# Zero to the Power of Zero

## Contents

## Examples

In light of some mathematicians' (or, indeed, muggles') reticence to define the zeroth power of zero as $1$, the following are examples of reasons why defining $0^0 = 1$ is a good idea.

### Empty Product

We can interpret $0^0$ as meaning:

*zero multiplied by itself zero times*.

Using product notation:

- $0^0 = \displaystyle \prod_{\large \text{false}} 0$

This is a vacuous product, so by definition should be equal to $1$.

### Binomial Theorem

Consider the real polynomial function:

- $y = \left({x + c}\right)^n$

for $n \in \N, c \in \R$.

By the binomial theorem, $y$ contains a term of the form:

- $\dbinom n n x^{n - n} c^n$

If we did not define $0^0 = 1$, $y$ would have a discontinuity at $x = 0$.

This would contradict the theorem that a polynomial is continuous on the entire real line.

### Cardinality of Mappings

By Cardinality of Set of All Mappings, the number of mappings from the empty set to the empty set should be given by:

\(\displaystyle \left \vert {\varnothing^\varnothing}\right \vert\) | \(=\) | \(\displaystyle \left \vert {\varnothing }\right \vert^\left \vert {\varnothing }\right \vert\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 0^0\) | Cardinality of Empty Set |

By Empty Mapping is Unique, there is exactly $1$ such mapping, demanding that $0^0 = 1$.

### Exponential of Zero

- $\exp 0 = 1$

From Power Series Expansion for Exponential Function

- $\exp x = \dfrac {x^0} {0!} + \dfrac {x^1} {1!} + \dfrac {x^2} {2!} + \cdots$

For these theorems to be consistent, it is necessary that:

- $\exp 0 = 1 = \dfrac {0^0} {0!} + 0 + 0 + \cdots$

which holds only if $0^0 = 1$.

### Derivatives

Consider the identity mapping:

- $I_{\mathbb F} \left({x}\right) = x$

where $\mathbb F \in \left\{ {\R, \C}\right\}$.

From Derivative of Identity Function:

- $\dfrac {\d I_{\mathbb F} } {\d x} = 1$

But $I_{\mathbb F} \left({x}\right) = x^1$ is also an order one polynomial.

By Power Rule for Derivatives:

- $\dfrac {\d I_{\mathbb F} } {\d x} = 1 x^0$

As $I_{\mathbb F}$ is differentiable at $0$, for these theorems to be consistent, we insist that $0^0 = 1$.

### As a Limit

Consider the real function:

- $y = x^x$

This function is well defined for $x > 0$.

It is not obvious whether or not the right hand limit:

- $\displaystyle \lim_{x \mathop \to 0^+} y$

exists.

If it does, it would be nice if:

- $\displaystyle \lim_{x \mathop \to 0^+} x^x = 0^0$

Indeed, by Limit of x to the x, we have:

- $\displaystyle \lim_{x \mathop \to 0^+} x^x = 1$

We see that defining $0^0 = 1$ allows $x^x$ to be right-continuous at $x = 0$.

## Historical Note

Whether $0^0 = 0$ or $0^0 = 1$ has been in question since the concept of zero was first raised as a concept.

From one point of view:

- $\forall a \in \R: a^0 = 1$

and so $0^0 = 1$.

From the other point of view:

- $\forall a \in \Z_{\ge 0}: 0^a = 0$

and so $0^0 = 0$.

## Sources

- 1986: David Wells:
*Curious and Interesting Numbers*... (previous) ... (next): $0$ Zero - 1997: David Wells:
*Curious and Interesting Numbers*(2nd ed.) ... (previous) ... (next): $0$ Zero