On 24.06.2016 04:36, Smoke Adams wrote:
....
You do realize that e^(-1/t)^t is a counter example?
e^(-1/t) -> 0 as t -> 0
t -> 0 as t -> 0
....
That's not a counterexample to anything I said. ^ is discontinuous at
(0,0) and indeed, you can force the limit to an arbitrary value by
choosing the two expressions the right way. That's clear.
but e^(-1/t)^t does not -> 1 as t-> 0, which is obvious since it/s 1/e.
So, We can define 0^0 = 1 and maybe that is better than 2, but it is
arbitrary in the sense that it's a definition. It may bear fruit but it
is not necessarily meaningful.
...
It's meaningful in those cases where you want to use 0^0, and otherwise
just don't use it.
Suppose a person is running some numerical simulation that happens to be
computing an a value for an equation that is essentially based on the
above.
Then the numerical simulation is inherently broken. a^b takes on any
arbitrary non-negative value in an arbitrary small region around (0,0)
and is undefined at many points in such a region. (BTW: It would be fine
with me if 0.0^^0.0 was NaN -- that's a completely different case than
the one at hand: pow on integers.)
... break the laws of physics by
arbitrarily defining something to be true when it is not.
...
Utter nonsense. (Also note that the 'laws of physics' typically give
rise to piecewise analytic functions, and if you only consider analytic
functions, 0 ^ 0 = 1 is actually the right answer.)
