# Difference between revisions of "Polymath15 test problem"

We are initially focusing attention on the following

Test problem For $t=y=0.4$, can one prove that $H_t(x+iy) \neq 0$ for all $x \geq 0$?

If we can show this, it is likely that (with the additional use of the argument principle, and some further information on the behaviour of $H_t(x+iy)$ at $y=0.4$) that one can show that $H_t(x+iy) \neq 0$ for all $y \geq 0.4$ as well. This would give a new upper bound

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

for the de Bruijn-Newman constant.

For very small values of $x$ we expect to be able to establish this by direct calculation of $H_t(x+iy)$. For medium or large values, the strategy is to use a suitable approximation

$H_t(x+iy) \approx A + B$

for some relatively easily computable quantities $A = A_t(x+iy), B = B_t(x+iy)$ (it may possibly be necessary to use a refined approximation $A+B-C$ instead). The quantity $B$ contains a non-zero main term $B_0$ which is expected to roughly dominate. To show $H_t(x+iy)$ is non-zero, it would suffice to show that

$\frac{|H_t - A - B|}{|B_0|} \lt \frac{|A + B|}{|B_0|}.$

Thus one will seek upper bounds on the error $\frac{|H_t - A - B|}{|B_0|}$ and lower bounds on $\frac{|A+B|}{|B_0|}$ for various ranges of $x$. Numerically it seems that the RHS stays above 0.4 as soon as $x$ is moderately large, while the LHS stays below 0.1, which looks promising for the rigorous arguments.

## Choices of approximation

There are a number of slightly different approximations we have used in previous discussion. The first approximation was $A+B$, where

$A := \frac{1}{8} \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \sum_{n=1}^N \frac{\exp(\frac{t}{16} \log^2 \frac{s+4}{2\pi n^2})}{n^s}$
$B := \frac{1}{8} \frac{s(s-1)}{2} \pi^{-(1-s)/2} \Gamma((1-s)/2) \sum_{n=1}^N \frac{\exp(\frac{t}{16} \log^2 \frac{5-s}{2\pi n^2})}{n^{1-s}}$
$B_0 := \frac{1}{8} \frac{s(s-1)}{2} \pi^{-(1-s)/2} \Gamma((1-s)/2) \exp( \frac{t}{16} \log^2 \frac{5-s}{2\pi} )$
$s := \frac{1-y+ix}{2}$
$N := \lfloor \sqrt{\frac{\mathrm{Im} s}{2\pi}} \rfloor = \lfloor \sqrt{\frac{x}{4\pi}} \rfloor.$

There is also the refinement $A+B-C$, where

$C:= \frac{1}{8} \exp(-\frac{t\pi^2}{64}) \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \frac{e^{-i\pi s} \Gamma(1-s)}{2\pi i} (2\pi i N)^{s-1} \Psi( \frac{s}{2\pi i N}-N )$
$\Psi(\alpha) := 2\pi \frac{\cos \pi(\frac{1}{2} \alpha^2 - \alpha - \frac{\pi}{8})}{\cos(\pi \alpha)} \exp( \frac{i \pi}{2} \alpha^2 - \frac{5 \pi i}{8}).$

The first approximation was modified slightly to $A'+B'$, where

$A' := \frac{2}{8} \pi^{-s/2} \sqrt{2\pi} \exp( (\frac{s+4}{2}-\frac{1}{2}) \log \frac{s+4}{2} - \frac{s+4}{2}) \sum_{n=1}^N \frac{\exp(\frac{t}{16} \log^2 \frac{s+4}{2\pi n^2})}{n^s}$
$B' := \frac{2}{8} \pi^{-(1-s)/2} \sqrt{2\pi} \exp( (\frac{5-s}{2}-\frac{1}{2}) \log \frac{5-s}{2} - \frac{5-s}{2}) \sum_{n=1}^N \frac{\exp(\frac{t}{16} \log^2 \frac{5-s}{2\pi n^2})}{n^{1-s}}$
$B'_0 := \frac{2}{8} \pi^{-(1-s)/2} \sqrt{2\pi} \exp( (\frac{5-s}{2}-\frac{1}{2}) \log \frac{5-s}{2} - \frac{5-s}{2}) \exp( \frac{t}{16} \log^2 \frac{5-s}{2\pi} )$
$s := \frac{1-y+ix}{2}$
$N := \lfloor \sqrt{\frac{\mathrm{Im} s}{2\pi}} \rfloor = \lfloor \sqrt{\frac{x}{4\pi}} \rfloor.$

