Polymath15 test problem

From Polymath Wiki
Jump to navigationJump to search

We are initially focusing attention on the following

Test problem For [math]\displaystyle{ t=y=0.4 }[/math], can one prove that [math]\displaystyle{ H_t(x+iy) \neq 0 }[/math] for all [math]\displaystyle{ x \geq 0 }[/math]?

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 [math]\displaystyle{ H_t(x+iy) }[/math] at [math]\displaystyle{ y=0.4 }[/math]) that one can show that [math]\displaystyle{ H_t(x+iy) \neq 0 }[/math] for all [math]\displaystyle{ y \geq 0.4 }[/math] as well. This would give a new upper bound

[math]\displaystyle{ \Lambda \leq 0.4 + \frac{1}{2} (0.4)^2 = 0.48 }[/math]

for the de Bruijn-Newman constant.

For very small values of [math]\displaystyle{ x }[/math] we expect to be able to establish this by direct calculation of [math]\displaystyle{ H_t(x+iy) }[/math]. For medium or large values, the strategy is to use a suitable approximation

[math]\displaystyle{ H_t(x+iy) \approx A + B }[/math]

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

[math]\displaystyle{ \frac{|H_t - A - B|}{|B_0|} \lt \frac{|A + B|}{|B_0|}. }[/math]

Thus one will seek upper bounds on the error [math]\displaystyle{ \frac{|H_t - A - B|}{|B_0|} }[/math] and lower bounds on [math]\displaystyle{ \frac{|A+B|}{|B_0|} }[/math] for various ranges of [math]\displaystyle{ x }[/math]. Numerically it seems that the RHS stays above 0.4 as soon as [math]\displaystyle{ x }[/math] 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

[math]\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} }[/math]
[math]\displaystyle{ 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}} }[/math]
[math]\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} ) }[/math]
[math]\displaystyle{ s := \frac{1-y+ix}{2} }[/math]
[math]\displaystyle{ N := \lfloor \frac{\mathrm{Im} s}{2\pi} \rfloor = \lfloor \frac{x}{4\pi} \rfloor. }[/math]

This was modified slightly to

[math]\displaystyle{ 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} }[/math]
[math]\displaystyle{ 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}} }[/math]
[math]\displaystyle{ 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} ) }[/math]
[math]\displaystyle{ s := \frac{1-y+ix}{2} }[/math]
[math]\displaystyle{ N := \lfloor \frac{\mathrm{Im} s}{2\pi} \rfloor = \lfloor \frac{x}{4\pi} \rfloor. }[/math]

In Effective bounds on H_t - second approach, a more refined approximation was introduced:

[math]\displaystyle{ 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}} }[/math]
[math]\displaystyle{ 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}} }[/math]
[math]\displaystyle{ 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})} }[/math]
[math]\displaystyle{ 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} ) }[/math]
[math]\displaystyle{ \alpha_1(s) := \frac{1}{2s} + \frac{1}{s-1} + \frac{1}{2} \log \frac{s}{2\pi} }[/math]
[math]\displaystyle{ N := \lfloor \sqrt{ \frac{T'}{2\pi}} \rfloor }[/math]
[math]\displaystyle{ T' := \frac{x}{2} - \frac{\pi t}{8}. }[/math]

Finally, a simplified approximation is

[math]\displaystyle{ 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}} }[/math]
[math]\displaystyle{ 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}} }[/math]
[math]\displaystyle{ 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} }[/math]
[math]\displaystyle{ N := \lfloor \sqrt{\frac{x}{4\pi}} \rfloor. }[/math]

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

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

Here some typical values of [math]\displaystyle{ B/B_0 }[/math] (note that [math]\displaystyle{ B/B_0 }[/math] and [math]\displaystyle{ B'/B'_0 }[/math] are identical):

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

Here some typical values of [math]\displaystyle{ A/B_0 }[/math], which seems to be about an order of magnitude smaller than [math]\displaystyle{ B/B_0 }[/math] in many cases:

