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3

[alemagno12]

3

lol.
But it's not valid.

I'm willing to dispute your statements that A(G,G) is smaller than your number.

Apparently you have no idea how large large numbers can actually be.

First of all, A(G) = A(G,G) = f_ω(G), and A^G(G) = f_ωⁿ(G) = f_ω+1(G), and your number is A^AG(G)(G) < A^AG(G)(A^G(G)) = f_ω+1(f_ω+1(G)) = f_ω+1²(G) < f_ω+1^G(G) = f_ω+2(G). Note how incredibly unexact those inequalities are. And since ω+2 is so very very much smaller than ψ(ψ_I(0)), your number is so incredibly much smaller than f_ψ(ψI(0))(G) you can't even imagine how hard it would be to even come up with this shit in the first place.

EDIT: Suddenly I know so much about ordinals. Thanks, thread.

EDIT2: Aaaaaaactually, after reading through the definition of fast-growing hierarchies again, my previous statements are all not really true. Since alemagno's number is dependent on the choice of the fast-growing hierarchy (which is defined by the choice of fundamental sequences of limit ordinals) and since both ψ(ψ_I(0)) and ω are limit ordinals, fψ(ψ_I(0))(G) could very well be smaller than f_ω+1(G), namely if you choose the first G elements of ω's and ψ(ψ_I(0))'s limit sequences to both be the first G integers, then A^AG(G)(G) > f_ω+1(G) > f_ω(G) = f_ω[n](G) = f_G(G) = f_{ψ(ψI(0))[n]}(G) = f_ψ(ψI(0))(G). So alemagno, you need to specify your fundamental sequences first before making such claims, because your number was not completely defined.

Noope.
Read again.
It's wrong.
ψ(ψ_I(0)) is FAR greater than ω.
And f_ω+1(G) beats your ''f_ψ(ψI(0))(G)''
The fundamental sequences for limit ordinals can produce larger values than fundamental sequences for smaller limit ordinals.
Really.

[FallenAngel]

Great, now I'm cross-eyed again.
I'm going to go back to my circles whose circumference is greater than πd and toot on my Gabriel's Horn.

Also, I realized a typo in what I said previously. I had many instances of f_ω+1, while I had meant to put f_ω+1.
Revising my previous statement...

I know how large numbers can get - I just wanted to pry some info out of the situation.
Going by the numbers that have been derived from the xkcd function, I have a new, larger number.
f_ω+1(f_ω+1(G)), my large ...thing made of nested Ackermann functions with the root of it all being A(G,G), is not close to your number, or f_ψ(ψI(0))(G).
Now, to take this information. Since f_ω+1(f_ω+1(G)) is approximately A(A(A(A(A(A(...(G,G)...), where the number of A() is equal to A(A(A(A(...(G,G)...), in which the number of A() is Graham's Number,  the two can technically be interchangeable - we're at large enough numbers where most differences between similar numbers can be ignored safely - A(G,G)-1 can be written as A(G,G) to no ill effect.
From this, my new number is f_ω+1(f_ω+1(f_ω+1(f_ω+1(f_ω+1(...(G)...), in which function the f_ω+1(...) are nested G taken to the xkcd hyperoperation taken to the xkcd hyperoperation times. Plus 1. Because one is my seventh-favorite number, and 7 is considered "lucky".

Also, you had previously stated that f_ω+1(G) is drastically lower than f_ψ(ψI(0))(G), but now you are saying that the latter is the smaller. Make up your mind.
And as a "I'm-tired-right-now" thing to put down, f_ω+(fw+G)(G). I'm admittedly sort of lost by now, but I'm rather confident that said number is a large number, and certain that it is larger than f_ω+1(G).

[alemagno12]

3

lol.
But it's not valid.

I'm willing to dispute your statements that A(G,G) is smaller than your number.

Apparently you have no idea how large large numbers can actually be.

First of all, A(G) = A(G,G) = f_ω(G), and A^G(G) = f_ωⁿ(G) = f_ω+1(G), and your number is A^AG(G)(G) < A^AG(G)(A^G(G)) = f_ω+1(f_ω+1(G)) = f_ω+1²(G) < f_ω+1^G(G) = f_ω+2(G). Note how incredibly unexact those inequalities are. And since ω+2 is so very very much smaller than ψ(ψ_I(0)), your number is so incredibly much smaller than f_ψ(ψI(0))(G) you can't even imagine how hard it would be to even come up with this shit in the first place.

EDIT: Suddenly I know so much about ordinals. Thanks, thread.

