Talk:Xi function

I think this is where we start hitting the ceiling as to where googology can go.

Next stop: infinity. FB100Z &bull; talk &bull; contribs 23:00, February 5, 2013 (UTC)

Whatsoever, infinity is not reachable, because it is not a number. There is also no largest finite number. Take something like \(\Xi(\Xi(\Sigma(100)))\) = n. To prove that isn't the largest, just double it. We know that for any n>0: \(2n > n\), so n isn't the largest. Ikosarakt1 (talk) 08:59, February 6, 2013 (UTC)


 * Yeah, but \(\Xi(\Xi(\Sigma(100)))\) is ugly. Without creating a salad ordinal, where can we go after \(\omega^\text{CK}_{\omega^\text{CK}_{\omega^\text{CK}_{._{._.}}}}\)? FB100Z &bull; talk &bull; contribs 19:52, February 6, 2013 (UTC)

Probably it is \(\omega_1\), the first uncountable ordinal. Is it limit of sequence of \(\omega^\text{CK}_{\omega^\text{CK}_{\omega^\text{CK}_{._{._.}}}}\)? Ikosarakt1 (talk) 22:02, February 6, 2013 (UTC)

I doubt it. Any countable sequence of countable ordinals has a countable limit ordinal. Just think about it. If every ordinal in a countable sequence of ordinals is countable, you can show that the limit ordinal is countable simply by diagonalizing across w^2. It's basically impossible to reach w1 via "w" just like its impossible to reach "w" via finite arithmetic. Sbiis Saibian (talk) 23:26, February 6, 2013 (UTC)


 * I don't think they're equal, but if they are, then the xi function is the end of googology. If not, what is there between \(\omega^\text{CK}_{\omega^\text{CK}_{\omega^\text{CK}_{._{._.}}}}\) and \(\omega_1\)? FB100Z &bull; talk &bull; contribs 23:24, February 6, 2013 (UTC)


 * perhaps there is no established ordinal notation after this point. That would be a perfect opportunity for googology to do something new and invent some Sbiis Saibian (talk) 23:28, February 6, 2013 (UTC)
 * So, then, we come to a Big Question:

What's between computability theory and infinity?


 * FB100Z &bull; talk &bull; contribs 23:31, February 6, 2013 (UTC)


 * I don't think that's a question that can be answered. Robert Munafo makes the point that if we went beyond formalism and computability theory, we'd by definition have nothing to define. There probably is no hard limit that we are going to bump into in googology, but ... that will still all be within a formalism that we can't escape, and that will place a certain fundamental limit on what kinds of finite numbers we can express. The catch is, we'll probably never be able to describe the size of that box without barry paradoxes or worse. So some finite numbers will probably always be inaccessible to us. We know they must exist, but that's about all we can say without being self-contradictory. If it's true that we can't escape computability theory then it proves the point I was trying to make with my site: that saying we can "continue indefinitely" is not exactly true. Not at least in the sense that there is no fundamental limit Sbiis Saibian (talk) 23:41, February 6, 2013 (UTC)

Higher-order xi functions
Any guesses about the growth rates of the order-2 xi function, etc.? FB100Z &bull; talk &bull; contribs 00:20, February 7, 2013 (UTC)

Maybe \(f_{\alpha \times 2}(n)\), where \(\alpha\) is the ordinal for the normal Xi function? It is by analogy with \(\Sigma(n)\) and \(\Sigma_2(n)\). Ikosarakt1 (talk) 20:56, February 8, 2013 (UTC)

Certainly big underestimate. Adam Goucher admited he was wrong when he first wrote about strength of \(\Sigma_2(n)\). It is actually \(\omega_2^{CK}\), well over \(\omega_1^{CK} \times 2\). My personal guess, without any analysis, is as follows: consider function \(\alpha \rightarrow \omega_\alpha^{CK}\). Strength of Xi function is smallest fixed point of that function (which is guaranteed, because it's a normal function). But there are other fixed points - actually infinitely many of them. Let \(f_\alpha \) be \(\alpha\)-th such fixed point. Then I think strength of \(\Xi_2\) function is smallest fixed point of that function. If we define higher order combinators, then strengths of corresponding Xi functions would be created according to Veblen hierarchy based on \(\alpha \rightarrow \omega_\alpha^{CK}\).

