This is the third “research” thread of the Polymath15 project to upper bound the de Bruijn-Newman constant ${\Lambda}$, 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 are making progress on the following test problem: can one show that ${H_t(x+iy) \neq 0}$ whenever ${t = 0.4}$, ${x \geq 0}$, and ${y \geq 0.4}$? This would imply that

$\displaystyle \Lambda \leq 0.4 + \frac{1}{2} (0.4)^2 = 0.48$

which would be the first quantitative improvement over the de Bruijn bound of ${\Lambda \leq 1/2}$ (or the Ki-Kim-Lee refinement of ${\Lambda < 1/2}$). Of course we can try to lower the two parameters of ${0.4}$ later on in the project, but this seems as good a place to start as any. One could also potentially try to use finer analysis of dynamics of zeroes to improve the bound ${\Lambda \leq 0.48}$ further, but this seems to be a less urgent task.

Probably the hardest case is ${y=0.4}$, as there is a good chance that one can then recover the ${y>0.4}$ case by a suitable use of the argument principle. Here we appear to have a workable Riemann-Siegel type formula that gives a tractable approximation for ${H_t}$. To describe this formula, first note that in the ${t=0}$ case we have

$\displaystyle H_0(z) = \frac{1}{8} \xi( \frac{1+iz}{2} )$

and the Riemann-Siegel formula gives

$\displaystyle \xi(s) = \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \sum_{n=1}^N \frac{1}{n^s}$

$\displaystyle + \frac{s(s-1)}{2} \pi^{-(1-s)/2} \Gamma((1-s)/2) \sum_{m=1}^M \frac{1}{m^{1-s}}$

$\displaystyle + \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \frac{e^{-i\pi s} \Gamma(1-s)}{2\pi i} \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1}\ dw$

for any natural numbers ${N,M}$, where ${C_M}$ is a contour from ${+\infty}$ to ${+\infty}$ that winds once anticlockwise around the zeroes ${e^{2\pi im}, |m| \leq M}$ of ${e^w-1}$ but does not wind around any other zeroes. A good choice of ${N,M}$ to use here is

$\displaystyle N=M=\lfloor \sqrt{\mathrm{Im}(s)/2\pi}\rfloor = \lfloor \sqrt{\mathrm{Re}(z)/4\pi} \rfloor. \ \ \ \ \ (1)$

In this case, a classical steepest descent computation (see wiki) yields the approximation

$\displaystyle \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1}\ dw \approx - (2\pi i M)^{s-1} \Psi( \frac{s}{2\pi i M} - N )$

where

$\displaystyle \Psi(\alpha) := 2\pi \frac{\cos \pi(\frac{1}{2}\alpha^2 - \alpha - \pi/8)}{\cos(\pi \alpha)} \exp( \frac{i\pi}{2} \alpha^2 - \frac{5\pi i}{8} ).$

Thus we have

$\displaystyle H_0(z) \approx A^{(0)} + B^{(0)} - C^{(0)}$

where

$\displaystyle A^{(0)} := \frac{1}{8} \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \sum_{n=1}^N \frac{1}{n^s}$

$\displaystyle B^{(0)} := \frac{1}{8} \frac{s(s-1)}{2} \pi^{-(1-s)/2} \Gamma((1-s)/2) \sum_{m=1}^M \frac{1}{m^{1-s}}$

$\displaystyle C^{(0)} := \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \frac{e^{-i\pi s} \Gamma(1-s)}{2\pi i} (2\pi i M)^{s-1} \Psi( \frac{s}{2\pi i M} - N )$

with ${s := \frac{1+iz}{2}}$ and ${N,M}$ given by (1).

Heuristically, we have derived (see wiki) the more general approximation

$\displaystyle H_t(z) \approx A + B - C$

for ${t>0}$ (and in particular for ${t=0.4}$), where

$\displaystyle A := \frac{1}{8} \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \sum_{n=1}^N \frac{\exp(\frac{t}{16} \log^2 \frac{s+4}{2\pi n^2} )}{n^s}$

$\displaystyle B := \frac{1}{8} \frac{s(s-1)}{2} \pi^{-(1-s)/2} \Gamma((1-s)/2) \sum_{m=1}^M \frac{\exp(\frac{t}{16} \log^2 \frac{5-s}{2\pi m^2} )}{m^{1-s}}$

$\displaystyle C := \exp(-\frac{t \pi^2}{64}) C^{(0)}.$

In practice it seems that the ${C}$ term is negligible once the real part ${x}$ of ${z}$ is moderately large, so one also has the approximation

$\displaystyle H_t(z) \approx A + B.$

For large ${x}$, and for fixed ${t,y>0}$, e.g. ${t=y=0.4}$, the sums ${A,B}$ converge fairly quickly (in fact the situation seems to be significantly better here than the much more intensively studied ${t=0}$ case), and we expect the first term

