real analysis - $0^0$ -- indeterminate, or $1$?

One of my teachers argued today that 0^0 = 1. However, WolframAlpha, intuition(?) and various other sources say otherwise... 0^0 doesn't really "mean" anything..

can anyone clear this up with some rigorous explanation?

Answer

Short answer: It depends on your convention and how you define exponents.

Long answer: There are a number of ways of defining exponents. Usually these definitions coincide, but this is not so for $0^0$: some definitions yield $0^0=1$ and some don't apply when both numbers are zero (leaving $0^0$ undefined).

For example, given nonnegative whole numbers $m$ and $n$, we can define $m^n$ to be the number of functions $A \to B$, where $A$ is a set of size $n$ and $B$ is a set of size $m$. This definition gives $0^0=1$ because the only set of size $0$ is the empty set $\varnothing$, and the only function $\varnothing \to \varnothing$ is the empty function.

However, an analyst might not want $0^0$ to be defined. Why? Becuase look at the limits of the following functions:
$$\lim_{x \to 0^+} 0^x = 0, \qquad \lim_{x \to 0} x^0 = 1, \qquad \lim_{x \to 0^+} (e^{-1/t^2})^{-t} = \infty$$
All three limits look like $0^0$. So when this is desired, you might want to leave $0^0$ undefined, so that it's a lack of definition rather than a discontinuity.

Typically this is resolved by:

• If you're in a discrete setting, e.g. considering sets, graphs, integers, and so on, then you should take $0^0=1$.


• If you're in a continuous setting, e.g. considering functions on the real line or complex plane, then you should take $0^0$ to be undefined.

Sometimes these situations overlap. For example, usually when you define functions by infinite series
$$f(x) = \sum_{n=0}^{\infty} a_nx^n$$

problems occur when you want to know the value of $f(0)$. It is normal in these cases to take $0^0=1$, so that $f(0)=a_0$; the reason being that we're considering what happens as $x \to 0$, and this corresponds with $\lim_{x \to 0} x^0 = 1$.