# Does there exist a power series which sends every algebraic number in its radius of convergence to a rational number?

Section 1.3 of https://swc-math.github.io/aws/2008/08BeukersNotesDraft.pdf claims there exists a non-constant power series $$f(x)$$ with positive radius of convergence $$\rho,$$ such that for any algebraic number $$a$$ with $$|a| < \rho,$$ the value $$f(a)$$ is rational. The author says a proof can be found in [], presumably meaning they meant to add a citation but forgot.

Where can I find a reference to this power series?

• I think this is relevant (but I'm in the middle of something else now and don't have time to give it much thought): Functions that take rationals to rationals Commented Aug 2 at 16:33
• @hardmath Thanks for catching that, edited! Commented Aug 2 at 16:36
• Also, $0$ would work, so perhaps you want to exclude constants. Commented Aug 2 at 16:57
• mathoverflow.net/a/434311/95172 looks like this works Commented Aug 2 at 17:03
• The algebraic numbers are dense in $\mathbb C$, but all rationals are in $\mathbb R$. The only analytic functions taking a nonempty open subset of $\mathbb C$ into $\mathbb R$ are constants. Commented Aug 2 at 17:14

Given any sequence $$\{x_n\}_{n\ge 1}$$ of distinct reals, I will define a non-constant entire function $$f$$ that takes $$S$$ into the rationals. This will be of the form $$f(z) = \sum_{n=1}^\infty c_n \prod_{k < n} (z - x_k) \tag 1$$ Note that for such a function, $$f(x_j) = \sum_{n=1}^{j} c_n \prod_{k < n} (x_j - x_k)$$, and given real numbers $$c_1, \ldots, c_{j-1}$$, there is a dense set of reals $$c_j$$ for which this will be rational. We just need to ensure that the series (1) converges uniformly on compact sets: this can be done by taking $$|c_n| \le b_n$$, where $$b_n > 0$$ is small enough that, say, $$b_n \prod_{k < n} |z - x_k| < 2^{-n}$$ for $$|z| \le n$$.
• Something is wrong in your equation at the start of the second paragraph, where $z$ only appears on the RHS. Commented Aug 3 at 0:44