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此文是根据本人在arXiv的一篇文章稍加改动而成。原文目前正在一个杂志做Peer Review。博文改成英文标题是因为有外国人来电垂询以前的一篇博文,他们认为以前那篇证明是正确的,但不知道为什么没有得到专家认可。对于目前这篇文章,虽然证明过程绝大部分人能看懂,但我希望国内的专业人士不要错过一个找错的机会:-)
其实,这篇文章经几个大咖瞄过一眼,好像没啥问题。本人以为唯一可能有问题的地方是Zeta为0时由(3)到(4)是否依然成立。为回答这个问题,我先打个比方,对于直线 y = 3x,斜率为3。那么,在 x = 0时,斜率是否还是3?答案是肯定的。因为在x = 0时,对两边求导数,斜率依然为3。因为Zeta是一个连续函数,pi指数函数也是,所以,对(3)的两边求导,就可得到(4),与Zeta是否为0无关。
Here is a really, really simple proof of the Riemann Hypothesis, which is based on a radical thinking:
Take the absolute value for both sides of the symmetrical reflection functional equation, one finds that
|pi(s/2)gamma(s/2)zeta(s)| = |pi((1-s)/2)gamma((1-s)/2)zeta(1-s)| (1)
对黎曼zeta函数的对称函数方程两边同时取绝对值,得出
|pi(s/2)gamma(s/2)zeta(s)| = |pi((1-s)/2)gamma((1-s)/2)zeta(1-s)|
Based on the fact that the gamma and the pi exponential functions have no zero, one can easily realize that a necessary condition for the Riemann zeta function to have nontrivial zeros is:
|zeta(s)| = |zeta(1-s)| (2)
根据gamma函数、pi指数函数都没有0这一特征,推出黎曼zeta函数有非简单0点的必要条件是黎曼zeta函数的绝对值必须对称相等:
|zeta(s)| = |zeta(1-s)|
It follows that 由此导出
|pi(s/2)||zeta(s)| = |pi((1-s)/2)||zeta(1-s)| (3)
from which one can easily find that 由此进一步导出
|pi(s/2)| = |pi((1-s)/2)|. (4)
pi指数函数的绝对值必须对称相等;
Solving this equality,
sigma = 1/2.
解pi指数函数的绝对值对称等式,得出黎曼zeta函数存在非简单0点的sigma数值为:sigma = 1/2。
It is evident that the Riemann Hypothesis is actually a necessary condition for the zeta function to have nontrivial zeros. Consequently, all the nontrivial zeros of the zeta function must have real part
sigma = 1/2.
由此可见,黎曼猜想其实是黎曼zeta函数有非简单0点的必要条件。也就是说,黎曼zeta函数的所有非简单0点的实部必须是
sigma = 1/2。
Since this proof is independent of the form of the zeta function, thus, the nontrivial zeros of the general L-functions must also have real part
sigma = 1/2。
由于这个证明过程不依赖黎曼zeta函数的具体表达式,因此,广义黎曼zeta函数的非简单0点也必须是:
sigma = 1/2。
这就是黎曼猜想的证明。详情参见下面的链接:Link to the original paper:
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