In Effective bounds on H_t - second approach, a more refined approximation $A^{eff} + B^{eff}$ was introduced:

$A^{eff} := \frac{1}{8} \exp( \frac{t}{4} \alpha_1(\frac{1-y+ix}{2})^2 ) H_{0,1}(\frac{1-y+ix}{2}) \sum_{n=1}^N \frac{1}{n^{\frac{1-y+ix}{2} + \frac{t \alpha_1(\frac{1-y+ix}{2})}{2} - \frac{t}{4} \log n}}$
$B^{eff} := \frac{1}{8} \exp( \frac{t}{4} \overline{\alpha_1(\frac{1+y+ix}{2})}^2 ) \overline{H_{0,1}(\frac{1+y+ix}{2})} \sum_{n=1}^N \frac{1}{n^{\frac{1+y-ix}{2} + \frac{t \overline{\alpha_1(\frac{1+y+ix}{2})}}{2} - \frac{t}{4} \log n}}$
$B^{eff}_0 := \frac{1}{8} \exp( \frac{t}{4} \overline{\alpha_1(\frac{1+y+ix}{2})}^2 ) \overline{H_{0,1}(\frac{1+y+ix}{2})}$
$H_{0,1}(s) := \frac{s (s-1)}{2} \pi^{-s/2} \sqrt{2\pi} \exp( (\frac{s}{2} - \frac{1}{2}) \log \frac{s}{2} - \frac{s}{2} )$
$\alpha_1(s) := \frac{1}{2s} + \frac{1}{s-1} + \frac{1}{2} \log \frac{s}{2\pi}$
$N := \lfloor \sqrt{ \frac{T'}{2\pi}} \rfloor$
$T' := \frac{x}{2} + \frac{\pi t}{8}.$

There is a refinement $A^{eff}+B^{eff}-C^{eff}$, where

$C^{eff} := \frac{1}{8} \exp( \frac{t\pi^2}{64}) \frac{s'(s'-1)}{2} (-1)^N ( \pi^{-s'/2} \Gamma(s'/2) a^{-\sigma} C_0(p) U + \pi^{-(1-s')/2} \Gamma((1-s')/2) a^{-(1-\sigma)} \overline{C_0(p)} \overline{U})$
$s' := \frac{1-y}{2} + iT' = \frac{1-y+ix}{2} + \frac{\pi i t}{8}$
$a := \sqrt{\frac{T'}{2\pi}}$
$p := 1 - 2(a-N)$
$\sigma := \mathrm{Re} s' = \frac{1-y}{2}$
$U := \exp( -i (\frac{T'}{2} \log \frac{T'}{2\pi} - \frac{T'}{2} - \frac{\pi}{8} ))$
$C_0(p) := \frac{ \exp( \pi i (p^2/2 + 3/8) )- i \sqrt{2} \cos(\pi p/2)}{2 \cos(\pi p)}.$

One can also replace $C^{eff}$ by the very slightly different quantity

$\tilde C^{eff} :=\frac{2 e^{-\pi i y/8}}{8} \exp( \frac{t\pi^2}{64}) (-1)^N \mathrm{Re}( H_{0,1}(iT') C_0(p) U e^{\pi i/8} ).$

Finally, a simplified approximation is $A^{toy} + B^{toy}$, where

$A^{toy} := B^{toy}_0 \exp(i ((\frac{x}{2} + \frac{\pi t}{8}) \log \frac{x}{4\pi} - \frac{x}{2} - \frac{\pi}{4} )) N^{-y} \sum_{n=1}^N \frac{1}{n^{\frac{1-y+ix}{2} + \frac{t}{4} \log \frac{N^2}{n} + \pi i t/8}}$
$B^{toy} := B^{toy}_0 \sum_{n=1}^N \frac{1}{n^{\frac{1+y-ix}{2} + \frac{t}{4} \log \frac{N^2}{n} - \pi i t/8}}$
$B^{toy}_0 := \frac{\sqrt{2}}{4} \pi^2 N^{\frac{7+y}{2}} \exp( i (-\frac{x}{4} \log \frac{x}{4\pi} + \frac{x}{4} + \frac{9-y}{8} \pi) + \frac{t}{16} (\log \frac{x}{4\pi} - \frac{\pi i}{2})^2 ) e^{-\pi x/8}$
$N := \lfloor \sqrt{\frac{x}{4\pi}} \rfloor.$

