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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?

edited Jan 11 2011 at 0:42

Sean Tilson
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asked Jan 11 2011 at 0:35

Joakim
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Limits in what sense? – Thierry Zell Jan 11 2011 at 0:39

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Are we overloading the word "closure"? – Igor Rivin Jan 11 2011 at 0:40

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Isn't $f(x)=sin(x)$ a limit of polynomials but not the solution to a real rational runction? – Sean Tilson Jan 11 2011 at 0:47

Point-wise limits with respect to the standard distance. – Joakim Jan 11 2011 at 0:47

quid · Answer 1 · 2011-01-11 00:52:45Z UTC

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.)

algebraic vis-a-vis analytic functions

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