Simply Beautiful Art's realm of calculus and analysis - 2017-12-08

« first day (275 days earlier) ← previous day next day → last day (124 days later) »

Simply Beautiful Art

12:07 AM

1

A: Create the slowest growing function you can in under 100 bytes

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

Simply Beautiful Art

Heyo

googology.wikia.com/wiki/Beklemishev%27s_worms

Interesting worm game

Didn't get to solving your thing @Nilknarf

You can post the answer

?

Which thing?

Simply Beautiful Art

23 hours ago, by Nilknarf

$$a_{n+1}=a_n+\sqrt{13\cdot 2^n-a_n^2}$$

@SimplyBeautifulArt not slow enough :P

Simply Beautiful Art

12:19 AM

@LeakyNun :P Yeah ik

Ah, there it is

Simply Beautiful Art

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
$$b_{n+1}=a_n-b_n$$

Simply Beautiful Art

oh

Sneaky

This can now be solved in a straightforward way

using...

Simply Beautiful Art

12:24 AM

math

INVARIANTS! woohoo!

Simply Beautiful Art

x'D

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

Simply Beautiful Art

yaya

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

Simply Beautiful Art

12:26 AM

Beautiful

Ah, yes

beautiful indeed

Simply Beautiful Art

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

...just kidding, it's really ugly. The algebra gets nasty.

Oh, no

Simply Beautiful Art

lol

I mean, no I haven't

Simply Beautiful Art

12:28 AM

Well, u should when you get the chance.

Where do you watch it?

Simply Beautiful Art

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

Simply Beautiful Art

amWhy is active in other chats

Typhon likely got himself in trouble

D:

sorry to hear that

but not necessarily surprised

Simply Beautiful Art

12:31 AM

yeah

oh, maybe he didn't get in trouble

"last seen 1 hr ago"

Simply Beautiful Art

but still likely on chat ban maybe

chat ban?

one can get banned just from chat?

Simply Beautiful Art

yeah

why?

would someone be banned from chat only?

Simply Beautiful Art

12:35 AM

For posting inappropriate chat messages

Or being bothersome to the mods

:P

Simply Beautiful Art

:P

darn

Simply Beautiful Art

tbh he was a pain

:(

welp

Simply Beautiful Art

12:36 AM

just saying

Hey, out of curiosity

do you follow politics at all?

Simply Beautiful Art

To the extent my macroeconomics class does

Haha

Simply Beautiful Art

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...

Simply Beautiful Art

12:38 AM

You'll probably be fine in it if you pay attention tbh

Hello @SimplyBeautifulArt

Simply Beautiful Art

:-)

@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

Simply Beautiful Art

Oh dear

warning heeded

Heh

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

Simply Beautiful Art

12:47 AM

oh dear

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

Simply Beautiful Art

We'll see

$$\alpha_{n+1}=\frac{2\alpha_n}{\alpha_n +n},\space\space\space \alpha_0=1$$

Looks nice enough :)

Simply Beautiful Art

:(

...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

Simply Beautiful Art

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))

Simply Beautiful Art

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

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

Simply Beautiful Art

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

Simply Beautiful Art

Which seems factorial-like

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

Simply Beautiful Art

12:50 AM

@LeakyNun x'D

Not yet

Yeahhh

it's guaranteed to terminate, but god knows when

@SimplyBeautifulArt is that to me?

Simply Beautiful Art

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

Simply Beautiful Art

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

12:52 AM

@SimplyBeautifulArt that's 'cause you're god

Simply Beautiful Art

Oh, tru

XD

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

Simply Beautiful Art

If the base isn't changing...

no, that isn't Goodstein!

Simply Beautiful Art

12:53 AM

It's not!

But!

Consider base n.

I'm not doing Goodstein at all

Simply Beautiful Art

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

Simply Beautiful Art

gn @Nilknarf

What is gn?

Simply Beautiful Art

12:57 AM

gud nite

Ah

you've predicted my schedule XD

Simply Beautiful Art

x'D

I have 3 minutes

Simply Beautiful Art

:P

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

Simply Beautiful Art

cya

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?

Simply Beautiful Art

@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

Simply Beautiful Art

@LeakyNun hm?

@SimplyBeautifulArt isn't it right?

Simply Beautiful Art

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 $\sin(x+\frac{\pi}{6}=\sin(\frac{\pi}{4})$

Simply Beautiful Art

Hm

Oops, you are right

github.com/kckennylau/ordinals/blob/master/ordinals.py

with addition :P

@SimplyBeautifulArt

Simply Beautiful Art

1:30 AM

:P @LeakyNun

1:44 AM

@SimplyBeautifulArt did you solve the equation?

Simply Beautiful Art

@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

Simply Beautiful Art

lol

Yeah don't worry about it

(omega+1)^3 = omega^3+omega+1

I'm not even sure if this is right

Simply Beautiful Art

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

Simply Beautiful Art

@WheatWizard what brings you to my realm?

2:08 AM

I saw it in the sidebar so I thought I would take a look.

Simply Beautiful Art

:P

Okay

It looks like you do more Differential Calculus than Ananlysis here.

Simply Beautiful Art

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

Simply Beautiful Art

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?

Simply Beautiful Art

@LeakyNun It just becomes (ω+1)^n

You can use w=ω

no, I want to be able to write it in normal form

Simply Beautiful Art

Normal form is ω^ω

but how do I figure that out?

Simply Beautiful Art

The limit of (ω+1)^n

2:16 AM

I know, but my computer doesn't know

Simply Beautiful Art

So?

It knows what (ω+1)^n is

we're going in circles

Simply Beautiful Art

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

Simply Beautiful Art

Ofc it is

But it's so much easier to go with the flow and the syntax

13 hours later…

3:16 PM

The hardest problem on the hardest test

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?

Simply Beautiful Art

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

Simply Beautiful Art

x'D

It's still gonna be a pain if d isn't a limit ordinal

it isn't

Simply Beautiful Art

And I just thought of a cool function

what is it?

Simply Beautiful Art

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 = ?

Simply Beautiful Art

@LeakyNun (ω^a)∙d + ω^b + ω^c

Assuming d is not a limit

if d is limit?

Simply Beautiful Art

If it is a limit, it'll just be (ω^a)∙d

(omega^a)*(omega^b+omega^c) = ?

Simply Beautiful Art

10:07 PM

c>0?

b>=c

c>0

Simply Beautiful Art

IIRC, a∙(b+c) = a∙b + a∙c

hmm

Simply Beautiful Art

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?

Simply Beautiful Art

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

Simply Beautiful Art

5

Q: Distributivity of ordinal arithmetic

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...

elementary-set-theory ordinals

thanks

Simply Beautiful Art

tio.run/##KypNqvz/…

tio.run/…

Perhaps I should remove exponentiation from the program...

Interesting... it jumps to $\psi_0(6)=178$ it seems

Simply Beautiful Art

10:47 PM

0

Q: Growth rate of the nth natural number not constructable with n steps of addition and multiplication

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...

discrete-mathematics asymptotics

I wonder how one would tackle this question

And whether or not there are better tags

« first day (275 days earlier) ← previous day next day → last day (124 days later) »

Transcript for

Dec '178

$Mathematics$ Simply Beautiful Art's realm of calcu…

where you may feel free to do your calculus

join this room
about this room

00:00

06:00

12:00

18:00

all times are UTC

$Mathematics$

site design / logo © 2024 Stack Exchange Inc; legal

mobile