Return to Dirichlet Beta Function

Let us return once more to the Dirichlet Beta function i.e.

L(χ,s)

= 1 − 3-s + 5-s − 7-s + 9-s − 11-s + 13-s − 15-s + 17-s − 19-s + 21-s − 23-s + 25-s − 27-s + ...

Only the odd natural numbers are involved here alternating as between positive and negative values.

So we have two distinct series i.e. 1, 5, 9, 13, 17, … and 3, 7, 11, 15, 19, … etc.

All of the odd numbers represent ½ of the natural numbers; then each of these two series of odd numbers represent ¼ of the natural numbers.

We can now attempt to estimate the frequency of factors of both of these series in two ways.

1. We can do so with respect to the cumulative total of factors up to n. The formula that can be used here to approximate the no. of factors involved = n(ln n − 1) 

2. We can alternatively use the cumulative total of zeta zeros up to t (where n = t/2π). The formula that can be used here to give an extremely accurate answer for  the frequency of zeros =   t/2π(ln t/2π − 1).  

In the following illustration I manually counted the cumulative frequency of factors (excluding 1) of the 1^st series 1, 5, 9, 13, 17, 21, … in blocks of 50 up to n = 300.

I then manually counted the cumulative frequency of all the natural nos. (except 1) again in blocks of 12.5 up to n/4 = 75.

This latter count of all natural number factors to n/4 then provided an estimate of the corresponding count of the odd factors in this first series.

For example there are 29 odd factors (belonging to the 1^st series) up to n = 50 and the corresponding count of natural no. factors (except 1) to n/4 = 12.5 is 23. So this computation involving all natural no. factors provides an estimate of the frequency of factors 5, 9, 13, 17, 21, …

We could also use the frequency of zeta zeros up to t/4 as an estimate.

As t = 2nπ, t/4 = nπ/2

So again to measure the frequency of factors 5, 9, 13, 17, 21, … up to 50 we calculate t/8π(ln t/8π − 1). In the first instance this serves as an accurate measurement of the frequency of zeros up to t/4 which then likewise serves as an approximation of the frequency of the odd factors in the series to n. And there are 19 zeta zeros to t/4 (i.e. 50π/2 = 25π) serving as a not so accurate estimate in this case. However as n increases the approximation steadily improves.

n	Total no. of factors 5, 9, 13,17, 21, …to n	Estimated total using natural no. factors to n/4		Zeta zero estimate to t/4
50	29	23	19
100	79	62	55
150	132	106	98
200	189	157	146
250	247	207	196
300	308	264	249

Now it is interesting to note that this 1^st series of odd numbers follows the distribution of the Riemann zeta zeros.

The second task is then to estimate the frequency of odd factors in the 2^nd series i.e. 3, 7, 11, 15, 19, …

Again I measured this in blocks of 50 up to n = 300. I first calculated the actual no. of factors (in the 2^nd series) up to n. I then used an estimate that is directly based on the Dirichlet distribution.

In this case I calculated all the even factors - not to n/4 - but rather to n, before then dividing the total by 4 to get an estimate for the actual no. of factors. This estimate turns out to be remarkably accurate.

For example the actual frequency of the odd factors 3, 7, 11, 15, 19, … to 50 = 39.

And the corresponding estimate using all factors (except 1) to 50 is also 39. And this is no coincidence as the table below indicates.

Also the estimate using the Dirichlet Beta zeros as an estimate is also very accurate in this case where we count all the zeros to t (where t = 2nπ) before dividing the total by the conductor of the function (i.e. N = 4).

n	Total no. of factors 3, 7, 11,15, 19, …to n	(Estimated total using natural no. factors to n)/4		Zeta zero estimate to t before division by 4
50	39	39	36
100	95	96	90
150	155	155	150
200	222	222	215
250	287	286	283
300	368	364	353

One further interesting fact is that the sum of all Riemann zeta zeros to 2t minus the sum of all zeta zeros to t = the sum of all Dirichlet Beta zeros to t.

For example there are 79 zeta zeros to 200 and 29 zeta zeros to 100. And the difference = 50 is the frequency of Dirichlet Beta zeros to 100.

Estimating Individual Zeros of Dirichlet L-Functions

Yesterday I showed how the frequency of zeros of all Dirichlet L-Functions (degree 1) is intimately related through the conductor to the corresponding frequency of (non-trivial) zeros of the Riemann zeta function.

However this does not imply that the individual zeros of these functions can be precisely calculated with reference to the individual zeros of the Riemann zeta function.

One must remember that a complementary nature connects the nature of the primes and the Riemann zeta zeros.

