Strings of digits and prime numbers

This weekend, I attended a conference about geometry and topology. Last night was the conference banquet and there I was sitting in a table at dinner time with a bunch of mathematicians. After some egg rolls and a wonton soup, and obvious topic made its appearance: prime numbers.

My friend Brian Streit was sitting next to me, and in the middle of the dinner he proposed this interesting and rather old question:

Given any string of digits, can one find a prime number ending in such a string? How about beginning in it?

My immediate reaction was to answer yes to both questions, as I recall seen them before in my early olympic training days.

That also made me remember a conjecture that was given to me few months ago by one of my friends back home, Rafa Martinez, who was claiming the following:

Given any string of digits, there exists a digit such that you can insert it in at a position in the original string such that the resulting number is prime.

This statement is in a sense much stronger than the previous one, and as strong as it seems, it turns out to be false, but thinking about it can make you discover some interesting properties about prime numbers and base representation of numbers.

When Rafa gave me this problem, he was so convinced of its validity and I was so convinced of its falsity, that I started looking for a counterexample. First, I started with a single digit prime number, then I start looking for two digit prime numbers that contained the first one. Then I looked for 3-digit prime numbers that contained the previous one and so on.

The first counterexample that I found by this method was 612113, but thinking of this led me to state the same question but in a different base, say $b$. Of course, the statement will still be false, but now, depending on $b$, the length of the counterexample would be different. For instance, if we are in base 2, a counterexample would be 1 (which is not prime), in base 3 we have for example 212, etc.

Basically, this phenomenon is due to the rate of growth of the prime numbers and how that can be related with the base in which numbers are written in.

Broadly speaking, we can make use of the prime number theorem to establish the rate of growth of prime numbers. It basically says, in simple words, that prime numbers grow in a logarithmic way, notice I'm not talking about the density of the primes, but about the size of a prime itself.

Hence, there is a high probability that if a number $p$ is prime, then "$ep$" is also a prime number, which, in some sense says that the most natural "base" to write prime numbers in is un "base $e$".

Of course $ep$ is not a prime number, as it is irrational, even more, what does "base $e$" mean anyway? The idea is to use this approach to find some appropriate bases for writing prime numbers. An appropriate base would be one such $b$ is close to a power of $e$. The first of these such powers is

$b=20\sim e^3=20.085536923187667740928529654581717896987907838554150144...$

The next one is

$b\sim e^8=2980.9579870417282747435920994528886737559679391328357022089...$

and so on.

Of course, changing base representation wont make the behavior of prime numbers different or anything like that, but maybe using this representations could lead to nice patterns on its representations.

After this little digression, I was then left with the original question of Brian, which I didn't think until today in the morning when I was discussing it about with other participant of the conference.

About the question that if you could find a prime number starting with a given string of digits (not ending in even or 5,0 of course) the simplest solution I could think was using Dirichlet's theorem on arithmetic progressions. Given a string of digits $a$, we look at the sequence

$a+k10^n$

with $k$ running over the positive integers and $n$ been the length of $a$. Hence Dirichlet's theorem asserts that there are in fact, infinitely many primes in this sequence, and hence, infinitely many numbers answering positively the question posed.

For the remaining question, my proof didn't come as quickly as the previous one, and it took me the entire last talk (which sadly I didn't pay attention to) to come up with a proof, aided with my good old friend Wikipedia.

I found this interesting fact about prime gaps due to Hoheisel, that states that

$p_{n+1}-p_n \leq p_n^\theta$ for $\theta$ less than 1

where $p_n$ is the $n$th prime number.

With the help of this, proving the statement is not really hard. Proceeding by contradiction, suppose that $a$ is a given string of digits, and that there is no prime number in between $a10^m$ and $(a+1)10^m$ for all $m$, hence this would imply that the relative gap of prime numbers would be of at least

$\frac{ a 10^m- (a+1) 10^m}{a 10^m}=\frac{1}{a} $

but by Hoheisel's result, as $n\to\infty$

$ \frac{p_{n+1}-p_n}{p_n}\to 0$

which leads to a contradiction.

