Mathworks (Not the main chat!) - 2017-10-24 (page 2 of 3)

Simply Beautiful Art

20:01

@chronolegends You should really get towards seeing the footnotes on my TREE(3) program though.

Secret

$\Omega[n,\Omega] =\omega, \psi(\omega), \psi^2(\omega), \psi^3(\omega), \psi^4(\omega),...$

$= \omega, \epsilon_{\omega},\epsilon_{\epsilon_{\omega}},\epsilon_{\epsilon_{\epsilon_{\omega}}},\epsilon_{\epsilon_{\epsilon_{\epsilon_{\omega}}}},...$

So: $\psi (\Omega) = \zeta_0$

Simply Beautiful Art

@Secret :P Who need's $\zeta_0$ when you have $\psi(\Omega)$? amirite?

chronolegends

i need it

Secret

It still bothers me that any OCF need the first uncountable ordinal to reach $\zeta_0$

Simply Beautiful Art

@chronolegends need what?

@Secret not any

chronolegends

20:02

z_0

Simply Beautiful Art

@chronolegends Oh.

@Secret The OCF in my program has $\psi(\Omega)>\Gamma_0$

Actually.... lemme calculate $\psi(\Omega)$ in my program real quick...

(brb...)

chronolegends

lol

@Secret it depends on the closure operations allowed

Simply Beautiful Art

In terms of the Veblen function

my program has $\psi(\Omega)=\varphi(1,1,0)$

@chronolegends what he said

chronolegends

psi(W) = psi(1,1,0) is a result of having Gamma_alpha in there ... (?)

Secret

That translates to : $\Gamma_0\text{ something } \epsilon_0$

Simply Beautiful Art

20:06

@chronolegends The second one is a phi.

chronolegends

right

Simply Beautiful Art

$\psi(\alpha\cdot\omega)=\Gamma_\alpha$ for small $\alpha$

So yeah, more or less.

Secret

ah nope. $\varphi (1,1,0)$ is $\Gamma_{\epsilon_0}$ I think...

Simply Beautiful Art

@Secret nope

chronolegends

psi(1,1,0) = limit to T_T_T....

i mean phi

dammit. that hiatus really set me back

Simply Beautiful Art

20:08

$\varphi(1,1,0)=1$-st fixed point of $x=\Gamma_x$

@chronolegends lots of tears were required to reach φ(1,0,0)

Btw, you know you can copy the symbols from the MathJax and it'll give you the unicode characters?

chronolegends

i need to get my ordinal-fu back to its prime

Secret

$\varphi (1,1,0)=\Gamma_{\Gamma_{\Gamma_{\ddots}}}$?

Simply Beautiful Art

lmao

chronolegends

and thanks i didnt know that

Simply Beautiful Art

@Secret yup

chronolegends

20:10

Yeah

Simply Beautiful Art

But back to our functions here

I'll explain mine later.

=P

Secret

I think I started to have some idea on what the second argument is doing: It is needed to vary the argument at the bottom of the nesting. Since we have $\Omega=\omega_1$ that second argument allow us to access the uncountably many fundemental sequence in increasing order to reach higher ordinals before $\Omega$

chronolegends

How do i make it show me the jax code?

Simply Beautiful Art

@chronolegends right click -> show math as -> TeX commands

Or refresh the page

Secret

I.e.
$\Omega[n,\alpha] =\omega, \psi(\alpha), \psi^2(\alpha), \psi^3(\alpha), \psi^4(\alpha),...$

chronolegends

20:12

Cool

Simply Beautiful Art

@Secret Noppie doodles

$\Omega[1,1]=\psi(1[0,1])=\psi(0)\ne\psi(1)$

Get's a little bit tricky

You'll never reach the expression $\Omega[1,1]$, but it was the quickest counterexample that came to mind.

chronolegends

0\ne\1

Simply Beautiful Art

@chronolegends Put dollar signs around it to make math

chronolegends

$0\ne\1$

I see

Simply Beautiful Art

Yeah, also no idea why you're doing a \1

chronolegends

20:15

$0\ne1$

Simply Beautiful Art

:P

Secret

Climbing the $\psi(\Omega)$:

Simply Beautiful Art

@Secret The second term should actually be $\psi(\alpha[0,\alpha])$

And the third term is not always $\psi^2(\alpha)$

In fact, $\psi(\Omega)$ is the only time we'll be able to use iteration notation so nicely.

