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11:01 PM
the fun stuff beings up to and above $\psi(\Omega^{\Omega^\Omega})$ , below we can use $\varphi(...)$
 
oops mistake
 
np :)
 
11:17 PM
@SimplyBeautifulArt: Did you call out for me?
 
Howdy
 
@chronolegends $$\psi(\Omega^{\Omega^\Omega}) =\varphi(1, \underbrace{0,\dots,0}_{\varphi(1, \underbrace{0,\dots,0}_{\varphi(1, \underbrace{0,\dots,0}_{\varphi(1, \underbrace{0,\dots,0}_{\vdots})})})})$$
 
hahahahaha, nice
 
11:32 PM
\begin{align}
\psi (\Omega)[n,\Omega] & = \psi^{n+1}(\omega) = \epsilon_{\ddots\epsilon_{\omega}}\\
\psi (\Omega +1)[n,\Omega +1] = \psi(\Omega +1[n,\Omega+1]) & = \psi(\Omega) = \zeta_0\\
\psi (\Omega +(m+1))[n,\Omega +(m+1)] = \psi(\Omega +(m+1)[n,\Omega+1]) & = \psi(\Omega+m) = \epsilon_{\zeta_0+m}\\
\psi (\Omega + \Omega)[n,\Omega + \Omega] = \psi(\Omega +\Omega[n,\Omega+\Omega]) & = [\psi(\Omega +]^{n+1}(\omega) = \zeta_1\\
\psi (\Omega2 + \Omega)[n,\Omega2 + \Omega] = \psi(\Omega2 +\Omega[n,\Omega2+\Omega]) & = [\psi(\Omega 2 +]^{n+1}(\omega) = \zeta_2\\
 
Oh dear.
@Secret I'd prefer to write "$\to$" instead of the last equality. You should be considering $n\to\omega$ there.
I assume you are asking "What's $\psi(\Omega^22)$ equal to?"
Before you just so far, consider the following pattern:
$\psi(x)=\varepsilon_x$
$\psi(\Omega(1+a)+b)=\varepsilon_{\zeta_a+b}$
$\psi(\Omega^2(1+a)+\Omega(1+b)+c)=\varepsilon_{ \zeta_{?+b}+c}$
 
try to avoid using minus signs when handling ordinals
 
@chronolegends Huh? The only minus signs I see are in relation to $n$, which should be a natural number.
Logically, the $?$ is a function of $a$, and it's probably the natural continuation of the sequence $\varepsilon,\zeta,?,\dots$
 
I know, there isn't a problem with the way he did but, its just like a general use advice
 
@SimplyBeautifulArt Is $\Omega^m$ where $m < \omega$ the m-th fixed point map in the Veblen hierarchy (i.e. $\eta_a$ for $m=2$)?
 
11:39 PM
@chronolegends >.> I feel like you say that every time you say something you want to take back.
 
I don't take it back
lol
 
@Secret Actually, you can just let $m<\Gamma_0$.
And yes, that is how it goes.
 
ok
 
Pretty nifty huh?
 
Into the Veblen hierarchy:
 
11:40 PM
But wait
 
He's diving into it
exciting
 
You still didn't answer my questions!!!
56 mins ago, by Simply Beautiful Art
(in terms of more natural numbers, $\omega$, $\Omega$ and $\psi$'s, but not reducing to anything else)
What did $\psi(\Omega)$ expand into?
What did $\psi(\Omega+\Omega)$ expand into?
What do you think $\psi(\Omega2+\Omega)$ will expand into?
Oh, well you did...
 
That's where I align all the equal signs
 
I meant to ask a different underlying question about all this before you moved on!
That is.
What effect do you think $\Omega$ has?
 
$\Omega$ tracks the type of fixed point mapping involved?
e.g. Given some fixed point map $X_y$, $\Omega^X y$ controls the map involved using $X$ and $y$
 
11:44 PM
Yup
And so...
It should almost feel natural how this ties into the Veblen function.
 
give him $\Gamma_0$ now
 
he's ready
 
lol, okay
@Secret Expand $\psi(\Omega^\Omega)$
@chronolegends See, if you had put those two lines in one I would've starred it.
 
hahaha
 
11:46 PM
:P
Knowledge from the future and your intuition:
$$\psi(\Omega^x)=\varphi_{1+x}(0), 0<x<\Omega$$
For most large enough $x$, this reduces down to
$$\psi(\Omega^x)=\varphi_x(0)$$
@chronolegends hope I'm not spoon-feeding him the answers.
 
No, thats good pacing
i'll throw in, $\Gamma_0 = \varphi_{..._{\varphi(0)}}$
 
@chronolegends Horrible syntax.
 
lol fixed
 
Horrendous syntax.
Nah mate.
 
11:52 PM
Those look like single argument Veblen functions
Like what is that
 
they are
$\varphi_{..._{\varphi(0)}(0)}(0)$ better?
 
@chronolegends BLEH
Use \ddots for diagonalized dots.
 
nice
k
 
Also
the number of φ's does not match the number of $(0)$'s
 

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