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12:03 AM
@SimplyBeautifulArt are you here?
 
@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
 
hi
 
12:28 AM
@LeakyNun Yup. I'm a blind bat
O_O
 
@SimplyBeautifulArt Wanna do another asymptotic expansion problem?
 
Big-O is giving me trouble :P
 
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
 
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
 
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?
$$\Sigma \tau \upsilon \pi \iota \delta \text{?}$$
 
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
 
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
hope your headache goes away :(
 
1:00 AM
cya
thnx
hey @LeakyNun
 
hi
 
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 $f(\alpha,n,k)$
please do ping me
 
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
So ocf's usually start with a set of ordinals
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
 
It's about my bedtime too
Wow
I hit an HNQ
weird
 
 
19 hours later…
9:12 PM
@SimplyBeautifulArt hi
 
10:00 PM
17
Q: Create the slowest growing function you can in under 100 bytes

PyRulezYour 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!
 
:P
 
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
 

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