Complementary L1 and L2 Functions Continued

In the last entry, I indicated how each individual term, in the product over primes expression of the L function, can be represented as a complementary L function (where the base and dimensional aspects of number are inverted).

I thereby refer to the former collective function as the L1 function and the latter individual function (representing a single term) as the L2 function.

So the (collective) L1 function is comprised of an infinite series of (individual) L2 functions. And once again these operate in a dynamic complementary manner with respect to each other.

So again in general terms (s > 1), with respect to the Riemann zeta function, the L1 function is customarily represented (in its sum over the positive integers form) as

1^{– s} + 2^{– s} + 3^{– s} + 4^{– s}  + …,

with the corresponding L2 function represented as

1 + (1/p^s)¹  + (1/p^s)²  + (1/p^s)³  + (1/p^s)⁴  + …,

where p is defined over all the primes.  

However it is not only each term of the product over primes that can be represented as an L2 function, but likewise - with minor adjustment - each term of the sum over natural numbers expression.

The 1st term (in the sum over positive integers expression) as 1 stands alone, as it were, without a corresponding L2 expression.

However every other individual term can then be expressed in a general manner as an L2 function as follows.

[1/{1 – 1/(k^s + 1)}] – 1 where k = 1, 2, 3, 4, …

= [{1 + {1/(k^s + 1)}¹ + {1/(k^s + 1)}² + {1/(k^s + 1)}³ + {1/(k^s + 1)}⁴ + …] – 1 = 1/k^s.

For example for ζ(2), where p = 2, the 2^nd term is,

[{1 + {1/(2² + 1)}¹ + {1/(2² + 1)}² + {1/(2² + 1)}³ + {1/(2² + 1)}⁴ + …] – 1,

= [1 + 1/5 + 1/5²  + 1/5³ + 1/5⁴  + …] – 1,

= 5/4 – 1  = 1/2²  = 1/4.

So just as the 1^st term 1, (i.e. + 1) is omitted in terms of a L2 expression, every other term includes the addition of + 1 , in its denominator expression.   

In this context it is fascinating to observe the case where both L1 and L2 functions, with respect to the Riemann zeta function in the analytically continued complex plane, diverge.

For the L1, with s = 1 (representing the dimensional aspect of number as exponent) we have the harmonic series

1/1¹ + 1/2¹ + 1/3¹ + 1/4¹ + … = ∞.

Then with the L2 when s = 1 (representing the corresponding base aspect of number) we have,

1/(1 – 1) = 1 + 1¹ + 1² + 1³  + …  = ∞.

Thus from a dynamic interactive perspective, there is both an L1 and L2 explanation for the one pole in the Riemann zeta function i.e. s = 1 (where the functions are not defined).

So for the L1 function this pole occurs where s (representing the dimensional aspect of number) = 1.

For the L2 function, this pole occurs in complementary fashion where s (representing the corresponding base aspect of number) = 1.