Polymath15 test problem: Difference between revisions

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:<math>U := \exp( -i (\frac{T'}{2} \log \frac{T'}{2\pi} - \frac{T'}{2} - \frac{\pi}{8} ))</math>
:<math>U := \exp( -i (\frac{T'}{2} \log \frac{T'}{2\pi} - \frac{T'}{2} - \frac{\pi}{8} ))</math>
:<math>C_0(p) := \frac{ \exp( \pi i (p^2/2 + 3/8) )- i \sqrt{2} \cos(\pi p/2)}{2 \cos(\pi p)}.</math>
:<math>C_0(p) := \frac{ \exp( \pi i (p^2/2 + 3/8) )- i \sqrt{2} \cos(\pi p/2)}{2 \cos(\pi p)}.</math>
One can also replace <math>C^{eff}</math> by the very slightly different quantity
:<math>\tilde C^{eff} :=\frac{2 e^{-\pi i y/8}}{8} \exp( \frac{t\pi^2}{64}) (-1)^N \mathrm{Re}( H_{0,1}(iT') C_0(p) U e^{\pi i/8} ).</math>


Finally, a simplified approximation is <math>A^{toy} + B^{toy}</math>, where
Finally, a simplified approximation is <math>A^{toy} + B^{toy}</math>, where
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* [https://drive.google.com/open?id=1YBIA5gRv2DUXX74MLwfn9F2J0QJBWUh_ N between 20 and 150] ([https://drive.google.com/open?id=1ZBX7jNGXhQZQ50t8UX4boTPW93dmRAun raw data]
* [https://drive.google.com/open?id=1YBIA5gRv2DUXX74MLwfn9F2J0QJBWUh_ N between 20 and 150] ([https://drive.google.com/open?id=1ZBX7jNGXhQZQ50t8UX4boTPW93dmRAun raw data]
* [https://drive.google.com/file/d/1NvEv-1R4KTEchWMbJCpZA6Uf1xLkiYWM/view N between 151 and 300]
* [https://drive.google.com/file/d/1NvEv-1R4KTEchWMbJCpZA6Uf1xLkiYWM/view N between 151 and 300]
* [https://drive.google.com/open?id=15Xf9GsaAzydl-39zyG9nei5aZyaHwsFe The (A+B)/B0 mesh data for N=300 to 20, y=0.45, t=0.4, c=0.065]
* [https://drive.google.com/open?id=1kK8tV2bRfACm1lUKRFIktca58ZV8L8U3  c=0.26 for N=7 to 19, y=0.4,t=0.4]
* https://drive.google.com/open?id=13_mzqvtaZCghmj7oAZtDRXnkb2zbxQH3 c=0.26 for N=19 to 7, y=0.45,t=0.4]


[https://github.com/km-git-acc/dbn_upper_bound/blob/master/dbn_upper_bound/python/research/mod_abbeff_lower_Nbounds.csv Here is a table of analytic lower bounds for <math>A^{eff}+B^{eff}/B^{eff}_0</math> for <math>3 \leq N \leq 2000</math>].
[https://github.com/km-git-acc/dbn_upper_bound/blob/master/dbn_upper_bound/python/research/mod_abbeff_lower_Nbounds.csv Here is a table of analytic lower bounds for <math>A^{eff}+B^{eff}/B^{eff}_0</math> for <math>3 \leq N \leq 2000</math>].
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See [[Controlling H_t-A-B/B_0]].
See [[Controlling H_t-A-B/B_0]].


