LOG#104. Primorial objects.

Posted on by September 9, 2019

clock-primorial

My post today will be discussing two ideas: the primorial and the paper “The product over all primes is 4\pi^2“ (2003).

The primorial is certain generalization of the factorial, but running on prime numbers. While the factorial is defined as

(1)   \begin{equation*} n!=n\cdot (n-1)\cdots 3\cdot \cdot 2\cdot 1\end{equation*}

the primorial is defined as follows

\boxed{p_n\#=\prod_{k=1}^np_k}

The first primorial numbers for n=1,2,\ldots are

2,6,30,210,2310,30030,510510,\ldots

We can also extend the notion of primorial to integer numbers

(2)   \begin{equation*} \boxed{n\#=\prod_{k=1}^{\pi (n)}p_k}\end{equation*}

where \pi (n) is the prime counting function. The first primorial numbers for integer numbers are

1,2,6,6,30,30,210,210,210,210,2310,\ldots

In fact, if you take the limit

\displaystyle{\lim_{n\rightarrow \infty}(p_n)^{1/p_n}=e}

since the Chebyshev function provides

\displaystyle{\lim_{x\rightarrow \infty}\dfrac{x}{\theta (x)}=1}

By the other hand, the factorial of infinity can be regularized

(3)   \begin{equation*} \infty ! =1\cdot 2\cdot 3\cdots =\sqrt{2\pi}\end{equation*}

The paper mentioned above provides a set of cool formulae related to infinite products of prime numbers powered to some number! The main result of the paper is that

(4)   \begin{equation*} \boxed{\prod_p p=4\pi^2}\end{equation*}

If you have an increasing sequence of numbers 0<\lambda_1\leq \lambda_2\leq \ldots, then we can define the regularized products thanks to the Riemann zeta function (this technique is called zeta function regularization procedure):

(5)   \begin{equation*} \boxed{\prod_{n=1}^\infty \lambda_n =e^{-\zeta'_\lambda (0)}}\end{equation*}

(6)   \begin{equation*} \boxed{\zeta_\lambda (s)=\sum_{n=1}^\infty \lambda_n^{-s}}\end{equation*}

and where the \lambda_n is some sequence of positive numbers. The Artin-Hasse exponential can be defined in the following way:

\boxed{\exp (X)=\prod_{n=1}^\infty (1-X^n)^{-\mu (n)/n}}

and there \mu (n) is the Möbius function. From this exponential, we can easily get that

(7)   \begin{equation*} \boxed{e^{p^{-s}}=\prod_{n=1}^\infty (1-p^{-ns})^{-\mu (n)/n}}\end{equation*}

Using the prime zeta function

(8)   \begin{equation*} \displaystyle{\mathcal{P}(s)= \sum_{p}\dfrac{1}{p^s}}\end{equation*}

we obtain

(9)   \begin{equation*} \displaystyle{e^{\mathcal{P}(s)}=\prod_p e^{p^{-s}}=\prod_p \prod_{n=1}^\infty (1-p^{-ns})^{-\mu (n)/n}}\end{equation*}

(10)   \begin{equation*} \displaystyle{e^{\mathcal{P}(s)}=\prod_{n=1}^\infty \prod_p (1-p^{-ns})^{-\mu (n)/n}=\prod_{n=1}^\infty \zeta (ns)^{\mu (n)/n}}\end{equation*}

Therefore

(11)   \begin{equation*} \boxed{e^{\mathcal{P}(s)}=\prod_{n=1}^\infty \zeta (ns)^{\mu (n)/n}}\end{equation*}

Now, if we remember that

(12)   \begin{equation*} \displaystyle{\sum_{n=1}^ \infty\dfrac{\mu (n)}{n^{s}}=\dfrac{1}{\zeta (s)}}\end{equation*}

and that

(13)   \begin{equation*} \displaystyle{\mathcal{P}'(s)=\sum_{n=1}^\infty \dfrac{\mu (n)}{n}\dfrac{n\zeta' (ns)}{\zeta (ns)}=\sum_{n=1}^\infty \mu (n)\dfrac{\zeta' (ns)}{\zeta (ns)}}\end{equation*}

(14)   \begin{equation*} \displaystyle{\mathcal{P}'(0)=\left(\sum_{n=1}^\infty \mu (n)\right) \dfrac{\zeta' (0)}{\zeta (0)}=\dfrac{\zeta' (0)}{\zeta (0)^2}=-2\log (2\pi)}\end{equation*}

and from this we get

(15)   \begin{equation*} \displaystyle{\prod_p p=e^{\mathcal{P}'(0)}=(2\pi)^2=4\pi^2}\end{equation*}

Q.E.D.

And similarly, it can be proved the beautiful formula

(16)   \begin{equation*} \boxed{\displaystyle{\prod_p p^s=(2\pi)^{2s}}}\end{equation*}

Moreover, using the Riemann zeta function

(17)   \begin{equation*} \displaystyle{\zeta (s)=\prod_p (1-p^{-s})^{-1}=\dfrac{\displaystyle{\prod_p p^s}}{\displaystyle{\prod_p (p^s-1)}}}\end{equation*}

we also have

(18)   \begin{equation*} \displaystyle{\boxed{\prod_p(p^s-1)=\dfrac{(2\pi)^{2s}}{\zeta (s)}}}\end{equation*}

In particular, e.g., we get

(19)   \begin{equation*} \displaystyle{\prod_p (p-1)=0}\end{equation*}

and

(20)   \begin{equation*} \displaystyle{\prod_p (p^2-1)=48\pi^2}\end{equation*}

The final part of the paper is a proof using a “more convergent” series of the previous “prime products”/products of prime numbers. It uses the series

(21)   \begin{equation*} \displaystyle{e^{\mathcal{P}(s,t)}=\prod_{n=1}^\infty \zeta (ns)^{\mu (n)/n^t}}\end{equation*}

and it converges \forall Re (s)>1, Re (t)>1. But then, the series

(22)   \begin{equation*} \displaystyle{\dfrac{\partial \mathcal{P}}{\partial s}(s,t)=\sum_{n=1}^\infty \dfrac{\mu (n)}{n^{t-1}}\dfrac{\zeta' (ns)}{\zeta (ns)}}\end{equation*}

converges \forall Re (t)>1, Re (s)> 0. Then, if t\in \mathbb{C}, and Re (t)>2, with s\longrightarrow 0 we have a limit

(23)   \begin{equation*} \displaystyle{L=\lim_{Re s>0}\lim _{s\rightarrow 0}\dfrac{\partial \mathcal{P}}{\partial s}(s,t)=\left(\sum_{n=1}^\infty \dfrac{\mu (n)}{n^{t-1}}\right)\dfrac{\zeta' (0)}{\zeta (0)}}\end{equation*}

and thus

(24)   \begin{equation*} L=\dfrac{\zeta ' (0)}{\zeta (t-1) \zeta (0)}\end{equation*}

Therefore, the meromorphic extension of this formula to the whole complex plane provides that

(25)   \begin{equation*} \displaystyle{\lim_{t\rightarrow 1} L=\dfrac{\zeta '(0)}{\zeta (0)^2}}\end{equation*}

as we wanted to prove (Q.E.D.).

Let the prime numbers and the primorial be with you!

transformers-optimus-prime

Optimus_Prime

View ratings
Rate this article
Liked it? Take a second to support amarashiki on Patreon!

Leave a Reply

Your email address will not be published.

This site uses Akismet to reduce spam. Learn how your comment data is processed.