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]
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] |
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 |
In the toy case, we have
- [math]\displaystyle{ \frac{|A^{toy}+B^{toy}|}{|B^{toy}_0|} \geq |\sum_{n=1}^N \frac{b_n}{n^s}| - |\sum_{n=1}^N \frac{a_n}{n^s}| }[/math]
where [math]\displaystyle{ b_n := \exp( \frac{t}{4} \log^2 n) }[/math], [math]\displaystyle{ a_n := (n/N)^{y} b_n }[/math], and [math]\displaystyle{ s := \frac{1+y+ix}{2} + \frac{t}{2} \log N + \frac{\pi i t}{8} }[/math]. For the effective approximation one has
- [math]\displaystyle{ \frac{|A^{eff}+B^{eff}|}{|B^{eff}_0|} \geq |\sum_{n=1}^N \frac{b_n}{n^s}| - |\sum_{n=1}^N \frac{a_n}{n^s}| \quad (2.1) }[/math]
where now [math]\displaystyle{ b_n := \exp( \frac{t}{4} \log^2 n) }[/math], [math]\displaystyle{ s := \frac{1+y+ix}{2} + \frac{t}{2} \alpha_1(\frac{1+y+ix}{2}) }[/math], and
- [math]\displaystyle{ a_n := |\frac{\exp( \frac{t}{4} \alpha_1(\frac{1-y+ix}{2})^2 ) H_{0,1}( \frac{1-y+ix}{2} )}{ \exp( \frac{t}{4} \alpha_1(\frac{1+y+ix}{2})^2 ) H_{0,1}( \frac{1+y+ix}{2} ) }| n^{y - \frac{t}{2} \alpha_1(\frac{1-y+ix}{2}) + \frac{t}{2} \alpha_1(\frac{1+y+ix}{2}) )} b_n. }[/math]
It is thus of interest to obtain lower bounds for expressions of the form
- [math]\displaystyle{ |\sum_{n=1}^N \frac{b_n}{n^s}| - |\sum_{n=1}^N \frac{a_n}{n^s}| }[/math]
in situations where [math]\displaystyle{ b_1=1 }[/math] is expected to be a dominant term.
From the triangle inequality one obtains the lower bound
- [math]\displaystyle{ |\sum_{n=1}^N \frac{b_n}{n^s}| - |\sum_{n=1}^N \frac{a_n}{n^s}| \geq 1 - |a_1| - \sum_{n=2}^N \frac{|a_n|+|b_n|}{n^\sigma} }[/math]
where [math]\displaystyle{ \sigma := \frac{1+y}{2} + \frac{t}{2} \log N }[/math] is the real part of [math]\displaystyle{ s }[/math]. There is a refinement:
Lemma 1 If [math]\displaystyle{ a_n,b_n }[/math] are real coefficients with [math]\displaystyle{ b_1 = 1 }[/math] and [math]\displaystyle{ 0 \leq a_1 \lt 1 }[/math] we have
- [math]\displaystyle{ |\sum_{n=1}^N \frac{b_n}{n^s}| - |\sum_{n=1}^N \frac{a_n}{n^s}| \geq 1 - a_1 - \sum_{n=2}^N \frac{\max( |b_n-a_n|, \frac{1-a_1}{1+a_1} |b_n+a_n|)}{n^\sigma}. }[/math]
Proof By a continuity argument we may assume without loss of generality that the left-hand side is positive, then we may write it as
- [math]\displaystyle{ |\sum_{n=1}^N \frac{b_n - e^{i\theta} a_n}{n^s}| }[/math]
for some phase [math]\displaystyle{ \theta }[/math]. By the triangle inequality, this is at least
- [math]\displaystyle{ |1 - e^{i\theta} a_1| - \sum_{n=2}^N \frac{|b_n - e^{i\theta} a_n|}{n^\sigma}. }[/math]
We factor out [math]\displaystyle{ |1 - e^{i\theta} a_1| }[/math], which is at least [math]\displaystyle{ 1-a_1 }[/math], to obtain the lower bound
- [math]\displaystyle{ (1-a_1) (1 - \sum_{n=2}^N \frac{|b_n - e^{i\theta} a_n| / |1 - e^{i\theta} a_1|}{n^\sigma}). }[/math]
By the cosine rule, we have
- [math]\displaystyle{ (|b_n - e^{i\theta} a_n| / |1 - e^{i\theta} a_1|)^2 = \frac{b_n^2 + a_n^2 - 2 a_n b_n \cos \theta}{1 + a_1^2 -2 a_1 \cos \theta}. }[/math]
This is a fractional linear function of [math]\displaystyle{ \cos \theta }[/math] with no poles in the range [math]\displaystyle{ [-1,1] }[/math] of [math]\displaystyle{ \cos \theta }[/math]. Thus this function is monotone on this range and attains its maximum at either [math]\displaystyle{ \cos \theta=+1 }[/math] or [math]\displaystyle{ \cos \theta = -1 }[/math]. We conclude that
- [math]\displaystyle{ \frac{|b_n - e^{i\theta} a_n|}{|1 - e^{i\theta} a_1|} \leq \max( \frac{|b_n-a_n|}{1-a_1}, \frac{|b_n+a_n|}{1+a_1} ) }[/math]
and the claim follows.