I particularly find really interesting this connections between prime distributions and base representations, it seems that we don't completely understand yet the interaction of these two, but definitely there is an intimate relation among them.

Ecuaciones cónicas y extensiones algebráicas

Hace un par de meses discutiamos con mi amigo José Carlos sobre ecuaciones diofantinas y de posibles métodos para poder resolverlas completamente sobre los enteros.

Este tipo de ecuaciones han sido sujeto de estudio intenso a través de la historia, siendo uno de los ejemplos más famosos el del último teorema de Fermat, dado un natural $n$, determinar las soluciones enteras de la ecuación

$x^n+y^n=z^n$

Como es bien sabido, gracias a Andrew Wiles, dicha ecuación no posee soluciones enteras a menos que $n=1$ o $n=2$.

Proponiendo una meta menos ambisiosa, consideremos una ecuación diofantina de segundo grado

$ax^2+bxy+cy^2=z$

La idea era caracterizar todos los enteros $z$ que pueden ser escritos de la forma $ax^2+bxy+cy^2$, y en el mejor de los casos, dar la forma de los enteros $x$ y $y$ que cumplen con el trabajo.

Con un poco de ejemplos numericos para ciertos valores de $a,b,c$, José Carlos conjeturó que las soluciones $x,y$ de la ecuación son multiplicativas, es decir, si

$ax_1^2+bx_1y_1+cy_1^2=z_1$

$ax_2^2+bx_2y_2+cy_2^2=z_2$

con $x_1,y_1$ y $x_2,y_2$ primos relativos, entonces existen $x_3,y_3$ tales que

$a(x_3)^2+b(x_3)(y_3)+c(y_3)^2=(z_1z_2)$

con afán de simplificar un poco la notacion, podemos denotar la ecuación como

$p(x,y)=z$

y si reescribimos la propiedad multiplicativa como

$p(x_1,y_1)\cdot p(x_2,y_2)=p(x_3,y_3)=z_1z_2$

Esto da una noción de un cierto tipo de norma en $Z^2$. Esto no es más que una extensión de la idea de los enteros gaussianos, en donde se hacen corresponder a los pares $(x,y)$ con $x+iy$, ahora, en este caso, es necesario trabajar con otra extensión algebraica de segundo orden distinta de $\mathbb{Z}\oplus i\mathbb{Z}$.

En pocas palabras, la idea es hacer corresponderle a cada par de enteros $(x,y)$ un numero $x+\alpha y$, donde $\alpha$ es un numero algebraico sobre $\mathbb{Z}$ de grado 2, y de tal manera que $p(x,y)$ sea una norma en dicha extensión algebráica.

Para esto, consideremos el conjunto $G_\alpha=\mathbb{Z}\oplus\alpha\mathbb{Z}$, con $\alpha$ un número algebráico de segundo orden con polinomio mínimo $p(x)=x^2+ex+f$. La idea es convertir $G_\alpha$ en algo que posea una estructura muy parecida a la de los numeros complejos, digamos, un anillo junto con una operación de conjugación.

En primer lugar, la multiplicación de dos elementos se puede hallar utilizando el hecho que $\alpha^2=-b-a\alpha$, lo cual da

$(x,y)(w,z)=(xw-byz,yw+xz-ayz)$

Ahora, es necesario tener un cirto tipo de $\alpha$-conjugado, es decir, un homomorfismo involutivo en $G_\alpha$. Esta conjugación queda totalmente determinada por su valor en $\alpha$. Si tenemos que $\overline{\alpha}=k+d\alpha$, utilizando la propiedad involutiva y la de ser homomorfismo, llegamos a que la única conjugación no trivial es $k=-e$ y $d=-1$.

