From: utzoo!decvax!duke!harpo!utah-cs!utah-gr!thomas
Newsgroups: net.math
Title: Re: x**x**x**x... : Where did I go wrong?
Article-I.D.: utah-gr.487
Posted: Sun Jul 18 23:33:01 1982
Received: Sat Jul 24 01:35:10 1982


You didn't go wrong.  The function x^(1/x) (x>=0) has the values given
below at specified points:

x		x^(1/x)
0		0
.1		1e-10
.5		.25
1		1
2		1.4142135623
2.7082818284	1.4446642425
2.7182818284	1.4446678610	[x = e, this is the maximum point]
2.7282818284	1.4446642866
3		1.4422495703
4		1.4142135623
10		1.2589254117
100		1.0471285480
(Thank goodness for bc!)

Or, the calculus approach:
	x^1/x = exp(log(x)/x)
	d/dx(x^1/x) = exp(log(x)/x) * (1 - log(x)) / x^2
		= (1 - log(x)) * x^(1/x - 2)
The important thing about this result is that x^(1/x - 2) is always
positive, thus the sign of the derivative is totally dependent on the
sign of (1 - log(x)).  This is 0 for x = e, positive for x < e and
negative for x > e.

The final question to be answered, then is
	"For what x,y (x != y) is x^(1/x) = y^(1/y)?"
or "when is x^y = y^x?".  It is "well known" that the only integer solutions
to this equation are the pairs (2,4) and (4,2) (for x and y positive).

=Spencer Thomas