We can also mollify the [math]\displaystyle{ a_n,b_n }[/math]:
Lemma 2 If [math]\displaystyle{ \lambda_1,\dots,\lambda_D }[/math] are complex numbers, then
- [math]\displaystyle{ |\sum_{d=1}^D \frac{\lambda_d}{d^s}| (|\sum_{n=1}^N \frac{b_n}{n^s}| - |\sum_{n=1}^N \frac{a_n}{n^s}|) = ( |\sum_{n=1}^{DN} \frac{\tilde b_n}{n^s}| - |\sum_{n=1}^{DN} \frac{\tilde a_n}{n^s}| ) }[/math]
where
- [math]\displaystyle{ \tilde a_n := \sum_{d=1}^D 1_{n \leq dN} 1_{d|n} \lambda_d a_{n/d} }[/math]
- [math]\displaystyle{ \tilde b_n := \sum_{d=1}^D 1_{n \leq dN} 1_{d|n} \lambda_d b_{n/d} }[/math]
Proof This is immediate from the Dirichlet convolution identities
- [math]\displaystyle{ (\sum_{d=1}^D \frac{\lambda_d}{d^s}) \sum_{n=1}^N \frac{a_n}{n^s} = \sum_{n=1}^N \frac{\tilde a_n}{n^s} }[/math]
and
- [math]\displaystyle{ (\sum_{d=1}^D \frac{\lambda_d}{d^s}) \sum_{n=1}^N \frac{b_n}{n^s} = \sum_{n=1}^N \frac{\tilde b_n}{n^s}. }[/math]
[math]\displaystyle{ \Box }[/math]
Combining the two lemmas, we see for instance that we can show [math]\displaystyle{ |\sum_{n=1}^N \frac{b_n}{n^s}| - |\sum_{n=1}^N \frac{a_n}{n^s}| \gt 0 }[/math] whenever can find [math]\displaystyle{ \lambda_1,\dots,\lambda_D }[/math] with [math]\displaystyle{ \lambda_1=1 }[/math] and
- [math]\displaystyle{ \sum_{n=2}^N \frac{\max( \frac{|\tilde b_n-\tilde a_n|}{1-a_1}, \frac{|\tilde b_n+ \tilde a_n|}{1+a_1})}{n^\sigma} \lt 1. }[/math]
A usable choice of mollifier seems to be the Euler products
- [math]\displaystyle{ \sum_{d=1}^D \frac{\lambda_d}{d^s} := \prod_{p \leq P} (1 - \frac{b_p}{p^s}) }[/math]
which are designed to kill off the first few [math]\displaystyle{ \tilde b_n }[/math] coefficients.
Analysing the toy model
With regards to the toy problem of showing [math]\displaystyle{ A^{toy}+B^{toy} }[/math] does not vanish, here are the least values of [math]\displaystyle{ N }[/math] for which this method works source source source source:
| [math]\displaystyle{ P }[/math] in Euler product | [math]\displaystyle{ N }[/math] using triangle inequality | [math]\displaystyle{ N }[/math] using Lemma 1 |
|---|---|---|
| 1 | 1391 | 1080 |
| 2 | 478 | 341 |
| 3 | 322 | 220 |
| 5 | 282 | 192 |
| 7 | 180 | |
| 11 | 176 |
Dropping the [math]\displaystyle{ \lambda_6 }[/math] term from the [math]\displaystyle{ P=3 }[/math] Euler factor worsens the 220 threshold slightly to 235 source.
Analysing the effective model
The differences between the toy model and the effective model are:
- The real part [math]\displaystyle{ \sigma }[/math] of [math]\displaystyle{ s }[/math] is now [math]\displaystyle{ \frac{1+y}{2} + \frac{t}{2} \mathrm{Re} \alpha_1(\frac{1+y+ix}{2}) }[/math] rather than [math]\displaystyle{ \frac{1+y}{2} + \frac{t}{2} \log N }[/math]. (The imaginary part of [math]\displaystyle{ s }[/math] also changes somewhat.)
- The coefficient [math]\displaystyle{ a_n }[/math] is now given by
- [math]\displaystyle{ a_n = \lambda n^{y + \frac{t}{2} (\alpha_1(\frac{1+y+ix}{2}) - \alpha_1(\frac{1-y+ix}{2}))} b_n }[/math]
rather than [math]\displaystyle{ a_n = N^{-y} n^y b_n }[/math], where
- [math]\displaystyle{ \lambda := |\frac{\exp( \frac{t}{4} \alpha_1(\frac{1-y+ix}{2})^2 H_{0,1}( \frac{1-y+ix}{2})}{\exp( \frac{t}{4} \alpha_1(\frac{1-y+ix}{2})^2 H_{0,1}( \frac{1-y+ix}{2})}|. }[/math]
Two complications arise here compared with the toy model: firstly, [math]\displaystyle{ \sigma,a_n }[/math] now depend on [math]\displaystyle{ x }[/math] and not just on [math]\displaystyle{ N }[/math], and secondly the [math]\displaystyle{ a_n }[/math] are not quite real-valued making it more difficult to apply Lemma 1.
However we have good estimates for [math]\displaystyle{ \sigma,a_n }[/math] that depend only on [math]\displaystyle{ N }[/math]. Note that
- [math]\displaystyle{ 2\pi N^2 \leq T' \lt 2\pi (N+1)^2 }[/math]
and hence
- [math]\displaystyle{ x_N \leq x \lt x_{N+1} }[/math]
where
- [math]\displaystyle{ x_N := 4\pi N^2 - \frac{\pi t}{4}. }[/math]
To control [math]\displaystyle{ \sigma }[/math], it suffices to obtain lower bounds because our criteria (both the triangle inequality and Lemma 1) become harder to satisfy when [math]\displaystyle{ \sigma }[/math] decreases. We compute
- [math]\displaystyle{ \sigma = \frac{1+y}{2} + \frac{t}{2} \mathrm{Re}(\frac{1}{1+y+ix} + \frac{2}{-1+y+ix} + \frac{1}{2} \log \frac{1+y+ix}{4\pi}) }[/math]
- [math]\displaystyle{ = \frac{1+y}{2} + \frac{t}{2} (\frac{1+y}{(1+y)^2+x^2} + \frac{-2+2y}{(-1+y)^2+x^2} + \frac{1}{2} \log \frac{|1+y+ix|}{4\pi}) }[/math]
- [math]\displaystyle{ \geq \frac{1+y}{2} + \frac{t}{2} (\frac{1+y}{(-1+y)^2+x^2} + \frac{-2+2y}{(-1+y)^2+x^2} + \frac{1}{2} \log \frac{x}{4\pi}) }[/math]
- [math]\displaystyle{ \geq \frac{1+y}{2} + \frac{t}{2} (\frac{3y-1}{(-1+y)^2+x^2} + \log N) }[/math]
- [math]\displaystyle{ \geq \frac{1+y}{2} + \frac{t}{2} \log N }[/math]
assuming that [math]\displaystyle{ y \geq 1/3 }[/math]. Hence we can actually just use the same value of [math]\displaystyle{ \sigma }[/math] as in the toy case.
