Talk: Slow-growing hierarchy

Revision as of 07:56, 24 February 2020

I've tried to create comparisons with BEAF and SGH, assuming that $g_{\varepsilon_0}(n) = n \uparrow\uparrow n$, $g_{\zeta_0}(n) \approx n \uparrow\uparrow\uparrow n$, $g_{\eta_0}(n) \approx n \uparrow\uparrow\uparrow\uparrow n$, etc. First two comparisons are correct, but I have doubts about $g_{\eta_0}(n) \approx n \uparrow\uparrow\uparrow\uparrow n$, although it seems reasonable. Just look at these comparisons past $\zeta_0$:

$g_{\zeta_0}(n) \approx n \uparrow\uparrow\uparrow n$

$g_{\zeta_0^{\zeta_0}}(n) \approx (n \uparrow\uparrow\uparrow n) \uparrow\uparrow 2$

$g_{\varepsilon_{\zeta_0+1}}(n) \approx (n \uparrow\uparrow\uparrow n) \uparrow\uparrow n$

$g_{\varepsilon_{\zeta_0+2}}(n) \approx ((n \uparrow\uparrow\uparrow n) \uparrow\uparrow n) \uparrow\uparrow n \approx (n \uparrow\uparrow\uparrow n) \uparrow\uparrow (2n)$ (using LAPL).

$g_{\varepsilon_{\zeta_0+3}}(n) \approx (n \uparrow\uparrow\uparrow n) \uparrow\uparrow (3n)$

$g_{\varepsilon_{\zeta_0+\omega}}(n) \approx (n \uparrow\uparrow\uparrow n) \uparrow\uparrow (n^2)$

$g_{\varepsilon_{\zeta_0+\omega^\omega}}(n) \approx (n \uparrow\uparrow\uparrow n) \uparrow\uparrow (n^n)$

$g_{\varepsilon_{\zeta_0+\varepsilon_0}}(n) \approx (n \uparrow\uparrow\uparrow n) \uparrow\uparrow (n \uparrow\uparrow n)$

$g_{\varepsilon_{\zeta_0 2}}(n) \approx (n \uparrow\uparrow\uparrow n) \uparrow\uparrow (n \uparrow\uparrow\uparrow n) \approx n \uparrow\uparrow\uparrow (n+1)$

$g_{\varepsilon_{\zeta_0 3}}(n) \approx n \uparrow\uparrow\uparrow (n+2)$

$g_{\varepsilon_{\zeta_0 \omega}}(n) \approx n \uparrow\uparrow\uparrow (2n)$

$g_{\varepsilon_{\zeta_0 \omega 2}}(n) \approx n \uparrow\uparrow\uparrow (3n)$

$g_{\varepsilon_{\zeta_0 \omega^2}}(n) \approx n \uparrow\uparrow\uparrow (n^2)$

$g_{\varepsilon_{\zeta_0 \omega^\omega}}(n) \approx n \uparrow\uparrow\uparrow (n^n)$

$g_{\varepsilon_{\zeta_0 \varepsilon_0}}(n) \approx n \uparrow\uparrow\uparrow (n \uparrow\uparrow n)$

$g_{\varepsilon_{\zeta_0^2}}(n) \approx n \uparrow\uparrow\uparrow (n \uparrow\uparrow\uparrow n)$

I'm not sure, but isn't it that $g_{\varepsilon_{\zeta_0^\omega}}(n) \approx n \uparrow\uparrow\uparrow\uparrow n$?

Ikosarakt1 (talk ^ contribs) 15:59, April 4, 2013 (UTC)

Other

Can add more? $Jiawhein$ ^{$a$$l$$t$} 11:00, April 21, 2013 (UTC)

I think things from SVO to $\theta(\Omega^\omega+1)$ are most likely to cause misunderstanding. Almost everybody thinks $g_{\theta(\Omega^\omega)}(n)\approx f_{\omega^\omega}(n)\approx\{n,n(1)2\}$. However, some people think $g_{\theta(\Omega^\omega+1)}(n)\approx f_{\omega^\omega+1}(n)$ and some think $g_{\theta(\Omega^\Omega+1)}(n)\approx f_{\omega^\omega+1}(n)$. That's why some people think SGH catches up FGH at $\psi(\Omega_\omega)$ but some think it LVO. {hyp<hyp··cos>cos} (talk) 01:50, October 10, 2013 (UTC)

