# How to define the $0^0$? [duplicate]

Possible Duplicate:
Zero to zero power

According to Wolfram Alpha:

$0^0$ is indeterminate.

According to google: $0^0=1$

According to my calculator: $0^0$ is undefined

Is there consensus regarding $0^0$? And what makes $0^0$ so problematic?

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Duplicate question. – Graphth Dec 15 '12 at 21:09
Google is all knowing. – Daniel Montealegre Dec 15 '12 at 21:13
@ashley 1 is never prime, though. It's a unit. – Potato Dec 15 '12 at 21:54
@Graphth Where is the duplicate? I searched it, but couldn't find. – Kasper Dec 15 '12 at 21:54

## marked as duplicate by mixedmath♦Dec 15 '12 at 23:57

This question will probably be closed as a duplicate, but here is the way I used to explain it to my students:

Since $x^0=1$ for all non-zero $x$, we would like to define $0^0$ to be 1. but ...

since $0^x = 0$ for all non-zero $x$, we would like to define $0^0$ to be 0.

The end result is that we can't have all the "rules" of indices playing nicely with each other if we decide to chose one of the above options, it might be better if we decided that $0^0$ should just be left as "undefined".

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+1: Best answer so far – Amr Dec 15 '12 at 21:20
Thanks! I met the problem of trying to explain this to students so many times, that I finally developed an argument that would convince even stubborn students as quickly and simply as possible :) – John Wordsworth Dec 15 '12 at 21:24
My vote goes to this ans. as well. – ashley Dec 15 '12 at 21:29
@OldJohn: It completely fails to convince me. Who cares about $0^x$? On the other hand $x^0$ must be $1$ always, or polynomials wouldn't work at $x=0$ ... – Henning Makholm Dec 15 '12 at 21:42
But there are also standard identities that fail unless $0^0 = 0$ ... – John Wordsworth Dec 16 '12 at 0:01

One way of looking at it is that there are two different exponentiation operators that are denoted $a^b$:

• Discrete (algebraic): when $b$ is an integer. Then the exponentiation is multiple multiplication or division. This can be used in any group. This operator occurs in Taylor series, the binomial expansion, when calculating the size of the set $A^B$ given sizes of $A$ and $B$ etc. For this operator, $0^0=1$, or in general for any element $a$, $a^0$ is the multiplicative identity.

• Continuous (analytic): when $b$ is real, and $a>0$. Then we define $a^b = \exp(b \ln a)$. Note that here $a$ must be positive. Here it is best to leave $0^0$ undefined, as otherwise the function will be discontinuous.

You can have several more variants. In a monoid, you can define exponentiation $a^n$ where $n$ is a nonnegative integer. In a semigroup, you can define exponentiation where $n$ is a positive integer. In complexes (or an algebraically closed field), you can define multi-valued $a^{p/q}$ for a rational exponent. Cardinals and ordinals have their own exponentiations. The "continuous" exponent can be extended to complex numbers: when $a>0$ then you can define $\exp(b \ln a)$. Yet another exponentiation on complex numbers is multi-valued $\exp(b \operatorname{Ln} a)$.

All those operations are different - they have different domains. Mathematicians are unusually sloppy about which exponentiation they are talking about and use context-dependent $a^b$. (Some programming languages have multiple exponentiation operators to deal with this problem.)

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 "Here it is best to leave $0^0$ undefined, as otherwise the function will be discontinuous." I don't understand this. Which function would be discontinuous ? – Kasper Dec 15 '12 at 22:02 @Kasper: If you extend the function $f:(0,\infty) \times \mathbb R \to \mathbb R$, $f(x,y)=x^y$ to the point $(0,0)$ it will no longer be continuous. – sdcvvc Dec 15 '12 at 22:02

In set theory, where everything is a set, $0$ is represented by the empty set. Exponentiation of sets $\alpha^\beta$, lets call them cardinalities and write $|\alpha|^{|\beta|}$, is defined to be the cardinality (number of elements) of all functions from $\beta \to \alpha$. If both $\alpha$ and $\beta$ are empty, then there is exactly one function $\varnothing \to \varnothing$, hence $0^0 = 1$.

Though this is just a convention, I like how it justifies $0^0 = 1$.

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 The question is tagged (real-analysis). So set theoretic justifications don't qualify here :) – Ralph Dec 15 '12 at 22:45 @Ralph : ) ${}{}{}$ – Matt N. Dec 16 '12 at 8:21

The consensus is that if you are going to adopt a convention defining $0^0$, then it should probably be $0^0=1$.

The problem is that there is good reason for basic arithmetic operations on real numbers are continuous, and the real and complex exponentiation operators cannot be continuous at $0^0$.