Next we control [math]\displaystyle{ \lambda }[/math]. Note that we can increase [math]\displaystyle{ \lambda }[/math] (thus multiplying [math]\displaystyle{ \sum_{n=1}^N \frac{a_n}{n^s} }[/math] by a quantity greater than 1) without affecting (2.1), so we just need upper bounds on [math]\displaystyle{ \lambda }[/math]. We may factor
- [math]\displaystyle{ \lambda = \exp( \frac{t}{4} \mathrm{Re} (\alpha_1(\frac{1-y+ix}{2})^2 - \alpha_1(\frac{1+y+ix}{2})^2) + \mathrm{Re}( f(\frac{1-y+ix}{2}) - f(\frac{1+y+ix}{2} ) ) }[/math]
where
- [math]\displaystyle{ f(s) := -\frac{s}{2} \log \pi + (\frac{s}{2} - \frac{1}{2}) \log \frac{s}{2} - \frac{s}{2}. }[/math]
By the mean value theorem, we have
- [math]\displaystyle{ \mathrm{Re} (\alpha_1(\frac{1-y+ix}{2})^2 - \alpha_1(\frac{1+y+ix}{2})^2) = -2 y \alpha_1(s') \alpha'_1(s') }[/math]
for some [math]\displaystyle{ s_1 }[/math] between [math]\displaystyle{ \frac{1-y+ix}{2} }[/math] and [math]\displaystyle{ \frac{1+iy}{2} }[/math]. We have
- [math]\displaystyle{ \alpha_1(s_1) = \frac{1}{2s_1} + \frac{1}{s_1-1} + \frac{1}{2} \log \frac{s_1}{2\pi} }[/math]
- [math]\displaystyle{ = O_{\leq}(\frac{1}{x}) + O_{\leq}(\frac{1}{x/2}) + \frac{1}{2} \log \frac{|s_1|}{2\pi} + O_{\leq}(\frac{\pi}{4}) }[/math]
- [math]\displaystyle{ = O_{\leq}( \frac{\pi}{4} + \frac{3}{x_N}) + \frac{1}{2} O_{\leq}^{\mathbf{R}}( \log \frac{|1+y+ix_{N+1}|}{4\pi} ) }[/math]
and
- [math]\displaystyle{ \alpha'_1(s_1) = -\frac{1}{2s_1^2} + \frac{1}{(s_1-1)^2} + \frac{1}{2s_1} }[/math]
- [math]\displaystyle{ = O_{\leq}(\frac{1}{x^2/2}) + O_{\leq}(\frac{1}{x^2/4}) + \frac{1}{2s_1} }[/math]
- [math]\displaystyle{ = O_{\leq}(\frac{6}{x_N^2}) + \frac{1}{2s_1} }[/math]
- [math]\displaystyle{ = O_{\leq}(\frac{6}{x_N^2}) + O_{\leq}( \frac{1}{x_N} ). }[/math]
Thus one has
- [math]\displaystyle{ \mathrm{Re} (\alpha_1(\frac{1-y+ix}{2})^2 - \alpha_1(\frac{1+y+ix}{2})^2) = 2y O_{\leq}( (\frac{\pi}{4} + \frac{3}{x_N}) (\frac{1}{x_N} + \frac{6}{x_N^2}) ) }[/math]
- [math]\displaystyle{ + 2y O_{\leq}( \log \frac{|1+y+ix_{N+1}|}{4\pi} (\frac{6}{x_N^2} + |\mathrm{Re} \frac{1}{2s'}|) ) }[/math]
Now we have
- [math]\displaystyle{ \mathrm{Re} \frac{1}{2s'} = \frac{\mathrm{Re}(s')}{2|s'|^2} }[/math]
- [math]\displaystyle{ \leq \frac{1+y}{x^2} }[/math]
- [math]\displaystyle{ \leq \frac{1+y}{x_N^2}; }[/math]
also
- [math]\displaystyle{ (\frac{\pi}{4} + \frac{3}{x_N}) (\frac{1}{x_N} + \frac{6}{x_N^2}) \leq \frac{\pi}{4} (1 + \frac{12/\pi}{x_N}) \frac{1}{x_N-6} }[/math]
- [math]\displaystyle{ \leq \frac{\pi}{4} ( \frac{1}{x_N-6} + \frac{12/\pi}{(x_N-6)^2} ) }[/math]
- [math]\displaystyle{ \leq \frac{\pi}{4} \frac{1}{x_N - 6 - 12/\pi}. }[/math]
We conclude that
- [math]\displaystyle{ \mathrm{Re} (\alpha_1(\frac{1-y+ix}{2})^2 - \alpha_1(\frac{1+y+ix}{2})^2) = O_{\leq}(\frac{\pi y}{2 (x_N - 6 - 12/\pi)} + \frac{2y(7+y)}{x_N^2} \log \frac{|1+y+ix_{N+1}|}{4\pi}). }[/math]
In a similar vein, from the mean value theorem we have
- [math]\displaystyle{ \mathrm{Re}( f(\frac{1-y+ix}{2}) - f(\frac{1+y+ix}{2} ) = -y \mathrm{Re} f'(s_2) }[/math]
for some [math]\displaystyle{ s_2 }[/math] between [math]\displaystyle{ \frac{1-y+ix}{2} }[/math] and [math]\displaystyle{ \frac{1+y+ix}{2} }[/math]. We have
- [math]\displaystyle{ \mathrm{Re} f'(s_2) = -\frac{1}{2} \log \pi + \frac{1}{2} \log \frac{|s_2|}{2} - \mathrm{Re} \frac{1}{2s_2} }[/math]
- [math]\displaystyle{ = \frac{1}{2} \log \frac{|s_2|}{2\pi} + O_{\leq}(\frac{\mathrm{Re}(s_2)}{2|s_2|^2}) }[/math]
- [math]\displaystyle{ \geq \log N + O_{\leq}(\frac{1+y}{x^2}) }[/math]
- [math]\displaystyle{ \geq \log N + O_{\leq}(\frac{1+y}{x_N^2}) }[/math]
and thus
