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I would like to ask the following question. This is related to one lemma that I need in a recent research on the arithmetic behavior of transcendental functions with integer coefficients.

Let $\alpha, \beta\in (-1,1)$, with $\alpha\neq \pm \beta$. I would like to prove that there exists a function $f\in \mathbb{Z}[[z]]$ analytic in the unit ball and such that $f(\beta)=0$ and $f(\alpha)\neq 0$.

I am able to prove this when $\alpha\cdot \beta<0$ and when $|\beta|>|\alpha|$.

Any help?

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    $\begingroup$ How did you prove it for the cases you mentioned? $\endgroup$
    – user78249
    Commented Nov 17, 2016 at 22:57
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    $\begingroup$ Is this a homework question? $\endgroup$
    – Anthony Quas
    Commented Nov 17, 2016 at 23:06
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    $\begingroup$ it could be, but it appears in order to prove one lemma in one of my recent works. $\endgroup$
    – Diego
    Commented Nov 17, 2016 at 23:14
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    $\begingroup$ @PietroMajer When an anonymous asks to prove a homework-level "lemma", the burden is not us to prove that it is a student trying to have their homeworks solved. $\endgroup$
    – Boris Bukh
    Commented Nov 18, 2016 at 2:17
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    $\begingroup$ @BorisBukh Of course the burden is not to us. On the other hand, if we raise a mistrust of somebody's purposes, it is generally acknowledged that we should at least provide a more solid reason than just smell. That's why (IMHO) it is better to stay on the question, objecting either its content or form, not the questioner's good faith. If a question appears too elementary to me, I kindly suggest a more suitable site. But note, what is trivial to you may be difficult to other professional mathematicians, just because it is not their field, and this is a situation where MO proves greatly useful. $\endgroup$
    – Pietro Majer
    Commented Nov 18, 2016 at 10:06

1 Answer 1

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We may assume w.l.o.g $0<\beta<1$. Write $\beta^{-1}$ in the $ \beta^{-2}$ expansion as: $$\beta^{-1}=\sum_{k=0}^\infty d_k\beta^{2k}$$ with integer digits $0\le d_k<\beta^{-2}$, and define $$f(x):=-1+\sum_{k=0}^\infty d_kx^{2k+1}$$ Then $f\in\mathbb{Z}[[x]]$ is analytic in the unit disk, strictly increasing on the interval $(-1,1)$, and $f(\beta)=0$.

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    $\begingroup$ @Diego $f$ is strictly increasing, so it has only one zero. $\endgroup$
    – Wojowu
    Commented Nov 18, 2016 at 11:12
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    $\begingroup$ Thanks @Pietro, it works. Another question, how to manage when $\alpha$ and $\beta$ are complex numbers inside the unit ball with $\alpha$ different of $\beta$ and $\overline{\beta}$? $\endgroup$
    – Diego
    Commented Nov 18, 2016 at 12:12
  • $\begingroup$ Good question! Making $f\in\mathbb{Z}[[x]]$ an univalent (slicht) map on the unit disk seems hard to me (but it should be known, and it is worth posting a new question). The sufficient condition on coefficients for univalence I know is Alexander's one: $n|a_n|$ be decreasing, but for $f\in\mathbb{Z}[[x]]$ it implies $f$ is a polynomial. On the other hand, de Branges' necessary condition $|a_n|\le n|a_1|$ leaves some hope. If univalence proves hard or impossible, one should try to build an $f$ depending on both $\alpha$ and $\beta$, not only on $\beta$, possibly not slicht. $\endgroup$
    – Pietro Majer
    Commented Nov 18, 2016 at 13:18
  • $\begingroup$ Thanks @PietroMajer. I already posed the other question. I think the better idea is to consider a construction depending on both, but I am afraid that I need to split such a construction in many cases depend on the their arguments (rational, irrational, linearly independent, etc). $\endgroup$
    – Diego
    Commented Nov 18, 2016 at 20:42
  • $\begingroup$ If you are ok with $f$ depending on both $\alpha$ and $\beta$, then I think the above construction can be repeated. "Complex base system" can be considered like in the real case. I think, given a base $\beta$, a large enough set of coefficients $C$ allows to represent any complex number, so you can make plenty of $f\in\mathbb{Z}[[z]]$ with bounded coefficients such that $f(\beta)=0$; then also getting $f(\alpha)\neq0$ should not be difficult. $\endgroup$
    – Pietro Majer
    Commented Nov 18, 2016 at 21:39

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