SearcharxivSearch

arXiv · 2312.00297

A note on Apery's constant is transcendental

Abstract

Beuker's [2] considers the following integral $$ \int_{0}^{1}\int_{0}^{1} \frac{-\log xy}{1-xy} P_n(x)P_n(y)\ dx dy$$If $d_n=\text{LCM}(1,2,...,n)$, then $$ 0<\frac{|A_n+B_n\zeta(3)|}{d_n^3}<2(\sqrt{2}-1)^{4n} \zeta(3) $$ for some $A_n,B_n\in\mathbb{Z}$. We establish that if Apery's constant is algebraic then the above inequality fails to be true. This proves that $\zeta(3)$ is

Explore related subjects

Keep this discovery

BibTeXRIS

Shekhar Suman. 2023-08-07. A note on Apery's constant is transcendental. https://arxiv.org/abs/2312.00297

Cite the original work for its findings. Save a collection to share your selection of sources.

KEEP EXPLORING

Related papers

Average Chord Lengths in a Triangle

Let $P$ be a point inside a triangle $T$. We consider the average length of the chords of $T$ through $P$, where the direction of the chord is chosen uniformly. An elementary formula is obtained in terms of the distances from $P$ to the sides and vertices of the triangle. Several classical triangle centers give especially simple specializations. For example, if $I$ is the incenter, then \[ M_T(I)=\frac{2r}{\pi} \log\left(\cot\frac A4\cot\frac B4\cot\frac C4\right). \] Our main result is the sharp inequality \[ M_T(P)\le \frac{p}{\pi\sqrt3}\log(2+\sqrt3), \] valid simultaneously for every triangle of perimeter $p$ and every interior point $P$. Thus, among all such pairs $(T,P)$, the largest possible average chord length occurs only when $T$ is equilateral and $P$ is its center. The proof is an elementary symmetrization argument. We close with brief remarks relating the problem to the radial center of a convex body, the electrostatic potential center of a triangle, and dual quermassintegrals.

math.GM

A Proof of Liu's Conjecture on the Fundamental Triangle Inequality

Let $a,b,c$ be the side lengths of a triangle, and let $R$ and $r$ denote its circumradius and inradius, respectively. We prove a conjecture of Liu stating that \[\sum_{\mathrm{cyc}} \left(\frac{a(b+c-a)}{bc}\right)^k \geq 2+\left(\frac{2r}{R}\right)^k,~~k>1, \] with the reverse inequality for $0<k<1$. The proof reduces the problem to three positive variables with fixed sum and product. We also determine the equality cases.

math.GM