# Are polynomial functions with fractional exponents transcendental?

I'm having trouble categorizing fractional--order systems, that means functions like

$$f(x) = K \cdot (1 + \sqrt{x})$$

or more generally, e.g.

$$g(x) = \dfrac{a_0 + a_1 x^{1/2} + a_2 x^{1} + a_3 x^{3/2}}{b_0 + b_1 x^{1/2} + b_2 x^{1} + b_3 x^{3/2}}$$

where the exponents are rational numbers $\gamma \in \mathbb{Q}$. All coefficients and $K$ are real numbers. $x$ is, in my case, complex.

I doubt they can still be refered as rational functions, can they? But on the other hand, are these functions transcendental functions?

The definition on Wikipedia is not entirely clear to me on that matter:

A transcendental function is an analytic function that does not satisfy a polynomial equation, in contrast to an algebraic function.1[2] In other words, a transcendental function "transcends" algebra in that it cannot be expressed in terms of a finite sequence of the algebraic operations of addition, multiplication, and root extraction.

Would the definition change, if we include real exponents $\gamma \in \mathbb{R}$?

And what happens in case of the implicit form:

$$h(x) = K \cdot \sqrt{1 + x}$$

I'm a little confused about all these definitions and I hope you can shed some light on my problem.

$$f(x)=1+\sqrt x$$ is algebraic because it satisfies the polynomial equation

$$(f(x)-1)^2-x=0.$$

Same goes with your function $g(x)$, though the polynomial will be of a much higher degree.

Transcendence begins with irrational exponents, as they would require a polynomial of infinite degree.

The confusion arises because it is quite unclear in which ring the coefficients of the polynomial are supposed to be. The notion of algebraicity can only be defined properly if it is clear what is meant by a polynomial equation. Visiting the French article on the same subject, we see that the coefficients are supposed to be polynomials over the ground field, which in this case are the complex numbers.

Based on that definition, I would say that the first function is indeed algebraic, because it satisfies a quadratic equation, as is easily seen. (Given that $K$ denotes a constant.)

The second function should also be algebraic, because one may use polynomials to counteract growing exponents in the denominator. (I don't see the equation immediately.)

• In general, transcendental means precisely "not algebraic". Once one settles for a polynomial ring, the definitions become clear. – AlgebraicsAnonymous Jul 16 '18 at 9:10
• I gave some more information about the other variables. – thewaywewalk Jul 16 '18 at 9:11