[FOM] Query on real functions

[Toby Ord]

On Tuesday, October 14, 2003, at 10:19  pm, JoeShipman at aol.com wrote:

> Can anyone identify a real-valued function f continuous on [1, 
> infinity) with the following two properties:
>
> 1) f eventually dominates any function in the sequence e^x, e^(e^x), 
> e^(e^(e^x)), ....
>
> 2) f is defined in some other way than by defining it first at all 
> integers and then interpolating

I have a tentative answer to this. I should, however, first introduce 
myself as a BPhil student (studying philosophy at Oxford) and perhaps a 
little rusty on such questions involving continuity in the reals. I 
have spent some time studying fast growing functions on the naturals 
and find them to be quite philosophically interesting, so am interested 
in the apparent inability to construct such things on the reals.

> Unfortunately, a nice way to do this for "tower" rather than "power" 
> is not apparent: 2T1=2, 2T2=4, 2T3=16, 2T4=65536, but what could 
> 2T(3.5) possibly be?

Take the following function:

q(x) = x;             if x is in [1,2)
q(x) = x^q(x-1);      if x is in [2, infinity)

We can then look at q(x) at some values of x around (say) 3, where d is 
to be thought of as some delta just greater than 0.

q(3+d) = (3+d)^(2+d)^(1+d)
q(3+0) = (3+0)^(2+0)^(1+0)
q(3-d) = (3-d)^(2-d)
        = (3-d)^(2-d)^(1+0)

It appears that around such natural numbers, where one might fear 
discontinuity, we find none. However, my powers at expanding out such 
exponentials are not great and I would be happy to be corrected if in 
error.

Note that while I believe this to be continuous, something special is 
definitely happening at the integer points and the function is 
presumably less uniform than you would like.

Incidentally, the obvious generalisation of this approach to 
Ackermann/tower type functions does not work. If the base is 2 as in 
your examples, it seems to remain continuous, giving 2T(3.5) = 
2^2^2^(1.5), but if the base is, say, 3 it is not continuous, jumping 
from 3^1.999 to 3^3.

Toby Ord.