Another Interesting Generalisation (2)

In the last entry, I mentioned the new general formula, which can be used to extend relationships as between sum over integers and product over primes expressions with respect to the Riemann zeta function i.e.

1/1^s + (1 + k₁)/2^s + (1 + k₂)/3^s + (1 + k₁)/4^s + (1 + k₃)/5^s + {(1 + k₁)(1 + k₂)}/6^s + …
= 1/{1 – (1 + k₁)/(2^s + k₁)} * 1/{1 – (1 + k₂)/(3^s + k₂)} * 1/{1 – (1 + k₃)/(5^s + k₃)} * …,
 
where k₁, k₂, k₃, … are rational numbers which can be either positive or negative.

In that entry, I demonstrated the - surely - interesting fact that when s is 2, 4, 6, …

and k₁ = k₂ = k_3, …  = 1, that the value of the two expressions (sum over the integers and product over the primes respectively) is a rational number.

In this case there is a clear relationship as between the values of the standard zeta function for s and 2s respectively.

So - again when s is a positive even integer - the value of the zeta expression (for both sum over the integers and product over the primes expression) is t₁π^s (where t₁ is a rational number).

Likewise the value of the expression when the dimension is 2s is t₂π^2s (where t₂ is a rational number).

Let the ratio t₂/t₁  = t_3.

Then the value of the extended zeta function where k₁ = k₂ = k_3, …  = 1, is t₃/t₁ i.e. t₂/(t₁)².

So when s = 2, ζ(2) = π²/6 and when s = 4, ζ(4) = π⁴/90.

So t₁ = 6 and t₂  = 90.

Therefore t₂/(t₁)²  = 90/36 = 5/2

And this the value of the extended expression where k₁ = k₂ = k_3, …  = 1.


Another interesting - if trivial - case arises when k₁ = k₂ = k_3, …  = – 1.

This leads to the elimination on both sides of all terms entailing k so that we are left with the identity 1 = 1.

This is useful to remind us that 1 precedes all subsequent relationships entailing prime factors. So 1 is not obtained from these relationships but rather serves as a necessary precondition for their use. 

Thus internally the very notion of a prime entails unit components which then have both a quantitative identity as (independent of each other) and a qualitative identity as (interdependent with other) respectively.

So far in the use of the extended general formula we have shown how to generate expressions where the numerator  > 1.

However by picking the value of k₁, k₂, k_3, …, appropriately we can likewise generate expressions where the denominator - rather than the numerator term -  increases.

For example when k₁ = – 1/2, k₂ = – 2/3, k₃ = – 4/5, …, we have 

1/1^s + (1 – 1/2)/2^s + (1 – 2/3)/3^s + (1 – 1/2)/4^s + (1 – 4/5)/5^s + {(1 – 1/2)(1 – 2/3)}/6^s + …

= 1/{1 – (1 – 1/2)/(2^s – 1/2)} * 1/{1 – (1 – 2/3)/(3^s – 2/3)} * 1/{1 – (1 – 4/5)/(5^s – 4/5)} * …,

i.e.

1/1^s + 1/(2.2^s) + 1/(3.3^s) + 1/(2.4^s) + 1/(5.5^s) + 1/(6.6^s) + …

= 1/1 – {(1/2)/(2^s – 1/2)} * 1/1 – {(1/3)/(3^s – 2/3)} * 1/1 – {(1/5)/(5^s – 4/5)} * …,

So for example, when s = 2

1/1² + 1/(2.2²) + 1/(3.3²) + 1/(2.4²) + 1/(5.5²) + 1/(6.6²) + …

= 1/{1 – [(1/2)/(2² – 1/2)]} * 1/{1 – [(1/3)/(3² – 2/3)]} * 1/{1 – [(1/5)/(5² – 4/5]} * …,

= 1/(1 – 1/7) * 1/(1 – 1/25) * 1/(1 – 1/121) * …= 7/6 * 25/24 * 121/120 * …

And likewise k₁, k₂, k_3, …, can be chosen so that both numerator and denominator are multiples of values that in the standard case.