A Proof of the Riemann Hypothesis Based on a New Expression of the Completed Zeta Function

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.

Proof of the functional equation for the Riemann zeta-function

In this article, we shall prove a result which enables us to transfer from finite to infinite Euler products. As an example, we give two new proofs of the infinite product for the sine function depending on certain decompositions. We shall then prove some equivalent expressions for the functional equation, i.e. the partial fraction expansion and the integral expression involving the generating function for Bernoulli numbers. The equivalence of the infinite product for the sine functions and the partial fraction expansion for the hyperbolic cotangent function leads to a new proof of the functional equation for the Riemann zeta function.

Γ-convergence of Onsager–Machlup functionals: II. Infinite product measures on Banach spaces

We derive Onsager–Machlup functionals for countable product measures on weighted ℓ p subspaces of the sequence space R N . Each measure in the product is a shifted and scaled copy of a reference probability measure on R that admits a sufficiently regular Lebesgue density. We study the equicoercivity and Γ-convergence of sequences of Onsager–Machlup functionals associated to convergent sequences of measures within this class. We use these results to establish analogous results for probability measures on separable Banach or Hilbert spaces, including Gaussian, Cauchy, and Besov measures with summability parameter 1 ⩽ p ⩽ 2. Together with part I of this paper, this provides a basis for analysis of the convergence of maximum a posteriori estimators in Bayesian inverse problems and most likely paths in transition path theory.

A Proof of the Riemann Hypothesis Based on a New Expression of the Completed Zeta Function

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.

A Proof of the Riemann Hypothesis Based on A New Expression of the Completed Zeta Function

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.

A Proof of the Riemann Hypothesis Based on A New Expression of the Completed Zeta Function

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.

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

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.

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

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.

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

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.