[https://github.com/km-git-acc/dbn_upper_bound/blob/master/dbn_upper_bound/python/research/bounded_normalized_E1_and_E2_and_E3_and_overall_error.csv Here is a table on bounds on error terms <math>E_1/B^{eff}_0, E_2/B^{eff}_0, E_3^*/B^{eff}_0</math> for N=3 to 2000]
[https://github.com/km-git-acc/dbn_upper_bound/blob/master/dbn_upper_bound/python/research/bounded_normalized_E1_and_E2_and_E3_and_overall_error.csv Here is a table on bounds on error terms <math>E_1/B^{eff}_0, E_2/B^{eff}_0, E_3^*/B^{eff}_0</math> for N=3 to 2000].  [https://github.com/km-git-acc/dbn_upper_bound/blob/master/dbn_upper_bound/python/research/e1_e2_e3_sharper_Nbound.csv Here is a table] with some sharpened estimates from the PDF writeup.


[https://ibb.co/n9Gw2x Here are some graphs] depicting <math>|H_t-A^{eff}-B^{eff}/B_0^{eff}|</math> and <math>E_1+E_2+E_3^*/|B_0^{eff}|</math> for <math>x \leq 400</math>.
[https://ibb.co/b7baZc Here is a graph] depicting <math>|H_t-A^{eff}-B^{eff}/B_0^{eff}|</math> and <math>E_1+E_2+E_3^*/|B_0^{eff}|</math> for <math>x \leq 1600</math>.


== Small values of x ==
== Small values of x ==
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* [https://pastebin.com/NkFKs3pB x=1000 to x=1300 with step size 0.1]
* [https://pastebin.com/NkFKs3pB x=1000 to x=1300 with step size 0.1]
* [https://pastebin.com/k7vC6e7n x=1300 to x=1600 with step size 0.1]
* [https://pastebin.com/k7vC6e7n x=1300 to x=1600 with step size 0.1]
* [https://gist.githubusercontent.com/p15-git-acc/3ada0ff0b9ec77e23cb7cace0dcb8691/raw/807e1b0a16356a9bbd2a5af872f71bc064830c38/gistfile1.txt x=20 to x=1000, adaptive mesh]


[https://ibb.co/fOroa7 Here are some snapshots of <math>H_t/B^{eff}_0</math>].
[https://ibb.co/fOroa7 Here are some snapshots of <math>H_t/B^{eff}_0</math>].


In this range we will need [[Bounding the derivative of H_t]] or [[Bounding the derivative of H_t - second approach]].
In this range we will need [[Bounding the derivative of H_t]] or [[Bounding the derivative of H_t - second approach]] or [[Bounding the derivative of H_t - third approach]].
 
Tables of <math>H'_t(x+iy)</math>:
 
* [https://drive.google.com/open?id=1mtzrJ_-hBMyt90gVzJdNCq4RU6Pz_dQQ x=0 to x=100 with step size 0.1]
* [https://drive.google.com/open?id=1wSZJgoPp9-6C8CJqLr5g3E1uhEBHKmTJ x=100 to x=200 with step size 0.1]
* [https://drive.google.com/open?id=1q3h0x8jSF0Z1KQ9iLx23JI1oCXfrx0RL x=200 to x=300 with step size 0.1]
 
[https://drive.google.com/open?id=1ge0TD5hvs1O6BKLAz34JmTqxSrguvjsJ Here is a table] of <math>x</math>, pari/gp prec, <math>H_{t}, H^{'}_{t}, |H_{t}|, |H^{'}_{t}|, \frac{|H_{t}|}{|B_{0}^{eff}|}, \frac{|H^{'}_{t}|}{|B_{0}^{eff}|}</math> for x=0 to x=30 with step size 0.01.
 
 
 
[https://drive.google.com/open?id=1855iryE-7uDDEyW7hXJ5-3Njomsz-mMH Here is a plot] of <math>H_t/B_0</math> for a rectangle <math> \{x+iy: 0 \leq x \leq 300; 0.4 \leq y \leq 0.45\}</math>.  Here is [https://drive.google.com/open?id=1oGQ4HfXlEiC5WUnWHOzAt5EJQlEg9SRb an adaptive mesh plot]; here is a [https://drive.google.com/open?id=12zkFFXBF7H6Mjd1KBmLhZw10io3J_wer closeup near the origin].
 