Fundamental sequences

Can you guys state exactly what fundamental sequences you are using? It's important that we are all on the same page. Deedlit11 (talk) 11:34, April 21, 2013 (UTC)

I used the same fundamental sequences as for fast-growing hierarchy. By the way, I made a mistake in my previous comparisons, now I improved it. Ikosarakt1 (talk ^ contribs) 11:41, April 21, 2013 (UTC)

Okay, but what fundamental sequences are those? It's rather important that we know the details. Deedlit11 (talk) 12:11, April 21, 2013 (UTC)

Here is the list of some of them that I believe to be reasonable:

$\omega = lim(1,2,3,\cdots)$

$\varepsilon_0 = lim(\omega,\omega^\omega,\omega^{\omega^\omega},\cdots)$

$\varepsilon_1 = lim(\varepsilon_0,\varepsilon_0^{\varepsilon_0},\varepsilon_0^{\varepsilon_0^{\varepsilon_0}},\cdots$

$\zeta_0 = lim(\varepsilon_0,\varepsilon_{\varepsilon_0},\varepsilon_{\varepsilon_{\varepsilon_0}},\cdots)$

$\zeta_1 = lim(\varepsilon_{\zeta_0+1},\varepsilon_{\varepsilon_{\zeta_0+1}},\varepsilon_{\varepsilon_{\varepsilon_{\zeta_0+1}}},\cdots)$

$\eta_0 = lim(\zeta_0,\zeta_{\zeta_0},\zeta_{\zeta_{\zeta_0}},\cdots)$

$\varphi(\omega,0) = lim(\varepsilon_0,\zeta_0,\eta_0,\cdots)$

$\varphi(1,0,0) = lim(\varepsilon_0,\varphi(\varepsilon_0,0),\varphi(\varphi(\varepsilon_0,0),0))$

$\varphi(1,0,0,0) = lim(\varphi(1,0,0),\varphi(\varphi(1,0,0),0,0),\varphi(\varphi(\varphi(1,0,0),0,0),0,0))$

$\vartheta(\Omega^\omega) = lim(\vartheta(\Omega) = \Gamma_0,\vartheta(\Omega^2),\vartheta(\Omega^3),\cdots)$

I believe that it is easy to fill the intermediate terms, for example $\zeta_6 = lim(\varepsilon_{\zeta_5},\varepsilon_{\varepsilon_{\zeta_5}},\varepsilon_{\varepsilon_{\varepsilon_{\zeta_5}}},\cdots)$, following this pattern.

Ikosarakt1 (talk ^ contribs) 13:19, April 21, 2013 (UTC)

Catching ordinal

Is the official opinion wrong about catching ordinal with SGH and FGH? It really turns out for me that it is LVO. Ikosarakt1 (talk ^ contribs) 11:39, April 21, 2013 (UTC)

I'm pretty sure they aren't wrong. There are papers on the subject available on JSTOR, if you have access. Some papers are here and here. Deedlit11 (talk) 12:30, April 21, 2013 (UTC)

We can notice the effect from replacing all omega into n's in FGH: it gives very good approximation in googological sense. Chris Bird determined that at LVO ordinal numbers in his array notation catches up the finite numbers: his separator $[1 [1 \neg 1 \neg 2] 2]$ has level LVO, and LVO is also equal to $\{\omega,\omega [1 [1 \neg 1 \neg 2] 2]\}$, when visualised in array notation. It means that $g_{\vartheta(\Omega^\Omega)}(n) \approx \{n,n [1 [1 \neg 1 \neg 2] 2] 2\}$, and $f_{\vartheta(\Omega^\Omega)}(n) \approx \{n,n [1 [1 \neg 1 \neg 2] 2] 2\}$, because Bird's array hierarchy grows roughly as fast as fast-growing one, and at LVO they are catched up. It means that catching ordinal for SGH and FGH is $\vartheta(\Omega^\Omega)$.

Also I can say that LVO is just first ordinal when SGH catches up FGH. After it, two hierarchies start to be very different again.

