω∙2+1))∙3,3)
= H((ω^((ω^2)∙2+ω∙2+2))∙2+(ω^((ω^2)∙2+ω∙2+1))∙2+(ω^((ω^2)∙2+ω∙2))∙3,3)
= H((ω^((ω^2)∙2+ω∙2+2))∙2+(ω^((ω^2)∙2+ω∙2+1))∙2+(ω^((ω^2)∙2+ω∙2))∙2+(ω^((ω^2)∙2+ω+3)),3)
= ...

user202729

Just post your solution.

Simply Beautiful Art

Yeah, it gets horribly messy

I mean, its way bigger than Graham's number.

And Graham's number cannot be written in the observable universe.

Anywho, I'll be back later.

Feel free to ponder how large these numbers are and what sorts of nightmares must await if my TREE(3) program involves crazy symbols.

And welcome to Googology :D

(Also, H(ω^x,n) = f_x(n) in the fast growing hierarchy)

user202729

Nov 7, 2017 15:10

Write a program to calculate those things seems easy enough, and yet the result is extremely large. Nice.

Simply Beautiful Art

googology.wikia.com/wiki/list_of_functions

You can also now understand how large some functions are.

Simply Beautiful Art

Nov 7, 2017 18:15

Hey @Mr.Xcoder

Mr. Xcoder

Hello @SimplyBeautifulArt

Simply Beautiful Art

U gud with the above?

Mr. Xcoder

I am reading the above rn.

Simply Beautiful Art

k

Mr. Xcoder

@SimplyBeautifulArt So $ε_0=\{ω\uparrow\uparrow 0, ω\uparrow\uparrow 1, ω\uparrow\uparrow 2, ω\uparrow\uparrow 3, ...\}$?

Simply Beautiful Art

Nov 7, 2017 18:19

@Mr.Xcoder yup

Mr. Xcoder

@SimplyBeautifulArt Ok, I think I got the above, but could you please give me another exercise with the Hardy hierarchy? I want to test my abilities first (also typing dot product is annoying on my keyboard, so do you mind if I use another symbol for that instead?)

Simply Beautiful Art

@Mr.Xcoder sure, I don't mind something like a period or something or implicit multiplication.

Try out H(ε_1,2)

Mr. Xcoder

Ok that's going to take a while

Simply Beautiful Art

Just stop when you feel comfortable.

Mr. Xcoder

ಠ_ಠ $ε_1 = \{1, ε_0, ε_0^{ε_0}, ε_0^{ε_0^{ε_0}}, ...\}$

These numbers are just out of control

:P

Simply Beautiful Art

Nov 7, 2017 18:25

that's the point :P

The goal: Write out everything that's sane, then go beyond. :D

Mr. Xcoder

Ok check the first part to make sure I am not dumb:

H(ε_1,2)=
H(ε_1[2],2)=
H(ω↑↑2,2)

Simply Beautiful Art

Nope

Mr. Xcoder

Oh. Where did I go wrong?

Simply Beautiful Art

3 hours ago, by Simply Beautiful Art

While we're at it, we may as well make
ε_1 = {1, ε_0, ε_0^ε_0, ε_0^ε_0^ε_0, ...}
ε_2 = {1, ε_1, ε_1^ε_1, ε_1^ε_1^ε_1, ...}
etc.

Mr. Xcoder

Ah...

Simply Beautiful Art

Nov 7, 2017 18:27

3 mins ago, by Mr. Xcoder

ಠ_ಠ $ε_1 = \{1, ε_0, ε_0^{ε_0}, ε_0^{ε_0^{ε_0}}, ...\}$

Mr. Xcoder

Yup I thought it was $0$, I am dumb

Simply Beautiful Art

:p

Mr. Xcoder

H(ε_1,2)=
H(ε_1[2],2)=
H(ε_0^ε_0,2)=
H(ε_0^(ε_0[2]),2)=
H(ε_0^(ω↑↑2),2)

@SimplyBeautifulArt Please check again, I am quite dumb as you can see.

