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2:01 AM
@SimplyBeautifulArt exponential is a can of worms
is it?
I mean, (omega+1)^2 = omega^2+omega+1
 
lol
Yeah don't worry about it
 
(omega+1)^3 = omega^3+omega+1
I'm not even sure if this is right
 
If you do it naturally as (ω+1)∙(ω+1) -> (ω+1)∙ω + (ω+1) = ...
Then it comes out naturally to the right result
So don't worry about it too much
Hey @WheatWizard
 
Hello
 
@WheatWizard what brings you to my realm?
 
2:08 AM
I saw it in the sidebar so I thought I would take a look.
 
It looks like you do more Differential Calculus than Ananlysis here.
 
Lol
It's what the people bring
@WheatWizard you interested in large finite numbers?
 
I really don't know much about them, but I am interested
 
And I think we were more along the lines of numerical analysis

 Ordinality?

Trying to understand extraordinarily large numbers.
You may be interested in that chat room @LeakyNun @WheatWizard
 
2:14 AM
@SimplyBeautifulArt yes, but how do i do it for (omega+1)^omega?
 
@LeakyNun It just becomes (ω+1)^n
You can use w=ω
 
no, I want to be able to write it in normal form
 
Normal form is ω^ω
 
but how do I figure that out?
 
The limit of (ω+1)^n
 
2:16 AM
I know, but my computer doesn't know
 
So?
It knows what (ω+1)^n is
 
we're going in circles
 
My point is that (ω+1)^n is close enough
Teaching it Cantor normal form is only going to drive you crazy
 
it's supposed to be decidable
 
Ofc it is
But it's so much easier to go with the flow and the syntax
 
 
13 hours later…
 
7 hours later…
9:46 PM
@SimplyBeautifulArt is it true that (omega^a+omega^b+omega^c)^d is just (omega^a)^d if a>=b>=c and d is limit ordinal?
 
9:57 PM
@LeakyNun Yes
 
exactly
20 hours ago, by Simply Beautiful Art
Teaching it Cantor normal form is only going to drive you crazy
so this is simply false :P
 
x'D
It's still gonna be a pain if d isn't a limit ordinal
 
it isn't
 
And I just thought of a cool function
 
what is it?
 
10:01 PM
$$C(\alpha)_0=\{0,\omega\}\\C(\alpha)_{n+1} =C(\alpha)_n\cup\{\gamma+\delta, \gamma\cdot\delta,\gamma^\delta, \psi_\eta(\zeta):\gamma,\delta, \eta,\zeta\in C(\alpha)_n ,\eta<\alpha,\zeta<\omega\}\\ \psi_\alpha(n)=\min\{\gamma \notin C(\alpha)_n\}$$
It defines a fast growing function more or less.
Intuition says that $\psi_\alpha(n)\sim f_\alpha(n)$ in the fast growing hierarchy
 
(omega^a+omega^b+omega^c)*d = ?
 
@LeakyNun (ω^a)∙d + ω^b + ω^c
Assuming d is not a limit
 
if d is limit?
 
If it is a limit, it'll just be (ω^a)∙d
 
(omega^a)*(omega^b+omega^c) = ?
 
10:07 PM
c>0?
 
b>=c
c>0
 
IIRC, a∙(b+c) = a∙b + a∙c
 
hmm
 
Trivially, $\psi_\alpha(0)=1$, since $1\notin\{0,ω\}$
$C(0)_1 = \{0,1,\omega,\dots\}$, where $\dots$ contains ordinals $>ω$
So $\psi_0(1)=2$
$C(0)_2= \{0,1,2,\omega,\dots\}$
So $\psi_0(2) = 3$
$C(0)_3= \{0, 1, 2, 3, 4, ω, \dots\}$
So $\psi_0(3) = 5$
$C(0)_4 = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 16, 27, 64, 256, ω, \dots\}$
So $\psi_0(4) = 10$?
 
@SimplyBeautifulArt could you prove it?
 
10:14 PM
Prove what?
$C(0)_5 = \{0, \dots, 25, 27, 28, 30, 35, 36, 48, 49, \dots\}$
So $\psi_0(5) = 26$
Too much, gotta write a program for it lol
 
@SimplyBeautifulArt prove that a(b+c) = ab+ac
 
5
Q: Distributivity of ordinal arithmetic

Math Student 020Let greek letters be ordinals. I want to prove $\alpha(\beta + \gamma) = \alpha\beta + \alpha\gamma$ by induction on $\gamma$ and I already know it holds true for $\gamma = \emptyset$ and $\gamma$ a successor ordinal. Let $\gamma$ be a limit ordinal. I found $$ \alpha(\beta + \gamma) = \alpha \cd...

 
thanks
 
Perhaps I should remove exponentiation from the program...
Interesting... it jumps to $\psi_0(6)=178$ it seems
 
10:47 PM
0
Q: Growth rate of the nth natural number not constructable with n steps of addition and multiplication

Simply Beautiful ArtWhile messing around with the idea of ordinal collapsing functions, I stumbled upon an interesting simple function: $$C(0)=\{0,1\}\\C(n+1)=C(n)\cup\{\gamma+\delta:\gamma,\delta\in C(n)\}\\\psi(n)=\min\{k\notin C(n),k>0\}$$ The explanation is simple. We start with $\{0,1\}$ and repeatedly add it...

I wonder how one would tackle this question
And whether or not there are better tags
 
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