# Hey, what’s that!?

Hey, long time without posting. Hope this will change drastically in the near future. In the meantime… can anybody tell me what’s that!? :)

# There’s music in the primes… (part I)

OK, now a new section at physicsnapkins, in which I will discuss a bit about my own research… Recently, Germán Sierra and I have submitted to the ArXiv a paper about the Riemann hypothesis, which you can see here. To be honest, the real expert in the field is Germán, my contribution is mostly technical. Anyway, I’ll try to convey here the basic ideas of the story… We’ll give a walk around the concepts, assuming only a freshman maths level.

It’s fairly well known that the sum of the reciprocal numbers diverges: $\sum_{n=1}^\infty 1/n\to\infty$. Euler found, using an amazing trick the sum of the inverse squares and, in fact, the sum of the inverse of any even power. This formula is simply amazing: $\sum 1/n^2 = \pi^2/6$, isn’t it? Now, Riemann defined the “zeta” function, for any possible exponent:

$\displaystyle{\zeta(z)=\sum_{n=1}^\infty {1\over n^z}}$

So, we know that $\zeta(1)=\infty$, $\zeta(2)=\pi^2/6$, and many other values. Riemann asked: what happens when z is complex? Complex function theory is funny… We know that z=1 is a singularity.  If you do a Taylor series around, say, around z=2, the radius of convergence is the distance to the nearest singularity, so R=1. But now, your function is well defined in a circle of center z=2 and radius R=1. This means that you can expand again the function in a Taylor series from any point within that circle. And, again, the radius of convergence will be the distance to the closest singularity. This procedure is called analytical continuation.

The series of circles show how to compute the analytical continuation of a complex function...

Well… in the case of the Riemann $\zeta$ function, the only singularity is z=1. Therefore, I can do the previous trick and… bypass it! Circle after circle, I can reach z=0 and get a value, which happens to be… -1/2. So, somehow, we can say that, if we had to give a value to the sum 1+1+1+1+…, it should be -1/2. Also, and even more amazing, $\zeta(-1)=\sum_{n=1}^\infty n=1+2+3+4+\cdots=-1/12$. Hey, that’s really pervert maths! Can this be useful for real life, i.e.: for physics. Well, it is used in string theory, to prove that you need (in the simplest bosonic case) dimension D=26… but that’s not true physics. Indeed, it’s needed for the computation of the Casimir effect. Maybe, I’ll devote a post to that someday. Anyway, this is the look of the Riemann zeta function in the complex plane:

The Riemann zeta function. Color hue denotes phase, and intensity denotes modulus. The white point at z=1 is the singularity.

Even more surprises… $1^2+2^2+3^2+\cdots=0$. In fact, it’s easy (ok, ok… it’s easy when you know how!) to prove that $\zeta(-2n)=0$ for all positive n. Those are called the trivial zeroes of the Riemann zeta function (amazing!)… So, what are the non-trivial ones? Riemann found a few zeroes which were not for negative even numbers. But all of them had something in common: their real part was 1/2. And here comes the Riemann hypothesis: maybe (maybe) all the non-trivial zeroes of the $\zeta$ function will have real part 1/2.

OK, I hear you say. I got it. But I still don’t get the fun about the title of the post, and why so much fuss about it. Here it comes…

Euler himself (all praise be given to him!) found an amazing relation, which I encourage you to prove by yourselves:

$\displaystyle{\zeta(z)=\sum_{n=1}^\infty {1\over n^z} = \prod_{p \hbox{ prime}} {1\over 1-p^{-z}}}$

Ahí comienza el link verdadero. Una pista para la demostración: expandimos el producto:

$\displaystyle{\prod_{p \hbox{ primes}} {1\over 1-p^{-z}} = {1\over 1-2^{-z}} {1\over 1-3^{-z}} {1\over 1-5^{-z}} \cdots}$

Wonderful. But $1/(1-x)$ can be easily recognized as the sum of a geometric series, right?

OK, in a few days, I’ll post the second part, explaining why there’s music in the primes, and how quantum mechanics might save the day…

# Log rules…

Here is a very simple problem which appeared with my first year calculus students. Consider the function

$f(x)=\log(x^2)$

Its domain is ${\mathbb R}-\{0\}$. But now, we may take the exponent down, using the rules of the logarithm…

$f(x)=2\log(x)$

and now, the domain is $(0,\infty)$… What happens??? Try to explain it: (a) without complex numbers, (b) with them…