Another Simple L-Function

This is another of the simpler L-functions (degree 1) on the LMFDB site, which I examined with a view to assessing common features with respect to all functions of this class.

So, L(χ, s) where χ is the Dirichlet character with label 7, 6 is given as,

1 + 1/2^s – 1/3^s + 1/4^s – 1/5^s – 1/6^s + 1/8^s + …

So with respect to the Label, 7 refers to the number by which we divide each successive natural number. Then when the remainder is 1, 2, or 4, then χ(n) = 1; when the remainder is 3, 5 or 6 then χ(n) = – 1; and when the remainder is zero, χ(n) = 0.

Therefore for each cycle of 7 successive natural numbers, we have 6 terms (with numbers exactly divisible by 7 omitted).

The corresponding product over primes expression (for this Dirichlet sum) is given as

(1 – 1/2^s) * (1 + 1/3^s) * (1 + 1/5^s) * (1 – 1/11^s) * …

In this case, when s is a positive odd integer, the value of both expressions = kπ^s/√7 (where k is a rational number).

So for example when s = 1,

1 + 1/2 – 1/3 + 1/4 – 1/5 – 1/6 + 1/8 + …

 = (1 – 1/2) * (1 + 1/3) * (1 + 1/5) * (1 – 1/11) * …

= π/√7

Then when s = 3,

1 + 1/2³ – 1/3³ + 1/4³ – 1/5³ – 1/6³ + 1/8³ + …

= (1 – 1/2³) * (1 + 1/3³) * (1 + 1/5³) * (1 – 1/11³) * …

= √7π³/75

And when s = 5,

1 + 1/2⁵ – 1/3⁵ + 1/4⁵ – 1/5⁵ – 1/6⁵ + 1/8⁵ + …

= (1 – 1/2⁵) * (1 + 1/3⁵) * (1 + 1/5⁵) * (1 – 1/11⁵) * …

= 64π⁵/(7203√7).

Again as always is the case, the above L-function, where s is a positive integer, it can be related to the Riemann zeta function (excluding those terms where 7 is a factor).

(1 + 1/2^s – 1/3^s + 1/4^s – 1/5^s – 1/6^s + 1/8^s + …)²/(1 + 1/2^2s + 1/3^2s + 1/4^2s + 1/5^2s – 1/6^2s + 1/8^2s + …)

= 1 + 2/2^s  – 2/3^s  + 2/4^s – 2/5^s – 4/6^s + 2/8^s + …

with a product over primes expression,

1/{1 – 2/(2^s + 1)} * 1/{1 + 2/(3^s + 1)} * 1/{1 + 2/(5^s + 1)} * 1/{1 – 2/(11^s + 1)} * …

And where s is a positive odd integer, the result is always a rational number.

It is important here to stress that the general relationship holds in all cases, where is a positive odd integer.

However in this case, where s is even, the result is an irrational number.

So for example in the simplest case where s = 1,

(1 + 1/2 – 1/3 + 1/4 – 1/5 – 1/6 + 1/8 + …)²/(1 + 1/2² + 1/3² + 1/4² + 1/5² – 1/6² + 1/8² + …)

= 1 + 2/2  – 2/3  + 2/4 – 2/5 – 4/6 + 2/8 + …,

With a product over primes expression,

1/{1 – 2/(2 + 1)} * 1/{1 + 2/(3 + 1)} * 1/{1 + 2/(5 + 1)} * 1/{1 – 2/(11 + 1)} * …

= {(π/√7)²}/²{(8π²/49)  = 7/8.

A Couple of More L-Functions

I looked at a couple of the other simpler L-functions on the LMFDB site to see if similar sort of patterns as demonstrated with the Dirichlet Beta Function exist.

For example,

L(χ, s) where χ is the Dirichlet character with label 8, 3 is given as,

1 + 1/3^s  – 1/5^s – 1/7^s + 1/9^s  + 1/11^s  – 1/13^s – 1/15^s  + …  

 = 1/(1 – 1/3^s) * 1/(1 + 1/5^s) * 1/(1 + 1/7^s) * 1/(1 – 1/11^s) * 1/(1 + 1/13^s) * …

So when on dividing each successive number n by 8, if remainder is an even number then χ(n) = 0; if remainder is 1 or 3, χ(n) = 1; if remainder is 5 or 7, χ(n) = – 1.     

When s is a positive odd integer, the above L-function has a value of the form kπ^s/√2

So for example, when s = 1,

1 + 1/3  – 1/5 – 1/7 + 1/9  + 1/11  – 1/13 – 1/15  + …  = √2π/4

It is interesting to contrast this with the Dirichlet Beta Function (when s = 1)

i.e. 1 – 1/3  + 1/5 – 1/7 + 1/9  – 1/11  + 1/13 – 1/15  + …  = π/4

Therefore,

(1 + 1/3  – 1/5 – 1/7 + 1/9  + …)/(1 – 1/3  + 1/5 – 1/7 + 1/9  –  …) = √2

Then when s = 3,

1 + 1/3³  – 1/5³ – 1/7³ + 1/9³  + 1/11³  – 1/13³ – 1/15³  + …  = 3π³/(64√2)

Again the above L-function, where s is a positive odd integer, can be related to the Riemann zeta function (excluding those terms where 2 is a factor).

