Consider functions $\mathbb{R}^n\to\mathbb{R}$. If the set of algebraic functions is the algebraic closure of the rationals, does this mean that limits of polynomials are algebraic? In particular, are real-analytic functions algebraic?
-
1$\begingroup$ Limits in what sense? $\endgroup$– Thierry ZellJan 11, 2011 at 0:39
-
2$\begingroup$ Are we overloading the word "closure"? $\endgroup$– Igor RivinJan 11, 2011 at 0:40
-
3$\begingroup$ Isn't $f(x)=sin(x)$ a limit of polynomials but not the solution to a real rational runction? $\endgroup$– Sean TilsonJan 11, 2011 at 0:47
-
$\begingroup$ Point-wise limits with respect to the standard distance. $\endgroup$– JoakimJan 11, 2011 at 0:47
1 Answer
Algebraic means that the function satisfies a polynomial equation (whose coefficients are also polynomial), see http://en.wikipedia.org/wiki/Algebraic_function.
There are real analytic functions that are not algebraic, for example exp(x), sin(x), cos(x), see http://en.wikipedia.org/wiki/Transcendental_function.
To recall an analogy: an algebraic number (over the rationals) is one that is the solution of a polynomial equation with rational coefficients.
Numbers that are not algebraic are called transcendental.
And, limits of sequences of algebraic numbers are not necessarily algebraic. (Every real number is the limit of a sequence of rationals.)