Ahora, podemos definir la norma en $G_\alpha$ como

$|(x,y)|^2=(x,y)\overline{(x,y)}$

lo cual, por puede escribirse como

$|(x,y)|^2=x^2-exy+fy^2$

con esto tenemos que es inmediata la observación dada por José Carlos, puesto que si $p(x,y)=x^2-exy+fy^2$, y si $p(x_1,y_1)=z_1$ y $p(x_2,y_2)=z_2$, entonces $p(x_3,y_3)=z_1z_2$, donde $x_3,y_3$ están dados por la multiplicación en $G_\alpha$

$(x_1,y_1)(x_2,y_2)=(x_3,y_3)$

Así, el problema inicial de resolver $p(x,y)=z$ se reduce a encontrar puntos con coordenadas enteras en el círculo de radio $z$ centrado en el origen en $G_\alpha$, donde $\alpha$ es una de las raices de $p(x,1)$.

Es importante observar que cualquier polinomio $ax^2+bxy+cy^2$ sustituirse por un polinomio de la forma $x^2-exy+fy^2$ sin afectar el problema inicial.

Projectivizations and orthogonal groups

Few days ago, I went to a summer school in algebraic geometry in Peru, and ithere was one course in projectivization of some special types of curves and properties of them.

One of the examples was doing the projectivization of a general conic over $\mathbb{C}^n$ and finding some nice properties that the resulting curves have.

The idea of the projectivization of a curve, is to grab the domain of definition and then do its compactification, for instance, in the case of a conic in $\mathbb{C}^2$, the result is to have a conic in a sphere (Riemann sphere). Algebraically, the idea is to have a curve defined as the zero set of a polynomial, for example, $p(x,y)=ax^2+bxy+cy^2=0$ with $(a,b,c)\in\mathbb{C}^3$, and then to extend the domain where the parameters are defined, i.e. $[a:b:c]\in\mathbb{P}(\mathbb{C}^2)$.

The speaker was talking about the cases in dimension 2 and 3, where the proyective spaces are $\mathbb{P}(\mathbb{C}^2)$ and $\mathbb{P}(\mathbb{C}^5)$ , and she defined a nice realization of the later one as the set of $3\times 3$ complex symmetric matrices, i.e.

$p(x,y,z)=a_1x^2+a_2xy+a_3xz+a_4y^2+a_5yz+a_6z^2$

$\mapsto \begin{pmatrix} a_1 & a_2 & a_3\\ a_2 & a_4 & a_5 \\ a_3 & a_4 & a_6 \end{pmatrix}$

This reminded me of the good old $O(3)$ with some differences. First of all, we have complex entries in our matrix, and second, the entries must satisfy and extra condition, we identify set of parameters with a common factor ($A\sim B$ iff $A=\lambda B$ for some $\lambda\in\mathbb{C}^x$), that is, the matrix defines the same curve up to a scalar multiple, which means that we need to scale it in some sense.

This two things can be fixed by considering $SO(3)$ instead of the whole $O(3)$ and then, by doing its complexification, $SO(3)\times_{\mathbb{R}}\mathbb{C}$.

So, my natural conjecture was that

$\mathbb{P}(S^n_2)\sim SO(n)\otimes_{\mathbb{R}}\mathbb{C}$

where $S^n_2$ is the space of homogeneous polynomials with degree 2 in $n$ variables and the isomorphism is as complex manifolds.

After discussing with a couple of people, I found that there were reasons to believe that this is true, since some topological properties of both spaces matched, but so far, I haven't found yet a formal proof of this fact.

In the case of this being true, it would turn out to be a really interesting property, since the object on the right is a Lie group, being isomorphic will imply that one can define a group structure in the set of conics, which is rather an interesting fact.

El calor y la redondez de los objetos

Una pregunta muy tonta aparentemente es la de ¿cómo saber si un objeto es plano o no?, sin embargo, al tratar de determinar si algo es plano o no, frecuentemente nos referimos a algún objeto patrón, el cual acordamos en decir que es plano, pero cómo saber si es plano en realidad, y más aún, ¿qué significa que algo sea plano?

Esta linea de pensamientos anterior puede resultar un poco sin sentido y muy abstracta, sin embargo al meditar sobre el asunto es posible darse cuenta del problema latente que se tiene en esta situación. Nuestra referencia de algo plano está dada principalmente por la superficie del suelo en donde vivimos, la cual, desafortunadamente, no es para nada plana.

