欧拉在他的论文《无穷级数的一些检视》(Various Observations about Infinite Series)中证明黎曼ζ函数的欧拉乘积公式,并于1737年由当时的科学院出版。[1][2]
黎曼ζ函数以欧拉乘积的方式可写成
![{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{s}}}=\prod _{p{\text{ prime}}}{\frac {1}{1-p^{-s}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eae0968ff29d511b6ac1e2a3463ba92313f94526)
而左方等于黎曼ζ函数:
![{\displaystyle \zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}=1+{\frac {1}{2^{s}}}+{\frac {1}{3^{s}}}+{\frac {1}{4^{s}}}+{\frac {1}{5^{s}}}+\ldots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/a448138f05d20838238bc2836b712e7e316ab163)
右方的乘积则扩展至所有质数p:
![{\displaystyle \prod _{p{\text{ prime}}}{\frac {1}{1-p^{-s}}}={\frac {1}{1-2^{-s}}}\cdot {\frac {1}{1-3^{-s}}}\cdot {\frac {1}{1-5^{-s}}}\cdot {\frac {1}{1-7^{-s}}}\cdots {\frac {1}{1-p^{-s}}}\cdots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/9f22da57dec5301a7a1535e8dec3e1f18aec0c82)
证明方法采用了埃拉托斯特尼筛法的概念,此筛法用于找寻出特定范围内的质数。
证明过程只需用到简单的代数概念,这亦是欧拉当初使用的证明方法。
(1)
(2)
从(1)式减去(2)式:
(3)
重复上面步骤:
(4)
从(3)式减去(4)式,可得:
![{\displaystyle \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}}}+\ldots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/7a37540ed47178a76e5c4f4bee9592c012fd86ae)
这次2和3的所有倍数项都被减去。可见右方的的倍数项可被筛去,不断重复以上步骤可得:
![{\displaystyle \ldots \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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a498cd38314b0411e963571bc2a18c0431eeb0b1)
左右两方除以所有括号项,我们得到:
![{\displaystyle \zeta (s)={\frac {1}{\left(1-{\frac {1}{2^{s}}}\right)\left(1-{\frac {1}{3^{s}}}\right)\left(1-{\frac {1}{5^{s}}}\right)\left(1-{\frac {1}{7^{s}}}\right)\left(1-{\frac {1}{11^{s}}}\right)\ldots }}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bb8ab931e451715ee7845bd5563c4c98d58586b7)
最后,公式可写成质数的无穷乘积:
![{\displaystyle \zeta (s)=\prod _{p{\text{ prime}}}{\frac {1}{1-p^{-s}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5969f6b5a81ab4a9e32dba75156443830601f7f8)
证毕。
为了使证明更严密,我们只需注意到当
,已筛的右方项趋向1,并遵从狄利克雷级数的收敛性。
特别情形
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从以上公式可推导出 ζ(1) 的有趣结果。
![{\displaystyle \ldots \left(1-{\frac {1}{11}}\right)\left(1-{\frac {1}{7}}\right)\left(1-{\frac {1}{5}}\right)\left(1-{\frac {1}{3}}\right)\left(1-{\frac {1}{2}}\right)\zeta (1)=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/81779f00fa170ebdd9c8bdd505cae3456db0addd)
可以写成,
![{\displaystyle \ldots \left({\frac {10}{11}}\right)\left({\frac {6}{7}}\right)\left({\frac {4}{5}}\right)\left({\frac {2}{3}}\right)\left({\frac {1}{2}}\right)\zeta (1)=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b3128ac0ea620cf1ba257d8372ea1113c4cc3716)
![{\displaystyle \left({\frac {\ldots \cdot 10\cdot 6\cdot 4\cdot 2\cdot 1}{\ldots \cdot 11\cdot 7\cdot 5\cdot 3\cdot 2}}\right)\zeta (1)=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5214d949ad56b06685b1103afd0aa003f2213e9b)
又知:
![{\displaystyle \zeta (1)=1+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}+{\frac {1}{5}}+\ldots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/0be0d6276277e9af46e7864c40988ed283980c79)
所以
![{\displaystyle 1+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}+{\frac {1}{5}}+\ldots ={\frac {2\cdot 3\cdot 5\cdot 7\cdot 11\cdot \ldots }{1\cdot 2\cdot 4\cdot 6\cdot 10\cdot \ldots }}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/14050961eea89847f373b8f52180ce35749e8b7a)
我们得知左式是调和级数,并发散至无穷大,故此右式的分子(质数阶乘)必定同样发散至无穷大。由此可以证明质数有无限多个。
参考资料[编辑]
- ^ O'Connor, J.J. and Robertson, E.F. A history of calculus. University of St Andrews. February 1996 [2007-08-07]. (原始内容存档于2007-07-15).
- ^ John Derbyshire (2003), chapter 7, "The Golden Key, and an Improved Prime Number Theorem"
- John Derbyshire, Prime Obsession: Bernhard Riemann and The Greatest Unsolved Problem in Mathematics, Joseph Henry Press, 2003, ISBN 978-0-309-08549-6