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.

My first post

July 11th, 2013 | Posted by in Opinion - (0 Comments)

My first post will be to explain why I have taken the work and time necessaries for this job.

I think that I can give my visitors a pure and sincere opinion of what I think and do about mathematics and I think I can help the very valuable students to gain faith and experience in their own researches and studies.

First of all I would like to say that the Three Primes Theorem that I have proved in the autumn of 2013 (South Hemisphere) is by no means a great theorem of Mathematics. It is not the Green Tao Theorem, nor the Twin Primes Conjecture and much less the Riemann Hypothesis.

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To the class and level of these great mathematicians I have not the slightest pretension to belong to.

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Nevertheless the theorem shares with them the fact that it is a result of the elusive prime numbers. Then it is a small grain of sand added to the Theory of Primes and thus to the science of Math.

And the second thing that I would like to say is that I am aware of each of the steps that my mind had taken in the construction and proof of this theorem. And it is on these steps that I shall write my next post.

This is (I think) the best I can give to my students : to explain them in detail these steps so they can try themselves to construct their own work.

Having this said I propose to take some time to write how I was able to prove this theorem.

The Three Primes Theorem

July 11th, 2013 | Posted by in - (Comments Off on The Three Primes Theorem)

A new proof of a theorem of Erdos and Turan.
Author: MARTIN MAULHARDT

ABSTRACT. A sequence of three consecutive primes is said to have the closer property if the difference between the first two of them is greater or equal than the difference between the second two. The first three such terns are (3,5,7), (7,11,13) and (13, 17, 19). In this paper we prove using only divergent series that there are infinitely many such terns of primes. This result was first found by P. Erdos and P. Turan in their 1948 paper "On some new questions on the distribution of prime numbers". But they used a quite advanced inequality about primes, namely that π(x) > c x / ln(x). We now conclude this result once again by completely different and simpler arguments and incidentally we find a new criteria of convergence of series.

INTRODUCTION

We first define the main concept of our work.

Definition 1: Let $p_1 < p_2 < p_3 <$ ... $< p_n<$ ... be the sequence of prime numbers.
Let $p_k$, $p_{k+1}$, $p_{k+2}$ be three consecutive terms of that sequence. Let the difference between the first two of them be: $d_k=p_{k+1}-p_k$ and the difference between the two second ones be $d_{k+1}=p_{k+2}-p_{k+1}$.

We say that the tern ($p_k, p_{k+1}, p_{k+2}$) has the closer property if $d_{k+1} \leq d_k$.

Three examples are (3,5,7), (7,11,13) and (13, 17, 19). In the first of these special cases we have $d_{k+1}=d_k=2$. In the second one we have $d_{k+1}=2$ and $d_k=4$. And finally in the third example we have $d_{k+1}=2$ and $d_k=4$.

A question naturally arises:
Are there infinitelly many such terns in the sequence of primes?

The answer is in the affirmative as we will prove in the next paragraphs.

We first need a new theorem on convergent series.

A NEW CRITERIA OF CONVERGENCE

Theorem 1.
Let $a_1,a_2,a_3,$ ..., $a_n,$ ... be any sequence of natural numbers ordered in increasing magnitude. Let the consecutive first difference between them be $d_n=a_{n+1}-a_n$ for n=1,2,3 .... If this differences form an increasing sequence then the series

$\sum\limits_{n=1}^{\infty}\frac{1}{a_n}$

is convergent.

Proof
We have by definition of $d_k$:
$d_1 = a_2 - a_1$

$d_2 = a_3 - a_2$

$d_3 = a_4 - a_3$

...

$d_n = a_{n+1} - a_n$

Because the $d_i$ are natural numbers and because they're increasing in magnitude we deduce the inequalities

$a_2=a_1+d_1\geq a_1+1$

$a_3=a_2+d_2 = a_1+d_1+d_2 \geq a_1 + 1 + 2$

...

