Saturday, October 31, 2009

On general derivatives

Long time ago I remember one of my former professors back home talking about a way of generalize the order of derivatives to real and then complex numbers. At the moment I was maybe in introduction to analysis or so, and I only did understand the meaning of the $n$th derivative, or at least, I knew how to calculate them.

Then, I came to graduate school and people where talking about fractional calculus, which in some sense is a generalization for the usual derivatives, providing a way of calculating the $\frac{p}{q}$ derivative of a function.

The purpose is to calculate $\frac{d^\alpha}{dt^\alpha}f(x)$ for $\alpha\in\mathbb{R}^+,\mathbb{R},\mathbb{C}$.

I was thinking of some sort of an easy and non-elaborated way of generalizing this idea, and it came to my mind the spectral theorem. If we define an operator $T=\frac{d}{dx}$ to be the derivative operator for a space of functions $H$ to itself, one would like to calculate $T^\alpha$ of a function, and a nice way of solving this would be to use functional calculus by means of the spectral theorem, which would say that

$T^\alpha(f)(x)=\int_{\sigma(T)}\lambda^\alpha dE(\lambda)(f)(x)$

where $\sigma(T)$ is the spectrum of the operator $T$ and $E(\lambda)$ is a partition of unity. This would provide a quick and dirty way of accomplish our mission, but we have one problem: $T$ is not a bounded operator, and we cannot quite use this same result.

Anyhow, trying to go further with this first approach, one has some classical results when applying this idea.

It is well known that the eigenfunctions of $T$ are exponentials given by the solutions of

$Tf-\lambda f=0$

subject to some boundary/initial conditions. Suppose that we work in the space $H=L^2$ with the usual inner product and the condition for the eigenfunctions to be $f(0)=1$. Then this result is the same as taking the Fourier transform of the function and passing the derivative along the integration sign

$T^\alpha (f)(x)=\int_{-\infty}^\infty \hat{f}(t) (2\pi i t)^\alpha e^{2\pi i xt}dt$

for $\alpha=0$ we have $T^0(f)(x)=f(x)$. When $\alpha=n\in\mathbb{N}$ we have the old result from Fourier analysis and in fact, we have that $T^n(f)(x)=\frac{d^n}{dx^n}f(x)$.

Also, for negative integers $\alpha=-n\in\mathbb{Z}^-$, this realization for the generalized derivative agrees with the result od Fourier transforms, having that $T^{-n}(f)(x)=F^{(n)}(x)$, where $F^{(n)}(x)$ is the $n$th antiderivative of $f(x)$.

But this method is not so true, since as we pointed out before, $T$ is not bounded in $L^2$, so the idea is to use some kind of spectral theorem for symmetric unbounded operators. This can be done by finding a self adjoint extension of $T$ and apply the spectral theorem to it or equivalently, to use the symmetric unbounded version of this result to $T$.

By this means, one can calculate the derivative of any complex order $\alpha$ of (in general) any $L^2$ function.

Thursday, October 15, 2009

Posible vs Probable

Hace unos días discutí con mi amigo Andrés sobre el post de como calcular la probabilidad de obtener un triángulo al partir un palillo en 3 trocitos. Él me preguntaba que pasaría si decidiéramos imponer la restricción que el triángulo obtenido fuera equilátero o isóceles.

Obviamente, dichos casos son posibles, sin embargo al calcular la probabilidad de ocurrencia, esta es cero.

Por ejemplo, en el caso de un triángulo equilátero, existe solamente una manera de obtener dicha configuración, esto es, cuando los 3 trocitos miden lo mismo. Sin embargo, el número total de posibilidades es infinita, por lo que podríamos decir que la probabilidad de obtener un triángulo equilátero es

$\frac{1}{\infty}=0$

Ahora bien, al calcular el número de triángulos isóceles que pueden ser obtenidos, este es infinito, sin embargo la probabilidad asociada sigue siendo cero. Este fenómeno puede ser explicado si utilizamos una interpretación geométrica de la probabilidad.

Así como vimos anteriormente la probabilidad se obtiene al dividir el área correspondiente a la región que representa los casos deseados sobre el área de los casos totales. En el caso de obtener un triángulo isóceles, el conjunto de casos favorables está dado por las tres rectas

$x=y \quad \cup\quad y=1-2x\quad \cup \quad 2y=1-x$

Puesto que este conjunto es unidimensional, su área es 0, y por lo tanto la probabilidad asociada a los triángulos isóceles es

$\frac{\text{area de las rectas}}{\text{area roja}}=\frac{0}{1/2}=0$

Esto quiere decir que dichos casos son posibles sin embargo no son probables. Pero ¿qué significa que algo sea posible pero no probable?

