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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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    Your last questions is answered here - http://math.stackexchange.com/questions/11150/zero-to-zero-power – Joe Tait Feb 05 '13 at 10:23

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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 05 '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 $1$. – hmakholm left over Monica Feb 05 '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 05 '13 at 11:36
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    @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. – hmakholm left over Monica Feb 05 '13 at 12:49
  • @user49685: First of all, $\lim_{x\rightarrow 0+}(x^{\frac{1}{x}})^x = \lim_{x\rightarrow 0+} x$. Secondly, you can't do that. For example, by your logic: $\lim_{x\rightarrow 0} e^{x+1} = \lim_{x\rightarrow 0} (e^x)^{\frac{x+1}{x}}$ would be equal to 1, since $\lim_{x\rightarrow 0} e^x = 1$. – Daniel Robert-Nicoud Feb 05 '13 at 12:50
  • @Henning Makholm: I don't quite get it, isn't the function $(x^{\frac{1}{x}})^x$ continuous for every positive $x$? – user49685 Feb 05 '13 at 13:16
  • @Daniel Robert-Nicoud: Nope, you can't interprete like that, since it'll be the indeterminate form $1^\infty$, and its limit is $e$, not 1. What I mean is: $(x^\frac{1}{x})^x$ is of the indeterminate form $0^0$, and its limit is 0, not 1, so $0^0$ can be anything, 0, or 1, or anything, you just need to choose the right expression. – user49685 Feb 05 '13 at 13:18
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    @user49685: $(x,y)\mapsto x^y$ is not continuous at $(0,0)$ -- therefore you cannot use limits to find what its value at the discontinuity point is or ought to be. That you can speak about a different function that can be made continuous doesn't in itself tell you anything about the exponentiation operator. – hmakholm left over Monica Feb 05 '13 at 13:21
  • @Henning Makholm: No, it's certainly not true, a function $f$ can be undefined at some value, say $a$, but it still have its limit at that very same value. Since when taking limit as $x \rightarrow a$, you only consider the value near $a$, but not $a$. Another example is $x^x$ is not defined at 0, but its limit as $x$ tends to 0 is, indeed, 1 – user49685 Feb 05 '13 at 13:23
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    @user49685: It is possible for a function to have a limit for $x$ some point, but that limit tells you nothing about the value of the function at that point -- unless you already know for independent reasons that the function is continuous. And $(x,y)\mapsto x^y$ is not continuous at $(0,0)$. – hmakholm left over Monica Feb 05 '13 at 13:26
  • @Henning Makholm: Ah, yes, I get your point; but what I'm doing is to point out another example, which is of the indeterminate form $0^0$, but tends to 0, instead of 1 (this case Daniel Robert-Nicoud has already given an example). Ofcourse, $0^0$ is undefined, but the indeterminate form $0^0$ can tend to several values, depending on the expression. – user49685 Feb 05 '13 at 13:30
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    @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. – hmakholm left over Monica Feb 05 '13 at 13:35