- [math]\displaystyle{ \lambda \leq N^{-y} \exp( \frac{\pi y}{2 (x_N - 6 - 12/\pi)} + \frac{2y(7+y)}{x_N^2} \log \frac{|1+y+ix_{N+1}|}{4\pi} + \frac{y(1+y)}{x_N^2} ) }[/math]
- [math]\displaystyle{ \leq e^\delta N^{-y} }[/math]
where
- [math]\displaystyle{ \delta := \frac{\pi y}{2 (x_N - 6 - \frac{14+2y}{\pi})} + \frac{2y(7+y)}{x_N^2} \log \frac{|1+y+ix_{N+1}|}{4\pi} }[/math]
Asymptotically we have
- [math]\displaystyle{ \delta = \frac{\pi y}{2 x_N} + O( \frac{\log x_N}{x_N^2} ) = O( \frac{1}{x_N} ). }[/math]
Now we control [math]\displaystyle{ \alpha_1(\frac{1+y+ix}{2}) - \alpha_1(\frac{1-y+ix}{2}) }[/math]. By the mean-value theorem we have
- [math]\displaystyle{ \alpha_1(\frac{1+y+ix}{2}) - \alpha_1(\frac{1-y+ix}{2}) = O_{\leq}( y |\alpha'_1(s_3)|) }[/math]
for some [math]\displaystyle{ s_3 }[/math] between [math]\displaystyle{ \frac{1+y+ix}{2} }[/math] and [math]\displaystyle{ \frac{1-y+ix}{2} }[/math]. As before we have
- [math]\displaystyle{ \alpha'_1(s_3) = -\frac{1}{2s_3^2} - \frac{1}{(s_3-1)^2} + \frac{1}{2s_3} }[/math]
- [math]\displaystyle{ = O_{\leq}( \frac{1}{x^2/2} + \frac{1}{x^2/4} + \frac{1}{x} ) }[/math]
- [math]\displaystyle{ = O_{\leq}( \frac{1}{x_N} + \frac{6}{x_N^2} ) }[/math]
- [math]\displaystyle{ = O_{\leq}( \frac{1}{x_N-6} ). }[/math]
We conclude that (after replacing [math]\displaystyle{ \lambda }[/math] with [math]\displaystyle{ e^\delta N^{-y} }[/math])
- [math]\displaystyle{ a_n = (n/N)^y \exp( \delta + O_{\leq}( \frac{t y \log n}{2(x_N-6)} ) ) b_n. }[/math]
The triangle inequality argument will thus give [math]\displaystyle{ A^{eff}+B^{eff} }[/math] non-zero as long as
- [math]\displaystyle{ \sum_{n=1}^N (1 + (n/N)^y \exp( \delta + \frac{t y \log n}{2(x_N-6)} ) ) \frac{b_n}{n^\sigma} \lt 2. }[/math]
The situation with using Lemma 1 is a bit more complicated because [math]\displaystyle{ a_n }[/math] is not quite real. We can write [math]\displaystyle{ a_n = e^\delta a_n^{toy} + O_{\leq}( e_n ) }[/math] where
- [math]\displaystyle{ a_n^{toy} := (n/N)^y b_n }[/math]
and
- [math]\displaystyle{ e_n := e^\delta (n/N)^y (\exp( \frac{t y \log n}{2(x_N-6)} ) - 1) b_n }[/math]
and then by Lemma 1 and the triangle inequality we can make [math]\displaystyle{ A^{eff}+B^{eff} }[/math] non-zero as long as
- [math]\displaystyle{ a_1^{toy} + \sum_{n=2}^N \frac{\max( |b_n-a_n^{toy}|, \frac{1-a_1^{toy}}{1+a_1^{toy}} |b_n + a_n^{toy}|}{n^\sigma} + \sum_{n=1}^N \frac{e_n}{n^\sigma} \lt 1. }[/math]
Controlling |H_t-A-B|/|B_0|
As computed in Effective bounds on H_t - second approach, there is an effective bound
- [math]\displaystyle{ |H_{eff} - A^{eff} - B^{eff}| \leq E_1 + E_2 + E_3 }[/math]
where
- [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{ E_1 := \frac{1}{8 (T - 3.33)} \exp( \frac{t}{4} \mathrm{Re} \alpha_1(\frac{1-y+ix}{2})^2 ) |H_{0,1}(\frac{1-y+ix}{2})| \epsilon'(\frac{1-y+ix}{2}) }[/math]
- [math]\displaystyle{ E_2 := \frac{1}{8 (T - 3.33)} \exp( \frac{t}{4} \mathrm{Re} \alpha_1(\frac{1+y+ix}{2})^2 ) |H_{0,1}(\frac{1+y+ix}{2})| \epsilon'(\frac{1+y+ix}{2}) }[/math]
- [math]\displaystyle{ E_3 := \frac{1}{8} \sqrt{\pi} \exp( -\frac{t \pi^2}{64} ) (T')^{3/2} e^{-\pi T/4} \int_{-\infty}^\infty v(\sigma) w(\sigma) f(\sigma)\ d\sigma }[/math]
- [math]\displaystyle{ \epsilon'(s) := \frac{1}{2} \sum_{n=1}^N \frac{1}{n^{\mathrm{Re}(s) + \frac{t \mathrm{Re} \alpha_1(s)}{2} - \frac{t}{4} \log n}} \exp(\frac{1}{2(T-3.33)} (\frac{t^2}{4} |\alpha_1(s) - \log n|^2 + \frac{1}{3} + t)) (\frac{t^2}{4} |\alpha_1(s) - \log n|^2 + \frac{1}{3} + t ) }[/math]
- [math]\displaystyle{ f(\sigma) := \frac{1}{2\sqrt{\pi t}} (e^{-(\sigma-(1-y)/2)^2/t} + e^{-(\sigma-(1+y)/2)^2/t}) \quad (4.1) }[/math]
