# What is 0^0 ?

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

$$x_n^{y_n} \to 0$$

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