Friday, October 12, 2018

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/ns, (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 n2s, 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 – 1/2s + 1/4s + 1/5s – 1/8s – 1/10s – 1/11s – 1/13s + 1/16s + 1/17s – 1/19s + 1/20s + 1/22s – 1/23s + 1/25s + …

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 – 1/2s + 1/4s + 1/5s – 1/8s – 1/10s – …)2/(1 + 1/22s + 1/42s + 1/52s + 1/82s + 1/102s – …)

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

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 = 2t, 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 2t, 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). 

Tuesday, September 25, 2018

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

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

In this case, when s is a positive odd integer, the value of both expressions = kπs/√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 + 1/2s – 1/3s + 1/4s – 1/5s – 1/6s + 1/8s + …)2/(1 + 1/22s + 1/32s + 1/42s + 1/52s – 1/62s + 1/82s + …)

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

with a product over primes expression,

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

Monday, September 17, 2018

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

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

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

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

with a product over primes expression,

1/{1 – 2/(3s + 1)} * 1/{1 + 2/(5s + 1)} * 1/{1 + 2/(7s + 1)} * 1//{1 – 2/(11s + 1)} *
1{1 + 2/(13s + 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 1/3s  – 1/5s + 1/7s + 1/9s  1/11s  – 1/13s + 1/15s  + … 

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

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

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

with a product over primes expression,

1/{1 + 2/(3s + 1)} * 1/{1 + (2/(5s + 1)} * 1/{1 – (2/(7s + 1)} * 1/{1 + 2/(11s + 1)} *
1/{1 + 2/(13s + 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.

Saturday, September 15, 2018

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

with a product over the primes expression as

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

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

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

with a product over the primes expression,

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

Friday, August 17, 2018

Fascinating Dirichlet Beta Function Relationships

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,

1/1s – 1/3s + 1/5s – 1/7s + 1/9s     = 1/(1 + 1/3s) *  1/(1 1/5s) * 1/(1 + 1/7s) * …

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 kπs, where k is a rational number.

So for example β(3) = 1 – 1/33 + 1/53 – 1/73 + 1/93    = π3/32 and

β(5) = 1 – 1/35 + 1/55 – 1/75 + 1/95    = 5π5/1536.


Now this function can be connected with the Riemann zeta function in a surprising manner.

For using a similar type general connection as illustrated in the last blog entry “More on Riemann Generalisation” we have

(1s – 1/3s + 1/5s – 1/7s + 1/9s  …)2/(12s + 1/32s + 1/52s + 1/72s + 1/92s +  …)

= 1/1s – 2/3s + 2/5s – 2/7s + 2/9s  2/11s + 2/13s 4/15s  + …)  = k (where k is a rational number). 

And this is another L1 function with corresponding product over the primes expression, i.e.

1/{1 + 2/(3s + 1)} *  1/{1 2/(1/5s + 1)}* 1/{1 + 2/(7s + 1)} * …

Thus again the value of the numerator is determined by the number of distinct prime factors in the corresponding denominator value. So with just one distinct prime factor in the denominator, the value of the numerator is 2; however with the denominator 15 for example, we now have two distinct prime factors, so the value of the corresponding numerator is 4 (i.e. 22).

So for example when s = 1, we obtain,

(11 – 1/31 + 1/51 – 1/71 + 1/91  …)2/(12 + 1/32 + 1/52 + 1/72 + 1/92 +  …)

= 1 – 2/3 + 2/5 – 2/7 + 2/9 2/11 + 2/13 4/15  + …  = (π/4)2/(π2/8) = 1/2.

And when s = 3, we obtain

(13 – 1/33 + 1/53 – 1/73 + 1/93  …)2/(16 + 1/36 + 1/56 + 1/76 + 1/96 +  …)

= 1 – 2/33 + 2/53 – 2/73 + 2/93 2/113 + 2/133 4/153  + … = (π3/32)2/(π6/960) = 15/16. 


In fact interestingly just as we have seen above,

(11 – 1/31 + 1/51 – 1/71 + 1/91  …)2/(12 + 1/32 + 1/52 + 1/72 + 1/92 +  …) = 1/2,

likewise,

(11 – 1/31 + 1/51 – 1/71 + 1/91  …)3/(13 – 1/33 + 1/53 – 1/73 + 1/93  …) = 1/2.


Here is another very interesting connection!

As we know,

12 + 1/32 + 1/52 + 1/72 + 1/92 +  … = 1/(1 – 1/32) * 1/(1 – 1/52) * 1/(1 – 1/72) * … = π2/8.

So, if we eliminate from the product over primes expression the term involving 3 as a prime, we must then eliminate in turn all terms with 3 as a factor in the in sum over integers expression.

Thus, 12 + 1/52 + 1/72 + 1/112 +    =  1/(1 – 1/52) * 1/(1 – 1/72) * 1/(1 – 1/112) * …  = π2/9.

And if we eliminate with respect to the corresponding Dirichlet beta function the term involving 3 as a prime, we must then eliminate in turn all terms with 3 as a factor in the in sum over integers expression.

Thus, 1 + 1/5 – 1/7 1/11 + …  = (1 + 1/5) *  1/(1 1/7) * 1/(1 + 1/11) * … = π/3.

and (1 + 1/5 – 1/7 1/11 +  …)2   = π2/9, where we consistently have two positive terms followed by two negative terms in the series and vice versa.