Here is a [https://pastebin.com/TiFk6CfF script] for verifying the absence of zeroes of <math>H_t</math> in a rectangle.  It can eliminate zeros in the rectangle <math>\{0 \leq x \leq 1000, 0.4 \leq y \leq 0.45\}</math> when t = 0.4.
 
 
== Large negative values of <math>t</math> ==
 
See also [[Second attempt at computing H_t(x) for negative t]].
 
We heuristically compute <math>H_t(x)</math> in the regime where <math>x</math> is large and <math>t</math> is large and negative with <math>|t|/x \asymp 1</math>.  We shall only be interested in the zeroes and so we discard any multiplicative factor which is non-zero: we write <math>X \sim Y</math> if X is equal (or approximately equal) to Y times something that is explicit and non-zero.
 
From equation (35) of the writeup we have
 
:<math>H_t(x) = \int_{\bf R} \frac{1}{8} \xi(\frac{1+ix}{2} + i |t|^{1/2} v) \frac{1}{\pi} e^{-v^2}\ dv \quad (3.1)</math>
:<math> \sim \int_{\bf R} \xi(\frac{1+ix}{2} + i |t|^{1/2} v) e^{-v^2}\ dv. \quad (3.2)</math>
 
To cancel off an exponential decay factor in the <math>\xi</math> function, it is convenient to shift the v variable by <math>\pi |t|^{1/2}/8</math>, thus
:<math> H_t(x) \sim \int_{\bf R} \xi(\frac{1+ix}{2} + i |t|^{1/2} v - \pi i |t|/8) e^{-(v - \pi |t|^{1/2}/8)^2}\ dv \quad (3.3)</math>
:<math> \sim \int_{\bf R} \xi(\frac{1+i\tilde x}{2} + i |t|^{1/2} v) e^{-v^2 + \pi |t|^{1/2} v / 4}\ dv \quad (3.4)</math>
where
:<math> \tilde x := x - \pi |t|/4 = x + \frac{\pi t}{4}. \quad (3.5)</math>
Now from the definition of <math>\xi</math> and the Stirling approximation we have
:<math> \xi(s) \sim M_0(s) \zeta(s)\quad (3.6)</math>
where <math>M_0</math> is defined in (6) of the writeup.  Thus
:<math> H_t(x) \sim \int_{\bf R} M_0(\frac{1+i\tilde x}{2} + i |t|^{1/2} v) \zeta(\frac{1+i\tilde x}{2} + i |t|^{1/2} v) e^{-v^2 + \pi |t|^{1/2} v / 4}\ dv.\quad (3.7)</math>
By Taylor expansion we have
:<math> M_0(\frac{1+i\tilde x}{2} + i |t|^{1/2} v) \sim M_0(\frac{1+i\tilde x}{2}) \exp( \alpha( \frac{1+i\tilde x}{2} ) i |t|^{1/2} v + \alpha'(\frac{1+i \tilde x}{2}) \frac{-|t| v^2}{2} )\quad (3.8)</math>
:<math> \sim \exp( \alpha( \frac{1+i\tilde x}{2} ) i |t|^{1/2} v + \alpha'(\frac{1+i \tilde x}{2}) \frac{-|t| v^2}{2} )\quad (3.9)</math>
where <math>\alpha</math> is defined in equation (8) of the writeup.  We have the approximations
:<math> \alpha(\frac{1+i\tilde x}{2} ) \approx \frac{1}{2} \log \frac{\tilde x}{4\pi} + \frac{i\pi}{4} \quad (3.10)</math>
and
:<math> \alpha'(\frac{1+i\tilde x}{2} ) \approx \frac{-i}{\tilde x} \quad (3.11)</math>
and hence
:<math> H_t(x) \sim \int_{\bf R} \exp( \frac{i |t|^{1/2} v}{2} \log \frac{\tilde x}{4\pi} - \pi |t|^{1/2} v/4 + i |t| v^2 / 2\tilde x) \zeta(\frac{1+i\tilde x}{2} + i |t|^{1/2} v) e^{-v^2 + \pi |t|^{1/2} v / 4}\ dv.\quad (3.12)</math>
