On the definition of algebraic/transcendental functions.

Question

On Wikipedia it says ``Formally, an analytic function $f(z)$ of one real or complex variable $z$ is transcendental if it is algebraically independent of that variable. This can be extended to functions of several variables.''

I have a real analytic function $f$ which is defined only on some closed and bounded interval $I$. Suppose there exists a polynomial $g(x, y)$ such that $$ g(x, f(x)) = 0 $$ on $I$. Does it then mean that $f$ is algebraic? I was not sure because this holds only on $I$... Any comments would be appreciated. Thank you.

Answer 1

To make the algebraic functions on $\Bbb{R}$ into a field (or any connected subset of $\Bbb{C}$), the easiest way is to choose a branch of $\log$, set $x^{1/n} = e^{\log(x)/n}$, then $f$ is algebraic iff around each $a \in \Bbb{R}$, $f(a+x) = F(x^{1/n})$ for some $F$ meromorphic algebraic at $0$, which means for some $r> 0$ and polynomials $c_n(z)$, for $0 < |z|< r$, $\sum_{n=0}^N c_n(z) F(z)^n = 0$.

Answer 2

An algebraic function is algebraic on its whole domain.

A characteristic property of algebraic functions is that the function values are algebraic for all algebraic places in the domain of the function.

$f$ is algebraic iff there exists a polynomial $g\in\mathbb{Q}[x,y]\setminus (\mathbb{Q}[x]\cup\mathbb{Q}[y])$ that is not a zero polynomial and if $g(x,f(x))=0$ on dom($f$).