SearcharxivSearch

arXiv · 2106.14607

Faulhaber's Formula, Odd Bernoulli Numbers, and the Method of Partial Sums

Abstract

Let ``Faulhaber's formula'' refer to an expression for the sum of powers of integers written with terms in n(n+1)/2. Initially, the author used Faulhaber's formula to explain why odd Bernoulli numbers are equal to zero. Next, Cereceda gave alternate proofs of that result and then proved the converse, if odd Bernoulli numbers are equal to zero then we can derive Faulhaber's formula. Here, the original author will give a new proof of the converse using the method of partial sums and mathematical induction.

Explore related subjects

Keep this discovery

BibTeXRIS

Ryan Zielinski. 2021-05-25. Faulhaber's Formula, Odd Bernoulli Numbers, and the Method of Partial Sums. https://arxiv.org/abs/2106.14607

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