En términos matemáticos, la pregunta es cómo saber si un objeto tiene curvatura $0$ o no. Una manera indirecta de aproximarse a la respuesta es por medio de la propagación del calor.

Por ejemplo, para determinar si una superficie es plana o no, bastaría con aplicarle una fuente de calor puntual y analizar las curvas isotérmicas, si éstas son círculos, entonces la superficie tiene curvatura constante, y en especial, si es $0$, la superficie es plana (un plano o un cilindro)

Para lograr determinar en sí si se trata de un plano o no, es posible analizar el flujo del calor a cortos tiempos. Básicamente, si el calor se propaga aproximadamente a una razón constante o menos, entonces la superficie es plana. En otras palabras, la forma más lenta en que el calor puede propagarse es en un medio plano, lo cual tiene sentido, puesto que al haber curvatura, existe más proximidad entre las partículas del material y por lo tanto el calor puede propagarse a mayor velocidad.

Esto puede obtenerse por medio de analizar la ecuación de propagación de calor, la cual en coordenadas locales puede escribirse como

$\Delta f(x,t) =\frac{\partial f(x,t)}{\partial t}$

donde $\Delta$ es el laplaciano en $\mathbb{R}^2$ y $x\in M$ una variedad diferencial compacta sin frontera de dimensión 2.

Unos de los métodos más comunes para resolver esta ecuación es por medio de la utilización del kernel o núcleo de calor $K(t,x,x')\in C^\infty\left(\mathbb{R}^+\times M \times M\right)$, el cual da una solución a la ecuación diferencial con condición inicial $f(x,0)=g(x)$

$f(x,t)=\int_M K(t,x,y)g(y)dy.$

Para tiempos cortos, ie. $t\to 0$, se tiene la expansión asintótica del kernel

$K(t,x,x)\sim t^{-1}\sum\limits_{k=0}^\infty a_k(x)t^{k}$

en donde los coeficientes del kernel de calor dependen unicamente de la variedad $M$, su métrica y sus derivadas.

Es un resultado conocido que el primer coeficiente $a_0$ es un múltiplo de la curvatura escalar de la variedad, y en nuestro caso, estamos interesados en que esta sea $0$, por lo tanto para que la superficie sea plana, debemos tener que para tiempos pequeños ($t\sim 0$)

$K(t,x,x)\sim a_1(x)+a_2(x)t+O(t^2)$

por otra parte, si la curvatura escalar no es cero, se tiene que la expansión del kernel es

$K(t,x,x)\sim a_0(x)t^{-1}+a_1(x)+a_2(x)t+O(t^2)$

cuyo término predominante a pequeños tiempos es $a_0$, el cual hace que la propagación del calor sea más rápida, del orden de $t^{-1}$.

Una consecuencia curiosa de este análisis es el hecho que el hecho de que una superficie sea plana o no depende de la rapidez del flujo de calor, en otras palabras, el que una superficie sea plana, no es solamente una propiedad de las dimensiones espaciales de dicho objeto, sino que también de la dimensión temporal en la que está inmerso.

Este hecho da indicios de una relación más intima entre las dimensiones espaciales y temporales, y que al final de cuentas, no están tan desligadas unas de otras.

Polygons and vertex orbits

Last Christmas, I went with my family to a road trip. I drove from Waco to New Orleans, a 8 hrs drive more or less, and when we reached there, I was a little tired. I laid down in the bed and stared at the roof of our room, and then, I started imaging a rubber ball bouncing all over a vertical cross section of our room, like a billiards ball, and then I wondered which kind of orbits could be ball have, depending of the incident angle of the first bounce.

After thinking a little, one can realize that for a rectangular cross section (my room's) this problem is not really hard, and the possible answers are quite few, depending on the ratio of the lengths of the rectangle. Then I though that a more interesting question would be when the cross section its a circle.

Basically two thing can happen, one is that the incident angle is such that the ball bounces only in finitely many places, and hence, the bouncing points make a periodic sequence on the circumference, and the other is that the bouncing points are dense in the circumference.