Here is a table comparing the size of the various main terms:

$x$ $B_0$ $B'_0$ $B^{eff}_0$ $B^{toy}_0$
$10^3$ $(3.4405 + 3.5443 i) \times 10^{-167}$ $(3.4204 + 3.5383 i) \times 10^{-167}$ $(3.4426 + 3.5411 i) \times 10^{-167}$ $(2.3040 + 2.3606 i) \times 10^{-167}$
$10^4$ $(-1.1843 - 7.7882 i) \times 10^{-1700}$ $(-1.1180 - 7.7888 i) \times 10^{-1700}$ $(-1.1185 - 7.7879 i) \times 10^{-1700}$ $(-1.1155 - 7.5753 i) \times 10^{-1700}$
$10^5$ $(-7.6133 + 2.5065 i) * 10^{-17047}$ $(-7.6134 + 2.5060 i) * 10^{-17047}$ $(-7.6134 + 2.5059 i) * 10^{-17047}$ $(-7.5483 + 2.4848 i) * 10^{-17047}$
$10^6$ $(-3.1615 - 7.7093 i) * 10^{-170537}$ $(-3.1676 - 7.7063 i) * 10^{-170537}$ $(-3.1646 - 7.7079 i) * 10^{-170537}$ $(-3.1590 - 7.6898 i) * 10^{-170537}$
$10^7$ $(2.1676 - 9.6330 i) * 10^{-1705458}$ $(2.1711 - 9.6236 i) * 10^{-1705458}$ $(2.1571 - 9.6329 i) * 10^{-1705458}$ $(2.2566 - 9.6000 i) * 10^{-1705458}$

Here some typical values of $B/B_0$ (note that $B/B_0$ and $B'/B'_0$ are identical):

$x$ $B/B_0$ $B'/B'_0$ $B^{eff}/B^{eff}_0$ $B^{toy}/B^{toy}_0$
$10^3$ $0.7722 + 0.6102 i$ $0.7722 + 0.6102 i$ $0.7733 + 0.6101 i$ $0.7626 + 0.6192 i$
$10^4$ $0.7434 - 0.0126 i$ $0.7434 - 0.0126 i$ $0.7434 - 0.0126 i$ $0.7434 - 0.0124 i$
$10^5$ $1.1218 - 0.3211 i$ $1.1218 - 0.3211 i$ $1.1218 - 0.3211 i$ $1.1219 - 0.3213 i$
$10^6$ $1.3956 - 0.5682 i$ $1.3956 - 0.5682 i$ $1.3955 - 0.5682 i$ $1.3956 - 0.5683 i$
$10^7$ $1.6400 + 0.0198 i$ $1.6400 + 0.0198 i$ $1.6401 + 0.0198 i$ $1.6400 - 0.0198 i$

Here some typical values of $A/B_0$, which seems to be about an order of magnitude smaller than $B/B_0$ in many cases:

$x$ $A/B_0$ $A'/B'_0$ $A^{eff}/B^{eff}_0$ $A^{toy}/B^{toy}_0$
$10^3$ $-0.3856 - 0.0997 i$ $-0.3857 - 0.0953 i$ $-0.3854 - 0.1002 i$ $-0.4036 - 0.0968 i$
$10^4$ $-0.2199 - 0.0034 i$ $-0.2199 - 0.0036 i$ $-0.2199 - 0.0033 i$ $-0.2208 - 0.0033 i$
$10^5$ $0.1543 + 0.1660 i$ $0.1543 + 0.1660 i$ $0.1543 + 0.1660 i$ $0.1544 + 0.1663 i$
$10^6$ $-0.1013 - 0.1887 i$ $-0.1010 - 0.1889 i$ $-0.1011 - 0.1890 i$ $-0.1012 - 0.1888 i$
$10^7$ $-0.1018 + 0.1135 i$ $-0.1022 + 0.1133 i$ $-0.1025 + 0.1128 i$ $-0.0986 + 0.1163 i$

