User blog:Kolmogrov142 Oracle Tuto X/Building up the fast growing hiearchy

According to my research into the fast-growing hierarchy, the fgh is a hierarchy which maps numerical or transfinite ordinals to functions.

Furthermore, I also understand how the rules work:

f_0(n) = n+1

f_a+1(n) = f_a-1^n(n) = f_a(f_a...(f_a(n))) where f_a is applied n times to n.

f_a(n) = f_a[n](n) iff (if and only if) a or alpha is a limit ordinal i.e. there is no imediate ordinal before it.

Now onto making pretty large numbers using the fast growing hiearchy.

f_0(0) = 1 since 0+1 = 1 and +1 denotes the most fundamental function of them all the successor function.

f_0(1) = 1+1 = 2.

f_1(1) = f_0^1(n) = f_0(1) = 2.

In fact any f_a>=0(1) =2 since 2 is the base number for this hiearchy and all further diagonalizations on the hiearchy will be based on the number 2.

for any n in the input of the function f_1(n) the output will always be equal to 2n.

f_2(n) can be seen as constantly doubling the input of f_1(n) n times therefore, f_2(n) = 2^n *n.

To give an example of that here is f_2(2).

f_2(2) = f_1(f_1(2)) = f_1(f_0(f_0(2))) = f_1(f_0(3)) = f_1(4) = 2*4 = 8 = 2^2*2.

Furthermore the fast growing hiearchy can be seen as representing a loose aproximation of Knuth up arrows.

as f_0(n) = n+1 which is below the range of Knuth up arrows.

f_1(n) = 2n which has multiplicative strength hence is still far to weak for Knuth up arrows.

f_2(n) = 2^n *n, roughly equals 2^n.

f_3(n) roughly equals 2^^n.

From this we can gather than f_n>=2(n) roughly equals 2^(n-1) n.

(^ = one up arrow).

The next most logical step would be to look at implanting ordinals into the fast growing hiearchy specifically omega (not to be confused with one of the founders of Time Lord society).

Omega or ω for short is the lowest transfinite ordinal and hence the lowest infinite number.

A transfinite number is a number which is infinite in the sense that it is larger than any finite number but not necasserily absolutely infinite.

When one applies omega to f _a(n) we find that omega diagonalizes over n to produce f_n(n).

This is because omega is the limit of the set of natural numbers {0,1,2,3,4,5,6...} or in slightly more elegant terms {n:nEN} set of all n such that n is an element of the set of natural numbers.

Omega is also the order type of the natural numbers since every natural number can be mapped to omega at least one time.

After f_omega(n) we reach a problem.

The problem is trying to figure out how to break past this limit trap in terms of omega since all of the finite numbers have been used up.

However due to the work of some pretty smart men and women a solution has been found in simply taking the successor of omega which ironically turns out to equal omega+1.

We then apply this new ordinal omega+1 to f_a(n) to find that f_omega+1(n) = f_omega ^n (n) = f_omega (f_omega ... (f_omega(n))).

Thus we may also understand that f_omega+n(n) = f_omega+(n-1) ^n (n) = f_omega+(n-1) applied n times to n.

Just as f_omega(n) = f_n(n) due to Cantors diagonal argument f_omega+omega(n) = f_omega+n(n) due to cantors diagonal argument.