Interesting Relationships Continued (2)

In the last entry, I considered the (infinite) reciprocal sum of the unique numbers associated with the polynomial equation xⁿ – kⁿ = 0, (where n = 2 and k prime) to find that the respective sums generate the successive terms of the product over primes expression for the (Riemann) Zeta 1 function i.e.
ζ₁(2), where s = 2, i.e.

4/3, 9/8, 25/24, …

And then when n = 3, the successive terms of ζ₁(3), where s = 3, are generated i.e.

8/7, 27/26, 125/124, …

However, when we now consider the respective reciprocal sums for xⁿ + kⁿ = 0 (where n = 2 and k prime) we generate the following terms,

4/5, 9/10, 25/26,…

And 4/3 * 9/8 * 25/24 * …   = π²/6, and

4/5 * 9/10 * 25/26 * …  = π²/15

Of course this means that multiplying the respective terms in each series,

16/15 * 81/80 * 625/624 * …  = π⁴/90 = ζ₁(4)

And in general, where k ranges over all the primes

Then (2ⁿ/2ⁿ – 1) * (3ⁿ/3ⁿ – 1) * (5ⁿ/5ⁿ – 1) * … = ζ₁(n),  when multiplied by

(2ⁿ/2ⁿ + 1) * (3ⁿ/3ⁿ + 1) * (5ⁿ/5ⁿ + 1) * … = ζ₁(2n).

However there is an easier of showing the true nature of the simple equation (x – k)ⁿ = 0 = 0

Let us initially - in relation to the this equation - consider the simple case where k = 1, 2, 3, … and n = 1

As we have already seen unique numbers associated with (x – 1)¹ = 0 are,

1, 1, 1, …

And of course the sum of reciprocals of these numbers diverges to infinity.

Then when (x – 2)¹ = 0, the unique numbers thereby associated are,

1, 2, 4, …

And the (infinite) sum of reciprocals of these numbers

= 1 + 1/2 + 1/4 + …    = 2. (i.e. 2/1)

And when (x – 3)¹ = 0,  the unique numbers associated are

1, 3, 9, … and the corresponding (infinite) sum of reciprocals =

1 + 1/3 + 1/9 + … = 3/2.

Thus in general terms, the (infinite) sum of reciprocals of the unique numbers associated with (x – k)¹ = 0, = k/k – 1).

Let us now again in relation to the alternative equation (x – k)ⁿ = 0, consider the reverse complementary situation where k = 1 and n = 1, 2, 3, …

Thus again we start with (x – 1)¹ = 0.

And as we have seen the infinite reciprocal sum of unique numbers associated with this equation = 1 + 1 + 1 + … (which diverges to infinity).

Then when k = 1 and n = 2 we obtain (x – 1)² = 0.

And the unique number sequence - as seen in previous entries - of this equation is the set of natural numbers 1, 2, 3, …

And the (infinite) sum of reciprocals of these numbers (i.e. the harmonic series = 1 + 1/2 + 1/3 + … (which diverges to infinity).

Then when k = 1 and n = 3 we obtain (x – 1)³ = 0.

And again as we have seen - in previous entries - the unique numbers associated with this equation are 1, 3, 6, …

And the (infinite) sum of reciprocals of these numbers = 1 + 1/3 + 1/6 + … = 2 (i.e. 2/1).

And taking one more example, when k = 1 and n = 4, we obtain (x – 1)⁴ = 0 with the unique numbers associated 1, 4, 10, …

And the (infinite) sum of reciprocals of these numbers = 1 + 1/4 + 1/10 + … = 3/2.

So in general, the (infinite) sum of reciprocals = (n – 1)/(n – 2).

Now I have referred before to the first formulation i.e. the (infinite) sum of reciprocals of unique number sequences associated with (x – k)¹ = 0 i.e. k/(k – 1) as the Zeta 2 function i.e. ζ₂(1/k).

Thus when k = 2, 1/k = 1/2;

So ζ₂(1/k) = 1 + (1/k)¹ + (1/k)² + (1/k)³ + …  = 1 + (1/2)¹ + (1/2)² + (1/2)³ + …  = 2.                

And the alternative formulation, i.e. the (infinite) sum of reciprocals of unique numbers associated with (x – 1)ⁿ is expressed as Alt ζ₂(n).

However as both k and (x – 1)ⁿ range over the same natural numbers 1, 2, 3, …

then (x – 1)ⁿ = (x – 1)^k and Alt ζ₂(n) = Alt ζ₂(k)

So ζ₂(1/k) = Alt ζ₂(k + 1).

Thus, just as there are two formulations of the Zeta 1 (Riemann) function as the sum over natural numbers and product over primes expressions, equally there are two formulations of the Zeta 2 function.

And the key to understanding the two Zeta 2 formulations is that they represent complementary notions of number (both in terms of base and dimensional formulations) which are analytic (quantitative) and holistic (qualitative) with respect to each other. In other words, number keeps switching as between both its particle (independent) and wave (interdependent) expressions.

And these two expressions can only be properly understood in a dynamic interactive manner.

So it is equally true that with respect to the macro formulation of the number system, that the sum over natural numbers and product over primes formulations represent complementary expressions that are analytic (quantitative) and holistic (qualitative) with respect to each other.

So for example in macro terms when s = 2, in the sum over natural numbers expression,

ζ₁(2) = 1 + 1/4 + 1/9 + …  = π²/6

However in the product over primes expression,

ζ₁(2) = 4/3 * 8/9 * 24/25 + …  = π²/6.  

And properly understood i.e. in dynamic interactive terms, these two expressions are analytic (quantitative) and holistic (qualitative) with respect to each other.

However equally in micro terms when k = 1/2²,

ζ₂(1/2²) = 1 + 1/2² + 1/2⁴ + … = 4/3

So we see how the 1^st term of the product over primes expression of the Zeta 1 represents a corresponding Zeta 2 expression.

And equally this is true of every other term of the Zeta 1, which can be expressed in Zeta 2 terms.  

And also each individual term of the Zeta 1 can be expressed in Alt Zeta 2 terms

So when k = 2² + 1, Alt ζ₂(5) = 1 + 1/5 + 1/15 + 1/35 + … = 4/3.

And equally, this is true of every other term of the Zeta 1, which can be expressed in Alt Zeta 2 terms.

This is also true - though slightly more convoluted - in relation to the sum over natural numbers expression of the Zeta 1.

For example the 2^nd term here = 1/4 = 5/4 – 1

So ζ₂{(1/(2² + 1)} = 1 + 1/5 + 1/5² + … = 5/4

Thus 1/4 as 2^nd term of ζ₁(2), = ζ₂{(1/(2² + 1)} – 1

And similar Zeta 2 type expressions are available for ever other term of Zeta 1 and by extension ζ₁(s) where s > 2.

Finally each term of Zeta 1 (sum over natural numbers expression) can be expressed in appropriate Alt Zeta 2 terms.

So Alt ζ₂ (2² + 2) = Alt ζ₂ (6)  = 1 + 1/6 + 1/21 + 1/56 + …  = 5/4.

Thus 1/4 = Alt ζ₂ (2² + 2) – 1.

And every other individual term of ζ₁(2) in a sum over natural numbers expression, and by extension ζ₁(s) where s > 2, can be given an Alt Zeta 2 formulation.

So a full understanding of Riemann’s Zeta function requires not only that the Zeta 1 function can be expressed in two different ways (which in dynamic terms are complementary), but that each individual term of the Zeta 1 (for both expressions) can equally be expressed in two different ways i.e. through the Zeta 2 and Alt Zeta 2 functions respectively.