Degree of Constant Function

Date: 11/08/2001 at 19:07:59
From: Masha Albrecht
Subject: Degree of constant function

Dear Dr Math,

I teach at Galileo High School in San Francisco and we teachers are 
stumped on this question:

First, we agree that F(x) = 1 is a polynomial function of degree 0. 
We tell the kids that this is true because this function is equivalent 
to the function F(x) = 1x^0.

My students recently pointed out that these functions are not 
equivalent. This is because F(x) = 1 has a domain over all the reals, 
but F(x) = 1x^0 has a discontinuity at x = 0.

Now we think F(x) = 1x^0 is not a polynomial function (because 
polynomials shouldn't have discontinuities), but F(x) = 1 is a 
polynomial. And F(x) = 1 still has degree 0 but for reasons we can't 
explain well.

Help!
Masha Albrecht

Date: 11/08/2001 at 23:15:15
From: Doctor Peterson
Subject: Re: Degree of constant function

Hi, Masha.

Have you seen the Dr. Math FAQ on 0^0?

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

I've never been quite happy with what that says, but you've convinced 
me that it's right!

Although 0^0 is properly considered indeterminate, it is for many 
purposes taken to be equal to 1. What you describe is a very good 
reason for doing so: it makes x^0 a continuous function, _always_ 
equal to 1. Having made that definition, there's no more trouble!

It's important to distinguish between a discontinuity like that of 
x^-1 at x = 0, where the function is actually undefined and the 
discontinuity is not removable, and one like x^0, where it is only 
indeterminate and can be removed by a proper definition. Even if there 
were no consensus on doing this with 0^0 in general, there would be no 
problem with calling a constant function a degree-zero polynomial, 
because the difficulty is so easily dealt with.

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