Ikosarakt1 (talk ^ contribs) 13:28, April 21, 2013 (UTC)

I think your result may indeed be right. We all know how SGH is sensitive to fundamental sequences. Bird used sequences based on collapsing function and both papers mentioned by Deedlit use sequences based on abstract tree-representations, which, while much stronger, may lead to diametrally different sequences. LittlePeng9 (talk) 14:31, April 21, 2013 (UTC)

I don't know about that. It's possible, but I don't think Ikosarakt's analysis is deep enough to say for sure that it is correct. It depends on a case by case analysis of succeeding ordinals using intuition to guide the way. I think what is needed is a rigorous proof. I understand this is hard to come by, but until we have one it is better to stick by the established results. Note that the Wainer paper lists a bunch of papers by a bunch of different authors that prove that F_{epsilon_0} is at the same level as H_{BHO}, and I don't think they all use tree ordinals - take this paper, for instance. Deedlit11 (talk) 15:02, April 21, 2013 (UTC)

I also personally doubt Ikosarakt's result, I just said it might be true. One big lack in his "proof" is that we don't know if "replace $\omega$'s with n's" extends that far into hierarchy. It may break near Feferman-Schutte ordinal, for example. By the way, I think you meant G_{BHO}, as Hardy hierarchy meets Wainer hierarchy at epsilon_0. LittlePeng9 (talk) 15:15, April 21, 2013 (UTC)

No matter what, we can always mangle the fundamental sequence so everything turns out okay.

I believe that under the standard definition of the Veblen hierarchy, $\Gamma_0$ is actually the first fixed point of $\alpha \mapsto \omega\ \{\alpha + 1\}\ \omega$, so it's just a hair away from expansion. Of course we can just change the Veblen function so it really does come out to $\omega\ \{\{1\}\}\ \omega$. FB100Z • talk • contribs 15:22, April 21, 2013 (UTC)

It's actually the Ackermann ordinal. 80.98.179.160 10:08, January 12, 2018 (UTC)

But why it is not related to Chris Bird's proved fact that $f_{\omega,\omega [1 [1 \neg 1 \neg 2] 2] 2}(n) = f_{\text{LVO}}(n) \approx \{n,n [1 [1 \neg 1 \neg 2] 2] 2\}$? Also, why you said that replacing all $\omega$'s to n's in SGH is not true after some ordinal? SGH must behave equally with any ordinals, since it is defined equally for all ordinals. I shall, of course, develop comparisons with SGH and Bird's array notation more carefully. Ikosarakt1 (talk ^ contribs) 19:08, April 21, 2013 (UTC)

FB100Z pointed out that $\Gamma_0$ isn't equal to $\omega\ \{\{1\}\}\ \omega$. It is in fact close, but it isn't exactly the same. Going even further difference may become much more significant. And when we reach level when we can't use Veblen hierarchy anymore, there is no single definition of fundamental sequence then, so $\omega$'s to n's may not work for all of them. LittlePeng9 (talk) 19:41, April 21, 2013 (UTC)

Fundamental sequences

Why the most googologists think that minor changes for fundamental sequences can significantly affect the growth rates? For example, we know that fund. sequence for $\epsilon_1$ can be defined in two significantly different ways, and both will lead to $f_{\epsilon_1}(n) \approx n \uparrow\uparrow (2n)$. Ikosarakt1 (talk ^ contribs) 14:21, July 2, 2013 (UTC)

I believe you heard result that $f_{\varepsilon_0}\approx g_{BH}$ under standard sequences. There are definitions of fundamental sequences based on tree representations such that $f_{\varepsilon_0}\approx g_{\varepsilon_0}$. LittlePeng9 (talk) 14:46, July 2, 2013 (UTC)

What exactly these sequences? I only read that they were defined, but no clear definitions I've seen so far. Ikosarakt1 (talk ^ contribs) 14:49, July 2, 2013 (UTC)

I think this is what you look for, but I can't check it right now. LittlePeng9 (talk) 14:56, July 2, 2013 (UTC)

