This is the second “research” thread of the Polymath15 project to upper bound the de Bruijn-Newman constant , continuing this previous thread. Discussion of the project of a non-research nature can continue for now in the existing proposal thread. Progress will be summarised at this Polymath wiki page.

We now have the following proposition (see this page for a proof sketch) that looks like it can give a numerically feasible approach to bound :

Proposition 1Suppose that one has parameters obeying the following properties:

- All the zeroes of with are real.
- There are no zeroes with in the region .
- There are no zeroes with and .
Then one has .

The first hypothesis is already known for up to about (we should find out exactly what we can reach here). Preliminary calculations suggest that we can obtain the third item provided that . The second hypothesis requires good numerical calculation for , to which we now turn.

The initial definition of is given by the formula

where

This formula has proven numerically computable to acceptable error up until about the first hundred zeroes of , but degrades after that, and so other exact or approximate formulae for are needed. One possible exact formula that could be useful is

where

and

and can be chosen arbitrarily. We are still trying to see if this can be implemented numerically to give better accuracy than the previous formula.

It seems particularly promising to develop a generalisation of the Riemann-Siegel approximate functional equation for . Preliminary computations suggest in particular that we have the approximation

where

Some very preliminary numerics suggest that this formula is reasonably accurate even for moderate values of , though further numerical verification is needed. As a proof of concept, one could take this approximation as exact for the purposes of seeing what ranges of one can feasibly compute with (and for extremely large values of , we will presumably have to introduce some version of the Odlyzko-Schönhage algorithm. Of course, to obtain a rigorous result, we will eventually need a rigorous version of this formula with explicit error bounds. It may also be necessary to add more terms to the approximation to reduce the size of the error.

Sujit Nair has kindly summarised for me the current state of affairs with the numerics as follows:

—

- We need a real milestone and work backward to set up intermediate goals. This will definitely help bring in focus!
- So far, we have some utilities to compute zeroes of with a nonlinear solver which uses roots of as an initial condition. The solver is a wrapper around MINPACK’s implementation of Powell’s method. There is some room for optimization. For example, we aren’t providing the solver with the analytical Jacobian which speeds up the computation and increases accuracy.
- We have some results in the output folder which contains the first 1000 roots of for some small values of , etc. They need some more organization and visualization.

We have a decent initial start but we have some ways to go. Moving forward, here is my proposition for some areas of focus. We should expand and prioritize after some open discussion.

- Short term Optimize the existing framework and target to have the first million zeros of (for a reasonable range of ) and the corresponding plots. With better engineering practice and discipline, I am confident we can get to a few tens of millions range. Some things which will help include parallelization, iterative approaches (using zeroes of to compute zeroes of ), etc.
- Medium term We need to explore better ways to represent the zeros and compute them. An analogy is the computation of Riemann zeroes up to height . It is computed by computing the sign changes of (page 119 of Edwards) and by exploiting the speed up with the Riemann-Siegel formulation (over Euler-Maclaurin). For larger values of , I am not sure the root solver based approach is going to work to understand the gaps between zeroes.
- Long term We also need a better understanding of the errors involved in the computation — truncation, hardware/software, etc.

## 102 comments

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2 February, 2018 at 9:01 pm

Terence TaoOne medium-term numerical goal which would be helpful would be to gather some data regarding the accuracy of the Riemann-Siegel type approximations. When , the error term in these approximations is about times the main term; hopefully a similar accuracy is present for as well. If we are lucky, that is enough accuracy that it will be justifiable (for a preliminary pass at least) to pretend that the Riemann-Siegel expansion is in fact exact, and then try to control zeroes of that expansion at large height.

3 February, 2018 at 4:45 am

KMI ran some quick tests using the original Riemann Siegel eqns (as implemented in Gourdon’s work), and checked the effect of using the first few error terms. As expected, the difference is significant and even the estimates of the smaller zeroes become much better. It would be great to have corresponding first few error terms for Ht.

Also, there seem to be practical considerations with the root finding algorithms using initial guesses. If there are many close roots in an interval, then it’s difficult to trivially control which root will the algorithm converge to. Which results in occasionally the algorithm converging to the same root from multiple nearby guesses, and missing a few in between. It can be dealt with by carefully and iteratively expanding the interval where a root should be searched for, till the algorithm finds a sign change.

But then it’s unclear whether this approach is less or more expensive than the original approach of simply evaluating H_t at close enough regularly spaced intervals and then counting the sign changes.

Will try out the counting approach between two large T heights for some t values.

3 February, 2018 at 4:49 am

KMOh, just noticed you have mentioned the root solving limitations in the last section.

3 February, 2018 at 12:17 am

meditationataePerhaps Gauss-Legendre quadrature, a method in the family of Gaussian

quadrature methods, could be useful in obtaining a few high-accuracy values of H_t(z), for example to validate other computation methods. A reference at Wikipedia: https://en.wikipedia.org/wiki/Gaussian_quadrature .

I experimented with trying to compute H_0 (2*z_100), where z_100 is the

imaginary part of the 100th zeta zero, using the first exact formula. Using 5120 intervals of equal lengths, a length of 1/1600 as it turns out, and a 100-point Gauss-Legendre method on each interval, the command in PARI/gp returned a numerical value of roughly 1 E-156 after 31 minutes. For comparison, H_0 (2*z_100 + 1) returns roughly 9.9 E-60, much larger than at 2*z_100; this is to be expected if the integration of the very oscillatory H_0 function is highly accurate, since 2*z_100 is a zero of H_0 .

I should add that I’ve only used Gaussian quadrature methods on a couple of occasions, and that it’s new to me…

3 February, 2018 at 8:25 am

meditationataeI find that the Gauss-Legendre method using a 400-point rule and 32 intervals of length 1/16 delivers the same accuracy as above in computing H_0(2*z_100), but in only 15 seconds. This includes the time spent initializing the tables of weights and abscissas for the 400-point Gauss-Legendre rule.

For illustration and reference, the PARI/gp command entered is copied below:

intnumgaussinit(400);

s = 0.0; x=2*z100 ;

for(J=0,32, s=s+intnumgauss(X=J/16,(J+1)/16, Phi(X)*cos(x*X),));

s

3 February, 2018 at 4:57 am

AnonymousA possible approach for such approximation to is to use its Taylor series in powers of

Where

Since , and is entire (in both and ), the Taylor coefficients (which are entire functions) are super exponentially decaying, so in order to estimate by truncating its Taylor series, the required number of terms should depend on the required precision (say), and and is expected to be roughly proportional to .

It seems that the asymptotic approximations of (via their defining integrals) should be similar to that of .

3 February, 2018 at 6:03 am

AnonymousFrom the integral representation of (or directly from the backwards heat equation for ), it follows that

,

So it seems that the asymptotic expansions of the coefficients can be computed recursively from that of .

3 February, 2018 at 9:43 am

AnonymousThis approach can be used for each given disk with an explicit analytic approximation of and with explicit (uniform) upper bound on the magnitude of the approximation error on the given disk, to derive for each (complex) an explicit analytic approximation to with an explicit (uniform) bound on the magnitude of the approximation error on the given disk as follows:

Using Taylor series expansion for , with the coefficients

It is natural to approximate them (replacing by ) and truncate the resulting series to terms, giving

where

Note that and the coefficients can be computed recursively (for ) using the same recursive relation of above.

