S.O.S. Mathematics CyberBoard

I came up with this on my own. I am not sure, maybe someone else has done it too. I do not know the answer. Came up with these while trying to think of functions that grow more quickly than busy beaver functions.

Which, "over the long run," grows more quickly?

1) Not sure how to write this as a function, but you will probably get the idea:

X1= (1!)
X2= (2!)^(2!)
X3= [(3!)^(3!)]^(3!)
X4= {[(4!)^(4!)]^(4!)}^(4!)= 1.097E19080

2) The first term is 3. Xn = (X(n-1))!. So:

X1= 3
X2= 3!=6
X3= 6!=720
X4= 720!=2.60E1746
X5= (X4)!

Also, does anyone know if either or both of these will, in the long run, grow faster than busy beaver functions? If you start working these out, it quickly becomes obvious that they grow at a tremendously fast rate. Also, has anyone seen either of these before, and if so, is there some practical use for them? Thanks and good luck!

Of course, if you wanted something that certainly must grow faster than busy beaver functions, just combine the two ideas:

X1= (2!)^(2!)=4
X2= {[(4!)^(4!)]^(4!)}^(4!)= 1.097E19080
X3= [(X2!)^(X2!)]^(X2!)... and so forth with a tower consisting of X2! levels.
X4= [(X3!)^(X3!)]^(X3!)... and so forth with a tower consisting of X3! levels.
Can anyone calculate the value of the 3rd term here? I would be very surprised if it is less than a googolplex (actually I think X3 probably makes a googolplex look tiny). Can anyone come up with a unique, faster growing "function" than this?

No! Your function must grow slower (eventually) than busy beavers because your function is computable (it is basically composing the factorial with the Knuth arrow).

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It is true that busy beaver functions are noncomputable, but that does not mean that they have an infinite value do they? Does some f(x)=infinity for some beaver function? Doesn't noncomputable just mean that the exact value can not be mathematically determined? Therefore, is it not possible for computable functions, such as the ones I came up with, to grow faster than noncomputable functions, such as busy beaver functions?

No! Infinity is not a number!

(or partial function ...) is computable means there is a finite state Turing machine (or equivalently a register machine which is usually easier to demonstrate) that will accept any natural number as input and compute f for that input (and in the case of partial functions, need machine halting for those input where f is defined and doesn't halt if f is not defined at the input).

Even if f is noncomputable, any truncation

is computable.

While you can cook up computable functions that beats the busy beavers function for the first N inputs, eventually the busy beavers function will win (see the construction in the other thread). That is why the answer to your initial question (does your function grow faster than the beavers function in the long run) is negative.

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Hi,
You have to find the maximum and minimum values of the first derivative. These values are where the function is increasing or decreasing the fastest.