We know that the growth rate of the slow-growing hierarchy is very sensitive to changes in the fundamental sequences mainly because of Andreas Weiermann, who wrote several papers on the subject. For example, you can read his paper "Sometimes Slow Growing is Fast Growing", or his paper "An Extremely Sharp Phase Transition Threshold for the Slow Growing Hierarchy". The latter is available via the internet, do a Google search. For an elementary proof that $f_{\varepsilon_0}\approx g_{BHO}$ using standard fundamental sequences, look at the paper "The Slow-growing and Gregorczyk Hierarchies" by S.S. Wainer, available on jstor.org - you just have to sign up, and then you can read up to three papers at a time. You can also read Wainer's paper "Slow Growing versus Fast Growing", also available on jstor.org, which proves that the Slow Growing Hierarchy first catches up to the Fast Growing Hierarchy at $\psi_0 (\Omega_\omega)$. Deedlit11 (talk) 16:11, July 2, 2013 (UTC)

Note that the result that $f_{\varepsilon_0}\approx g_{\varepsilon_0}$ uses Cichon's normed based fundamental sequences. The paper that uses tree ordinals ("Slow Growing versus Fast Growing") proves the standard results, that $f_{\varepsilon_0}\approx g_{BHO}$ and $f_{\psi_0 (\Omega_\omega)}\approx g_{\psi_0 (\Omega_\omega)}$. Deedlit11 (talk) 16:11, July 2, 2013 (UTC)

1 rule SGH

There's a variant of FGH with 1 basic rule instead of 3:

\(g_{\alpha+m}(n) = g_{\alpha[m+n]}(n)</math>

It works if we set 0[n] = 0. Ikosarakt1 (talk ^ contribs) 09:25, June 21, 2014 (UTC)

Not quite right, I think you need:

Failed to parse (syntax error): {\displaystyle g_{\alpha+m}(n) = g_{\alpha[n]}(n) + m\) :Also, you still need the second rule \(g_0(n) = 0} . With just the one rule, you can only set g equal to another expression with g, so you will keep evaluating forever. Deedlit11 (talk) 11:15, June 21, 2014 (UTC)

Okay, thanks. I'm pretty satisfied with reducing it to 2 rules. By the way, we can make the same thing with Hardy Hierarchy, using \(H_{\alpha+m}(n) = H_{\alpha[n]}(n+m)</math>. Ikosarakt1 (talk ^ contribs) 05:50, June 24, 2014 (UTC)

That should be \(H_{\alpha+m}(n) = H_{\alpha[n+m]}(n+m)</math>. Deedlit11 (talk) 12:58, June 25, 2014 (UTC)

OBJECTION!

SGH goes beyond X^^X & n.

Go to the FGH talk page, section What happened?

Why are values here defined beyond X^^X & n?

Should they be not well-defined?

-- A Large Number Googologist -- 19:54, October 11, 2014 (UTC)

They are well-defined, see this site. They say that $\omega\uparrow^n\omega=(n+1)\varphi0$ iff n<ω, $n\varphi0$ if n=ω, Γ0 if n=ω+1. Also, this site gives an example about {ω,2,2,2}={ω,ω,1,2}. 80.98.179.160 10:03, January 12, 2018 (UTC)

Ordinals written by omega

\(g_{\gamma}(\omega)={\gamma}</math>,right?

In other words, I want to say that SGH shows how big the ordinal is by omega which is replaced by x. --Nayuta Ito (talk) 10:07, October 25, 2014 (UTC)

SGH isn't really defined for transfinite ordinals. You can suggest a definition and then we can talk. LittlePeng9 (talk) 10:15, October 25, 2014 (UTC)

For example, \(g_{\omega^{\omega}}(x)=x^x</math>. Then, what should \(g_{\omega^{\omega}}(\omega)</math> be? It must be

\omega^{\omega}

. I can't imagine other definitions for ordinals.--Nayuta Ito (talk) 01:22, October 26, 2014 (UTC)

No, the equation $g_{\omega^{\omega}}(x)=x^x$ is only valid for nonnegative integers x. The domain of every function in the hierarchy is defined to be $\mathbb{N} \mapsto \mathbb{N}$, and therefore ordinals in general are not valid arguments. If you want to modify the hierarchy to accept ordinals, then that's okay but you have to define what you're doing. it's vel time 01:27, October 26, 2014 (UTC)

What I want to say is from ordinal to ordinal. This may work:

g_{\gamma }({\alpha })

is a limit of \(g_{\gamma}({\alpha}[0]),g_{\gamma}({\alpha}[1]),g_{\gamma}({\alpha}[2])\cdots</math>--Nayuta Ito (talk) 03:58, October 26, 2014 (UTC)

