I would define x^x = e^(x*ln(x)), for any values of x for which the operations have meaning, since this expression obeys the properties we expect for exponentials and logarithms: x^x = e^ln(x^x) = e^(x*ln(x)).
There's no problem with e^u when u is negative or fractional, but the domain for ln(x) in its simplest form is (0,infinity): ln(0) is undefined. The definition can be extended to negative arguments, or indeed to all non-zero complex arguments, cf. http://mathworld.wolfram.com/NaturalLogarithm.html
But for arguments not on the positive real axis, it is now a multivalued function (with complex output values):
ln z = ln |z| + i arg(z), where arg(z) is the angle at which z lies from the origin of the complex plane, and is defined up to an additive multiple of 2*pi. Or we can arbitrarily restrict the treatment to the principal value, which amounts to selecting one of the infinite number of options of arg(z).
No difficulty is posed by multiplying any selected choice, including the principle value, of ln(x) by x.
Finally we can use e^(x + i y) = e^x e^(iy) = e^x (cos(y) + i sin(y)).
With z = -x for positive x,
lnz = ln(-x) = ln |-x| + i arg(-x) = ln(x) + i (pi + 2n pi) => principal value: ln(x) + i pi
z*lnz = (-x)*(ln(x) + i (2n+1) pi x) = -x*ln(x) - x*(i (2n+1) pi x)
z^z = e^(z*lnz) = e^(-x*ln(x) - x*( i (2n+1) pi x)) = e^(-x*ln(x)) * e^(-i(2n+1) pi x)
= (1/(x^x)) * (cos((2n+1) pi x) - i sin((2n+1) pi x)) => principal value: (cos(pi x) - i sin(pi x))/(x^x).
If you take the multivalue output of arg(-x), letting n be any integer, even the number of distinct output values of (cos((2n+1) pi x) - i sin((2n+1) pi x)) depends on the choice of x, giving an integer number of values for rational x, and an infinite number for irrational x, seemingly a meaningless choice, but defining a circle about the negative x axis in the visualization.
But the principal value looks unambiguous, assuming a consistent definition of the principal value on the negative real axis in the complex plane), and we see immediately that the limit approaches 0 as x -> -infinity, and also understand the spiraling about the negative x axis in the 3-d image.
This interpretation of x^x for negative x avoids the problems JD highlighted, and pinpoints one type of ambiguity which is resolved by an unambiguous definition of the principal value of arg(-x). And we do get the 3-d visualization shown in the original video clip.
Curious indeed.
Points awarded for data visualization.
Points deducted for disastrous mathematical unsophistication.
Holy wars are fought over the meaning of b^c when the base (b) is negative and the exponent (c) is not an integer.
As always, I do not wish to take sides in holy wars, but we can perhaps take note of some of the possibilities. As a /subset/ of the problem, consider the case where the exponent is a rational number. We write c = p/q in lowest terms, so that p and q are relatively prime integers.
So the first question is, what do we mean by y = b^(p/q) ???
It could mean:
-- Find all y such that y^q = b^p.
There will be q different elements in the solution-set.
Multiply any element by a qth root of unity to find another.
-- Find some y such that y^q = b^p.
That is, choose any element from the aforementioned solution-set.
-- Find the principal y such that y^q = b^p
based on some specified set of preferences.
Again, I don't want to take sides in holy wars, but here is one possible way of choosing the preferred "principal" value (PV).
Note that b is still negative.
*) If p and q are both odd, then b^p is negative and we can
choose PV(y) to be a negative real number.
*) If p is even and q is odd, then b^p is positive and we can
choose PV(y) to be a positive real number.
*) If p is odd and q is even, then b^p is negative and every
y is a nontrivial complex number.
*) We can't have both p and q even.
This is just one possibility. For an arbitrary piece of software operating on an arbitrary floating-point number, it is not the least bit obvious what it will choose as "the" principal value.
If we generalize to the case where the exponent (c) is irrational, then I have no idea how to make sense of b^c when b is negative.
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Now fix p=1000 and consider the sequence q=1001, 1003, 1005, et cetera. Let's set x=-p/q and take a look at
y = (x)^(x)
= (-p/q)^(-p/q)
= (-q/p)^(p/q)
Since p is even, the aforementioned PV(y) is a positive real number in all these cases. That means PV(y) is a positive real number at infinitely many places on the x-interval from -1 to 0. It's a positive real number at 500 different places just between -1 and -0.5.
This is not what is shown in the video. Not even close.
If the presenter had chosen slightly different x-values, and/or chosen a different software package, the results would have been wildly different.
If the presenter had considered all elements of the solution- set (not just some ill-defined principal value) then the results would have been a spectacular mess.
Bottom line: Just because some number crawled out of a software library doesn't mean it's the whole story.