scholarly journals A Proof of the Riemann Hypothesis Based on MacLaurin Expansion and Hadamard Product of the Completed Zeta Function

Author(s):  
Weicun Zhang

The basic idea is to expand the completed zeta function $\xi(s)$ in MacLaurin series (infinite polynomial), which can be further expressed as infinite product (Hadamard product) by conjugate complex roots. Finally, the functional equation $\xi(s)=\xi(1-s)$ leads to $(s-\alpha_i)^2 = (1-s-\alpha_i)^2, i \in \mathbb{N}$ with solution $\alpha_i= \frac{1}{2}, i \in \mathbb{N}$, where $\alpha_i$ are the real parts of the zeros of $\xi(s)$, i.e., $s_i =\alpha_i\pm j\beta_i, i\in \mathbb{N}$. Therefore, a proof of the Riemann Hypothesis is achieved.

Author(s):  
Weicun Zhang

The basic idea is to expand the completed zeta function $\xi(s)$ in MacLaurin series (infinite polynomial), which can be further expressed as infinite product (Hadamard product) by conjugate complex roots $\alpha_i\pm j\beta_i, i\in \mathbb{N}$. Then, according to the functional equation $\xi(s)=\xi(1-s)$, we have $$\xi(0)\prod_{i=1}^{\infty}\frac{\beta_i^2}{\alpha_i^2+\beta_i^2}\Big{(}1+\frac{(s-\alpha_i)^2}{\beta_i^2}\Big{)} =\xi(0)\prod_{i=1}^{\infty}\frac{\beta_i^2}{\alpha_i^2+\beta_i^2}\Big{(}1+\frac{(1-s-\alpha_i)^2}{\beta_i^2}\Big{)}$$ which is equivalent to $$(s-\alpha_i)^2 = (1-s-\alpha_i)^2, i \in \mathbb{N}$$ with solution $\alpha_i= \frac{1}{2}, i\in \mathbb{N}$. Therefore, a proof of the Riemann Hypothesis can be achieved.


Author(s):  
Weicun Zhang

The basic idea is to expand the completed zeta function $\xi(s)$ in MacLaurin series (infinite polynomial), which can be further expressed as infinite product by conjugate complex roots. Then, according to Lemma 3, Lemma 4, and Lemma 5, the functional equation $\xi(s)=\xi(1-s)$ leads to $(s-\alpha_i)^2 = (1-s-\alpha_i)^2$ with solution $\alpha_i= \frac{1}{2}$, where $\alpha_i$ are the real parts of the zeros of $\xi(s)$, i.e., $s_i =\alpha_i\pm j\beta_i, i\in \mathbb{N}$. Thus a proof of the Riemann Hypothesis is achieved.


Author(s):  
Weicun Zhang

The completed zeta function $\xi(s)$ is expanded in MacLaurin series (infinite polynomial), which can be further expressed as infinite product (Hadamard product) by its complex conjugate zeros $\alpha_i\pm j\beta_i, \beta_i\neq 0, i\in \mathbb{N}$. Then, according to the functional equation $\xi(s)=\xi(1-s)$, we have $$\xi(0)\prod_{i=1}^{\infty}\frac{\beta_i^2}{\alpha_i^2+\beta_i^2}\Big{(}1+\frac{(s-\alpha_i)^2}{\beta_i^2}\Big{)} =\xi(0)\prod_{i=1}^{\infty}\frac{\beta_i^2}{\alpha_i^2+\beta_i^2}\Big{(}1+\frac{(1-s-\alpha_i)^2}{\beta_i^2}\Big{)}$$ which, by Lemma 3 and Corollary 1, is equivalent to $$(s-\alpha_i)^2 = (1-s-\alpha_i)^2, i \in \mathbb{N}$$ with solution $\alpha_i= \frac{1}{2}, i\in \mathbb{N}$. Thus, a proof of the Riemann Hypothesis can be achieved.


Author(s):  
Weicun Zhang

The completed zeta function $\xi(s)$ is expanded in MacLaurin series (infinite polynomial), which can be further expressed as infinite product (Hadamard product) of quadratic factors by its complex conjugate zeros $\alpha_i\pm j\beta_i, \beta_i\neq 0, i\in \mathbb{N}$ ($\mathbb{N}$ is the set of natural numbers, from $1$ to infinity). Then, according to the functional equation $\xi(s)=\xi(1-s)$, we have $$\xi(0)\prod_{i=1}^{\infty}\frac{\beta_i^2}{\alpha_i^2+\beta_i^2}\Big{(}1+\frac{(s-\alpha_i)^2}{\beta_i^2}\Big{)} =\xi(0)\prod_{i=1}^{\infty}\frac{\beta_i^2}{\alpha_i^2+\beta_i^2}\Big{(}1+\frac{(1-s-\alpha_i)^2}{\beta_i^2}\Big{)}$$ which, by Lemma 3 and Corollary 1, is equivalent to $$(s-\alpha_i)^2 = (1-s-\alpha_i)^2, i \in \mathbb{N}$$ with solution $\alpha_i= \frac{1}{2}, i\in \mathbb{N}$ (another solution $s=\frac{1}{2}$ is invalid due to obvious contradiction). Thus, a proof of the Riemann Hypothesis is achieved.


