Polynomials: A Matter of Degrees

Last time we examined why polynomials are defined as they are. This time, let’s look at some tricky aspects of the concept of “degree”, mostly involving something being zero.

Recall that the degree of a polynomial is the highest degree of any of its terms (that is, the degree of the leading term); and the degree of a term is the exponent (a non-negative integer) on the variable (or, when there are more than one variable, the sum of the exponents in the term).

But what is the degree when there is no exponent, namely a constant term? This will be uncertain, then obvious, and then uncertain again!

Degree of a constant term

Here is a question from 2002:

Degree of a Constant

Why is the degree of a constant zero?

We have just finished studying that any number to the zero power is one.

It makes sense that the degree of a monomial like 2x is one, but it doesn't make sense to me that in a constant like 5, the degree is zero.

Joy is referring to the degree of a constant term, such as the 5 in \(x^2+2x+5\). The first term has degree 2, the second 1 because \(2x=2x^1\); but where does the degree of zero come from?

Doctor Ian answered:

Hi Joy,

It looks like you mostly figured it out yourself. Each term in a polynomial is the product of a constant (which may be 1 or 0) and a variable raised to an exponent, e.g., 

  3x^4 + 0x^3 + 1x^2 + 5x^1 + 7x^0

For convenience, we leave out terms where the coefficient is zero, 

  3x^4 + 1x^2 + 5x^1 + 7x^0

and we leave multiplication by 1 implied, 

  3x^4 + x^2 + 5x^1 + 7x^0

We abbreviate x^1 as just x, 

  3x^4 + x^2 + 5x + 7x^0

and we simplify x^0 to 1... which is left implied (see above):

  3x^4 + x^2 + 5x + 7

So coefficients, exponents, and even entire terms can be omitted when they do nothing, but we can still think of them as existing. The constant term is one whose exponent (when written out) is zero. And that’s the degree.

So it's convenient to write a polynomial this way, but it's much cleaner for the definition to require no special cases:

          n
         ---
         \       i
  p(x) = /    a x
         ---   i
         i=0

If you have seen summation notation, this is just another way to write a sum like he started with: $$\sum_{i=0}^n a_ix^i=a_0x^0+a_1x^1+a_2x^2+\dots+a_nx^n$$ This says to add up terms \(a_ix^i\) while \(i\) varies from 0 through n, which is the degree of the polynomial (as long as \(a_n\ne0\)).

When n=0, we get

      0
   a x  = a
    0      0

When you look at it this way, a constant is just a very short polynomial. This is nice, because you can just have one set of rules for dealing with polynomials of any degree, and constants are covered by it.

This, by the way, is a common theme of math: When there is a special case, we make definitions so that we can include it with the others. When we say that each term is a product of a coefficient and some number of variables, that includes the coefficient being 1, or the number of variables being 0 (the degree), just as we saw that a sum of terms can be only one term.

But isn’t that indeterminate?

The previous year, though, we had a question from a teacher who questioned this, for a good reason:

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.

The issue is one we briefly touched on in Zero Divided By Zero: Undefined and Indeterminate. We call the form \(0^0\) indeterminate, meaning that a limit \(\displaystyle\lim_{x\to a}f(x)^{g(x)}\) where both functions approach 0 may take different values depending on how each of them does so. So normally, and particularly in calculus, \(0^0\) is considered undefined, just as \(\frac{0}{0}\) is; we need to use special techniques, such as L’Hôpital’s rule, rather than just substituting and calling it 1. So isn’t \(x^0\) undefined for \(x=0\)?

I answered:

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!

Here is a quote from that FAQ:

Other than the times when we want it to be indeterminate, 0^0 = 1 seems to be the most useful choice for 0^0 . This convention allows us to extend definitions in different areas of mathematics that would otherwise require treating 0 as a special case. Notice that 0^0 is a discontinuity of the function f(x,y) = x^y, because no matter what number you assign to 0^0, you can’t make x^y continuous at (0,0), since the limit along the line x=0 is 0, and the limit along the line y=0 is 1.

This means that depending on the context where 0^0 occurs, you might wish to substitute it with 1, indeterminate or undefined/nonexistent.

