Analytic functions with algebraic Taylor coefficients at some point. - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T07:13:40Z http://mathoverflow.net/feeds/question/73560 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/73560/analytic-functions-with-algebraic-taylor-coefficients-at-some-point Analytic functions with algebraic Taylor coefficients at some point. wood 2011-08-24T12:13:31Z 2011-08-26T04:10:49Z <p>This question just came to my mind when reading the question</p> <p><a href="http://mathoverflow.net/questions/73553/when-may-function-meromorphic-be-expanded-as-power-series-with-coefficients-of" rel="nofollow">http://mathoverflow.net/questions/73553/when-may-function-meromorphic-be-expanded-as-power-series-with-coefficients-of</a></p> <p>Suppose $f$ is an analytic function on some open subset $U \subseteq \mathbb{C}$. Are there sufficient or necessary conditions to be put on $f$ that $f$ has the following property. There exist a point $\zeta \in U$ such that the Taylor expansion of $f$ around $\zeta$ has only rational (algebraic) coefficients.</p> <p>More generally, let us call the set of functions with this property $\mathcal{R}(U)$. It is easy to see that $\mathcal{R}(U)$ is dense in the set of all analytic functions with respect to the topology given by locally uniform convergence, since all polynomials with rational coefficients are contained in $\mathcal{R}(U)$ and are dense. But of course $\mathcal{R}(U)$ does not contain all analytic functions since for example $z^2+\pi$ or constant functions are not all contained. Also $\mathcal{R}(U)$ is uncountable, since $z-\zeta$ for any $\zeta \in \mathbb{C}$ satisfies the property. </p> <p>So what can be said about $\mathcal{R}(U)$? Is it in some sense interesting?</p> http://mathoverflow.net/questions/73560/analytic-functions-with-algebraic-taylor-coefficients-at-some-point/73630#73630 Answer by Peter Luthy for Analytic functions with algebraic Taylor coefficients at some point. Peter Luthy 2011-08-25T04:20:37Z 2011-08-26T04:10:49Z <p>Suppose you have a polynomial $p(z)=a_nz^n+a_{n-1}z^{n-1}+...+a_0$. Suppose that $p$ has the property you desire, which is equivalent to all of the derivatives of $p$ being simultaneously rational at some point $c$. In particular, this means $a_n$ must be rational, and so we may assume wlog that $a_n=1$. </p> <p>If $a_{n-1}\in\mathbb{Q}$, then clearly $c$ must be rational, in which case it is immediate that all the $a_i$ are rational and we are done.</p> <p>So, suppose that $a_{n-1}$ is not rational. Then by looking at the $(n-1)$th derivative at $c$, it must be that $a_{n-1}+nc$ is a rational number. Thus $a_{n-1}$ and $c$ must either both be algebraic or both be in the algebraic numbers adjoin $a_{n-1}$. Looking at the higher derivatives, one sees that all the terms are polynomially related, and so all the $a_i$ must all lie in the algebraic closure of the algebraic numbers adjoin $a_{n-1}$.</p> <p>This necessary condition guarantees that almost none of the polynomials are in $\mathcal{R}(\mathbb{C})$ and, as a consequence, almost no analytic function is in there either. This is despite them being dense with respect to the topology you mentioned.</p> <p>For a necessary condition, one can again see that it comes down to the value of $a_{n-1}$. One can "easily" iteratively generate a list of exact polynomial relations between $a_{n-1}$ and each of $a_i$ with $i\le n-2$. This isn't exactly a pleasant relation, however, since the relations are different for different degrees and also one ends up with complicated polynomials. For fourth- and lower-degree polynomials, this allows you to get exact relations in terms of radicals. However, for higher degree polynomials, this problem is decidedly more complicated because finding roots even for 5th degree polynomials is not universally possible. It might be that the polynomials in question are all of a particularly nice flavor which happen to be solvable nicely by radicals, but I don't know enough about that kind of thing. My guess is it is likely a hopeless venture.</p>