L-functions et alia

One important L-function - closely related to the Riemann zeta function - is known as Dirichlet’s beta function (also Catalan’s beta function) with its L-series (i.e. L1) sum over the integers and product over the primes expressions as follows 1/1 s – 1/3 s + 1/5 s – 1/7 s + 1/9 s –   …    = 1/(1 + 1/3 s ) *   1/(1 – 1/5 s ) * 1/(1 + 1/7 s ) * … Then in the simplest case when s = 1, we have, 1 – 1/3 + 1/5 – 1/7 + 1/9 –   …   =   3/4 * 5/4 * 7/8 * …         = π/4. So the well-known Leibniz formula for π represents a special case of the Dirichlet beta function i.e. β(s), for s = 1. And when s = 1, 3, 5, …   β(s), results in a value of the form kπ s , where k is a rational number. So for example β(3) = 1 – 1/3 3 + 1/5 3 – 1/7 3 + 1/9 3 –   …   = π 3 /32 and β(5) = 1 – 1/3 5 + 1/5 5 – 1/7 5 + 1/9 5 –   …   = 5π 5 /1536. Now this function can be connected with the Riemann zeta function in a surprising manner. For using a similar type ge

In an earlier entry “ Intertwining L1and L2 Functions ” I showed that (1/1 2 + 1/2 2   + 1/3 2 + 1/4 2 + …) 2 /(1/1 4 + 1/2 4   + 1/3 4 + 1/4 4 + …)     =  (π 2 /6) 2 /(π 4 /90)   = 90/36 = 5/2. And   that this can be written as a Dirichlet L-series i.e. 1/1 2 + 2/2 2 + 2/3 2 + 2/4 2 + 2/5 2 + 4/6 2 + 2/7 2 + 2/8 2 + 2/9 2 + 4/10 2 + … = 5/2 This in turn makes use of our generalised Riemann zeta function formula, 1/1 s + (1 + k 1 )/2 s + (1 + k 2 )/3 s + (1 + k 1 )/4 s + (1 + k 3 )/5 s + {(1 + k 1 )(1 + k 2 )}/6 s + … = 1/{1 – (1 + k 1 )/(2 s + k 1 )} * 1/{1 – (1 + k 2 )/(3 s + k 2 )} * 1/{1 – (1 + k 3 )/(5 s + k 3 )} * …,   where k 1 , k 2 , k 3 , …    = 1. So ζ(2) 2 / ζ(4) = 1/1 2 + 2/2 2 + 2/3 2 + 2/4 2 + 2/5 2 + 4/6 2 + 2/7 2 + 2/8 2 + … = 5/2 And the result can be generalised so that when s is a positive even integer, (1/1 s + 1/2 s   + 1/3 s + 1/4 s + …) 2 /(1/1 2s + 1/2 2s   + 1/3 2s + 1/4 2s + …)   =

Imagine we are presented with the first of the Riemann zeros i.e. 1/2 + 14.134725 …, (or equally 1/2 –   14.134725 …), without knowing the origin of the value and asked to find a solution to the following infinite equation, k 1 – (1 /2 + 14.134725 …) + k 2 – (1 /2 + 14.134725 …) + k 3 – (1 /2 + 14.134725 …) + k 4 – (1 /2 + 14.134725 …) + …    = 0, where k 1, k 2, k 3, k 4 , … are complex numbers, then by lucky chance we might be able to find that when k 1, k 2, k 3, k 4 , … represent the natural numbers 1, 2, 3, 4 , … respectively, that the value of the equation = 0. And imagine in turn if we were presented in turn with all of the known Riemann zeros and asked to solve the same equation for each one individually, we might then find that in every case when k 1, k 2, k 3, k 4 , …, represent the natural numbers 1, 2, 3, 4 , … respectively, that again the value of the equation = 0. So this would indeed be remarkable! For on the one hand we would have a list of known