Then $g_{\omega^{\omega}}(\omega) = \lim\{0^0, 1^1, 2^2, 3^3, \ldots\} = \omega$. it's vel time 04:47, October 26, 2014 (UTC)

This is what I want to say: If the ordinal Failed to parse (syntax error): {\displaystyle \gamma\) is written with omegas, \(g_{\gamma}(x)} is what you get when you change all omegas to x. If it is not written with omegas, write with omegas and put them x. --Nayuta Ito (talk) 09:02, October 26, 2014 (UTC)

For larger ordinals it's often not even clear how to write this number "with omegas". For example, how would one write $\varepsilon_1$ using omegas? There is a dispute between users if it should be $\omega\uparrow\uparrow\omega+1,\omega\uparrow\uparrow\omega2,\omega\uparrow\uparrow\omega^2$ or maybe something else. LittlePeng9 (talk) 09:12, October 26, 2014 (UTC)

ε1 is actually ω↑↑↓3=(ω↑↑ω)↑↑ω. 80.98.179.160 10:05, January 12, 2018 (UTC)

Even worse, it's dependent on the FS system you use. There are some definitions of FS where $f_{\omega^\omega}(x) \neq x^x$. In the Wainer hierarchy, yes, you can "substitute $x$ for $\omega$" but the Wainer hierarchy is only defined up to $\varepsilon_0$. it's vel time 10:06, October 26, 2014 (UTC)

In most cases $f_{\omega^\omega}(x) \neq x^x$ is true.... Wythagoras (talk) 16:20, October 29, 2014 (UTC)

I don't think there is a notion of "most" for these things but I agree LittlePeng9 (talk) 16:39, October 29, 2014 (UTC)

That's not really true -- if we're talking about all valid FS systems, the vast majority will not respect that equation. It's just that we're very accustomed to the Wainer hierarchy. it's vel time 18:58, October 29, 2014 (UTC)

Another thing is that we can express $g_\gamma(n)$ in many different ways, e.g. $g_{\omega2}(n)=2n$, so we would also need to have $g_{\omega2}(\omega)=2\omega=\omega$. Even worse, we would sometimes have expressions which are undefined among ordinals, like $g_2(n)=n+2-n$, but expression $\omega+2-\omega$ doesn't have an agreed-upon definition. LittlePeng9 (talk) 12:04, October 26, 2014 (UTC)

So... g_a(w) has no static definition at all? -- A Large Number Googologist -- 19:46, October 27, 2014 (UTC)

Not really - g_a(w) can have many different definitions, just g_a(n) for finite n. It all depends on how we choose fundamental sequences, and how we deal with transfinite arguments. LittlePeng9 (talk) 19:48, October 27, 2014 (UTC)

ok... -- A Large Number Googologist -- 20:06, October 27, 2014 (UTC)

accurate comparisons

Given that $\varepsilon_n = {\varepsilon_{n - 1}} \uparrow\uparrow {\varepsilon_{n - 1}}$ or ${\varepsilon_{n - 1}} \uparrow\uparrow\uparrow 2$

these comparisons can be made accurate, up to a point.

$g_{\varepsilon_0}(n)=n \uparrow\uparrow n$ or \(n\uparrow\uparrow\uparrow 2</math>,

$g_{\varepsilon_1}(n)=g_{\varepsilon_0}(n)\uparrow\uparrow\uparrow 2$ or<math>(n\uparrow\uparrow\uparrow 2)\uparrow\uparrow\uparrow 2\) $g_{\varepsilon_2}(n) =g_{\varepsilon_1}(n)\uparrow\uparrow\uparrow 2$

$g_{\varepsilon_m}(n) =g_{\varepsilon_{m - 1}}(n)\uparrow\uparrow\uparrow 2$

Next, define the following notation:

$x[0] = x \uparrow\uparrow\uparrow 2$

$x[y] = x[{y - 1}]\uparrow\uparrow\uparrow 2$ if $x > 0$

basically a modified version of Steinhaus-Moser notation such that:

it starts at 0 instead of 3
base is <math>^xx\) instead of $x^x$

This means that:

$g_{\varepsilon_m}(n) = n[m]$

$g_{\varepsilon_{\omega}}(n) = n[n]$

$g_{\varepsilon_{\omega^m}}(n) = n[n^m]$

$g_{\varepsilon_{\varepsilon_0}}(n) = n[n[n]]$

$g_{\varepsilon_{\varepsilon_{\varepsilon_0}}}(n) = n[n[n[n]]]$

...

