BEAF visualization? FB100Z • talk • contribs 17:02, April 4, 2013 (UTC)
Since BEAF ends at \(\vartheta((\Omega^\Omega)\omega)\), BHO can't be visualised in that system. However, in Bird's array notation it can be represented as \(\{\omega,\omega [1 [1 [1 \ldots [1 \backslash_{\omega} 1 \backslash_{\omega} 2] \ldots 2] 2] 2] 2\}\) (with \(\omega\) pairs of brackets). Ikosarakt1 (talk ^ contribs) 19:09, April 4, 2013 (UTC)
So, bird array grows faster than BEAF? Jiawhien (talk) 10:43, May 30, 2013 (UTC)
- Shortly, yes. Ikosarakt1 (talk ^ contribs) 10:52, May 30, 2013 (UTC)
What does the ε function here mean? Is εΩ+1=ω^ω^...ω^(εΩ+1)(with ω ω's), which is equal to φ(1,Ω+1), or εΩ+1=εΩ^εΩ^...^εΩ=Ω^Ω^...^Ω(with ω Ω's), which is equal to θ1(1,0)? {hyp<hyp··cos>cos} (talk) 15:24, July 18, 2013 (UTC)
One variant of ε function, used for definition of BHO, uses equation εα+1 = αααα... (with ω α's) Ikosarakt1 (talk ^ contribs) 15:42, July 18, 2013 (UTC)
εΩ+1=ω^ω^...ω^(Ω+1)=Ω^Ω^...^Ω (yes, I know, abuse of notation) LittlePeng9 (talk) 16:20, July 18, 2013 (UTC)
Both expressions represent the same ordinal, the (Ω+1)st fixed point of a -> ω^a, or the first fixed point larger than Ω. To see this, let x be that fixed point. Obviously Ω+1 < ω^(Ω+1), so Ω+1 < x. Also, if a < x, then ω^a < ω^x = x. So Ω+1, ω^(Ω+1), ω^ω^(Ω+1), ... are all less than x, and therefore y = lim (Ω+1, ω^(Ω+1), ω^ω^(Ω+1), ...) <= x. On the other hand, ω^y = lim (ω^(Ω+1), ω^ω^(Ω+1), ω^ω^ω^(Ω+1) ...) = y, so y is a fixed point of a -> ω^a and therefore y = x.
Similarly, if a,b < x then a^b < x. Proof: First, we will prove that if a,b < x, then a + b < x. First, x must be a limit ordinal, as if x = y + 1, then ω^x = ω^(y+1) = ω^y * ω = y * ω > y + 1 = x. Now, suppose max(a,b) = a. Then if a,b < x, then a+1 < x, and so x > ω^(a+1) = a * ω > a * 2 > a + b. Similarly for max(a,b) = b. Finally, having proven a,b < x => a + b < x, we have a^b <= (ω^ω^a)^(ω^b) = ω^ω^(a+b) < ω^ω^x = x.
So, since Ω < x, we have Ω^Ω, Ω^Ω^Ω, Ω^Ω^Ω^Ω, ... < x. So z = lim (Ω, Ω^Ω, Ω^Ω^Ω, ...) <= x. On the other hand, ω^z = lim (ω^Ω, ω^Ω^Ω, ω^Ω^Ω, ...) <= lim (Ω^Ω, Ω^Ω^Ω, Ω^Ω^Ω^Ω, ...) = z <= ω^z. So z is a fixed point of a -> ω^a, and so z = x.
Thus, εΩ+1 is both lim (ω^(Ω+1), ω^ω^(Ω+1), ω^ω^ω^(Ω+1) ...) and lim (Ω, Ω^Ω, Ω^Ω^Ω, ...). Deedlit11 (talk) 16:31, July 18, 2013 (UTC)
I can show that sequence with ω's (later: A, other sequence is B) is upper bounded by BHO, but I'm not sure about lower bound. For upper bound, we can see that ω^(Ω+1) = ω^Ω*Ω = Ω*Ω = Ω^2 < Ω^Ω. So A[1] is lower than B[2]. Then we get A[n] = w^(A[n-1]) and B[n] = Ω^(B[n-1]). By obvious reasons, A grows no faster than B. Ikosarakt1 (talk ^ contribs) 18:40, July 18, 2013 (UTC)
- Your A[2]=ω^(Ω+1)=ω^Ω*ω=Ω*ω (not ω^Ω*Ω), A[3]=ω^(Ω*ω)=(ω^Ω)^ω=Ω^ω>B[1], and A[4]=ω^Ω^ω=ω^(Ω*Ω*Ω*…*Ω)=(ω^Ω)^(Ω*Ω*…*Ω)=Ω^Ω^ω>B[2], A[5]=ω^Ω^Ω^ω=Ω^Ω^Ω^ω>B[3].