@chronolegends Say, do you program by any chance?

chronolegends

Not outside of minor golfing

@Secret have you skimmed the wikipedia article on ordinal collapsse?

Simply Beautiful Art

@chronolegends Probably

chronolegends

20:19

Ordinal collapsing function

In mathematical logic and set theory, an ordinal collapsing function (or projection function) is a technique for defining (notations for) certain recursive large countable ordinals, whose principle is to give names to certain ordinals much larger than the one being defined, perhaps even large cardinals (though they can be replaced with recursively large ordinals at the cost of extra technical difficulty), and then “collapse” them down to a system of notations for the sought-after ordinal. For this reason, ordinal collapsing functions are described as an impredicative manner of naming ordinals....

Simply Beautiful Art

But Secret has also found all those technical definitions a bit too much.

Secret

@chronolegends I have, but the non recursive version is so annoying to follow. This one is MUCH BETTER

Simply Beautiful Art

lol

I'll have to agree though, I like this recursive form much more.

Secret

I have two rules in maths life:

1. Higher dimensions and columns makes things easy
2. Operators makes things easy

Simply Beautiful Art

You can actually make an ordinal collapsing function in a program by asking it to repeat a set of operations infinitely and find the smallest ordinal that's not inside of it @chronolegends

chronolegends

20:20

um.

okay

Simply Beautiful Art

Would probably result in a program that never terminates.

If you're to do it like the definitions prescribe

23

Q: Output the simplified Goodstein sequence

A number is in base-b simplified Goodstein form if it is written as b + b + ... + b + c, 0 < c ≤ b The simplified Goodstein sequence of a number starts with writing the number in base-1 simplified Goodstein form, then replacing all 1's with 2's and subtracting 1. Rewrite the result in base-2...

@chronolegends This challenge might be easy enough and still of interest to you

Anyways...

@Secret So can you see that $\psi(\Omega+1)=~^\omega(\psi(\Omega)) =\varepsilon_{\zeta_0+1}$

chronolegends

$\psi(\Omega+\alpha)=~^\omega(\psi(\Omega)) =\varepsilon_{\zeta_0+\alpha}$

um, exclude the middle part

Simply Beautiful Art

:P

19 mins ago, by chronolegends

i need to get my ordinal-fu back to its prime

You really do.

chronolegends

\psi(\Omega+1) =\varepsilon_{\zeta_0+1}

$\psi(\Omega+\alpha) = \varepsilon_{\zeta_0+\alpha}$

ok better

so then we

Simply Beautiful Art

But yeah, $\psi(\Omega+\alpha)=\varepsilon_{\zeta_0+\alpha}$ for reasonably small $\alpha$.

Wonder where @Secret went.

@chronolegends and I totally cut you off! xD Sorry.

chronolegends

20:31

$\psi(\Omega+\psi(\Omega)) =\varepsilon_{\zeta_0+\varepsilon_{\zeta_0}}$

Simply Beautiful Art

@chronolegends D:

Stop stealing my light man

chronolegends

i meant $ \psi(\Omega+\psi(\Omega)) =\varepsilon_{\zeta_0+\varepsilon_{\zeta_0+1}} $

but since limit ordinals additively left cancel from the left, we have

effectively $\psi(\Omega+\psi(\Omega)) =\varepsilon_{\varepsilon_{\zeta_0+1}} $

Secret

\begin{align}
\psi (\Omega)[n,\Omega] & = \psi^{n+1} (\omega) = \zeta_0\\
\psi (\Omega+1)[n,\Omega] & = {}^n\psi(\Omega) = \epsilon_{\zeta_0+1}\\
\psi (\Omega+m)[n,\Omega] & = {}^{n^m}\psi(\Omega) = \epsilon_{\zeta_0+m}\\
\psi (\Omega (m+1))[n,\Omega] & = \psi (\Omega m+ \Omega [n,\Omega]) = \psi (\Omega m + \psi^{n+1}(\omega)) = \psi(\Omega m + \zeta_0) = \text{I need to do it on paper}
\end{align}

chronolegends

you can take the torch @SimplyBeautifulArt to $\zeta_1$ if you want :p

Simply Beautiful Art

@Secret Whoa whoa whoa

chronolegends

20:37

not so fast cowboy :D

Simply Beautiful Art

We don't jump from $\Omega+m$ to $\Omega m$

Dats a bit jump

Back down to $\Omega2$

Or rather, $\Omega+\Omega$

$$\psi(\Omega2)=~?$$

Do this one.