But again, this is only my guess, but with some effort I think I could prove that it is lower bound LittlePeng9 (talk) 21:40, February 9, 2013 (UTC)
 * Stop it, you're making me dizzy! :) FB100Z &bull; talk &bull; contribs 00:59, February 10, 2013 (UTC)

Define the Veblen function with \(\varphi^\text{CK}_0(\alpha) = \omega^\text{CK}_\alpha\). That makes Goucher's ordinal \(\varphi^\text{CK}_1(0)\), and the ordinal you just defined is \(\varphi^\text{CK}_2(0)\). So, assuming your guess is correct, it seems natural that further xi functions would be \(\Xi_m(n) = O(f_{\varphi^\text{CK}_m(0)}(n))\). Eep. FB100Z &bull; talk &bull; contribs 17:11, February 21, 2013 (UTC)
 * Okay, challenge problem: devise a function with a growth rate of \(\Gamma^\text{CK}_0\) (the supremum of this "upper Veblen hierarchy"). FB100Z &bull; talk &bull; contribs 17:15, February 21, 2013 (UTC)
 * In every \(\Xi_m(n)\) we are allowed to use oracles with indexes up to and including m. But to define function with ordinal \(\varphi^\text{CK}_\omega(0)\) we need to alter this a bit. Going to hypercomputability makes weird things to ordinal properties. Namely, limit of admissibles isn't always admissible (e.g. \(\omega_\omega^\text{CK}\) isn't admissible), similarly there is no \(\Omega_\omega\) combinator. But it's not a problem - if we allow every possible \(\Omega_n\) combinator to appear corresponding \(\Xi_\omega(n)\) will be what we want. This is what we'll do for limit ordinals. Standard oracles will work for succesors (i.e. \(\Omega_{\omega+1}\) will do). Using this we can ad hoc create function you wished. But it isn't what you expected, is it? Don't worry, right now I'm trying to find way to make function reaching \(\Gamma^\text{CK}_0\), and even further! (Have we already reached point at which we'll call it hyper-hypercomputability?) LittlePeng9 (talk) 18:47, February 21, 2013 (UTC)

Values
We know some exact values and lower bounds for \(\Sigma(n)\), why not make it for \(\Xi(n)\)? At least, what is the value of \(\Xi(1)\)? (I believe it is not so unimaginably large). Ikosarakt1 (talk ^ contribs) 20:12, May 18, 2013 (UTC)
 * For Xi(1) we must have only one combinator. Single combinator can't beta reduce, so Xi(1)=1. If we have 2 combinators, either left one is I and tree reduces to single combinator, or we can't reduce. So Xi(2)=2. Similarily Xi(3)=3. I think also Xi(4)=4, although I didn't check all possibilities. LittlePeng9 (talk) 21:41, May 18, 2013 (UTC)
 * Good, but maybe at least \(\Xi(10)\) is already extremely large? Ikosarakt1 (talk ^ contribs) 21:45, May 18, 2013 (UTC)


 * I doubt it. I think we need more combinators to make good use of oracle, so for Xi(10) we don't have to take much care about it. LittlePeng9 (talk) 07:24, May 19, 2013 (UTC)


 * For \(\Xi(4)\), we observe that reducing I and K always reduce the length of the string, and reducing S will keep the string at the same length, unless z has length greater than 1. But for a length 4 string you can't have z greater than 1, so \(\Xi(4) = 4\). This also reduces the possibilities for \(\Xi(5)\), so perhaps we can prove that \(\Xi(5) = 6\). Deedlit11 (talk) 09:44, May 19, 2013 (UTC)


 * Now I see that \(\Xi(n)\) doesn't "explodes" so fast. Ikosarakt1 (talk ^ contribs) 09:43, May 19, 2013 (UTC)


 * Suppose leftmost combinator in n-combinator tree is I, K or \(\Omega\). For every case, after single reduction we have smaller tree which will leave same output. So, if we want to make an improvement from smaller trees, leftmost combinator must be S. But then, to make proper reduction, we need tree to have height at least 4 (including root). It makes things much easier for small trees. LittlePeng9 (talk) 10:15, May 19, 2013 (UTC)


 * But \(\Xi(10^6)\) more than for example BB(TREE(3)) or BB(BB(..BB(1)...))? Konkhra (talk) 11:20, May 19, 2013 (UTC)
 * \(\Xi(10^6)\) is very likely to be larger than BB(TREE(3)) or BB(BB(..BB(1)...)). By the way, BB(BB(..BB(1)...)) = 1 because BB(1) = 1, BB(BB(1)) = BB(1) = 1, etc. -- I want more clouds! 11:35, May 19, 2013 (UTC)
 * Xi is the current winner. $Jiawhein$\(a\)\(l\)\(t\) 12:17, May 19, 2013 (UTC)


 * Rayo's function is way way way WAY faster growing. LittlePeng9 (talk) 12:36, May 19, 2013 (UTC)


 * I can confirm that \(\Xi(5)=6\), and there are 160 trees reaching this value - trees of form S(Sx)ab or Sxa(bc) where a,b,c are any combinators and x is either S or \(\Omega\). 6-combinator trees are smallest which can explode to infinity, so analysis there may be harder. I found \(\Xi(6)\geq 9\). LittlePeng9 (talk) 13:16, May 19, 2013 (UTC)

It's interesting how the really fast-growing functions often have extremely slow starts. --Ixfd64 (talk) 17:46, May 19, 2013 (UTC)