The bound follows from Cauchy estimates for the coefficients and (as derivatives of ), giving

Where is an upper bound on in the given disk which is upper bounded by .

Remark: It seems that for each given , one may take the disk center , but the disk radius and the number of terms should be optimized (to minimize ).

3 February, 2018 at 9:55 am

AnonymousCorrection: in the expressions above for the coefficients , the repeated derivatives should be

3 February, 2018 at 10:33 am

AnonymousI just realized that unfortunately, Cauchy derivatives estimates used for the bound are not sufficient (the second sum diverges). The approximant may (hopefully) still be useful, but it seems that a better local estimates for the of magnitude of the derivatives of are required for a rigorous bound.

3 February, 2018 at 3:22 pm

meditationataeMichael Rubinstein has a gzipped file with the first 35,161,820 zeros of the Riemann zeta function, as described in the Readme at: http://oto.math.uwaterloo.ca/~mrubinst/L_function_public/ZEROS/Readme.txt . The zipped file zeros_0001_35161820.gz is in the directory: http://oto.math.uwaterloo.ca/~mrubinst/L_function_public/ZEROS/DEGREE_1/ .

He’s also the author of “l-calc”, an L-function calculator, described in

“Software” at http://oto.math.uwaterloo.ca/~mrubinst/L_function_public/L.html

In the past, I’ve succeeded in building lcalc from source code using “make” in Linux, but today the “build” process encountered a bug/error.

3 February, 2018 at 5:23 pm

David Bernier (@doubledeckerpot)I changed one line in the Makefile, so as to use the System/Distribution C++ compiler originating from Red Hat. With that modification, the build process using the “make” command completed without a hitch, and as administrator/root, I installed lcalc.

lcalc is very fast: for roughly 12 significant figures, it computes one million Riemann zeta zeros in just over 5 minutes. I’m now running lcalc to compute the first 10 million zeros.

The command and output at the prompt for 1 million zeros were:

$ time lcalc –zeros=1000000 > /dev/null

real 5m8.578s

user 4m55.946s

sys 0m0.758s

4 February, 2018 at 7:14 am

David Bernier (@doubledeckerpot)The command:

$ time lcalc -zeros=10000000 > /dev/null ,

which computes the first 10,000,000 zeros of zeta, then redirects the output to “trash” instead of displaying it on the terminal, took about 2 hours to finish.

3 February, 2018 at 8:47 pm

Sujit NairTerry, the right side of doesn’t make sense if .

It took some fiddling around (and going over the asymptotics page) for me to realize that .

Maybe you should define before the expression for .

3 February, 2018 at 9:18 pm

Terence TaoActually to evaluate one has to evaluate at two values of , and (see the equation before the definition of ). So I would prefer to think of as an independent variable rather than as a quantity defined in terms of and .

3 February, 2018 at 9:27 pm

Sujit NairIf that is the case, shouldn’t the summation limit in be in terms of . Right now, it is ?

I am trying to recover Riemann-Siegel for and it is stumping me.

3 February, 2018 at 9:42 pm

Sujit NairAh. I think I got it.

So you mean in the summation limit on the right side?

3 February, 2018 at 9:50 pm

Sujit NairActually, this won’t work. For example, computing would require computing . In this case, .

4 February, 2018 at 8:39 am

David Bernier (@doubledeckerpot)In my PARI/gp code for , I put where your expression had : https://terrytao.wordpress.com/2018/01/27/polymath15-first-thread-computing-h_t-asymptotics-and-dynamics-of-zeroes/#comment-491988 .

Hopefully, this substitution was correct for the purpose of computing .

4 February, 2018 at 8:56 am

Terence TaoYes, one should have instead of there. I’ll edit the text accordingly.

4 February, 2018 at 5:56 am

KMOn running the approx functional eqn (Ht_AFE for brevity) for t=0 between heights T=50000 to 100000 (H_0 has 34517 actual zeros in this range), we get these results

T range———–detected—not—–actual—-% not detected

50000-60000——6613——58—–6671——-0.87%

60000-70000——6773——30—–6803——-0.44%

70000-80000——6869——50—–6919——-0.72%

80000-90000——6969——48—–7017——-0.68%

90000-100000—-7053——-54—-7107——–0.76%

Total—————34277——240—34517——-0.70%

Whenever zeros were not detected they came in pairs, for eg.

Ht_AFE root–Actual root

54677.865—-54677.8535

54680.645—-54680.66596

——————54681.9331

——————54682.27354

54684.085—-54684.04412

54685.165—-54685.18586

Instead near the undetected ones, Ht_AFE came close to the origin and flipped back forming a local minima or maxima, which I think is due to these zeros being close by. In a plot of actual H_0, its likely the function just crosses the origin before going back.

The computations are being done using a multiprecision library mpmath using 30 significant digits, hence I checked if increasing the precision in this region to 100 digits would help. Using a bisection solver

Solving Ht_AFE between 54679.0 and 54681 -> 54680.642507114… as expected

Solving Ht_AFE between 54681.0 and 54683 -> 54681.0000000000…

Solving Ht_AFE between 54680.65 and 54684.08 -> 54680.650000….

the last two instances showing the solver sticks to the endpoint of the given interval. Similar behavior was observed at another such pair.

I will now run the same exercise at T>2000000, and check the behavior there. As indicated by the AFE formula, Its likely that beyond a certain height such pairs will be either rare or non-existent, as Ht_AFE simulates H_t with higher precision.

4 February, 2018 at 10:02 pm

Sujit NairKM, mpmath is a good find!

I assume you are still using the root solver to detect, right?

5 February, 2018 at 1:14 am

AnonymousFrom a numerical viewpoint, If there is a cluster of closely packed zeros of a function , a fast local root solver based on Newton-Raphson algorithm () is expected to show such erratic behavior (because the increment can be very large if is very close to a local maximizer of on the real axis – inside the zeros cluster).

A possible solution is to limit the increment magnitude

to be at most (say), and then to verify that (e.g. by dividing the increment by a suitable power of ). The resulting monotonicity of ensures convergence of to a local minimizer of (on the real axis). If complex increments are allowed, then (obviously) if is analytic, any local minimizer of must be a zero of .

5 February, 2018 at 8:39 am

AnonymousI think using root solving approaches with closely packed zeroes is feasible if done carefully, as you laid out. What also should be considered while doing this is that the zero density gradually increases with T, and the gaps between zeroes can sometimes be much smaller than average gaps in that region.

5 February, 2018 at 5:23 am

KMNo, this is based on evaluating the function for each z with step size of 0.01, and then counting the sign changes..the intervals have to become finer for larger heights (I am using 0.002 for the T>2 mn exercise). Currently the zero location is taken as the midpoint of an interval where a sign change took place.

The bisection solver was only used to investigate further where a zero pair had been missed. I think such solvers in general can also be used to find the exact roots in the intervals with detected sign changes (thus improving upon the midpoint estimate). The counting method and the solver approach are then complementary.

5 February, 2018 at 8:06 am

Sujit NairGot it. Is there a branch for this on github? I didn’t see any after a “git fetch”.