I was, until writing this, a little unsatisfied with saying that \(0^0=1\). Context makes a difference.

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!

So we define \(0^0=1\) so that polynomials make sense!

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 the discontinuity 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.

In other words, we can ignore the problem in the context of polynomials, by making an appropriate definition. So that’s what we do!

And what if the constant is zero?

There’s always one more level to take it to, as this 2003 question did:

Degree of Zero

Dear Dr. Math,

I just wanted to know what the degree of zero was.

Thank you.

Now we’re moving from the degree of a constant term, to the degree of a specific constant term, namely 0. Doctor Rick took the challenge, referring first to those two answers above:

Hi, Kristi.

That's an interesting question. I assume you're asking for the degree of the function f(x) = 0 regarded as a polynomial. 

In general, a constant function is regarded as a polynomial of degree zero, as we discuss here:

   Degree of a Constant
   http://mathforum.org/library/drmath/view/61845.html 

   Degree of Constant Function
   http://mathforum.org/library/drmath/view/54602.html 

This is true because a constant such as 2 can be regarded as 2*x^0, and the degree of a polynomial is the highest power of the variable that has a non-zero coefficient.

Are you starting to see the problem?

I presume this is why you asked the question - there is no term in f(x) = 0 that has a non-zero coefficient. So do we say that this function has no degree?

To me, yes, this is the most sensible answer. Think about degree another way: a polynomial has (at most) as many zeros as its degree. (It has exactly as many zeros as its degree if we count the multiplicities of the zeros, for instance, x^3-10x^2+33x-36 = (x-3)^2(x-4) has 3 zeros: 3, of multiplicity 2, and 4, of multiplicity 1.)

A non-zero constant function like f(x) = 5 has zero zeros, in keeping with its degree of zero. But f(x) = 0 has an infinite number of zeros: every value of x is a zero of the function. Thus it makes sense to say that the degree of f(x) = 0 is undefined.

Another way to look at this would be that, if you were given 0 as a term (rather than, as discussed here, the entire function), you would want to ask, does this term represent \(0\) as a constant term, or \(0x\), a linear term, or \(0x^2\), a quadratic term, or what? It could equally well be any of those. So we can’t decide what its degree is; it is, in a way, indeterminate.

The zero polynomial is so special that we can’t assign it a degree. If we could, it would be expected to have that many zeros. At best, then, we could call its degree infinite. But it seems better to say “undefined”.

A polynomial with degree 1, but 3 terms?

A recent question involves a different sort of issue with terms and degrees, so I’ll close with it. This came from Za in February:

I’ve learnt that a linear polynomial is a polynomial with degree as 1 and that it only consists of 2 terms.

But “√2 + 2 + y” is a polynomial with degree 1 even though it consists of 3 terms. So, is it a linear polynomial? If not, why?

A linear polynomial has degree 1, and typically looks like $$ax + b,$$ where a and b are the coefficients.

It can have at most two terms, whose degrees will be 1 and 0. So what is happening in Za’s example? $$\sqrt{2}+2+y$$ Terms are the things we add; and we’re adding three things, not just two!

Doctor Fenton answered:

Hi Za,

Yes, that is a linear polynomial, because the sum 2+√2 is one number (approximately 3.414213562…), so it counts as a single term.  A linear polynomial (in the variable y) has the form

a₁y + a_{0 ,}

where a₁ and a₀ are numbers.  Those numbers might be complicated combinations of other numbers, but they are still a single number, so a₁y is a single term, and the constant a₀ is a single term.

This is something I would never have thought to teach: When we talk about coefficients in a polynomial, and say they can be any kind of number, that includes numbers that can only be written (exactly) as a sum of simpler numbers. In Za’s example, this happens because the constant term is a surd (radical sum); but it also happens when coefficients are complex numbers, like this: $$(1+i)x^2-2ix+(11-3i)$$ We just don’t happen to do that very often!

Note my use of parentheses to write the coefficients. To display Za’s linear polynomial as a sum of two distinct terms, we can similarly write $$y+(2+\sqrt{2})$$