Room for Simply Beautiful Art and Astyx

next day → last day (16 days later) »

Astyx

21:00

What's up ?

Simply Beautiful Art

@Astyx You busy?

Astyx

Not really

Nearly bedtime, but I'm not doing much

Simply Beautiful Art

a.k.a. want to try and see why I want to use that bounded thing I made?

Astyx

Sure, remind me what it is

Simply Beautiful Art

f(x+a) < g(x) < f(x+b) for some a,b and all x>x0, then f and g are bounded by each other.

Astyx

21:03

Right, that's fair enough

Simply Beautiful Art

Do you know of ordinals?

Astyx

A little

Simply Beautiful Art

Well,
$H_0(n)=n$
$H_\alpha(n)=H_{\alpha-1}(n+1)$ if $\alpha$ is a successor ordinal.
$H_\alpha(n)=H_{\alpha[n]}(n)$ if $\alpha$ is a limit ordinal

So $H_k(n)=n+k$

Letting $\omega[n]=n$,
$H_\omega(n)=H_n(n)=2n$

$H_{\omega+k}(n)=H_\omega(n+k)=2n+2k$

$H_{\omega2}(n)=H_{\omega+n}=4n$

$H_{\omega k}(n)=2^kn$

$H_{\omega^2}(n)=2^nn\approx2^n$

$H_{\omega^2+\omega}(n)\approx2^{2n}$

$H_{\omega^2+\omega k}(n)\approx2^{2^kn}$

$H_{\omega^22}(n)\approx2^{2^n}$

@Astyx You understand how this works?

Astyx

Give me 5 minutes :)

You're defining $H_k(n)$ recursively right

Simply Beautiful Art

Sort of yeah

Usually $k$ is reserved for things less than $\omega$

Astyx

21:19

Okay I think I follow (without going too much into the details)

Simply Beautiful Art

So you think you know what $H_{\omega^3}(n)\approx?$

Astyx

Something like $2^{2^{2^{n}}}$ ?

Simply Beautiful Art

$H_{\omega^22+\omega}(n)\approx2^{2^{2n}}$

Astyx

Let me think

Was it just a typo from my part or is what I wrote still wrong ?

Simply Beautiful Art

You are still wrong.
$H_{\omega^23}(n)\approx2^{2^{2^n}}$

Notice the jump from $H_\omega$ to $H_{\omega^2}$

Astyx

21:23

So $2^{2^{2^nn}}$ ?

Simply Beautiful Art

$H_{\omega^23}(n)\approx2^{2^{2^nn}}$

Astyx

Tell me what it is ?

Simply Beautiful Art

$H_{\omega^3}(n)\approx\underbrace{2^{2^{2^{\dots}}}}_n$

$H_{\omega^2}(n)\approx\underbrace{2\times2\times2\times\dots\times2}_n$

Astyx

Oh yeah right

Simply Beautiful Art

$H_{\omega^4}(n)\approx\underbrace{H_{\omega^3}(H_{\omega^3}(H_{\omega^3}(\dots)‌))}_n$

etc.

$H_{\omega^\omega}(n)=H_{\omega^n}(n)$

Astyx

21:28

Huh, cool

So what's the point ?

Simply Beautiful Art

So $\varepsilon_0=\omega^{\omega^{\dots}}$

And I want to show that $H_{\varepsilon_0}(n)\approx f_{\varepsilon_0}(n)$

where $f$ is the fast growing hierarchy

Astyx

$f$ is what ?

And what does that indice mean for $f$ ?

Simply Beautiful Art

$f_0(n)=n+1$
$f_\alpha(n)=\underbrace{f_{\alpha-1}(f_{\alpha-1}(\dots f_{\alpha-1}(n)\dots))}_n$
$f_\alpha(n)=f_{\alpha[n]}(n)$

Astyx

So same as $H$ except it's $n+1$

Simply Beautiful Art

And initial condition

Clearly, $f_a(n)>H_a(n)$

Astyx

21:33

Huh ?

Simply Beautiful Art

$H_0(n)=n<n+1=f_0(n)$

Astyx

Why "And initial condition" ? That's what I meant ? Or am I missing something ?

Simply Beautiful Art

Oh, nvm then

Astyx

Simply Beautiful Art

Never mind lol

Astyx

21:36

Nevermind what ?

Simply Beautiful Art

never mind my saying about initial conditions

Astyx

Oh right

Go on :)

Simply Beautiful Art

But I wanted to say $H_\alpha(n)\approx f_\alpha(n)$ for some very large $\alpha$

Particularly, when $\alpha=\varepsilon_0$

But as you can see, both these functions will grow insanely fast

Too fast for me to apply normal $\approx,\sim\mathcal O,\Omega,etc$

Astyx

What would both $H_{\epsilon_0}$ and $f_{\epsilon_0}$ look like ?

(Or is there no way to write them properly using basic operations ?)

Simply Beautiful Art

I can't write them using things other than $f$ and $H$

Astyx

21:42

Fair enough

I'm not too confortable with ordinals, especially infinite tetration of them

This'll get some time for me to get used to it

I have to go for today

I'll let you know if I make progress and come up with ideas

Bye

Simply Beautiful Art

@Astyx Bye!

next day → last day (16 days later) »

all rooms

Transcript for

May '1711

May12

$Mathematics$ Room for Simply Beautiful Art and Astyx

join this room
about this room

00:00

06:00

12:00

18:00

all times are UTC

$Mathematics$