The solution, IMO, is to honestly recognize that there are multiple exponentiation operators. All of the cases where the convention $0^0 = 1$ is useful are discrete: e.g. in a power series, where we are interested in monomials with integer exponents. The needs we have for discrete exponents are very different from the needs we have for continuous exponents.

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The answer is: The meaning is context-sensitive. This is surprising only if you assume that mathematical terms are context-insensitive. They are not. See x^y by Sam Derbyshire.

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$\forall x \not= 0, x^0 = 1$

$\forall x \not= 0, 0^x = 0$

Those are either definitions or conventions chosen to extend formulas (just like you chose $0!=1$).

We can't have both functions $x\mapsto 0^x$ and $x\mapsto x^0$ continuous at $x=0$ no matter how we define $0^0$

Continuous functions are functions that commute with limit, ie $\lim f( x_n) = f( \lim x_n)$

And in this case it doesn't work for at least one of the cases.

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$0^0=1$, and "$0^0$" is an indeterminate form.

The fact that it's a well defined expression in no way conflicts with the fact that it's an indeterminate form.

$0^0=1$ because it's an empty product. Multiplying by no number is the same as multiplying by $1$; therefore when one multiplies by no number, the product is $1$.

It's indeterminate because one can let the pair $(x,y)$ approach $(0,0)$ along a path that makes the limit of $x^y$ equal to $5$ or to $1$ or to $\infty$, or to any of infinitely many other values.

If one approaches $(0,0)$ along any path that remains between two lines of positive slope, then the limit is $1$.

If $0^0$ were not equal to $1$, then the familiar expansion $$e^z= \frac{z^0}{0!} + \frac{z^1}{1!} + \frac{z^2}{z!} + \cdots$$ would fail when $z=0$, since the first term is $\dfrac{0^0}{0!}$.

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Define $f(x)=x^x$ and compute $\lim_{x\to 0} f(x)$.

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Shouldn't you take "$x \to 0+$" ? – Ralph Dec 15 '12 at 21:16
@Ralph : Yes. That's a suggestion of a definition ; it is not universal. – Patrick Da Silva Dec 15 '12 at 21:26

It is undefined. It can be 1 and 0, depending on how you define exponentiation.

Exemple where $0^0 = 1$ Define:

$x^n := 1 *x *x ... x$

when n = 0

$x^0 = 1$

$0^0 = 1$

But consider following definition

$x^n :=lim _{z->n} +x^z$

$0^0 = lim_{z->0} + 0^z =lim_{z->0} + 0 = 0$

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$0^0$ can be described as:

\begin{align} \lim\limits_{x \to 0^+} x^x = e^{\lim\limits_{x \to 0^+}x\cdot\log(x)} = \left|t = \dfrac{1}{x}\right| & = e^{\lim\limits_{t \to+\infty}\dfrac{\log{\frac{1}{t}}}{t}} \\ = \left|\text{Using L'Hospital's rule.}\right| & = e^{\lim\limits_{t \to+\infty}\dfrac{1}{t}} \\ = \exp\left(\dfrac{1}{\lim\limits_{t \to+\infty} t}\right) & = e^{0} = 1 \end{align}

For this case he answer is 1.

And! The $0^0$ can be described as:

$$\lim\limits_{x\to 0^+}\left(\dfrac{\cos{x}}{x}\right)^{\dfrac{\cos{x}}{x}} = \infty$$ and from the other side:

$$\lim\limits_{x\to 0^-}\left(\dfrac{\cos{x}}{x}\right)^{\dfrac{\cos{x}}{x}} = 0$$

For this case the two-sided limit didn't exist.

I think $0^0$ cannot be defined in existing calculus.

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The choice of the function $x^x$ to get $0^0$ is completely arbitrary. – Andrea Mori Dec 15 '12 at 21:24
@Andrew Mori : the answer says "$0^0$ CAN be described as..." so there's nothing wrong with his answer. He doesn't say it is a universal definition or whatever. – Patrick Da Silva Dec 15 '12 at 21:27
@PatrickDaSilva : ah, ok. Then $0^0=-\sqrt{\pi}$ – Andrea Mori Dec 15 '12 at 21:34
$0^0$ cannot be described in any of those ways, unless you already know that $(x,y)\mapsto x^y$ is continuous such that it preserves limits. And we well know that it actually isn't continuous, so there is no reason at all to think any of those limits necessarily has anything to do with $0^0$ itself. – Henning Makholm Dec 15 '12 at 21:45
I think the problem with using $x^x$ as an argument to $0^0=1$ is that the same reasoning will tell that $0/0$ should be 1, which seems rather unpleasant. – sdcvvc Dec 15 '12 at 21:45