Degree of Constant FunctionDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/