Polymath15 test problem
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+B }[/math], where
- [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]
There is also the refinement [math]\displaystyle{ A+B-C }[/math], where
- [math]\displaystyle{ 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 ) }[/math]
- [math]\displaystyle{ \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}). }[/math]
The first approximation was modified slightly to [math]\displaystyle{ A'+B' }[/math], where
- [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 [math]\displaystyle{ A^{eff} + B^{eff} }[/math] 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}_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})} }[/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]
There is a refinement [math]\displaystyle{ A^{eff}+B^{eff}-C^{eff} }[/math], where
- [math]\displaystyle{ 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}) }[/math]
- [math]\displaystyle{ s' := \frac{1-y}{2} + iT' = \frac{1-y+ix}{2} + \frac{\pi i t}{8} }[/math]
- [math]\displaystyle{ a := \sqrt{\frac{T'}{2\pi}} }[/math]
- [math]\displaystyle{ p := 1 - 2(a-N) }[/math]
- [math]\displaystyle{ \sigma := \mathrm{Re} s' = \frac{1-y}{2} }[/math]
- [math]\displaystyle{ U := \exp( -i (\frac{T'}{2} \log \frac{T'}{2\pi} - \frac{T'}{2} - \frac{\pi}{8} )) }[/math]
- [math]\displaystyle{ C_0(p) := \frac{ \exp( \pi i (p^2/2 + 3/8) )- i \sqrt{2} \cos(\pi p/2)}{2 \cos(\pi p)}. }[/math]
One can also replace [math]\displaystyle{ C^{eff} }[/math] by the very slightly different quantity
- [math]\displaystyle{ \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} ). }[/math]
Finally, a simplified approximation is [math]\displaystyle{ A^{toy} + B^{toy} }[/math], where
- [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] |
Here some typical values of [math]\displaystyle{ C/B_0 }[/math], which is significantly smaller than either [math]\displaystyle{ A/B_0 }[/math] or [math]\displaystyle{ B/B_0 }[/math]:
| [math]\displaystyle{ x }[/math] | [math]\displaystyle{ C/B_0 }[/math] | [math]\displaystyle{ C^{eff}/B^{eff}_0 }[/math] |
|---|---|---|
| [math]\displaystyle{ 10^3 }[/math] | [math]\displaystyle{ -0.1183 + 0.0697i }[/math] | [math]\displaystyle{ -0.0581 + 0.0823 i }[/math] |
| [math]\displaystyle{ 10^4 }[/math] | [math]\displaystyle{ -0.0001 - 0.0184 i }[/math] | [math]\displaystyle{ -0.0001 - 0.0172 i }[/math] |
| [math]\displaystyle{ 10^5 }[/math] | [math]\displaystyle{ -0.0033 - 0.0005i }[/math] | [math]\displaystyle{ -0.0031 - 0.0005i }[/math] |
| [math]\displaystyle{ 10^6 }[/math] | [math]\displaystyle{ -0.0001 - 0.0006 i }[/math] | [math]\displaystyle{ -0.0001 - 0.0006 i }[/math] |
| [math]\displaystyle{ 10^7 }[/math] | [math]\displaystyle{ -0.0000 - 0.0001 i }[/math] | [math]\displaystyle{ -0.0000 - 0.0001 i }[/math] |
Some values of [math]\displaystyle{ H_t }[/math] and its approximations at small values of [math]\displaystyle{ x }[/math] source source:
| [math]\displaystyle{ x }[/math] | [math]\displaystyle{ H_t }[/math] | [math]\displaystyle{ A+B }[/math] | [math]\displaystyle{ A'+B' }[/math] | [math]\displaystyle{ A^{eff}+B^{eff} }[/math] | [math]\displaystyle{ A^{toy}+B^{toy} }[/math] | [math]\displaystyle{ A+B-C }[/math] | [math]\displaystyle{ A^{eff}+B^{eff}-C^{eff} }[/math] |
|---|---|---|---|---|---|---|---|
| [math]\displaystyle{ 10 }[/math] | [math]\displaystyle{ (3.442 - 0.168 i) \times 10^{-2} }[/math] | 0 | 0 | 0 | N/A | N/A | [math]\displaystyle{ (3.501 - 0.316 i) \times 10^{-2} }[/math] |