 (1 + 1/3^s  – 1/5^s – 1/7^s + 1/9^s  + …)²/(1 + 1/3^2s  + 1/5^2s + 1/7^2s + 1/9^2s  + …)

= 1 + 2/3^s  – 2/5^s – 2/7^s + 2/9^s  + 2/11^s  – 2/13^s – 4/15^s …,

with a product over primes expression,

1/{1 – 2/(3^s + 1)} * 1/{1 + 2/(5^s + 1)} * 1/{1 + 2/(7^s + 1)} * 1//{1 – 2/(11^s + 1)} *

1{1 + 2/(13^s + 1)} …

And this will always result in a rational number.

So for example when s = 1,

(1 + 1/3  – 1/5 – 1/7 + 1/9  + …)²/(1 + 1/3²  + 1/5² + 1/7² + 1/9²  + …)

 

= 1 + 2/3  – 2/5 – 2/7 + 2/9  + 2/11  – 2/13 – 4/15  + …,

with a product over primes expression,

1/{1 – 2/(3¹ + 1)} * 1/{1 + 2/(5¹ + 1)} * 1/{1 – 2/(7¹ + 1)} * 1/{1 + 2/(11¹ + 1)} *

1/{1 – 2/(13¹ + 1)} …  

= 4/2 * 6/8 * 8/6 * 12/14 * 14/12 * …

= 1.

So this is in its own way remarkable i.e. that the L-function,

1 + 2/3  – 2/5 – 2/7 + 2/9 + 2/11  – 2/13 – 4/15  + …  = 1 

And in brief when s = 3,

1 + 1/3³  – 1/5³ – 1/7³ + 1/9³  + …)²/(1 + 1/3⁶  + 1/5⁶ + 1/7⁶ + 1/9⁶  + …)

= 1 + 2/3³  – 2/5³ – 2/7³ + 2/9³  + 2/11³  – 2/13³ – 4/15³…

= 135/128.

Then, L(χ, s) where χ is the Dirichlet character with label 8, 5 is given as,

1 – 1/3^s  – 1/5^s + 1/7^s + 1/9^s  – 1/11^s  – 1/13^s + 1/15^s  + … 

 = 1/(1 + 1/3^s) * 1/(1 + 1/5^s) * 1/(1 – 1/7^s) * 1/(1 + 1/11^s) * 1/(1 + 1/13^s) * …

So here when on dividing each successive number n by 8, if remainder is an even number then χ(n) = 0; if remainder is 1 or 7, χ(n) = 1; if remainder is 3 or 5, χ(n) = – 1.    

When s is a positive even integer, the above L-function has a value of the form kπ^s/√2

So for example, when s = 2, the Dirichlet series expansion is

1 – 1/3²  – 1/5² + 1/7² + 1/9² – 1/11²  – 1/13² + 1/15²  + … 

= π²/(8√2)

And when s = 4, the Dirichlet series expansion is

1 – 1/3⁴  – 1/5⁴ + 1/7⁴ + 1/9⁴ – 1/11⁴  – 1/13⁴ + 1/15⁴ + … 

= 11π⁴/(768√2)

Again the above L-function, where s is a positive even integer, can be related to the Riemann zeta function (excluding those terms where 2 is a factor).

In this case,

(1 – 1/3^s  – 1/5^s + 1/7^s + 1/9^s  – …)²/(1 + 1/3^2s  + 1/5^2s + 1/7^2s + 1/9^2s + …)

= 1 – 2/3^s  – 2/5^s + 2/7^s + 2/9^s  – 2/11^s – 2/13^s + 4/15^s +  …,

with a product over primes expression,

1/{1 + 2/(3^s + 1)} * 1/{1 + (2/(5^s + 1)} * 1/{1 – (2/(7^s + 1)} * 1/{1 + 2/(11^s + 1)} *

1/{1 + 2/(13^s + 1)} * …

Furthermore, the answer will be a rational fraction.

So when for example s = 2,

(1 – 1/3²  – 1/5² + 1/7² + 1/9²  – …)² /(1 + 1/3⁴  + 1/5⁴ + 1/7⁴ + 1/9⁴ + …)

= 1 – 2/3² – 2/5² + 2/7² + 2/9² – 2/11² – 2/13² + 4/15² +  …,

with a product over primes expression,

1/{1 + 2/(3² + 1)} * 1/{1 + (2/(5² + 1)} * 1/{1 – (2/(7² + 1)} * 1/{1 + 2/(11² + 1)} *

1/{1 + 2/(13² + 1)} * … 

= 10/12 * 26/28 * 50/48 * 122/124 * 170/172 * …

= 3/4.

