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Riemann

11 May 2014

\begin{align} \zeta(s) &= 1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\frac{1}{5^s}+\frac{1}{6^s}+\cdots \\ \ &= \sum_{n=1}^\infty \frac{1}{n^s} \\ \ &= \sum_{n=1}^\infty n^{-s} \end{align}

埃拉托色尼筛法

\begin{align} \frac{1}{2^s}\zeta(s) &= \frac{1}{2^s}+\frac{1}{4^s}+\frac{1}{6^s}+\frac{1}{8^s}+\frac{1}{10^s}+\cdots \\ \left(1 - \frac{1}{2^s}\right) \zeta(s) &= 1 + \frac{1}{3^s}+\frac{1}{5^s}+\frac{1}{7^s}+\frac{1}{9^s}+\frac{1}{11^s}+\cdots \\ \frac{1}{3^s}\left(1 - \frac{1}{2^s}\right) \zeta(s) &= \frac{1}{3^s}+\frac{1}{9^s}+\frac{1}{15^s}+\frac{1}{21^s}+\frac{1}{27^s}+\cdots \\ \left(1-\frac{1}{3^s}\right) \left(1 - \frac{1}{2^s}\right) \zeta(s) &= 1 + \frac{1}{5^s}+\frac{1}{7^s}+\frac{1}{11^s}+\frac{1}{13^s}+\frac{1}{17^s}+\cdots \\ \end{align}

直到某个大素数,比如997,得到:

\begin{align} \left(1-\frac{1}{997^s}\right) \left(1-\frac{1}{991^s}\right) + \cdots + \left(1-\frac{1}{5^s}\right) \left(1-\frac{1}{3^s}\right) \left(1 - \frac{1}{2^s}\right) \zeta(s) &= 1 + \frac{1}{1009^s}+\frac{1}{1013^s}+\frac{1}{1019^s}+\frac{1}{1021^s} + \cdots \\ \end{align}

重复下去:

\[ \cdots \left(1-\frac{1}{13^s}\right) \left(1-\frac{1}{11^s}\right) \left(1-\frac{1}{7^s}\right) \left(1-\frac{1}{5^s}\right) \left(1-\frac{1}{3^s}\right) \left(1 - \frac{1}{2^s}\right) \zeta(s) = 1 \\ \]

金钥匙,欧拉积公式:

\[ \zeta(s) = \prod_p (1-p^{-s})^{-1}\]