
Re: Leaving 0^0 undefined  A numbertheoretic rationale
Posted:
Sep 13, 2013 11:21 AM


On Friday, September 13, 2013 4:56:07 AM UTC4, Peter Percival wrote: > Dan Christensen wrote: > > > > > Playing around with a calculator, I found the following: > > > > > > If x=0 and y is a very small positive real number, then we have > > > x^y=0. Shifting x just slightly into the positive suddenly results in > > > x^y being very close to 1. > > >
Put another way: For very small positive values of x and y, x^y can be very close to 1. If x actually becomes 0, then x^y drops to 0. If both become zero, the result is ambiguous (see above).
> > > Nothing earthshattering here, but it seems to reinforce my > > > recommendation that 0^0 ought to be left undefined for any realworld > > > applications. f(x,y)=x^y simply behaves too strangely close to the > > > origin. > > > > > > Comments? > > > > You've changed the question, haven't you? Didn't it start out as what > > is 0^0 in the natural numbers? > > Now it seems you're talking about reals >
Yes.
> (and even the realworld, shudder). For x^y in the reals you might want > > to look at this picture: > > https://en.wikipedia.org/wiki/File:X%5Ey.png. >
Nice.
> Note the text below it: "3D plot of z = abs(x)^y showing how different > > limits are reached along different curves approaching (0,0)." >
Using a calculator, I get
0.000 001 ^ 0.000 001 = 0.999 986 (rounded to 6 decimal places)
Moving the base value x closer to 0, y is still very close to 1
0.000 000 1 ^ 0.000 001 = 0.999 984
0.000 000 01 ^ 0.000 001 = 0.999 982
At x=0, x^y suddenly becomes 0
0 ^ 0.000001 = 0
I'm sure it's more complicated than this, but to play it safe, you should probably leave 0^0 undefined, even in the reals.
Dan Download my DC Proof 2.0 at http://www.dcproof.com