And when s = 4,

(1 – 1/3⁴  – 1/5⁴ + 1/7⁴ + 1/9⁴  – …)² /(1 + 1/3⁸  + 1/5⁸ + 1/7⁸ + 1/9⁸ + …)

= 1 – 2/3⁴ – 2/5⁴ + 2/7⁴ + 2/9⁴– 2/11⁴ – 2/13⁴ + 4/15⁴ +  …,

with a product over primes expression,

1/{1 + 2/(3⁴ + 1)} * 1/{1 + (2/(5⁴ + 1)} * 1/{1 – (2/(7⁴ + 1)} * 1/{1 + 2/(11⁴ + 1)} *

1/{1 + 2/(13⁴ + 1)} * … 

= 82/84 * 626/628 * 2402/2400 * 14642/14644 * 28562/28564 * …

= 4235/4352.

A Simple L-Function

Perhaps the simplest L-function at the LMFDB site (besides the Riemann zeta function) is,

L(χ, s) where χ is the Dirichlet character with label 3, 2, which is given as

1 – 1/2^s + 1/4^s – 1/5^s + 1/7^s – 1/8^s + 1/10^s – 1/11^s + …,

with a product over the primes expression as

1/(1 + 1/2^s) * 1/(1 + 1/5^s) * 1/(1 – 1/7^s) * 1/(1 + 1/11^s) * …

So when on dividing each successive number n by 3, if remainder is 0, then χ(n) = 0; if remainder is 1, χ(n) = 1; if remainder is 2, χ(n) = – 1.  

When s is a positive odd integer the above L-function, results in a simple expression of the form kπ^s/√3.

For example when s = 1,

1 – 1/2 + 1/4 – 1/5 + 1/7 – 1/8 + 1/10 – 1/11 + …  = 2/3 * 5/6 * 7/6 * 11/12 * … 

= π/(3√3).

And when s = 3,

1 – 1/2³ + 1/4³ – 1/5³ + 1/7³ – 1/8³ + 1/10³ – 1/11³ + …

= 8/9 * 125/126 * 343/342 * 1331/1332 * …

= 4π³/(81√3).

And when s = 5,

1 – 1/2⁵ + 1/4⁵ – 1/5⁵ + 1/7⁵ – 1/8⁵ + 1/10⁵ – 1/11⁵ + …

= 32/33 * 3125/3126 * 16807/16806 * 161051/161052 * …

= 4π⁵/(729√3).

An interesting connection can be made here with the Riemann zeta function (excluding all terms where 3 is a factor).

So when s is a positive odd integer,

(1 – 1/2^s + 1/4^s – 1/5^s + 1/7^s – 1/8^s + …)²/(1 + 1/2^2s + 1/4^2s + 1/5^2s + 1/7^2s + 1/8^2s + …)

= 1 – 2/2^s + 2/4^s – 2/5^s + 2/7^s – 2/8^s + 4/10^s – …,

with a product over the primes expression,

= 1/{1 + 2/(2^s +1)} * 1/{1 + 2/(5^s +1)} * 1/{1 – 2/(7^s +1)} * 1/{1 – 2/(11^s +1)} * …

And this result is a rational number.

So when s = 1,

(1 – 1/2 + 1/4 – 1/5 + 1/7 – 1/8 + …)²/(1 + 1/2² + 1/4² + 1/5² + 1/7² + 1/8² + …)

= 1 – 2/2 + 2/4 – 2/5 + 2/7 – 2/8 + 4/10 – …

with a product over primes expression,

1/{1 + 2/(2 + 1)} * 1/{1 + 2/(5 + 1)} * 1/{1 – 2/(7 + 1)} * 1/{1 – 2/(11 + 1)} * …

=  3/5 * 6/8 * 8/6 * 10/12 * 14/16 * 18/20 * …

= 1/4.

And when s = 3,

(1 – 1/2³ + 1/4³ – 1/5³ + 1/7³ – 1/8³ + …)²/(1 + 1/2⁶ + 1/4⁶ + 1/5⁶+ 1/7⁶ + 1/8⁶ + …)

= 1 – 2/2³ + 2/4³ – 2/5³ + 2/7³ – 2/8³ + 4/10³ – …,

with a product over primes expression,

1/{1 + 2/(2³  + 1)} * 1/{1 + 2/(5³  + 1)} * 1/{1 – 2/(7³  + 1)} * 1/{1 – 2/(11³  + 1)} * …

= 9/11 * 126/128 * 344/342 * 1332/1330 *

= 10/13.

Finally briefly when s = 5 

(1 – 1/2⁵ + 1/4⁵ – 1/5⁵ + 1/7⁵ – 1/8⁵ + …)²/(1 + 1/2¹⁰ + 1/4¹⁰+ 1/5¹⁰+ 1/7¹⁰ + 1/8¹⁰ + …)

= 1 – 2/2⁵ + 2/4⁵ – 2/5⁵ + 2/7⁵ – 2/8⁵ + 4/10⁵ – …  = 6930/7381.