This is the Realm of Simply Beautiful Art

DHMO

4:09 AM

@AkivaWeinberger que pasa aqui?

Akiva Weinberger

@DHMO Estamos pensando en códigos (no seguro si es la palabra correcta) que nos dan números grandes

Computer codes

DHMO

@AkivaWeinberger what is the restraint?

Akiva Weinberger

Less than 256 characters, I think?

DHMO

you know, 256 chars can produce huge numbers.

much bigger than you can ever imagine.

Akiva Weinberger

2 hours ago, by Simply Beautiful Art

Would you like to try and make a larger number than S.C.B. Nisman and I in under 256 characters with the restriction of no infinity and no calling in constants (or else you'd just call in some really big number and +1 it)

DHMO

4:12 AM

simply 9**9**9**9**9**9**9 is large as heaven

in which language?

Akiva Weinberger

@DHMO Mine above is much larger, pretty sure. For looping a function into itself

He did Ruby, I did Python

repl.it/languages

^Simulates a lot of them

DHMO

@AkivaWeinberger eval("9**"*9**9**9**9**9+"9") have fun computing this

Akiva Weinberger

Ha.

My code actually counts up to a ridiculously large number one by one, so it's as slow as possible essentially

since the only tools are "add one" and recursion

DHMO

I see

can you roughly estimate which number is bigger? @AkivaWeinberger

Akiva Weinberger

I'mma guess mine, but I really have no idea

TheGreatDuck

4:19 AM

hi

either of you two aware of the reasoning behind en.wikipedia.org/wiki/…

specifically the value for a and how it plays into finding the nTH prime?

Akiva Weinberger

I like the quote they have

> "Any one of these formulas (or any similar one) would attain a different status [i.e. different from being considered useless] if the exact value of the number α ... could be expressed independently of the primes. There seems no likelihood of this, but it cannot be ruled out as entirely impossible."

TheGreatDuck

...

Akiva Weinberger

Footnote 35.

TheGreatDuck

im just wanting to know how the second and third equations are reasoned

@AkivaWeinberger i know, but it's annoying how people keep quoting it...

:p

i have nothing more to say regarding that quote

Akiva Weinberger

@TheGreatDuck I have no idea.

TheGreatDuck

4:24 AM

hmm

maybe I'll ask on wikipedia how to find the article cited

that would probably explain why

(i hope)

Akiva Weinberger

Try plugging in r=10?

TheGreatDuck

lol

Akiva Weinberger

Plug in r=10, see what $\alpha$ is, plug it into the second thing

TheGreatDuck

lol

note the fact that a defines the sequence of primes which defines a

Akiva Weinberger

Seems to be $.200300005000000700000001100\dots$

or something like that

Ah, I see.

TheGreatDuck

4:26 AM

:p

but good point

you manually calculated it

Akiva Weinberger

They're just trying to mine the $n^2$-eth digit of the number

TheGreatDuck

oh

that's it?

Akiva Weinberger

Like, if you multiply by $10^9$ and take the floor, you get:

DHMO

@AkivaWeinberger i=99;eval"i**=i;"*9**9**9**9**9**9

Akiva Weinberger

$200300005$

If, on the other hand, you multiply by $10^4$, take the floor, and multiply by $10^5$, you get:

$200300000$

Subtracting gets you $5$.

TheGreatDuck

4:28 AM

ah

Akiva Weinberger

The formula they have is this idea generalized.

TheGreatDuck

i thought I had a sum yielding a

but perhaps my precision wasn't good enough

Akiva Weinberger

(I plugged in $n=3$.)

TheGreatDuck

after about 7 or 11

it gave nothing but 0

was honestly wondering if it was a crackpot equation

Akiva Weinberger

Seems legit.

TheGreatDuck

4:28 AM

and absolute garbage

in thats its false

ah

cause I can compute a with a normal infinite series

Akiva Weinberger

It could only fail if the primes grow faster than $r^{2n-1}$, which they very much do not.

TheGreatDuck

granted, it's an infinite series of 3-4 sub-series

Akiva Weinberger

So the equation seems to be true.

TheGreatDuck

@AkivaWeinberger proven or based on riemann hypothesis?

Akiva Weinberger

No, just formalize the argument I gave above.

I showed that with $r=10$ and $n=3$, it works

TheGreatDuck

4:30 AM

"It could only fail if the primes grow faster than $r^{2n-1}$, which they very much do not." ^proven or based on riemann hypothesis?

Akiva Weinberger

No, it's known that there's always a prime between $n$ and $2n$

There are other lower bounds on $\pi(n)$ also

TheGreatDuck

oh

wait a second then

doesn't that imply that there is a prime between consecutive squares?

