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Consider the set $\mathbb{Z}_P[[X]]=\left\{ \sum_{n=0}^\infty a_n X^n \in \mathbb{Z}[[X]]\mid \exists k \in \mathbb{N_0}\colon (a_n)_{n\in\mathbb{N}_0} \in \mathcal{O}(n^k)\right\}$. It is easy to show that the elements of $\mathbb{Z}_P[[X]]$ have radius of convergence equal to 1, except for the polynomials.

I am interested in the set $N_P =\left\{ \beta \in (-1,1) \mid \exists f(X) \in \mathbb{Z}_P[[X]] \colon f(\beta) = 0\right\}$. We have $\mathbb{A} \cap (-1,1) \subseteq N_P$, were $\mathbb{A}$ are the algebraic numbers.

Question: What is known about the set $N_P$? Do we have $N_P = \mathbb{A} \cap (-1,1)$, or $N_P = (-1,1)$, or neither?

Some things to note:

  • If we consider uniformly bounded coefficents – that is, we replace $\mathcal{O}(n^k)$ above with $\mathcal{O}(1)$ – the functions we get are all rational functions (see for example this paper by Borwein et al.). The set of corresponding zeros is thus equal to $\mathbb{A} \cap (-1,1)$. Edit: I may have misunderstood the paper. I will follow up on this. Edit 2: I definitely misunderstood the paper. The functions are not in general rational functions.
  • The following theorem holds: For $\gamma \in (-1,1)$ there is a power series with integer coefficients $g(X) \in \mathbb{Z}[[X]]$ such that $g(\gamma) = 0$ (see for example this math.stackexchange question). The question is whether one can construct $g$ in such a way that the coefficients are polynomially bounded.
Vincent
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1 Answers1

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The set $N_P$ is in fact equal to $(-1, 1)$, even if we restrict ourselves to uniformly bounded coefficients:

Theorem: Let $\gamma \in (-1,1)$. Then there exists a sequence $(a_n)_{n \in \mathbb{N}_0}$ of integers and $M > 0$ such that:

  1. $\forall n \in \mathbb{N}_0\colon |a_n| \leq M$
  2. $\exists n \in \mathbb{N}_0\colon a_n \neq 0$
  3. $\sum_{n=0}^\infty a_n \gamma^n = 0$.

Proof Without loss of generality $\gamma > 0$, otherwise just multiply the coefficients by $(-1)^n$. Let $\beta = \frac{1}{\gamma} > 1$, $N = \lceil\beta\rceil$.

Consider the $\beta$-expansion of $N$: $$ N = \sum_{n=-\infty}^k d_n \beta^n $$ where $k$ and $d_n$ are integers and $0 \leq d_n < \beta$ for all $n \in \{k, k-1, k-2, \ldots\}$. Notice that there must be some $n \neq 0$ with $d_n \neq 0$. Otherwise we would have $N = d_0$, which is impossible, since $N \geq \beta > d_0$.

Rearranging, we see that we get the non-trivial representation $$ \begin{aligned} 0 =& \left(\sum_{n=-\infty}^{-1} d_n \beta^n\right) + (d_0-N) + \left(\sum_{n=1}^{k} d_n \beta^n\right)\\ =& \left(\sum_{n=1}^{k} d_n \gamma^{-n}\right)+ (d_0-N) + \left(\sum_{n=1}^{\infty} d_{-n} \gamma^n\right). \end{aligned} $$ Upon multiplication with $\gamma^k$, we see that $$ 0 = \left(\sum_{n=0}^{k-1} d_{k-n} \gamma^n\right) + (d_0-N)\gamma^k + \left(\sum_{n=k+1}^{\infty} d_{k-n} \gamma^{n}\right). $$ Since the coefficients are bounded by $\max(\beta, |d_0-N|)$, the proof is complete.

Vincent
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