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 := \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

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 improves the 220 threshold slightly to 213 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' }[/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') = \frac{1}{2s'} + \frac{1}{s'-1} + \frac{1}{2} \log \frac{s'}{2\pi} }[/math]

[math]\displaystyle{ = O_{\leq}(\frac{1}{x}) + O_{\leq}(\frac{1}{x/2}) + \frac{1}{2} \log \frac{|s'|}{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') = -\frac{1}{2(s')^2} + \frac{1}{(s'-1)^2} + \frac{1}{2s'} }[/math]

[math]\displaystyle{ = O_{\leq}(\frac{1}{x^2/2}) + O_{\leq(\frac{1/x^2/4}) + \frac{1}{2s'} }[/math]

[math]\displaystyle{ = O_{\leq}(\frac{6}{x_N^2}) + \frac{1}{2s'} }[/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'') }[/math]

for some [math]\displaystyle{ s'' }[/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'') = -\frac{1}{2} \log \pi + \frac{1}{2} \log \frac{|s''|}{2} - \mathrm{Re} \frac{1}{2s''} }[/math]

[math]\displaystyle{ = \frac{1}{2} \log \frac{|s''|}{2\pi} + O_{\leq}(\frac{\mathm{Re}(s'')}{2|s''|^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''')|) }[/math]

for some [math]\displaystyle{ s''' }[/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''') = -\frac{1}{2(s''')^2} - \frac{1}{(s'''-1)^2} + \frac{1}{2s'''} }[/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)} ) ) b_n \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 attempt], 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

[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

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]

...