$\displaystyle B_0 := \frac{1}{8} \frac{s(s-1)}{2} \pi^{-(1-s)/2} \Gamma((1-s)/2) \exp( \frac{t}{16} \log^2 \frac{5-s}{2\pi} )$

of the ${B}$ series to dominate. Indeed, analytically we know that ${\frac{A+B-C}{B_0} \rightarrow 1}$ (or ${\frac{A+B}{B_0} \rightarrow 1}$) as ${x \rightarrow \infty}$ (holding ${y}$ fixed), and it should also be provable that ${\frac{H_t}{B_0} \rightarrow 1}$ as well. Numerically with ${t=y=0.4}$, it seems in fact that ${\frac{A+B-C}{B_0}}$ (or ${\frac{A+B}{B_0}}$) stay within a distance of about ${1/2}$ of ${1}$ once ${x}$ is moderately large (e.g. ${x \geq 2 \times 10^5}$). This raises the hope that one can solve the toy problem of showing ${H_t(x+iy) \neq 0}$ for ${t=y=0.4}$ by numerically controlling ${H_t(x+iy) / B_0}$ for small ${x}$ (e.g. ${x \leq 2 \times 10^5}$), numerically controlling ${(A+B)/B_0}$ and analytically bounding the error ${(H_t - A - B)/B_0}$ for medium ${x}$ (e.g. ${2 \times 10^5 \leq x \leq 10^7}$), and analytically bounding both ${(A+B)/B_0}$ and ${(H_t-A-B)/B_0}$ for large ${x}$ (e.g. ${x \geq 10^7}$). (These numbers ${2 \times 10^5}$ and ${10^7}$ are arbitrarily chosen here and may end up being optimised to something else as the computations become clearer.)

Thus, we now have four largely independent tasks (for suitable ranges of “small”, “medium”, and “large” ${x}$):

1. Numerically computing ${H_t(x+iy) / B_0}$ for small ${x}$ (with enough accuracy to verify that there are no zeroes)
2. Numerically computing ${(A+B)/B_0}$ for medium ${x}$ (with enough accuracy to keep it away from zero)
3. Analytically bounding ${(A+B)/B_0}$ for large ${x}$ (with enough accuracy to keep it away from zero); and
4. Analytically bounding ${(H_t - A - B)/B_0}$ for medium and large ${x}$ (with a bound that is better than the bound away from zero in the previous two tasks).

Note that tasks 2 and 3 do not directly require any further understanding of the function ${H_t}$.

Below we will give a progress report on the numeric and analytic sides of these tasks.

— 1. Numerics report (contributed by Sujit Nair) —

There is some progress on the code side but not at the pace I was hoping. Here are a few things which happened (rather, mistakes which were taken care of).

1. We got rid of code which wasn’t being used. For example, @dhjpolymath computed ${H_t}$ based on an old version but only realized it after the fact.
2. We implemented tests to catch human/numerical bugs before a computation starts. Again, we lost some numerical cycles but moving forward these can be avoided.
3. David got set up on GitHub and he is able to compare his output (in C) with the Python code. That is helping a lot.

Two areas which were worked on were

1. Computing ${H_t}$ and zeroes for ${t}$ around ${0.4}$
2. Computing quantities like ${(A+B-C)/B_0}$, ${(A+B)/B_0}$, ${C/B_0}$, etc. with the goal of understanding the zero free regions.

Some observations for ${t=0.4}$, ${y=0.4}$, ${x \in ( 10^4, 10^7)}$ include:

• ${(A+B) / B_0}$ does seem to avoid the negative real axis
• ${|(A+B) / B0| > 0.4}$ (based on the oscillations and trends in the plots)
• ${|C/B_0|}$ seems to be settling around ${10^{-4}}$ range.

See the figure below. The top plot is on the complex plane and the bottom plot is the absolute value. The code to play with this is here.

— 2. Analysis report —

The Riemann-Siegel formula and some manipulations (see wiki) give ${H_0 = A^{(0)} + B^{(0)} - \tilde C^{(0)}}$, where

$\displaystyle A^{(0)} = \frac{2}{8} \sum_{n=1}^N \int_C \exp( \frac{s+4}{2} u - e^u - \frac{s}{2} \log(\pi n^2) )\ du$

$\displaystyle - \frac{3}{8} \sum_{n=1}^N \int_C \exp( \frac{s+2}{2} u - e^u - \frac{s}{2} \log(\pi n^2) )\ du$

$\displaystyle B^{(0)} = \frac{2}{8} \sum_{m=1}^M \int_{\overline{C}} \exp( \frac{5-s}{2} u - e^u - \frac{1-s}{2} \log(\pi m^2) )\ du$