$g_{\zeta_0}(n) = n[n[n[n[...[n[n[n[n]]]]...]]]]$ with n nested brackets

etc.

Of course, when you have to develop separate notation to express such values accurately, is there even a point?

Regardless, putting this out there to show that the true equivalent of the values can be expressed.

ArtismScrub (talk) 01:58, November 7, 2017 (UTC)

About Slow-growing hierarchy, ordinals and nonrecursive ordinals

Is there an ordinal $\alpha$, that $g_\alpha(n)>f_\alpha(n)$? And is $g_{\omega_1^\text{CK}}$ well defined? 80.98.179.160 10:51, December 18, 2017 (UTC)

To the latter question: if you fix some choice of fundamental sequences for $\omega_1^\text{CK}$ and all recursive limit ordinals, then yes. LittlePeng9 (talk) 19:55, December 18, 2017 (UTC)

For the former question, we will have $f_\alpha(n) > g_\alpha(n)$ for all $\alpha$ and $n > 1$, regardless of how the fundamental sequences are defined. First, one can prove pretty easily that $f_\alpha(n) \ge n+1$ for all $\alpha$ and $n$. Then, we prove the main statement by transfinite induction. First, $f_0(n) = n+1 > 0 = g_0(n)$. Then for successor ordinals we have $f_{\alpha+1}(n) = f_\alpha^n(n) \ge f_\alpha(n) + n - 1 \ge f_\alpha(n) + 1 > g_\alpha(n)+1 = g_{\alpha+1}(n)$. Finally, for $\alpha$ a limit ordinal, we have $f_\alpha(n) = f_{\alpha[n]}(n) > g_{\alpha[n]}(n) = g_\alpha(n)$. Deedlit11 (talk) 08:03, December 19, 2017 (UTC)

So this means that:

$g_{\omega_1^\text{CK}}$ is well-defined,
there's no x that $g_x>^*f_x$. 80.98.179.160 09:57, January 12, 2018 (UTC)

Array-of limit is catching ordinal?

Although BEAF is very ambiguously defined at that level, in the interpretation I most agree with, the ordinal where the slow-growing hierarchy catches the fast-growing hierarchy is the same as the limit of the array-of operator. In the SGH, the SVO corresponds to linear arrays in the SGH and linear-array arrays in the FGH. Similarly, the Bachmann-Howard ordinal corresponds to tetrational arrays in SGH, while in the FGH it corresponds to tetrational-array arrays in my interpretation. In both cases, this is an offset of just one application of the array-of operator, which would disappear (in terms of the growth rate) when diagonalizing. Allam(2^^n mod 10^6 for n >= 8) (talk) 03:40, February 24, 2020 (UTC)

The notion of "catching ordinal" is known to be ill-defined. It is not a predicate on an ordinal, but is a predicate of a system of fundamental sequences.

p-adic 06:07, February 24, 2020 (UTC)