5 February, 2018 at 8:43 am

AnonymousWell these were some adhoc loops, repeatedly calling the AFE function and recording the output on my PC. I will upload whatever I have as a separate branch.

5 February, 2018 at 8:07 am

AnonymousPlatt’s recent paper (Math. comp. 86(2017), 2449-2467)

https://doi.org/10.1090/mcom/3198

Describes an efficient (using FFT) algorithm for the computation of more than first Zeta zeros with (rigorous) absolute precision of . Is it possible to extend this algorithm for a large-scale, high-precision computation of zeros?

5 February, 2018 at 8:56 am

AnonymousThanks for the link. Well having a FFT version to compute the AFE at many simultaneous points would be cool. The single point evaluations have become quite slow at larger heights.

4 February, 2018 at 5:52 pm

David Bernier (@doubledeckerpot)I’m trying to set up a numerical integration expression for the second exact formula in the post for , where is the first non-trivial zero of zeta.

I tried the same expression with realprecision = 150, and with realprecision = 200. I get respectively roughly 4 e-29 and 1 e-38.

This puzzles me, and I’m wondering if it could be that the integrand takes on large values, but I haven’t looked into that.

For reference, the PARI/gp code, along with the output for realprecision = 200 digits are copied below.

z=2*z1;

t=0.0;

thet=Pi/8 – 1/(4*z);

K=0.0;

for(Y=1, 80, b=z-9*I;

beta=Pi*Y^2;

K=K+2*Pi*Pi*(Y^4)*sum(J=0,64, intnumgauss(X=J/16, (J+1)/16, exp( t*(X+I*thet)^2 – beta*exp(4*(X+I*thet)) + I*b*(X+I*thet)), ) ) ;

b = z-5*I;

K=K – 3*Pi*(Y^2)*sum(J=0,64, intnumgauss(X=J/16, (J+1)/16, exp( t*(X+I*thet)^2 – beta*exp(4*(X+I*thet)) + I*b*(X+I*thet)), ) ) );

print( real((K+conj(K)))/2)

Output:

1.08 E-38 // realprecision = 200 digits

4 February, 2018 at 7:57 pm

David Bernier (@doubledeckerpot)I used the first formula for after equation (2) [ 1st thread ] to evaluate numerically .

Among the (numerical) expressions in the first thread springing from (2) to (3), it’s the one with value closest to zero.

Phi(X) = sum(Y=1, 50, (2*Pi*Pi*(Y^4)*exp(9*X) -3*Pi*(Y^2)*exp(5*X))*exp(-Pi*(Y^2)*exp(4*X)))

intnumgaussinit(800); // 800-point Gauss-Legendre method

realprecision = 154 significant digits

For the PARI/gp numerical integration code:

z=2*z1;

t=0.0;

thet=Pi/8 – 1/(4*z);

K=0.0;

for(J=0, 256 , K=K+intnumgauss(X=J/128, (J+1)/128, Phi(X+I*thet)*exp(I*z*(X+I*thet)) ));

print( real((K+conj(K)))/2)

output:

1.809 E-127

5 February, 2018 at 1:33 am

David Bernier (@doubledeckerpot)It seems at first glance as though the real/imaginary parts of the integrand might decay more rapidly when than when …

Below, there is some data on the real part of the integrand for the choice :

z=2*z1;

t=0.0;

thet=0.0;

for(J=0, 16 ,X=J/16.0;

K=real(Phi(X+I*thet)*exp(I*z*(X+I*thet)) );

print(X,” “, K))

0.00000000000000000000 0.4466969004671234441

0.06250000000000000000 -0.07506099482654958609

0.1250000000000000000 -0.2247326399152970553

0.1875000000000000000 0.05928707622660076264

0.2500000000000000000 0.02137269899984563796

0.3125000000000000000 -0.004075283488938568051

0.3750000000000000000 -0.0001521228758848166709

0.4375000000000000000 1.281421125067103168 E-5

0.5000000000000000000 3.364341936999641644 E-10

0.5625000000000000000 -3.299802452045272008 E-10

0.6250000000000000000 4.775197762416083978 E-14

0.6875000000000000000 3.527315493478114852 E-18

0.7500000000000000000 -4.571153262716285137 E-24

0.8125000000000000000 -1.053342402347743622 E-31

0.8750000000000000000 3.104515713884114793 E-41

0.9375000000000000000 1.738363069214473369 E-54

1.000000000000000000 -5.101940498958935872 E-70

? z

%101 = 28.26945028346938758

5 February, 2018 at 8:05 am

Sujit NairKM, David,

Here is some feedback from Fredrik Johansson (main author of mpmath) to compute with arbitrary precision.

——————-

I was going to recommend implementing the function H_t using Arb (http://arblib.org/). This will be a bit more complicated to code for than mpmath, but it should be much faster, and it will allow obtaining rigorous error bounds.

I recently added code for rigorous numerical integration which is described here

This is using a rigorous, adaptive version of Gauss-Legendre quadrature which should be quite efficient for the type of integrand defining H_t, although there could be some subtleties in practice.

5 February, 2018 at 8:59 am

AnonymousThanks, will try it out.

5 February, 2018 at 9:22 am

David Bernier (@doubledeckerpot)Thanks, this is encouraging news.

5 February, 2018 at 11:23 am

David Bernier (@doubledeckerpot)An adaptive version of Gauss-Legendre quadrature seems particularly appropriate, at least for , since the integrand decreases extremely rapidly on e.g. the interval [0,2],

cf.:

https://terrytao.wordpress.com/2018/02/02/polymath15-second-thread-generalising-the-riemann-siegel-approximate-functional-equation/#comment-492141 .

5 February, 2018 at 9:45 am

David Bernier (@doubledeckerpot)There are two references regarding the Riemann-Siegel formula that I’d like to mention as possibly relevant: the first is Chapter 7 of Edwards’ book, and the second is Chapter 3 of Glen Pugh’s Master’s Thesis at the University of British Columbia, available from the link https://web.viu.ca/pughg/thesis.d/masters.thesis.pdf .

5 February, 2018 at 11:50 am

Terence TaoThanks for this! I am in the process of extending the Riemann-Siegel formula to and trying to compute the next term in the asymptotic expansion. I now see that the main term should only have an accuracy of (I had previously thought it was ) and so a first correction term will probably be needed (which would improve the accuracy to a much more tolerable ). It is unfortunately a bit of a mess though and I have already caught a number of sign errors etc. in my calculations, but Pugh’s thesis will be a good cross-check. (I have some partial computations on the wiki at http://michaelnielsen.org/polymath1/index.php?title=Asymptotics_of_H_t right now, but it is not yet completed.)

5 February, 2018 at 1:49 pm

Gil KalaiA side qiestion. Looking at the (impressive) asymptotic expansions and error estimations I wonder if this part could also be automated (a la Zeilberger-Wilf etc.). and not just the numerical work.

6 February, 2018 at 12:24 am

KThis sounds very interesting, could you please elaborate a bit more on this?

5 February, 2018 at 3:37 pm

Terence TaoOK, my preliminary computations suggest that a refined asymptotic expansion for is

[EDIT: This should be ] where

where , are integers with (for instance one can set if one wishes), one uses the standard branch of the logarithm throughout, and is the function

This refines the previous approximation, which was essentially . Roughly speaking, the first term should be of size , the second term should be of size (and become dominant once is significantly larger than ), and the final term should be of size , with the error being of the order of (plus perhaps another error of size that becomes relevant when is much larger than ).

