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From: tet@uvaee.UUCP (Thomas E. Tkacik)
Newsgroups: net.math
Subject: Re: Beyond Exponentiation
Message-ID: <172@uvaee.UUCP>
Date: Wed, 30-Jan-85 13:44:33 EST
Article-I.D.: uvaee.172
Posted: Wed Jan 30 13:44:33 1985
Date-Received: Fri, 1-Feb-85 00:48:50 EST
References: <186@ihnet.UUCP>
Organization: EE Dept., U of Virginia, Charlottesville
Lines: 82

> <  where no function has gone before  >
> 
> what is the next function in the following sequence?
> Allow me to use '^' for exponentiation.
> 	y = 2
> 	y = 2*x
> 	y = 2^x
> 	???

K. A. Dahlke poses the problem of the function 

        f(x) = 2~x = 2^ (2^ (2^ ... x times ...) ) ).

What is f(2.5)?  How is this function defined for nonintegers?

The function f(x) has the recurrence relation

        f(x) = 2^f(x-1)

thus we need only define g(x)=f(x) only over the interval (0,1)
and give f(0)=1.
This, along with the recurrence relation will then define f(x)
for all real numbers.

Ignoring for the momemt that we want continuity for the function and its
derivatives, we can assume any function we wish for g(x).

Let us pick  g(x) = 0.   If we do then  f(x) = 0  for any noninteger,
but will be what we want for all integers.  Thus f(2.5)=0.

The next step is to assume continuity at integer values only from the left.
The function    g(x) = f(0) = 1  over the interval  (0,1).
Now  f(x) = f(int(x)) for all x, and   f(2.5) = f(2) = 4.

Now impose the restriction that f(x) be continuous over all x.
Let
        g(x) = c0 + c1*x   
We can find c0 and c1 by forcing   g(0) = f(0)   and  g(1) = f(1).
Solving, we get  c0 = 1  and  c1 = 1.
        g(x) = 1 + x
With this g(x),  f(x) will be continuous everywhere, and we
find f(2.5) from the recurrence relation.
        f(0.5) = 1.5; f(1.5) = 2^f(0.5) = 2.82; 
and f(2.5) = 2^f(1.5) = 7.1

Next impose the resrtiction that f'(x) be continuous everywhere.
Now
        g(x) = c0 + c1*x + c2*x^2
We now need to define the first derivative of f(x).
        f'(x) = log(2) * f(x) * f'(x-1).
However, this is a function of f(x) (which we do not yet know), so using
        f(x) = 2^f(x-1)
we get
        f'(x) = log(2) * 2^f(x-1) * f'(x-1)
g(x) is found by setting
        g(0) = f(0);  g(1) = f(1);  and g'(1) = f'(x) = log(2) *2^g(0) * g'(0).
Solving we get
        c0 = 1;   c1 = 1/(log(2)+0.5);   c2 = (log(2)-0.5)/(log(2)+0.5).
                  c1 = 0.838             c2 = 0.162
Again we find f(2.5) from f(0.5),
        f(0.5) = 1.46;  f(1.5) = 2.75;   f(2.5) = 6.73.

This is as far as I got with this approach.  To find the exact solution
we set
             infinity
               ---        i
        g(x) = >     c * x
               ---    i
               i=0
    
and solve for the c's  by setting  g(0)=f(0), g(1)=f(1) and
forcing all of the derivatives of g(1) equal to the derivatives of f(1).
This might be difficult as it appears that the higher order derivatives
of f(x) get more difficult to calculate.

When this is done, the power series of g(x) will then define f(x) for
all x, not just over the interval (0,1).

----

Tom Tkacik    mcnc!ncsu!uvacs!uvaee!tet
Jeff Labuz                ...!uvaee!jl