So each prime is characterised by the maximum amount of individual uniqueness as it were consistent with the ordered nature of the natural numbers.

Therefore though it is indeed possible to predict with a progressively greater relative degree of accuracy the frequency of primes up to a given number, in absolute terms the deviation as between the actual and predicted number of primes tends to increase.

However by contrast each zeta zero is characterised by the maximum amount of ordered sociability (between the primes and natural numbers) consistent with each zero maintaining its individual uniqueness at an individual level.

Therefore it is possible to predict the number of zeta zeros up to a given magnitude not only accurately in relative terms but likewise also in an absolute manner.

Thus the very nature of the zeta zeros is to keep reconciling the individual uniqueness of the zeros at a local with the maximum amount of order possible at a collective level.

Thus deviations at a local level keep getting cancelled out in terms of maintaining this order at the collective level.

However this entails that though we can calculate almost exactly in absolute terms the frequency of Riemann zeros and the zeros of other Dirichlet functions up to any given level, because of the uniqueness of each zero, this does not apply so well at the individual level.

However even here, from knowledge of the individual zeta zeros, we can however, through knowledge of the conductor, make a reasonable attempt to approximate the individual zeros of the other Dirichlet functions.

For example to illustrate I made an estimate of the first 10 zeros of the Dirchlet Beta Function.

As we have seen the Dirichlet Beta Function as a conductor of 4. So starting with the 3^rd Riemann zero I estimated each 4^th zero up to the 39th zero (giving 10 in all).

Then I divided each these zeros by 4 to provide an estimate of the corresponding zero of the Dirichlet Beta Function

  

Ordinal number of zeta zero	Zeta zero (correct to 3 decimal places)	Zeta zero/4	First 10 Dirichlet Beta Function zeros
3	25.011	6.253	6.001
7	40.919	10.230	10.264
11	52.970	13.243	13.002
15	65.113	16.278	16.353
19	75.705	18.926	18.285
23	84.735	21.184	21.445
27	94.651	23.663	23.274
31	103.726	25.932	25.726
35	111.875	27.969	28.358
39	121.370	30.343	29.662

The estimates here, though not exact are however quite good with some local deviation however in evidence. However there is a certain degree of arbitrariness in the manner that I started with the 3^rd Riemann zero in obtaining my estimates.

Perhaps the most unbiased approach is to obtain a mean or median within the range of conductor values.

So in this case this would imply getting an average of each successive group of 4 Riemann zeta values (before then dividing by 4) or alternatively the average of the 2^nd and 3^rd values within each group of before dividing by 4. Though the estimates would not be quite as good as above they would still be reasonably accurate. And this approach could then be generalised to approximate the individual zeros for any Dirichlet L-function.   

Calculating Frequency of Zeros for Dirichlet L-Functions

Some time ago I was interested to find computations of the Zeros for the Dirichlet Beta Function by Anthony Lander (the first 320 zeros).

This table of zeros enabled me to establish a close connection with the better known (non-trivial) zeros of the Riemann Zeta Function.

Now again the Dirichlet Beta function is given as

L(χ, s) = 1/1^s – 1/3^s + 1/5^s – 1/7^s + …   = 1/(1 + 1/3^s) *  1/(1 – 1/5^s) * 1/(1 + 1/7^s) * …, where χ is the Dirichlet character and s the power associated with each term

Now using the established terminology this represents a Dirichlet L-function of degree 1.

So when on dividing each successive number n by 4, if number is even then χ(n) = 0; if remainder is 1, χ(n) = 1; if remainder is 3, χ(n) = – 1.  

So the conductor N, which is the number by which n is divided, in this case is 4.

Now this conductor is directly relevant in terms of establishing the relationship of zeros in the Dirichlet Beta Function and zeros in the Riemann zeta function respectively.

Quite simply if the total frequency of zeros in the Riemann Zeta function up to n = k, the corresponding frequency of the zeros in the Dirichlet Beta function up to n/4 = k/4.

For example there are 649 zeros for the Riemann zeta function up to 1000.

So here n = 1000 and k = 649.

Therefore this entails that the corresponding number of zeros in the Dirichlet Beta

Function to 250 (i.e. n/4) = 649/4 = 162 (to the nearest integer).

And Lander’s table does indeed show exactly 162 Dirichlet Beta zeros up to 250.

Again there are 79 zeros for the Riemann zeta function up to 200.

This entails that the frequency of Dirichlet Beta zeros up to 50 = 79/4 = 20 (to the nearest integer).

And again Lander’s table shows exactly 20 such zeros to 50.

So there seems a pretty exact correspondence here as between the frequency of both sets of zeros.

