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12:07 AM
1
A: Create the slowest growing function you can in under 100 bytes

Simply Beautiful ArtRuby, 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
You can post the answer
 
?
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
$$b_{n+1}=a_n-b_n$$
 
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
with adblock on
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
 
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
or the sine 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
 
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
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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