Which is actually $H(ε_0^{(ω^ω)},2)$

Right?

Simply Beautiful Art

@Mr.Xcoder Nah, its good

@Mr.Xcoder Yup

Mr. Xcoder

4 hours ago, by Simply Beautiful Art

Now we can get into some syntax.

Posting for easy access ^

@SimplyBeautifulArt I arrived to $H(ε_0^{ω\cdot ω},2)$. What should I do now?

Nvm...

Simply Beautiful Art

Nov 7, 2017 18:37

:P

Mr. Xcoder

@SimplyBeautifulArt Is the result a (relatively) small integer?

Also I need another check (especially the last steps), and I think I overcomplicated this:

H(ε_1,2)=
H(ε_1[2],2)=
H(ε_0^ε_0,2)=
H(ε_0^(ε_0[2]),2)=
H(ε_0^(ω^ω),2)=
H(ε_0^(ω^ω[2]),2)=
H(ε_0^(ω^2),2)=
H(ε_0^(ω*ω),2)=
H(ε_0^(ω*ω[2]),2)=
H(ε_0^(ω+ω),2)=
H(ε_0^(ω+2),2)=
H(ε_0^4,2)=

Simply Beautiful Art

@Mr.Xcoder Definitely not

@Mr.Xcoder Almost.

H(ε_0^(ω+2),2)

ε_0^(ω+2) = a^(b+1)

Mr. Xcoder

So this was right: H(ε_0^(ω+2),2)?

Simply Beautiful Art

Yup

You need to pull out the exponent into multiplication.

Mr. Xcoder

Wait can I just transfer the +2 to the ,2 part?

NVM ^...

Simply Beautiful Art

Nov 7, 2017 18:41

No

Mr. Xcoder

@SimplyBeautifulArt Huh?

Simply Beautiful Art

@Mr.Xcoder a^(b+1) = (a^b)a

Mr. Xcoder

Yeah but... my brain is slow

Simply Beautiful Art

lol

ε_0^(ω+2) = (ε_0^(ω+1))ε_0

Now we can apply (ab)[n] = a(b[n])

Hey @RobertFrost

Mr. Xcoder

Wait I can handle it.

Simply Beautiful Art

Nov 7, 2017 18:44

We be learning 'bout ordinals and how to make stupidly big numbers with them.

Mr. Xcoder

H((ε_0^(ω+1))ε_0,2)=
H((ε_0^(ω+1))ε_0[2],2)=
H((ε_0^(ω+1))ω*ω,2)

@SimplyBeautifulArt ^?

And then that again...

@SimplyBeautifulArt I have this. Won't we just loop like this forever?

Simply Beautiful Art

@Mr.Xcoder ε_0[2] = ω^ω

Sorry, I'm in class, so I may be slow to respond.

And no

It will end eventually

@Mr.Xcoder On each step, the ordinal gets smaller and smaller...

Mr. Xcoder

@SimplyBeautifulArt Where did you pull ε_0[2] from?

Simply Beautiful Art

4 hours ago, by Simply Beautiful Art

ε_0 = {1, ω, ω^ω, ω^ω^ω, ω^ω^ω^ω, ...}
(Note that a^b^c = a^(b^c))

Mr. Xcoder

@SimplyBeautifulArt Yes but where does ε_0[2] appear in our situation?

@SimplyBeautifulArt No problem

Simply Beautiful Art

Nov 7, 2017 18:55

@Mr.Xcoder ((ε_0^(ω+1))ε_0)[2] = (ε_0^(ω+1))(ε_0[2])

Mr. Xcoder

Also I assume we can change the order since multiplication is commutative.

Also I have to say you are a very cool professor

Simply Beautiful Art

@Mr.Xcoder No

Definitely no

Mr. Xcoder

Why?

Simply Beautiful Art

@Mr.Xcoder I'll also have to mention I'm not even 18 yet.

Mr. Xcoder

Still

Simply Beautiful Art

Nov 7, 2017 18:58

@Mr.Xcoder Imagine ω∙2 versus 2∙ω.

