Re: Zeroes and infinities

[Thorsten Siebenborn <7_born@xxxxxx>]

Nick Maclaren schrieb:
> Thorsten Siebenborn <7_born@xxxxxx> wrote:
> > What exactly yields the example of one case 0^0 == 1 in regard of
> > the question what NaN^0 should be ? And 0^0 == 1 is not just a
> > "feeling", it fulfills the valuable binominal theorem in case of 0.
>
> Er, what exactly do you mean?

?

I mean (x+y)^n = sum((n over k)x^(n-k)*y^k,k,0,n)

for all x,y element of R/C and n = N0 (which is
the binominal theorem)

So (1-1)^0 = (0 over 0) * 1^0 * -1^0 = 1 so 0^0 => 1

Or the definition of exp as taylor series:

exp(x) = sum(x^n / n!, n, 0, infinity)

       = 0^0 / 0! = 1

Yes, I know you don't like it because it has no
consistent limit and while defining the value on a
singular point to "heal" different limits is
mathematically not a problem, the approximation of
numbers with limited precision like
pow(0.0000....,0.00000...) may hide a different and
more correct result.

Please, no discussion about 0^0 = 1. Too much
ink and venom has been spilled over this topic.

If you knew this and just wanted to know what I tried
to say: I want to say that even a huge number of
similar examples with different values doesn't say
anything about the result of a particular value.
If I say 0^0 = 1, 1^0 = 1, 2^0
= 1 etc. etc. it says nothing about the value of NaN^0.

HTH
Thorsten