Also, no skipping steps

The intuition from $\psi(\Omega)$ may be your downfall, pressing you to make jumps that are not valid.

chronolegends

holds @Secret 's hand

Secret

$\psi (\Omega 2)[n,\Omega] = \psi (\Omega +\Omega)[n,\Omega] = \psi (\Omega + \Omega [n, \Omega]) = \psi (\Omega + \psi (\Omega [n-1,\Omega]))$

Simply Beautiful Art

@Secret yeup.

R.I.P. because it's not being very nice for iteration notation

Secret

$=\psi (\Omega + \psi^2 (\Omega [n-2,\Omega])) = \psi (\Omega + \psi^3 (\Omega [n-3,\Omega])) = \cdots = \psi (\Omega + \psi^{n+1}(\omega)) = \psi (\Omega + \zeta_0) = ???$

Simply Beautiful Art

20:42

Nope

Oh

x.x you messed up the $z$

You should be having $\Omega2[n,\Omega2]$

Not $\Omega2[n,\Omega]$

Secret

$$\psi (\Omega 2)[n,\Omega 2] = \psi (\Omega +\Omega)[n,\Omega 2] = \psi (\Omega + \Omega [n, \Omega 2]) = \psi ( \Omega + \psi ( \Omega 2 [n-1,\Omega 2])) = \psi (\Omega + \psi (\Omega +\Omega [n-1,\Omega 2])) = \psi (\Omega + \psi (\Omega +\psi (\Omega [n-2,\Omega 2]))) = \cdots = [\psi (\Omega + ]^{n+1} (\omega) = [\psi (\Omega + ]^{n} (\psi (\Omega+\zeta_0)) = ???$$

chronolegends

no, tripped again

$\psi(\Omega2)=~?$

Simply Beautiful Art

@chronolegends what?

@Secret well, you shouldn't be getting any $\zeta_0$ at the end

rest looks fine to me.

So we more or less are looking at:

$$[\psi(\Omega+]^n\omega)$$

What a broken notation lol

chronolegends

$\psi(\Omega2)=\psi(\Omega+\Omega)=\psi(\Omega+0),\psi(\Omega+\psi(\Omega+0)), \psi(\Omega+\psi(\Omega+\psi(\Omega+0))) ...$

Simply Beautiful Art

@chronolegends For some reason I made it $\omega$'s instead of zeroes because that's how my program went.

$$\underbrace{\psi(\Omega +\psi(\Omega +\psi(\Omega +\dots}_n\omega\dots)))$$

But we can agree on this expression more or less

Maybe there's $n+1$, but we get the idea.

Secret

20:53

$$\psi (\Omega 2)[n,\Omega 2] = \psi (\Omega +\Omega)[n,\Omega 2] = \psi (\Omega + \Omega [n, \Omega 2]) = \psi ( \Omega + \psi ( \Omega 2 [n-1,\Omega 2])) = \psi (\Omega + \psi (\Omega +\Omega [n-1,\Omega 2])) = \psi (\Omega + \psi (\Omega +\psi (\Omega 2[n-2,\Omega 2]))) = \cdots = [\psi (\Omega + ]^{n+1} (\omega) = ???$$

Simply Beautiful Art

Now, recall that we said that:

chronolegends

$ = \zeta_0, \epsilon_{\zeta_0+1},\epsilon_{\epsilon_{\zeta_0+1}}, ... = \zeta_1$

Simply Beautiful Art

$$\psi(\Omega+\alpha)=\varepsilon_{\zeta_0+\alpha}$$

@chronolegends jeez man, I'm trying to teach what I wrote, not that normal stuff

chronolegends

:D

brb sorry lol trying to warm up back to a working standard

Simply Beautiful Art

My FS are non-standard FYI @chronolegends

chronolegends

20:55

i know i'm cramping your style in the process, i apologize :p

Simply Beautiful Art

@Secret Anyways, we get:$$\psi(\Omega2) =\varepsilon_{\zeta_0 +\varepsilon_{\zeta_0 +\varepsilon_{\zeta_0 +\varepsilon_{\zeta_0 +\varepsilon_\ddots}}}}$$

Secret

That's even worse then continued fractions

Simply Beautiful Art

X'D

Realize that $\zeta_0$ is smaller than the limit ordinal to it's right.

