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A Proof Of The Riemann Hypothesis Based On A New Expression Of The Completed Zeta Function

Weicun Zhang

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Source: Crossref

Published: Oct 12, 2022

DOI: 10.20944/preprints202108.0146.v22

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Based on the Hadamard product ξ(s)=ξ(0)ρ(1sρ)\xi(s)= \xi(0)\prod_{\rho}(1-\frac{s}{\rho}), a new expression of ξ(s)\xi(s) is obtained by paring ρ\rho and ρˉ\bar{\rho} ξ(s)=ξ(0)i=1(1sρi)(1sρˉi)=ξ(0)i=1(βi2αi2+βi2+(sαi)2αi2+βi2)\xi(s)= \xi(0)\prod_{i=1}^{\infty}(1-\frac{s}{\rho_i})(1-\frac{s}{\bar{\rho}_i})=\xi(0)\prod_{i=1}^{\infty}\Big{(}\frac{\beta_i^2}{\alpha_i^2+\beta_i^2}+\frac{(s-\alpha_i)^2}{\alpha_i^2+\beta_i^2}\Big{)} where ξ(0)=12\xi(0)=\frac{1}{2}, ρi=αi+jβi\rho_i=\alpha_i+j\beta_i and ρˉi=αijβi\bar{\rho}_i=\alpha_i-j\beta_i are the complex conjugate zeros of ξ(s)\xi(s), 0<αi<10<\alpha_i<1 and βi0\beta_i\neq 0 are real numbers, βi\beta_i are in order of increasing βi|\beta_i|, i.e., β1β2β3|\beta_1|\leq|\beta_2|\leq|\beta_3|\leq \cdots.\\ Then, by the functional equation ξ(s)=ξ(1s)\xi(s)=\xi(1-s), we have ξ(0)i=1(βi2αi2+βi2+(sαi)2αi2+βi2)=ξ(0)i=1(βi2αi2+βi2+(1sαi)2αi2+βi2)\xi(0)\prod_{i=1}^{\infty}\Big{(}\frac{\beta_i^2}{\alpha_i^2+\beta_i^2}+\frac{(s-\alpha_i)^2}{\alpha_i^2+\beta_i^2}\Big{)} =\xi(0)\prod_{i=1}^{\infty}\Big{(}\frac{\beta_i^2}{\alpha_i^2+\beta_i^2}+\frac{(1-s-\alpha_i)^2}{\alpha_i^2+\beta_i^2}\Big{)} i.e., i=1(1+(sαi)2βi2)=i=1(1+(1sαi)2βi2)\prod_{i=1}^{\infty}\Big{(}1+\frac{(s-\alpha_i)^2}{\beta_i^2}\Big{)}=\prod_{i=1}^{\infty}\Big{(}1+\frac{(1-s-\alpha_i)^2}{\beta_i^2}\Big{)} which, by Lemma 3, is equivalent to αi=12,i=1,2,3,,\alpha_i= \frac{1}{2}, i =1, 2, 3, \cdots, \infty Thus, we conclude that the Riemann Hypothesis is true.

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