User:Jrsousa2

About me

I'm a SAS programmer working mostly in the insurance industry. I've graduated from the University of Sao Paulo with a bachelor's degree in Pure Math (1995), and another in Statistics (1997). I didn't have to pay a dime, cause superior education in Brazil is free, if you are admitted through vestibular.

Math as a hobby

When it comes to math, I’m not interested in the rigorous treatment of math, I think it gets too much after a while, its gets boring, too much yada yada.

On here I will post links to some of my papers, even the most ridiculous ones, starting with the harmonic numbers.

The writing is not professional, no academic advisor after all — besides, thank God I don’t depend on academia for a living, it must be a nightmare, but with plenty of time to do nothing but type equations on a computer I’m sure it’d be neat. But the underlying ideas are somewhat interesting:
Generalized Harmonic Progression
On the Limits of a Generalized Harmonic Progression
An Exact Formula for the Prime Counting Function

The last one has a somewhat underwhelming logic behind it. Who would’ve thought that a formula for prime numbers could be obtained so easily, it’s almost like cheating.

Last but not least, let's pray for peace, all kinds of peace. We need a lighter world, not a heavier one. So many bad things going on right now in the world.

New formulae for Harmonic Numbers

In 2018, my first paper was released with a new formula for the harmonic number. It utilizes the Taylor series expansion of $\sin {\pi x}$ as a way to create a power series for $1/x$ which only holds for integer $x$, since $1/x$ is not analytic at 0. From there, the harmonic number is obtained via Lagrange's trigonometric identities and Faulhaber's formula for the sum of the powers of the first $n$ positive integers:

$H_{n}={\frac {1}{2n}}+\pi \int _{0}^{1}(1-u)\left(1-\cos {2\pi nu}\right)\cot {\pi u}\,du$

The paper also provides a generalization of the above formula for the so called generalized harmonic numbers (further defined later on in this page), through the employment of Bernoulli numbers:

$H_{2i}(n)={\frac {1}{2n^{2i}}}-{\frac {(-1)^{i}(2\pi )^{2i}}{2}}\int _{0}^{1}\sum _{j=0}^{i}{\frac {B_{2j}\left(2-2^{2j}\right)(1-u)^{2i-2j}}{(2j)!(2i-2j)!}}\sin {2\pi nu}\cot {\pi u}\,du$

$H_{2i+1}(n)={\frac {1}{2n^{2i+1}}}+{\frac {(-1)^{i}(2\pi )^{2i+1}}{2}}\int _{0}^{1}\sum _{j=0}^{i}{\frac {B_{2j}\left(2-2^{2j}\right)(1-u)^{2i+1-2j}}{(2j)!(2i+1-2j)!}}\left(1-\cos {2\pi nu}\right)\cot {\pi u}\,du$