| [math]\displaystyle{ 30 }[/math] | [math]\displaystyle{ (-1.000 - 0.071 i) \times 10^{-4} }[/math] | [math]\displaystyle{ (-0.650 - 0.188 i) \times 10^{-4} }[/math] | [math]\displaystyle{ (-0.211 - 0.192 i) \times 10^{-4} }[/math] | [math]\displaystyle{ (-0.670 - 0.114 i) \times 10^{-4} }[/math] | [math]\displaystyle{ (-0.136 + 0.021 i) \times 10^{-4} }[/math] | [math]\displaystyle{ (-1.227 - 0.058 i) \times 10^{-4} }[/math] | [math]\displaystyle{ (-1.032 - 0.066 i) \times 10^{-4} }[/math] |
| [math]\displaystyle{ 100 }[/math] | [math]\displaystyle{ (6.702 + 3.134 i) \times 10^{-16} }[/math] | [math]\displaystyle{ (2.890 + 3.667 i) \times 10^{-16} }[/math] | [math]\displaystyle{ (2.338 + 3.742 i) \times 10^{-16} }[/math] | [math]\displaystyle{ (2.955 + 3.650 i) \times 10^{-16} }[/math] | [math]\displaystyle{ (0.959 + 0.871 i) \times 10^{-16} }[/math] | [math]\displaystyle{ (6.158 + 12.226 i) \times 10^{-16} }[/math] | [math]\displaystyle{ (6.763 + 3.074 i) \times 10^{-16} }[/math] |
| [math]\displaystyle{ 300 }[/math] | [math]\displaystyle{ (-4.016 - 1.401 i) \times 10^{-49} }[/math] | [math]\displaystyle{ (-5.808 - 1.140 i) \times 10^{-49} }[/math] | [math]\displaystyle{ (-5.586 - 1.228 i) \times 10^{-49} }[/math] | [math]\displaystyle{ (-5.824 - 1.129 i) \times 10^{-49} }[/math] | [math]\displaystyle{ (-2.677 - 0.327 i) \times 10^{-49} }[/math] | [math]\displaystyle{ (-3.346 + 6.818 i) \times 10^{-49} }[/math] | [math]\displaystyle{ (-4.032 - 1.408 i) \times 10^{-49} }[/math] |
| [math]\displaystyle{ 1000 }[/math] | [math]\displaystyle{ (0.015 + 3.051 i) \times 10^{-167} }[/math] | [math]\displaystyle{ (-0.479 + 3.126 i) \times 10^{-167} }[/math] | [math]\displaystyle{ (-0.516 + 3.135 i) \times 10^{-167} }[/math] | [math]\displaystyle{ (-0.474 + 3.124 i) \times 10^{-167} }[/math] | [math]\displaystyle{ (-0.406 + 2.051 i) \times 10^{-167} }[/math] | [math]\displaystyle{ (0.175 + 3.306 i) \times 10^{-167} }[/math] | [math]\displaystyle{ (0.017 + 3.047 i) \times 10^{-167} }[/math] |
| [math]\displaystyle{ 3000 }[/math] | [math]\displaystyle{ (-1.144+ 1.5702 i) 10^{-507} }[/math] | [math]\displaystyle{ (-1.039+ 1.5534 i) 10^{-507} }[/math] | [math]\displaystyle{ (-1.039+ 1.5552 i) 10^{-507} }[/math] | [math]\displaystyle{ (-1.038+ 1.5535 i) 10^{-507} }[/math] | [math]\displaystyle{ (-0.925+ 1.3933 i) 10^{-507} }[/math] | [math]\displaystyle{ (-1.155+ 1.5686 i) 10^{-507} }[/math] | [math]\displaystyle{ (-1.144+ 1.5701 i) 10^{-507} }[/math] |
| [math]\displaystyle{ 10000 }[/math] | [math]\displaystyle{ (-0.558 - 4.088 i) \times 10^{-1700} }[/math] | [math]\displaystyle{ (-0.692 - 4.067 i) \times 10^{-1700} }[/math] | [math]\displaystyle{ (-0.687 - 4.067 i) \times 10^{-1700} }[/math] | [math]\displaystyle{ (-0.692 - 4.066 i) \times 10^{-1700} }[/math] | [math]\displaystyle{ (-0.673 - 3.948 i) \times 10^{-1700} }[/math] | [math]\displaystyle{ (-0.548 - 4.089 i) \times 10^{-1700} }[/math] | [math]\displaystyle{ (-0.558 - 4.088 i) \times 10^{-1700} }[/math] |
| [math]\displaystyle{ 30000 }[/math] | [math]\displaystyle{ (3.160 - 6.737) \times 10^{-5110} }[/math] | [math]\displaystyle{ (3.065 - 6.722) \times 10^{-5100} }[/math] | [math]\displaystyle{ (3.066 - 6.722) \times 10^{-5100} }[/math] | [math]\displaystyle{ (3.065 - 6.722) \times 10^{-5100} }[/math] | [math]\displaystyle{ (2.853 - 6.286) \times 10^{-5100} }[/math] | [math]\displaystyle{ (3.170 - 6.733) \times 10^{-5100} }[/math] | [math]\displaystyle{ (3.160 - 6.737) \times 10^{-5100} }[/math] |
Controlling |A+B|/|B_0|
See Controlling A+B/B_0.
