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In the sum

$$\sum_{n=1}^N \left(\{n \alpha\}-\frac{1}{2}\right)$$

where $\{x\}$ indicates the fractional part of $x$ and $\alpha$ is an irrational number, Koksma inequality suggests an order of $\log^2 N $ if we use a weak bound for the discrepancy (Kuipers and Niederreiter, Chapter 2, "Special sequences" section) and a stronger one using the bound of $\log N (\log \log N)^{1+\varepsilon}$ that is valid for almost all $ \alpha$ (Khintchine). Some details on this are reported in this previous question.

I wonder what can we say about the order of higher-degree sums, as

$$\sum_{n=1}^N\left( \{n \alpha\}^2-\frac{1}{3}\right)$$ $$\sum_{n=1}^N \left(\{n \alpha\}^3-\frac{1}{4}\right)$$

generalizing to any exponent $k$. Should we use Erdös-Turan inequality to estimate the discrepancy?

I am not searching particularly strong bounds, but only a way to get a formal proof for even weak O terms. After estensive search, I did not find anything on these higher-degree sequences in the literature.

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  • $\begingroup$ Surely you get the same bounds: all of these boil down to sum f({nt}) where f is of the form 1_J(x)-|J| for an interval J. $\endgroup$ Commented Nov 1, 2019 at 0:16
  • $\begingroup$ Having said this, is there some problem with your question? Your summands all appear to be 0 or -1. $\endgroup$ Commented Nov 1, 2019 at 0:17
  • $\begingroup$ Sorry... I misinterpreted the question - I interpreted $[\cdot]$ as the integer part, but I think you just mean a parenthesis here. So I don't know the answer to your question - I still expect it to have the same growth rate, but I don't know a proof. $\endgroup$ Commented Nov 1, 2019 at 7:29
  • $\begingroup$ Ok, I changed the square brackets into round ones to avoid misinterpretations. $\endgroup$
    – Anatoly
    Commented Nov 1, 2019 at 8:47

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