Mr. Xcoder

@SimplyBeautifulArt ε_0^(ω+1))(ε_0[2])=ε_0^(ω+1))(ω^ω). And in H(ε_0^(ω+1))(ω^ω),2)=H((ε_0^(ω+1))(ω^ω))[2],2)=H(ε_0^(ω+1))(ω^ω[2]),2)=H(ε_0^(ω‌+1))(ω^2),2) right?

@SimplyBeautifulArt You're like at IMO level :D?

Simply Beautiful Art

ω∙2 = ω+ω = {ω, ω+1, ω+2, ω+3, ...}
2∙ω = {0, 2, 4, 6, 8, ...}

They are entirely different.

Mr. Xcoder

I understand.

Simply Beautiful Art

@Mr.Xcoder :P

@Mr.Xcoder Yup

Mr. Xcoder

@SimplyBeautifulArt I am not joking, I am asking.

Simply Beautiful Art

Nov 7, 2017 19:01

@Mr.Xcoder I'm not actually entirely sure what IMO level is.

Mr. Xcoder

Do you want to/Did you compete in IMO (international mathematics olympiad)?

Simply Beautiful Art

If you mean university level mathematics, sure, but this usually isn't the sort of thing you'd find in the average university.

@Mr.Xcoder No, I don't do olympiads. Those don't really teach you 'new mathematics', which is what I'd like.

Mr. Xcoder

I understand, just being curious.

Simply Beautiful Art

np

Mr. Xcoder

I have:

H((ε_0^(ω+1))ε_0,2)=
H((ε_0^(ω+1))ε_0[2],2)=
H((ε_0^(ω+1))ω^ω,2)=
H((ε_0^(ω+1))ω^2,2)=
H((ε_0^(ω+1))ω^2,2)=
H((ε_0^(ω+1))ω*ω,2)=
H((ε_0^(ω+1))ω*2,2)=

Simply Beautiful Art

Nov 7, 2017 19:04

Looks good

Mr. Xcoder

For now. I have to go eat and will be back in ~5'

Simply Beautiful Art

(using 2 instead of 3 simplifies things so much)

@Mr.Xcoder cya

Mr. Xcoder

@SimplyBeautifulArt Anyway, I am not sure on the steps to follow, so you can leave me a hint while I'm gone.

Simply Beautiful Art

Sure. ω^2 = ω∙ω.

ω∙2 = ω...?

(Think about what x∙2 is in general.)

Mr. Xcoder

Nov 7, 2017 19:17

i am back

@SimplyBeautifulArt Umm, not sure what you mean by that

Oh of course $ω∙2 = ω+ω$

Simply Beautiful Art

:P

Mr. Xcoder

H((ε_0^(ω+1))(ω+ω),2)=
H((ε_0^(ω+1))(ω+ω[2]),2)=
H((ε_0^(ω+1))(ω+2),2)=

Simply Beautiful Art

Yup

Mr. Xcoder

Now what?

@SimplyBeautifulArt TBH I have no idea how to continue.

Simply Beautiful Art

Nov 7, 2017 19:37

Well

(ε_0^(ω+1))(ω+2) = (ε_0^(ω+1))(ω+1) + ε_0^(ω+1)

Mr. Xcoder

Oh well

Simply Beautiful Art

x'D

It sure is a nightmare

Mr. Xcoder

Rght

Simply Beautiful Art

Well

Time to go bigger?

Mr. Xcoder

@SimplyBeautifulArt No it's a daydream.

@SimplyBeautifulArt Yes.

Simply Beautiful Art

Nov 7, 2017 19:40

Well

ε_(x+1) = {1, ε_x, ε_x^ε_x, ε_x^ε_x^ε_x, ...}

ε_x = {ε_(x_0), ε_(x_1), ε_(x_2), ε_(x_3), ...}

Mr. Xcoder

Wait, I am havibg trouble with the internet connection. Ping you when I move back to my mac, currently on mobile

Simply Beautiful Art

@Mr.Xcoder I'm about to go, and I won't be back for about 3.5 hours.

