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Does there exist an algebraic number $\alpha$ such that $$\left|\frac{\alpha^n+\alpha^n_1}{n!}\right|\sim_{n\to+\infty}\frac1{(n!)^2}$$ where $\alpha_1$ is a conjugate of $\alpha$? Obviously $\alpha$ can not be a rational number.

Thanks in advance for any answer.

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  • $\begingroup$ Sorry, I forgot something in the question. I edit it. But anyway your argument still works. Thanks $\endgroup$
    – joaopa
    Dec 31, 2019 at 0:25

1 Answer 1

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For any fixed nonzero complex numbers $z_1,\dotsc,z_m$, there are infinitely many $n$'s such that the arguments of $z_1^n,\dotsc,z_m^n$ all lie in $[-\pi/4,\pi/4]$. This follows from Dirichlet's theorem on simultaneous diophantine approximation. For such $n$'s, $$|z_1^n+\dotsb+z_m^n|\geq\Re(z_1^n+\dotsb+z_m^n)\geq\frac{|z_1|^n+\dotsb+|z_m|^n}{\sqrt{2}}.$$ In particular, the left hand side cannot be asymptotically $1/n!$, because the right hand side is exponentially small at worst.

In short, there is no $\alpha$ satisfying the requirements.

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  • $\begingroup$ Nice answer for this question. I have a one more that is related. You could be interested in it :D $\endgroup$
    – joaopa
    Dec 31, 2019 at 3:23
  • $\begingroup$ @joaopa: Please ask your related question in a separate post. I will look at it as time and mood permits :-) $\endgroup$
    – GH from MO
    Dec 31, 2019 at 3:57

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