## miércoles, 21 de junio de 2017

### Edmund Landau, the irrepressible spoiled child of mathematics

... Landau was an instance of that uncommon phenomenon, the scion of a wealthy family who yet had a powerful work ethic and a record of great achievement in a non-commercial field. Landau's mother Johanna, née Jacoby, came from a rich banking family. His father was a Professor of Gynecology in Berlin, with a successful practice. Landau Senior was also a keen supporter of Jewish causes. The family home was at Pariser Platz 6a, in the most elegant quarter of Berlin, close to the Branderburg Gate. Edmund was appointed to a professorship at Göttingen in 1909. When people asked for directions to his house, he would reply "You can't miss it. It's the finest house in town." He followed his father's (and Hadamard's) interest in Zionism, helping to establish the Hebrew University of Jerusalem and giving the first math lecture there, in Hebrew, shortly after the university opened in April 1925.

Landau was something of a character--this was a great age for mathematical characters--and there are apocrypha about him rivaling those of Hilbert and Hardy. Perhaps the best-known story is his remark about Emmy Noether, a colleague at Göttingen. Noether was mannish and very plain. Asked if she was not an instance of a great female mathematician, Landau replied: "I can testify that Emmy is a great mathematician, but that she is a female, I cannot swear." His work ethic was legendary. It is said that when one of his junior lecturers was in hospital, recuperating from a serious illness, Landau climbed a ladder and pushed a huge folder of work through the poor man's window. According to Littlewood, Landau simply did not know what it was like to be tired...

References
J. Derbyshire, Prime obsession: B. Riemann and the greatest unsolved problem in mathematics. Published by Plume (a member of Penguin Group), USA, 2003, pp. 230-231.

## lunes, 5 de junio de 2017

### Revisitando la demostración de Erdös del postulado de Bertrand

Al releer recientemente el artículo donde Erdös expone su prueba del postulado de Bertrand, caí en la cuenta de que el buen Paul sí mencionó en ese trabajo que su prueba daba también una cota inferior para el número de primos en los intervalos $(n,2n]$ (donde $n \in \mathbb{N}$) y que dicha cota es prácticamente la que predice el teorema de los números primos.

Recordemos que lo que Erdös hace en su artículo es acotar inferior y superiormente los coeficientes binomiales $c_{n}:=\binom{2n}{n}$ y comparar entre sí sendas estimaciones. La estimación inferior es $$\frac{4^{n}}{2n} \leq \binom{2n}{n}$$ y la obtiene de la identidad $\binom{2n}{0} + \ldots + \binom{2n}{2n} = 2^{2n} = 4^{n}$. La estimación superior la obtiene al estudiar de una manera muy astuta la descomposición en números primos de $\binom{2n}{n}$: en efecto, del teorema fundamental de la aritmética, de la fórmula Legendre (cf. A. M. Legendre, Essai sur la théorie des nombres. Seconde édition, 1808, págs. 8-10; de acuerdo con W. Narkiewicz, la atribución a de Polignac (1826-1863) y/o a Chebyshev (1821-1894) de este resultado es incorrecta) y de la desigualdad de Erdös-Kalmár se llega a que \begin{eqnarray*} \binom{2n}{n} &\leq & \prod_{p \leq \sqrt{2n}} p^{\alpha_{p}(c_{n})} \prod_{\sqrt{2n} < p \leq \frac{2}{3}n} p \prod_{n < p \leq 2n} p\\ &\leq& \prod_{p \leq \sqrt{2n}} (2n) \cdot 4^{\frac{2}{3}n} \cdot \prod_{n < p \leq 2n} p\\ &\leq& (2n)^{\sqrt{2n}} \cdot 4^{\frac{2}{3}n} \cdot \prod_{n < p \leq 2n} p \end{eqnarray*} para cada número natural $n \geq 3$. La conexión con el estudio de los números primos en $(n,2n]$ se acaba de hacer más que patente en este momento, ¿cierto?

De ambas estimaciones se desprende que si $n\geq 3$ entonces \begin{eqnarray*} 4^{\frac{n}{3}} \leq (2n)^{1+\sqrt{2n}} \prod_{n < p \leq 2n} p. \end{eqnarray*} De esto y de la desigualdad $$(2n)^{1+\sqrt{2n}} < 2^{\frac{n}{2}},$$ la cual es válida para todo número natural $n$ suficientemente grande (lo que en este caso quiere decir, en números redondos, siempre que $n > 22 \, 620$), se obtiene que $$\prod_{n < p \leq 2n} p > 2^{\frac{n}{6}}$$ si $n$ es suficientemente grande: ergo, para cada $n$ así de grande se cumple que $$(2n)^{\pi(2n)-\pi(n)} > 2^{\frac{n}{6}},$$ o equivalentemente que \begin{eqnarray*} \pi(2n)-\pi(n) > \frac{\log 2}{6} \cdot \frac{n}{\log 2n}. \end{eqnarray*} En resumidas cuentas: ¡la formulación clásica del postulado de Bertrand es sumamente conservadora!

## miércoles, 26 de abril de 2017

### A play on the interplay 'twixt primes and polynomials

In what follows, we shall denote the set of positive prime numbers by $\mathbf{P}$.

Act I. It is more or less well-known that there does not exist a non-constant polynomial $f \in \mathbb{Z}[x]$ such that $f(n) \in \mathbf{P}$ for every $n \in \mathbb{N}$. This can be proven by reductio ad absurdum: if $f(n) \in \mathbf{P}$ for every $n \in \mathbb{N}$ and $f(1) =: p$, then $p \mid f(1+kp)$ for every $k \in \mathbb{N}$; it follows that at least one of the equations $f(x)=p$ or $f(x)=-p$ has more solutions than $\deg(f)$, Q. E. A. This result is typically attributed to Christian Goldbach: W. Narkiewicz, on page 25 of his "The Development of Prime Number Theory", even mentions that it can be found in a letter from Goldbach to Euler written on September 28th, 1743. Luckily for us, Springer-Verlag published two years ago a translation into English of the correspondence of L. Euler with C. Goldbach edited and commented by F. Lemmermeyer and M. Mattmüller.

