Testing for L-Functions

In investigating the simpler Dirichlet L-functions (where numerator of each term is 1) an interesting pattern appears to universally hold.

                            _∞

Thus if L(s, χ) = ∑ χⁿ/n^s, (where again the numerator of each term = 1),

                                 ⁿ⁼¹

when the square of this expression is divided by the sum of the terms of the corresponding Riemann type zeta function (i.e. where all the terms of the original series are now positive) with the denominator of each term n^2s, then a new L-function is always generated.

To illustrate, I have taken, from the LFMDB site the example of L(s, χ), where χ is the Dirichlet character with Label (21.20). 

The Dirichlet L-series for this function is given by

L(s, χ) = 1 – 1/2^s + 1/4^s + 1/5^s – 1/8^s – 1/10^s – 1/11^s – 1/13^s + 1/16^s + 1/17^s – 1/19^s + 1/20^s + 1/22^s – 1/23^s + 1/25^s + …

So when a number is divisible by either 3 or 7 (which are the prime factors of 21)

χ(n) = 0;

if on division by 21, a number leaves a remainder of 1, 4, 5, 16, 17 or 20, then 

χ(n) = 1;

and finally if on division by 21, a number leaves a remainder of 2, 8, 10, 11, 13, or 19, then

χ(n) = – 1;

Therefore because the numerator of each term in this function is 1, then

(1 – 1/2^s + 1/4^s + 1/5^s – 1/8^s – 1/10^s – …)²/(1 + 1/2^2s + 1/4^2s + 1/5^2s + 1/8^2s + 1/10^2s – …)

= 1 – 2/2^s + 2/4^s + 2/5^s – 2/8^s – 4/10^s – …

So the form of this function (with respect to the signs of each term) is the same as the original L-function.

However the numerator is now governed by the number of distinct prime factors contained in the denominator part of each term.

So the numerator = 2^t, where t represents the number of (distinct) factors in the corresponding denominator part of the term.

This in fact provides a ready means of determining whether a given Dirichlet type series does indeed constitute an L-function (with a corresponding Euler product expression).

Therefore if we square this series and divide by the sum of corresponding terms of the Riemann function (where s in the original series is replaced by 2s in the Riemann), when a new function of the same form with respect to sign as the original is generated (where again the numerator of each term is 2^t, with t representing the number of distinct factors in the corresponding denominator part of the term), then the original Dirichlet series is indeed an L-function (of degree 1).