Wednesday, November 29, 2017

La paradoja de pi=4

Hace unos días, Sergio Lopez-Permouth posteó un meme sobre por qué $\pi=4$


Este es un ejemplo interesante de por qué algunos procesos no son completamente continuos cuando realizamos un límite. Un resultado similar muestra que $\sqrt{2}=2$,



En ambos casos el problema principal ocurre al querer aproximar la longitud de arco por medio de segmentos infinitesimales verticales y horizontales. En este caso tenemos que este no es un sistema lineal, sino que cuadrático. Es decir,

$$ds\neq dx+dy\,,$$

sino que

$$ds^2=dx^2+dy^2\,.$$

En otras palabras, siempre tendremos triángulos infinitesimales y es por esto que se obtiene una aparente paradoja visual.

Dada que esta relación sí se cumple en orden cuadrático, es de esperar que el resultado sea verdadero, no acerca de los perímetros, sino sobre las áreas, como bien lo sugirió Mario Blanco. Esto es cierto dado que al remover los cuadrados se obtienen una suma de Riemann correspondiente al área del círculo. 

Si nos enfocamos en un cuadrante de la figura, por ejemplo

$$0\leq x\leq 1/2\,,y=\sqrt{1/4-x^2}\,, $$

tenemos que estamos realizando una suma de Riemann para la región entre

$$x=1/2\,, y=1/2\,, y=\sqrt{1/4-x^2}\,.$$

Es de notar que luego de la segunda etapa, los cuadrados obtenidos en la $n$-ésima etapa no tendrán todos el mismo lado. Por lo tanto es un tanto complejo escribir una serie que describa la suma de Riemann, sin embargo, sabemos que los lados de los cuadrados tienden a 0 cuando $n$ tiende a infinito, y esto asegura la existencia del límite y que sea igual a $1/4-\pi/4$.

Inicialmente quise escribir la serie, pero no me gustaría tener 20 páginas de ecuaciones, hay series más bonitas que describen $\pi$.


Wednesday, January 25, 2017

The voodoo behind 1+2+3+4+...=-1/12

A couple days ago, my friend Adolfo posted a rant on Facebook about the famous (or infamous) series

$$\sum_{n=1}^\infty n=-\frac{1}{12}\,.$$

His comment was not about the result itself, but about the little effort made by mathematicians to clarify this voodoo.


This result is very counter-intuitive as it states that if you add all positive integers, you will not only get a negative quantity, but also a fraction! I remember speaking about this result a couple years ago with a friend who was skeptical about it and he was telling me that it shouldn't be true since if he grabbed a calculator and start adding 1+2+3+4+..., he wouldn't approach a negative number, on the contrary, it would become bigger and bigger!


The trick lies in realizing we are talking about a mathematical object called a series, which is essentially different that any usual finite sum. One thing that can help to see that series are different than just adding a finite number of terms is that sometimes the order in which we add makes a difference! This is called Riemann rearrangement theorem, and states that if a series is conditionally convergent, then we can permute its terms to make it converge to any real number, or to make it diverge.

There are several concepts that play important roles here. One of them is the idea of dealing with an infinite number of things. It is not natural to operate with an infinite number of objects and a major problem is that we cannot directly apply an algorithm to compute this, as one of the key things of algorithms is that they must end, in other words, they must have a finite number of steps. 

Hence, the traditional addition algorithm that we use in everyday life (or at least that our calculators use), cannot be applied directly to infinite series. Then the relevant question regarding 

$$\sum_{n=1}^\infty n$$

is not how much is it? but rather, what is it?


The traditional approach for computing infinite series is by means of sequences of partial sums. This means that since we don't know what does it mean to add an infinite number of things, we approach infinity in the potential infinity sense, which establishes that infinity is the ability of taking an increasing sequence of big numbers forever. In other words, we take a big number of terms, we add them, and we assume that this result should somehow be close to the value of the series. Then we take a bigger amount of terms, and a bigger, and we keep doing this. Eventually, if the results obtained from the finite sums tend to cluster around a value, we say that that value is the result of the infinite series. 

It is important to remark that when we write

$$\sum_{n=1}^\infty\frac{1}{2^n}=1\,,$$

we really don't mean equal in the same sense we mean 1+1=2. This first equal sign really means

$$\lim_{k\to\infty} s_i=1\,,$$

where 

$$ s_i=\frac{1}{2}+\frac{1}{4}+\dots+\frac{1}{2^i}\,.$$

Of course writing $\sum 1/2^n=1$ is shorter and conveys the same idea, or at least that's what lazy mathematicians think.  Again, the equal sign used here should not be thought as a comparison between two objects but rather as assigning a value to the series. 

This assignment is very intuitive and we could say is very natural to think that this is a reasonable way to approach defining the meaning of an infinite series. But this it's not the only option. Deeper questions arise when this method of partial sums don't provide an answer, as it is the example of $\sum n$. The easy way out is simply to say there is no answer and the series diverges. But just as it happened with the equation $x^2=-1$, mathematicians saw an opportunity to extend the theory and also assign values to series for which the partial sums method is not enough.


It is possible to use the idea of actual infinity instead of potential infinity in order to approach these objects. An actual infinity approach considers having all infinite terms at the same time, as opposed to just a never-ending source of terms. Euler and Leibniz first started to develop ideas around divergent series and one of their key insights was to look at the meaning of a sum rather than its value.

A very interesting example of this occurs with the series

$$\sum_{n=0}^\infty (-1)^n\,.$$

This is an alternating sum for which the traditional approach of partial sums gives no answer (the partial sums jump from 0 to 1 and back forever).  But if we look a more abstract realization of the series, we can make it correspond to a geometric series, 

$$\sum_{n=0}^\infty r^n\,.$$

This is another representation of the function $1/(1-r)$ inside the unit circle. With this we can make the conceptual assignment of $r=-1$ to this series and make the correspondence

$$\sum_{n=0}^\infty (-1)^n\mapsto \frac{1}{2}\,.$$

Sometimes we abuse of the notation and we prefer to write it down as $\sum(-1)^n=1/2$, but we have to stress the fact that "=" does not represent a comparison between objects but rather a correspondence

Something similar happens with the series $\sum n$. One of the most used methods involve the famous Riemann Zeta function. This is defined as

$$\zeta(s)=\sum_{n=1}^\infty \frac{1}{n^s}\,,$$

for complex numbers $s$ with $\Re(s)$ bigger than 1. It is possible to extend the definition of the Zeta function to the entire complex plane using the reflection formula

$$\frac{\zeta(1-s)}{\zeta(s)}=\frac{2\Gamma(s)}{(2\pi)^s}\cos\left(\frac{\pi s}{2}\right)\,.$$

Using the series representation of the Zeta function suggest the correspondence

$$\sum_{n=1}^\infty n\mapsto\zeta(-1)\,.$$

Hence, we can assign the value of 

$$\zeta(-1)=\zeta(2)\frac{2\Gamma(s)}{(2\pi)^s}\cos\left(\frac{\pi s}{2}\right)=-\frac{1}{12}$$

to our series. Assign, not evaluate.