The two factors of <math>\exp( \pi |t|^{1/2} v/4 ) </math> cancel.  If we now write
:<math>N := \sqrt{\frac{\tilde x}{4\pi}}\quad (3.13)</math>
and
:<math>u := |t|/N^2 = 4\pi |t|/\tilde x,\quad (3.14)</math>
we conclude that
:<math> H_t(x) \sim \int_{\bf R} \exp( i |t|^{1/2} v \log N + i u v^2 / 8 \pi) \zeta(\frac{1+i\tilde x}{2} + i |t|^{1/2} v) e^{-v^2}\ dv.\quad (3.15)</math>
If we formally write <math>\zeta(s) = \sum_n \frac{1}{n^s}</math> (ignoring convergence issues) we obtain
:<math> H_t(x) \sim \sum_n \int_{\bf R} \exp( i |t|^{1/2} v \log N + i u v^2 / 8 \pi) n^{-\frac{1+i\tilde x}{2} - i |t|^{1/2} v} e^{-v^2}\ dv\quad (3.16)</math>
:<math> \sim \sum_n \int_{\bf R} \exp( - i |t|^{1/2} v \log \frac{n}{N} + i u v^2 / 8 \pi -\frac{1+i\tilde x}{2} \log \frac{n}{N} ) e^{-v^2}\ dv\quad (3.17)</math>
We can compute the <math>v</math> integral to obtain
:<math> H_t(x) \sim \sum_n \exp( - \frac{|t| \log^2 \frac{n}{N}}{4 (1 - iu / 8 \pi)} -\frac{1+i\tilde x}{2} \log \frac{n}{N}).\quad (3.18)</math>
Using the Taylor approximation
:<math> \log \frac{n}{N} \approx \frac{n-N}{N} - \frac{(n-N)^2}{2N^2} \quad (3.19)</math>
and dropping some small terms, we obtain
:<math> H_t(x) \sim \sum_n \exp( - \frac{|t| (n-N)^2}{4 N^2 (1 - iu/8\pi)} -\frac{i\tilde x}{2} \frac{n-N}{N} + \frac{i \tilde x (n-N)^2}{4N^2} ).\quad (3.20)</math>
Writing <math>\tilde x = 4\pi N^2</math> and <math>|t| = u N^2</math>, this becomes
:<math> H_t(x) \sim \sum_n \exp( - \frac{2\pi u (n-N)^2}{8\pi - iu} -2 \pi i N(n-N) + \pi i (n-N)^2 ).\quad (3.21)</math>
Writing
:<math> N(n-N) = \frac{1}{2} n^2 - \frac{1}{2} (n-N)^2 - \frac{1}{2} N^2 \quad (3.22)</math>
we thus have
:<math> H_t(x) \sim \sum_n \exp( - \frac{2 \pi u (n-N)^2}{8 \pi - iu} - \pi i n^2 + 2 \pi i (n-N)^2 )\quad (3.23)</math>
:<math> \sim \sum_n \exp( \frac{16 \pi^2 i (n-N)^2}{8 \pi - iu} ) e^{\pi i n}\quad (3.24)</math>
:<math> \sim \theta_{01}( \frac{16 \pi N}{8\pi - iu}, \frac{16 \pi}{8\pi - iu} )\quad (3.25)</math>
where <math>\theta_{01}</math> is the theta function defined in [https://en.wikipedia.org/wiki/Theta_function#Auxiliary_functions this Wikipedia page].  Using the Jacobi identity we then have
:<math> H_t(x) \sim \theta_{10}(N, \frac{iu - 8\pi}{16 \pi} )\quad (3.26)</math>
:<math> \sim \theta( N + \frac{1}{2} \frac{iu - 8\pi}{16 \pi}, \frac{iu - 8\pi}{16 \pi})\quad (3.27)</math>
:<math> \sim \sum_n \exp( - \pi i n(n+1) / 2 ) e^{2\pi i (n+1/2) N} e^{-u n(n+1)/16}.\quad (3.28)</math>
As a sanity check, one can verify that the RHS is real-valued, just as <math>H_t(x)</math> is (by the functional equation).
 