Mesh evaluations of [math]\displaystyle{ A^{eff}+B^{eff}/B^{eff}_0 }[/math] in the ranges
- N between 11 and 19
- N between 20 and 150 (raw data
- N between 151 and 300
- The (A+B)/B0 mesh data for N=300 to 20, y=0.45, t=0.4, c=0.065
- c=0.26 for N=7 to 19, y=0.4,t=0.4
- https://drive.google.com/open?id=13_mzqvtaZCghmj7oAZtDRXnkb2zbxQH3 c=0.26 for N=19 to 7, y=0.45,t=0.4]
Controlling |H_t-A-B|/|B_0|
Here is a table on bounds on error terms [math]\displaystyle{ E_1/B^{eff}_0, E_2/B^{eff}_0, E_3^*/B^{eff}_0 }[/math] for N=3 to 2000. Here is a table with some sharpened estimates from the PDF writeup.
Here is a graph depicting [math]\displaystyle{ |H_t-A^{eff}-B^{eff}/B_0^{eff}| }[/math] and [math]\displaystyle{ E_1+E_2+E_3^*/|B_0^{eff}| }[/math] for [math]\displaystyle{ x \leq 1600 }[/math].
Small values of x
Tables of [math]\displaystyle{ H_t(x+iy) }[/math] for small values of [math]\displaystyle{ x }[/math]:
- x=0 to x=300 with step size 0.1
- x=200 to x=600 with step size 0.1
- x=600 to x=1000 with step size 0.1
- x=1000 to x=1300 with step size 0.1
- x=1300 to x=1600 with step size 0.1
- x=20 to x=1000, adaptive mesh
Here are some snapshots of [math]\displaystyle{ H_t/B^{eff}_0 }[/math].
In this range we will need Bounding the derivative of H_t or Bounding the derivative of H_t - second approach or Bounding the derivative of H_t - third approach.
Tables of [math]\displaystyle{ H'_t(x+iy) }[/math]:
Here is a table of [math]\displaystyle{ x }[/math], pari/gp prec, [math]\displaystyle{ H_{t}, H^{'}_{t}, |H_{t}|, |H^{'}_{t}|, \frac{|H_{t}|}{|B_{0}^{eff}|}, \frac{|H^{'}_{t}|}{|B_{0}^{eff}|} }[/math] for x=0 to x=30 with step size 0.01.
Here is a plot of [math]\displaystyle{ H_t/B_0 }[/math] for a rectangle [math]\displaystyle{ \{x+iy: 0 \leq x \leq 300; 0.4 \leq y \leq 0.45\} }[/math]. Here is an adaptive mesh plot; here is a closeup near the origin.
Here is a script for verifying the absence of zeroes of [math]\displaystyle{ H_t }[/math] in a rectangle. It can eliminate zeros in the rectangle [math]\displaystyle{ \{0 \leq x \leq 1000, 0.4 \leq y \leq 0.45\} }[/math] when t = 0.4.