Here some typical values of $C/B_0$, which is significantly smaller than either $A/B_0$ or $B/B_0$:

$x$ $C/B_0$ $C^{eff}/B^{eff}_0$
$10^3$ $-0.1183 + 0.0697i$ $-0.0581 + 0.0823 i$
$10^4$ $-0.0001 - 0.0184 i$ $-0.0001 - 0.0172 i$
$10^5$ $-0.0033 - 0.0005i$ $-0.0031 - 0.0005i$
$10^6$ $-0.0001 - 0.0006 i$ $-0.0001 - 0.0006 i$
$10^7$ $-0.0000 - 0.0001 i$ $-0.0000 - 0.0001 i$

Some values of $H_t$ and its approximations at small values of $x$ source source:

$x$ $H_t$ $A+B$ $A'+B'$ $A^{eff}+B^{eff}$ $A^{toy}+B^{toy}$ $A+B-C$ $A^{eff}+B^{eff}-C^{eff}$
$10$ $(3.442 - 0.168 i) \times 10^{-2}$ 0 0 0 N/A N/A $(3.501 - 0.316 i) \times 10^{-2}$
$30$ $(-1.000 - 0.071 i) \times 10^{-4}$ $(-0.650 - 0.188 i) \times 10^{-4}$ $(-0.211 - 0.192 i) \times 10^{-4}$ $(-0.670 - 0.114 i) \times 10^{-4}$ $(-0.136 + 0.021 i) \times 10^{-4}$ $(-1.227 - 0.058 i) \times 10^{-4}$ $(-1.032 - 0.066 i) \times 10^{-4}$
$100$ $(6.702 + 3.134 i) \times 10^{-16}$ $(2.890 + 3.667 i) \times 10^{-16}$ $(2.338 + 3.742 i) \times 10^{-16}$ $(2.955 + 3.650 i) \times 10^{-16}$ $(0.959 + 0.871 i) \times 10^{-16}$ $(6.158 + 12.226 i) \times 10^{-16}$ $(6.763 + 3.074 i) \times 10^{-16}$
$300$ $(-4.016 - 1.401 i) \times 10^{-49}$ $(-5.808 - 1.140 i) \times 10^{-49}$ $(-5.586 - 1.228 i) \times 10^{-49}$ $(-5.824 - 1.129 i) \times 10^{-49}$ $(-2.677 - 0.327 i) \times 10^{-49}$ $(-3.346 + 6.818 i) \times 10^{-49}$ $(-4.032 - 1.408 i) \times 10^{-49}$
$1000$ $(0.015 + 3.051 i) \times 10^{-167}$ $(-0.479 + 3.126 i) \times 10^{-167}$ $(-0.516 + 3.135 i) \times 10^{-167}$ $(-0.474 + 3.124 i) \times 10^{-167}$ $(-0.406 + 2.051 i) \times 10^{-167}$ $(0.175 + 3.306 i) \times 10^{-167}$ $(0.017 + 3.047 i) \times 10^{-167}$
$3000$ $(-1.144+ 1.5702 i) 10^{-507}$ $(-1.039+ 1.5534 i) 10^{-507}$ $(-1.039+ 1.5552 i) 10^{-507}$ $(-1.038+ 1.5535 i) 10^{-507}$ $(-0.925+ 1.3933 i) 10^{-507}$ $(-1.155+ 1.5686 i) 10^{-507}$ $(-1.144+ 1.5701 i) 10^{-507}$
$10000$ $(-0.558 - 4.088 i) \times 10^{-1700}$ $(-0.692 - 4.067 i) \times 10^{-1700}$ $(-0.687 - 4.067 i) \times 10^{-1700}$ $(-0.692 - 4.066 i) \times 10^{-1700}$ $(-0.673 - 3.948 i) \times 10^{-1700}$ $(-0.548 - 4.089 i) \times 10^{-1700}$ $(-0.558 - 4.088 i) \times 10^{-1700}$
$30000$ $(3.160 - 6.737) \times 10^{-5110}$ $(3.065 - 6.722) \times 10^{-5100}$ $(3.066 - 6.722) \times 10^{-5100}$ $(3.065 - 6.722) \times 10^{-5100}$ $(2.853 - 6.286) \times 10^{-5100}$ $(3.170 - 6.733) \times 10^{-5100}$ $(3.160 - 6.737) \times 10^{-5100}$

