12:07 AM
1

Ruby, 100 bytes, score = fω³-1(n) Basically borrowed from my Largest Number Printable, here's my program: ->n{k=0;f=->a,b=k,c=k{k.times{k*=c>0?f[a,b,c-1]:b>0?f[a,b-1]:a<0?k:f[a-1]}};k+=1 while f[k]<n;k} Try it online Basically computes the inverse of fω³(n) in the fast growing hierarchy. Th...

Updated to grow a bit slower

Hiya

Heyo
Interesting worm game
Didn't get to solving your thing @Nilknarf

?
Which thing?

23 hours ago, by Nilknarf
$$a_{n+1}=a_n+\sqrt{13\cdot 2^n-a_n^2}$$

@SimplyBeautifulArt not slow enough :P

12:19 AM
@LeakyNun :P Yeah ik

Ah, there it is

Need golfier language...
But that stuff is hard for me lol

Ok, lemme try and find my answer
Define $b_n$ as
$$a_n^2+b_n^2=13\cdot 2^n$$
so that
$$a_{n+1}=a_n+b_n$$
It can also be shown that
$$b_{n+1}=a_n-b_n$$

oh
Sneaky

This can now be solved in a straightforward way
using...

12:24 AM
math

INVARIANTS! woohoo!

Namely, the invariants
$$\mu(x,y)=x+(-1+\sqrt 2)y$$
$$\nu(x,y)=x+(-1-\sqrt 2)y$$
Ok, I lied, they aren't really invariants

But they satisfy
$$\mu(a_n,b_n)=\sqrt 2 ^n\mu(a_0,b_0)$$
$$\nu(a_n,b_n)=(-\sqrt 2) ^n\nu(a_0,b_0)$$

12:26 AM
Beautiful

Ah, yes
beautiful indeed

Have you watched any of "No game no life"?

...just kidding, it's really ugly. The algebra gets nasty.
Oh, no

I mean, no I haven't

12:28 AM
Well, u should when you get the chance.

Where do you watch it?

I watch it on crunchyroll
Cuz I don't wanna watch ads

ha, ofc
Whatever happened to amWhy? She doesn't seem to be around anymore
and typhon

amWhy is active in other chats
Typhon likely got himself in trouble

D:
sorry to hear that
but not necessarily surprised

12:31 AM
yeah

oh, maybe he didn't get in trouble
"last seen 1 hr ago"

but still likely on chat ban maybe

chat ban?
one can get banned just from chat?

why?
would someone be banned from chat only?

12:35 AM
For posting inappropriate chat messages
Or being bothersome to the mods

:P

darn

tbh he was a pain

:(
welp

12:36 AM
just saying

Hey, out of curiosity
do you follow politics at all?

To the extent my macroeconomics class does

Haha

Hello and welcome to my realm @MrAP

Is that class hard? Some seniors in my calc class seem to struggle with it... but then again, they also struggle with calc, so...

12:38 AM
You'll probably be fine in it if you pay attention tbh

Hello @SimplyBeautifulArt

@Simply Earlier today I was looking at the asymptotics of
$$a_{n+1}=\text{lg}(a_n^n+1)$$
It's very gross, don't try it

Oh dear
warning heeded

Heh
I've got another one to try, if you're up for it

12:47 AM
oh dear

I haven't tried it myself yet though, so idk if it is messy

We'll see

$$\alpha_{n+1}=\frac{2\alpha_n}{\alpha_n +n},\space\space\space \alpha_0=1$$
Looks nice enough :)

...at least, it looks nice compared to the other one. XD
Should be pretty easy if you get each term in terms of only the previous term

12:48 AM
X'D
Yeah
You actually get a weird looking fraction if you do it one way...

import copy

class ord:
def check(val):
if type(val) == list:
return all(ord.check(x) for x in val)
return False
def le(a, b):
if a == []:
return True
if b == []:
return False
if a[0] == b[0]:
return ord.le(a[1:],b[1:])
if ord.le(a[0],b[0]):
return True
return False
def str(a):
if a == []:
return "0"
if len(a) == 1:
return "\u03c9^"+ord.str(a[0])
return "("+"+".join(ord.str([x]) for x in a)+")"
def call(a,n):
assert(type(n)==int and n >= 0)
if a == []:
d=ord([[[[]]]])
print(d)
print(d(3))
print(d(3)(3))
print(d(3)(3)(3))
print(d(3)(3)(3)(3))
print(d(3)(3)(3)(3)(3))
print()
print(d(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3))
print()
print(d(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3)(3))

$$\frac1{\alpha_{n+1}}= \frac12+\frac n{2\alpha_n}$$

ω^ω^ω
ω^ω^(1+1+1)
ω^(ω^(1+1)+ω^(1+1)+ω^(1+1))
ω^(ω^(1+1)+ω^(1+1)+ω+ω+ω)
ω^(ω^(1+1)+ω^(1+1)+ω+ω+1+1+1)
ω^(ω^(1+1)+ω^(1+1)+ω+ω+1+1)+ω^(ω^(1+1)+ω^(1+1)+ω+ω+1+1)+ω^(ω^(1+1)+ω^(1+1)+ω+ω+1+1)