Large negative values of [math]\displaystyle{ t }[/math]
We heuristically compute [math]\displaystyle{ H_t(x) }[/math] in the regime where [math]\displaystyle{ x }[/math] is large and [math]\displaystyle{ t }[/math] is large and negative with [math]\displaystyle{ |t|/x \asymp 1 }[/math]. We shall only be interested in the zeroes and so we discard any multiplicative factor which is non-zero: we write [math]\displaystyle{ X \sim Y }[/math] 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
- [math]\displaystyle{ 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 \quad (3.1) }[/math]
- [math]\displaystyle{ \sim \int_{\bf R} \xi(\frac{1+ix}{2} + i |t|^{1/2} v) e^{-v^2}\ dv. \quad (3.2) }[/math]
To cancel off an exponential decay factor in the [math]\displaystyle{ \xi }[/math] function, it is convenient to shift the v variable by [math]\displaystyle{ \pi |t|^{1/2}/8 }[/math], thus
- [math]\displaystyle{ 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 \quad (3.3) }[/math]
- [math]\displaystyle{ \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 \quad (3.4) }[/math]
where
- [math]\displaystyle{ \tilde x := x - \pi |t|/4 = x + \frac{\pi t}{4}. \quad (3.5) }[/math]
Now from the definition of [math]\displaystyle{ \xi }[/math] and the Stirling approximation we have
- [math]\displaystyle{ \xi(s) \sim M_0(s) \zeta(s)\quad (3.6) }[/math]
where [math]\displaystyle{ M_0 }[/math] is defined in (6) of the writeup. Thus
- [math]\displaystyle{ 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.\quad (3.7) }[/math]
By Taylor expansion we have
- [math]\displaystyle{ 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} )\quad (3.8) }[/math]
- [math]\displaystyle{ \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} )\quad (3.9) }[/math]
where [math]\displaystyle{ \alpha }[/math] is defined in equation (8) of the writeup. We have the approximations
- [math]\displaystyle{ \alpha(\frac{1+i\tilde x}{2} ) \approx \frac{1}{2} \log \frac{\tilde x}{4\pi} + \frac{i\pi}{4} \quad (3.10) }[/math]
and
- [math]\displaystyle{ \alpha'(\frac{1+i\tilde x}{2} ) \approx \frac{-i}{\tilde x} \quad (3.11) }[/math]
and hence
- [math]\displaystyle{ 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.\quad (3.12) }[/math]
The two factors of [math]\displaystyle{ \exp( \pi |t|^{1/2} v/4 ) }[/math] cancel. If we now write
- [math]\displaystyle{ N := \sqrt{\frac{\tilde x}{4\pi}}\quad (3.13) }[/math]
and
- [math]\displaystyle{ u := |t|/N^2 = 4\pi |t|/\tilde x,\quad (3.14) }[/math]
we conclude that
- [math]\displaystyle{ 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.\quad (3.15) }[/math]
If we formally write [math]\displaystyle{ \zeta(s) = \sum_n \frac{1}{n^s} }[/math] (ignoring convergence issues) we obtain
- [math]\displaystyle{ 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\quad (3.16) }[/math]
- [math]\displaystyle{ \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\quad (3.17) }[/math]
We can compute the [math]\displaystyle{ v }[/math] integral to obtain
- [math]\displaystyle{ 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}).\quad (3.18) }[/math]
Using the Taylor approximation
- [math]\displaystyle{ \log \frac{n}{N} \approx \frac{n-N}{N} - \frac{(n-N)^2}{2N^2} \quad (3.19) }[/math]
and dropping some small terms, we obtain
- [math]\displaystyle{ 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} ).\quad (3.20) }[/math]
Writing [math]\displaystyle{ \tilde x = 4\pi N^2 }[/math] and [math]\displaystyle{ |t| = u N^2 }[/math], this becomes
- [math]\displaystyle{ 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 ).\quad (3.21) }[/math]
Writing
- [math]\displaystyle{ N(n-N) = \frac{1}{2} n^2 - \frac{1}{2} (n-N)^2 - \frac{1}{2} N^2 \quad (3.22) }[/math]
we thus have
- [math]\displaystyle{ 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 )\quad (3.23) }[/math]
- [math]\displaystyle{ \sim \sum_n \exp( \frac{16 \pi^2 i (n-N)^2}{8 \pi - iu} ) e^{\pi i n}\quad (3.24) }[/math]
- [math]\displaystyle{ \sim \theta_{01}( \frac{16 \pi N}{8\pi - iu}, \frac{16 \pi}{8\pi - iu} )\quad (3.25) }[/math]
where [math]\displaystyle{ \theta_{01} }[/math] is the theta function defined in this Wikipedia page. Using the Jacobi identity we then have
- [math]\displaystyle{ H_t(x) \sim \theta_{10}(N, \frac{iu - 8\pi}{16 \pi} )\quad (3.26) }[/math]
- [math]\displaystyle{ \sim \theta( N + \frac{1}{2} \frac{iu - 8\pi}{16 \pi}, \frac{iu - 8\pi}{16 \pi})\quad (3.27) }[/math]
- [math]\displaystyle{ \sim \sum_n \exp( - \pi i n(n+1) / 2 ) e^{2\pi i (n+1/2) N} e^{-u n(n+1)/16}.\quad (3.28) }[/math]
As a sanity check, one can verify that the RHS is real-valued, just as [math]\displaystyle{ H_t(x) }[/math] is (by the functional equation).
