English

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

Number Theory 2023-03-13 v1

Abstract

For a given transcendental number ξ\xi and for any polynomial P(X)=:λ0++λkXkZ[X]P(X)=: \lambda_0+\cdots+\lambda_k X^k \in \mathbb{Z}[X], we know that P(ξ)0. P(\xi) \neq 0. Let k1k \geq 1 and ω(k,H)\omega (k, H) be the infimum of the numbers r>0r > 0 satisfying the estimate λ0+λ1ξ+λ2ξ2++λkξk>1Hr, \left|\lambda_0+\lambda_1 \xi+\lambda_2 \xi^{2}+ \ldots +\lambda_k\xi^{k}\right| > \frac{1}{H^r}, for all (λ0,,λk)TZk+1{0}(\lambda_0, \ldots ,\lambda_k)^T \in \mathbb{Z}^{k+1}\setminus\{\overline{0}\} with max1ik{λi}H\max_{1\le i\le k} \{|\lambda_i|\} \le H. Any function greater than or equal to ω(k,H)\omega (k, H) is a {\it transcendence measure of ξ\xi}. In this article, we find out a transcendence measure of e1/n e^{1/n} which improves a result proved by Mahler(\cite{Mahler}) in 1975.

Keywords

Cite

@article{arxiv.2303.05542,
  title  = {Transcendence measure of $e^{1/n}$},
  author = {Marta Dujella and Anne-Maria Ernvall-Hytönen and Linda Frey and Bidisha Roy},
  journal= {arXiv preprint arXiv:2303.05542},
  year   = {2023}
}

Comments

This collaboration has come up due to the wonderful programme: Women in Numbers Europe 4