User:Vel!/Kleene's O

Observe as FB100Z blunders through Kleene's O.


 * \(\mathcal{O}(0) = 0\)
 * \(\mathcal{O}(2^0) = \mathcal{O}(1) = 1\)
 * \(\mathcal{O}(2^1) = \mathcal{O}(2) = 2\)
 * \(\mathcal{O}(2^2) = 3\)
 * \(\mathcal{O}(2^{2^2}) = 4\)
 * \(\mathcal{O}(2 \uparrow\uparrow n) = n + 1\)
 * \(0 <_O 1 <_O < 2 <_O < 2^2 <_O 2^{2^2} <_O \ldots\)
 * \(\varphi_0(n) := 2 \uparrow\uparrow n\) (say)
 * Since \(\varphi_0(n)\) is a notation and \(\varphi_0(n) <_O \varphi_0(n + 1)\), \(\mathcal{O}(3 \cdot 5^0) = \mathcal{O}(3)\) exists.
 * \(\mathcal{O}(3) = \sup\{\mathcal{O}(0), \mathcal{O}(1), \mathcal{O}(2), \mathcal{O}(2^2), \ldots\} = \sup\{0, 1, 2, 3, \ldots\} = \omega\)
 * \(0, 1, 2, 2^2, 2^{2^2}, \ldots <_O 3\)
 * \(\mathcal{O}(3) = \omega\)
 * \(\mathcal{O}(2^3) = \omega + 1\)
 * \(\mathcal{O}(2^{2^3}) = \omega + 2\)
 * \(\mathcal{O}(2^{2^{2^3}}) = \omega + 3\)
 * \(\mathcal{O}([2 \uparrow\uparrow n]^3) = \omega + n\), and you can probably figure out what \([a \uparrow\uparrow b]^c\) means
 * \(3 <_O 2^3 <_O < 2 <_O < 2^2 <_O 2^{2^2} <_O \ldots\)
 * \(\varphi_1(n) := [2 \uparrow\uparrow n]^3\) (say &mdash; I'm providing contrived definitions for the sake of simplicity)
 * All \(\varphi_0(n)\) are notations and \(\varphi_1(n) <_O \varphi_1(n + 1)\), so \(\mathcal{O}(3 \cdot 5)\) exists.
 * \(\mathcal{O}(3 \cdot 5) = \sup\{\mathcal{O}(3), \mathcal{O}(2^3), \mathcal{O}(2^{2^3}), \mathcal{O}(2^{2^{2^3}}), \ldots\} = \sup\{\omega, \omega + 1, \omega + 2, \ldots\} = \omega \times 2\)
 * \(3, 2^3, 2^{2^3}, 2^{2^{2^3}}, \ldots <_O 3 \cdot 5\)