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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.
 
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