|
|
以下是：转载计算黎曼假设非平凡零点的式子，否定黎曼假设的人能否定这个计算式吗？如果不能否定它，就说黎曼假设是不成立的，既然不能破旧，自然不能立新。仅凭自己“推理”得出结论的人是不是有点狂妄自大，就像许多编辑一样根本就不看，因为这压根就不是他们认为的数论，简直就是胡说八道。
否定黎曼假设的人为什么就不看看计算黎曼假设非平凡零点的公式到底是哪里错了还是对的呢？如果不能否定原定猜想的依据，而得出的与猜想相反的结论都是垃圾，不值得一看！
The Riemann hypothesis posits that all the non-trivial zeros of the zeta function ζ (shown below) on the critical strip bounded by Re(s) = 0 and Re(s) = 1 will always be at the critical line Re(s)=1/2: - ζ(s)=∑ 1/ns=1+1/2s+1/3s+1/4s+1/5s+…(1)n=1， This has been found to be true for the 1st. 1013non-trivial zeros.
How to calculate non-trivial zeros of the Riemann zeta function：
(1)One can calculate nontrivial zeros of ζ(s) along the line s=1/2+t∈C using the algorithm below.
Newton−Raphson Algorithm for ζ(s): Given an initial tk∈R, iterative solutions tk+1 converge to non-trivial zeros of ζ(s),
2i
tk+1=tk -—————————————————————
16itk Γ′(1/4+itk/2)2ζ′(1/2+itk)
——— + loge(π)-———————————
1+4t2k Γ(1/4+itk/2)ζ(1/2+itk)
Note that itk is the imaginary part of the root s=1/2+itk, with ζ(1/2+itk)≈0.
(2)Basically, naming s=σ+it, the idea is to use the function
ξ(s)=Γ(s/2)π−s/2(s−1)ζ(s)
because it's real-valued on the critical line t=1/2, hence you'll find a zero whenever ξ(1/2+it) changes sign. There are various method to do that, a very nice introduction can be found in Edwards' book ``Riemann's Zeta Function", see for example Section 6.5.
The infinite number of non-trivial zeros are arranged mirror-symmetrically to the real axis. So if a non-trivial zero in the following table has the imaginary part , then the too complex conjugate number is also a non-trivial zero of the Riemann zeta function (denotes the imaginary unit ).
|
|
IP卡
狗仔卡
发表于 2023-1-8 10:43
显身卡
楼主