nt.number theory - Bounds on $sum {n ((alpha n))}$ where $((x))$ is the sawtooth function

Here is an approach which may give some better estimates for particular values of $alpha$:

$$sum_{i= 1}^N i((ialpha)) = sum_{i=1}^N sum_{j=i}^N ((j alpha)) = sum_{i=1}^N sum_{k=0}^{N-i}((Nalpha -kalpha))$$ .

So, if you can estimate

$$sup_{substack{xin (0,1) \ M le N}} bigg|sum_{k=0}^{M} ((x-kalpha))bigg|,$$

then you can crudely multiply by $N$ to get an estimate for $|sum i((ialpha))|$.

Specifically, my guess is that for quadratic irrationals $alpha$, there is an upper bound for

$$bigg|sum_{k=0}^M ((x-kalpha))bigg|$$

which is $O (log M)$, which would give you a bound of $O(N log N)$, and more generally that there is a bound in terms of the coefficients of the simple continued fraction for $alpha$, so that if those are bounded, then you still get $O(N log N)$.

For the particular value $phi = (sqrt5 + 1)/2$, $sum_{i=0}^{M} ((i phi))$ has logarithmic growth $c + (5sqrt5 - 11)/4 log_phi M$ (achieved at indices in the sequences A064831 (+) and A059840 (-)), which suggests that $sup sum_{i=0}^M ((x-iphi))$ also has logarithmic growth, which would give an $N log N$ bound for the sum.

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In the opposite direction, for all $alpha notin frac 12mathbb Z$,

$$limsup bigg(log_N bigg|sum_{i=0}^N i((ialpha))bigg|bigg) ge 1$$ since there are terms proportional to $N$.

The sum can be greater than $N^{2-epsilon}$ infinitely often by choosing $alpha$ so that it is extremely well approximated by infinitely many rational numbers. When $alpha$ is very closely approximated by $p/q$, then for $N$ a small multiple of $q$ (where "small" is relative to how well $p/q$ approximates $alpha$), about $1/q$ of the terms can be moved past integers with a small perturbation of $alpha$ to $alpha'$, which causes a jump of about $N^2/q$ in the sum. So, either the sum for $alpha$ or $alpha'$ is large. We can choose a sequence $p_n/q_n$ which converges to an $alpha$ which produces large sums infinitely often, so that for these $alpha,$

$$limsup bigg(log_N bigg|sum_{i=0}^N i((ialpha))bigg|bigg) = 2$$ .

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