Akiva Weinberger

How

TheGreatDuck

isn't n to 2n a smaller bound?

Akiva Weinberger

How do we get a prime between, say, 100 and 121

We know there's a prime between 50 and 100, or between 60 and 120, but that doesn't seem helpful

TheGreatDuck

4:33 AM

nvmd

i was thinking n to n*n

@Ramanujan was looking at yer work. good job. en.wikipedia.org/wiki/…

Nisman

4:50 AM

Isnt there like a known "Most increasing function ever" ?

So you code it and done ?

TheGreatDuck

@Nisman um... what?

Nisman

I was talking about the biggest possible number , the challenge they were doing

TheGreatDuck

i can prove that there isn't a largest number by induction

Nisman

Is there like a mathematical function that is known for being the most fastest growing function ever ??

TheGreatDuck

first, I claim that there is an integer larger than 1 which is 2

then I assert that n+1 is always greater than n

Nisman

4:52 AM

@TheGreatDuck Yeah but we cant do recursion infinitely

Akiva Weinberger

@Nisman No. There are functions that grow faster than any computable function, but those are noncomputable.

TheGreatDuck

@Nisman induction works infinitely. When did I ever mention recursion?

Akiva Weinberger

If there were a fastest function, add $1$ to it. Or add $n$ to it. Also, if you have infinite sequence $f_k$ of functions, each faster than the last, you can get one faster than all of those by doing $n\mapsto f_n(n)$.

Nisman

@AkivaWeinberger yeah yeah I know what you mean

But you have a great starting point from those functions

TheGreatDuck

@AkivaWeinberger grows faster

@Nisman let's say f(x) grows the fastest

then 2*f(x) grows twice as a fast

Nisman

4:55 AM

If I have $f(x) = 2^n $ and $ g(x) = 2+2 $
At some point g will have the same values as f

Akiva Weinberger

I mean, for any finite $k$, there's a fastest function that can be definable in Python in $\le k$ characters, I guess

since there are finitely many codes with $\le k$ characters.

Nisman

But F will be much faster to compute than g, and also it grows much faster

TheGreatDuck

@Nisman those are both constant functions, stupid!

Nisman

So from there I have a great starting point

TheGreatDuck

what do you mean by "grows fastest?"

do you mean purely the derivative f'(x) is greatest?

rather than the actual values of f(x)?

Nisman

4:56 AM

f(f(f(f(x)))) > g(g(g(g(x)))))

You know what I mean with grows faster

Akiva Weinberger

The function $n\mapsto{}$"the largest number definable in $\le n$ characters in Python" grows faster than all computable functions, and is uncomputable.

(Pretty sure.)

TheGreatDuck

@Nisman no I don't

Akiva Weinberger

@TheGreatDuck I assume functions from $\Bbb N\to\Bbb N$

TheGreatDuck

f(f(f(f(x)))) = f(x) = 2^n

Nisman

@TheGreatDuck What grows faster $n^n$ or $n!$ ??

TheGreatDuck

4:57 AM

neither

they're constants

Nisman

...

Akiva Weinberger

but $f$ grows faster than $g$ ("$f$ dominates $g$") if there's an N such that for all $n>N$ we have $f(n)>g(n)$. That is, its eventually larger.

TheGreatDuck

f(x) = n^n is a constant function

Akiva Weinberger

There's a word for if $f$ and $g$ grow at roughly the same rate (neither dominates the other), but I forget.

TheGreatDuck

you're referring to the variables as x, but yet your function depends on this mysterious 'n'

i can only presume n to be a constant

Nisman

4:59 AM

@TheGreatDuck I can tell you $f(x) = x^x $ grows faster than $ g(x) = x! $ because for any $ n \in N $, f(n) > g(n)

TheGreatDuck

@Nisman x! doesn't grow. It's a discrete function. Do you mean gamma, which is the continuous version of x!?

Akiva Weinberger

Derp, never mind

TheGreatDuck

i mean, I know what you mean but x! doesn't grow.

Akiva Weinberger

@TheGreatDuck Discrete functions grow

The formal notion you want here is eventual domination, I'm pretty sure

where eventually $f$ is always larger than $g$

TheGreatDuck

a function that grows faster than all other functions?

it's impossible

Akiva Weinberger

5:00 AM

Yup.

