Rational numbers with small denominators in short intervals

We use bounds on bilinear forms with Kloosterman fractions and improve the error term in the asymptotic formula of Balazard and Martin (2023) on the average value of the smallest denominators of rational numbers in short intervals.


Introduction
Given integer N 1, and j = 1, . . ., N , we denote by q j (N) the smallest integer q such that for some a we have a q ∈ j − 1 N , j N .
The bound on the error term in (1.1) is based on the classical bound of Kloosterman sums, see, for example, [9,Corollary 11.12].
Here, we use bounds on bilinear forms with Kloosterman fractions due to Duke, Friedlander and Iwaniec [6] and improve the error term in the asymptotic formula (1.1) as follows.

Preliminary reductions
As usual, we use the expressions U ≪ V and U = O(V ) to mean |U| cV for some constant c > 0 which throughout this paper is absolute.
We first recall the following expression for T 11 (N) given in [2, Section 5.3]: with the Bernoulli function where {u} is the fractional part of a real u, the inversion r −1 in the fractional part {Nr −1 /s} is computed modulo s and R s is a certain sequence of positive integers, satisfying (we refer to [2] for an exact definition, which is not important for our argument).
It is more convenient for us to work with the function which coincides with B 1 (u) for all u ∈ Z.

Vaaler polynomials
By a result of Vaaler [14], see also [8,Theorem A.6] we have the following approximation to ψ(u).Lemma 3.1.For any integer H 1 there is a trigonometric polynomial for coefficients a h ∈ [0, 1] and such that

Bilinear forms with Kloosterman fractions
For an integer q , let e(z) = exp(2πz).Here we collect some estimates on bilinear form with exponantials e (hr −1 /s) where, as before, r −1 in the argument is computed modulo s.
For U 1 we aslo also use u ∼ U to indicate U u < 2U .We start with recalling the following bound of Duke, Friedlander and Iwaniec [6, Theorem 1].
of complex numbers, an nonzero integer K and real positive R and S we have 1) , Next, given two sequences of complex numbers and an integer h, for S 1 we define the bilinear form Note that in the sums B K (S; R, α, β) the range of summation over r depends on s and hence Lemma 4.1 does not directly apply.
We observe that for Note that one can also derive (4.2) via [6, Lemma 8] and partial summation.
In fact using the bound (4.2) for S N 2/3 and the trivial bound in our argument below, one recovers the asymptotic formula (1.1).However using some other bounds we achieve a stronger result.We also remark that for us only the choice of α = {α r } ∞ r=1 satisfying (4.1) matter.However we present the below results for a more general α (but still they admit even more general forms).
Using Lemma 4.1 together with the standard completing technique, see, for example, [9, Section 12.2], we derive our main technical tool.
Proof.Note that Using the orthogonality of exponential functions, we write α r e(−ur/R)β s e Kr −1 /s Rs t=1 e(ut/R).
It remains to observe that for each u = 0, . . ., R − 1 the bound of Lemma 4.1 applies to the inner sum and implies 1) .
Recalling (4.3), this now simplifies as .However these bounds do not seem to improve our main result.

Proof of Theorem 1.1
As we have noticed in Section 2, it is only enough to estimate T 11 (N), as we borrow the bounds on T 12 (N) and T 2 (N) from [2].Furthermore, we see from (2.9) and (2.11) that it is enough to estimate U(N) given by (2.10).
We note that it is important to observe that the sum defining ψ H (u) in Lemma 3.1 does not contain the term with h = 0, while the sum on the right hand side of (3.1) does.Hence, for any integer H 1, by Lemma 3.1 we have r e hNr −1 /s .Note that R s 1 implies s ≪ N .Therefore, partitioning the corresponding summation over s into dyadic intervals, we see that there is some integer S with V (N, S) N 1+o(1) S 1/2 H 1/10 N 13/10+o (1) .