1:32 AM
Meanwhile I need to figure out how to make Asaf more weird, cause past experience told me that I often have trouble explaining what I am doing to non weird people
I don't understand what you are doing, or why you expect the limit to be an open set. But in any case, writing $\lim_{k\to\aleph_0}$ is making me cringe. — Asaf Karagila 7 hours ago
2:13 AM
2 days ago, by Simply Beautiful Art
$$[x,y,z] = \begin{cases} x+1,& y\cdot z=0\\ \sup\{[[x,y,a],b,x]: a<z, b<y\} ,&\rm else\end{cases}$$
1 hour later…
4:19 AM
$\sup (n_4 < \omega : \varphi (n_4,\sup(n_3 : < \omega : \sup (n_2 < \omega : \sup (n_1 < \omega : n_1)+n_2)n_3)))$
after this point, I am going to omit all $n_i < \omega$ and they will be implied by the nesting of the sups
$\sup([\varphi(\cdot ,\sup(\varphi(,\sup(\sup(n_1)+n_2)n_3))] \circ^{n_5} \sup(\varphi(n_4,\sup(\sup(n_1)+n_2)n_3)))$
$\sup (\varphi \binom {n_6}{\sup([\varphi(\cdot ,\sup(\varphi(,\sup(\sup(n_1)+n_2)n_3))] \circ^{n_5} \sup(\varphi(n_4,\sup(\sup(n_1)+n_2)n_3)))})$
$\sup ([\varphi \binom{1}{\cdot}]\circ^{n_7} \sup(\varphi \binom {n_6}{\sup([\varphi(\cdot ,\sup(\varphi(,\sup(\sup(n_1)+n_2)n_3))] \circ^{n_5} \sup(\varphi(n_4,\sup(\sup(n_1)+n_2)n_3)))}))$
$\sup (\varphi_{n_8}\binom{1}{\sup ([\varphi \binom{1}{\cdot}]\circ^{n_7} \sup(\varphi \binom {n_6}{\sup([\varphi(\cdot ,\sup(\varphi(,\sup(\sup(n_1)+n_2)n_3))] \circ^{n_5} \sup(\varphi(n_4,\sup(\sup(n_1)+n_2)n_3)))}))}))$
$\sup (\varphi_{\cdot}\binom{1}{\sup (\varphi_{n_8}\binom{1}{\sup ([\varphi \binom{1}{\cdot}]\circ^{n_7} \sup(\varphi \binom {n_6}{\sup([\varphi(\cdot ,\sup(\varphi(,\sup(\sup(n_1)+n_2)n_3))] \circ^{n_5} \sup(\varphi(n_4,\sup(\sup(n_1)+n_2)n_3)))}))}))}) \circ^{n_9} \sup (\varphi_{n_8}\binom{1}{\sup ([\varphi \binom{1}{\cdot}]\circ^{n_7} \sup(\varphi \binom {n_6}{\sup([\varphi(\cdot ,\sup(\varphi(,\sup(\sup(n_1)+n_2)n_3))] \circ^{n_5} \sup(\varphi(n_4,\sup(\sup(n_1)+n_2)n_3)))}))}))$
$0, \omega, \omega 2, \omega^2, \psi (\Omega^{\omega} + \omega^2), \psi (\Omega^{\Omega} + \psi (\Omega^{\omega} + \omega^2)), \psi (\Omega^{\Omega^{\omega}}+\psi (\Omega^{\Omega} + \psi (\Omega^{\omega} + \omega^2))), \psi (\Omega^{\Omega^{\Omega}}+\psi (\Omega^{\Omega^{\omega}}+\psi (\Omega^{\Omega} + \psi (\Omega^{\omega} + \omega^2))), $
actually, now that I get a better feel of the pattern, I can rewrite this mess more concisely in terms of $\psi$
The function grows like this:
$0,\omega,\omega 2, \omega^2, \psi (\Omega + \omega^2), \psi (\Omega^{\omega} + \psi (\Omega + \omega^2)), \psi (\Omega^{\Omega^{\omega}}+ \psi (\Omega^{\omega} + \psi (\Omega + \omega^2))$ and so on...
$0,\omega,\omega 2, \omega^2, \psi (\Omega + \omega^2), \psi (\Omega^{\omega} + \psi (\Omega + \omega^2)), \psi (\Omega^{\Omega^{\omega}}+ \psi (\Omega^{\omega} + \psi (\Omega + \omega^2))$ and so on...
$f(n) = \begin{cases} n+1, \text{n is a sup or n is 0}\\ \sup (something (n)), \text{otherwise}\end{cases}$
$f(3) = \sup (\omega^{\cdot} \circ^n f(2)) = \sup (\omega,\omega^{\omega+1},\omega^{\omega^{\omega +1}},...) = \epsilon_0 \prod_{n < \omega}{}^{n}\omega$
$f(7) = \sup (\varphi (\cdot, f(6)) \circ^n f(6)) = \Gamma_0 = \varphi (1,0,0) = \psi (\Omega^{\Omega})$
$f(9) = \sup (\varphi \binom{n}{f(8)}) = \varphi \binom{\omega}{1} = SVO = \psi (\Omega^{\Omega^{\omega}})$
6:04 AM
We can prove $\sup (f(n))$ is computable by induction since for all $n < \omega$, $f(n)$ is a finite sequence of computable function and successors, thus the result is computable. Then for $\omega = \sup (n)$, since all finite cases are computable, and the $\omega$th case is just a union of them all, it follows $f(\omega)$ should be computable, even if we don't have a way to write it down
6:37 AM
While each $g_n$ is computable hence terminate in finite steps, the whole sequence is countable, thus by definition is the same as not terminating. Therefore $f(\omega)$ will run forever unless given countable steps. Thus $f(\omega)$ is uncomputable
But the top one seems more suitable as it grows faster and $g_{n}$ is required to be a function that grows faster than the one in the previous stage
In fact, even a tower of $\omega_{\cdot}^{CK}$ is not fast enough,as after finite number of steps, it will get stuck
To be fast enough so that it does not get stuck in finite number of steps, we need something like $f(\omega_1^{CK})$. This is well defined as having $\omega_1^{CK}$, we can collapse it down to obtain all the countable ordinals (which will be given by some finite sequence of computable functions)
The reason I can get to $\omega_1^{CK}$ despite using only computable functions is because it is the function hierarchy $\{g_0,g_1,g_2,..\}$ is uncomputable.
Suppose we instead define the hierarchy to be one which adds two levels of $\Omega$ into the $\psi$ for each n, then the nth stage will be the notation $\psi ({}^{n}\Omega)$ which is computable and thus supremum can easily bring it to the $\omega$th level with the notation $\psi ({}^{\omega}\Omega) = BHO$ and be computable (thus never reaching $\omega_1^{CK})$
4
I was wondering whether there exists a (computable) sequence of numbers, for which it can be proven that no closed form can exist. Edit: By closed form I mean an expression involving only a constant number of elementary functions. So something like a sum can not occur in the expression.
1
It is always easy to forge recurrence relations. E.g. $$a_{n+1}=2a_n+\dfrac{1}{a_n}, a_0=1$$ But it is always hard to find the general closed form expression. And it is even harder to prove that there is no such expression. Another example is here. And I highly suspect sequences like $$a_n = \...
15 hours later…
11:41 PM
2 days ago, by Simply Beautiful Art
$$[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}$$
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Dec '179
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Simply Beautiful Art's realm of calcu…
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