- [math]\displaystyle{ w(\sigma) := (1 + \frac{\sigma^2}{(T'_0)^2})^{1/2} (1 + \frac{(1-\sigma)^2}{(T'_0)^2})^{1/2} \exp( \frac{(\sigma-1)_+}{4} \log (1 + \frac{\sigma^2}{(T'_0)^2}) + (\frac{T'_0}{2} \arctan \frac{\sigma}{T'_0} - \frac{\sigma}{2}) 1_{\sigma \lt 0} + \frac{1}{12(T'_0 - 0.33)}) }[/math]
- [math]\displaystyle{ v(\sigma) := 1 + (0.400 \frac{9^\sigma}{a_0} + 0.346 \frac{2^{3\sigma/2}}{a_0^2}) 1_{\sigma \geq 0} + (9/10)^{\lceil -\sigma \rceil} \sum_{1 \leq k \leq 4-\sigma} (1.1)^k \frac{\Gamma(k/2)}{a_0^k} 1_{\sigma \lt 0} }[/math]
- [math]\displaystyle{ a_0 := \sqrt{\frac{T'_0}{2\pi}} }[/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]
- [math]\displaystyle{ T'_0 := T_0 + \frac{\pi t}{8} }[/math]
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 source
| [math]\displaystyle{ x }[/math] | [math]\displaystyle{ \frac{|H_t-(A+B-C)|}{|B_0|} }[/math] | [math]\displaystyle{ \frac{|H_t-(A^{eff}+B^{eff}-C^{eff})|}{|B_0^{eff}|} }[/math] |
|---|---|---|
| 160 | 0.06993270565802375041 | 0.009155667752 |
| 320 | 0.006716674125965016299 | 0.0005529962481 |
| 480 | 0.005332893070605698501 | 0.0004966282128 |
| 640 | 0.003363431256036816251 | 0.0004482768972 |
| 800 | 0.1548144749150572349 | 0.002686105466 |
| 960 | 0.03009229958121352990 | 0.001270168744 |
| 1120 | 0.004507664238680722472 | 0.0009957229500 |
| 1280 | 0.002283591962997851167 | 0.0007024411378 |
| 1440 | 0.01553727684468691873 | 0.0007000473085 |
| 1600 | 0.001778051951547709718 | 0.0004882487218 |
| 1760 | 0.02763769444052338578 | 0.0002518910919 |
| 1920 | 0.002108779890256530964 | 0.0008378989413 |
| 2080 | 0.02746770886040058927 | 0.0004924765754 |
| 2240 | 0.001567020041379128455 | 0.0001171320991 |
| 2400 | 0.01801417530687959747 | 0.0002443802551 |
| 2560 | 0.001359561117436848149 | 0.0004569058755 |
| 2720 | 0.008503327577240081269 | 0.0006355966221 |
| 2880 | 0.001089253262122934826 | 0.0008864917365 |
| 3040 | 0.003004181560093288747 | 0.00004326840265 |
| 3200 | 0.02931455383125538672 | 0.0003598521453 |
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 |
In practice [math]\displaystyle{ E_1/B^{eff}_0 }[/math] is smaller than [math]\displaystyle{ E_2/B^{eff}_0 }[/math], which is mostly dominated by the first term in the sum which is close to [math]\displaystyle{ \frac{t^2}{16 x} \log^2 \frac{x}{4\pi} }[/math]:
| [math]\displaystyle{ x }[/math] | [math]\displaystyle{ E_1 / B^{eff}_0 }[/math] | [math]\displaystyle{ E_2 / B^{eff}_0 }[/math] | [math]\displaystyle{ \frac{t^2}{16x} \log^2 \frac{x}{4\pi} }[/math] |
|---|---|---|---|
| 10^3 | [math]\displaystyle{ 1.389 \times 10^{-3} }[/math] | [math]\displaystyle{ 2.341 \times 10^{-3} }[/math] | [math]\displaystyle{ 1.915 \times 10^{-4} }[/math] |
| 10^4 | [math]\displaystyle{ 1.438 \times 10^{-4} }[/math] | [math]\displaystyle{ 3.156 \times 10^{-4} }[/math] | [math]\displaystyle{ 4.461 \times 10^{-5} }[/math] |
| 10^5 | [math]\displaystyle{ 1.118 \times 10^{-5} }[/math] | [math]\displaystyle{ 3.574 \times 10^{-5} }[/math] | [math]\displaystyle{ 8.067 \times 10^{-6} }[/math] |
| 10^6 | [math]\displaystyle{ 7.328 \times 10^{-7} }[/math] | [math]\displaystyle{ 3.850 \times 10^{-6} }[/math] | [math]\displaystyle{ 1.273 \times 10^{-6} }[/math] |
| 10^7 | [math]\displaystyle{ 4.414 \times 10^{-8} }[/math] | [math]\displaystyle{ 4.197 \times 10^{-7} }[/math] | [math]\displaystyle{ 1.846 \times 10^{-7} }[/math] |
Estimation of [math]\displaystyle{ E_1,E_2 }[/math]
...
Estimation of [math]\displaystyle{ E_3 }[/math]
Here we assume that [math]\displaystyle{ T_0 \geq 100 }[/math], which implies also [math]\displaystyle{ T'_0 \geq 100 }[/math].
We first bound [math]\displaystyle{ w }[/math] by a Gaussian type quantity.
