Definition for the complex exponential function

Question

We define the exponential function as $\exp(z)=\sum\limits_{j=0}^\infty= \frac{z^j}{j!}$ for all $z\in \mathbb{C}$. Lets now compute $\exp(0)$, then we would have to calculate $0^0$ which undefined. But for all other $j$ $z^j=0$ and as we know that $\exp(0)=1$ this leaves $0^0=1$. So is it useful to define zero to the zeroth power as $1$? But the again there are surely cases where it could be useful to define it as $0$.

Now my main question is:

Are we actually allowed to use the above defintion or any other equation where one has to consider $z^i$ for all $z \in \mathbb{C}$ and $i$ ranging over $\mathbb{Z}$ including $0$?

Answer 1

Yours is a fair question.

Graham, Knuth, and Patashnik, in their Concrete Mathematics, suggest that we should just define $0^0=1$ for a quite similar reason. They say this on page 162 of the second edition:

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Reference 220 is Two notes on notation by Knuth.

Answer 2

$0^0$ isn't undefined. It's $1$. (If $n, m$ are nonnegative integers, then $n^m$ is the number of functions from a set of size $m$ to a set of size $n$. When $n = m = 0$ you're asking for the number of functions from the empty set to itself, and there's exactly one such function: the empty function.)

When people say that $0^0$ is an indeterminate form, they don't mean it's undefined, they just mean no more and no less than that if $f(x), g(x)$ are functions such that $\lim_{x \to r} f(x) = \lim_{x \to r} g(x) = 0$, that is not enough information to determine the value of $\lim_{x \to r} f(x)^{g(x)}$. Said another way, the exponential function $x^y$, regarded as a function of two nonnegative reals (so it's always well-defined), fails to be continuous at $(x, y) = (0, 0)$ no matter what value it's assigned there. But there are good reasons to assign the value $1$ anyway.