That is, $\alpha+\beta=\beta$ if $\beta$ is a limit ordinal and $\beta>\alpha$

So it reduces down to$$\psi(\Omega2)=\varepsilon_{ \varepsilon_{ \varepsilon_{ \varepsilon_{ \ddots_{\zeta_0+\omega}}}}}$$

Anyways, you should recognize this as $\zeta_1$.

Secret

I always thought $\zeta_1$ is $\zeta_{\zeta_{\zeta_{\cdots}}}(\zeta_0)$

Simply Beautiful Art

@Secret Uh, that doesn't even make sense?

Secret

21:07

subscript towers of $\zeta_0$

Simply Beautiful Art

$\zeta_{x+1}$ is basically infinitely many $\varepsilon$'s with a $\zeta_x$ at the bottom (but with a +1 or something to avoid collapsing due to fixed-points)

@Secret That's $\eta_0$.

$\varepsilon$ enumerates over exponentiation.
$\zeta$ enumerates over $\varepsilon$'s.
$\eta$ enumerates over $\zeta$'s.

But it all works out much much nicer once you develop a good understanding of OCF's and polynomials of $\Omega$'s.

chronolegends

@Secret $\zeta_{\zeta_...} = \eta_0$

Simply Beautiful Art

@chronolegends >.> I already said that mate.

chronolegends

i was like agreeing with u on that

Simply Beautiful Art

>.>

chronolegends

21:15

btw if there is infinitely many ε's then there can't be a ζx at the bottom :p

since there is no bottom :D

Simply Beautiful Art

:P

@Secret Done for the day, or moving on?

Secret

Tidying stuff in my brain atm

Simply Beautiful Art

Fair warning, its getting to the point where we may want to switch to the Veblen function.

chronolegends

take it easy bro

one ordinal at a time :D

Simply Beautiful Art

@chronolegends Too slow man. Never would've reached $\psi(\Omega)$ at that rate. Sometimes you gotta jump and ask questions afterwards.

chronolegends

21:23

You are the better teacher so i concede you might be right

so should we jump to $\psi(\Omega^2)$

grin

Simply Beautiful Art

@chronolegends >:(

chronolegends

$\psi(\Omega3) = \psi(\Omega2+\Omega) = \psi(\Omega2), \psi(\Omega2+\psi(\Omega2)), \psi(\Omega2+\psi(\Omega2+\psi(\Omega2))) ...$

$= \zeta_1 , \epsilon_{\zeta_1+1}, \epsilon_{..._{\zeta_1+1}} = \zeta_2$

Simply Beautiful Art

@chronolegends Nah

In here, it shall be:
$\psi(\Omega2+\omega),\psi(\Omega2+\psi(\Omega2+\omega)),\dots$

chronolegends

$\psi(\Omega( 1+\alpha)) = \zeta_\alpha$

Simply Beautiful Art

Uh no

Parenthesis sir.

chronolegends

21:37

there ;D

Simply Beautiful Art

=P

chronolegends

ah feels good doing the baby steps again. thanks for inviting me to the chat lol, brb

Simply Beautiful Art

@chronolegends I really like how you can interpret the Veblen function as $\psi(P(\Omega))$, where $P(\Omega)$ is a polynomial of $\Omega$'s with countable coefficients.

