$0^0 = 1$ or indeterminate?

Question

Is $0^0 = 1$ or indeterminate? I've heard a number of conflicting answers. If it was indeterminate, how would

$$e^x = \sum_{n = 0}^\infty \frac{x^n}{n!}$$

make sense since $e^0 = 1$?

It is one and indeterminate. Indeterminate doesn't mean undefined, rather it means not continuously defined. — Thomas Andrews, May 31 '13 at 01:50

Jared · Answer 1 · 2013-05-31T01:59:53.237

1

The sum you've written is short hand for

$$e^x=1+x+\frac{1}{2}x^2+\frac{1}{6}x^3+\ldots$$

You'll notice that I wrote $1$ and not $x^0$, so there is no problem with the equation $e^0=1$.

You are right, however, to say that $0^0$ is indeterminate. In fact, for any $a\in\mathbb{R}$, notice the following:

$$\lim_{x\to 0}a(x^2)^{(x^2)}=a$$

edited May 31 '13 at 01:59

answered May 31 '13 at 01:26

Jared

31,451

1

This implicitly assumes that $x^0=1$, and in particular $0^0=1$. Somewhat circular for my taste... – Asaf Karagila May 31 '13 at 01:29

score 0 · Answer 2 · answered May 31 '13 at 01:26

0

It really depends on what the context is. There are very good reasons to define $0^0 = 1$ like the one you just mentioned. Refer to here for a list of arguments for and against.

answered May 31 '13 at 01:26

Cameron Williams

29,432

$0^0 = 1$ or indeterminate?

2 Answers2