ω^(ω^(1+1)+ω^(1+1)+ω+ω+1+1)+ω^(ω^(1+1)+ω^(1+1)+ω+ω+1+1)+ω^(ω^(1+1)+ω^(1+1)+ω+ω+1)+ω^(ω^(1+1)+ω^(1+1)+ω+ω+1)+ω^(ω^(1+1)+ω^(1+1)+ω+ω)+ω^(ω^(1+1)+ω^(1+1)+ω+ω)+ω^(ω^(1+1)+ω^(1+1)+ω+1+1)+ω^(ω^(1+1)+ω^(1+1)+ω+1+1)+ω^(ω^(1+1)+ω^(1+1)+ω+1)+ω^(ω^(1+1)+ω^(1+1)+ω+1)+ω^(ω^(1+1)+ω^(1+1)+ω)+ω^(ω^(1+1)+ω^(1+1)+ω)+ω^(ω^(1+1)+ω^(1+1)+1+1)+ω^(ω^(1+1)+ω^(1+1)+1+1)+ω^(ω^(1+1)+ω^(1+1)+1)+ω^(ω^(1+1)+ω^(1+1)+1)+ω^(ω^(1+1)+ω^(1+1)…

AAAHHHHH!!!

Sorry for spamming

12:49 AM
@SimplyBeautifulArt Ooh, that is very helpful

So it really comes down to$$\beta_n=\frac12+\frac n2\beta_n$$

@SimplyBeautifulArt do you get what I'm doing?

That's much nicer looking

Which seems factorial-like

theorem: "finite" doesn't mean "small"

12:50 AM
@LeakyNun x'D
Not yet

Yeahhh

it's guaranteed to terminate, but god knows when
@SimplyBeautifulArt is that to me?

Yeah I get it

Theorem 1: you have no idea how big finite is
Theorem 2: you have no idea how big countable is
@SimplyBeautifulArt nice

@LeakyNun not true, I also know when it'll terminate

12:52 AM
@SimplyBeautifulArt that's 'cause you're god

Oh, tru

XD

seriously, it takes that many steps for it to become a successor ordinal

If the base isn't changing...

no, that isn't Goodstein!

12:53 AM
It's not!
But!
Consider base n.

I'm not doing Goodstein at all

A constant k terminates in k steps
Consider ω
it terminates in n steps more or less
ω+k in n+k steps
ω+ω in n+n steps
ω^ω in n^n steps
etc
ez

hmm
nice

gn @Nilknarf

What is gn?

12:57 AM
gud nite

Ah
you've predicted my schedule XD

I have 3 minutes

and I intend to use them XD

12:58 AM
Can someone please help me solve this: $\cos x +\sqrt{3} \sin x =\sqrt{2}$

Gah, gotta go
Try dividing both sides by $2$ and using the cosine addition formula

I did

Argh, gotta go

Using cosine addition formula i got $x=\frac{\pi}{12}-2n\pi$ and $x=\frac{7\pi}{12}-2n\pi$
Using sine addition formula i got $x=n\pi-\frac{5\pi}{12}$ and $x=n\pi-\frac{\pi}{12}$

1:04 AM
@SimplyBeautifulArt are you here?

@MrAP $$\sin(x+y)=\sin(y)\cos(x) + \cos(y)\sin(x)$$
Yeah I'm here lol

to everyone else, sorry for the spam:
import copy

class ord:
def check(val):
if type(val) == list:
return all(ord.check(x) for x in val)
return False
def le(a, b):
if a == []:
return True
if b == []:
return False
if a[0] == b[0]:
return ord.le(a[1:],b[1:])
if ord.le(a[0],b[0]):
return True
return False
def str(a):
if a == []:
return "0"
if len(a) == 1:
return "\u03c9^"+ord.str(a[0])
return "("+"+".join(ord.str([x]) for x in a)+")"
def call(a,n):
assert(type(n)==int and n >= 0)
if a == []:
a = ord([],[])
n = 7
print("fgh(%s,%d) = %d" % (a,n,fgh(a,n)))
fgh(1+1,7) = 896
don't expect it to work for fgh(3,3)
a = ord([[]])
n = 2
print("fgh(%s,%d) = %d" % (a,n,fgh(a,n)))
fgh(ω,2) = 8

@LeakyNun hm?

@SimplyBeautifulArt isn't it right?

Oh it is
Sorry, my brain is in sgh and hh mode
slow-growing and Hardy
@MrAP And by dividing by 2, as suggested, and plugging in $y=\pi/6$, we get:$$\sin(x+\pi/6) = \sin(\pi/3)$$

1:14 AM
i got $\sin(x+\frac{\pi}{6}=\sin(\frac{\pi}{4})$

Hm
Oops, you are right

@SimplyBeautifulArt

1:30 AM
:P @LeakyNun

1:44 AM
@SimplyBeautifulArt did you solve the equation?

@MrAP I'm too tired, and it seems you've got it close enough, which leads me to believe you probably did the rest right

Okay.

2:01 AM
@SimplyBeautifulArt exponential is a can of worms
is it?
I mean, (omega+1)^2 = omega^2+omega+1

lol

(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.

:P
Okay

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

Let 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 $$\alpha(\beta + \gamma) = \alpha \cd... thanks Perhaps I should remove exponentiation from the program... Interesting... it jumps to \psi_0(6)=178 it seems 10:47 PM 0 While messing around with the idea of ordinal collapsing functions, I stumbled upon an interesting simple function:$$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