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I've had something that's been bugging me, and I tried research and asked my math teacher. None had sufficient answers. The concept of $0$ is that when $0$ goes to any exponent except for $0$, it becomes $0$. For example,

$0^3 = 0$, but $0^0 =$ undefined

However, the proof that $0^0$ is undefined is shown thus: $0^x$ ... (divided by) = $0^{(x-x)} = 0^0$ = undefined $0^x$

You can apply this to any exponent though, such as: $0^6$ ... = $0^6 = 0$ and $0^3 = 0$, so this expression is equal to $0/0$, which should be $0^3$ undefined, right?

Am I doing something wrong here? Please help! Gil

Gil Keidar
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    If $x>0,$ then $x^0 = 1$ but $0^x = 0.$ There is nothing consistent left for $0^0$ – Will Jagy Jul 29 '15 at 03:00
  • @WillJagy 's comment hits the nail on the head. It shows why we can't define $0^0$ in a universally acceptable manner. +1. But--I point out that the usual way of writing Taylor series assumes $0^0=1$ (think about the constant term); in older texts, the constant term is often placed outside of the summation sign, in recognition of this problem. These days it seems rare (to me, anyway). I should say that, in many instances, that convention is specifically commented on by authors using the more abbreviated form, so they understand the implicit abuse of notation. That's just for convenience. – MPW Jul 29 '15 at 03:42

2 Answers2

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To clarify one thing (Michael already posted a great answer). You don't prove something is undefined. Undefined is not some mystical object. Undefined means we have not defined it.

You can prove something can't be defined in a way consistent with other math. I can define $0^0$ anyway I like. I define $0^0=54$. There, now it is defined, but it is not consistent with other things.

I won't add anymore as Michael already added enough.

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One should not say "equals undefined"; one should say "is undefined". The is the "is" of predication, not the "is" of equality.

$0^0$ is indeterminate in the sense that if $f(x)$ and $g(x)$ both approach $0$ as $x\to a$ then $f(x)^{g(x)}$ could approach any positive number or $0$ or $\infty$ depending on which functions $f$ and $g$ are.

But in some contexts it is important that $0^0$ be taken to be $1$. One of those contexts is in the identity $$ e^z = \sum_{n=0}^\infty \frac{z^n}{n!}. $$ This fails to hold when $z=0$ unless $0^0=1$, since the first term of the series is then $\dfrac{0^0}{0!}$.

In combinatorial enumeration it is also important in some contexts that $0^0$ is $1$, for the same reason $2^0$ is $1$: if you multiply by something $0$ times, it's the same as multiplying by $1$. That is also one way of looking at the reason why $0!$ is $1$.

Michael Hardy
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    Sorry I downvoted because $0^0$ is not indeterminate. Rather we can only say that some limit has an indeterminate form, which due to abuse of notation people might call it a "$0^0$" form. Whether limits have an indeterminate form has nothing to do with the value of $0^0$, and in fact it only makes sense to define $0^0 = 1$. My point is that nothing implies any indeterminacy of $0^0$, so if you rephrase your answer I will remove my downvote. – user21820 Jul 29 '15 at 04:30
  • @user21820 : I wrote "is indeterminate in the sense that${},\ldots,{}$." ${}\qquad{}$ – Michael Hardy Jul 29 '15 at 04:47
  • I know, but people who read it don't read it properly, which is why there is such a great confusion, like on the other thread where people keep conflating its value and its being used as a label for limit forms. – user21820 Jul 29 '15 at 05:17