What is the value of

\(0^0 \)

Well, we usually define that as 1. Simple, nice, and useful. So, by convention,

\(0^0 = 1 \)

Let us call this convention (*). We do require it to be able to write down a power series (or a polynomial) in the form

\(\sum_{k=0}^\infty a_k x^k \)

and to apply it for x=0. The result is

\(a_0 \cdot 0^0 = a_0\)

as expected. Otherwise, we would have to write

\(a_0 + \sum_{k=1}^\infty a_k x^k \)

all the time.

Why is this so difficult that many sites and net postings have problems? First of all, trying to derive the convention from

\(x^y = \exp(y \log(x)) \)

is not meaningful, since this equation is only defined for x,y>0. If we set

\(\log(0) = -\infty \)

using extended reals, we face the problem to define zero times infinity. Some areas of mathematics use the convention that this product is 0, which agrees with our convention (*). Note that this convention cannot be used to compute limits. See below for the problem with limits.

If we extend the definition of the powers to integer y in such a way that we have

\(x^{y_1+y_2}= x^{y_1} \cdot x^{y_2} \)

we get

\(0^0 = 0^{0+0} = 0^0 \cdot 0^0 \)

and from that we can only deduce that it is either 0 or 1. In fact, both definitions work for that rule. But since

\(0^y \)

is not defined for negative integers y, we can just pick between the two values, and we pick 1, because it is more useful.

Some people argue using limits of

\(f(x,y) = x^y \)

defined for x,y>0. But this function simply is not continuous in (0,0). The fact, that the limit can take any value between 0 and 1 does not mean that we cannot define it in (0,0). We just do, and note that it is not continuous. There is nothing wrong with that.

So, use the convention (*) and be happy. You can use it in the complex plain, or anywhere, where multiplication makes sense.

Exercise 1: Proof that setting

\(0^0 = 0 \)

works, if we are only interested in satisfying the usual exponential rules.

Exercise 2: Proof that

\(x_n^{y_n} \to 1 \)

if we set

\(y_n = a x_n \)

for any 0<a<1.

Exercise 3: Setting

\(y_n=1/n, \quad x_n=\exp(-1/y_n^2) \)

show

\((x_n,y_n) \to (0,0) \)

and

\(x_n^{y_n} \to 0 \)