12:03 AM
@SimplyBeautifulArt are you here?

maybe

@SimplyBeautifulArt ping me next time you reply

12:25 AM
:41587665

@SimplyBeautifulArt Hi

@SimplyBeautifulArt ... I didn't receive that ping lol

@LeakyNun Hello to you too

12:28 AM
@LeakyNun Yup. I'm a blind bat
O_O

@SimplyBeautifulArt Wanna do another asymptotic expansion problem?

Oh sure

Big-O is giving me trouble :P

1 message moved to trash

Okay, here comes the original problem:
$$\lambda_{n+1}=\lambda_n+\sqrt{5^n-\lambda_n}$$
$$\lambda_0=0$$
But I've reduced it a bit

12:31 AM
oh dear
Why're all your things so weird

Heh heh, don't worry. It's not as ugly as it looks
...at least, I think it isn't.
It reduces to
$$L_{n+1}=\frac{1+2\sqrt{L_n-L_n^2}}{5}$$
where $$L_n=\frac{\lambda_n^2}{5^n}$$

So if I can approximate $L_n$ within $O(5^n)$, then I can get a good asymptotic expansion for $\lambda_n$

$\lambda_1=1$

Yah

12:33 AM
@SimplyBeautifulArt teach me anything about ordinals lol

$\lambda_2=3$

Uh huh...

@LeakyNun lol I mean I mostly do recursion stuffs

Whoops whoops!

12:34 AM
@SimplyBeautifulArt you still know a lot of stuff

It should be $$\lambda_{n+1}=\lambda_n+\sqrt{5^n-\lambda_n^2}$$

I was amazed by how you could have a transfinite (countable) hierarchy on $\Bbb N \to \Bbb N$ functions

Hm
K

But yeah, your initial values still fit
since $0^2=0$ and $1^2=1$

12:35 AM
$f_{\omega+1}(3) = f_{f_{f_3(3)}(f_3(3))}(f_{f_3(3)}(f_3(3)))$

It appears to approximately double per step
@LeakyNun :P

Note that
$$L_{\infty}=\frac{7+2\sqrt{5}}{29}$$

@SimplyBeautifulArt I'm looking to define $f:\varepsilon_0 \times \Bbb N \times \Bbb N \to \Bbb N$ instead to make it more rigorous
of course I would also like to define the cofinal sequence $\lambda : \varepsilon_0 \times \Bbb N \to \varepsilon_0$ more rigorously

Sorry, perhaps I should wait

Lol, don't mind @Nilk

12:38 AM
You're getting pegged with questions from two people XD
Ok, if you say so
So
Since $y=\sqrt x$ is concave down

@Nilknarf seems like $\lambda_n\in\mathcal O(2.23606797749979^n)$

and increasing, and positive

@LeakyNun $\varepsilon_0\times\Bbb N\times\Bbb N$???

then
$$0\le \sqrt{x+\delta}-\sqrt x\le \delta\frac{d}{dx}\sqrt x$$
correct?

:| ._. |: .-.
@Nilknarf probably

12:39 AM
so

Got a headache actually, so I'm not thinking much

@SimplyBeautifulArt afterall you need iteration

$$\sqrt x\le \sqrt{x+\delta}\le \frac{\delta}{2\sqrt x}+\sqrt x$$
Oh, sorry
I'll back off :P

Hm
I don't think $\lambda_n$ should be bounded.

No, it's not
But $L_n$ is, and $\lambda_n=\sqrt{5^n L_n}$
$L_n$ is easier to work with, since it is a simple iteration
and there is no additional term in terms of $n$

12:41 AM
oh
Ah
So $\lambda_n\in\mathcal O(5^{n/2})$
@LeakyNun could you be more specific?

Hm
Ah, yeah
Correct
But I want
$$\lambda_n=5^{n/2}+\mathcal O(\text{something})$$

@SimplyBeautifulArt well, we're supposed to have $f(\beta,n,1) = f(\lambda(\beta,n),n,1)$ where $\beta$ is a limit ordinal

no?

Oops
@LeakyNun wait why are there 3 arguments?

