Is a real power series that maps rationals to rationals defined by a rational function? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T02:08:57Z http://mathoverflow.net/feeds/question/42460 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/42460/is-a-real-power-series-that-maps-rationals-to-rationals-defined-by-a-rational-fun Is a real power series that maps rationals to rationals defined by a rational function? SJR 2010-10-17T05:41:14Z 2010-10-17T17:34:29Z <p>Suppose that the function $p(x)$ is defined on an open subset $U$ of $\mathbb{R}$ by a power series with real coefficients. Suppose, further, that $p$ maps rationals to rationals. Must $p$ be defined on $U$ by a rational function? </p> http://mathoverflow.net/questions/42460/is-a-real-power-series-that-maps-rationals-to-rationals-defined-by-a-rational-fun/42464#42464 Answer by Pietro Majer for Is a real power series that maps rationals to rationals defined by a rational function? Pietro Majer 2010-10-17T06:47:32Z 2010-10-17T17:34:29Z <p>You can map $\mathbb{Q}$ in itself by plenty of entire functions that are not rational functions. Let fix an enumeration of the rationals, $\mathbb{Q}=\{q _ j\ : j=1,2,\dots \}$. Consider a series $$f(x):=\sum_{n=1}^\infty\ \epsilon _n \ \prod _{j=1}^n (x-q _j)$$ If $\epsilon _n$ is a sequence of rationals converging to 0 with sufficient velocity, the series converges uniformly on bounded sets to an entire function.</p> <p>(edit) <strong>rmk.</strong> of course with some more small care we can even make an entire function $f(x)=\sum_{n=0}^\infty p _n(x)$ invertible over $\mathbb{R}$, and a bijection between two assigned countably infinite dense subsets $A$ and $B$. Start the series with the identity $p_0(x)=x$, then add inductively only odd degree polynomials $p _n$, that vanish on the (finitely many) already settled points, and do not destroy the invertibility on $\mathbb{R}$ (say, keeping all partial sums of the series with derivative greater than $1/2$). The bijectivity between $A$ and $B$ is to be ensured by a standard ping-pong argument.</p> http://mathoverflow.net/questions/42460/is-a-real-power-series-that-maps-rationals-to-rationals-defined-by-a-rational-fun/42465#42465 Answer by Lavender Honey for Is a real power series that maps rationals to rationals defined by a rational function? Lavender Honey 2010-10-17T06:49:35Z 2010-10-17T06:49:35Z <p>No. In fact, $p(x)$ can be a complex analytic function with rational coefficients that takes any algebraic number $\alpha$ in an element of $\mathbb{Q}(\alpha)$. (And everywhere analytic functions are not rational unless they are polynomials).</p> <p>The algebraic numbers are countable, so one can find a countable sequence of polynomials $q_1(x), q_2(x), \ldots \in \mathbb{Q}[x]$ such that every algebraic number is a root of $q_n(x)$ for some $n$. Suppose that the degree of $q_i(x)$ is $a_i$, and choose integers $b_i$ such that $$b_{n+1} > b_{n} + a_1 + a_2 + \ldots + a_n.$$</p> <p>Then consider the formal power series:</p> <p>$$p(x) = \sum_{n=0}^{\infty} c_n x^{b_n} \left( \prod_{i=0}^{n} q_i(x) \right),$$</p> <p>By the construction of $b_n$, the coefficient of $x^k$ for $k = b_n$ to $b_{n+1} -1$ in $p(x)$ is the coefficient of $x^k$ in $c_n x^{b_n} \prod_{i=1}^{n} q_i(x)$. Hence, choosing the $c_n$ to be appropriately small rational numbers, one can ensure that the coefficients of $p(x)$ decrease sufficiently rapidly and thus guarantee that $p(x)$ is analytic.</p> <p>On the other hand, clearly $p(\alpha) \in \mathbb{Q}[\alpha]$ for every (algebraic) $\alpha$, because then the sum above will be a finite sum.</p> <p>With a slight modification one can even guarantee that the same property holds for all derivatives of $p(x)$.</p> <p>I learnt this fun argument from the always entertaining Alf van der Poorten (who sadly died recently).</p>