7 月他在母校北京大学的北京国际数学研究中心 (BICMR) 有一个系列的学术报告: Distribution of Prime Numbers and the Riemann Zeta Function I, II, III. 这个报告分三场, 原定时间是 July 8, 10, 15,  2014 16:00-17:00, 地点是镜春园 78 号院的 77201 室.

BICMR 官网上这个报告的 Abstract 是这么写的:

The distribution of prime numbers is one of the most important subjects in number theory.

There are many interesting problems in this field. It may not be difficult to understand the problems themselves, but the solutions are extremely difficult.

In this series of talks we will describe the application of certain analytic tools to the distribution of prime numbers. In particular, the role played by the Riemann zeta function will be discussed. We will also describe some early and current researches on the Riemann Hypothesis.

These talks are open to everyone in the major of mathematics, including undergraduate students.

Yitang Zhang at BICMR :Distribution of Prime Numbers and the Riemann Zeta Function

8 日下午 4 点, 田刚现身. 因为人比较多, 改为在镜春园82号甲乙丙楼的中心报告厅进行. 主持人刘若川是 1999 年的 IMO 金牌(他本来也是 1998 年中国国家队的队员).

$\zeta(2k)=\sum_{n=1}^\infty\frac1{n^{2k}}=(-1)^{k+1}\frac{(2\pi)^{2k}B_{2k}}{2(2k)!}$

10 日下午 4 点的第二场, 依旧在镜春园82号甲乙丙楼的中心报告厅. 不过, 15 日的一场会在镜春园 78 号院的 77201 室, 16:30 开始.

15 日下午 4:30 的最后一场, 要深入一点. 田刚坐在教室最后一排, 刘若川, 许晨阳坐在教室左边的走廊.张大师谈到有 Goldston, Pintz and Yildirim 的工作, 说他自己最大的贡献是把 $$c$$ 改进为 $$\dfrac14+\dfrac1{1168}$$.

$$a_0=0,a_1=1,a_2=2,a_3=4,a_4=5,\dotsc.$$

$$\varlimsup_{n\rightarrow\infty}d_n= \infty.$$

$\begin{cases}x+1\equiv p_1\pmod{p_1^2},\\ x+2\equiv p_2\pmod{p_2^2},\\ \cdots\\x+n\equiv p_n\pmod{p_n^2}.\end{cases}$

$S:=\{a_0,a_1,a_2,a_3,\dotsc\}.$

$$S$$ 的密率是 $$0$$. 事实上, 我们有精确得多的结果.

$$\lim_{n\rightarrow\infty}\frac{A(n)\sqrt{\log n}}n=1.$$

$$L_{1}(s)=\sum_{n\,\text{odd}}(-1)^\frac{n-1}2 n^{-s}=\prod_{p\in\Pi_1}\dfrac{1}{1-p^{-s}}\prod_{p\in\Pi_3}\dfrac1{1+p^{-s}}.$$

$$L_{0}(s)=\sum_{n\,\text{odd}}n^{-s}=\prod_{p\in\Pi_1}\dfrac1{1-p^{-s}}\prod_{p\in\Pi_3}\dfrac1{1-p^{-s}}=(1-2^{-s})\zeta(s),$$

$$\log L_0+\log L_1=2\sum_{p\in\Pi_1}p^{-s}+O(1),$$

$$\log L_0-\log L_1=2\sum_{p\in\Pi_3}p^{-s}+O(1).$$

$$L_1$$ 是交换级数, 因而收敛, 有界, 并且当 $$s\geqslant1$$ 时, 其上下界与 $$s$$ 无关. 于是

$$\sum_{p\in\Pi_1}p^{-s}=\frac12\log\left[(1-2^{-s})\zeta (s)\right]+O(1),$$

$$\sum_{p\in\Pi_3}p^{-s}=\frac12\log\left[(1-2^{-s})\zeta (s)\right]+O(1).$$

$$\prod_{p\in\Pi_1}\frac1{1-p^{-s}}=\sqrt{(1-2^{-s})\zeta (s)}(1+O(1)),$$

$$\prod_{p\in\Pi_3}\frac1{1-p^{-s}}=\sqrt{(1-2^{-s})\zeta (s)}(1+O(1)).$$

$$\sum_{n=1}^\infty a_n^{-s}=\frac1{1-2^{-s}}\prod_{p\in\Pi_1}\frac1{1-p^{-s}}\prod_{p\in\Pi_3}\frac1{1-p^{-2s}}=\sqrt{\zeta (s)\zeta (2s)}(1+O(1)).$$

$$\sum_{n=1}^\infty a_n^{-s}=\sqrt{\zeta(s)}(1+O(1)).$$

$$\sum_{n=1}^\infty a_{n}^{-s}=\sum_{n=1}^\infty\chi(n)n^{-s},$$

$$\chi(n)=\begin{cases}\,1,\quad\,n\in S\\0,\quad \text{otherwise}\end{cases}$$

$A(x)=\sum_{n\leqslant x}\chi(n),$

$$\sqrt{\zeta (s)}+O(1)=s\int_{0}^\infty A(x)x^{-(s+1)}dx.$$

$$\frac1{\sqrt{s-1}}+O(1)=\int_{0}^\infty A(x)x^{-(s+1)}dx.$$

$$A(x)\sim\frac{x}{\sqrt{\log x}}.$$

$\{\frac{s_1}{s_2}|s_1,s_2\in \Bbb S\}$

$g(x)=a_n, h(x)=a_{n+1}(a_n\leqslant x<a_{n+1},n=0,1,2,\dotsc.),$

$\dfrac{\dfrac{g(np)}{h(nq)}}{\dfrac{h(np)}{g(nq)}}<\dfrac{\dfrac{g(np)}{g(nq)}}{\dfrac pq}<\dfrac{\dfrac{h(np)}{g(nq)}}{\dfrac{g(np)}{h(nq)}},$

$\lim_{n\rightarrow\infty}\frac{g(np)}{g(nq)}=\frac pq.$

$$\sum_{n=1}^\infty\frac1{a_n}$$

$\sum_{p\equiv b\pmod a\atop (a,b)=1}\frac 1p=\infty$