Zeta 2 Formulation of Euler Product Formula

In yesterday's blog entry I commented on the fact that the Zeta 1 and Zeta 2 (non-trivial) zeros represent the holistic counterparts to the Type 1 and Type 2 aspects of the number system respectively.

Putting it more precisely, the Zeta 1 zeros represent the (qualitative)  holistic counterpart to the cardinal aspect of the number system, whereas the Zeta 2 zeros represent the corresponding (quantitative) holistic counterpart to the ordinal aspect of the number system.

It must be clearly remembered in this context that both the cardinal and ordinal aspects of the number system enjoy a relative identity which are - in direct terms - quantitative and qualitative with respect to each other.

We saw how the ordinal identity implies qualitative interdependence in the ordered relationship of numbers with each other.

The Zeta 2 zeros then consisted of giving a new quantitative numerical identity (in circular terms) to the corresponding qualitative notions of 1st, 2nd, 3rd etc. And these new numbers (with respect to the n roots of 1) have then a holistic identity with respect to their combined sum = o.

Now again, one of the great blind spots of Conventional Mathematics resides in the failure to recognise that ordinal aspect of numbers clearly implies a qualitative - rather than quantitative - identity. This failure leads to the consistent reduction of qualitative to quantitative meaning with respect to the interpretation of number! 


By contrast, the cardinal identity of number implies quantitative independence in the individual nature of each number. So the prime numbers 2, 3, 5 etc. are independent in this locally confined sense. All the natural numbers (except 1) therefore depend for their identity on the prime numbers (as the independent building blocks of the number system). 

The Zeta 1 zeros thereby consist of giving a new qualitative numerical identity (in linear imaginary terms) to the corresponding shared identity of the primes with the natural number system.

It is important in this context to recognise that the imaginary number line in fact is the appropriate way of expressing meaning, which by nature is inherently qualitative and holistic, in an independent local manner. So in this sense each of the Zeta 1 zeros has an individual local identity on the imaginary number line! 

Therefore, if we are then to properly decode the collective nature of the Zeta 1 zeros , we must interpret them in a (holistic) qualitative - rather than (analytic) quantitative - manner.

So the holistic nature of the zeros resides in the combined manner in which they can be used to ultimately demonstrate the perfect (interdependent) order of the primes with respect to the natural number system. And this relationship, which is intimately connected with the distribution of non-trivial zeros, is strictly of a qualitative - rather than quantitative - nature!


I mentioned in my last blog that the Zeta 1 and Zeta 2 zeros are ultimately totally complementary with each other. Put another way, the full appreciation of the cardinal aspect of the number system implies corresponding full appreciation of the ordinal;  from the other direction, full appreciation of the ordinal aspect likewise implies full appreciation of the cardinal.

So the Zeta 1 ultimately imply the Zeta 2; and the Zeta 2 in turn ultimately imply the Zeta 1 zeros. And in turn the Zeta 1 Function therefore implies the Zeta 2 and the Zeta 2 the Zeta 1 Function respectively.

With this in mind, I sought recently to demonstrate that the Euler Product Formula (that beautifully relates the additive properties of the natural numbers to the multiplicative properties of the primes) can be equally fully expressed in terms of the Zeta 2 Function.

At present it is exclusively associated with the Zeta 1 Function!

So in the traditional manner of expression for example Euler’s Product Formula for the Zeta 1 Function is given as,

    _∞

   ∑ 1/n^s   = ∏ 1/(1 – p^–s)  = ζ₁(s)

  ^{n = 1                p}

Thus for example when s = 2 in the Zeta 1 expression,

ζ₁(2) = 1 + 1/4 + 1/9  + 1/16 +…..  = 4/3 * 9/8 * 25/24 * 49/48 * ….    =  π²/6

However this equally can be expressed in terms of the Zeta 2 Function! 

Once again,

ζ₂(s) = 1 + s¹  + s² + s³ + s⁴……   = 1/(1 – s).

Thus when s = 1/n,

ζ₂(1/n) = 1 + (1/n)¹  + (1/n)² + (1/n)³ + (1/n)⁴……

= 1/(1 - 1/n) = n/(n – 1)

Therefore when n = 2,

ζ₂(1/2) = 1 + 1/2 + 1/4 + 1/8 + ….  = 2 /(2 – 1) = 2

Thus,

ζ₂(1/2) – 1 = 1

Then when n = 3

ζ₂(1/3) = 1 + 1/3 + 1/9 + 1/27 + ….  = 3/(3 – 1) = 3/2 

Thus,

ζ₂(1/3) – 1 = 1/2

Then when n = 4

ζ₂(1/4) = 1 + 1/4 + 1/16 + 1/64 +…… =  4/(4 – 1) = 4/3

Thus,

ζ₂(1/4) – 1 = 1/3

Thus summing over the natural numbers i.e. n = 2, 3, 4,….

 _∞

∑{ζ₂(1/n) – 1} =  1 + 1/2 + 1/3 + ………

^{n = 2}  

Therefore,

_∞

∑{ζ₂(1/n) – 1}^s =  1^s + ( 1/2)^s + ( 1/3)^s + ………

^{n = 2}

So when s = 2,

  _∞
  ∑{ζ₂(1/n) – 1}^{2
n = 2}

=  1² + ( 1/2)² + ( 1/3)² + ……… = π²/6

Likewise when p = 2

ζ₂(1/p)  = ζ₂(1/2)

Then when s = 2

ζ₂(1/2)²   =  ζ₂(1/4)   

ζ₂(1/4)  = 1 + 1/4 + 1/16 + 1/64 + ...  1 = 1/(1 –1/4) = 4/3  

Then when p = 3

ζ₂(1/p)  = ζ₂(1/3)

So again with s = 2,

ζ₂(1/3)²   =  ζ₂(1/9)    

ζ₂(1/9)  = 1 + 1/9 + 1/81 + 1/729 + ... = 1/(1 – 1/9) = 9/8

Finally to illustrate with p = 5 and s = 2  

ζ₂(1/5)²   =  ζ₂(1/25)  = 1 + 1/25 + 1/625 + …. = 1/(1 – 1/25)  = 25/24  

Therefore multiplying over all the primes when s = 2,

  _∞

 ∏{ζ₂(1/p)²} = 4/3 * 9/8 * 25/24 * …… = π²/6

  ^p

So we have now demonstrated the famous Euler Product Formula solely in terms of the Zeta 2 expression i.e.,

                                                       

  ∑{ζ₂(1/n) – 1}^s   =    ∏{ζ₂(1/p)^s}
  ^{n = 2                                          p}
                                   

Therefore in terms of both Zeta 1 and Zeta 2 Functions, the Euler Product Formula can be expressed:

                                                                              

∑ 1/n^s   = ∏ 1/(1 – p^–s)  = ζ₁(s)  =   ∑{ζ₂(1/n) – 1}^s =   ∏{ζ₂(1/p)^s}
^{n = 1                p                                         n = 2                                          p}