@@ Line 120: / Line 120: @@
 There's a variant of FGH with 1 basic rule instead of 3:
-<math>g_{\alpha+m}(n) = g_{\alpha[m+n]}(n)</math>
+\(g_{\alpha+m}(n) = g_{\alpha[m+n]}(n)</math>
 It works if we set 0[n] = 0. [[User:Ikosarakt1|Ikosarakt1]] ([[User_talk:Ikosarakt1|talk]] ^ [[Special:Contributions/Ikosarakt1|contribs]]) 09:25, June 21, 2014 (UTC)
@@ Line 126: / Line 126: @@
 :Not quite right, I think you need:
-:<math>g_{\alpha+m}(n) = g_{\alpha[n]}(n) + m</math>
+:<math>g_{\alpha+m}(n) = g_{\alpha[n]}(n) + m\)
-:Also, you still need the second rule <math>g_0(n) = 0</math>. With just the one rule, you can only set g equal to another expression with g, so you will keep evaluating forever. [[User:Deedlit11|Deedlit11]] ([[User talk:Deedlit11|talk]]) 11:15, June 21, 2014 (UTC)
+:Also, you still need the second rule \(g_0(n) = 0</math>. With just the one rule, you can only set g equal to another expression with g, so you will keep evaluating forever. [[User:Deedlit11|Deedlit11]] ([[User talk:Deedlit11|talk]]) 11:15, June 21, 2014 (UTC)
-::Okay, thanks. I'm pretty satisfied with reducing it to 2 rules. By the way, we can make the same thing with Hardy Hierarchy, using <math>H_{\alpha+m}(n) = H_{\alpha[n]}(n+m)</math>. [[User:Ikosarakt1|Ikosarakt1]] ([[User_talk:Ikosarakt1|talk]] ^ [[Special:Contributions/Ikosarakt1|contribs]]) 05:50, June 24, 2014 (UTC)
+::Okay, thanks. I'm pretty satisfied with reducing it to 2 rules. By the way, we can make the same thing with Hardy Hierarchy, using \(H_{\alpha+m}(n) = H_{\alpha[n]}(n+m)</math>. [[User:Ikosarakt1|Ikosarakt1]] ([[User_talk:Ikosarakt1|talk]] ^ [[Special:Contributions/Ikosarakt1|contribs]]) 05:50, June 24, 2014 (UTC)
-:::That should be <math>H_{\alpha+m}(n) = H_{\alpha[n+m]}(n+m)</math>. [[User:Deedlit11|Deedlit11]] ([[User talk:Deedlit11|talk]]) 12:58, June 25, 2014 (UTC)
+:::That should be \(H_{\alpha+m}(n) = H_{\alpha[n+m]}(n+m)</math>. [[User:Deedlit11|Deedlit11]] ([[User talk:Deedlit11|talk]]) 12:58, June 25, 2014 (UTC)
 ==OBJECTION!==
@@ Line 148: / Line 148: @@
 ==Ordinals written by omega==
-<math>g_{\gamma}(\omega)={\gamma}</math>,right?
+\(g_{\gamma}(\omega)={\gamma}</math>,right?
 In other words, I want to say that SGH shows how big the ordinal is by omega which is replaced by x.
@@ Line 155: / Line 155: @@
 : SGH isn't really defined for transfinite ordinals. You can suggest a definition and then we can talk. [[User:LittlePeng9|LittlePeng9]] ([[User talk:LittlePeng9|talk]]) 10:15, October 25, 2014 (UTC)
-::For example, <math>g_{\omega^{\omega}}(x)=x^x</math>. Then, what should <math>g_{\omega^{\omega}}(\omega)</math> be? It must be <math>\omega^{\omega}</math>. I can't imagine other definitions for ordinals.--[[User:Nayuta Ito|Nayuta Ito]] ([[User talk:Nayuta Ito|talk]]) 01:22, October 26, 2014 (UTC)
+::For example, \(g_{\omega^{\omega}}(x)=x^x</math>. Then, what should \(g_{\omega^{\omega}}(\omega)</math> be? It must be <math>\omega^{\omega}</math>. I can't imagine other definitions for ordinals.--[[User:Nayuta Ito|Nayuta Ito]] ([[User talk:Nayuta Ito|talk]]) 01:22, October 26, 2014 (UTC)
 :::No, the equation \(g_{\omega^{\omega}}(x)=x^x\) is only valid for nonnegative integers x. The domain of every function in the hierarchy is defined to be \(\mathbb{N} \mapsto \mathbb{N}\), and therefore ordinals in general are not valid arguments. If you want to modify the hierarchy to accept ordinals, then that's okay but you have to define what you're doing. [[User:Vel!|it's]] [[User talk:Vel!|vel]] [[Special:Contributions/Vel!|time]] 01:27, October 26, 2014 (UTC)