Secret

ugh I am trying to rederive $\psi (\epsilon_{\omega})$ using that supremum rule so I can see why it is acting an epsilon

Simply Beautiful Art

@Secret ew, don't write it like dat.

$$\psi(\psi(\omega))$$

No $\Omega$'s in this expression, so it's safe to use [n] instead of [n,z]

\begin{align}\psi(\psi(\omega))[n]&\to \psi(\psi(\omega)[n])\\&\to\psi(\psi(\omega[n]))\\&\to\psi(\psi(n))\end{align}

Secret

21:42

Now how to apply the supremum rule and thus conclude that it is $\epsilon_{\epsilon_{\omega}}$. I am always losing track...

Simply Beautiful Art

Uh...

Well it just follows from $\varepsilon_\alpha=\sup\{\varepsilon_\beta, \beta<\alpha\}$ for all $\alpha\in\Bbb{Lim}$

The rule aligns with the rules of the OCF is all you need to really verify.

chronolegends

@SimplyBeautifulArt true, that principle also applies to extending up to and above hyper inaccessibles

but replacing psi with the appropiate function

Simply Beautiful Art

Yeah ofc

chronolegends

and phi too :D

Simply Beautiful Art

@Secret I like to imagine $\psi$ as multiple different functions. You have the function $\psi(x)$, then $\psi(\Omega+x)$, then $\psi(\Omega2+x)$, ...

Then maybe a generalized $\psi(\Omega y+x)$

And you'll notice something peculiar here....

@Secret How would you describe the nature of what the $\Omega$ is doing so far?

Secret

21:54

It keeps nesting $\zeta_0$ terms at increasingly higher sophistication controlled by $y$

Simply Beautiful Art

@Secret Nope. What does $\Omega$ do in general.

$\psi(\Omega)$ never nested any $\zeta_0$'s, for example.

Secret

Other than to get past $\zeta_0$ I have no idea yet

Simply Beautiful Art

What did $\psi(\Omega)$ expand into?

(in terms of more $\psi$'s, but not reducing to anything else)

What did $\psi(\Omega+\Omega)$ expand into?

What do you think $\psi(\Omega2+\Omega)$ will expand into?

Remember. Only write your expansions in terms of $0,1,2,\dots,\omega,\Omega,\psi,+,\times$, and if necessary, exponentiation.

@chronolegends Say, have you been on Discord?

*::Patiently waits for the messages to stop sliding to the right::*

chronolegends

no i haven't

Simply Beautiful Art

Okay, just checking

chronolegends

22:02

i'm too ashamed by my lack of practice to go back

Simply Beautiful Art

lol

Hm

@chronolegends you know how to write TREE(n)'s fgh approximation in terms of these OCF's?

chronolegends

which one ? :p

Simply Beautiful Art

$ϑ(Ω^ωω)+1$

chronolegends

Hold on, i think i may have a way if i just look it up

Simply Beautiful Art

I'm pretty confident it should translate directly into $\psi$ where that $\vartheta$ is.

But I just wanna check in on it.

chronolegends

22:07

$\theta(\alpha, \beta) = \vartheta (\Omega * \alpha + \beta)$

$\theta (\alpha, \beta) = \psi (\Omega^{\alpha} (1 + \beta))$

so um. it should follow that

Secret

What is $\sup (\epsilon_n +1 | n < \omega)$?

chronolegends

$ \vartheta (\Omega * \alpha + \beta) = \psi (\Omega^{\alpha} (1 + \beta)) $

Simply Beautiful Art

Ah, okay

chronolegends

@Secret $\epsilon_\omega$

Simply Beautiful Art

@Secret $\varepsilon_n<\varepsilon_n+1<\varepsilon_{n+1}$

Secret

22:10

Hmm...

Simply Beautiful Art

What do the upper and lower bounds go to as $n\to\omega$?

Secret

$\omega$

Simply Beautiful Art

Gonna go eat.

cya

Secret

$\psi(\psi(\omega)) = \sup ({}^{\omega}\psi (\psi (\omega)[n,\omega]|n < \omega)) = \sup ({}^{\omega}\psi (\psi (n)|n < \omega)) = .........................$

chronolegends

$\psi(\psi(\omega)) = \epsilon_{\epsilon_\omega} $

Secret

22:13

no way I can churn a $\epsilon_{\epsilon_{\omega}}$ from here... using just the sup

chronolegends

I don't know about @SimplyBeautifulArt 's method but

Secret

How can you show that with the $\psi$ fully unravelled into sups?

