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We all know that there are algebraic numbers that can't be expressed by radical. For example the real root of the equation $x^5-x+1=0$ (which is near $-1.16$) is algebraic but can't be expressed by radicals.

We say that a function $f(x)$ is algebraic if there exists a polynomial $p(x,y)$ with integer coefficients such that $p\left(x,f(x)\right)=0$. Obviously, polynomial functions, rational functions and irrational functions are algebraic. Is there an algebraic function (defined by a series or by an integral) that isn't polynomial, rational or irrational?

user5402
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  • If $f$ does not need to be continuous consider $f(x) = \sqrt{1/2}$ on $\mathbb{Q}$ and $f(x)=-\sqrt{1/2}$ otherwise. Then $2f^2-1=0$. – Loreno Heer Sep 12 '15 at 17:14
  • @LorenoHeer $f$ doesn't doesn't need to be continuous (although it's better if it was) but $p\not\equiv 0$. – user5402 Sep 12 '15 at 17:17
  • In the example i gave $p(x,y)=2y^2-1$ – Loreno Heer Sep 12 '15 at 17:18
  • Let $\mathbb{E}$ be the set of polyomial, rational and irrational functions and let $A_1,\cdots,A_n$ be a partition of a subset $B$ of $\mathbb{R}$. Let $f_1,\cdots,f_n\in\mathbb{E}$ and let $f$ be defined on $B$ by $f(x)=f_i(x)$ over $A_i$ ($i\in[1,n]\cap\mathbb{N}$). I'm not looking for functions defined this way which merely are restrictions of functions of $\mathbb{E}$ to subsets of $\mathbb{R}$. – user5402 Sep 12 '15 at 17:23
  • We say that a function $f(x)$ is algebraic if there exists a non-zero polynomial $p(x,y)$ with integer coefficients such that $p\left(x,f(x)\right)=0$ for all $x$. – jjagmath Aug 03 '22 at 22:00

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Rational functions and root functions have a function term that is a radical expression. That are the explicit algebraic functions.
But just as there are algebraic numbers that cannot be represented by radicals, there are algebraic functions (the implicit algebraic functions) whose function term cannot be represented by a radical, see e. g. the algebraic function $f$ that is determined by the irreducible algebraic equation

$$f(x)^5-f(x)+1=0.$$

The function can be represented by series or integrals.
see e.g. Is a polynomial equation of degree $\ge 5$ not solvable by any way?

IV_
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