We have
- [math]\displaystyle{ 1 + \frac{\sigma^2}{(T'_0)^2} \leq \exp( \frac{\sigma^2}{(T'_0)^2}) }[/math]
and
- [math]\displaystyle{ 1 + \frac{(1-\sigma)^2}{(T'_0)^2} \leq \exp( \frac{(1-\sigma)^2}{(T'_0)^2}) }[/math]
and thus
- [math]\displaystyle{ ( 1 + \frac{\sigma^2}{(T'_0)^2} )^{1/2} (1 + \frac{(1-\sigma)^2}{(T'_0)^2})^{1/2} \leq \exp( \frac{1}{2} \frac{\sigma^2}{(T'_0)^2} + \frac{1}{2} \frac{(1-\sigma)^2}{(T'_0)^2} ) }[/math]
- [math]\displaystyle{ = \exp( \frac{(\sigma-1/2)^2}{(T'_0)^2} + \frac{1}{4 (T'_0)^2} ). }[/math]
Next, from calculus one can verify the bounds
- [math]\displaystyle{ \log(1+x^2) \leq 1.479 \sqrt{x} }[/math]
and
- [math]\displaystyle{ x - \mathrm{arctan}(x) \leq 0.230 x^2 }[/math]
for any [math]\displaystyle{ x \geq 0 }[/math], and hence
- [math]\displaystyle{ \frac{(\sigma-1)_+}{4} \log (1 + \frac{\sigma^2}{(T'_0)^2}) \leq \frac{1}{4} 1.479 \frac{\sigma(\sigma-1)}{T'_0} 1_{\sigma \geq 1} }[/math]
- [math]\displaystyle{ \leq 0.37 \frac{(\sigma-1/2)^2}{T'_0} 1_{\sigma \geq 1} }[/math]
and
- [math]\displaystyle{ (\frac{T'_0}{2} \arctan \frac{\sigma}{T'_0} - \frac{\sigma}{2}) 1_{\sigma \lt 0} \leq \frac{T'_0}{2} 1_{\sigma\lt 0} 0.230 (\frac{|\sigma|}{T'_0})^2 }[/math]
- [math]\displaystyle{ \leq 0.115 \frac{(\sigma-1/2)^2}{T'_0} 1_{\sigma \lt 0}. }[/math]
We conclude that
- [math]\displaystyle{ w(\sigma) \leq \exp( \frac{(\sigma-1/2)^2}{(T'_0)^2} + \frac{1}{4 (T'_0)^2} + 0.37 \frac{(\sigma-1/2)^2}{T'_0} + \frac{1}{12(T'_0 - 0.33)}) }[/math]
- [math]\displaystyle{ \leq \exp( 0.37 \frac{(\sigma-1/2)^2}{T'_0-2.71} + \frac{1}{12(T'_0 - 3.33)}). }[/math]
Now we work on [math]\displaystyle{ \nu }[/math]. Observe that if [math]\displaystyle{ k \leq \frac{T'_0}{2.42 \pi} = \frac{a_0^2}{1.21} }[/math] then
- [math]\displaystyle{ (1.1)^{k+2} \frac{\Gamma(\frac{k+2}{2})}{a_0^{k+2}} = \frac{1.21 k}{2 a_0^2} \frac{\Gamma(\frac{k}{2})}{a_0^k} \leq \frac{1}{2} (1.1)^{k} \frac{\Gamma(\frac{k}{2})}{a_0^{k}}, }[/math]
and hence
- [math]\displaystyle{ \sum_{2 \leq k \leq \frac{T'_0}{2.24 \pi}; k\ \mathrm{even}} (1.1)^{k} \frac{\Gamma(\frac{k}{2})}{a_0^{k}} \leq 2 (1.1)^2 \frac{\Gamma(\frac{2}{2})}{a_0^2} = \frac{2.42 \sqrt{\pi}}{a_0^2} }[/math]
and similarly
- [math]\displaystyle{ \sum_{3 \leq k \leq \frac{T'_0}{2.42 \pi}; k\ \mathrm{odd}} (1.1)^{k} \frac{\Gamma(\frac{k}{2})}{a_0^{k}} \leq 2 (1.1)^3 \frac{\Gamma(\frac{3}{2})}{a_0^2} = \frac{1.331}{a_0^3} }[/math]
and hence
- [math]\displaystyle{ \sum_{1 \leq k \leq \frac{T'_0}{2.42 \pi}} (1.1)^{k} \frac{\Gamma(\frac{k}{2})}{a_0^{k}} \leq \frac{1.1 \sqrt{\pi}}{a_0} + \frac{2.42}{a_0^2} + \frac{1.331 \sqrt{\pi}}{a_0^3} }[/math]
- [math]\displaystyle{ \leq \frac{1.1 \sqrt{\pi}}{a_0 - 1.25}; }[/math]
also
- [math]\displaystyle{ (0.400 \frac{9^\sigma}{a_0} + 0.346 \frac{2^{3\sigma/2}}{a_0^2})1_{\sigma \geq 0} \leq 0.400 \times 9^\sigma (\frac{1}{a_0} + 0.865 \frac{1}{a_0^2}) }[/math]
- [math]\displaystyle{ \leq 0.4 \frac{9^\sigma}{a_0 - 0.865} }[/math]
and hence (bounding [math]\displaystyle{ (0.9)^{\lceil -\sigma \rceil} \leq \frac{1}{1.1} }[/math])
- [math]\displaystyle{ v(\sigma) \leq 1 + 0.400 \frac{9^\sigma}{a_0-0.865} + \frac{\sqrt{\pi}}{a_0-1.25} + \sum_{\frac{T'_0}{2.42 \pi} \lt k \leq 4-\sigma} \frac{(1.1)^{k-1}}{a_0^k} \Gamma(k/2). }[/math]
We conclude (using Fubini's theorem) that
- [math]\displaystyle{ \int_{-\infty}^\infty vwf(\sigma)\ d\sigma \leq \exp(\frac{1}{12(T'_0 - 3.33)}) ( (1 + \frac{\sqrt{\pi}}{a_0-1.25}) \int_{-\infty}^\infty f(\sigma) \exp( 0.37 \frac{(\sigma-1/2)^2}{T'_0-2.71} )\ d\sigma }[/math]
