Why Does 0^0 = 1 and Not Undefined?

Date: 11/30/2007 at 15:31:42
From: Jesse
Subject: 0^0

Your proof for why x^0 = 1 uses a law which breaks down at x = 0. 
Then in your definition for 0^0 you side significantly in favor of 
0^0 = 1 based on your rule for x^0 = 1 (which was based on a law that
breaks down at 0).  Based on what I've read I would side in favor of 
undefined.  Are there any more conclusive reasons for siding with 
0^0 = 1?

Date: 11/30/2007 at 23:10:49
From: Doctor Peterson
Subject: Re: 0^0

Hi, Jesse.

I presume you are referring to our FAQs:

  N to 0 power
    http://mathforum.org/dr.math/faq/faq.number.to.0power.html 

  0 to 0 power
    http://mathforum.org/dr.math/faq/faq.0.to.0.power.html 

You can also find a restatement of the 0^0 issue here:

  Proof That 0/0 = 1 Based on x^0 Equaling 1?
    http://www.mathforum.org/library/drmath/view/69917.html 

Ultimately, the answer really depends on your context, as both 0^0
discussions above recognize to different extents.  It HAS to be taken
as an indeterminate form in calculus, because different limits that
reduce to 0^0 have different values; we can't just say, "Oh, 0^0 is
defined as 1, so that's the limit".  But in many specific formulas or
types of equations, taking 0^0 as 1 is necessary in order to write the
formula without exceptions.

For me, the clincher was when I realized that I've always talked about
the constant term in a polynomial as the zero degree term, and yet if
you write

  ax^2 + bx + c = ax^2 + bx^1 + cx^0

and don't take 0^0 as 1, you've changed the polynomial, which is
defined for all x, into something that is undefined for x=0!  I've
never intended to do that, and never even thought about it, until a
student pointed it out.  Now I'm a believer: 0^0 not only must be, but
simply IS, taken as 1 in many cases in ordinary algebra.

But that doesn't mean we can automatically assume that it makes sense
to define it that way in any context we come across.  We have to keep
our eyes open, and determine whether a new context is one in which
this definition fits.

If you have any further questions, feel free to write back.


- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/