We know that some real numbers (actually, most of them) are not algebraic and the proof of this fact is beautiful: algebraic numbers, like polynomials with integer coefficients, are countable, contrary to $\mathbb R$, hence there should be some non algebraic numbers.

I was wondering if any real number $\alpha$ is "analytic", in the sense that it exists a power series $\displaystyle{ \sum_{n\geq 0} a_n(x-a)^n }$ of positive radius, with $a_n\in{\mathbb Q}$, and $a_n\in{\mathbb Q}$ for any $n\geq 0$, such that $\alpha$ is a root of $\displaystyle{ \sum_{n\geq 0} a_n(x-a)^n }$.

Of course, the above proof for non algebraic number does not work, since power series are uncountable, but I would be surprised that the answer is no. In that case, is there a simple example and does it change something if we consider Laurent series instead of power series?

Thanks in advance.


Without loss of generality, we may assume $|\alpha| < 1$.

We can construct the desired Taylor series

$$ f(x) = \sum_{n=0}^{+\infty} a_n x^n $$

as follows. Let $f_k(x)$ be the polynomial

$$ f_k(x) = \sum_{n=0}^{k} a_n x^n $$

  • Choose $a_0$ so that $0 < a_0 < \alpha$
  • Choose $a_k$ so that $|f_{k-1}(\alpha) + a_k \alpha^k| < \alpha^{k+1} $ and $|a_k| < 1$.

The radius of convergence of the resulting series will be at least $1$, and since $|f_k(\alpha)| < \alpha^{k+1}$, we have $f(\alpha) = 0$.

  • $\begingroup$ (I know this is picky, but you should show such sequence can be constructed.) $\endgroup$ – Pedro Tamaroff Mar 5 '14 at 4:45
  • $\begingroup$ The thing that makes it obvious to me is that $|f_{k-1}(\alpha)| < \alpha^k$. Thus, we can make any smaller value by adding a multiple of $\alpha^k$ with coefficient less than 1. $\endgroup$ – user14972 Mar 5 '14 at 4:48

I add my own answer, although I prefer Hurkyl's one:

By taking the decimal expansion of $\alpha$, there exists a power series $\displaystyle{f(x)=\sum_{n\geqslant 0}a_nx^n}$ and $a_0\in\mathbb N$ and $a_k\in\{0,1,\dots,9\}$ for $k\geqslant 1$ (hence with radius at least $1$) such that $\alpha=f(\frac{1}{10})$.

We have $\displaystyle{f'(\frac{1}{10})=\sum_{n\geqslant 1}n\frac{a_n}{10^{n-1}}>0}$. By the Inverse Function Theorem, there exists an analytic function $g$ defined on a neighborhood of $\alpha$ such that $g(\alpha)=\frac{1}{10}$ and $g$ is the inverse of $f$ (on a domain small enough).

Since $g$ satisfies, $g\circ f(x)=x$, it is not difficult to check that $g$ has also rational coefficients. So $g-\frac{1}{10}$ is the function we are looking for.


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