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Let $\alpha$ be an algebraic integer of modulus 1, and $ R_\alpha z=\alpha z$. Is $$\lim_{n\to\infty}\frac{\log|\sum_{k=1}^n \Re R_\alpha^k z|}{\log n}=\frac12$$ for all $z\in S^1$?

Birkhoff's ergodic theorem works for all $z\in S^1 $ since $R_\alpha$ is uniquely ergodic. Does the CLT? (I only need $\sum_{k=1}^n\Re( R_\alpha^k z)\asymp c_n(\alpha)\sqrt n$ with $c_n(\alpha)=o(\sqrt n)$ , not the full power of the CLT.)

ANSWER. $$ \sum_{k=1}^{n} \text{Re}(\alpha^k)=\sum_{k=1}^{n}T_j(\cos\text{arg}\alpha) \asymp n, $$ where $T_j(t)=\cos\arccos jt$ is the $j$th Chebysh"ev polynomial. The lower bound comes from $\alpha$ being badly approximable by Liouville/Roth theorems.

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  • $\begingroup$ The sum is o(n) but it does not imply that the limit is 0. Example $\log(\sqrt{n})/\log(n) = 1/2$. Yet, in our example, the sums are bounded, and can be negative and the logarithm can be undefined. $\endgroup$ Nov 24, 2023 at 11:56
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    $\begingroup$ So. Almost all $\alpha$ on the unit circle have this property, but do all algebraic integers on the unit circle have the property? It reminds me of the (open) question of whether algebraic irrational numbers must be normal. $\endgroup$ Nov 24, 2023 at 12:21
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    $\begingroup$ At the very least you need to exclude roots of unity. $\endgroup$ Nov 24, 2023 at 13:29
  • $\begingroup$ @GeraldEdgar Birkhoff's ergodic theorem works for all $x$ since $R_\alpha$ is uniquely ergodic. Does the CLT? $\endgroup$ Nov 25, 2023 at 10:09
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    $\begingroup$ @Nikita This is not very helpful. Your answer came after another contradictory answer, and you're not saying what you think is wrong with it. Just to give some empirical evidence: your sum of real parts of powers of alpha is just a sum of cos(nx) (as you wrote). From Mathematica: Sum[Cos[kSqrt[2]],{k,10000}] = -0.939269... Sum[Cos[kSqrt[2]],{k,100000}] =-0.401409... Sum[Cos[k*Sqrt[2]],{k,1000000}] = 0.218388... It surely seems that these are not growing like n. (I'm aware that this is just one example, but try your favorite and see what happens!) $\endgroup$ Dec 1, 2023 at 21:38

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I am surprised by the question. The sums are bounded if $\alpha \ne 1$, since for every $n \ge 1$, $$\Big|\sum_{k=1}^n \Re R_\alpha^k z \Big| = \Big|\Re \Big(\sum_{k=1}^n R_\alpha^k z \Big) \Big| = \Big| \Re \Big(\frac{\alpha-\alpha^{n+1}}{1-\alpha}z\Big) \Big| \le \Big|\frac{\alpha-\alpha^{n+1}}{1-\alpha}z\Big| \le \frac{2}{|1-\alpha|}.$$ The log of this quantity is bounded above, so it can not increase like $(1/2)\log(n)$. But it is not bounded below, it can takes large negative values for the integers $n$ such that $\alpha^n$ is close to $1$.

ADDENDUM : one can check that $$\limsup_{n \to +\infty} \log\Big|\sum_{k=1}^n \Re R_\alpha^k z \Big| \times \frac{1}{\log(n)} = 0,$$ whereas $$\liminf_{n \to +\infty} \log\Big|\sum_{k=1}^n \Re R_\alpha^k z \Big| \times \frac{1}{\log(n)} \le -1.$$ Indeed, setting $\alpha = e^{i\theta}$ with $\theta \in \mathbb{R}$, Dirichlet principle ensures that $|\alpha^n-1| = |e^{in\theta}-1| \le \mathrm{dist}(n\theta,\mathbb{Z}) < 1/n$ for infinitely many $n$. Actually, the liminf above is $-1$ for almost every $\alpha$.

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  • $\begingroup$ The question should at least be reformulated, since the logarithm of a possibly negative number is not well-defined. If the true question is « Does the CLT [work for all $z$] ?», please give a correct statement of what it means. $\endgroup$ Nov 25, 2023 at 12:39
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    $\begingroup$ I don't understand the weird responses you're giving. Christophe pointed out that your sums are bounded in $\mathbb{C}$, so there's no version of log where they can grow like $1/2 \log n$. The whole talk about log of negative number seems immaterial. What am I missing? $\endgroup$ Nov 25, 2023 at 13:54
  • $\begingroup$ $\Re z$ stands for the real part of $z$. Log is well defined. $\endgroup$ Nov 28, 2023 at 11:08
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    $\begingroup$ I still don't understand. The sums you work with are bounded (in $n$). So their real parts are also bounded, so your numerator is the logarithm of a bounded quantity. Therefore, it is bounded from above, and so dividing by $\log n$ cannot approach $1/2$. Can you either describe what you think is wrong (maybe we're both making a mistake!) or accept his answer? The whole removal of your previous comments and changing the question and adding a large bounty and asking for an authoritative reference is strange given that it's a one-line proof which has already been given. $\endgroup$ Nov 28, 2023 at 14:11
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    $\begingroup$ @NikitaSidorov The question has been corrected a bit. I modified my answer accordingly. $\endgroup$ Nov 29, 2023 at 12:07

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