Is Google too cool for school?

A tale about a bunch of nothing

What does 0^0 equal? WolframAlpha says its is an indeterminate form. But Google calculator insists the answer is 1 -- no ifs, ands or buts.

Google's result comes without explanation, meaning that some kid may end up with a dud answer on his homework assignment (which is what he gets for not paying attention in class).

This writer put the question to a mathematician, who writes:

"You can easily show that Lim_{x->0} x^x=1, but that's not relevant.

"The question is whether the two-variable limit Lim_{(x,y)->(0,0)} x^y

exists, and it doesn't. So, you are free to adopt whatever convention

you want for 0^0, and some people apparently have argued for 0^0=1,

but whatever convention you adopt, the function f(x,y)=x^y will be

discontinuous at (0,0). It is important to realize this, because

otherwise when taking limits, and seeing that you have one of the form

0^0 you might think it is 1, but this is false. It is an interesting

exercise to construct a path g(t) into (0,0) such that the limit of

x^y along that path is any given positive real number."

He sent me this result from another mathematician

http://math.stackexchange.com/questions/11150/zero-to-zero-power

who uses a similar argument, with the caveat that from one set theoretic perspective as well as from a combinatorial math perspective, one does indeed arrive at 0^0 = 1.

In most cases, however, one would think it makes more sense to leave 0^0 as indeterminate.

In line with the quotation above, consider this:

x^0 = (x^1)(x^-1) = x/x, which holds for all x other than 0.

However, if we write (lim x --> 0) x^0 = x/x, we may use l'Hopital's rule to arrive at D(x/x) = 1. Or in other words the limit of x^0 = 1. That limit point however is not analytical for x = 0. The limit is outside the range, and so it is a discontinuous point. We are free to accept the limit, while understanding that it is not analytical. Calculus permits indeterminate forms such as 0/0 if handled with care (usually via l'Hopital's rule).

One more point: at least one expert in complex variables regards division by 0 as implying "blowing up" to infinity, though infinities are generally excluded from the realm of numbers, unless one is talking about Cantor's cardinals.

Note added Nov. 28, 2001:

=

So the binomial expansion formula can be used to justify 0^0 = 1 because a = 0 either requires that 0^0 = 1 or that (0 + b)^x not be represented by the binomial expansion formula.


Here's the Google calculator graphic:

0^0 = 1 More about calculator.