Let $\alpha$ be an irrational number. Denote by $\mu(\alpha)$ its irrationality measure. Can one say anything about $\mu(\alpha^n)$ for every $n\in\mathbb N$?

Even more, one knows that $\mu(e)=2$. Can one say anything for $\mu(e^{p/q})$ for $\frac pq\in\mathbb Q^*$?

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    $\begingroup$ Reguarding your second question: it is known that, for all positive integers $k$, $\mu(e^{2/k})=2$. I do not know if this is also true for other rational powers of $e$. $\endgroup$ – Manuel Norman Oct 14 at 17:09
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    $\begingroup$ One can say that if there exist good rational approximations to $\alpha$ then there also exist good rational approximations to $\alpha^n$. But the converse is presumably not true - maybe the square root of Liouville's number already gives a counterexample. $\endgroup$ – Will Sawin Oct 14 at 17:25

Let $\alpha$ be irrational. There are two cases: $\alpha$ can be either algebraic or transcendental. Products of algebraic numbers are algebraic, while rational powers of transcendental numbers are transcendantal. Hence, for all positive integer $n$, $\alpha^n$ is algebraic in the first case, and transcendental in the second one. Roth proved that algebraic irrational numbers all have irrationality measure $2$. Instead, little is known about transcendental numbers. We can only say, in general, that the irrationality measure is $\geq 2$. Thus, by the previous discussion, we can conclude that:

$\alpha$ irrational algebraic $\Rightarrow$ $\mu(\alpha^n)=2$ for all integers $n \geq 1$ (in fact, this also holds for all nonzero rationals $n$, since roots of algebraic numbers are algebraic).

$\alpha$ transcendental $\Rightarrow$ $\mu(\alpha^n) \geq 2$ for all integers $n \geq 1$ (in fact, as before, this also holds for all nonzero rationals $n$).

I think that the actual value of the irrationality measure of a power of a transcendental number highly depends on the particular case. However, it is worth recalling that almost (in the sense of Lebesgue measure) all irrational numbers have irrationality measure $2$.

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    $\begingroup$ Thank you for the case $\alpha$ irationnal. In the transcendental case, at least can one obtain a upper bound of $\mu(\alpha^n)$ depending on $\mu(\alpha)$? $\endgroup$ – joaopa Oct 14 at 19:03
  • $\begingroup$ That's an interesting question. I think that it might be possible. Maybe an upper bound can be derived using the limit formula involving the convergents of the simple continued fractions (of the powers, in this case) found by Sondow. This is just an idea - I have not checked if it works $\endgroup$ – Manuel Norman Oct 14 at 19:54

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