## Controlling |A+B|/|B_0|

Mesh evaluations of $A^{eff}+B^{eff}/B^{eff}_0$ in the ranges

## Controlling |H_t-A-B|/|B_0|

Here is a table on bounds on error terms $E_1/B^{eff}_0, E_2/B^{eff}_0, E_3^*/B^{eff}_0$ for N=3 to 2000. Here is a table with some sharpened estimates from the PDF writeup.

Here is a graph depicting $|H_t-A^{eff}-B^{eff}/B_0^{eff}|$ and $E_1+E_2+E_3^*/|B_0^{eff}|$ for $x \leq 1600$.

## Small values of x

Tables of $H_t(x+iy)$ for small values of $x$:

Tables of $H'_t(x+iy)$:

Here is a table of $x$, pari/gp prec, $H_{t}, H^{'}_{t}, |H_{t}|, |H^{'}_{t}|, \frac{|H_{t}|}{|B_{0}^{eff}|}, \frac{|H^{'}_{t}|}{|B_{0}^{eff}|}$ for x=0 to x=30 with step size 0.01.

Here is a plot of $H_t/B_0$ for a rectangle $\{x+iy: 0 \leq x \leq 300; 0.4 \leq y \leq 0.45\}$. Here is an adaptive mesh plot; here is a closeup near the origin.

Here is a script for verifying the absence of zeroes of $H_t$ in a rectangle. It can eliminate zeros in the rectangle $\{0 \leq x \leq 1000, 0.4 \leq y \leq 0.45\}$ when t = 0.4.

## Large negative values of $t$

We heuristically compute $H_t(x)$ in the regime where $x$ is large and $t$ is large and negative with $|t|/x \asymp 1$. We shall only be interested in the zeroes and so we discard any multiplicative factor which is non-zero: we write $X \sim Y$ if X is equal (or approximately equal) to Y times something that is explicit and non-zero.

From equation (35) of the writeup we have

$H_t(x) = \int_{\bf R} \frac{1}{8} \xi(\frac{1+ix}{2} + i |t|^{1/2} v) \frac{1}{\pi} e^{-v^2}\ dv$
$\sim \int_{\bf R} \xi(\frac{1+ix}{2} + i |t|^{1/2} v) e^{-v^2}\ dv.$

To cancel off an exponential decay factor in the $\xi$ function, it is convenient to shift the v variable by $\pi |t|^{1/2}/8$, thus

$H_t(x) \sim \int_{\bf R} \xi(\frac{1+ix}{2} + i |t|^{1/2} v - \pi i |t|/8) e^{-(v - \pi |t|^{1/2}/8)^2}\ dv$
$\sim \int_{\bf R} \xi(\frac{1+i\tilde x}{2} + i |t|^{1/2} v) e^{-v^2 + \pi |t|^{1/2} v / 4}\ dv$

where

$\tilde x := x - \pi |t|/4 = x + \frac{\pi t}{4}.$

Now from the definition of $\xi$ and the Stirling approximation we have

$\xi(s) \sim M_0(s) \zeta(s)$

where $M_0$ is defined in (6) of the writeup. Thus

$H_t(x) \sim \int_{\bf R} M_0(\frac{1+i\tilde x}{2} + i |t|^{1/2} v) \zeta(\frac{1+i\tilde x}{2} + i |t|^{1/2} v) e^{-v^2 + \pi |t|^{1/2} v / 4}\ dv.$

By Taylor expansion we have

$M_0(\frac{1+i\tilde x}{2} + i |t|^{1/2} v) \sim M_0(\frac{1+i\tilde x}{2}) \exp( \alpha( \frac{1+i\tilde x}{2} ) i |t|^{1/2} v + \alpha'(\frac{1+i \tilde x}{2}) \frac{-|t| v^2}{2} )$
$\sim \exp( \alpha( \frac{1+i\tilde x}{2} ) i |t|^{1/2} v + \alpha'(\frac{1+i \tilde x}{2}) \frac{-|t| v^2}{2} )$

where $\alpha$ is defined in equation (8) of the writeup. We have the approximations