The function looks strange, but can also be defined as the unique solution to the functional equations

and is also given by a contour integral

where is for instance the contour for .

The formula is messy, but can be simplified a little (up to lower order terms) by using the Stirling approximation and collecting some terms. An interesting feature is that one can modify and by without making significant changes to the final approximation, although some of the individual terms in that approximation would change. (This is one way to check that all the signs etc. in the above approximation are accurate.) If one sets and then one essentially recovers the classical Riemann-Siegel asymptotics (after extracting the main term of the remainder and discarding the rest).

I will try to write up a rigorous proof of this approximation with an error term (using notation with unspecified constants for now, since putting in explicit constants, while ultimately necessary, will make the derivation far more messier). (I started the computations on the wiki but they are unfinished and not properly organised, I will probably end up rewriting it.)

5 February, 2018 at 4:31 pm

AnonymousAnother function is already used in the definition of .

[Good point; I’ve renamed it for now. -T.]6 February, 2018 at 6:16 am

AnonymousIn the sub-page on asymptotics, the letter denotes also the imaginary part of .

[Changed to – T.]5 February, 2018 at 7:31 pm

David Bernier (@doubledeckerpot)I started writing the various formulae , mainly as functions of , in PARI/gp.

It finds a sign change in near , , which looks questionable.

I figure I’ll take another look at these formulae later.

Function definitions:

ga =

(S)->S*(S-1)*exp(-S*log(Pi)/2)*gamma(S/2)/16

gb =

(S)->S*(S-1)*exp((S-1)*log(Pi)/2)*gamma((1-S)/2)/16

M =

(S)->floor(sqrt(imag(S)/(2*Pi)))

A =

(S)->ga(S)*sum(Y=1,M(S),exp((t/16)*(log((S+4)/(2*Pi*(Y^2)))^2))*exp(-S*log(Y)))

B =

(S)->gb(S)*sum(Y=1,M(S),exp((t/16)*(log((5-S)/(2*Pi*Y^2))^2))*exp((S-1)*log(Y)))

Phi =

(Z)->2*Pi*(cos(Pi*((Z^2)/2-Z-1/8))/cos(Pi*Z))*exp(I*(Pi/2)*(Z^2)-(5/8)*Pi*I)

C =

(S)->ga(S)*gamma(1-S)*(exp(-I*Pi*S)/(2*Pi*I))*exp((S-1)*log(2*Pi*I*M(S)))*exp(-t*(Pi^2)/64)*Phi(S/(2*Pi*M(S)*I)-M(S))

E =

(S)->A(S)+B(S)+C(S)

H =

(Z)->E(I*real(Z)/2+(1-imag(Z))/2)

5 February, 2018 at 8:53 pm

David Bernier (@doubledeckerpot)Could there be a sign error in ?

(for my functions based on Terry’s calculations), i.e.

Changing from to improves accuracy,

at least some of the time :

E3(S) = A(S)+B(S)-C(S)

H3(Z) = E3(I*real(Z)/2+(1-imag(Z))/2)

Calculations:

? abs(zeta(1/2+I*74919.075161121))

= 1.44654472644 E-9

? abs(H3(2*74919.075161121))/abs(H(2*74919.075161121))

= 0.000834063509455

I also checked that two zeros of that were missed in [440, 444] via sign changes in the old formula, without remainder, are found when using the H3 above, near twice the zeta zeros: 220.7 and 221.45 .

6 February, 2018 at 7:15 am

AnonymousIs there a Riemann-Siegel integral formula for every $H_t$?http://mathworld.wolfram.com/Riemann-SiegelIntegralFormula.html

8 February, 2018 at 7:58 pm

David Bernier (@doubledeckerpot)Glen Pugh writes on one of his web pages:

“Gram’s Law is the statement that zeros of Z(t) alternate with the Gram points”, references at: https://web.viu.ca/pughg/RiemannZeta/RiemannZetaLong.html and https://en.wikipedia.org/wiki/Riemann%E2%80%93Siegel_theta_function .

For , could there be an analogue of Gram’s Law and/or an analogue of the Riemann-Siegel function ?

12 February, 2018 at 3:25 pm

David Bernier (@doubledeckerpot)Could there be an analogue for the of the Hardy function ?

The function can be used with Euler-Maclaurin summation to get very accurate values of the zeta zeroes:

https://web.viu.ca/pughg/RiemannZeta/RiemannZetaLong.html#ZetaFunction

Also, Edwards Section 6.4 of Chapter 6.

5 February, 2018 at 4:55 pm

AnonymousDear Terry,

There are smoothed approximate functional equations that are exact and provide much better error terms (for instance one can select Gaussian kernels e^{-x^2} in the smooth weights). See for instance the method of proof in https://arxiv.org/pdf/math/0111312.pdf (the argument is folklore). Cutting such a smoothed approximate functional equation at N = \sqrt{T} \log T gives an excellent error term at the price of a slightly longer computation (by a factor of \log T). Is there a particular reason why the same technique wouldn’t go through in the case of H_t ?

5 February, 2018 at 6:41 pm

Terence TaoThere are likely formulae like this – the exact formula in the third display of this blog post is something of this sort, actually (particularly if one applies a smooth splitting of the contour from to rather than a rough one), but I think these formulae will be slower to compute, even if they are slightly more compact in form. The error term in the Riemann-Siegel formula has an asymptotic expansion whose first few terms decay extremely fast (each term in the expansion is about the size of the previous one), and are far easier to compute than the additional terms one would need in the smoothed formula.

One way to think about it is this: Poisson summation suggests that any summation of in the range can be converted to a summation of in the range , up to some tractable Gamma factors. Naively, the former sum takes about terms to compute and the latter takes to compute. One can split the Poisson summation in various ways, but the most efficient thing we can really do is to sum directly for up to and then apply Poisson summation beyond that range, for computations; the rough cutoff introduces an error term but we fortunately have a good asymptotic expansion for it. A smooth cutoff removes this small error term, but at the cost of placing more of the summation on the inefficient side of the Poisson summation formula, and does not seem to be a win in practice (certainly none of the numerical literature on RH verification use these sorts of formulae). Here by the way is a place where numerics diverges from theory, since when computational complexity (or implied constants) are not an issue it is certainly advantageous to smooth sums whenever possible.

6 February, 2018 at 1:40 am

David Bernier (@doubledeckerpot)I decided to count zeta zeros via sign changes in the real part of a function I defined in PARI/gp that I called H3, based on Terry Tao’s Riemann-Siegel formula calculations.

It’s set to look for zeros of zeta with imaginary part below 16,000.

The output is being written to a file, so that I can use it later.

The command that is now running is copied below, and already up to height 4000.0, the count of zeta zeros is complete.

last = 1.0;

count=0;

for(J=1000,1600000,

new = real(H3(J/50.0));

if( (new*last)<0,

count=count+1;

write(zeros_to_16k, count," ", J/100.0));

last = new);

6 February, 2018 at 3:10 am

David Bernier (@doubledeckerpot)The agreement with tables of zeta zeros using H3, with instead of , is excellent. It seems likely either I made a mistake in defining , or there was a mistake (sign error) in the complex equations…

H3 locates 17423 zeta zeros up to height 16000.0, and tables of zeta zeros say there are in fact 17425. It appears H3 didn’t locate zeros 5229.198557199 and 5229.241811259.

