Rational polynomial functions.

Question

What does it mean to have a rational polynomial function like quadratic (2 power), cubic (3 power)? I am aware that an n power polynomial function has n roots (real and lateral), so would a 3.6 power polynomial have 3.6 roots? Why is the power space discrete?

Answer 1

First of all, in addition to the "lateral" question, there is another terminology problem in your question. "Rational polynomial function", or just "rational function" already has a different standard meaning. A "Rational polynomial function" is the ratio of two polynomials.

But obviously, you are referring to functions of the form $$f(x) = a_0x^{r_0} + a_1x^{r_1} + \dots + a_nx^{r_n}$$ where the powers $r_i$ are rational numbers. If there is some standard terminology for such functions, I am not aware of it. "Polynomial" should not be used, as it exclusively means functions calculable using only a finite number of additions and multiplications. Let's call them rational power functions.

All the rules about polynomial roots do not extend to rational power functions. As Peter points out, there is no such thing as "3.6 roots". Roots are values of $x$ that satisfy the equation $f(x) = 0$, and when you talk about the number of them, you are counting, and so will only have cardinals (counting numbers): $0, 1, 2, \dots$. You could possibly have infinitely many, so you could refer to infinite cardinals such as $\aleph_0, \aleph_1, \dots$ or even $c$ (the cardinality of the real numbers, whose place in the aleph scheme is indeterminant). Such numbers as "3.6" and "-1" are not counting numbers, so cannot be the count of the number of roots.

Further, the fact that the number of roots of a complex polynomial, counted by multiplicity, is the degree of the polynomial is an artifact of the nature of polynomials. It does not apply to other functions, including rational power functions. The properties used to prove this for polynomials simply do not hold in general.

And another problem: if $r$ is not an integer (whether it is rational or irrational), then $x^r$ is only well-defined for non-negative real $x$. If $x < 0$ or is not real, then there are multiple possible values that it could take on, and no "obvious choice" that should be taken. We even see some of this when $x$ is positive and real: We know that $2^{1/2}$ should satisfy $\left(2^{1/2}\right)^2 = 2$, but there are actually two numbers that satisfy this: $\sqrt 2$ and $-\sqrt 2$. Which should we choose to be the value of $2^{1/2}$? In this case, one of the possibilities stands out: the positive one. So that is the one we choose. For $x^{1/3}$, it gets even harder: there are there 3 choices: $\sqrt[3] x, \sqrt[3] x\left(\frac{-1 + i\sqrt 3}2\right), \sqrt[3] x\left(\frac{-1 - i\sqrt 3}2\right)$. Again, the real value stands out, so we choose that one.

But when $x$ is negative, or not real. This breaks down. What is $(-4)^{1/4}$? In this case there are four possibilities: $1 + i, -1 + i. 1 - i, -1 - i$. We might go for $1 + i$ automatically, but depending on what we are doing, this is not always the best choice. It may put a discontinuity exactly where we need to look. And that is the problem. Rational powers always have a discontinuity when extended to the complex numbers. As you circle $0$, when you get back to where you started, you find that the function value differs from the one you started with. It will now be a different root than the one you started with. You end up having to pick where this discontinuity occurs, but no one choice stands out.