# Explicit numbers with square root cancellation in Weyl's exponential sum

I'm interested in examples of real numbers $\alpha$ where we have

$$\left| \sum_{n=1}^N \mathrm e(\alpha n) \right| \ll N^{1/2}$$

or perhaps with the weaker estimate with the right side replaced with $C_{\epsilon} N^{1/2+\epsilon}$.

For example is this known for any classes of number, such as badly approximate numbers? Is there a formal relationship between Diophantine estimates in the form of Roth's theorem and exponential sum estimates? I assume this is well-known, pointers to the literature will be very helpful.

• @AlexM. Rather $e^{i\alpha n}$. This is quite a standard notation in the Analytic Number Theory :-) – fedja May 14 '18 at 18:36
• The standard notation in analytic number theory is $e(x) = e^{2 \pi i x}$. – Matt Young May 14 '18 at 19:05
• Considering instead the sum with $n$ ranging from $0$ to $N-1$ (which is off by at most $2$ from your original sum), the absolute value of this sum is $|\sin(\pi N\alpha)/\sin(\pi\alpha)|$, which is $O(1)$, with the implicit constant depending on $\alpha$ (but not on $N$). Does this answer your question? – Seva May 14 '18 at 19:19
• @Seva, the sum from 1 to $N$ is just $e(\alpha)$ times the sum from 0 to $N-1$, so either way you get the same absolute value, no? – Gerry Myerson May 14 '18 at 23:10
• @Gerry Myerson: absolutely; the two sums are off by at most 2, their absolute values are the same. – Seva May 15 '18 at 5:21

Yes, such exponential sums can be estimated using the continued fraction coefficients of $\alpha$. There is a field of mathematics called uniform distribution theory which is concerned with the distribution of sequences in the unit interval. The degree of uniformity is measured by something called discrepancy. The infinite sequence of fractional parts of $n \alpha$ is called Kronecker sequence. There are explicit estimates for the discrepancy of Kronecker sequences in terms of continued fraction coefficients.
To answer your particular question, if $\alpha$ is badly approximable then the continued fractions coefficients are bounded, which implies that the discrepancy is of order $(\log N)/N$, which implies that the exponential sum is of order $\log N$.