The Inverse Naturals and The Euler Product

 

Mathematical Conundrum Series

This post is part of a series of posts that should be read in order:

Mathematical Conundrum Series

• Introduction
• Foundations - The Counting Numbers
• Foundations - The Fundamental Theorem of Arithmetic
• The Fundamental Theorem of Arithmetic - Part 2
• The Inverse Naturals and The Euler Product
• The Conundrum
• The Conundrum - Free Will and Choice

In this post I am going to look at the inverse natural numbers, and the easiest way to do that is to look at Leonhard Euler's 'Theorem 7', also known as Euler’s Product Formula, from his 1744 published work 'Variae observationes circa series infinitas' Several Remarks on Infinite Series.

Reference: E072en.pdf

Theorem 7 of Euler's work concludes that the harmonic series is equal to the infinite product "where the numerators are all the prime numbers and the denominators are the numerators less one unit". 

$${\displaystyle 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\cdots =\frac{2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19\cdot 23\cdots }{1\cdot 2\cdot 4\cdot 6\cdot 10\cdot 12\cdot 16\cdot 18\cdot 22\cdots }}$$

Euler's infinite product is also, like in the natural numbers, an infinite product of prime geometric power series. How do I know that?

$${\displaystyle \frac{2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19\cdot 23\cdots }{1\cdot 2\cdot 4\cdot 6\cdot 10\cdot 12\cdot 16\cdot 18\cdot 22\cdots }}$$

Each term of the Euler product can be written as a prime geometric power series.

The first term: $$\begin{array}{rl}\frac{2}{1}& =1+1\\ & =1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\dots \end{array}$$
The second term: $$\begin{array}{rl}\frac{3}{2}& =1+\frac{1}{2}\\ & =1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\dots \end{array}$$
The entire Euler product can be written as a product of prime power series. $$\left(\begin{array}{c}\\ 1\\ +\\ 2^{-1}\\ +\\ 2^{-2}\\ +\\ 2^{-3}\\ +\\ 2^{-4}\\ +\\ 2^{-5}\\ ⋮\\ +\\ 2^{-n}\\ \end{array}\right)\times \left(\begin{array}{c}\\ 1\\ +\\ 3^{-1}\\ +\\ 3^{-2}\\ +\\ 3^{-3}\\ +\\ 3^{-4}\\ +\\ 3^{-5}\\ ⋮\\ +\\ 3^{-n}\\ \end{array}\right)\times \left(\begin{array}{c}\\ 1\\ +\\ 5^{-1}\\ +\\ 5^{-2}\\ +\\ 5^{-3}\\ +\\ 5^{-4}\\ +\\ 5^{-5}\\ ⋮\\ +\\ 5^{-n}\\ \end{array}\right)\times \left(\begin{array}{c}\\ 1\\ +\\ 7^{-1}\\ +\\ 7^{-2}\\ +\\ 7^{-3}\\ +\\ 7^{-4}\\ +\\ 7^{-5}\\ ⋮\\ +\\ 7^{-n}\\ \end{array}\right)\times \left(\begin{array}{c}\\ 1\\ +\\ {11}^{-1}\\ +\\ {11}^{-2}\\ +\\ {11}^{-3}\\ +\\ {11}^{-4}\\ +\\ {11}^{-5}\\ ⋮\\ +\\ {11}^{-n}\\ \end{array}\right)\times \left(\begin{array}{c}\\ 1\\ +\\ {13}^{-1}\\ +\\ {13}^{-2}\\ +\\ {13}^{-3}\\ +\\ {13}^{-4}\\ +\\ {13}^{-5}\\ ⋮\\ +\\ {13}^{-n}\\ \end{array}\right)\times \left(\begin{array}{c}\\ 1\\ +\\ {17}^{-1}\\ +\\ {17}^{-2}\\ +\\ {17}^{-3}\\ +\\ {17}^{-4}\\ +\\ {17}^{-5}\\ ⋮\\ +\\ {17}^{-n}\\ \end{array}\right)\times \left(\begin{array}{c}\\ 1\\ +\\ {19}^{-1}\\ +\\ {19}^{-2}\\ +\\ {19}^{-3}\\ +\\ {19}^{-4}\\ +\\ {19}^{-5}\\ ⋮\\ +\\ {19}^{-n}\\ \end{array}\right)\times \cdots$$

For a modern reference, you can checkout Wofram's Mathworld reference on the Euler product.

Can we model the factorization of the inverse naturals using this infinite product of inverse prime geometric power series?

Yes.

The infinite product of prime geometric power series in the naturals and the infinite product of inverse prime geometric power series are similar. Factorization in the naturals is in the numerator, and in the inverse naturals we have the exact same factorization but in the denominator. Outside of that, the models are identical.

The visual model of tracing distinct/unique multiplicative paths through the infinite product of inverse prime geometric power series.

The infinite multiplicative sequences model:

$$\begin{array}{cl}\frac{1}{n}& \text{sequence}\\ 1& 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...\\ \frac{1}{2}& 2^{-1},1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...\\ \frac{1}{3}& 1,3^{-1},1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...\\ \frac{1}{4}& 2^{-2},1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...\\ \frac{1}{5}& 1,1,5^{-1},1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...\\ \frac{1}{6}& 2^{-1},3^{-1},1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...\\ \frac{1}{7}& 1,1,1,7^{-1},1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...\\ \frac{1}{8}& 2^{-3},1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...\\ \frac{1}{9}& 1,3^{-2},1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...\\ \frac{1}{10}& 2^{-1},1,5^{-1},1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...\\ \frac{1}{11}& 1,1,1,1,{11}^{-1},1,1,1,1,1,1,1,1,1,1,1,1,1,...\\ \frac{1}{12}& 2^{-2},3^{-1},1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...\\ \frac{1}{13}& 1,1,1,1,1,{13}^{-1},1,1,1,1,1,1,1,1,1,1,1,1,...\end{array}$$

So, as can be seen here, both the infinite product of prime geometric power series in the naturals and the infinite product of inverse prime geometric power series can model the factorization of their respective sets. 

Note: I keep saying factorization in these posts regarding the infinite prime geometric power series products in the naturals and inverse naturals, but here I am not really working with factorization. Instead of taking the numbers in the respective sets and determining their factors, here I am starting with the actual creators, the mathematical objects that create/generate the members of the respective sets, the naturals and the inverse naturals. I am literally performing a reverse process. I am demonstrating that the prime geometric power series creates the infinite canonical representation of the factors of each member of each set.

Next up, 'The Conundrum'.