EDIT2: Aaaaaaactually, after reading through the definition of fast-growing hierarchies again, my previous statements are all not really true. Since alemagno's number is dependent on the choice of the fast-growing hierarchy (which is defined by the choice of fundamental sequences of limit ordinals) and since both ψ(ψ_I(0)) and ω are limit ordinals, fψ(ψ_I(0))(G) could very well be smaller than f_ω+1(G), namely if you choose the first G elements of ω's and ψ(ψ_I(0))'s limit sequences to both be the first G integers, then A^AG(G)(G) > f_ω+1(G) > f_ω(G) = f_ω[n](G) = f_G(G) = f_{ψ(ψI(0))[n]}(G) = f_ψ(ψI(0))(G). So alemagno, you need to specify your fundamental sequences first before making such claims, because your number was not completely defined.

Noope.
Read again.
It's wrong.
ψ(ψ_I(0)) is FAR greater than ω.
And f_ω+1(G) beats your ''f_ψ(ψI(0))(G)''
The fundamental sequences for limit ordinals can produce larger values than fundamental sequences for smaller limit ordinals.
Really.

You were the one who said the latter, by the way. Does that mean I win?

I mean in the user's sense.

[FallenAngel]

I agree.
...what?

I mean, I often make statements that make no real sense. Typically such statements are on purpose, but sometimes they aren't.
But that is one heck of a statement.
I'm going to assume you're discussing ontological inertia and continue on with my day.

[alemagno12]

So game over?

[FallenAngel]

Technically no, if we stretch the rules a tad.
We can certainly get larger. Very easily, I might add.
Just the creativity would slowly boil away until we're doing the higher mathematical version of x+1.

[Putnam]

non-computables?

you know a lot of us here have probably read Gödel, Escher, Bach and know how this will end (or rather, how it won't)

so let's do this baby, the fun ends (or starts) now

Define Α(x) as "f[0](x)->f[1](x)->f[2](x)...f[n](x)", where "f" is the set of all functions put forward in this thread (not including Α(x)) sorted in order of growth speed (smallest to largest) and -> is Conway's chained arrow notation. This is retroactive and proactive, so any functions put forward later will add in to this.

Define Ω_A(x) as A(A(A...A(x)...)), where the number of "A"s nested is equal to the maximum number of "A"s put forward plus one, with a minimum of one.

Define NiB as the sum of all numbers put forward in this thread, not including the particular NiB defined here (but including any other number with a similar meta definition as NiB, but only as they were originally defined, so no infinite shenanigans). NiB is also retroactive and proactive.

My number is Ω_A(NiB), nested NiB times.

This is now a "who can get more meta" contest

[Putnam]

"The universe" is vague.

[Putnam]

See, now it's not vague, because you gave a term And yeah, that's a very nice, all-encompassing definition. Trying to one-up it will just result in it getting "bigger".

[MagmaMcFry]

-

Noope.
Read again.
It's wrong.
ψ(ψ_I(0)) is FAR greater than ω.
And f_ω+1(G) beats your ''f_ψ(ψI(0))(G)''
The fundamental sequences for limit ordinals can produce larger values than fundamental sequences for smaller limit ordinals.
Really.

My point is that they don't necessarily have to, because the definition of f depends on the choice of all your fundamental sequences (which need not be related at all), and since the only requirement for a fundamental sequence for a limit ordinal is that it strictly increases and converges to that limit ordinal, you can very well choose such sequences for ω and ψ(ψ_I(0)) such that the first few elements are the same, e.g. ψ(ψ_I(0))[G] = G and ω[G] = G.

Now remember the definition of fast-growing hierarchy, which says that if α is a limit ordinal, f_α(n) is defined as f_α[n](n). So in this case, f_ω+1(G) = f_ω^G(G) > f_ω(G) = f_ω[G](G) = f_G(G) = f_{ψ(ψI(0))[G]}(G) = f_ψ(ψI(0))(G).

Now as I said before, different choices of fundamental sequences lead to different fast-growing hierarchies, and it may very well be possible (for appropriate choice of the fundamental sequences) for your number to be larger than Fallen's. But until you provide a choice of fundamental sequence for all the limit ordinal numbers up to ψ(ψ_I(0)), your number is not even well-defined, as opposed to Fallen's perfectly well-defined A^AG(G)(G).

My number is Ω_A(NiB), nested NiB times.

Well, I hereby define a function Z that takes an integer and adds the imaginary unit. Now your definition of A(x) is broken forever, because chained arrows don't operate on complex numbers. So there.

[alemagno12]

Ok, i'm going to see if your entries are valid.

