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I've read in several places that defining $0^0=1$ is convenient in several (primarily discrete) settings. One argument on Wikipedia in favor of this definition was the need of a special case for the product rule $\frac{d}{dx} x^n = nx^{n-1}$ for $n=1$ at $x=0$.

Regardless of which definition, if any, is proper, is this not an argument to the contrary? Or I'm missing something? $nx^{n-1}$ for $n=1$ at $x=0$ is $1*0^0 = 1*1 = 1$ contradicts $\frac{d}{dx} 0^1 = \frac{d}{dx} 0 = 0$.

I feel as though there's something just incredibly obvious I'm missing here. How can this be an argument in favor of defining $0^0=1$?

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Also, IMO it's far better to think of defining a discrete exponentiation operation which is separate from the ordinary real exponentiation operation, than thinking of extending the real exponentiation operator to have a value at $0^0$. – Hurkyl Dec 6 '12 at 6:31
Look at the pithy responses here (especially how the mathematician replies):… – Amzoti Dec 6 '12 at 6:32
Sorry, revised the power rule. The problem is still the same, though. Isn't this a problem for $0^0=1$? – William Stagner Dec 6 '12 at 6:38
The value of $\frac{d}{dx} f(x)$ at $x=a$ is not $\frac{d}{dx} f(a)$. – Hurkyl Dec 6 '12 at 6:40
You're absolutely right. I feel embarrassed now... Thank you! – William Stagner Dec 6 '12 at 6:45
up vote 4 down vote accepted

There is actually a fairly good motivation for this definition, but it requires set theory, and a bit of torturing of definitions.

In set theory, given two sets $X,Y$, the set $X^Y$ is defined to be the set of all maps from $Y$ to $X$. Let $|X|$ be the cardinality ("size") of $X$ as usual. Then $|X^Y|$ = $|X|^{|Y|}$ if $|X|$ and $|Y|$ are positive integers. But if $|X|=|Y|=0$, so $X$ and $Y$ are both empty sets $\emptyset$, then there is one map; from the empty $Y$ to the empty $X$! In other words, $|\emptyset^\emptyset|=|\emptyset|^{|\emptyset|}=0^0=1$.

Now in order for this motivation to work fully, to have to declare that only the empty set can map to the empty set. Then $|X^\emptyset|=1$ for all $X$, because there is only the one map from $\emptyset$ to $\emptyset\in X$. Further, $|\emptyset^Y|=0$ if $Y\ne\emptyset$, because then there is no empty set to map from.

Like I said, a bit of torture, but it works!

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Review the summary discussion on this topic here:

Things tend to get messy at times!

Regards -A

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