The Half-Plane of Convergence of Prime Zeta Function

This post contains a brief proof of Conj. 3 posted on Lee’s Conjecture. To avoid confusion I used x as real variable, and s = \sigma + it as complex.

I proposed a conjecture as follows:

Conj. 3:
Let \mathbb{P} be a set of primes. The supremum of x such that \displaystyle \sum_{p \in \mathbb{P}} \frac{1}{p^x} diverges is x = 1.

Let us define a Prime Zeta Function as follows: \displaystyle P(s) = \sum_{p \in \mathbb{P}} \frac{1}{p^s}.
In fact, the conjecture turned out to be an easy comparison series problem.

Consider the original Riemann Zeta function, that is defined as \displaystyle \zeta(s) = \sum_{n \in \mathbb{N}} \frac{1}{n^s}. It was already proven that \zeta(s) is defined on a half plane of s such that \sigma > 1. The domain was extended over all complex plane except s = 1 via Analytic Continuation in Riemann’s paper.

Since the conjecture is asking the supremum of x, let us only consider the case of real numbers. Obviously, \zeta(x) \geq P(x) for x > 1, as each term in the summand of P(x) is also included in \zeta(x), and every term in \zeta(x) is positive. Thus, as long as \zeta(x) is defined, so is P(x). Hence, the domain of P(x) includes a half real line x > 1. Therefore, the supremum of x where P(x) diverges must be x = 1; the conjecture is true.

While studying Introduction to Analytic Number Theory by T. Apostol, I encountered a following theorem. (1976 Edition, pp.233)

Theorem 11.8:
If [a Dirichlet] series \sum f(n)n^{-s} converges for s = \sigma_0 + it_0, then it also converges for all s with \sigma > \sigma_0. If it diverges for s = \sigma_0+it_0, then it also diverges for all s with \sigma < \sigma_0


f(n) presented in the theorem is an arithmetical function. Consider a prime detecting function \rho (n) such that \rho(n) = 1 for n being prime, and \rho(n) = 0 for n being composite. Then, the Dirichlet series \sum \rho(n)n^{-s} = P(s). Therefore, we may apply Theorem 11.8 to deduce the following fact.

We already showed that for any real number x > 1, P(x) converges. Take arbitrarily small \epsilon > 0, then P(1+\epsilon) converges. Thus, by Theorem 11.8, for any s with \sigma > 1+\epsilon, P(s) also converges. Letting \epsilon \to 0, we can conclude $P(s)$ is defined for all \sigma > 1. Thus, the domain of P(s) is now extended to the half plane s where \sigma > 1.

Then, we can come up with a few new problems.

  1. Can we extend the domain of P(s) via Analytic Continuation?
  2. As we defined Prime Zeta Function, let us define Composite Zeta Function, that is \displaystyle C(s) = \sum_{n \in \mathbb{N} \setminus \mathbb{P}} \frac{1}{n^s}. Will the Composite Zeta Function have the same domain?
  3. Does Prime Zeta Function have functional equation as Riemann Zeta Function? If so, what are the trivial zeroes and nontrivial zeroes of P(s)?
The Half-Plane of Convergence of Prime Zeta Function

Constructing a set with arbitrary strict Hausdorff dimension

Here’s the link to Lee’s Challenge #1:

Consider a non-negative real number \beta; if \beta \in \mathbb{Z}, then constructing a set of strict Hausdorff dimension \beta is indeed trivial.

Suppose 0 < \beta < 1. Then, we may apply following theorem.

Theorem 1:
Suppose S_1, \dots, S_m are m separated similarities of a self-similar set F with the common ration r that satisfies 0 < r < 1. Then F has Hausdorff dimension equal to \log m / \log (1/r).

Theorem 1 suggests that Cantor set has Hausdorff dimension of \log 2 / \log 3. The proof of that Cantor set has a strict Hausdorff dimension is left to readers.

Let us define a \mathcal{C}_\delta as a Cantor set which was removed relative \delta from each interval for each stage of construction instead of 1/3. Then, Theorem 1 suggests the Hausdorff dimension of \mathcal{C}_\delta is \log 2 / \log (\frac{1-\delta}{2}). Since 0 < \delta < 1, we may choose appropriate \delta to construct \mathcal{C}_\delta of strict Hausdorff dimension \beta.

For \beta > 1, we may use the idea of n-dimensional Cantor dust. n-dimensional Cantor dust is constructed by n-cartesian product of Cantor sets. The n-dimensional Cantor dust has a strict Hausdorff dimension of n \log 2 / \log 3, and the proof is left to readers. (Key Idea: Use Theorem 1)

Therefore, for any \beta > 1, we may choose a natural number n satisfying \beta < n < \beta+1, and find appropriate \delta to render the strict Hausdorff dimension of \mathcal{C}_\delta to be \beta / n.

Thus, we may construct a set with arbitrary Hausdorff dimension.

