弗里德曼-伊万内茨双和猜想的反例
Counterexamples of Friedlander--Iwaniec dual sums conjecture
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中文总结 AI 辅助
本文通过构造A(s)=B(s)=ζ(s)^m，m≥4，给出弗里德曼-伊万内茨双和猜想的反例，推翻了该猜想的预测。
中文摘要 AI 辅助
令a(n)和b(n)为算术级数，并设A(s)=∑_{n≥1}a(n)n^{-s}，B(s)=∑_{n≥1}b(n)n^{-s}为两个相关的狄利克雷级数，它们由某种函数方程相关联。令m为函数方程的解析次数。对于x>0和正整数N，弗里德曼和伊万内茨（2005）定义了尖锐截断的非线性双和$$\mathcal B_{\ell,D}(x,N):= \sum_{\substack{n\in\mathbb N\\\\\\ n\le N}} b(n)n^{-\beta_m} \cos\left( 2\pi m\left(\frac{nx}{D}\right)^{1/m} +\frac{\pi\ell}{4} \right),$$其中D≥1为导数，β_m:=(m+1)/(2m)，而ℓ=m-3-2k由函数方程的阿基米德权重k确定。他们的猜想1预测，对于每一个ε>0，$$\mathcal B_{\ell,D}(x,N) \ll_{\varepsilon,\boldsymbol\kappa} (DNx)^\varepsilon,$$在变量x和N上一致成立，其中次数、导数和阿基米德数据固定。我们给出了对此预测的反例，其中A(s)=B(s)=ζ(s)^m，m≥4，其中ζ(s):=∑_{n≥1}n^{-s}（Re s>1）为黎曼ζ函数。
英文摘要
Let $a(n)$ and $b(n)$ be arithmetic sequences, and $$A(s)=\sum_{n\ge1}a(n)n^{-s}, \qquad B(s)=\sum_{n\ge1}b(n)n^{-s},$$ be the two Dirichlet series related by a certain functional equation. Let $m$ be the \emph{analytic degree} of the functional equation. For $x>0$ and a positive integer $N$, Friedlander and Iwaniec (2005) define the sharply truncated nonlinear dual sum $$\mathcal B_{\ell,D}(x,N) := \sum_{\substack{n\in\mathbb N\\ n\le N}} b(n)n^{-β_m} \cos\left( 2πm\left(\frac{nx}{D}\right)^{1/m} +\frac{π\ell}{4} \right),$$ where $D\ge1$ is the conductor, $β_m:=\frac{m+1}{2m}$, and $\ell=m-3-2k$ is determined by the archimedean weight $k$ of the functional equation. Their Conjecture 1 predicts that, for every $\varepsilon>0$, $$\mathcal B_{\ell,D}(x,N) \ll_{\varepsilon,\boldsymbolκ} (DNx)^\varepsilon,$$ uniformly in the variables $x$ and $N$, with the degree, conductor, and archimedean datum fixed. We give counterexamples to this prediction with $$A(s)=B(s)=ζ(s)^m,\; m\geq 4$$ where $$ζ(s):=\sum_{n\ge1}n^{-s} \qquad(\operatorname{Re}s>1)$$ is the Riemann zeta function.