- [math]\displaystyle{ + \frac{0.4}{a_0-0.865} \int_{-\infty}^\infty 9^\sigma f(\sigma) \exp( 0.37 \frac{(\sigma-1/2)^2}{T'_0-2.71} )\ d\sigma }[/math]
- [math]\displaystyle{ + \sum_{k \gt \frac{T'_0}{2.42\pi}} \frac{(1.1)^{k-1}}{a_0^k} \Gamma(k/2) \int_{-\infty}^{4-k} f(\sigma) \exp( 0.37 \frac{(\sigma-1/2)^2}{T'_0-2.71} )\ d\sigma.) }[/math]
Now we estimate the integrals appearing in the right-hand side. By symmetry we have
- [math]\displaystyle{ \int_{-\infty}^\infty f(\sigma) \exp( 0.37 \frac{(\sigma-1/2)^2}{T'_0-2.71} )\ d\sigma = \frac{1}{\sqrt{\pi t}} \int_{-\infty}^\infty \exp( - \frac{(\sigma - (1+y)/2)^2}{t} + 0.37 \frac{(\sigma-1/2)^2}{T'_0-2.71} )\ d\sigma. }[/math]
- [math]\displaystyle{ = \frac{1}{\sqrt{\pi t}} \int_{-\infty}^\infty \exp( - \frac{\sigma^2}{t} + 0.37 \frac{(\sigma-y/2)^2}{T'_0-2.71} )\ d\sigma. }[/math]
Using the Gaussian identity
- [math]\displaystyle{ \int_{-\infty}^\infty \exp( - (a\sigma^2 + b \sigma + c) )\ d\sigma = \sqrt{\pi} a^{-1/2} \exp( - c + \frac{b^2}{4a} ), }[/math]
valid for any [math]\displaystyle{ a,b,c }[/math] with [math]\displaystyle{ a }[/math] positive, we can write the above expression as
- [math]\displaystyle{ (1 - \frac{0.37 t}{T'_0 - 2.71})^{-1/2} \exp( \frac{0.37 y^2}{4 (T'_0 - 2.71 - 0.37 t)} ). }[/math]
Similarly, since [math]\displaystyle{ 9^\sigma }[/math] is larger for [math]\displaystyle{ \sigma \geq 1/2 }[/math] than for [math]\displaystyle{ \sigma \lt 1/2 }[/math], we have
- [math]\displaystyle{ \int_{-\infty}^\infty 9^\sigma f(\sigma) \exp( 0.37 \frac{(\sigma-1/2)^2}{T'_0-2.71} )\ d\sigma \leq \frac{1}{\sqrt{\pi t}} \int_{-\infty}^\infty 9^\sigma \exp( - \frac{(\sigma - (1+y)/2)^2}{t} + 0.37 \frac{(\sigma-1/2)^2}{T'_0-2.71} )\ d\sigma. }[/math]
- [math]\displaystyle{ = \frac{3^{1+y}}{\sqrt{\pi t}} \int_{-\infty}^\infty \exp( - \frac{\sigma^2}{t} + 0.37 \frac{(\sigma-y/2)^2}{T'_0-2.71} + \sigma \log 9)\ d\sigma. }[/math]
- [math]\displaystyle{ = 3^{1+y} (1 - \frac{0.37 t}{T'_0 - 2.71})^{-1/2} \exp( 0.37 \frac{y^2}{4 (T'_0-2.71)} + \frac{b^2}{4 (\frac{1}{t} - \frac{0.37}{T'_0-2.71})} ) }[/math]
where
- [math]\displaystyle{ b := - \log 9 + 0.37 \frac{y}{T'_0 - 2.71}. }[/math]
If [math]\displaystyle{ T'_0 \geq 100 }[/math] and [math]\displaystyle{ y \leq 1/2 }[/math] then [math]\displaystyle{ |b| \leq \log 9 }[/math], thus the above integral is at most
- [math]\displaystyle{ = 3^{1+y} (1 - \frac{0.37 t}{T'_0 - 2.71})^{-1/2} \exp( 0.37 \frac{y^2}{4 (T'_0-2.71)} + \frac{t \log^2 9}{4 (1 - \frac{0.37 t}{T'_0-2.71})} ). }[/math]
Now we consider the integral
- [math]\displaystyle{ \int_{-\infty}^{4-k} f(\sigma) \exp( 0.37 \frac{(\sigma-1/2)^2}{T'_0-2.71} )\ d\sigma. }[/math]
If we assume that [math]\displaystyle{ T_0 \geq 100 }[/math], then [math]\displaystyle{ 4-k \leq 4 - \frac{100}{2.42 \pi} \leq -9 }[/math] is negative, so this expression is at most
- [math]\displaystyle{ \leq \frac{1}{\sqrt{\pi t}} \int_{-\infty}^{4-k} \exp( - \frac{(\sigma - (1-y)/2)^2}{t} + 0.37 \frac{(\sigma-1/2)^2}{T'_0-2.71} )\ d\sigma }[/math]
- [math]\displaystyle{ \leq \frac{1}{\sqrt{\pi t}} \int_{-\infty}^{4-k} \exp( - \sigma^2 (\frac{1}{t} - \frac{0.37}{T'_0-2.71}) )\ d\sigma. }[/math]
With [math]\displaystyle{ t \leq 0.4 }[/math] and [math]\displaystyle{ T'_0 \geq 100 }[/math], one can verify numerically that
- [math]\displaystyle{ \frac{1}{t} - \frac{0.37}{T'_0-2.71} \geq 2 + \frac{1}{2} \log t }[/math]
and so (since [math]\displaystyle{ \sigma^2 \geq 1 }[/math]) one can bound the above by
- [math]\displaystyle{ \leq \frac{1}{\sqrt{\pi}} \int_{-\infty}^{4-k} \exp( - 2 \sigma^2 )\ d\sigma }[/math]