Through re-scaling, this appears as two “very near misses” with respect to sign changes:

? for(J=1045820, 1045860, print(J/200.0,” “, real(H3(J/100.0))*(10^1778)))

[…]

5229.18000000 0.0191439033097

5229.18500000 0.0151796159916

5229.19000000 0.0117431220497

5229.19500000 0.00883338972137

5229.20000000 0.00644920191919

5229.20500000 0.00458915818597

5229.21000000 0.00325167668798

5229.21500000 0.00243499624578

5229.22000000 0.00213717840158

5229.22500000 0.00235610952246

5229.23000000 0.00308950293842

5229.23500000 0.00433490111440

5229.24000000 0.00608967785543

5229.24500000 0.00835104054402

5229.25000000 0.0111160324088

[…]

6 February, 2018 at 5:40 am

David Bernier (@doubledeckerpot)The 17,425th zeta zero, as determined by a sign change in the real part of H3, is off by only 0.000074 .

6 February, 2018 at 8:45 am

Terence TaoIt is entirely plausible that I made a sign error. I will not have time to do this until much later today, but I suspect that I managed to forget about the minus sign in the formula for the remainder R(s) in the Riemann-Siegel formula https://en.wikipedia.org/wiki/Riemann%E2%80%93Siegel_formula .

6 February, 2018 at 9:28 am

David Bernier (@doubledeckerpot)Ok, thanks! I’m computing another dataset of zeta zeros up to height 64000 (using H3), and writing the output to a file. I’m not sure what the best tests might be for “quality control”, i.e. that the data is what it should be theoretically.

6 February, 2018 at 11:23 am

KMI too have started a new search for H_0 zeroes with the new eqn (between heights 50k to 100k), and the initial results show a dramatic improvement.

If by quality control you mean number of zeroes, there are 38379 H_0 zeroes from 0 to 64000, so it can be compared against that. Also with the zeroes estimated to two decimal places, the average gap between the estimated and actual zeroes near the 50k mark is close to 0.005.

6 February, 2018 at 4:51 pm

David Bernier (@doubledeckerpot)The program found, via sign changes in the real part of “H3”, 83816 of the 83818 non-trivial zeta zeros with imaginary part between 0 and 64,000 .

I plan to use something close to the secant method , https://en.wikipedia.org/wiki/Secant_method , to locate with higher precision the sign changes in the real part of “H3” …

6 February, 2018 at 6:40 pm

David Bernier (@doubledeckerpot)The approximate zeta zero, using with remainder term, at 63999.4809 (via sign change) is off by 0.0004 from the actual zeta zero at 63999.4813 .

6 February, 2018 at 7:00 pm

KMSo the new eqn has detected all sign changes and hence all 15142 zeroes between T=50k and 70k (in terms of zeta zeroes, T between 25k to 35k), hence I have stopped the benchmarking exercise on my side.

Now running computations for t=0.5 and T>2 million. As predicted by theory, we should start seeing both a decrease in average zero spacing and standard deviation of zero spacings, as the zeroes drift downward and become more regularly spaced.

At this point, we can start dividing t ranges between ourselves. For eg. I am planning to explore and generate statistics for t between 0.48 and 0.5 for a start.

7 February, 2018 at 3:46 am

David Bernier (@doubledeckerpot)For zeta zeroes with heights between 32k and 64k, I looked at the errors in the approximations using with the remainder term, as compared to tables of zeta zeroes.

The approximate zeta zeroes are based on where the real part of with remainder, changes sign. My program finds that the average error is 0.000113 .

7 February, 2018 at 5:14 am

David Bernier (@doubledeckerpot)Sujit, KM,

I’ll send Sujit my Github handle when I find it…

I’m not a Python programmer, but I can try to run your programs, to do a share of the zeros computations.

8 February, 2018 at 10:18 am

KMBy the way, checking some statistics for t=0.5 and upto first 30000 zeroes

T N_t(T) # detected zeroes of H_t

10000 4519.665 4521

20000 10142.321 10143

30000 16181.355 16182

pretty good match so far.

Also I just realized the (t/16)log(T/4PI) factor, and hence the drift is quite small. So even at T=10^15, we can expect just 1 more zero at t=0.5 compared to at t=0.

Also, as you had noticed, at t=0 there were 2 zeroes undetected with the ABC eqn at 2*5229.198 and 2*5229.242, but it seems at t=0.5, ABC has detected the corresponding zeroes for those ones too.

n H_0.5 zero zeta zero H_0 zero

4765 10457.4445 5229.1985 10458.3971

4766 10458.9976 5229.2418 10458.4836

8 February, 2018 at 3:37 pm

David Bernier (@doubledeckerpot)https://kommonmann.wordpress.com/ :

That’s interesting what you have about . At some point, we might try doing an exact count of zeroes of in a rectangle, something like a numerical contour integral, as far as I know, similar to what was done before “Turing’s trick” was found.

It could be something to think about.

6 February, 2018 at 2:33 pm

Terence TaoI have located the sign error (I had reversed the orientation of a contour). The correct approximation is indeed . Thanks for pointing it out! I am working on replacing the wiki page with corrected (and more logically structured) asymptotics.

[Edit: wiki page now replaced. -T]8 February, 2018 at 8:47 am

meditationataeI put together in a script file h3zeros16k.gp the PARI/gp commands and function definitions for the zeta zeros with height below 16000, based on the code above. Unfortunately, the script won’t run: it prints nothing.

But the examples scripts from PARI/gp Headquarters do work, for example lucas.gp to perform Lucas-Lehmer tests.

GP script files can be compiled with the gp2c compiler package. This worked for the example GP script lucas.gp .

6 February, 2018 at 7:59 am

PGniewekDear Terry, a minor nitpick: there appears to be an error in the second equation below the box containing Proposition 1 in this post: the argument of the rightmost exponential function is and I think it should be . Cheers.

[Corrected, thanks – T.]6 February, 2018 at 10:53 pm

Terence TaoHere is a very crude computation that nevertheless indicates that it should be possible to obtain a reasonable value of for which one can hope to establish the zero-free property whenever and for usable choices of .

I’ll pick and ; this will lead to a bound which is better than 0.5. I’ll pretend that the A+B approximation is exact, thus

with . To make non-zero, it suffices to make the first term in the second sum larger in magnitude than the sum of all the other magnitudes put together. Cancelling some factors, and using the approximation , this amounts to asking that

We approximate as and similarly with replaced by , so we reduce to showing that

In order to do a cheap computation, I crudely bounded , and then extended the sum to all , giving the criterion

.

Setting , it seems that we can make this criterion for or so, which is only barely within range of what is numerically verified for RH. One can probably do better if one doesn't bound things quite as crudely. It's also possible that when , the sum for is quite rapidly convergent, and one could numerically verify the absence of zeroes for for quite a large range of , e.g. , and then use something like the above argument to handle all larger values of .

