Polymath15 test problem: Difference between revisions

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


Finally, a simplified approximation is
Finally, a simplified approximation is

Revision as of 09:17, 25 February 2018

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 \sqrt{\frac{\mathrm{Im} s}{2\pi}} \rfloor = \lfloor \sqrt{\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 \sqrt{\frac{\mathrm{Im} s}{2\pi}} \rfloor = \lfloor \sqrt{\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

The [math]\displaystyle{ E_3 }[/math] error dominates the other two source:

[math]\displaystyle{ x }[/math] [math]\displaystyle{ \frac{E_3}{E_1+E_2} }[/math]
10000 9.11
15000 14.97
20000 19.26
50000 32.39
100000 42.99
[math]\displaystyle{ 10^7 }[/math] 87.23

[math]\displaystyle{ A+B-C }[/math] is a good approximation to [math]\displaystyle{ H_t }[/math] source

[math]\displaystyle{ x }[/math] [math]\displaystyle{ \frac{|H_t-(A+B-C)|}{|B_0|} }[/math]
160 0.06993270565802375041
320 0.006716674125965016299
480 0.005332893070605698501
640 0.003363431256036816251
800 0.1548144749150572349
960 0.03009229958121352990
1120 0.004507664238680722472
1280 0.002283591962997851167
1440 0.01553727684468691873
1600 0.001778051951547709718
1760 0.02763769444052338578
1920 0.002108779890256530964
2080 0.02746770886040058927
2240 0.001567020041379128455
2400 0.01801417530687959747
2560 0.001359561117436848149
2720 0.008503327577240081269
2880 0.001089253262122934826
3040 0.003004181560093288747
3200 0.02931455383125538672

A closer look at the "spike" in error near [math]\displaystyle{ x=800 \approx 256 \pi \approx 804 }[/math]:

[math]\displaystyle{ x }[/math] [math]\displaystyle{ \frac{|H_t-(A+B-C)|}{|B_0|} }[/math]
622.035345 0.003667321
631.460123 0.004268055
640.884901 0.003284407
650.309679 0.004453589
659.734457 0.003872174
669.159235 0.005048162
678.584013 0.005009254
688.008791 0.007418686
697.433569 0.007464541
706.858347 0.010692337
716.283125 0.012938629
725.707903 0.017830524
735.132681 0.022428596
744.557459 0.030907876
753.982237 0.040060298
763.407015 0.053652069
772.831793 0.071092824
782.256571 0.094081856
791.681349 0.123108726
801.106127 0.159299234
810.530905 0.002870724

...