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 can write

[math]\displaystyle{ \frac{A^{eff}+B^{eff}}{B^{eff}_0} = \sum_{n=1}^N \frac{b_n}{n^{s_B}} + \lambda \sum_{n=1}^N \frac{b_n}{n^{s_A}} }[/math]

where now [math]\displaystyle{ b_n := \exp( \frac{t}{4} \log^2 n) }[/math], [math]\displaystyle{ s_B := \frac{1+y-ix}{2} + \frac{t}{2} \alpha_1(\frac{1+y-ix}{2}) }[/math], [math]\displaystyle{ s_A := \frac{1-y+ix}{2} + \frac{t}{2} \alpha_1(\frac{1-y+ix}{2}) }[/math], and

[math]\displaystyle{ \lambda := \frac{\exp( \frac{t}{4} \alpha_1(\frac{1-y+ix}{2})^2 ) H_{0,1}( \frac{1-y+ix}{2} )}{ \overline{\exp( \frac{t}{4} \alpha_1(\frac{1+y+ix}{2})^2 ) H_{0,1}( \frac{1+y+ix}{2} )} }. }[/math]

In particular 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 [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 := |\lambda| 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 source:

[math]\displaystyle{ P }[/math] in Euler product	[math]\displaystyle{ N }[/math] using triangle inequality	[math]\displaystyle{ N }[/math] using Lemma 1	[math]\displaystyle{ N }[/math] using Lemma 1 and a custom mollifier
1	1391	1080
2	478	341	336
3	322	220	212
5	282	192	182
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. However, other custom mollifiers do work (see above table).

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{4 y(1+y)}{((-1+y)^2+x^2)((1+y)^2+x^2)} - \frac{1+3y}{(-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{1-3y-\frac{4y(1+y)}{x^2}}{(-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 \frac{1}{3} + \frac{4y(1+y)}{3x^2} }[/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} \quad (1) }[/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]

Large [math]\displaystyle{ x }[/math] case

When [math]\displaystyle{ N \geq 2000 }[/math] and [math]\displaystyle{ t=y=0.4 }[/math], we use the cruder lower bound

[math]\displaystyle{ |\sum_{n=1}^N \frac{b_n}{n^s}| - |\sum_{n=1}^N \frac{a_n}{n^s}| \geq 2 - \sum_{n=1}^N \frac{|b_n|}{n^\sigma} - \sum_{n=1}^N \frac{|a_n|}{n^\sigma} }[/math]

[math]\displaystyle{ \geq 2 - \sum_{n=1}^N \frac{1}{n^{0.7 + 0.1 \log \frac{N^2}{n}}} - e^\delta \exp( \frac{t y \log N}{2(x_N-6)} ) N^{-0.4} \sum_{n=1}^N \frac{1}{n^{0.3 + 0.1 \log \frac{N^2}{n}}} }[/math]

[math]\displaystyle{ \geq 2 - 1.706 - e^\delta \exp( \frac{0.08 \log N}{4\pi N^2 - 6.315} ) N^{-0.4} \times 3.469 }[/math]

using the estimates from Estimating a sum. We can bound

[math]\displaystyle{ \delta := \frac{0.2 \pi}{4\pi N^2 - 6.315 - \frac{14.8}{\pi}} + \frac{5.6}{(4\pi N^2 - 0.315)^2} \log |\frac{1.4}{4\pi}+i (N+1)^2| }[/math]

which for [math]\displaystyle{ N \geq 2000 }[/math] may be bounded by [math]\displaystyle{ 1.26 \times 10^{-8} }[/math]. One may similarly bound [math]\displaystyle{ \frac{0.08 \log N}{4\pi N^2 - 6.315} }[/math] by [math]\displaystyle{ 1.21 \times 10^{-8} }[/math], so we obtain the lower bound

[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}| \geq 2 - 1.706 - 0.166 = 0.128. }[/math]

Derivative bounds

We have

[math]\displaystyle{ \frac{A^{eff}+B^{eff}}{B^{eff}_0} = \sum_{n=1}^N \frac{b_n}{n^{s_B}} + \lambda \sum_{n=1}^N \frac{b_n}{n^{s_A}} }[/math]

where [math]\displaystyle{ b_n := \exp( \frac{t}{4} \log^2 n) }[/math], [math]\displaystyle{ s_B := \frac{1+y-ix}{2} + \frac{t}{2} \alpha_1(\frac{1+y-ix}{2}) }[/math], [math]\displaystyle{ s_A := \frac{1-y+ix}{2} + \frac{t}{2} \alpha_1(\frac{1-y+ix}{2}) }[/math], and

[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]

Differentiating in [math]\displaystyle{ x }[/math], we obtain

[math]\displaystyle{ \frac{d}{dx} \frac{A^{eff}+B^{eff}}{B^{eff}_0} = \sum_{n=1}^N \frac{-b_n \log n}{n^{s_B}} \frac{d}{dx}(s_B) + \lambda \sum_{n=1}^N \frac{b_n}{n^{s_A}} (-\log n \frac{d}{dx}(s_A) + \frac{d}{dx} \log \lambda) }[/math]

and hence

[math]\displaystyle{ |\frac{d}{dx} \frac{A^{eff}+B^{eff}}{B^{eff}_0}| \leq \sum_{n=1}^N \frac{b_n \log n}{n^{\mathrm{Re}(s_B)}} |\frac{d}{dx}(s_B)| + |\lambda| \sum_{n=1}^N \frac{b_n}{n^{\mathrm{Re}(s_A)}} |-\log n \frac{d}{dx}(s_A) + \frac{d}{dx} \log \lambda|. }[/math]