Well, if we stretch the rules to include non-computable values, then...

I will start off with f_ℵ1(G).

First off, ℵ₁ is NOT an ordinal, it's a cardinal.
Also, if we consider ℵ₁ to be ω₁ (first uncountable ordinal), then we have a problem here, because ω₁ does not have a fundamental sequence.
And i'm using ω₁ because ω₁ has cardinality ℵ₁, if you were wondering.
So invalid.

non-computables?

you know a lot of us here have probably read Gödel, Escher, Bach and know how this will end (or rather, how it won't)

so let's do this baby, the fun ends (or starts) now

Define Α(x) as "f[0](x)->f[1](x)->f[2](x)...f[n](x)", where "f" is the set of all functions put forward in this thread (not including Α(x)) sorted in order of growth speed (smallest to largest) and -> is Conway's chained arrow notation. This is retroactive and proactive, so any functions put forward later will add in to this.

Define Ω_A(x) as A(A(A...A(x)...)), where the number of "A"s nested is equal to the maximum number of "A"s put forward plus one, with a minimum of one.

Define NiB as the sum of all numbers put forward in this thread, not including the particular NiB defined here (but including any other number with a similar meta definition as NiB, but only as they were originally defined, so no infinite shenanigans). NiB is also retroactive and proactive.

My number is Ω_A(NiB), nested NiB times.

This is now a "who can get more meta" contest

Congratulations, you have just breaked the rules of the game!

non-computables?

you know a lot of us here have probably read Gödel, Escher, Bach and know how this will end (or rather, how it won't)

so let's do this baby, the fun ends (or starts) now

Define Α(x) as "f[0](x)->f[1](x)->f[2](x)...f[n](x)", where "f" is the set of all functions put forward in this thread (not including Α(x)) sorted in order of growth speed (smallest to largest) and -> is Conway's chained arrow notation. This is retroactive and proactive, so any functions put forward later will add in to this.

Define Ω_A(x) as A(A(A...A(x)...)), where the number of "A"s nested is equal to the maximum number of "A"s put forward plus one, with a minimum of one.

Define NiB as the sum of all numbers put forward in this thread, not including the particular NiB defined here (but including any other number with a similar meta definition as NiB, but only as they were originally defined, so no infinite shenanigans). NiB is also retroactive and proactive.

My number is Ω_A(NiB), nested NiB times.

This is now a "who can get more meta" contest

Hehe. Well, let's get truly meta:

j: V -> V

There we go. Note- this is in ZK, not ZKC. j is a cardinal, and V is the universe. We don't know whether such a cardinal can exist, but if it does, it would be the supremum of cardinals- as in, there can not be any greater cardinal.

Is this finite?
Are finite numbers allowed?
Are cardinals allowed?
Are ordinals allowed?
Are cardinals infinite?
Are ordinals infinite?

To quote the Wikipedia page on why the number I just mentioned can't exist in ZKC (Kunen's Inconsistency):

"There is no non-trivial elementary embedding of the universe V into itself."

So, mathematical universe.

Uhh, i see.
...
Like there are no other posts, i present you the Q:
The Q is the largest number definable with every function posted in this thread as of the date of posting on this thread and n, where n is the largest possible computable number.

[FallenAngel]

I vaguely recall stating that, if you used functions defined prior in the thread, you must at the very least mention them by name.

[alemagno12]

I vaguely recall stating that, if you used functions defined prior in the thread, you must at the very least mention them by name.

Then

non-computables?

you know a lot of us here have probably read Gödel, Escher, Bach and know how this will end (or rather, how it won't)

so let's do this baby, the fun ends (or starts) now

Define Α(x) as "f[0](x)->f[1](x)->f[2](x)...f[n](x)", where "f" is the set of all functions put forward in this thread (not including Α(x)) sorted in order of growth speed (smallest to largest) and -> is Conway's chained arrow notation. This is retroactive and proactive, so any functions put forward later will add in to this.

Define Ω_A(x) as A(A(A...A(x)...)), where the number of "A"s nested is equal to the maximum number of "A"s put forward plus one, with a minimum of one.

Define NiB as the sum of all numbers put forward in this thread, not including the particular NiB defined here (but including any other number with a similar meta definition as NiB, but only as they were originally defined, so no infinite shenanigans). NiB is also retroactive and proactive.

My number is Ω_A(NiB), nested NiB times.

This is now a "who can get more meta" contest

is not valid.

[FallenAngel]

I wasn't referring directly to you, I was just reminding everyone.
I have an unusually simple trump card that is crazy enough it could work.