## A SPANISH LANGUAGE BIOGRAPHY OF ERDOS

January 5th, 2014 | Posted by in Uncategorized - (0 Comments)

A SPANISH LANGUAGE BIOGRAPHY OF PAUL ERDOS

In the first post of this new year 2014 I leave to my students in Argentina and to every Spanish language interested a new and original biography of the great Paul Erdos. It is written in Spanish but I will soon post it translated into English. Hope you enjoy it!

erdosbio

## The simplest proof of the Fundamental Theorem of Algebra

September 16th, 2013 | Posted by in Uncategorized - (0 Comments)

This item is dedicated to my students of Mathematical Analysis 3 at Facultad de Ingenieria, University of Buenos Aires. First some background. We have seen in our first two classes how to solve quadratic and cubic equations with coefficients in C. The roots of the cubic equations are to be found by means of Tartaglia-Cardano' s formula. Because our course is not about algebra but about Analysis we will not study the solution of the quartic equation which is performed by means of Ludovico Ferrari's formula. We have also mentioned that there is no "algebraic formula" that gives the roots of the quintic, a fact proved by Galois, Abel and Ruffini. So, how can you be sure that any polinomial has a root in C if you can't effectively show it? The answer is the Fundamental Theorem of Algebra proved by Gauss in 1799 although its proof was completed by Ostrowski in 1920. The next proof is an existence non constructive one, which means that we will prove that there is a root but we will make no attempt to show which actually is it.

Fundamental Theorem of Algebra Let p(z) = $a_n z^n + a_{n-1} z^{n-1} + \dots\ + a_1 z + a_0$ be any polinomial with the $a_{i} \in C$ and $a_{n} \neq\ 0$. Then there is a $z_{0} \in C$ such that $p(z_0) = 0$.

Proof. We may assume p(z) real for z real. If this is not the case consider the polinomial q(z) = $p(z) \overline{p}(z)$ ($\overline{p}(z)$ being the same polynomial than $p(z)$ but with its coefficients conjugated)   which is real for z real and the contradiction will be met by this last polinomial q(z) if not with the original p(z).

Since p(z) does not change sign because it has no roots then the integral :

$\int_{0}^{2\pi} \frac{1}{p(2cos\theta)} d\theta$

is either positive or negative but not zero.

If we introduce $z = e^{(i\theta)}$ for $0 \leq \theta \leq 2\pi$ then this integral become :

$\frac{1}{i}\oint \frac{1}{z p(z+z^{-1})} dz$$\frac{1}{i}\oint \frac{z^{n-1}}{z^{n} p(z+z^{-1})} dz$.

Now, the integrand is analitic for $z\neq 0$ and for $z = 0$ we have that the denominator is $a_{n} \neq 0$.

Because the integrand is analytic, by Cauchy theorem the integral is zero. A contradiction. Thus, the Fundamental Theorem of Algebra is true.

Note : This beautiful proof was given by N.C.Ankeny, an american mathematician who was a fellow at Princeton and the Institute for Advance Studies. He specialized in Number Theory and he wrote a book on game theory and gambling.

## A COMMENT FROM FAN

August 12th, 2013 | Posted by in Uncategorized - (0 Comments)

On August the 5 th I have posted on Tao's blog a link to my proof of the three consecutive non increasing inequalities between the differences of consecutive primes. This, I thought, was a question raised by Erdos and Turan in their 1948 paper "On some new questions on the distribution of prime numbers". On August 11 th Fan answer my post with the following information : Erdos and Turan were asking for decreasing inequalities (not non increasing ones). This information was already at the end of my work but Fan pointed out that with non decreasing inequality my results can be obtained from the prime number theorem. In the end the value of my work is the "elementary" proof of these non increasing inequalities without use of the prime number theorem. Thanks again to Fan.