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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?

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    $\begingroup$ 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 $\endgroup$ Commented Aug 2 at 16:33
  • $\begingroup$ @hardmath Thanks for catching that, edited! $\endgroup$ Commented Aug 2 at 16:36
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    $\begingroup$ Also, $0$ would work, so perhaps you want to exclude constants. $\endgroup$ Commented Aug 2 at 16:57
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    $\begingroup$ mathoverflow.net/a/434311/95172 looks like this works $\endgroup$
    – QC_QAOA
    Commented Aug 2 at 17:03
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    $\begingroup$ 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. $\endgroup$ Commented Aug 2 at 17:14

1 Answer 1

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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 $|z| \le n$.

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  • $\begingroup$ Something is wrong in your equation at the start of the second paragraph, where $z$ only appears on the RHS. $\endgroup$ Commented Aug 3 at 0:44
  • $\begingroup$ Thanks (fixed it) $\endgroup$ Commented Aug 4 at 6:33

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