$a_n\geq a_1+1+2+3$ + … + (n-1) = $a_1+\frac {n(n-1)}{2}$

Taking reciprocals and summing over n, we obtain:

$\sum\limits_{n=1}^{\infty}\frac{1}{a_n}$ = $\frac {1}{a_1} + \sum\limits_{n=2}^{\infty}\frac{1}{a_n} \leq 1+ \sum\limits_{n=2}^{\infty}\frac{1}{a_1+\frac {n(n-1)}{2}} \leq$

$\leq 1+ \sum\limits_{n=2}^{\infty}\frac{1}{\frac {n(n-1)}{2}} = 1+ 2 \sum\limits_{n=2}^{\infty}\frac{1}{n(n-1)}<\infty$

which is the desired result.

Examples. A trivial example is the sequence 1,2,4,8,..., $2^n$ ....The consecutive differences are precisely the same numbers 1,2,4,8, ..., $2^n$ ... which form an increasing sequence. Then the trivial result that the series $\sum\limits_{n=1}^{\infty}\frac{1}{2^n}$ is convergent.

A less trivial example is the sequence $F_n$ of the Fibonacci numbers. This sequence is 1, 1, 2, 3, 5, 8, 13, 21, 34, ... and the difference between the consecutive terms is the similar sequence 0, 1, 2, 2, 3, 5, 8, 13 ....
This differences form an increasing sequence, then by application of Theorem 1 $\sum\limits_{n=1}^{\infty}\frac{1}{F_n} <\infty$.

Remark. Theorem 1 remains true if we replace the condition that the difference between the two terms form an increasing sequence by the weaker condition that they form an increasing sequence starting at some point $n_0$.

MAIN RESULT

Recall that three consecutive prime numbers $p_k, p_{k+1}, p_{k+2}$ have the closer property if the differences $d_k$ and $d_{k+1}$ satisfy the condition $d_{k+1} \leq d_k$, an example being 4111, 4127, 4129.

With this refreshed in our minds we now return to our main result.

Theorem 2 (The Three Primes Theorem).

Let the sequence $p_1, p_2,$ ... , $p_n$, ... be the sequence of prime numbers in increasing order. Then there are infinitely many terns of consecutive primes

$p_k, p_{k+1}, p_{k+2}$

such that their consecutive differences satisfy the closer condition, that is,

$d_{k+1} \leq d_k$

where $d_{k+1} = p_{k+2} - p_{k+1}$ and $d_k = p_{k+1} - p_k$.

Proof. We prove this theorem by Reductio ad absurdum.

Suppose, to get a contradiction, that there are only a finite number of terns satisfying the closer property. Let this last tern be

$p_{j-2}, p_{j-1},p_j$

Discarding from the original sequence of prime numbers all primes p $\leq p_j$ (which are finite in number), we obtain the sequence of all the remaining primes p $> p_j$, that is:

$p_{j+1}, p_{j+2}, p_{j+3}$, ..., $p_n$, ...

The differences between two consecutive primes of this last sequence must form an increasing sequence. If this last condition is not met, then we have a tern of three consecutive primes $p_r, p_{r+1}, p_{r+2}$ satisfying the closer condition, contradicting the fact that the last tern was

$p_{j-2}, p_{j-1},p_j$

Thus, the sequence $p_{j+1}, p_{j+2}, p_{j+3}$, ..., $p_n$, ... satisfies the condition of Theorem 1, that is, their differences form an increasing sequence of natural numbers; then the series of the reciprocals of this sequence is convergent. But the sequence of the reciprocals of the primes, that is:

$\sum\limits_{i=1}^{\infty}\frac{1}{p_i}$

is divergent (Euler, 1737). This is the contradiction we were searching for and this completes the proof of our Theorem.

REFERENCES

[1] I. Niven, H. Zuckerman, H. Montgomery, An introduction to the theory of numbers, 1991 John Wiley \& Sons, Inc.

[2] G.H. Hardy and E.M. Wright, An introduction to the theory of numbers, Clarendon Press (Oxford), 1979

[3] L.E. Dickson, History of the theory of numbers, Carnegie Institution of Washington, New York 1950

SPECIAL THANKS TO MY FRIEND AND MENTOR N.H.GUERSENZVAIG FOR REVISING THIS WORK.