Does 0^0 equal 1, or is it undefined?

1. May 15, 2007

Loren Booda

Does 0^0 equal one, or is it undefined?

2. May 16, 2007

chroot

Staff Emeritus
It's an indeterminate form. (It's many-valued; not the same as undefined.)

- Warren

3. May 16, 2007

Loren Booda

Is the limit of sin(x), as x approaches infinity, also "indeterminate"?

4. May 16, 2007

Kuno

It does not exist because the values oscillate between 1 and -1.

5. May 16, 2007

fopc

There are three interpretations (each one depends on context):

0^0 = 1, or indeterminate form, or undefined

The first is the set-theoretic interpretation. Justification?

Consider f:A->B. The set of all such functions is denoted B^A. In this context, {}^{} would represent the set of
functions f:{}->{}, and 0^0 would represent the number of functions in this set. There is only one such function.
(This is not the only justification for this particular interpretation, i.e., 0^0 = 1.)

It's a strange post. A cursory preliminary investigation would turn up something like the following:

http://en.wikipedia.org/wiki/Exponentiation#Zero_to_the_zero_power

Last edited: May 16, 2007
6. May 16, 2007

christianjb

In combinatorics it's usually convention to use 0^0=1. This makes some problems easier (less book-keeping).

Plot y^x and x^y and see what you get as y->0. The answer clearly depends on the direction of approach.

7. May 16, 2007

D H

Staff Emeritus
Just to add one very important interpretation of when 0^0 is defined to be 1: Power series.

In power series, the form 0^0 must be taken to mean 1. If not, we would have to write $\exp(x) = 1 + \sum_{n=1}^{\infty} x^n/n!$ rather than $\exp(x) = \sum_{n=0}^{\infty} x^n/n!$

8. May 16, 2007

Gib Z

And for the binomial theorem:

$$(1+x)^n = \sum_{k = 0}^n \binom{n}{k} x^k$$

Its not valid for x=0 except when defined 0^0 = 1

9. May 18, 2007

ZioX

We really only say indeterminate when we're talking about limits. It's either undefined, or defined arbitrarily depending on context.