7 February, 2018 at 6:31 am

AnonymousIn proposition 1 [of the wiki page on asymptotics], the definition of (which appears in both sides) and the exponent in the LHS, is not sufficiently clear.

7 February, 2018 at 8:20 am

Terence TaoI am not quite sure what you are referring to here. There is a definition of the complex exponential , but this is not a definition of itself, which is a dummy variable of integration.

7 February, 2018 at 8:36 am

AnonymousI was confused because the letter in the exponent of appears so tiny and unrecognizable (even under magnification) as the letter .

7 February, 2018 at 8:45 am

AnonymousThe letter in the exponent seems to be distorted(!) in that wiki page (but not here).

7 February, 2018 at 8:29 am

Terence TaoSet for the main term of . A plausible objective would be to show that for and and , the quantity does not vanish for all larger than some reasonable threshold . This is equivalent to not vanishing; if I did not make any computational errors, this function can be written explicitly as

In the limit this quantity converges to ; the question is how fast it converges and at what point it stays a significant distance away from zero (so that we would be able to replace by the full and still get a zero free region). This looks like something that could be amenable to numerics, given that the sums here should converge reasonably quickly once is moderately large. (In particular, comparative plots of , , and for smallish values of could be quite revealing as to the feasibility of this strategy.)

[Edited to restore some missing denominators of . -T]7 February, 2018 at 9:07 am

David Bernier (@doubledeckerpot)I see that the LHS of the inequality, for , has , but that the RHS for 2<= n <=N has an extra as in: …

7 February, 2018 at 1:25 pm

Terence TaoOops, good catch! I think I have restored all the missing denominators of now. (Also I had a which should also have been a .)

8 February, 2018 at 8:52 am

KMSome results on (A+B-C)/B0 and (A+B)/B0 computations for t=0.4, y=0.4. Two exercises were run,

one with x range with step size 1.0,

and the other a shorter one with single point evaluations with x = powers of 10

val below refers to the modulus |(A+B)/B0|

x range——minval–maxval–avgval–stdevval

upto 100k—0.179—4.491—1.077—0.455

100k-200k—0.488—3.339—1.053—0.361

200k-300k—0.508—3.049—1.047—0.335

300k-400k—0.517—2.989—1.043—0.321

400k-500k—0.535—2.826—1.041—0.310

500k-600k—0.529—2.757—1.039—0.303

600k-700k—0.548—2.728—1.038—0.296

more granular upto x=6000 (min value at x=29)

x range——minval–maxval–avgval–stdevval

upto 1000—0.179—4.074—1.219—0.782

1000-2000—0.352—4.403—1.164—0.712

2000-3000—0.352—4.050—1.145—0.671

3000-4000—0.338—4.174—1.134—0.640

4000-5000—0.386—4.491—1.128—0.615

5000-6000—0.377—4.327—1.120—0.599

x+yi————-(A+B-C)/B0—-(A+B)/B0

10^2 + 0.4i—-0.80-0.35i—-0.25-0.18i

10^3 + 0.4i—-0.50+0.44i—-0.39+0.51i

10^4 + 0.4i—-0.52+0.01i—-0.52-0.01i

10^5 + 0.4i—-1.28-0.15i—-1.28-0.16i

10^6 + 0.4i—-1.29-0.76i—-1.29-0.76i

10^7 + 0.4i—-1.54+0.13i—-1.54+0.13i

10^8 + 0.4i—-1.47-0.09i—-1.47-0.09i

10^9 + 0.4i—-1.23+0.18i—-1.23+0.18i

10^10 + 0.4i—-0.86+0.07i—-0.86+0.07i

10^11 + 0.4i—-1.09-0.07i—-1.09-0.07i

10^12 + 0.4i—-0.996+0.09i—0.996+0.09i

10^13 + 0.4i—-0.953+0.045i–0.953+0.045i

It seems (A+B)/B0 starts staying away from zero after a very short range, but converges to 1 quite slowly.

9 February, 2018 at 9:53 am

Terence TaoThis is very promising numerical data! It suggests for instance that the C error term is largely irrelevant for or so. One could now hope to establish that for all and by doing a numerical verification up to, say, , and then for , showing that (which seems plausible from the numerical data, and can hopefully be proven by some combination of numerics and analysis) and that (e.g. by showing and ). If not only stays away from the origin, but also from the negative real axis, then the argument principle (and Rouche’s theorem) should also preclude zeroes in the region as well (one may also have to control what goes on at some large value of , e.g. , but I think this should be very easy).

9 February, 2018 at 10:39 am

AnonymousSince for any and , it is not clear why larger values of should be considered?

10 February, 2018 at 10:32 am

Terence TaoTo apply the argument principle to establish a zero-free region such as , it is not enough to know that the function is non-zero on the boundary of the region; one also must control the winding number. For very large (I conservatively put , but one can probably do much better) one can approximate quite well by whose contribution to the winding number is very well controlled (coming almost entirely from the gamma function factor). One may still have to control for say on the sides, but probably we can take small enough that this verification is easy, and for large enough perhaps the analysis estimates would suffice.

10 February, 2018 at 7:55 pm

KMSome results for t=0.4, y=0.4 for Real((A+B)/B0) and |C/B0|. Every point n+yi was evaluated, where n is a natural number (showing results till n = 0.7 million)

While it may be possible for (A+B)/B0 and C/B0 to have different behavior between these points, I think that’s unlikely since n+0.4i is effectively random for H_t, and the sample size of observations is large.

It seems Real((A+B)/B0) stays positive beyond a small height and afterwards becomes comfortably above 0.4 and keeps moving towards 1. This reflects also in |(A+B)/B0|

n————–minval—maxval

upto 1000—(0.09)—4.06

1000-2000—0.02—–4.33

2000-3000—0.15—–3.99

3000-4000—0.19—–3.99

4000-5000—0.34—–4.48

5000-6000—0.33—–4.33

6000-7000—0.35—–3.88

———————————

upto 100k—(0.09)—4.48

100k-200k—0.48—–3.32

200k-300k—0.50—–3.00

300k-400k—0.52—–2.97

400k-500k—0.53—–2.82

500k-600k—0.53—–2.75

600k-700k—0.55—–2.72

|C/B0| keeps moving towards 0 and is less than 0.1 beyond a small height.

n————–minval—maxval

upto 1000—0.058—2.125

1000-2000—0.036—0.160

2000-3000—0.028—0.095

3000-4000—0.023—0.071

4000-5000—0.020—0.060

5000-6000—0.018—0.052

6000-7000—0.016—0.045

——————————————-

upto 100k—0.002236—2.124627

100k-200k—0.001262—0.005360

200k-300k—0.000888—0.003030

300k-400k—0.000699—0.002158

400k-500k—0.000572—0.001681

500k-600k—0.000485—0.001382

600k-700k—0.000424—0.001176

As a contrast, I also ran the same exercise for t=0.05 and y=0.4. It seems for t=0.05 to show similar behavior as t=0.4, we will have to go to much larger heights as expected.

|(A+B)/B0|

n————–minval—maxval

upto 100k—0.092—9.231

100k-200k—0.172—10.001

200k-300k—0.160—10.461

300k-400k—0.185—10.177

400k-500k—0.158—9.925

500k-600k—0.183—9.826

600k-700k—0.188—9.928

Real((A+B)/B0)

n————–minval—maxval

upto 100k—(1.24)—8.89

100k-200k—(1.29)—9.98

200k-300k—(1.35)—10.34

300k-400k—(1.17)—10.16

400k-500k—(1.22)—9.91

500k-600k—(0.94)—9.64

600k-700k—(1.23)—9.85

|C/B0|

upto 100k—0.0130—2.2336

100k-200k—0.0097—0.0315

200k-300k—0.0082—0.0236

300k-400k—0.0073—0.0200

400k-500k—0.0067—0.0177

500k-600k—0.0061—0.0161

600k-700k—0.0058—0.0149

Computing H_t independently of the ABC approach at such large heights (for eg. using the integral) is quite tricky and expensive, as shown by David in some of the comments below. We may need a Euler Maclaurin type formula, which was being attempted in one of the threads some time back.

