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Let $\{x\}$ be the fractional part of $x$, i.e. $\{x\}=x-[x]$, where $[x]$ is the biggest integer $\leq x$.

The question might be well known but I don't know where to look for: Assume $\alpha$ is an irrational number. Then does the sequence $\{\alpha\}$, $\{2\alpha\}$, $\{4\alpha\}$, $\{8\alpha\}$, ... distribute uniformly on $[0, 1]$?

In fact I was checking if the power series $z+z^2+z^4+z^8+...$ has bounded partial sum when $z=e^{2\pi i \alpha}$ and $\alpha$ is irrational. However even if we do have uniform distribution I am still not sure the partial sum is bounded.

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    $\begingroup$ Write out $\alpha$ in binary. As long as it doesn't terminate and isn't periodic, it's irrational. But that leaves a lot of room for distributions other than uniform. $\endgroup$
    – Gerry Myerson
    Commented Oct 20, 2018 at 4:55
  • $\begingroup$ Gerry, thanks! That is great. Now it might be $z+z^2+z^4+...$ equals infinity everywhere on the unit circle... $\endgroup$
    – Yu Ding
    Commented Oct 20, 2018 at 5:40
  • $\begingroup$ See mathworld.wolfram.com/NaturalBoundary.html $\endgroup$
    – Gerry Myerson
    Commented Oct 20, 2018 at 5:46

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This sequence is uniformly distributed if and only if $\alpha$ is normal in base 2. It is known that almost all irrational numbers (in the sense of Lebesgue measure) are normal, but the "usual" irrational numbers like $\sqrt{2},\log 2,\pi,e\ldots$ are not known to be normal.

Regarding the partial sums, Weyl's criterion says that uniform distribution is equivalent to $\frac{1}{n} \sum_{k=1}^n e^{2\pi i m 2^k \alpha} \to 0$ for every integer $m>0$. But there are cases where the sequence is uniformly distributed and the partial sums are unbounded (e.g. $u_n = \alpha n^2$ with $\alpha $ irrational I think). If you think of $u_n $ as a sequence of independent uniform random variables in $[0,1] $ then the partial sums $S_n $ define a planar random walk with unit step so $|S_n|$ typically grows like $\sqrt{n} $.

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    $\begingroup$ Uniform distribution of $(a_k)\subset\mathbb{R}/\mathbb{Z}$ is not equivalent to $\sum_{k=0}^n e^{2\pi i a_k}=o(n)$. A counterexample is when $a_k=\lfloor k/2\rfloor\bmod 1$. For equidistribution one needs that $\sum_{k=0}^n e^{2\pi i ma_k}=o(n)$ for any positive integer $m$. $\endgroup$
    – GH from MO
    Commented Oct 20, 2018 at 8:35
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    $\begingroup$ @GH from MO Sorry for misremembering Weyl's criterion, now corrected, thanks. I guess it's been too long I studied these things... $\endgroup$
    – François Brunault
    Commented Oct 20, 2018 at 10:26

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