Mr. Xcoder

@SimplyBeautifulArt Ok then, see yiu tomorrow then, I’ll most definitely sleep

Thanks a lot for this wonderful lesson! O/

Simply Beautiful Art

In syntax,
ε_0[n] = ω^^n
ε_(x+1)[n] = (ε_x)^^n
(ε_x)[n] = ε_(x[n])

Mr. Xcoder

Ok

Simply Beautiful Art

Nov 7, 2017 19:43

No problem. Hopefully you can imagine things like ε_(ε_0)

This

Mr. Xcoder

I can't think 4D :D

Simply Beautiful Art

This is still smaller than TREE(3).

;)

@user202729 Oh yeah, FYI, this is all less than TREE(3)

Mr. Xcoder

@SimplyBeautifulArt How do you know? Do you compare them using the fast growing hierarchy?

Simply Beautiful Art

But if you stick to it, I'll explain all the way up to TREE(3) and quite a ways beyond.

Mr. Xcoder

I will definitely stick to it.

Simply Beautiful Art

Nov 7, 2017 19:46

@Mr.Xcoder TREE has an ordinal of approximately ψ(Ω^((Ω^ω)ω))

So... yeah

Cya!

Mr. Xcoder

Ω_Ω Cya!

Robert Frost

Nov 7, 2017 20:03

@SimplyBeautifulArt hi, i only dropped in briefly earlier

Mr. Xcoder

Nov 7, 2017 20:40

@SimplyBeautifulArt For when you come back, can $$H(ε_0^{ω+1}(ω+1)+ε_0^{ω+1},2)$$ be worked on further? If so, how? I tried splitting, all the regular rules but I didn't succeed coming up with a method that simplifies it.

Simply Beautiful Art

Nov 7, 2017 23:06

@Mr.Xcoder You need to use (a+b)[n] = a+(b[n]) and a^(b+1) = (a^b)•a.

Simply Beautiful Art

Nov 7, 2017 23:59

What we've done so far:
H(0,n) = n
H(x+1,n) = H(x,n+1)
H(x,n) = H(x[n],n)
(a+b)[n] = a+(b[n])
(a∙b)[n] = a∙(b[n])
(a^b)[n] = a^(b[n])
a∙(b+1) = a∙b+a
a^(b+1) = (a^b)∙a
ω[n] = n
ε_0[n] = ω^^n
ε_(x+1)[n] = (ε_x)^^n
(ε_x)[n] = ε_(x[n])

2

Further expansions:
H(ε_0^(ω+1)(ω+1)+ε_0^(ω+1),2)
= H(ε_0^(ω+1)(ω+1)+(ε_0^ω)ε_0,2)
= H(ε_0^(ω+1)(ω+1)+(ε_0^ω)(ω^ω),2)
= H(ε_0^(ω+1)(ω+1)+(ε_0^ω)(ω^2),2)
= H(ε_0^(ω+1)(ω+1)+(ε_0^ω)(ω2),2)
= H(ε_0^(ω+1)(ω+1)+(ε_0^ω)(ω+2),2)
= H(ε_0^(ω+1)(ω+1)+(ε_0^ω)(ω+1)+ε_0^2,2)
= H(ε_0^(ω+1)(ω+1)+(ε_0^ω)(ω+1)+(ε_0)(ω^ω),2)
= H(ε_0^(ω+1)(ω+1)+(ε_0^ω)(ω+1)+(ε_0)(ω^2),2)
= H(ε_0^(ω+1)(ω+1)+(ε_0^ω)(ω+1)+(ε_0)(ω2),2)
= H(ε_0^(ω+1)(ω+1)+(ε_0^ω)(ω+1)+(ε_0)(ω+2),2)
= H(ε_0^(ω+1)(ω+1)+(ε_0^ω)(ω+1)+(ε_0)(ω+1)+ω^ω,2)
= ...

$Mathematics$ Simply Beautiful Art's realm of calcu…

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