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Zero to zero power

Suppose that $0^n$ where $n$ is any natural number (or non-negative real number.). What would be the result of this calculation? Also, what would $0^0$ be calculated as?

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marked as duplicate by dtldarek, Amr, Davide Giraudo, Per Manne, draks ... Feb 5 '13 at 11:38

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Your last questions is answered here - – Joe Tait Feb 5 '13 at 10:23

As you may already know: $\alpha ^ {m - n} = \dfrac{\alpha^m}{\alpha^n}, \mbox{ for } m > n$.

We then use that identity, and try to generalize it into cases where $m = n$, and $m < n$. So, for $m = n$, we have:

$\alpha^0 = \alpha ^ {m - m} = \dfrac{\alpha^m}{\alpha^m} = 1$

The identity above is true iff the denominator is not 0, or in other words $\alpha ^ m \neq 0$, or $\alpha \neq 0$.

So $\alpha ^ 0 = 1, \forall \alpha \neq 0$; and $0 ^ 0$ is undefined.

For negative powers, we have:

$\alpha^{-n} = \alpha^{0 - n} = \dfrac{\alpha^0}{\alpha^n} = \dfrac{1}{\alpha^n}$

This is again, valid iff the denominator is not 0, hence $\alpha \neq 0$. So,

$\alpha^{-n} = \dfrac{1}{\alpha^n}, \forall \alpha \neq 0$.

In conclusion, we have:

  • For $n > 0$, $0^n = \underbrace{0.0.0....0}{n \mbox{ times}} = 0$.
  • For $n \le 0$, $0^n$ is undefined.

You can also think about this as:

  • $\alpha^0 = 1, \forall \alpha \neq 0$.
  • $0^n = 0, \forall n > 0$.

So what's $0^0$, is it 1 (as in the first case, since it has the form $\alpha^0$), or is it 0 (as in the second case)? So it's undefined. Of course, this is not very rigorous.

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You can see $0^0$ as $\lim_{x\rightarrow 0+}x^x$, in this case you get $0^0 = 1$. – Daniel Robert-Nicoud Feb 5 '13 at 11:12
See the existing question 11150 about $0^0$. In most contexts it is not undefined, but quite definitely defined to mean $1$. – Henning Makholm Feb 5 '13 at 11:24
@Daniel Robert-Nicoud: But, if you take the lim of $\lim\limits_{x \rightarrow 0^+} {(x^{\frac{1}{x}})}^x$, this is clearly $0^0$, since $x^{\frac{1}{x}} \xrightarrow{x \rightarrow 0^+} 0$, but $\lim\limits_{x \rightarrow 0^+} {(x^{\frac{1}{x}})}^x = \lim\limits_{x \rightarrow 0^+} x ^ {x.\frac{1}{x}} = \lim\limits_{x \rightarrow 0^+} x ^ 1 = 0$ – user49685 Feb 5 '13 at 11:36
@user49685: Such limit arguments don't tell you anything unless you already know that the function $(x,y)\mapsto x^y$ is continuous everywhere you need it to be. And it isn't. – Henning Makholm Feb 5 '13 at 12:49
@user49685: $0^0$ doesn't "tend"; it contains no variables at all. The expression $f(x)^{g(x)}$ can tend to something, and if $f(x)\to 0$ and $g(x)\to 0$, then the limit of $f(x)^{g^(x)}$ can be called indeterminate. But there can't be any tending without a variable to vary. – Henning Makholm Feb 5 '13 at 13:35

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