Act II. Several years ago, while perusing a very interesting article on primes in arithmetic progressions by M. R. Murty, I learned the notion of prime divisor of a polynomial: if $p \in \mathbf{P}$ and $f \in \mathbb{Z}[x]$, we say that $p$ is a prime divisor of $f$ if $p \mid f(n)$ for some $n \in \mathbb{Z}$. According to Murty, the basic theorem on the set of prime divisors of a non-constant $f \in \mathbb{Z}[x]$ can be traced back (at least) to a 1912 paper of I. Schur. The theorem can be proven emulating the celebrated proof of Eucl. IX-20.

Theorem. If $f$ is a non-constant polynomial of integer coefficients, then its set of prime divisors is infinite.

Proof. If $f(0)=0$, then every $p \in \mathbf{P}$ is a prime divisor of $f$. If $f(0) = c \neq 0$, then $f$ has at least one prime divisor as it can take on the values $\pm 1$ only finitely many times. Given any finite set $\mathcal{P}_{k}:=\{p_{1}, \ldots, p_{k}\}$ of prime divisors of $f$, we are to show that we can always find another prime divisor of $f$ which does not belong to $\mathcal{P}_{k}$. Let $A := p_{1} \cdots p_{k}$ and consider the equality $f(Acx)=cg(x)$ where $g \in \mathbb{Z}[x]$ is a polynomial of the form $1+c_{1}x+c_{2}x^{2}+\cdots$ where every $c_{i}$ is a multiple of $A$. Since $g$ is a non-constant polynomial whose constant term is different from $0$, $g$ has at least one prime divisor $p$. Clearly enough, this prime number $p$ is also prime divisor of $f$ which does not belong to $\mathcal{P}_{k}$. Q.E.D.

Act III. Resorting to the ideas in the previous paragraphs plus Dirichlet's glorious theorem on primes in arithmetic progressions, we are going to determine all the non-constant polynomials $f \in \mathbb{Z}[x]$ such that $f(\mathbf{P}) \subseteq \mathbf{P}$.

- If $f$ is one such polynomial and $f(0)=0$, then $f(x)=xg(x)$ for some some $g \in \mathbb{Z}[x]$. Given that $p \cdot g(p) =f(p) \in \mathbf{P}$ for every $p \in \mathbf{P}$, it follows that $g(p) = \pm 1$ for every $p \in \mathbf{P}$; therefore, in this case we obtain that either $g(x)=1$ and $f(x) =x$ or $g(x)=-1$ and $f(x)=-x$.
- Let us assume now that $f$ is one such polynomial and $f(0)=c \neq 0$. By the above theorem, we may fix a prime divisor $q$ of $f$ which is greater than $|c|$. If $n \in \mathbb{Z}$ is a witness of the fact that $q$ is a prime divisor of $f$, then $q \nmid n$. Thus, if $p_{1} < p_{2} < p_{3} < \ldots$ are all the positive primes in the arithmetic progression whose first term is $n$ and whose common difference is $q$, we have that $f(p_{i}) \equiv f(n) \equiv 0 \pmod{q}$ for every $i \in \mathbb{N}$, which is decidedly absurd because $f$ can assume the values $\pm q$ only finitely many times.

Hence, $f(x)=x$ is the only non-constant polynomial with integer coefficients which sends $\mathbf{P}$ to one of its subsets.

THE END.

## martes, 18 de abril de 2017

### Nunca hubo milagro

NUNCA HUBO MILAGRO

Banach y Tarski se encontraban gesticulando y argumentando, en el mismo cubículo, frente a un inmenso pizarrón verde, cuando demostraban el teorema que sería conocido a la postre como la Paradoja de Banach-Tarski: dada una bola sólida en $\mathbb{R}^{3}$, existe una descomposición de esta en un número finito de subconjuntos disjuntos que se pueden juntar otramente para producir dos copias idénticas a la bola original. Justo cuando terminaron la prueba, ambos callaron y se miraron muy contentos. Tarski hizo una pequeña aspiración y retuvo el aire un instante hasta que finalmente, abstraído, le dijo a Banach: "Ahora sabemos cómo fue que Cristo multiplicó los peces y el pan".

Autor: Enrique Ruiz.

Postdata. Leí este cuento por vez primera en 2016: no obstante, debo de confesar que estuve aguardando su aparición desde aproximadamente el primer semestre de 2004 pues fue más o menos in illo tempore que el Prof. Vulfrano T. me comentó que la multiplicación de los panes y los peces se podía conectar con el Axioma de Elección.

## sábado, 7 de enero de 2017

### Para la reflexión

«It is related of the Socratic philosopher Aristippus (c. 435 – c. 356 BCE) that, being shipwrecked and cast ashore on the coast of the Rhodians, he observed geometrical figures drawn thereon, and cried out to his companions: "Let us be of good cheer, for I see the traces of man." With that he made for the city of Rhodes, and went straight to the gymnasium. There he fell to discussing philosophical subjects, and presents were bestowed upon him, so that he could not only fit himself out, but could also provide those who accompanied him with clothing and all other necessaries of life. When his companions wished to return to their country, and asked him what message he wished them to carry home, he bade them say this: that children ought to be provided with property and resources of a kind that could swim with them even out of a shipwreck...»

(Vitruvio en De architectura [Libro VI])