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Topic: Leaving 0^0 undefined -- A number-theoretic rationale
Replies: 48   Last Post: Sep 15, 2013 1:06 PM

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 Dan Christensen Posts: 8,160 Registered: 7/9/08
Re: Leaving 0^0 undefined -- A number-theoretic rationale
Posted: Sep 13, 2013 11:21 AM

On Friday, September 13, 2013 4:56:07 AM UTC-4, 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 earth-shattering here, but it seems to reinforce my
>
> > recommendation that 0^0 ought to be left undefined for any real-world
>
> > applications. f(x,y)=x^y simply behaves too strangely close to the
>
> > origin.
>
> >
>
>
>
>
> 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 real-world, 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