$\alpha(\frac{1+i\tilde x}{2} ) \approx \frac{1}{2} \log \frac{\tilde x}{4\pi} + \frac{i\pi}{4}$

and

$\alpha'(\frac{1+i\tilde x}{2} ) \approx \frac{-i}{\tilde x}$

and hence

$H_t(x) \sim \int_{\bf R} \exp( \frac{i |t|^{1/2} v}{2} \log \frac{\tilde x}{4\pi} - \pi |t|^{1/2} v/4 + i |t| v^2 / 2\tilde x) \zeta(\frac{1+i\tilde x}{2} + i |t|^{1/2} v) e^{-v^2 + \pi |t|^{1/2} v / 4}\ dv.$

The two factors of $\exp( \pi |t|^{1/2} v/4 )$ cancel. If we now write

$N := \sqrt{\frac{\tilde x}{4\pi}}$

and

$u := |t|/N^2 = 4\pi |t|/\tilde x,$

we conclude that

$H_t(x) \sim \int_{\bf R} \exp( i |t|^{1/2} v \log N + i u v^2 / 8 \pi) \zeta(\frac{1+i\tilde x}{2} + i |t|^{1/2} v) e^{-v^2}\ dv.$

If we formally write $\zeta(s) = \sum_n \frac{1}{n^s}$ (ignoring convergence issues) we obtain

$H_t(x) \sim \sum_n \int_{\bf R} \exp( i |t|^{1/2} v \log N + i u v^2 / 8 \pi) n^{-\frac{1+i\tilde x}{2} - i |t|^{1/2} v} e^{-v^2}\ dv$
$\sim \sum_n \int_{\bf R} \exp( - i |t|^{1/2} v \log \frac{n}{N} + i u v^2 / 8 \pi -\frac{1+i\tilde x}{2} \log \frac{n}{N} ) e^{-v^2}\ dv$

We can compute the $v$ integral to obtain

$H_t(x) \sim \sum_n \exp( - \frac{|t| \log^2 \frac{n}{N}}{4 (1 - iu / 8 \pi)} -\frac{1+i\tilde x}{2} \log \frac{n}{N}).$

Using the Taylor approximation

$\log \frac{n}{N} \approx \frac{n-N}{N} - \frac{(n-N)^2}{2N^2}$

and dropping some small terms, we obtain

$H_t(x) \sim \sum_n \exp( - \frac{|t| (n-N)^2}{4 N^2 (1 - iu/8\pi)} -\frac{i\tilde x}{2} \frac{n-N}{N} + \frac{i \tilde x (n-N)^2}{4N^2} ).$

Writing $\tilde x = 4\pi N^2$ and $|t| = u N^2$, this becomes

$H_t(x) \sim \sum_n \exp( - \frac{2\pi u (n-N)^2}{8\pi - iu} -2 \pi i N(n-N) + \pi i (n-N)^2 ).$

Writing

$N(n-N) = \frac{1}{2} n^2 - \frac{1}{2} (n-N)^2 - \frac{1}{2} N^2$

we thus have

$H_t(x) \sim \sum_n \exp( - \frac{2 \pi u (n-N)^2}{8 \pi - iu} - \pi i n^2 + 2 \pi i (n-N)^2 )$
$\sim \sum_n \exp( \frac{16 \pi^2 i (n-N)^2}{8 \pi - iu} ) e( \pi i n )$
$\sim \theta_{01}( \frac{16 \pi N}{8\pi - iu}, \frac{16 \pi}{8\pi - iu} )$

where $\theta_{01}$ is the theta function defined in this Wikipedia page. Using the Jacobi identity we then have

$H_t(x) \sim \theta_{01}(-N, \frac{iu - 8\pi}{16 \pi} )$
$= \sum_n (-1)^n \exp( - \pi i n^2 / 2 ) e^{-2\pi i n N} e^{-u n^2/16}.$