The square of the Zeta Function

[galoisjr]

It has been determined that the square of zeta can be written in terms of the divisor function.

$${\zeta ^2}(s) = \sum\limits_{n = 1}^\infty {\frac{{d(n)}}{{{n^s}}}}$$

Being a first semester student in complex variables, I have only recently started looking at zeta. But I have deduced an alternate and equivalent representation of the square of zeta. I must say that I am ignorant to the value of its importance, but I must say that in my opinion it is a beautiful series, and a very interesting representation.

From the Fundamental Theorem of Arithmetic, we know that each natural number n can be written as an unique product of primes. A consequence of this is that we can view the primes to be constant throughout the natural numbers, then we can write each n as an infinite product of primes in which the sequences of powers are unique to its respective value of n. In other words for each n,

$$n = p_0^{{\alpha _0}} \cdot \cdot \cdot p_n^{{\alpha _n}}$$

Although I know that it is unconventional to assume that 1 is a prime number, but I do also know that there have been a handful great mathematicians in history who have done so. For the sake of this argument, we can further define p_0=1 and alpha_0=1. So for example, since all primes are assumed in this constant sense, 6={1,1,1,0,...}(where alpha_0=1, alpha_1=1, alpha_2=1, and for n>2 the sequence vanishes).

Using this in the square of zeta we can instead sum over the prime factorizations of numbers and we obtain

$${\zeta ^2}(s) = \sum\limits_{n = 1}^\infty {\frac{{d(n)}}{{{n^s}}}} = 1 + \sum\limits_{n = 1}^\infty {\frac{{{\alpha _0} + {\alpha _n}}}{{p_n^{{\alpha _n}s}}} + \sum\limits_{n = 0}^\infty {\frac{{1 + {\alpha _0} + ... + {\alpha _n}}}{{{{\left( {p_0^{{\alpha _0}} \cdot \cdot \cdot p_n^{{\alpha _n}}} \right)}^s}}}} }$$

In which the first term has denominators that are powers of a single prime and the second term has more than 1 prime in its prime factorization (excluding 1). We can see that for the first term with denominator p^k will have numerator k+1, and for the second term each denominator will have a unique prime factorization, and thus, the numerator will be the sum of the powers + 1.

 

[galoisjr]

I would conjecture that for the nth power of zeta, as n grows each type of prime factorization will have distinct formulas for their respective numerators. But in each nth power the numerator over each prime would be n. And also for each term containing a denominator with multiple prime factorization types, such as p_1^k*p_2*p_3, will have a numerator that is the product of the respective numerators for each type, the previous example would have numerator=(numerator of p^k denominator)*(numerator of the term containing exactly 2 distinct primes in its denominator).

 

[galoisjr]

So I have actually found out something a little more interesting and that is pretty much that my previous conjecture was true but with an added twist.

Let's look at

$${\zeta ^n}(s) = {\left( {\sum\limits_{n = 1}^\infty {\frac{1}{{{n^s}}}} } \right)^n}$$

After expanding you can see that each prime denominator always has the power in its numerator. What is more interesting is that for each term which contains a power of a single prime in its denominator, if this power is a proper divisor of n then the numerator is not a multiple of the numerators of smaller powers as it was in the square of zeta. In other words, each prime power dividing n has a distinctly weighted numerator, and there is no obvious closed form for these terms unlike the zeta squared case. But the terms with composite denominators act the same as before except there is a unique product for each. Here is an example:

$$\begin{array}{l} {\zeta ^{12}}(s) = 1 + \frac{{12}}{{{2^s}}} + \frac{{12}}{{{3^s}}} + \frac{{78}}{{{4^s}}} + \frac{{12}}{{{5^s}}} + \frac{{144}}{{{6^s}}} + \frac{{12}}{{{7^s}}} + \frac{{366}}{{{8^s}}} + \frac{{78}}{{{9^s}}} + \frac{{144}}{{{{10}^s}}} + \frac{{12}}{{{{11}^s}}} + \frac{{936}}{{{{12}^s}}}\\ 12 \cdot 12 = 144\\ 12 \cdot 78 = 936\\ 12 \cdot 12 \ne 78\\ 12 \cdot 78 \ne 366 \end{array}$$

I only did it up to 1/12^s because it is all that could fit on my page and computing this by hand is pretty tedious. But this obviously assigns a weight to not only the primes but EACH prime factorization type.