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,

1/1^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 – 1/3³ + 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

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

= 1/1^s – 2/3^s + 2/5^s – 2/7^s + 2/9^s –  2/11^s + 2/13^s – 4/15^s  + …)  = 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/(3^s + 1)} *  1/{1 – 2/(1/5^s + 1)}* 1/{1 + 2/(7^s + 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 + 1/5^s – 1/7^s – 1/11^s + 1/13^s + …)²/(1 + 1/5^2s + 1/7^2s + 1/11^2s + 1/13^2s + …)

=  1 + 2/5^s – 2/7^s – 2/11^s + 2/13^s + …, 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.

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,

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

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

(1/1^s + 1/2^s  + 1/3^s + 1/4^s + …)²/(1/1^2s + 1/2^2s  + 1/3^2s + 1/4^2s + …)  = k (where k is a rational number).

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

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

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 = 2^t, 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 + 2^s)} * 1/{1 – 2/(1 + 3^s)} * 1/{1 – 2/(1 + 5^s)} * …

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 + …].

Dynamic Appreciation of Nature of Riemann Hypothesis

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,

k₁^{– (1/2 + 14.134725 …)} + k₂^{– (1/2 + 14.134725 …)} + k₃^{– (1/2 + 14.134725 …)} + k₄^{– (1/2 + 14.134725 …)} + …   = 0,

where k_1, k_2, k_3, k₄, … are complex numbers, then by lucky chance we might be able to find that when k_1, k_2, k_3, k₄, … 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, k₄, …, 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 exponents (i.e. dimensional numbers) that all lie on the same imaginary line (through 1/2).

Then on the other hand we would have an infinite series i.e. the natural numbers all on the real line (through 0) that provide in each case a solution to the equation.

However there would still remain a significant problem in that we would not be able to guarantee in any case that perhaps that another set of solutions for k_1, k_2, k_3, k₄, …  (with some or all not necessarily not lying on the real line) might likewise provide a solution to the equation.

Also we could not guarantee that for some future imaginary value for the exponent (not presently known) of a known zero, that is placed on the same imaginary line through 1/2, that k_1, k_2, k_3, k₄, …, would again always represent the natural numbers 1, 2, 3, 4 , … in providing a solution to the equation.

This would of course be a problem if a zero were in fact to lie off the imaginary line (through 1/2).

Therefore we could not guarantee that all the imaginary parts (of known zeros) would be associated with the natural numbers.

In other words another unique set of numbers for k_1, k_2, k_3, k₄, … would be associated with a zero where the imaginary part is off the line through 1/2 (shared by all other zeros).

Now of course customarily in looking at this issue we start with the natural numbers and then attempt to derive the solution as a common shared (negative) exponent.

So therefore when we have

1^–s + 2^–s + 3^–s + 4^–s + … = 0,

The first solution for s, i.e. the first Riemann zero is 1/2 + 14.134725 … (or alternatively

1/2 –  14.134725 …).

However as we know, there is - in the absence of proof of the Riemann Hypothesis - no guarantee that a future zero might lie off the imaginary line.

However when looked at correctly i.e. from a dynamic interactive perspective, there is no reason to necessarily start with the second formulation of the equation, with the natural numbers as given data (where we solve for the common dimensional value).

Equally we could start from the first formulation of the equation where each dimensional value is given (and we solve for an infinite set of associated numbers).

When expressed in a psychological manner what this entails is as follows.

Because of the strong rational conscious bias of present Mathematics we take the existence of the natural numbers as necessarily all existing on the real number line as a given unquestioned assumption.

Though we are not explicitly aware of this crucial fact, when we then solve for the set of exponents (as the Riemann zeros of the equation) we are implicitly looking for the unconscious basis that is consistent with our assumption of the real numbers (all lying on the real line).

We can therefore only correctly assume that all natural numbers lie on the real line if the corresponding set of Riemann zeros lie on the same imaginary line (through 1/2).

Expressed equally in a slightly different manner, for the analytic (quantitative) aspect of mathematical understanding to be consistent (that all real numbers exist on the same real number line) with the holistic (qualitative) aspect of mathematical understanding, all the Riemann zeros must correspondingly lie on the same imaginary line (though 1/2).

Or again in an equivalent manner, (conscious) rational interpretation of number relationships should be consistent with (unconscious) intuitive appreciation of such relationships. So consistency with respect to both aspects  - which are necessarily distinct from each other - are equally important.   

As we know all composite natural numbers represent the product of unique prime factors.

So 6 in common understanding  = 2 * 3 Now customarily we express the composite.

However if one considers fro example a table with width 2 metres and length 3 metres, the area of the table = 6 square metres.

Thus in multiplying the two numbers as well as the quantitative change in units, there is also a qualitative change in the nature of the units (i.e. from 1-dimesnional to 2 –dimensional).

However remarkably this crucial fact is ignored in the conventional interpretation of multiplication. So the quantitative transformation in units aspect is solely considered with the corresponding qualitative change (in the nature of units) simply reduced in a linear quantitative manner.

So strictly speaking we are not entitled to make the assumption that composite numbers lie on the same line (as the primes from which they are derived) without in turn justifying the validity of reducing the qualitative aspect (of dimensional change) to the quantitative aspect (of homogeneous linear measurements).

And perhaps surprisingly the condition for making this reduced assumption is that the Riemann zeros - which represent in an indirect quantitative manner the holistic (qualitative) nature of the natural number system - all lie on the same imaginary line (through  1/2). 

Indeed in holistic terms the imaginary line represents the indirect rational means of representing the qualitative aspect of the number system!

So again we can only make the assumption that all the natural numbers lie on the (1-dimensional) real line, if in turn all the Riemann zeros lie on the same imaginary line.

Equally from the complementary opposite perspective we can only assume that all the Riemann zeros lie on the same imaginary line (through 1/2) if the natural numbers (as products of the primes) all lie on the real line (through 0). 

So from the dynamic interact perspective the Riemann Hypothesis, which already assumes the consistency of the real number line, cannot be proven (or strictly disproven).

We can say however say that the Riemann Hypothesis (in the assumption that all the Riemann zeros lie on the imaginary line through 1/2) is consistent with the corresponding assumption of all the natural numbers lying on the same line.

Likewise we can say that the assumption that all the natural numbers lie in the same real line is consistent with the truth of the Riemann Hypothesis.

However we cannot strictly prove these propositions. Thus their acceptance ultimately requires faith in the consistency of the overall number system.