In fact this correspondence seems to universally apply to all Dirichlet L-functions of degree 1.

So therefore in general terms once again if the frequency of zeta zeros up to n is k, the frequency of the alternative set of zeros up to n/N (where N is the conductor) is k/N.

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

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

So the conductor here is 7.

Now frequency (k) of zeta zeros to n = 210 is 85.

Therefore the corresponding frequency of this alternative function with conductor 7 to 210/7 = 30 is 85/7 = 12 (to the nearest integer).

And in the list of the first few zeros of this function at the LMFDB, there are indeed exactly 12 zeros up to 30.

As I have said this seems a universal feature with respect to all Dirichlet L-functions of degree 1.

In fact this remains the case even where imaginary or complex numbers are involved.

For example the following is the Dirichlet L-function with label 5.3 from the LMFDB

L(χ,s)  = 1 − i·2^-s + i·3^-s − 4^-s + 6^-s − i·7^-s + i·8^-s − 9^-s + 11^-s − i·12^-s + i·13^-s − 14^-s + 16^-s − i·17^-s + i·18^-s − 19^-s + ⋯

So the conductor here is 5.

Therefore if n = 200, the frequency of zeros of this function to n/N i.e. 200/5 = 40 is given as k/N = 79/5 = 16 (correct to nearest integer).

And once again in the list of the first few zeros of this function there are exactly 16 up to 40.

And the following the Dirichlet L-function with label 7.5 from the LMFDB

L(χ,s) =

1 + (−0.5 − 0.866i)2-s + (0.5 − 0.866i)3-s + (−0.5 + 0.866i)4-s + (0.5 + 0.866i)5-s − 6-s +  8-s + (−0.5 − 0.866i)9-s + (0.5 − 0.866i)10-s + (−0.5 + 0.866i)11-s + (0.5 + 0.866i)12-s −  13-s + 15-s + (−0.5 − 0.866i)16-s + (0.5 − 0.866i)17-s + (−0.5 + 0.866i)18-s + (0.5 + 0.866i)19-s + ⋯

So the conductor here is 7.

As we have seen there are 85 zeros to 210. Therefore once again we would expect up to 210/7 = 30, 85/7 = 12 zeros. And the table of first few zeros at LMFDB does indeed show exactly 12 zeros for this function up to 30.

As we have seen before the formula for calculating zeta zeros up to a given number is remarkably accurate (even in absolute terms)

So therefore if the frequency of zeros to t is given as  t/2π(ln t/2π − 1) then the corresponding frequency of zeros of an alternative Dirichlet function of degree 1 (with conductor N) 
= t/2πN((ln t/2π − 1).

Using Zeta Zeros to Count Factors (2)

We have seen in the last entry, with respect to the Riemann Zeta Function, how to calculate the accumulated frequency of odd factors up to a certain number n.

This frequency as we have seen approximates closely the corresponding frequency of all factors (odd and even) to n/2.

So for example, if n = 300, the accumulated frequency of all the odd factors of the natural numbers up to 300, approximates closely the corresponding frequency of all factors (odd and even) to 150. 

Is implies therefore that to approximate the accumulated frequency of the even factors to 300, we can thereby count the frequency of all factors from 151 to 300.

Now the accumulated frequency of all odd factors to 300 is 680 (excluding as in all cases the trivial case where 1 is a factor). And if we count all factors (odd and even to 150) we get 629. So the relative accuracy of our result here is 92.5%. So the all factor underestimates the odd factor total in this case. But the approximation steadily increases towards 100% (in relative terms) as we progressively increase the value of n.

Likewise the accumulated frequency of all even factors to 300 is 803. However if we now count the frequency of all factors (odd and even) from 151 to 300 we get 854.

So the relative accuracy of our result is here 94% with the all factor now overestimating the frequency of even factors. However, once again the approximation steadily increases towards 100% as we progressively increase the value of n.

Alternatively we could say that if t = 2πn the frequency of zeros to t/2 = πn approximates the corresponding accumulated frequency of all odd number factors to n.

Also the frequency of zeros t/2 to t approximates the corresponding frequency of all even number factors to n.

Illustrating further, as we saw in the last entry, the accumulated frequency of factors up to n (not containing 3 or multiples of 3) can be measured by counting all factors to 2n/3.

Therefore again where n = 300, the accumulated frequency of factors (not containing 3 or multiples of 3) to 300 can be approximated by counting all factors up to 200.

This likewise implies that the accumulated frequency of those factors (comprising 3 or multiples of 3) can be approximated by counting all factors from 201 to 300.   

The cumulative frequency of factors (other than those containing 3 or multiples of 3) to 300 = 1001. And the cumulative frequency of all factors to 200 = 913.

