Nicolas Nagel

Peter Kritzer, Nicolas Nagel, Friedrich Pillichshammer

We investigate the periodic $L_2$-discrepancy of infinite sequences $§_d$ in $[0,1)^d$ and its analytic counterpart, the diaphony. We prove that infinite order-2 digital sequences over $\mathbb{F}_2$ attain the optimal order $L_{2,N}^{\rm per}(§_d) \le C_d (\log N)^{d/2}/N$ for all $N \in \mathbb{N}\setminus \{1\}$, matching known lower bounds for infinitely many $N \in \mathbb{N}$. This confirms the conjectured optimality of order-2 constructions. By this result, we improve upon previously known constructions using order-5 digital sequences, and reduce the underlying dimension for the interlacing construction from $5d$ to $2d$, significantly improving practicality. We establish our bounds within a broader framework of quasi-Monte Carlo integration for periodic Besov spaces $S_{p,q}^rB(\mathbb{T}^d)$ with dominating mixed smoothness $r \in (1/p,2)$, where $p,q\in [1,\infty]$. Rules based on infinite order-2 digital sequences yield worst-case errors of order $(\log N)^{(d-1)(1-1/q)} / N^{ \min(r,1)}$ for $r \not=1$, and $(\log N)^{d(1-1/q)}/N$ for $r=1$, for all $N \in \mathbb{N}\setminus\{1\}$, while preserving extensibility in $N$.

Melia Haase, Nicolas Nagel

We consider the asymptotics of sums of the form $$ \frac1{F_n^σ} \sum_{m = 1}^{F_n-1} \frac{f(m/F_n)}{\left|{\sin(πm/F_n)}\right|^σ} \frac{f(F_{n-1}m/F_n)}{\left|{\sin(πF_{n-1}m/F_n)}\right|^σ} $$ where $(F_n)_{n \in \mathbb N} = (1, 1, 2, 3, 5, 8, 13, \dots)$ are the Fibonacci numbers. Such sums appear, for example, in the context of discrepancy theory and numerical integration methods reformulated as energy minimization problems. We show that for parameters $σ> 1$ and a large class of functions $f$ the above sum behaves asymptotically like $$ C n + D + O\left((1-\varepsilon)^{n}\right) $$ for some constants $C$ and $D$. These constants can be given via infinite series connected to the Dedekind zeta function over the algebraic number field $\mathbb Q(\sqrt5)$. In special cases we even observe simple closed-form expressions for such sums as above, explicitly proving that $$ \sum_{m=1}^{F_n-1} \frac1{\sin(πm/F_n)^2} \frac1{\sin(πF_{n-1} m/F_n)^2} = \frac{4n}{75} F_{2n} - \frac{17}{225}F_n^2 - (-1)^n \frac2{15} - \frac19. $$