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Transcendence measure of $e^{1/n}$. (arXiv:2303.05542v1 [math.NT]) http://arxiv.org/abs/2303.05542

Transcendence measure of $e^{1/n}$

For a given transcendental number $ξ$ and for any polynomial $P(X)=: λ_0+\cdots+λ_k X^k \in \mathbb{Z}[X]$, we know that $ P(ξ) \neq 0.$ Let $k \geq 1$ and $ω(k, H)$ be the infimum of the numbers $r > 0$ satisfying the estimate $$ \left|λ_0+λ_1 ξ+λ_2 ξ^{2}+ \ldots +λ_kξ^{k}\right| > \frac{1}{H^r}, $$ for all $(λ_0, \ldots ,λ_k)^T \in \mathbb{Z}^{k+1}\setminus\{\overline{0}\}$ with $\max_{1\le i\le k} \{|λ_i|\} \le H$. Any function greater than or equal to $ω (k, H)$ is a {\it transcendence measure of $ξ$}. In this article, we find out a transcendence measure of $ e^{1/n}$ which improves a result proved by Mahler(\cite{Mahler}) in 1975.

arxiv.org
March 13, 2023 at 3:10 AM · · feed2toot · 0 · 0 · 0
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