chronolegends

We established earlier that for $\alpha < \Omega$ , $\psi(\alpha) = \epsilon_\alpha$

Secret

I am not very certain, I think I don't really understand how to establish that

chronolegends

Alright, in the closure operations we allow + x and ^

and for c_0 we allow 0 , 1 and w (using the wikipedia definition)

so the smallest ordinal we can't write as a finite expression using those is $\epsilon_0$

thus $\psi(0)=\epsilon_0$ , next since $\epsilon_0 has been constructed, it can be used, so the next ordinal we can't write as a finite expression is now

$\epsilon_0^{\epsilon_0^...} = \epsilon_1$

so $\psi(1) = \epsilon_1$

Good so far? :D

Secret

22:20

yup

chronolegends

well, to figure out what $\psi(\omega)$ we consider all the $\epsilon_n$ for $n<\omega$ are constructed, so

$ \epsilon_0^{\epsilon_0^...} = \epsilon_1 $ is constructed, $ \epsilon_1^{\epsilon_1^...} = \epsilon_2 $ is constructed, $ \epsilon_{(1+n)}^{\epsilon_{(1+n)}^...} = \epsilon_n $ is constructed ...

so the smallest ordinal not in the set c(ω) is clearly $\epsilon_\omega$

ergo $\psi(\omega) = \epsilon_\omega$

Simply Beautiful Art

I'm back.

chronolegends

*hides his ordinals beneath the table *

Simply Beautiful Art

@chronolegends You really need to get back into your ordinal-fu

chronolegends

i wasn't doing anything i swear

Simply Beautiful Art

22:28

13 mins ago, by chronolegends

We established earlier that for $\alpha < \Omega$ , $\psi(\alpha) = \epsilon_\alpha$

Pft.

What's this?

chronolegends

lies

Secret

17 mins ago, by Secret

$\psi(\psi(\omega)) = \sup ({}^{\omega}\psi (\psi (\omega)[n,\omega]|n < \omega)) = \sup ({}^{\omega}\psi (\psi (n)|n < \omega)) = .........................$

BSOD

Simply Beautiful Art

You really believe that $\psi(\alpha)$ outputs every epsilon number from $\varepsilon_0$ to $\Omega$?

@Secret :o new acronym.

chronolegends

not up to but below

Secret

Blue Screen of Death

A stop error, better known as a Blue Screen of Death (also known as a blue screen or BSoD) is an error screen displayed on a Windows computer system after a fatal system error, also known as a system crash: when the operating system reaches a condition where it can no longer operate safely. == History == BSoDs have been present in Windows NT 3.1 (the first version of the Windows NT family, released in 1993) and all Windows operating systems released afterwards. (See History of Microsoft Windows.) BSoDs can be caused by poorly written device drivers or malfunctioning hardware, such as faulty memory...

My brain is BSOD in trying to unravel $\psi(\psi(\omega))$

Simply Beautiful Art

22:30

@Secret :o

Secret

I am trying to justify why it is $\epsilon_{\epsilon_{\omega}}$

Simply Beautiful Art

@chronolegends So you believe it outputs all of those ordinals?

@Secret Easier to use induction

chronolegends

Yeah

I hope i was of some help @Secret

Simply Beautiful Art

Suppose that $\psi(y)=\varepsilon_y$ for every $y<x$.

@chronolegends :| You were using C.

Secret finds the C(..) to be a bit confusing.

Hence why I'm using this approach.

Back to the induction

chronolegends

But you haven't introduced me to your inductive ways

Simply Beautiful Art

22:33

Suppose that $\psi(y)=\varepsilon_y$ for every $y<x$.

chronolegends

so i don't know any better :P

Simply Beautiful Art

Suppose that $x=y+1$ for one of those $y$'s.

Prove that $\psi(x)=~^\omega(\psi(y))$ for that $x=y+1$.

Otherwise, $x$ is a limit (and not involving $\Omega$'s).

Then prove that $\psi(x)=\sup\{\psi(y):y<x\}$

chronolegends

me ?

Simply Beautiful Art

(not really a question. These are all from definitions.)

(No not you you silly)

(Hello @Hans)

@Secret If you prove those statements, you'll have proven that $\psi(x)=\varepsilon_x$ for every $x$ not involving $\Omega$'s.

