| For x>0, (x)^x is monotonically increasing - it goes up very fast, without
limit. At x=0 it has the value 1. For x<0 the function is complex
(= u + v*sqrt(-1)) and multivalued - very difficult to describe.
There are a lot of other functions that have better intuitive behavior for
a demo. How about F(x) = x^4 - x^2 + 1 ?
Lynn Yarbrough
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| Let f(x) = x^x. Now df(x)/dx = x^x * (log(x)+1) = f(x) * (log(x) + 1).
Setting this to zero gives the extrema, which occur when f(x) = 0, or
when x = 1/e (e is the base of natural logarithms).
When 0 < x <= 1/e, 1 > f(x) >= (1/e)^(1/e) = e^(-e^-1) = 0.692200627....
When 1/e <= x < inf, (1/e)^(1/e) <= f(x) < inf.
lim f(x) = 1, and (more difficultly), lim f(x) = 1.
x->0+ x->0-
For negative values of x, we have:
x^x = |(-x)^x| * (-1)^x
= |(-x)^x| * (cos(x*pi*(2*k+1)) + i*sin(x*pi*(2*k+1))),
where k is an integer.
So f(x) may be multi-valued. Indeed, if we look closely, we see that f(x)
is multi-valued even for x > 0; it's just convention that we usually take
the real value when raising a positive real to a real. For x > 0, we have:
x^x = |x^x| * (cos(x*pi*2*k) + i*sin(x*pi*2*k)),
where k is an integer.
We could consider just the *magnitude* of f(x) for x < 0. That is, just
|f(x)| = | |(-x)^x| * (cos(x*pi*(2*k+1)) + i*sin(x*pi*(2*k+1))) |
= |(-x)^x|
and find extrema, bounds, and limits for *that*. It's left as an exercise
for the interested reader.
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