So the relative accuracy of this latter estimate = 91.2% which is an underestimate but gradually increases towards 100% (in relative terms) as n increases.

And the corresponding cumulative frequency of factors (that are 3 or multiples of 3) to 300 = 479. And the cumulative frequency of all factors from 201 to 300 = 570.

So the relative accuracy of this latter estimate is 84% which is an overestimate but which again will approach 100% (in relative terms) as n progressively increases.

Alternatively we could say that if t = 2πn the frequency of zeros to t/3 = (2π/3)n approximates the corresponding accumulated frequency of all number factors (that do not include 3 as a divisor) to n.

Also the frequency of zeros 2t/3 to t approximates the corresponding frequency of all number factors (that include 3 as a divisor) to n.

Then we also saw in the last entry that if we exclude all factors containing either 2 or 3 as factors that the cumulative frequency of all factors to n/3 gives a good approximation.

Therefore again with n = 300 the cumulative frequency of all factors to 100 provides an estimate of those factors to 300 (that exclude 2 or 3 as divisors).

The cumulative frequency of all factors (excluding 2 or 3 as divisors) to 300 = 403.

And the corresponding cumulative frequency of all factors to 100 = 382.

Therefore the relative accuracy is here 94.8% and approaches ever closer to 100% as n becomes progressively larger.

We can also approximate the cumulative frequency of all those factors (with 2 or 3 as divisors) to 300 through the corresponding calculation of all factors between 101 and 300.

Now the cumulative frequency of those factors (with 2 or 3 as divisors) to 300 = 1080.

And the cumulative frequency of all factors between 101 and 300 = 1101.

So the relative accuracy is here 98.1% and approaches ever closer to 100% as n progressively increases.

Alternatively we could say that if t = 2πn the frequency of zeros to 2t/3 = (4π/3)n approximates the corresponding accumulated frequency of all number factors (that do not include 2 or 3 as a divisor) to n.

Also the frequency of zeros t/3 to t approximates the corresponding frequency of all number factors (that include 2 or 3 as a divisor) to n.

In principle these ideas can be extended to any combination of factors.

For example say we want to approximate the accumulation of all factors (excluding 3, 7 or 11 as divisors) to n

Now the exclusion of factors containing 3 = 1/3 of divisors.

Then 1/3 of numbers divisible by 7 are also divisible by 3 so fractions divisible by both 3 and 7 = 1/3 + 2/3(1/7) = 7/21 + 2/ 21 = 9/21 = 3/7.

Then the fraction of numbers divisible by 3, 7 or 11 = 3/7 + 1/11(4/7)

= 33/77 + 4/77 = 37/77.

Therefore the fraction of natural numbers not divisible by 3, 7 or 11 = 40/77.

So to approximate the frequency of factors to n (that do not have 3, 7 or 11) as divisors, one accumulates all factors to (40/77)n.

Thus in the convenient case where n = 77, one would count the factors of all numbers to 40 = 118.

And the actual number of such factors to 77 = 131. So the relative accuracy is 90.1% Then to approximate those factors with 3, 7 or 11 as divisors, one would count all factors from 41 to 77 = 153.  And the actual number of such factors = 140. So the relative accuracy = 93.6%, which would improve towards 100% as n becomes progressively larger.

And as before one can use zeta zeros to approximate the answer in both cases.

So in the former case we would count the zeta zeros to 80π i.e. 251.3 = 109 with a relative accuracy = 83.2%.

Then in the latter case we would count the zeros from 80π to 154π, i.e. from 251.3 to 483.8 i.e. 261 – 109 = 152. Therefore the relative accuracy here = 99.3%.

Using Zeta Zeros to Count Factors (1)

As we have seen there is a close relationship as between the cumulative frequency of factors up to a given number n and the corresponding frequency of zeta zeros to a given number t (where n = t/2π).

As an approximation, the formula n(ln n – 1) can be used to approximate the accumulated total of factors to n, with t/2π(ln t/2π – 1) used to approximate the frequency of zeta zeros to t. And the latter formula for zeta zeros proves an especially good approximation!

So for example the cumulative frequency of factors to n = 100 (excluding 1 as a factor) is 382. And the number of trivial zeros to t = 200π (i.e. where n = t/2π) = 361.

So the percentage accuracy here is 94.5% and this increases towards 100% as n becomes progressively larger)

The question then arises as to whether this relationship can be extended to situations where certain factors are omitted.

For example if we omit all even numbers (where 2 is a factor) can we still establish a connection as between the cumulative number of odd factors to n and trivial zeros to a related number?

And the answer here is yes!