Secret

$\psi (y+1)[n,z] = \sup ({}^{\omega}\psi (y+1[n,z])) = \sup ({}^{\omega}\psi (y))$

$\psi (x) [n,z] = \sup({}^{\omega}\psi(x[n,z])) = \sup ({}^{\omega} \psi (x[n])) = \sup ({}^{\omega} \psi (y_n))$

chronolegends

22:39

@SimplyBeautifulArt regarding your earlier question, i pm'd deedlit for help, in terms of psi it's $\psi(\Omega^{(\Omega^\omega\omega)})$

Simply Beautiful Art

@chronolegends M'kay

Oh, a full nother exponent thingy

chronolegends

Looks much baller doesn't it

Secret

and I guess we get the base case from $\epsilon_0$

chronolegends

join the dark c(...) side

Simply Beautiful Art

@chronolegends :|

Honestly much better over here.

Secret

22:42

Climbing the $\Omega$ (Again!)

chronolegends

yay

Simply Beautiful Art

Well

Secret

Use $\psi (\Omega +1)[n,\Omega +1]$ or $\psi (\Omega +1)[n,\Omega]$?

chronolegends

Can you link me to your recursive psi definition

Simply Beautiful Art

(in terms of more natural numbers, $\omega$, $\Omega$ and $\psi$'s, but not reducing to anything else)
What did $\psi(\Omega)$ expand into?
What did $\psi(\Omega+\Omega)$ expand into?
What do you think $\psi(\Omega2+\Omega)$ will expand into?

@chronolegends Uh, its kinda a mess.

chronolegends

22:45

iight

Simply Beautiful Art

You can kinda find it on the TREE(3) program

chronolegends

I'll give @Secret my cheaty way of dealing with $\Omega$

Simply Beautiful Art

(just replace a(↑b)c with addition, multiplication, and exponentiation)

@chronolegends did you know Deedlit is online?

Secret

Steps problem: $\psi (\Omega +1)[n,\Omega +1] = \psi(\Omega +1[n,\Omega+1])$ or $\psi (\Omega +1)[n,\Omega +1] = {}^{\omega}\psi(\Omega)[n,\Omega+1])$?

chronolegends

I didn't

Simply Beautiful Art

22:48

@Secret $\psi(Ω+(1[n,Ω+1]))$

@chronolegends he's in my realm.

Secret

but that will become $\psi (\Omega+0)$?

Simply Beautiful Art

@Secret $=ψ(Ω)=\zeta_0$, yes.

$\psi(x+1)$ is supposed to reduce straight down to $\psi(x)$, and then you take an infinite tower.

Secret

But that will mean $\psi (\Omega+m) = \psi (\Omega)$?

Simply Beautiful Art

No

chronolegends

@Secret Whenever you see a $*\Omega$ in an expression where * is some operation, make a sequence, the first step is $*0$, the second step is $*first step$, the third step is $*second step$ and so on

Simply Beautiful Art

22:51

You said it yourself, $(m+1)[n,z]=m+(1[n,z])=m$

It just drops to the next lower natural number.

@chronolegends \# and \text{...}

chronolegends

fixied

Secret

ok I see

Simply Beautiful Art

@chronolegends >:( No, the first step is $\omega$

chronolegends

Alright :D

Simply Beautiful Art

Anyways, I'm gonna go now.

We spent a lot of hours talking today lol

Don't you have, like, idk, a life to get back to @Secret ?

chronolegends

22:54

again, thanks for ropeing me back in

:)

Simply Beautiful Art

I'm sick at home more or less, so I have an excuse :P

@chronolegends anytime mate

chronolegends

feels good annoying you on a regular basis

Secret

I can push a bit more, but only a bit more, then I must get back to my chemistry

Also I did not sleep so far...

Simply Beautiful Art

@chronolegends -__-

chronolegends

lol jk @SimplyBeautifulArt i hope you feel better soon , friend :)

Simply Beautiful Art

22:55

:P

chronolegends

@Secret I can help you with psi stuff but, i only know the usual c(...) style

Transcript for

$Mathematics$ Mathworks (Not the main chat!)