- [math]\displaystyle{ \leq \frac{1}{\sqrt{\pi}} \exp( - 2 (k - 4)^2 ) \frac{1}{4 (k - 4)} }[/math]
and so the contribution to [math]\displaystyle{ \int_{-\infty}^\infty vwf(\sigma)\ d\sigma) }[/math] is at most
- [math]\displaystyle{ \frac{1}{4 (\frac{T'_0}{2.42\pi} - 4) \sqrt{\pi}} \sum_{k \gt \frac{T'_0}{2.42\pi}} c_k }[/math]
where
- [math]\displaystyle{ c_k := \frac{(1.1)^{k-1}}{a_0^k} \Gamma(k/2) \exp( - 2(k-4)^2 ). }[/math]
Observe that
- [math]\displaystyle{ c_{k+2}/c_k = \frac{(1.1)^2}{a_0^2} \frac{k}{2} \exp( - 4 (k+5) ) }[/math]
and this can be shown to be less than [math]\displaystyle{ 1/2 }[/math] if [math]\displaystyle{ T_0 \geq 100 }[/math], and [math]\displaystyle{ k \gt \frac{T'_0}{2.42 \pi} }[/math]. Thus
- [math]\displaystyle{ \sum_{k \gt \frac{T'_0}{2.42\pi}} c_k \leq 4 \sup_{\frac{T'_0}{2.42\pi} \lt k \leq \frac{T'_0}{2.42\pi}+2} a_k }[/math]
- [math]\displaystyle{ \leq 4 (\frac{1.1}{a_0})^{\frac{T'_0}{2.42\pi}} \Gamma( \frac{T'_0}{4.84\pi}+1 ) \exp( - 4 (\frac{T'_0}{2.42\pi}-4)^2 ). }[/math]
Putting all this together, we obtain
- [math]\displaystyle{ \int_{-\infty}^\infty vwf(\sigma)\ d\sigma \leq \exp(\frac{1}{12(T'_0 - 3.33)}) (1 - \frac{0.37 t}{T'_0 - 2.71})^{-1/2} \exp( \frac{0.37 y^2}{4 (T'_0 - 2.71 - 0.37 t)} ) \times }[/math]
- [math]\displaystyle{ (1 + \frac{\sqrt{\pi}}{a_0-1.25} + \frac{0.4}{a_0-0.85} 3^{1+y} \exp( \frac{t \log^2 9}{4 (1 - \frac{0.37}{T'_0-2.71})} ) ) }[/math]
- [math]\displaystyle{ + \varepsilon }[/math]
where [math]\displaystyle{ \varepsilon }[/math] is the exponentially small quantity
- [math]\displaystyle{ \varepsilon := \exp(\frac{1}{12(T'_0 - 3.33)}) \frac{1}{(\frac{T'_0}{2.42\pi} - 4) \sqrt{\pi}} (\frac{1.1}{a_0})^{\frac{T'_0}{2.42\pi}} \Gamma( \frac{T'_0}{4.84\pi}+1 ) \exp( - 4 (\frac{T'_0}{2.42\pi}-4)^2 ) }[/math]
which looks fearsome but is extremely negligible in practice. For instance, one can check that
- [math]\displaystyle{ \varepsilon \leq \frac{10^{-10}}{a_0^2} \leq 0.4 (\frac{1}{a_0-0.85} - \frac{1}{a_0-1.25}) }[/math]
whenever [math]\displaystyle{ T_0 \geq 100 }[/math], and hence
- [math]\displaystyle{ \int_{-\infty}^\infty vwf(\sigma)\ d\sigma \leq (1 - \frac{0.37 t}{T'_0 - 2.71})^{-1/2} \exp( \frac{1}{12(T'_0 - 3.33)} + \frac{0.37 y^2}{4 (T'_0 - 2.71 - 0.37 t)} ) \times }[/math]
- [math]\displaystyle{ (1 + \frac{\sqrt{\pi}}{a_0-1.25} + \frac{0.4}{a_0-1.25} 3^{1+y} \exp( \frac{t \log^2 9}{4 (1 - \frac{0.37}{T'_0-2.71})} ) ). }[/math]
To clean this up, we write
- [math]\displaystyle{ 1 - \frac{0.37 t}{T'_0 - 2.71} = \exp( O_{\leq}( \frac{0.37 t}{T'_0 - 2.71 - 0.37 t} ) }[/math]
and note that [math]\displaystyle{ T'_0 - 2.71 - 0.37t \geq T'_0 - 3.33 }[/math] to obtain
- [math]\displaystyle{ \int_{-\infty}^\infty vwf(\sigma)\ d\sigma \leq \exp( \frac{6 \times 0.37t + 1 + 3 \times 0.37 y^2}{12(T'_0 - 3.33)}) \times }[/math]
- [math]\displaystyle{ (1 + \frac{1}{a_0-1.25} (\sqrt{\pi} + 1.2 \times 3^y \exp( \frac{t \log^2 9}{4 (1 - \frac{0.37 t}{T'_0-2.71})} ) ). }[/math]
We bound [math]\displaystyle{ (6 \times 0.37t + 1 + 3 \times 0.37 y^2)/12 \leq 0.181 }[/math] and [math]\displaystyle{ 1.2 \times 3^y \exp( \frac{t \log^2 9}{4(1 - \frac{0.37 t}{T'_0-2.71}) \leq 5.15 }[/math] for [math]\displaystyle{ y \leq 1/2 }[/math], thus
- [math]\displaystyle{ \int_{-\infty}^\infty vwf(\sigma)\ d\sigma \leq \exp( \frac{0.181}{T'_0 - 3.33}) (1 + \frac{5.15}{a_0-1.25}). }[/math]
We conclude that
- [math]\displaystyle{ E_3 \leq \frac{1}{8} \sqrt{\pi} \exp( - \frac{t\pi^2}{64}) (T'_0)^{3/2} e^{-\pi T_0/4} \exp( \frac{0.181}{T'_0 - 3.33}) (1 + \frac{5.15}{a_0-1.25}). }[/math]