 ::::What I want to say is from ordinal to ordinal. This may work:
-::::<math>g_{\gamma}({\alpha})</math> is a limit of <math>g_{\gamma}({\alpha}[0]),g_{\gamma}({\alpha}[1]),g_{\gamma}({\alpha}[2])\cdots</math>--[[User:Nayuta Ito|Nayuta Ito]] ([[User talk:Nayuta Ito|talk]]) 03:58, October 26, 2014 (UTC)
+::::<math>g_{\gamma}({\alpha})</math> is a limit of \(g_{\gamma}({\alpha}[0]),g_{\gamma}({\alpha}[1]),g_{\gamma}({\alpha}[2])\cdots</math>--[[User:Nayuta Ito|Nayuta Ito]] ([[User talk:Nayuta Ito|talk]]) 03:58, October 26, 2014 (UTC)
 :::::Then \(g_{\omega^{\omega}}(\omega) = \lim\{0^0, 1^1, 2^2, 3^3, \ldots\} = \omega\). [[User:Vel!|it's]] [[User talk:Vel!|vel]] [[Special:Contributions/Vel!|time]] 04:47, October 26, 2014 (UTC)
-::::::This is what I want to say: If the ordinal <math>\gamma</math> is written with omegas, <math>g_{\gamma}(x)</math> is what you get when you change all omegas to x. If it is not written with omegas, write with omegas and put them x. --[[User:Nayuta Ito|Nayuta Ito]] ([[User talk:Nayuta Ito|talk]]) 09:02, October 26, 2014 (UTC)
+::::::This is what I want to say: If the ordinal <math>\gamma\) is written with omegas, \(g_{\gamma}(x)</math> is what you get when you change all omegas to x. If it is not written with omegas, write with omegas and put them x. --[[User:Nayuta Ito|Nayuta Ito]] ([[User talk:Nayuta Ito|talk]]) 09:02, October 26, 2014 (UTC)
 :::::::For larger ordinals it's often not even clear how to write this number "with omegas". For example, how would one write \(\varepsilon_1\) using omegas? There is a dispute between users if it should be \(\omega\uparrow\uparrow\omega+1,\omega\uparrow\uparrow\omega2,\omega\uparrow\uparrow\omega^2\) or maybe something else. [[User:LittlePeng9|LittlePeng9]] ([[User talk:LittlePeng9|talk]]) 09:12, October 26, 2014 (UTC)
 :::::::ε1 is actually ω↑↑↓3=(ω↑↑ω)↑↑ω. [[Special:Contributions/80.98.179.160|80.98.179.160]] 10:05, January 12, 2018 (UTC)
@@ Line 180: / Line 180: @@
 <span style="font-weight:normal;">these comparisons can be made accurate, up to a point.</span>
-<math>g_{\varepsilon_0}(n)=n \uparrow\uparrow n</math> or <math>n\uparrow\uparrow\uparrow 2</math>,
+\(g_{\varepsilon_0}(n)=n \uparrow\uparrow n\) or \(n\uparrow\uparrow\uparrow 2</math>,
-<math>g_{\varepsilon_1}(n)=g_{\varepsilon_0}(n)\uparrow\uparrow\uparrow 2</math> or<math>(n\uparrow\uparrow\uparrow 2)\uparrow\uparrow\uparrow 2</math>
+\(g_{\varepsilon_1}(n)=g_{\varepsilon_0}(n)\uparrow\uparrow\uparrow 2\) or<math>(n\uparrow\uparrow\uparrow 2)\uparrow\uparrow\uparrow 2\)
-<math>g_{\varepsilon_2}(n) =g_{\varepsilon_1}(n)\uparrow\uparrow\uparrow 2</math>
+\(g_{\varepsilon_2}(n) =g_{\varepsilon_1}(n)\uparrow\uparrow\uparrow 2\)
-<math>g_{\varepsilon_m}(n) =g_{\varepsilon_{m - 1}}(n)\uparrow\uparrow\uparrow 2</math>
+\(g_{\varepsilon_m}(n) =g_{\varepsilon_{m - 1}}(n)\uparrow\uparrow\uparrow 2\)
 <span style="font-weight:normal;">Next, define the following notation:</span>
@@ Line 195: / Line 195: @@
 <span style="font-weight:normal;">basically a modified version of Steinhaus-Moser notation such that:</span>
 *<span style="font-weight:normal;">it starts at 0 instead of 3</span>
-*<span style="font-weight:normal;">base is <math>^xx</math> instead of \(x^x\)</span>
+*<span style="font-weight:normal;">base is <math>^xx\) instead of \(x^x\)</span>
 <span style="font-weight:normal;">This means that:</span>