$\displaystyle - \frac{3}{8} \sum_{m=1}^M \int_C \exp( \frac{3-s}{2} u - e^u - \frac{1-s}{2} \log(\pi m^2) )\ du$

$\displaystyle \tilde C^{(0)} := -\frac{2}{8} \sum_{n=0}^\infty \frac{e^{-i\pi s/2} e^{i\pi s n}}{2^s \pi^{1/2}} \int_{\overline{C}} \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1} \exp( \frac{5-s}{2} u - e^u)\ du dw$

$\displaystyle +\frac{3}{8} \sum_{n=0}^\infty \frac{e^{-i\pi s/2} e^{i\pi s n}}{2^s \pi^{1/2}} \int_{\overline{C}} \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1} \exp( \frac{3-s}{2} u - e^u)\ du dw$

where ${C}$ is a contour that goes from ${+i\infty}$ to ${+\infty}$ staying a bounded distance away from the upper imaginary and right real axes, and ${\overline{C}}$ is the complex conjugate of ${C}$. (In each of these sums, it is the first term that should dominate, with the second one being about ${O(1/x)}$ as large.) One can then evolve by the heat flow to obtain ${H_t = \tilde A + \tilde B - \tilde C}$, where

$\displaystyle \tilde A := \frac{2}{8} \sum_{n=1}^N \int_C \exp( \frac{s+4}{2} u - e^u - \frac{s}{2} \log(\pi n^2) + \frac{t}{16} (u - \log(\pi n^2))^2)\ du$

$\displaystyle - \frac{3}{8} \sum_{n=1}^N \int_C \exp( \frac{s+2}{2} u - e^u - \frac{s}{2} \log(\pi n^2) + \frac{t}{16} (u - \log(\pi n^2))^2)\ du$

$\displaystyle \tilde B := \frac{2}{8} \sum_{m=1}^M \int_{\overline{C}} \exp( \frac{5-s}{2} u - e^u - \frac{1-s}{2} \log(\pi m^2) + \frac{t}{16} (u - \log(\pi m^2))^2)\ du$

$\displaystyle - \frac{3}{8} \sum_{m=1}^M \int_C \exp( \frac{3-s}{2} u - e^u - \frac{1-s}{2} \log(\pi m^2) + \frac{t}{16} (u - \log(\pi m^2))^2)\ du$

$\displaystyle \tilde C := -\frac{2}{8} \sum_{n=0}^\infty \frac{e^{-i\pi s/2} e^{i\pi s n}}{2^s \pi^{1/2}} \int_{\overline{C}} \int_{C_M}$

$\displaystyle \frac{w^{s-1} e^{-Nw}}{e^w-1} \exp( \frac{5-s}{2} u - e^u + \frac{t}{4} (i \pi(n-1/2) + \log \frac{w}{2\sqrt{\pi}} - \frac{u}{2})^2) \ du dw$

$\displaystyle +\frac{3}{8} \sum_{n=0}^\infty \frac{e^{-i\pi s/2} e^{i\pi s n}}{2^s \pi^{1/2}} \int_{\overline{C}} \int_{C_M}$

$\displaystyle \frac{w^{s-1} e^{-Nw}}{e^w-1} \exp( \frac{3-s}{2} u - e^u + \frac{t}{4} (i \pi(n-1/2) + \log \frac{w}{2\sqrt{\pi}} - \frac{u}{2})^2)\ du dw.$

Steepest descent heuristics then predict that ${\tilde A \approx A}$, ${\tilde B \approx B}$, and ${\tilde C \approx C}$. For the purposes of this project, we will need effective error estimates here, with explicit error terms.

A start has been made towards this goal at this wiki page. Firstly there is a “effective Laplace method” lemma that gives effective bounds on integrals of the form ${\int_I e^{\phi(x)} \psi(x)\ dx}$ if the real part of ${\phi(x)}$ is either monotone with large derivative, or has a critical point and is decreasing on both sides of that critical point. In principle, all one has to do is manipulate expressions such as ${\tilde A - A}$, ${\tilde B - B}$, ${\tilde C - C}$ by change of variables, contour shifting and integration by parts until it is of the form to which the above lemma can be profitably applied. As one may imagine though the computations are messy, particularly for the ${\tilde C}$ term. As a warm up, I have begun by trying to estimate integrals of the form

$\displaystyle \int_C \exp( s (1+u-e^u) + \frac{t}{16} (u+b)^2 )\ du$

for smallish complex numbers ${b}$, as these sorts of integrals appear in the form of ${\tilde A, \tilde B, \tilde C}$. As of this time of writing, there are effective bounds for the ${b=0}$ case, and I am currently working on extending them to the ${b \neq 0}$ case, which should give enough control to approximate ${\tilde A - A}$ and ${\tilde B-B}$. The most complicated task will be that of upper bounding ${\tilde C}$, but it also looks eventually doable.