For instance, if the initial angle happens to be $\pi/2$, we are going to have only 2 bouncing points. Suppose that we have an initial angle $a$ and we are working with a unit circle, so that the arc length is the same as the angle value. If we look look up for periodic points, we require that $na=m\pi$, for some $n,m\in\mathbb{Z}^+$. That means that $a= \frac{m}{n}\pi$, and with out loss of generality, we can require $m/n \leq 1$. Then, one can draw the path of the ball as follows:

• Step 1: Start drawing the path at vertex $k=1$
• Step 2: Draw a line from the vertex $k$ to the vertex $k+m$ (sums are taken modulo $n$)
• Step 3: Go to step 2 until you hit vertex 1

For instance, in the case $n=5$, we will have two different patterns

The orange path is obtained for $m=1$ and $m=4$, and the black one, for $m=2$ and $m=3$. In general, $m$ and $n-m$ will lead to the same orbit, but made in different directions (clockwise and counterclockwise).

So in the remaining case, when $a$ is not a rational multiple of $\pi$, we have that the bouncing points are dense in the circumference.

This result is not surprising, as this picture is related with the problem of wrapping a line around a torus, and it is well know that this wrapping is dense if it has an irrational slope, and periodic otherwise.

When glazed in this perspective, it turns out that the original problem, with the rectangle, and the one with the circle are exactly the same, since a torus is just the complex plane modulo a rectangle ( a lattice ), so at the end, my more interesting problem turned out to be as simple as the original one. It is quite interesting how apparently different problems became just two perspectives of the same phenomenon, one could seem totally boring and the other one, completely attractive and challenging. Some would say, beauty is in the eye of the beholder, but sometimes what happens is that differences are in the eye of the beholder.

Representación integral de la función máximo entero

Hace un par de semanas fui a un congreso/seminario sobre aplicaciones de la matemática en la industria y el impacto de minorias en estados unidos en la ciencia. Principalmente se habló sobre oportunidades de trabajo y aplicaciones de distintas áreas de la matemática en problemas actuales, como control en cambio climático, modelación de epidemias, etc. La verdad, no fue muy matemático que digamos puesto que fue más orientado a opciones de comenzar una carrera en la industria y no en la academia, sin embargo, hubo una charla muy interesante sobre funciones zeta, teoría de la representación y matrices aleatorias que capturó mi atención.

Durante su charla, el expositor tocó brevemente el tema de la hipótesis de Riemann, y para darnos la idea de la dificultad del problema, ofreció un premio de $100 si alguien era capaz de demostrar que el primer cero de la función zeta de Riemann tenía parte real igual a 1/2.

Como todo buen estudiante de postgrado y en permanente carecia de fondos, me llamó la atención el intentar este problema. El tiempo era limitado, puesto que al día siguiente se terminaba el congreso, así que durante el resto de las charlas me dedique a tratar de resolver el problema.

En la tarde, fuí a hablarle al conferencista para mostrarle mis avances y las ideas que llevaba, a lo cual no prestó mucha atención en el momento, así que decidí seguir intentando un poco mas en la noche. Al día siguiente fui a consultarle nuevamente, y esta vez me dijo que iba en buen camino, pero que no se recordaba muy bien de como iba la prueba, él mismo lo habia intentado hacía unos 15 años. Al final de cuentas, no logré probarlo, para alivió economico de él y frustración mia, sin embargo, me llevó a descurbir un par de cosas interesantes.

Estaba jugando un poco con representaciones integrales de la función zeta de Riemann, y se me ocurrió probar el viejo truco de utilizar el argumento principal de Cauchy. Definí una función $F$ que me permitía, por medio de una integral de contorno, contar el número de ceros de la función zeta en un rectángulo de altura variable $\sigma$, y puesto que la gráfica $F$ es una gráfica en escalera, intenté hacer algo similar con la función máximo entero.

La idea es trabajar con una función que se anule en todos los enteros, y un buen ejemplo de eso es $f(s)=\frac{1}{\pi}\cot(\pi s)$.