12:45 AM
@SimplyBeautifulArt because $f(\alpha+1,n,1) = f(\alpha,n,n)$

$$\mathcal O(1)\text{?}$$
Ah, I have two starred messages
I'm so proud XD

wait what are we doing? @LeakyNun
@Nilknarf Lol

@SimplyBeautifulArt fast-growing hierarchy

one moment brb

$$\mathcal O \kappa$$

12:49 AM
@Nilknarf we're stupid
Hey @mick
$$\lambda_n\approx\sqrt{L_\infty}5^{n/2}$$

Right...
and?
$$\Sigma \tau \upsilon \pi \iota \delta \text{?}$$

And...

drum roll...

$$\lambda_n=\sqrt{L_\infty}5^{n/2}+\mathcal O(1)$$

Naw
Surely we can do better than $\mathcal O(1)$

12:52 AM
Yaw
lol

XD
Would you believe me if I told you
that I found a closed-form
for

Probably

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

$a_0=2$

12:55 AM
Wait, I wanna tackle it

Hehe
Have fun
It's yucky

$$a_{n+1}^2-2a_{n+1}a_n+2a_n^2=13\cdot2^n$$
:|

Ok?

Looks horrible

It would be better

12:56 AM
Sh!

if you wrote
$$a_{n+1}^2=(a_n+\sqrt{13\cdot 2^n-a_n^2})^2$$
...and maybe did a little bit of sequence substitution?

Ok, no more hints, I promise

Welp
Too many radicals and squared terms

sigh

12:58 AM
Did enough thinking for now
x'D

Haha

I'll try later

I have a big chem test tomorrow

heh heh, it's a standardized test
don't worry
blah, gtg
cya

1:00 AM
cya
thnx
hey @LeakyNun

So, could you explain your function more for me?

sure
it's essentially the fast-growing hierarchy though

Well it's weird
And my brain isn't on straight

I just write down the index and the iteration as an argument

1:07 AM
So the first argument is the ordinal
The second and third arguments do what?

basically $f_\alpha^k(n)$ becomes $f(\alpha,n,k)$

Ah
That's fine @LeakyNun
Okay
Hm, what was the question again?

now how do we define $\lambda$ rigorously

36 mins ago, by Leaky Nun
of course I would also like to define the cofinal sequence $\lambda : \varepsilon_0 \times \Bbb N \to \varepsilon_0$ more rigorously
Uh
You mean like $\lambda[0]=1$ and $\lambda[n+1]=\omega^{\lambda[n]}$?

hmm
well
not that
31 mins ago, by Leaky Nun
@SimplyBeautifulArt well, we're supposed to have $f(\beta,n,1) = f(\lambda(\beta,n),n,1)$ where $\beta$ is a limit ordinal

1:15 AM
Oh
So like...
$$\lambda(\varepsilon_0,n)=\begin{cases}1,&n=0\\ \omega^{\lambda(\varepsilon_0,n-1)},&n>0\end{cases}$$
?

but $\beta$ can be other ordinals
actually
$\beta$ should be $\in \varepsilon_0$

:| Would you like to change that to $\beta\in\Gamma_0$?
Because I have all-encompassing fundamental sequences for all of those ordinals.

oh ok
sure

$$\lambda(\alpha,n)=\begin{cases}\alpha-1,& \alpha\in\omega\\ n,&\alpha=\omega \\ x,&\alpha=[x,y,z],y\cdot z=0\\ [[x,y, \lambda(z,n)], \lambda(y,n), x], & \alpha=[x,y,z], y\cdot z\ne0\end{cases}$$

what is $[]$?

1:22 AM
$$[x,y,z] = \begin{cases} x+1,& y\cdot z=0\\ \sup\{[[x,y,a],b,x]: a<z, b<y\} ,&\rm else\end{cases}$$
Fairly fancy notation
It encompasses all ordinals from $0$ to $\Gamma_0$.

interesting

I also ignored the case $\lambda(0,n)$.

what is $\varepsilon$?
@SimplyBeautifulArt oh you didn't: obviously $0-1=0$ :P

$$\varepsilon_x=[\omega,6,1+x]$$

where does $6$ come from

1:24 AM
No wait
My mistake
The 6 means that $\varepsilon$ is on the order of the 6th hyperoperation in my notation
$$\zeta_x=[\omega,6,1+x]$$

interesting

etc.
$[a,b,c]$ is really weird when $b=4,5$ though
$$[a,0,b]=a+1$$
$$[a,1,b]=a+2\cdot b$$