TheGreatDuck

the growth could be represented by the derivative

Akiva Weinberger

Or by the finite difference $f(n+1)-f(n)$

TheGreatDuck

and then you would be back to square one

@AkivaWeinberger derivative makes a larger point

for f(x) to grow faster than g(x) it would mean f'(x) > g'(x)

but then f'(x) grows faster than g'(x) or equal with it

so f''(x) >= g''(x)

and so on

eventually, the derivative will loop

or become undefined

since g(x) is all functions

Akiva Weinberger

Eh, I like the version with discrete functions better

TheGreatDuck

we can presume it to differentiate into some balance or trend to 0 with repeated differentiation

honestly

Akiva Weinberger

5:04 AM

The derivatives of functions could be faster than the functions themselves

Take $xe^x$

TheGreatDuck

i think such a function would need infinity

Akiva Weinberger

Its $n$th derivative is $ne^x+xe^x$

TheGreatDuck

however

infinity * x <= infinity * infinity * x * x

Nisman

You got off track here... but ok

TheGreatDuck

not at all

you asked for a function that grows faster than all other functions

Nisman

5:05 AM

Yeah, then you told me that didnt existed

TheGreatDuck

therefore, it's derivative function is greater than all other functions

Nisman

And I asked a function that grew extremely quick

TheGreatDuck

therefore there must be a function which is greater than all other functions

Akiva Weinberger

The Goodstein function I defined a while ago grows faster than all functions that PA can prove are defined everywhere

TheGreatDuck

therefore, there must be a number greater than all other numbers

Nisman

5:06 AM

Like for example, in your example @AkivaWeinberger, why instead of adding 1 to y you potentiate it ?

TheGreatDuck

infinity doesn't exist

Akiva Weinberger

but PA can't prove that the Goodstein function is defined everywhere

TheGreatDuck

therefore...

it is all nonexistant

Akiva Weinberger

@Nisman Which example

Nisman

The one with the 1 and 2

Which by the way is still running, with y = 167772160

TheGreatDuck

5:07 AM

@Nisman let's say f(x) is faster than most. c*f(x) for integer c > 1 is faster than f(x)

Akiva Weinberger

Oh, wow

You mean, why don't I exponentiate it?

That's a good idea...

...but it's still huge without that, I guess

Nisman

Yeah sorry, I guess its obvious that english is not my mother language

But it takes an eternity to compute haha

TheGreatDuck

native language

what about an infinite power tower?

can it be computed

Akiva Weinberger

What's your native language, if you don't mind my asking?

Nisman

@TheGreatDuck thanks

@TheGreatDuck Nope

TheGreatDuck

5:09 AM

here's an interesting thought

Nisman

@AkivaWeinberger Spanish, I'm from Argentina

Akiva Weinberger

Ahh

TheGreatDuck

i can prove all integers are finite

wanna know how?

Akiva Weinberger

I'm going to Argentina in the summer

to learn Spanish better

Nisman

Actually you know me

I just noticed

Akiva Weinberger

5:09 AM

?

Nisman

I'm logged in the other account

TheGreatDuck

first, we assert that 1 is finite

Nisman

I'm Maks

TheGreatDuck

then if n is finite then n+1 is finite

Akiva Weinberger

Ohh

Nisman

5:10 AM

I have two accounts, god knows why, and I dont remember which is which

TheGreatDuck

induction

Nisman

I think, its because I didnt know

TheGreatDuck

all integers >= 1 are of finite value

Nisman

My first account in stackexchange

Was in the blender forum

Akiva Weinberger

¿Vivés en Buenos Aires?

Nisman

5:10 AM

And I guess I thought that every forum was different and I created a different account for stackoverflow

And then I discovered that they were the same thing

Akiva Weinberger

(I'm not so familiar with the vos form. I've been learning tú and usted. But it doesn't sound so hard.)

Nisman

@AkivaWeinberger Was that google translate ? I live in Cordoba

Akiva Weinberger

@Nisman No.

Nisman

"Los del interior" which means, the ones who dont live in Buenos Aires, are not really fond of people living in buenos aires

Akiva Weinberger

Pensaba que Córdoba está en España...

Nisman

5:12 AM

Where are you from ?

Akiva Weinberger

¿Hay un otro Córdoba?

NYC

Nisman

Tambien hay un Cordoba en España

Akiva Weinberger

Ah, okey

Nisman

We got the city of Cordoba

And the "provincia" of Cordoba

Which is like a state

Are you students ?

Or already graduates ??

Akiva Weinberger

I'm a high school student

Nisman

5:16 AM

What are you studying ?

Oh, high school ?

TheGreatDuck

he's studying everything

XD

Nisman

Yeah hahaha I got confused with college

What about you @TheGreatDuck ?