Author(s):  
Weicun Zhang

The completed zeta function $\xi(s)$ is expanded in MacLaurin series (infinite polynomial), which can be further expressed as infinite product (Hadamard product) by its complex conjugate zeros $\alpha_i\pm j\beta_i, i\in \mathbb{N}$. Then, according to the functional equation $\xi(s)=\xi(1-s)$, we have $$\xi(0)\prod_{i=1}^{\infty}\frac{\beta_i^2}{\alpha_i^2+\beta_i^2}\Big{(}1+\frac{(s-\alpha_i)^2}{\beta_i^2}\Big{)} =\xi(0)\prod_{i=1}^{\infty}\frac{\beta_i^2}{\alpha_i^2+\beta_i^2}\Big{(}1+\frac{(1-s-\alpha_i)^2}{\beta_i^2}\Big{)}$$ which, by Lemma 3 and Corollary 1, is equivalent to $$(s-\alpha_i)^2 = (1-s-\alpha_i)^2, i \in \mathbb{N}$$ with solution $\alpha_i= \frac{1}{2}, i\in \mathbb{N}$. Thus, a proof of the Riemann Hypothesis can be achieved.


Author(s):  
Weicun Zhang

The completed zeta function $\xi(s)$ is expanded in MacLaurin series (infinite polynomial), which can be further expressed as infinite product (Hadamard product) of quadratic factors by its complex conjugate zeros $\rho_i=\alpha_i +j\beta_i, \bar{\rho}_i=\alpha_i-j\beta_i, 0<\alpha_i<1, \beta_i\neq 0, i\in \mathbb{N}$ are natural numbers from 1 to infinity, $\rho_i$ are in order of increasing $|\rho_i|=\sqrt{\alpha_i^2+\beta_i^2}$, i.e., $|\rho_1|<|\rho_2|\leq|\rho_3|\leq |\rho_4|, \cdots$, together with $\beta_1<\beta_2\leq\beta_3\leq\beta_4, \cdots$. Then, according to the functional equation $\xi(s)=\xi(1-s)$, we have $$\xi(0)\prod_{i\in \mathbb{N}}\Big{(}\frac{\beta_i^2}{\alpha_i^2+\beta_i^2}+\frac{(s-\alpha_i)^2}{\alpha_i^2+\beta_i^2}\Big{)} =\xi(0)\prod_{i\in \mathbb{N}}\Big{(}\frac{\beta_i^2}{\alpha_i^2+\beta_i^2}+\frac{(1-s-\alpha_i)^2}{\alpha_i^2+\beta_i^2}\Big{)}$$ which, by Lemma 3, is equivalent to $$(s-\alpha_i)^2 = (1-s-\alpha_i)^2, i \in \mathbb{N}, \text{from 1 to infinity.}$$ with only valid solution $\alpha_i= \frac{1}{2}$ (another solution $s=\frac{1}{2}$ is invalid due to obvious contradiction). Thus, a proof of the Riemann Hypothesis is achieved.


Author(s):  
Weicun Zhang

The completed zeta function $\xi(s)$ is expanded in MacLaurin series (infinite polynomial), which can be further expressed as infinite product (Hadamard product) of quadratic factors by its complex conjugate zeros $\alpha_i\pm j\beta_i, \beta_i\neq 0, i\in \mathbb{N}$ are natural numbers, from $1$ to infinity, $\mathbb{N}$ is the set of natural numbers. Then, according to the functional equation $\xi(s)=\xi(1-s)$, we have $$\xi(0)\prod_{i=1}^{\infty}\frac{\beta_i^2}{\alpha_i^2+\beta_i^2}\Big{(}1+\frac{(s-\alpha_i)^2}{\beta_i^2}\Big{)} =\xi(0)\prod_{i=1}^{\infty}\frac{\beta_i^2}{\alpha_i^2+\beta_i^2}\Big{(}1+\frac{(1-s-\alpha_i)^2}{\beta_i^2}\Big{)}$$ which, by Lemma 3 and Corollary 1, is equivalent to $$(s-\alpha_i)^2 = (1-s-\alpha_i)^2, i \in \mathbb{N}$$ with solution $\alpha_i= \frac{1}{2}, i\in \mathbb{N}$ (another solution $s=\frac{1}{2}$ is invalid due to obvious contradiction). Thus, a proof of the Riemann Hypothesis is achieved.


2020 ◽  
Vol 31 (10) ◽  
pp. 2050082
Author(s):  
Hau-Wen Huang

Let [Formula: see text] denote a connected [Formula: see text]-regular undirected graph of finite order [Formula: see text]. The graph [Formula: see text] is called Ramanujan whenever [Formula: see text] for all nontrivial eigenvalues [Formula: see text] of [Formula: see text]. We consider the variant [Formula: see text] of the Ihara Zeta function [Formula: see text] of [Formula: see text] defined by [Formula: see text] The function [Formula: see text] satisfies the functional equation [Formula: see text]. Let [Formula: see text] denote the number sequence given by [Formula: see text] In this paper, we establish the equivalence of the following statements: (i) [Formula: see text] is Ramanujan; (ii) [Formula: see text] for all [Formula: see text]; (iii) [Formula: see text] for infinitely many even [Formula: see text]. Furthermore, we derive the Hasse–Weil bound for the Ramanujan graphs.


Author(s):  
Fayez Alhargan

By unraveling a persistent misconception in the zeta Hadamard product expansion, and employing the zeta functional equation, a concise proof of the Riemann Hypothesis is presented, which conclusively demonstrate that the Riemann Hypothesis is true.


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