Therefore,

12 + 1/52 + 1/72 + 1/112 + 1/132 +     = (1 + 1/5 – 1/7 1/11 + 1/13 + …)2.      

And as (1 + 1/5s – 1/7s 1/11s + 1/13s…)2/(1 + 1/52s + 1/72s + 1/112s + 1/132s)

1 + 2/5s – 2/7s 2/11s + 2/13s…, this implies that when s = 1,

(1 + 1/5 – 1/7 1/11 + 1/13 )2/(12 + 1/52 + 1/72 + 1/112 + 1/132 + …)

1 + 2/5 – 2/7 2/11 + 2/13…    = 1.

So we have the interesting case here of where the value of this L-function (for s = 1) = 1.

Tuesday, August 7, 2018

More on Riemann Generalisation (1)

In an earlier entry “Intertwining L1and L2 Functions” I showed that,

(1/12 + 1/22  + 1/32 + 1/42 + …)2/(1/14 + 1/24  + 1/34 + 1/44 + …)  

= (π2/6)2/(π4/90)  = 90/36 = 5/2.

And  that this can be written as a Dirichlet L-series i.e.

1/12 + 2/22 + 2/32 + 2/42 + 2/52 + 4/62 + 2/72 + 2/82 + 2/92 + 4/102 + … = 5/2

This in turn makes use of our generalised Riemann zeta function formula,

1/1s + (1 + k1)/2s + (1 + k2)/3s + (1 + k1)/4s + (1 + k3)/5s + {(1 + k1)(1 + k2)}/6s + …

= 1/{1 – (1 + k1)/(2s + k1)} * 1/{1 – (1 + k2)/(3s + k2)} * 1/{1 – (1 + k3)/(5s + k3)} * …,
 
where k1, k2, k3, …   = 1.

So ζ(2)2/ζ(4) = 1/12 + 2/22 + 2/32 + 2/42 + 2/52 + 4/62 + 2/72 + 2/82 + … = 5/2

And the result can be generalised so that when s is a positive even integer,

(1/1s + 1/2s  + 1/3s + 1/4s + …)2/(1/12s + 1/22s  + 1/32s + 1/42s + …)  = k (where k is a rational number).

And this can be expressed as an L1 series (with exponent s) i.e.

1/1s + 2/2s  + 2/3s + 2/4s + 2/5s + 4/6s + 2/7s + …

Here the value of numerator is determined by the number of distinct prime factors contained by each corresponding number in the denominator with the numerator = 2t, where t represents the number of distinct prime factors.

So 2, 3, 4 and 5 respectively each contain one distinct prime factor. So the numerator is 21 = 2.
However 6 then contains two distinct prime factors so the numerator in this term = 22 = 4.

And the corresponding general product over primes expression for this new L-series is,

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


Thus to illustrate once again when s = 4, we have

(1/14 + 1/24  + 1/34 + 1/44 + …)2/(1/18 + 1/28  + 1/38 + 1/48 + …)  = (π4/90)2/(π8/9450)

= (π8/8100)/(π8/9450)  = 9450/8100 = 7/6

And this is represented by the new L-series (with exponent 4) i.e.

1/14 + 2/24  + 2/34 + 2/44 + 2/54 + 4/64 + 2/74 + …,

with a corresponding product over primes expression,

1/{1 – 2/(1 + 24)} * 1/{1 – 2/(1 + 34)} * 1/{1 – 2/(1 + 54)} * …


= 1/(1 – 2/17) * (1 – 2/82) * (1 – 2/626) * …

= 17/15 * 82/80 * 626/624 * …

And again each of the individual terms (for both the sum over the integers and product over the primes expressions) can then be expressed by a corresponding L2 series.

For example the first term of the product over primes expression i.e. 17/15

= 1 + (2/17)1 + (2/17)2  + (2/17)3  + …  = 17/15


And once more an unlimited number of possible variations can be made on this basic result through removing certain prime numbers (in the product over primes expressions) and then the corresponding natural numbers where these are factors in the sum over the integers expression.

So for example if we remove 2 (in product over primes expression) and then all corresponding even numbers (in the sum over integers expression) we have

 (1/14 + 1/34  + 1/54 + 1/74 + …)2/(1/18 + 1/38  + 1/58 + 1/78 + …)  = (π4/96)2/(255π8/9450 * 256) 

= 35/34

And this is represented by the new L-series expression,

1/14 + 2/34  + 2/54 + 2/74 + 2/94 + 2/114 + 2/134 + 4/154 + …,

with a corresponding product over primes expression

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

= 1/(1 – 2/82) * (1 – 2/626) * (1 – 2/2402) * …

= 82/80 * 626/624 *  2402/2400 * …  = 35/34

And just as this new L1 series represents a quotient of the two original L1 series,

in complementary fashion, each individual terms can be expressed as a product of two original L2 series

So for example 82/80 = 82/81 * 81/80

= [1 + {1/1 + 34}1 + {1/1 + 34}2 + {1/1 + 34}3 + …] * [1 + (1/34}1 + (1/ 34)2 + (1/34)3 + …].