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From: eklhad@ihnet.UUCP (K. A. Dahlke)
Newsgroups: net.math
Subject: Beyond Exponentiation
Message-ID: <186@ihnet.UUCP>
Date: Sun, 27-Jan-85 19:44:17 EST
Article-I.D.: ihnet.186
Posted: Sun Jan 27 19:44:17 1985
Date-Received: Tue, 29-Jan-85 05:33:20 EST
Organization: AT&T Bell Labs, Naperville, IL
Lines: 35

<  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
	???
Each operation evolve from an iteration of the previous (originally).
Consider a new operator '~', such that  a~b is equivalent to:
	a^ (a^ (a^ (a^ ... b times ... ) ) ).
Note, the parentheses are important, since the previous operator
(exponentiation) is not commutative.
The alternative:   a^a^a^a^... b times,
could simply be written   a ^ (a^(b - 1)).
Now consider the function  y = 2~x,  assuming y(0) = 1.
The function increases amazingly, taking on the values:
	1	2	4	16	65536	2^65536	huge	wow
I am having trouble with this function though.
What is  y(2.5)?  I am assuming, of course, continuity and infinite
differentiability and other nice things.
Perhaps I could come up with a differential equation which produces
the appropriate values for the integers.
If this function is f(x), then  f(x) = 2^f(x-1).
Differentiating,  df(x) = log(2) * f(x) * df(x-1).
If we substitute df (x-1) with the right side(using x-1 instead of x),
A pattern begins to develop.
df(x) = log(2)^x * f(x)*f(x-1)*f(x-2)*...
I find all this quite interesting, but i haven't been able
to go much further with it.
Any help would be appreciated.
what is  2~2.5?
-- 

Karl Dahlke    ihnp4!ihnet!eklhad