Constructing a set with arbitrary strict Hausdorff dimension

Constructing a polynomial with arbitrary Mahler Measure

Let p(z) \in \mathbb{C}[z], and p(z) = a(z-\alpha_1)(z-\alpha_2)\cdots(z-\alpha_n).

Here is a definition of Mahler measure.

M: \mathbb{C}[z] \to \mathbb{R} \text{, with } M(p) = |a| \prod_{i=1}^n \max\{1, |\alpha_i|\}

Lehmer conjectured,

Lehmer’s Mahler-measure Conjecture:
There exists a real number \mu such that \forall p(x) \in \mathbb{Z}[x], M(p) = 1 or 1 < \mu \leq M(p) . Lehmer suggested

M(p_\mu) = \mu = 1.176280818...

where p_\mu is given as

p_\mu = x^{10} - x^9 + x^7 - x^6 + x^5 - x^4 + x^3 - x^2 + 1.

Now, let us “try” to construct a polynomial p(x), such that 1 < M(p) < \mu.

First, consider 1 < \alpha < \mu. Hence, (x-\alpha)(x-\frac{1}{\alpha}) = x^2 - (\alpha+\frac{1}{\alpha}) + 1; however, \alpha + \frac{1}{\alpha} \not \in \mathbb{Z}, so we need to calibrate the polynomial with more terms.


p_N(x) = (x^2-(\alpha+\frac{1}{\alpha})x + 1) \prod_{i=1}^N (x- \beta_i)(x-\overline{\beta_i})

where \forall i, \beta_i, \overline{\beta_i} \in \mathbb{C} and \beta_i \overline{\beta_i} = {|\beta_i|}^2 = 1.

Thus, (x-\beta_i)(x-\overline{\beta_i}) = x^2-2\Re(\beta_i)x+1.

Let us substitute (\alpha+\frac{1}{\alpha}) by b_0 and 2\Re(\beta_i) by b_i; then, we may rewrite the polynomial p(x) as

p_{N}(x) = \prod_{i=0}^N (x^2-b_ix+1)

Here are the first few cases;

n = 0, then p_0(x) = x^2-b_0x+1

n = 1, then p_1(x) = x^4 - (b_0+b_1) x^3 + (b_0b_1+2)x^2 - (b_0+b_1)x + 1

n = 2, then p_2(x) = x^6 - (b_0+b_1+b_2)x^5 + (b_0b_1+b_0b_2+b_1b_2+3)x^4-(2b_0+2b_1+2b_2+b_0b_1b_2)x^3 + (b_0b_1+b_0b_2+b_1b_2+3)x^2 - (b_0+b_1+b_2)x + 1

Obviously, for given n, x^k and x^{2n-k} share the same coefficient.

Since the polynomial is getting messier, we need a new notation; let \pi_k^n denote the sum of product of k-combination from set of \{b_i\}_{i=0}^{n-1}. For instance, \pi_2^3 = b_0b_1 + b_0b_2 + b_1b_2.

Now, in stead of writing down the whole family of polynomials into a table. Let us define a matrix X where each entry x_{k,j}, (the k-th row and j-th column) represents the coefficient of x^{j-1} from p_{k-1}(x). The entry is left blank if the coefficient is 0, just to avoid confusion.

Featured image

The coefficients seem pretty random though, there is a specific pattern lurking.

 Consider k \leq 2i. Then, x^k term in p_i(x) is only determined by x^{k-2}, x^{k-1} and x^k terms from p_{i-1}(x).

For example, consider x_{5,5}; it is 10 + 3\pi_2^5 + \pi_4^5. According to the idea presented above, this term is created by

(4+\pi_2^4)x^2 \cdot x^2 + (-3\pi_1^4 - \pi_3^4)x^3 \cdot (-b_4) x + (6 + 2\pi_2^4 + \pi_4^4)x^4 \cdot 1

Since the degree of x is all 4, let us not consider about the x^4 for now. Then,

4 + 6 = 10

\pi_2^4 + (-b_4)(-3\pi_1^4) + 2\pi_2^4 = 3\pi_2^4 + 3b_4\pi_1^3 = 3 \pi_2^5


(-b_4)(-\pi_3^4) + \pi_4^4 = \pi_4^5.

Thus, it nicely transforms to the term, as desired.

With this nice property, I induced a following equation.

Lee Equation:

 Featured image

As we move back to the matrix, let us assume that p_{n-1}(x) has Mahler measure less than \mu. Then, every term along the n-th row is composed of integers; obviously it leads to the conclusion that for all k \leq n, \pi_k^n are integers.

If Lehmer’s conjecture turns out to be true, then it implies for every n, there exists no sets of \{b_i\}_{i=0}^{n-1} that renders \pi_k^n \in \mathbb{Z} for every k.

Constructing a polynomial with arbitrary Mahler Measure

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