Number of roots of a polynomial of non-integer degree How many roots does a polynomial-like function of degree $n$ have if $n$ is a rational or an irrational number?
 A: You mean something like $x^{\sqrt 7}+4x^2+x^{1/2}$? That's not usually called a polynomial at all -- it is implicit in the word "polynomial" that all of the exponents must be constant nonnegative integers.
A sum of terms of this form will generally only make sense for positive $x$. The number of positive real roots can still be bounded using Descartes' rule of signs if you sort the terms by decreasing exponents. In particular there cannot be more positive roots than there are terms.
(To see that this is true for rational exponents, substitute $y=x^{1/m}$ where $m$ is a common multiple of all the denominators, which makes the result an actual polynomial in $y$, with the same sign changes. By continuity, it then also has to be true for irrational exponents. If there are negative exponents, multiply through by a large enough power of $x$, which doesn't change the number of positive roots either).
A: A "generalized polynomial" with rational exponents can be turned to an ordinary polynomial by the change of variable $x=t^g$, where $g$ is the least common multiple of the denominators of all exponents.
Example:
$$4x^2-2\sqrt[3]x+3x^{4/7}=0$$
becomes
$$4t^{42}-2t^7+3t^{12}=0$$
which has $42$ (possibly complex) roots in $t$.
Things are less easy for irrational exponents as you need to choose some principal branch when taking the exponential ($z^r:=|z|^re^{ir\arg(z)}$), and a complex number then has a single irrational power (while it has $n$ $n^{th}$ roots). The fundamental theorem of algebra does not hold anymore, and usually you limit yourself to real positive values of the unknown $x$.
My guess (just a guess) is that for $x>0$, as every term is monotonous, the number of roots can lie between $0$ and the number of terms minus $1$.