$[\omega,1,1]=\sup\{[[\omega,1,0],0,\omega]\} = [\omega+1,0,\omega] = \omega+2$

Yup
$[x,0,\dots]=x+1$
Yup
Just as I claim
$$[a,2,b]\approx a\cdot b$$
Kinda weird

interesting

1:28 AM
The approximation would be like saying that $[a,1,b]\approx a+b$
$$[a,3,b]\approx a^b$$
$$[a,4,b] \approx a^{a^b}$$
And from there I don't have closed forms, but I can make some evaluations.
All of this basically 1/6 SOAP notation
1/3 SOAP includes expressions such as $[\omega,\omega,\omega]$
i.e. infinite hyperoperator/middle number
But this choice of $\lambda$ is very abnormal

hmm
what is the fixed point of $f:n \mapsto [n,n,n]$? :P

There is no fixed point

why not?
is it not increasing?

$[x,x,x]$ is strictly greater than $[x,0,0]=x+1$
So you can't have $x=[x,x,x]$
But nesting it limits off to $\Gamma_0$.

is it not continuous?

1:32 AM
wdym?

is $f$ increasing and continuous?

Uh, yeah?

then it has a fixed point by some theorem

but you say it has no fixed point

1:33 AM
I meant that we didn't have any $x=[x,x,x]$

but we must have $x=[x,x,x]$ for some $x$

What's the fixed point of $f:x\mapsto x+1$?

it isn't continuous

Oh
then my notation isn't continuous

is $x \mapsto [\omega,x,x]$ continuous?

1:35 AM
Nope

how can I make it continuous?

You can't
Notation is specifically designed not to be continuous or have fixed points
I use it to program big numbers mate
I can't have fixed points cuz that always ruins growth

hmm
@SimplyBeautifulArt teach me more lol

lol
In a bit maybe
Okay
wat u want more :'(
The next that I know are ordinal collapsing functions
@LeakyNun

1:52 AM
go on

oh dear
they repeatedly apply operations over the ordinals
and return a big result
in a game lol

sorry, I'll be back tomorrow

@SimplyBeautifulArt I'll try to read your messages, in case you want to leave them here

Wow
I hit an HNQ
weird

19 hours later…
9:12 PM
@SimplyBeautifulArt hi

10:00 PM
17

Your job is to create the slowest growing function you can in no more than 100 bytes. Your program will take as input a nonnegative integer, and output a nonnegative integer. Let's call your program P. It must meet the these two criterion: Its source code must be less than or equal to 100 byt...

beat it with your gamma_0 function :P

2 hours later…
11:32 PM
@LeakyNun HA!

Oh, you are asking me to beat it?
That sir, I shall try to do.
Shoot
Gamma_0 program is too long
Oh right, I'm doing inverse function.
Gah
Too long
->n{r=->a{b,c,d=a;b ?a==b ?a-1:a==a-[0]?b:[[b,c,r[d]],c-1,b]:n};(h=[],n,k=1;(k+=1;h=r[h])until h==0)until k>n;k}
12 bytes too long
Nvm, it doesn't even work

11:53 PM
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 "1"
if len(a) == 1:
return "\u03c9^"+ord.str(a[0])
return "+".join(ord.str(x) for x in a)
def __init__(self, *val):
val = list(val)
assert(ord.check(val))
self.val = []
a=ord([],[[]],[])
b=ord([[]],[],[])
print(a)
print(b)
print(a<=b)
print(a<b)

I'm building ordinals smaller than epsilon_0

Python
Yeup

I do belong to the Python camp

11:57 PM
correction:
	def __init__(self, *val):
val = list(val)
assert(ord.check(val))
self.val = []
curr_max = 0
for a in val:
if curr_max != 0 and ord.le(a,curr_max):
self.val.append(curr_max)
if a != curr_max:
curr_max = 0
curr_max = a
self.val.append(val[-1])
would you like to test it out?
in particular the CNF

Nah I'm fine
I'm doing my own stuffs rn

alright