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.

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!