11 February, 2018 at 11:11 am

KMEvaluating H_t(x+iy) for t=0.4 and y=0.8,1.2,…4.4 upto height 50k (for integer x) gives these results,

min value of Real((A+B)/B0)

—–x range

y—-upto10k-10k20k-20k30k-30k40k-40k50k

———————————————————

0.8—-0.19—–0.52—–0.55—–0.55—–0.57

1.2—-0.38—–0.60—–0.61—–0.62—–0.64

1.6—-0.47—–0.65—–0.67—–0.67—–0.69

2.0—-0.55—–0.70—–0.72—–0.72—–0.73

2.4—-0.61—–0.74—–0.76—–0.76—–0.77

2.8—-0.67—–0.78—–0.79—–0.80—–0.80

3.2—-0.71—–0.81—–0.82—–0.82—–0.83

3.6—-0.74—–0.84—–0.85—–0.85—–0.85

4.0—-0.77—–0.86—–0.87—–0.87—–0.87

4.4—-0.80—–0.88—–0.89—–0.89—–0.89

min value of |H/B0|

——x range

y—-upto10k-10k20k-20k30k-30k40k-40k50k

———————————————————

0.8—–0.18—–0.52—–0.55—–0.56—–0.58

1.2—–0.15—–0.60—–0.62—–0.62—–0.64

1.6—–0.20—–0.66—–0.67—–0.68—–0.69

2.0—–0.35—–0.70—–0.72—–0.72—–0.73

2.4—–0.48—–0.75—–0.76—–0.76—–0.77

2.8—–0.59—–0.78—–0.79—–0.80—–0.80

3.2—–0.67—–0.81—–0.82—–0.83—–0.83

3.6—–0.74—–0.84—–0.85—–0.85—–0.85

4.0—–0.78—–0.86—–0.87—–0.87—–0.87

4.4—–0.82—–0.88—–0.89—–0.89—–0.89

As expected, the behavior seen at t=0.4,y=0.4 becomes more pronounced as y is increased, and the above values converge to 1 faster. Also, Real((A+B)/B0) never went negative in this exercise.

7 February, 2018 at 7:31 am

AnonymousIn the wiki page on asymptotics (just above “approximation for “), it seems that in the exact expression of , the second sum (unlike the first sum) is independent of .

[Corrected, thanks – T.]8 February, 2018 at 1:14 am

AnonymousIs it possible (by obtaining more information on the dynamics of ) to improve the upper bound for in proposition 1 above (e.g. by reducing the factor in this bound for certain values of and )?

8 February, 2018 at 7:00 pm

KMFor t=0.5, we see evidence of the zeros getting quite regularly spaced compared to the the zeros for t=0, even while the average zero gap remains almost the same.

T range—# of——–avg—-avg——-stdev—stdev

————-H_0.5—–zero—zero——zero—–zero

————-zeroes—-gap—-gap——gap——gap

————————H_0.5—-H_0—–H_0.5—-H_0

—————————————————————

upto 10k–4521—–2.21—–2.21—–0.65—–1.00

10k-20k—5622—–1.78—–1.78—–0.23—–0.71

20k-30k—6039—–1.66—–1.66—–0.18—–0.66

30k-40k—6310—–1.58—–1.58—–0.16—–0.63

9 February, 2018 at 12:40 am

Doug StollI just found these blogs today. For all of you looking to confirm the Ho zeros here is potentially helpful hint. For Riemann zeta zeros one can use the following for a very “quick and rather good” approximation to the kth zero. I derived and tested this myself several months ago but have not published it nor have I seen it anywhere.

zz[k] ~ 2e*Pi*A/(ProdLog[A]) +Pi/k – (4.7159-0.3538loglog(k+1))/log(k+1)

Where A = (8k-7)/8e and ProdLog [A] = W[A] whichever notation you are more used to.

It works extremely well even out to 10^24 as I compared it to the published sets of zeros. It is not perfect as it presumes monotonically decreasing spacing but it is close.

So I would expect you to look at 2X this value +/- a little. It would be very informative to see if there is a “consistent” decrement or a “simple function” from 2X towards the origin the further up the i-axis one goes.

And a caution, be careful of your step size exactly for the reason the zeros still will crowd together. You should use a decreasing test step size. To get a feel for the stats of the first 10^13 zeta zeros and particularly the ones that get very close together (and therefore have an impact on the dB-N constant I refer you to:

http://numbers.computation.free.fr/Constants/Miscellaneous/zetazeroscompute.html –

particularly to published samples of zeros explored out to 10^24. Notice that the smallest spacing in your current range of interest (to 10^12) is 0.0001008 at T = 18,403,609,310.0854 (60,917,681,409th zeta zero). The smallest you should find in your test out to 160000 should be the 97,484th zero at 73,235.205353

I have heuristically found an estimate (unpublished) over the range of zz’s found to date that the minimum spacing for zeta zero’s is larger than:

min gap > 10^{[loglog[T]^3/2 + Pi/2]}

and the maximum spacing is:

max gap < 2/3*loglog[T]^3/2+1)*expected gap where the expected gap can be found from above as zz[k+1]-zz[k].

This is all extremely exciting if you improve Odlyzko's 1999 value for lamda's lower bound of -2.7*10^-9. This is feeling like it could be 0. What an interesting find that would be. Prove RH and show it is just barely true as expected.

9 February, 2018 at 5:03 am

Michael RuxtonDidn’t this thread start with Rodgers and Tao proving lambda >= 0?

9 February, 2018 at 9:15 am

Doug StollNo – proving it was non-negative – but allowing for 0,

9 February, 2018 at 7:04 am

David Bernier (@doubledeckerpot)I’d recommend taking a look at the Proposal post written by Terry Tao here: https://terrytao.wordpress.com/2018/01/24/polymath-proposal-upper-bounding-the-de-bruijn-newman-constant/ .

10 February, 2018 at 8:25 am

David Bernier (@doubledeckerpot)Using an 800-point Gauss-Legendre method,

I can accurately evaluate the integral

for H_0(2*z200), where z200 is the 200th zero of Riemann zeta.

By optimizing over the parameters, I got it down to

14 seconds.

The real precision was set to 140 digits, because

there is a lot of cancellation, and the tail

oscillations are extremely tiny.

Below, x6 := 2*z200 :

? intnumgaussinit(800); // 40 seconds to initialize 800-point method

time = 41,428 ms.