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

Controlling |A+B|/|B_0|

Some numerical data on [math]\displaystyle{ |A+B/B_0| }[/math] source and also [math]\displaystyle{ \mathrm{Re} \frac{A+B}{B_0} }[/math] source, using a step size of 1 for [math]\displaystyle{ x }[/math], suggesting that this ratio tends to oscillate roughly between 0.5 and 3 for medium values of [math]\displaystyle{ x }[/math]:

range of [math]\displaystyle{ x }[/math] minimum value max value average value standard deviation min real part max real part
0-1000 0.179 4.074 1.219 0.782 -0.09 4.06
1000-2000 0.352 4.403 1.164 0.712 0.02 4.43
2000-3000 0.352 4.050 1.145 0.671 0.15 3.99
3000-4000 0.338 4.174 1.134 0.640 0.34 4.48
4000-5000 0.386 4.491 1.128 0.615 0.33 4.33
5000-6000 0.377 4.327 1.120 0.599 0.377 4.327
[math]\displaystyle{ 1-10^5 }[/math] 0.179 4.491 1.077 0.455 -0.09 4.48
[math]\displaystyle{ 10^5-2 \times 10^5 }[/math] 0.488 3.339 1.053 0.361 0.48 3.32
[math]\displaystyle{ 2 \times 10^5-3 \times 10^5 }[/math] 0.508 3.049 1.047 0.335 0.50 3.00
[math]\displaystyle{ 3 \times 10^5-4 \times 10^5 }[/math] 0.517 2.989 1.043 0.321 0.52 2.97
[math]\displaystyle{ 4 \times 10^5-5 \times 10^5 }[/math] 0.535 2.826 1.041 0.310 0.53 2.82
[math]\displaystyle{ 5 \times 10^5-6 \times 10^5 }[/math] 0.529 2.757 1.039 0.303 0.53 2.75
[math]\displaystyle{ 6 \times 10^5-7 \times 10^5 }[/math] 0.548 2.728 1.038 0.296 0.55 2.72

Here is a computation on the magnitude [math]\displaystyle{ |\frac{d}{dx}(B'/B'_0)| }[/math] of the derivative of [math]\displaystyle{ B'/B'_0 }[/math], sampled at steps of 1 in [math]\displaystyle{ x }[/math] source, together with a crude upper bound coming from the triangle inequality source, to give some indication of the oscillation:

range of [math]\displaystyle{ T=x/2 }[/math] max value average value standard deviation triangle inequality bound
0-1000 1.04 0.33 0.19
1000-2000 1.25 0.39 0.24
2000-3000 1.31 0.39 0.25
3000-4000 1.39 0.38 0.27
4000-5000 1.64 0.37 0.26
5000-6000 1.60 0.36 0.27
6000-7000 1.61 0.36 0.26
7000-8000 1.55 0.36 0.27
8000-9000 1.65 0.34 0.26
9000-10000 1.47 0.34 0.26
[math]\displaystyle{ 1-10^5 }[/math] 1.78 0.28 0.23 2.341
[math]\displaystyle{ 10^5-2 \times 10^5 }[/math] 1.66 0.22 0.18 2.299
[math]\displaystyle{ 2 \times 10^5-3 \times 10^5 }[/math] 1.55 0.20 0.17 2.195
[math]\displaystyle{ 3 \times 10^5-4 \times 10^5 }[/math] 1.53 0.19 0.16 2.109
[math]\displaystyle{ 4 \times 10^5-5 \times 10^5 }[/math] 1.31 0.18 0.15 2.039
[math]\displaystyle{ 5 \times 10^5-6 \times 10^5 }[/math] 1.34 0.18 0.14
[math]\displaystyle{ 6 \times 10^5-7 \times 10^5 }[/math] 1.33 0.17 0.14


...

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

Comparison between [math]\displaystyle{ H^{eff} = A^{eff}+B^{eff} }[/math], [math]\displaystyle{ A'+B' }[/math], and the effective error bound [math]\displaystyle{ E_1+E_2+E_3 }[/math] on [math]\displaystyle{ H - H^{eff} }[/math] at some points of [math]\displaystyle{ x }[/math] source:

[math]\displaystyle{ x }[/math] [math]\displaystyle{ |H^{eff}/B'_0| }[/math] [math]\displaystyle{ |(A'+B')/B'_0| }[/math] [math]\displaystyle{ |(H^{eff}-(A'+B'))/B'_0| }[/math] [math]\displaystyle{ |(H^{eff}-(A'+B'))/B'_0| + |(E_1+E_2+E_3)/B'_0| }[/math]
10000 0.52 0.52 0.0006 0.039
12131 1.28 1.28 0.0004 0.033
15256 0.97 0.97 0.0003 0.027 18432 0.68 0.68 0.0003 0.023
20567 0.98 0.98 0.0004 0.022
30654 1.93 1.93 0.0004 0.016

...