Recall from above that [math]\displaystyle{ \alpha'_1(s) = O_{\leq}( \frac{1}{x_N-6} ) }[/math]. Thus we have

[math]\displaystyle{ \frac{d}{dx} s_B = \frac{-i}{2} - \frac{it}{4} \alpha'_1( \frac{1+y-ix}{2} ) }[/math]

[math]\displaystyle{ = O_{\leq}( \frac{1}{2} + \frac{t}{4(x_N-6)} ). }[/math]

Similarly

[math]\displaystyle{ \frac{d}{dx} s_A = \frac{i}{2} + O_{\leq}( \frac{t}{4(x_N-6)} ). }[/math]

Finally, writing [math]\displaystyle{ s := \frac{1-y+ix}{2} }[/math], we have

[math]\displaystyle{ \log \lambda = \frac{t}{4} (\alpha_1(s)^2 - \alpha_1(1-s)^2) + (-\frac{s}{2} \log \pi + (\frac{s}{2}-\frac{1}{2}) \log \frac{s}{2} - \frac{s}{2}) - (-\frac{1-s}{2} \log \pi + (\frac{1-s}{2}-\frac{1}{2}) \log \frac{1-s}{2} - \frac{1-s}{2}) }[/math]

and hence

[math]\displaystyle{ -\frac{i}{2} \log n + \frac{d}{dx} \log \lambda = -\frac{i}{2} \log n +\frac{i}{2} \frac{d}{ds} \log \lambda }[/math]

[math]\displaystyle{ = \frac{it}{4} (\alpha_1(s) \alpha'_1(s) + \alpha_1(1-s) \alpha'_1(1-s)) + \frac{i}{4} \log \frac{s}{2\pi n} - \frac{i}{4s} + \frac{i}{4} \log \frac{1-s}{2\pi n} - \frac{i}{4(1-s)} }[/math]

[math]\displaystyle{ = O_{\leq}( \frac{t}{4(x_N-6)} (|\alpha_1(s)| + |\alpha_1(1-s)| + \frac{1}{4} \log \frac{|1-y+ix| |1+y-ix|}{16\pi^2 n^2} + \frac{1}{4|s| |1-s|} ); }[/math]

bounding

[math]\displaystyle{ |\alpha_1(s)| \leq \frac{1}{2} \log \frac{|1-y+ix|}{4\pi} + \frac{3}{2x_N} }[/math]

and

[math]\displaystyle{ |\alpha_1(1-s)| \leq \frac{1}{2} \log \frac{|1+y+ix|}{4\pi} + \frac{3}{2x_N} }[/math]

we thus have

[math]\displaystyle{ |\frac{i}{2} \log n + \frac{d}{dx} \log \lambda| \leq \frac{1}{4} (1 + \frac{t}{2(x_N-6)}) \log \frac{|1-y+ix| |1+y-ix|}{16\pi^2 n^2} + \frac{3t}{4 x_N(x_N-6)} + \frac{1}{4 x_N^2} }[/math]

[math]\displaystyle{ \leq \frac{1}{4} (1 + \frac{t}{2(x_N-6)}) \log \frac{|1-y+ix_{N+1}| |1+y-ix_{N+1}|}{16\pi^2} + \frac{3t+1}{4 x_N(x_N-6)}. }[/math]

Also, as established above, we have

[math]\displaystyle{ \mathrm{Re}(s_B) \geq \frac{1+y}{2} + \frac{t}{2} (\frac{3y-1}{x_{N+1}^2} + \log N) }[/math]

(for [math]\displaystyle{ y \geq 1/3 + 4y(1+y)/3x^2 }[/math]) and a similar argument gives

[math]\displaystyle{ \mathrm{Re}(s_A) \geq \frac{1-y}{2} + \frac{t}{2} (\frac{2-3y}{x_{N}^2} + \log N). }[/math]

We conclude that

[math]\displaystyle{ |\frac{d}{dx} \frac{A^{eff}+B^{eff}}{B^{eff}_0}| \leq (1 + \frac{t}{2(x_N-6)}) \sum_{n=1}^N \frac{b_n}{n^{\frac{1+y}{2} + \frac{t}{2} (\frac{3y-1}{x_{N+1}^2} + \log N)}} \frac{\log n}{2} + e^\delta N^{-y} \sum_{n=1}^N \frac{b_n}{n^{\frac{1-y}{2} + \frac{t}{2} (\frac{2-3y}{x_{N}^2} + \log N)}} (\frac{t}{4(x_N-6)} \log n + \frac{1}{4} \log \frac{|1-y+ix_{N+1}| |1+y-ix_{N+1}|}{16\pi^2 n^2} + \frac{3t+1}{4 x_N(x_N-6)}) }[/math]

where [math]\displaystyle{ b_n = \exp( \frac{t}{4} \log^2 n) }[/math], [math]\displaystyle{ x_N = 4\pi N^2 - \frac{\pi t}{4} }[/math], and [math]\displaystyle{ \delta }[/math] was defined in (1) above.