A Dynamical Approach to Riemann Zeta Function

Here is a definition of the Riemann zeta function.

Def 1.1: The Riemann zeta function \zeta(s) is defined as

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

The function is defined on the whole complex plane but s = 1. Riemann proved that the zeta function has a symmetric property, which is

\zeta(s) = 2^s \pi^{s-1} \sin(\frac{\pi s}{2}) \Gamma(1-s) \zeta(1-s)   (1)

Originally, \zeta is defined on a complex plane to itself, but for now, let us limit the domain and the range of the function to real numbers in order to create a dynamical system. The first task is to look for fixed points of \zeta.

Thm 1.1: There are infinitely many fixed points of \zeta on s < 1.

Pf. of Thm 1.1

The functional equation (1) can be re-written into

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

The Taylor series of e^z is

e^z = \sum_{k=0}^\infty \frac{(z)^k}{k!}

Since e^z converges on the entire complex plane, the sequence of the terms \frac{(2\pi)^k}{k!} converges to zero. Hence,

\text{As } s \in \mathbb{N}, s \to \infty, \hspace{0.5cm} \frac{(s-1)!}{(2\pi)^{s-1}} = \Gamma(s) (2\pi)^{1-s} \to \infty

Since \zeta(s) asymptotically approaches to 1, 2(2\pi)^{-s}\Gamma(s)\zeta(s) still diverges. However, \cos\left(\frac{\pi s}{2}\right) forces the whole equation to return to zero for every 2 units. Therefore, it can be concluded that y = \zeta(1-s) crosses y = s infinitely many times. Thus, there are infinitely many fixed points along s < 1.

Featured image

As the graph above suggests, there are infinitely many periodic points along the line s < -20, and there are two other fixed points which are s = -.295905005575214... and s = 1.83377265168027... numerically. Before we go further, here I would like to introduce a pair of terminologies;

Def. 1.2: Let p be a periodic point of period n under a function f.

If |(f^n)'(p)| < 1, then p is an attracting periodic point (an attractor).
If |(f^n)'(p)| > 1, then p is a repelling periodic point (a repellor).

And, here is a proposition regarding the newly introduced terminologies.

Prop. 1.1: Let p be an attractor. Then, there is an open interval U about p such that if x \in U, then

\lim_{n\to \infty} f^n(x) = p. [1]

Prop. 1.2: Let p be a repellor. Then, there is an open interval U of p such that, if x \in U\setminus \{p\}, then there exists k > 0 such that f^k(x) \not \in U. [2]

All the periodic points along s<-20 and s=1.834... are repellors. Plus, all the points on s > 1 besides the periodic point are attracted to a cycle of prime period 2, which is (1, \infty).

The only fixed point left, s = -.296... , is the only attractor among the fixed points. If s=0 was entered in the dynamical system \zeta^n(s) for an arbitarily large n converged to s = -.296...; thus I hoped to find the distribution of s = -.296... along \zeta^n: \mathbb{C} \to \mathbb{C} for arbitrarily large n in order to observe the distribution of zeros of Riemann zeta function. However, this approach bears two crucial flaws.

The first flaw is that s=-.296... is indeed a strong attractor that sucks up all the real numbers on s < 1 to itself after sufficiently many re-iterations of \zeta(s); in other words, non-trivial zeros are not the only s \in \mathbb{C} which converge to the constant.

I assumed that my plan would be somehow feasible if I could define a \zeta on dynamical system from a set A \subset \mathbb{C} to itself. \zeta: A \to A. However, we need \forall s \in A, \Re(s) \geq 1/2, yet a non-trivial zero will send \zeta (s) = 0 \not \in A; thus it is impossible to find such A.

Quite disappointingly yet expectedly, the dynamical approach seems to be pretty futile in figuring out the trivial zeros of \zeta.

1. Devaney, Robert L. An Introduction to Chaotic Dynamical Systems, 2nd Edition: Westview, 2003. Print.
2. IBID.

A Dynamical Approach to Riemann Zeta Function

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s