? s=0.0;for(J=0,16, s=s+intnumgauss(X=J/12,(J+1)/12, Phi(X)*cos(x6*X),)); s

time = 13,778 ms.

%95 = 3.07144223057202094576[…] E-141 // 14 seconds elapsed time

? x6

%96 = 792.76370844600000000000[…]00000000000000000000

? zeta(1/2+I*x6/2)

time = 92 ms.

%97 = -7.0548531809481797311234[…] E-10 + 7.029082149944[…] E-10*I

? \p

realprecision = 154 significant digits (140 digits displayed)

Phi(X):= sum(Y=1,50,(2*Pi*Pi*(Y^4)*exp(9*X)-3*Pi*(Y^2)*exp(5*X))*exp(-Pi*(Y^2)*exp(4*X)))

10 February, 2018 at 10:21 am

David Bernier (@doubledeckerpot)For z400, I can get the time down to 38 seconds

to compute accurately H_0(2*z400).

It seems pretty clear that the main factor in the time taken

is the increasing real precision needed, when

passing from 200th to 400th and higher.

The table below is a first approximation

to the time needed to compute accurately H_0(2*z_k) using the integral,

where z_k is the k’th zero of zeta:

k=1000 time=2.36861 minutes

k=2000 time=6.42653 minutes

k=3000 time=11.5225 minutes

k=4000 time=17.4365 minutes

k=5000 time=24.0442 minutes

k=6000 time=31.2631 minutes

k=7000 time=39.0333 minutes

k=8000 time=47.3089 minutes

k=9000 time=56.0535 minutes

k=10000 time=65.2369 minutes

========================================

Commands to PARI/gp and output:

? z400=679.742197883;

? \p 38

realprecision = 38 significant digits

? zeta(1/2+I*z400)

%153 = -1.269246632[…] E-9 + 4.28823209915[…] E-9*I

? \p 240

realprecision = 250 significant digits (240 digits displayed)

? z400=679.742197883;

? x8=2*z400;

? s=0.0;for(J=0,14, s=s+intnumgauss(X=J/11,(J+1)/11, Phi(X)*cos(x8*X),)); s

%157 = 7.88880866256[…] E-237

? s=0.0;for(J=0,14, s=s+intnumgauss(X=J/11,(J+1)/11, Phi(X)*cos((x8+1)*X),)); s

%158 = 5.28032674733[…] E-228

? ##

*** last result computed in 37,923 ms.

10 February, 2018 at 12:26 pm

David Bernier (@doubledeckerpot)For z800, the 800th non-trivial zeta zero, the time to compute

the integral for H_0(2*z800) grows to almost 3 minutes.

Below, the assumption is that time is proportional

to . Then beyond the 10000th zeta

zero, the time to compute

using the integral formula becomes prohibitive:

k=1000 time=4.80261 minutes

k=2000 time=22.2359 minutes

k=3000 time=54.4995 minutes

k=4000 time=102.951 minutes

k=5000 time=168.617 minutes

k=6000 time=252.331 minutes

k=7000 time=354.805 minutes

k=8000 time=476.661 minutes

k=9000 time=618.455 minutes

k=10000 time=780.688 minutes

================================================

Commands to PARI/gp and output for z800:

? z800=1183.712775296;

? x10=2*z800;

? \p 410

? #

timer = 1 (on)

? s=0.0;for(J=0,15, s=s+intnumgauss(X=J/11,(J+1)/11, Phi(X)*cos((x10)*X),));s

time = 2min, 55,979 ms.

%179 = 7.096[…] E-410

10 February, 2018 at 1:50 pm

AnonymousIn the wiki page on the dynamics of simple zeros, the factor in the RHS of the result should be .

By the way, the simplifying assumption (for the Hadamard factorization) that has no zero at the origin, is not needed (since it leads to the same final result.)

[Corrected, thanks – T.]11 February, 2018 at 12:06 am

aAn attempt from UChicago on Hilbert’s eighth https://arxiv.org/pdf/1708.02653.pdf http://faculty.chicagobooth.edu/nicholas.polson/research/polson-hilbert-8.pdf.

11 February, 2018 at 2:05 am

AnonymousIn the arxiv paper, the function (also known as Digamma), defined by (21) as the logarithmic derivative of the Gamma function, should have its poles at the poles of (i.e. the nonpositive integers). But by (63), these poles should coincide with the zeros of which are known to have quasi-random properties. So there must be an error in the derivation of (63).

6 April, 2018 at 6:28 am

AnonymousGreat catch! But I asked Polson about it and apparently (63) was not intended to be the digamma function — it was just an unfortunate choice of notation. The 7th version (April 1st) on arXiv has $\psi_{\alpha}$ instead of $\psi$, but still this is apparently defining a new function, and is not claimed to be the digamma function.

11 February, 2018 at 2:44 am

David Bernier (@doubledeckerpot)There’s an expository paper by Andrew Booker that discusses Turing’s method and it’s use in the verification of the Riemann Hypothesis to some height. It would be nice if Turing’s method could be generalized to the functions . Here’s a link to Andrew Booker’s paper:

http://www.homepages.ucl.ac.uk/~ucahipe/turing1.pdf .

11 February, 2018 at 8:51 am

AnonymousThe ODE dynamics of the simple zeros of for is based on the fact that (in addition of obeying the Bacwards heat equation) is an even entire function of order 1 (allowing simple Hadamard factorization). Is of order 1 also for some values?

11 February, 2018 at 10:58 am

Terence Taois of order 1 for all (see the Ki-Kim-Lee paper).

11 February, 2018 at 10:57 am

Bob RobbinsWanna work on the Robbinsian method for finding new primes? I found the pattern.

12 February, 2018 at 5:53 am

zgtzsjWould you like to share it?

11 February, 2018 at 6:56 pm

KMFor t>0, a simplified form of Gram’s type law works quite well at larger T heights for quickly detecting roots. It takes advantage of the fact that zero locations get more predictable as t or T increase.

Using this I was able to compute for both t=0.4 and t=0.5, 85031 zeroes each between T = 500k and 600k. N_t estimate for both is around 85030.67

If after finding the first few zeroes in a region using the classical approach we see that that the stdev(zero gaps)/avg_zero_gap metric is less than a certain value, for eg. 10%, which would imply an interval of avg_gap/2 would cover 5 standard deviations, then such a method can be used. The zero gaps are not seen to follow a normal distribution and have a fatter tail to the right hand side, but the fatter tail shrinks as t or T are increased.

The pseudoalgo works in a 3 step loop,

1) Take a known root z_(t,i-1) and form a search interval

[z_(t,i-1) + avg_expected_gap/2, z_(t,i-1) + 3*avg_expected_gap/2]

2) Evaluate H_t at the interval endpoints and detect if there is a sign change

3) If a sign change exists, locate the root z_(t,i) using an interval solver

Locating z_(t,i) and using that to form the next search interval is necessary. For eg. we can’t use [z_(t,i-1) + 3*avg_gap/2, z_(t,i-1) + 5*avg_gap/2] and so on, since if there is a series of smaller or larger gaps, predetermined search intervals fail to detect all the roots. Also avg_expected_gap can be changed at every step, for eg. using [T-0.9T]/[N_t(T) – N_t(0.9T)]

12 February, 2018 at 3:08 pm

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