Matemáticamente, que algo sea posible significa que el conjunto de sucesos deseados en el espacio muestral sea no vacio. Por otra parte, que algo sea probable significa que la probabilidad asociada sea no nula. Resulta un poco dificil de convencerse al principio que esto pueda ocurrir, sin embargo al ver la probabilidad como la medida de un conjunto no suena tan artificial este concepto.

Si se tiene un conjunto $M$ una medida es una función que asigna a una familia de subconjuntos de $M$ un número real no negativo llamado medida. Por ejemplo, en la recta real, la medida mas usual es la de Lebesgue, la cual asigna a cada intervalo $[a,b]$ la medida $b-a$.

Este dilema se reduce a obtener subconjuntos no vacíos en $M$ cuya medida sea 0. El el caso de los reales, conjuntos discretos o conjuntos de cantor son buenos ejemplos de casos posibles mas no probables.

Luego de pensar un poco, no es difícil convencerse que si el espacio muestreal es un subconjunto con dimensión euclideana igual a $n$, cualquier conjunto no vacío con dimensión menor que $n$ será posible pero no probable, como el caso anterior, en el que el espacio muestreal tiene dimensión 2 (triángulo rojo) y el conjunto estudiado tiene dimensión menor, dimensión 1 para el caso de triángulos isóceles y dimensión 0 para el caso del equilátero.

Así, la unica manera de obtener probabilidad no nula es al comparar conjuntos de la misma dimensión, esto en el caso de espacios euclideanos, como fue el resultado en el caso anterior en donde comparamos la medida (área) de ambos conjuntos (triágulo rojo y triángulo azul)

Por lo tanto se puede tener algo posible pero no probable.

Monday, October 5, 2009

Elementary symmetric functions of the first natural numbers

About a month ago, my good friends Esteban and José Carlos were working in a really interesting problem in number theory involving harmonic series of sequences of something that they called the Esteban primes for $p_0$.

They were (or are, I hope) trying to find out if a particular series involving this sequences of primes, converges or not. They told me about the problem, and as always, I got really interested by this strange problem relating primes and analysis.

After working out a little bit the series, one can show that it is equivalent to prove that some infinite product converges to a nonzero value. Working with these expressions, I ended up looking at polynomials of the form

$p(x)=\prod_{k=1}^n (p_k-x)$

where the $p_k$ are primes. So, at first, I tried to not work with primes directly, but with the natural numbers $1,2,3,\dots,n$ so the polynomial will be

$p(x)=\prod_{k=1}^n (k-x)$

Expanding out this polynomial gives, by Cardano equations,

$p(x)=\sum_{k=0}^n (-1)^k s_{n-k} x^{k}$

were the $s_k$ are the elementary functions for $1,2,3\dots,n$, that are given by the sum of all possible $k$ products of these numbers, so for instance, $s_n=n!$, $s_1=1+2+\dots+n$ and $s_0=1$.

Therefore, evaluating $p(1)$ we have that $p(1)=\sum_{k=0}^n (-1)^k s_{n-k}=0$ and hence, the sum of the even elementary functions is equal to the sum of the odd ones.

For example, $n=3$ gives

$s_0=1$
$s_1=1+2+3=6$
$s_2=1\cdot 2+1\cdot 3+2\cdot 3=11$
$s_3=1\cdot 2\cdot 3=6$

and the result is

$s_0+s_2=s_1+s_3=12$

This reminds me of the property of the Binomial coefficients and must be somehow related, because these count the number of subsets with cardinality $k$, that is, for each subset of size $k$, we correspond to it a 1, then we add this results and that is $\binom{n}{k}$, while for the symmetric functions, for each subset of size $k$, we correspond to it the product of its elements, and then add over all subsets, and that gives $s_k$.

In other words, let $S=\{1,2,3,\dots,n\}$ and let $f$ be a function defined on the subsets of $S$.

If $f(A)=1$, for all $A\subset S$, we have that

$\binom{n}{k}=\sum_{A\subset S, |A|=k}f(A)$

on the other hand, if we define $f(A)=\prod_{k\in A} k$, then

$s_k=\sum_{A\subset S, |A|=k}f(A)$

so both forms have the same structure. I don't know what is the underlying property of such $f$ functions, so that the sum of the evens equals the sum of the odd ones, but I find it really interesting.