In fact is this case (with once again the trivial case of 1 as a factor excluded from consideration) the odd numbers i.e. 3, 5, 7, 9, 11, … will represent half of the divisors that can be used as factors in generating natural numbers.

So in the former case, where both even and odd divisors are used as factors n = t/2π.

Now where the number of divisors (as odd numbers) that can be used = ½ (of total of divisors), n = t/(2π * ½) = t/π.

In other words if we accumulate the total number of odd factors used in generating all numbers to n, this will be approximated closely by the corresponding number of zeta (non-trivial) zeros to nπ.

In the table below, I show the cumulative number of odd factors of all numbers up to n = 500 (starting with 50 and increasing in uniform blocks of 50).

Then I also show the corresponding number of trivial zeros to 500π (starting with 50π and increasing in uniform blocks of 50π).

And as one can see the approximation steadily improves as we ascend both number scales..

For example up to n = 100, we have 172 cumulative odd factors involved. And the corresponding number of zeta zeros to t = 100π = 145 with a relative accuracy here of 84.3%.

And when n = 500, we have 1262 odd factors and 1132 zeta zeros to 500π. So the relative accuracy has now improved to 89.54%. And this relative accuracy slowly increases towards 100% as n becomes progressively larger.

n	Odd factors to n	Zeta zeros to t (n = t/π)
50	70	55
100	172	145
150	290	249
200	413	361
250	541	479
300	680	602
350	818	729
400	963	860
450	1111	994
500	1262	1130

If we now for example were to exclude all numbers with 3 as a factors this would mean that 2/3 of all possible divisors would be used i.e. 2, 4, 5, 7, 8, 10, 11, …

And we would then in turn consider the cumulative factors of all natural numbers (excluding those with 3 is a factor).

To illustrate I will now accumulate the factors (excluding those with 3 as a factor) up to 20. Remember all numbers must be considered as some though divisible by 3 will contain other factor(s) not divisible by 3. For example, 15 though divisible by 3, contains 5 as a factor!

So 1 as a trivial factor is excluded in this case; 2 contains one factor, i.e. 2 (1);

3 is excluded; 4 contains two factors, i.e. 2 and 4 (2); 5 contains one factor, i.e. 5 (1); 6 is excluded; 7 contains one factor, i.e. 7 (1); 8 contains 3 factors i.e. 2, 4 and 8 (3); 9 is excluded; 10 contains three factors i.e. 2, 5 and 10 (3). 11 contains 1 factor, i.e. 11 (1); 12 is excluded; 13 contains one factor, i.e. 13 (1); 14 contains 3 factors, i.e. 2, 4 and 7 (3); 15 contains one factor i.e. 5 (1); 16 contains 4 factors i.e. 2, 4, 8 and 16 (4); 17 contains one factor i.e. 17; 18 is excluded; 19 contains one factor i.e. 19; 20 contains 5 factors i.e. 2, 4, 5, 10 and 20.

So the cumulative no. of factors to 20

= 1 + 2 + 1 + 1 + 3 + 3 + 1 + 1 + 3 + 1 + 4 + 1 + 1 + 5 = 28.  

As 2/3 of all divisors are involved in this case n = t/(2π * 2/3) = t /(4π/3)

Therefore t = 4nπ/3 = (80 * π)/3 = 84 (rounded to nearest integer).

And the number of zeta zeros to 84 = 22 (which already is a fairly reasonable approximation).

Therefore in general if k = percentage of total divisors involved in accumulating factors of numbers (containing these divisors) up to n, then n = t/(2π * k) where frequency of zeta zeros to t (on imaginary scale) is calculated.

Alternatively t = 2nπk 

Finally to illustrate this further, imagine we exclude from consideration all numbers containing either 2 or 3 as a factor.

Now, 1/2 of all divisors contain 2 as a factor and 1/3 contain 3. However as half of the factors containing 3 as a factor also contain 2, those divisors containing either 2 or 3 as factors =  1/2 + 1/6 = 2/3.

Therefore the remaining divisors i.e. 5, 7, 11, 13, … = 1/3 of total divisors.

So  with k = 1/3, if we accumulate the factors of all numbers (other than those divisible by 2 or 3 up to n) then t = 2nπ/3

The accumulated total of factors of numbers (neither divisible by 2 or 3) up to 100 = 97.

And the number of zeta zeros to 200π/3 = 209.43 = 81. So the percentage accuracy here = 85.5%

Then the accumulated total number of factors (neither divisible by 2 or 3) to 200 = 241.

And the number of zeta zeros to 400π/3 = 418.86 = 213. And the percentage accuracy has now improved to 88.4%.