Además, recordando que el argumento principal de Cauchy establece que

$\frac{1}{2\pi i}\int_\gamma F(s)ds=\sum_z Res (F(s))_{s=z}$

donde la suma corre sobre los polos de $F$ encerrados dentro de la curva $\gamma$.

Con estas dos cosas, tenemos que

$\frac{1}{2i}\int_\gamma \frac{1}{\cot (\pi s)}ds=\sum_z Res(F(s))_{s=z}$

si hacemos una curva que contenga a todos los enteros positivos, tenemos una representación para la función maximo entero, así que podemos definir $\gamma$ como el rectángulo con vértices opuestos $(\sigma, \epsilon)$ y $(1/2, -\epsilon)$, donde el contorno se toma en la dirección antihoraria.

Por lo tanto, podemos definir

$M(\sigma)=\frac{1}{2i}\int_{\gamma_\sigma}\frac{1}{\cot(\pi s)}ds$

Esta función suma los residuos de la funcion $F(s)=\frac{\pi}{\cot(\pi s)}$ en cada entero entre 1/2 y $\sigma$. No es dificil de probar que el residuo de $F(s)$ en $s=n\in\mathbb{Z}^+$ es 1, así que

$M(\sigma)=\sum_{n=1}^{[\sigma]} 1=[\sigma]$

Siguiendo esta idea, tambien es posible hallar una representación integral para la suma de los primeros $[\sigma]$ positivos

$\frac{1}{2i}\int_{\gamma_\sigma}\frac{s}{\cot(\pi s)}ds=\sum_z Res\left(\frac{\pi s}{\cot(\pi s)}\right)_{s=z}$

$=\sum_{n=1}^{[\sigma]}n=\frac{[\sigma]([\sigma]+1)}{2}$

En forma general, la representación integral de la suma de las primeras $[\sigma]$ $k$-potencias de positivos está dada por

$\sum_{n=1}^{[\sigma]} n^k=\frac{1}{2i}\int_{\gamma_{\sigma}}\frac{s^k}{\cot(\pi s)}ds$

para toda $\sigma>1/2$ y todo $k\in\mathbb{C}$.

Esta es una bonita conección entre funciones número teóricas y análisis, muchas veces el contar con una representación analítica resulta muy conveniente al estudiar matemáticas discretas.

The game of life

A couple of weeks ago, I was reading some lecture notes on game theory and I came across a really neat game.

After discussing the very basics of game theory and decision making theory, the author of the lectures gives an exercise which I found really interesting and enjoyable, at the point that I went ahead and gave it as a quiz to my business calculus class.

To my surprise, most of my class got the right answer, which was truly a grateful feeling. The game is really simple so anybody can understand it, but in my opinion, it represents many aspects of real life.

It is as follows:

Every student is to write down a real $x_i$ number in between 0 and 10 inclusively. After doing so, one computes the mean $\bar{x}$ of all of the students' bets and each student's grade is given by

$10-\left|x_i-\frac{2}{3}\bar{x}\right|$

This might look a really simple task, and by no means a game at all, but it is a game of strategy and common sense.

Our desire as a students is to maximize our grade, but that depends on the average choice of the class, which might complicate a bit the analysis of a best strategy to pick our $x_i$.

It is not hard to see that a global best strategy is to pick $x_i=0$, as if everybody is a good and logical player, having all bets equal to $0$ would give each student's grade to be $10$, which is the best possible.

So, our personal best strategy should be to pick $0$, but in real life, not all players are good thinkers or really logical, so at the end of the day, our best strategy won't give us the best out come possible.

In a sense, we can think of this game as rewarding you if you somehow think average, and most of the times, the average thinking is not precisely the most wise and logical.

By the way the game was set up, we can see that it neither rewards the average thinking as much as someone that was 2/3 away from it. If you are 2/3 away from the average, you'll get full credit, and this is somehow what happens in real life. Usually not the average people get the best outcome nor the people that plays the best, but people that are in between.

This gives a really good example that in most occasions, your outcome does not depend only on your own strategy, but also in someone else's strategy, and that making the best decisions and taking the best choices does not guarantee your success.