@AkivaWeinberger You know quite a lot of maths to be a high school student

Akiva Weinberger

Thank you

Nisman

What year are you on?

Akiva Weinberger

Argentina es en la misma "time zone" que NYC, ¿sí?

12:17am?

@Nisman 11th grade, junior year

Así que me voy a dormir

y pienso que tal vez querás dormir también

(Sorry for bad grammar)

Nisman

5:18 AM

@AkivaWeinberger We are 2 hours ahead

2:18am here

Akiva Weinberger

Oh

Then you definitely should sleep also

Good night

Nisman

@AkivaWeinberger Is that supposed to be the last high school year ?

Akiva Weinberger

Second-to-last

Nisman

@AkivaWeinberger Actually it was perfect

Akiva Weinberger

Last year is 12th grade, senior year

TheGreatDuck

5:19 AM

@Nisman college. Math and Computer Science

Akiva Weinberger

@Nisman I wasn't sure if I wanted querás or querés or querrías or...

Nisman

@TheGreatDuck I'm doing computer science too

what year @TheGreatDuck ?

TheGreatDuck

3

Nisman

@AkivaWeinberger its querrás

TheGreatDuck

what classes are you taking?

Nisman

5:20 AM

With two "r"

@TheGreatDuck well now that we're studying the same, I guess I could get some help from you ( ͡° ͜ʖ ͡°)

Akiva Weinberger

Thanks

TheGreatDuck

what are you taking right now?

Nisman

@TheGreatDuck In about ten days I have Mathematical Analysis II and Algebra test

Do you know any spanish ?

TheGreatDuck

no

i used to

but it's not the sort of language im interested in if you get the joke.

XD

Nisman

imgur.com/MzmMX2Q

Maybe you get some names

So, how good are you in math ? Cause I could use some help

TheGreatDuck

5:22 AM

very very good

Nisman

@TheGreatDuck LOL, not a great syntax

TheGreatDuck

im doing quite well in all my classes

Nisman

Do you remember anything about multiple variable functions and level curves and taylor series, etc ?

TheGreatDuck

but i've never heard of math analysis 2... :/

oh!

multivariate calculus?

Nisman

Yep

Are you familiar with the stewarts book ?

TheGreatDuck

5:23 AM

taylor series we never really focused on

idk. I don't think that's the one we used

the one we had was calc 1-3

15 chapters

Nisman

What about algorithms ?

Like differentiate programs ?

@TheGreatDuck Do you know the author ??

TheGreatDuck

algorithms? that's vague.

do you mean an algorithms class?

as in CS?

differentiate programs. Not sure what you mean by that.

@Nisman well level curves are really just the identification of the graph of some equation of x and y

it's basically just a way to crudely graph a 3d function onto paper

Nisman

5:56 AM

Yeah, I have to recheck how to do them

No one knows what algorithm differentiation is

@TheGreatDuck en.wikipedia.org/wiki/Program_derivation

S.C.B.

9:14 AM

I don't know how you count letters, but in the traditional method this seems to be about 251 letters. repl.it/F7Lz/1.

I meant repl.it/F7Lz/2

S.C.B.

12:12 PM

@SimplyBeautifulArt I cannot compare the two.

Simply Beautiful Art

12:38 PM

I am here!

Give me a few hours to read everything

@S.C.B. Aw man, you left me

@AkivaWeinberger @Nisman So, how's it coming along?

@S.C.B. Well, I took a look at your code, and I do believe you will be smaller than my numbers

@AkivaWeinberger I can't see how your number will grow very fast. In comparison to mine

@Nisman Did you make a number yet?

S.C.B.

12:53 PM

What is your number?

@SimplyBeautifulArt

Simply Beautiful Art

repl.it/F7MS/4

^ My current version

S.C.B.

So why do you think your number will be larger? I can't tell by looking at your code.

Simply Beautiful Art

I do believe jumps from 256 to larger than Graham's number by increasing the d or e arguments to 1

which is why I think it exceeds your number easily

S.C.B.

Can you tell me what your function is? I'm not experienced enough with code.

Simply Beautiful Art

So, I didn't give it a real name yet

but basically, let's consider the case of f(a,b,c,0,0) first

basically, f(a,b,c,0,0)>$a\uparrow^cb$

Knuth's uparrow notation

then, once we get c=0

and if d>0

then d drops one

so f(a,b,c,0,0)=a

then d drops one

then take f(a,a,a,d-1,0)

When d drops to 0

e drops one

then take f(a,a,a,a,e-1)

Clearly, it outgrows Graham's number very quickly

Transcript for

$Mathematics$ This is the Realm of Simply Beautiful…