Search arXiv⌕ Search
Home/Archive/arXiv

arXiv · math/0702605

Synthesizing Sums

Elemer E Rosinger
View original source ↗Read source document ↗Export metadata ↓

Abstract

Polynomial functions $f : \mathbb{N}_+ \longrightarrow \mathbb{N}_+$ are studied for which sums of arbitrary length $f (1) + f (2) + f (3) + >... + f (n)$, with $n \in \mathbb{N}_+$, can be expressed by polynomial functions $g : \mathbb{N}_+ \longrightarrow \mathbb{N}_+$ which involve a bounded number of operations, thus not depending on $n$. Open problems and extensions are presented.

Explore related subjects

math.GMmath.AC

Keep this discovery

BibTeXRIS

Elemer E Rosinger. 2007-02-21. Synthesizing Sums. https://arxiv.org/abs/math/0702605

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

KEEP EXPLORING

Related papers

arXiv2026

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↗
arXiv2026

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↗