- A[n] is lower bounded by B[n-2]. {hyp<hyp··cos>cos} (talk) 02:37, July 19, 2013 (UTC)
BEAF visualization[]
With more recent discoveries regarding the strength of BEAF, has a visualization been found? FB100Z • talk • contribs
- I don't see how ordinals past\(\varepsilon_0\) can be expressed in BEAF and\(\omega\)'s. X-structure arithmetic differs from\(\omega\)-arithmetic because, for example, X^^(X+1) = X^(X^^X)\(\neq\) X^^X while replacing X to\(\omega\)'s gives another result. Ikosarakt1 (talk ^ contribs) 16:25, June 2, 2014 (UTC)
Equation[]
What's wrong with the first ordinal statistifying the equation \(\vartheta(\Omega^\alpha) = \vartheta(\alpha)\)? Wythagoras (talk) 05:20, June 2, 2014 (UTC)
- I don't think there is anything wrong. LittlePeng9 (talk) 05:28, June 2, 2014 (UTC)
- If you say "the first ordinal satisfying the equation \(\vartheta(\Omega^\alpha) = \theta(\alpha)\)", it would mean that \(\alpha=\varepsilon_{\Omega+1}\). Therefore, BHO is equal to \(\vartheta(\varepsilon_{\Omega+1})\). King2218 (talk) 05:47, June 2, 2014 (UTC)
epsilon sub Omega plus one[]
Edit: Disregard all of this section since i am incapable of reading
Consider the function PT(a,b) = power tower of a, b high
Is it reasonable that for countable ordinals α=>ζ0, the set described by Union (from n=1 to infinity): PT(α,n) (lets call this set PT(α) for this talk section) call is similar to a fundamental sequence for e_{α+1} ?
I would think so, since even if the members are different from the usual definition of phi fundamental sequences which is α+1, ω^(α+1), ω^ω^(α+1), ... it can be seen to converge to the same limit. (i mean, fundamental sequences for e_{α+1}, not α) even though PT(α)[n]> e_{α+1}[n] which doesn't matter
Now my problem is, for uncountable ordinals, for example Ω, it does not seem the case that PT(Ω,n) converges to the same limit as the sequence for e{Ω+1}
for example psi(Ω*ω or ω^(Ω+1)) = Z_ω while psi(Ω^Ω) is Γ_0 (using this definition of psi [1] )
the implication is that the notation ε_{Ω+1} is not appropiate for "first fixed point of α -> Ω^α"
EDIT: I just read deedlit's earlier comment on this page, please disregard all of this as i shamefully walk away slowly Chronolegends (talk) 17:17, February 25, 2017 (UTC)
Extended phi functions[]
In the hierarchy of higher-order Veblen functions that I have devised, is this equal to \(\varphi_\omega(0)\), or the limit of \(\varphi_1(1)\),
\(\varphi_2(1)\), \(\varphi_3(1)\), ...? Allam(2^^n mod 10^6 for n >= 8) (talk) 18:09, March 5, 2020 (UTC)
- No. Your function is just ill-defined. Couldn't you read any comments given by others? Then why are you continuing to ask questions? If you are not interested in feedbacks, it is non-sense for you to ask others. Many of your questions has already been answered by others. Are you just kidding?
- p-adic 22:06, March 5, 2020 (UTC)
- In the interpretation of my function that I most agree with, phi_1(1, 1) is equal to theta(W^(W+1)), which leads to phi_1(2, 0) = theta(W^(W*2)), phi_1(1, 0, 0) = theta(W^W^2), phi_1(w, 0, 0) = theta(W^W^(W^2*w)), phi_2(3) or phi_1(1, 0, 0, 0) = theta(W^W^3). Continuing with this logic, we find phi_w(0), which is the supremum of phi_1(1), phi_2(1), phi_3(1), ..., is indeed the BHO. Allam(2^^n mod 10^6 for n >= 8) (talk) 01:32, March 6, 2020 (UTC)
- It is non-sense to state "In the interpretation of my function that I most agree with" without fixing a definition. The question is non-sense, because for example, you can say "In the interpretation of my function that I most agree with, φ_1(1,1 is equal to PTO(KPM), which leads to φ_1(2,0) = PTO(Stability). Continuing with this logic, we find φ_ω(0), which is the supremum of φ_1(1), φ_φ_1(1)(1), φ_φ_φ_1(1)(1)(1)(1), ..., is indeed PTO(ZFC)." How meaningless to ask others "Is PTO(ZFC) equal to φ_ω(0) in my function defined only in my mind?" As I taught you so many times, as long as your function is undefined, listing finitely many values does not characterise your function.
- p-adic 05:27, March 6, 2020 (UTC)
- I gave the definition for my extended Veblen function hierarchy on my user talk page. Allam(2^^n mod 10^6 for n >= 8) (talk) 16:37, March 6, 2020 (UTC)
- \(\varphi_1(1, 0, 0)\) is the first fixed point of \(\alpha \mapsto \varphi_1(\alpha, 0)\), and \(\varphi_2(1, 0)\) is the first fixed point of \(\alpha \mapsto \varphi_2(\alpha)\). Allam(2^^n mod 10^6 for n >= 8) (talk) 16:50, March 6, 2020 (UTC)
- It is obviously ill-defined. For example, you used ω-1 and Ω-1 in the definition of φ_1(ω,ω) and φ_1(Ω,Ω). I am not saying that it would be well-defined when you defined them in another way. There are many errors. I think that you are not even ready for creating a new ordinal function.
- Also, this is a talk page for BHO, but not for your ill-defined works. Please continue the argument at your talk page.
- p-adic 23:17, March 6, 2020 (UTC)