User blog comment:MachineGunSuper/The X Hierarchy/@comment-30754445-20171221101805

The last number on that page is at most level ω+2 of the fast growing hierarchy (around Class 14-15 on my scale).

I'll quickly add that you haven't done any actual (ω+2)-level recursion. The only reason it gets this large is that Knuth-arrows (which are already at level fω(n) themselves) have been explicitly used in the definition of Xε0 (n).

The biggest problem is that all the steps repeat the exact same thing (ordinary powers in the first half, knuth arrows in the second). Repeating the same function over and over can't get you very far. The trick is, to always repeat the strongest part you've defined so far.

So once you've defined something as strong as 10^^x (for example: 1₵x), you shouldn't bother with power towers anymore. Instead you should repeat the strongest function you have (which in this case is 1₵x).

So we could, say, write a(x) = 1 ₵ (1 ₵ (1 ₵ (1 ₵ (...x)))) with x ₵'s.

And then, instead of making power towers of a(x) or "₵ towers" of a(x), the next step should be repeating a itself:

b(x) = a(a(a(...(x)...))) with x a's.

( b(x) would be comparable to your q_x)

And then we do it again:

c(x) = b(b(b(...(x)...))) with x b's

d(x) = c(c(c(...(x)...))) with x c's

And so on.

Every new function here will get you a brand new shiney "+1" to the ordinal strength of your function, 'but only if you make sure to repeat the strongest step each time''. '''So if ₵ corresponds to f3, then

a(x) ~ f4(x)

b(x) ~ f5(x)

c(x) ~ f6(x)

d(x) ~ f7(x)

Now to level ω:

Level ω is what you get when you streamline the above approach and give a number to each and every single function in the above list. Since we already know that each level is created by repeating the previous one, we could do it like this:

A1(n) = a(n)

Am+1(n) = Am(Am(Am(...(n)...))) with n Am's

If you look really carefully at the above two lines, you'll see that:

A1(n) = a(n)

A2(n) = b(n)

A3(n) = c(n)

A4(n) = d(n)

and so on.

And now, we can write an ω-level function:

B(n) = An(n)

Now, since An(n) is level (n+1) on the FGH, so is B(n). In other words, B(n) goes through all the previous levels. So B(1000) is comparable to f1000(10). And B(googol) is comparable to fgoogol(10). This is why we call it "level ω": Because it is stronger than any finite level function.

So, how do we progress from here? Again, we repeat 'the strongest thing we've developed so far''. '''So we don't bother with power towers. We don't even bother with any specific An function because B(n) trumps them all.

So what should we repeat? The B function, of-course. We could write:

C(n) = B(B(B(....(n)...)))) with n B's

This, by the way, would be on par with Graham's number (Graham ~ C(64)). The C function leaves knuth arrows completely in the dust.

And now what? Well, since C is already stronger than knuth arrows, there's little point in using them now, is there? We should, again, repeat the strongest thing we have, which is C:

D(n) = C(C(C(...(n)...)))) with n C's

and again (repeating D):

E(n) = D(D(D(...(n)...)))) with n D's

and again

F(n) = E(E(E(...(n)...)))) with n E's

and so on.

Once again, every repetition gives us a shiny new +1 to our ordinal strength so:

B(n) ~ fω(n)

C(n) ~ fω+1(n)

D(n) ~ fω+2(n)

E(n) ~ fω+3(n)

F(n) ~ fω+4(n)

and so on.

If you make sure to always repeat the strongest part of your notation at each step, this can get you to fω+n(n) for an arbitrary n.

This is pretty strong, all things considered. F(2) is already far far bigger than anything on your X-hierarchy page.

But of-course, we don't have to stop here. We can use the "ω-trick" again and streamline all the capital letter functions into a brand new, stronger, function:

X0(n) = B(n)

Xm+1(n) = Xm(Xm(Xm(...(n)...))) with n Xm's

A careful look at the above definition will show that:

X0(n) = B(n)

X1(n) = C(n)

X2(n) = D(n)

X3(n) = E(n)

X4(n) = F(n)

which is, indeed, the thing we wanted to do.

So now we can write:

Y(n) = Xn(n). And since Xn(n) is at level ω+n, then Y(n) is at level ω+n. So Y(1000) ~ fω+1000(10) and Y(googol) ~  fω+googol(10). Since Y is stronger than any level of the form ω+k (for finite k) we can say that it is at level ω+ω = ω×2.

Now, once you fully understand how the above process works and are able to reconstruct it on your own, the next two challanges are:

(1) To properly repeat this entire process starting with Y (hint: the first step is to repeat the most powerful thing we have now, which is Y. In the end you should have a new infinite family of functions similar to a(x),b(x),c(x),d(x) and a new 'mega-function' Z(x) that goes through them all). Note that this is pretty difficult to get right on your first try, so don't get discouraged if you don't succeed immediately. Think of it as a long term goal (and try to do as much work as you can by yourself, before asking for help. That's the best way to learn).

(2) A real toughie: You might have noticed that we can repeat this entire process as many times as we wish. Find a way to streamline that repeation into a single function T(x). You should have something like T(1)=a(x), T(2)=B(x), T(3)=Z(x) and so on (warning: this is very very difficult to do without gaining some experience first. Don't even try this one before completing the first challange).

So, how strong will Z(n) and T(n) be? Well, we already know that every repetition of the entire process gets us another ultra-shiny "ω" to the function strength. So Z would correspond to ω×2+ω=ω×3, and T would correspond to ω+ω+ω+ω+.... = ω×ω = ω2.

And in my "psi levels" system, Z(n) corresponds to class 17 while T(n) corresponds to a nice and round class 20 (for smallish n>10).