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The following expressions in examples aren't polynomial expressions:

$$2x^2-5x+(3/x)$$ $$9- \sqrt x$$

Neither they are rational expressions, I've been just told that by book. but then what do you call them?

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Functions?${}{}$ – Asaf Karagila Oct 9 '11 at 18:12
The first one is a rational function: $$2x^2-5x+3/x=\frac{2x^3-5x^2+3}{x}$$ – SL2 Oct 9 '11 at 18:13
I'm not sure what the context is, but under many understandings the first one could be called rational. – Mark Bennet Oct 9 '11 at 18:14

The two examples you gave are both algebraic functions, i.e. functions that satisfy a polynomial equation whose coefficients are rational functions. For example, $$f=9-\sqrt{x}$$ satisfies the polynomial equation $$(y-9)^2-x=y^2-18y+(81-x)=0,$$ i.e. $y=f$ is a root of this polynomial, and the coefficients of the polynomial are the rational functions $1$, $-18$, and $81-x$. However, as has been pointed out in the comments, $$g=2x^2-5x+\tfrac{3}{x}$$ is itself a rational function, so there is a particularly simple polynomial it satisfies (specifically, a linear polynomial) $$y-(2x^2-5x+\tfrac{3}{x})=0.$$

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the rational function $2x^2-5x+3/x$ satisfies that equation but that's not what you want to say, right? – Rayleigh Oct 9 '11 at 20:05
No, it is what I want to say. What do you think I should be saying? – Zev Chonoles Oct 9 '11 at 20:17
I up-voted your answer, since it is the right answer to the question, but I think I'd like to second Rayleigh's complaint about your examples. Following your example, we can conclude that log x and sin(x) and erf(x) are algebraic functions, since they satisfy y - log x = 0 and y - exp(x) = 0 and y - erf(x) = 0. I think you want the equation they satisfy to be polynomials, which yours is not. – ziggurism Oct 10 '11 at 0:36
@ziggurism: Thank you for your upvote! Actually, we can't conclude that, because $\log$ and $\exp$ and $\text{erf}$ aren't rational functions, so the expressions $y-\log(x)$ etc. are not polynomials with coefficients that are rational functions, so the fact that $\log(x)$ etc. satisfy them does not imply that they are algebraic. – Zev Chonoles Oct 10 '11 at 0:56
Yes, I see. Your definition was an algebraic function is one who satisfies a polynomial equation whose coefficients are rational functions. In that case, my previous comment was incorrect, and what you said is correct. A rational function satisfies a trivial polynomial. I started to write further that I thought it would be more natural to define an algebraic function as one satisfying a polynomial with real (or perhaps even integer) coefficents, in analogy with algebraic numbers. So $9- \sqrt x$ is the inverse of $(9 - y)^2$. But this doesn't do the job. You can't make $1/x$ this way. – ziggurism Oct 10 '11 at 1:30

I suppose you could call them Puiseux polynomials by analogy with Puiseux series, though I'm not sure anyone has ever done so.

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You can call them Puiseux series! – Mariano Suárez-Alvarez Oct 10 '11 at 3:57
@Mariano, sure, but then you could call polynomials "power series," and you'd be losing information if you did. By the way, I have found that the computer algebra program SAGE actually uses the term, "Puiseux polynomials," see – Gerry Myerson Oct 10 '11 at 5:00

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