[[Category:Polymath15]]

Latest revision as of 22:26, 1 January 2019

We are initially focusing attention on the following

Test problem For [math]\displaystyle{ t=y=0.4 }[/math], can one prove that [math]\displaystyle{ H_t(x+iy) \neq 0 }[/math] for all [math]\displaystyle{ x \geq 0 }[/math]?

If we can show this, it is likely that (with the additional use of the argument principle, and some further information on the behaviour of [math]\displaystyle{ H_t(x+iy) }[/math] at [math]\displaystyle{ y=0.4 }[/math]) that one can show that [math]\displaystyle{ H_t(x+iy) \neq 0 }[/math] for all [math]\displaystyle{ y \geq 0.4 }[/math] as well. This would give a new upper bound

[math]\displaystyle{ \Lambda \leq 0.4 + \frac{1}{2} (0.4)^2 = 0.48 }[/math]

for the de Bruijn-Newman constant.

For very small values of [math]\displaystyle{ x }[/math] we expect to be able to establish this by direct calculation of [math]\displaystyle{ H_t(x+iy) }[/math]. For medium or large values, the strategy is to use a suitable approximation

[math]\displaystyle{ H_t(x+iy) \approx A + B }[/math]

for some relatively easily computable quantities [math]\displaystyle{ A = A_t(x+iy), B = B_t(x+iy) }[/math] (it may possibly be necessary to use a refined approximation [math]\displaystyle{ A+B-C }[/math] instead). The quantity [math]\displaystyle{ B }[/math] contains a non-zero main term [math]\displaystyle{ B_0 }[/math] which is expected to roughly dominate. To show [math]\displaystyle{ H_t(x+iy) }[/math] is non-zero, it would suffice to show that

[math]\displaystyle{ \frac{|H_t - A - B|}{|B_0|} \lt \frac{|A + B|}{|B_0|}. }[/math]

Thus one will seek upper bounds on the error [math]\displaystyle{ \frac{|H_t - A - B|}{|B_0|} }[/math] and lower bounds on [math]\displaystyle{ \frac{|A+B|}{|B_0|} }[/math] for various ranges of [math]\displaystyle{ x }[/math]. Numerically it seems that the RHS stays above 0.4 as soon as [math]\displaystyle{ x }[/math] is moderately large, while the LHS stays below 0.1, which looks promising for the rigorous arguments.

Choices of approximation

There are a number of slightly different approximations we have used in previous discussion. The first approximation was [math]\displaystyle{ A+B }[/math], where

[math]\displaystyle{ A := \frac{1}{8} \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \sum_{n=1}^N \frac{\exp(\frac{t}{16} \log^2 \frac{s+4}{2\pi n^2})}{n^s} }[/math]
[math]\displaystyle{ B := \frac{1}{8} \frac{s(s-1)}{2} \pi^{-(1-s)/2} \Gamma((1-s)/2) \sum_{n=1}^N \frac{\exp(\frac{t}{16} \log^2 \frac{5-s}{2\pi n^2})}{n^{1-s}} }[/math]
[math]\displaystyle{ B_0 := \frac{1}{8} \frac{s(s-1)}{2} \pi^{-(1-s)/2} \Gamma((1-s)/2) \exp( \frac{t}{16} \log^2 \frac{5-s}{2\pi} ) }[/math]
[math]\displaystyle{ s := \frac{1-y+ix}{2} }[/math]
[math]\displaystyle{ N := \lfloor \sqrt{\frac{\mathrm{Im} s}{2\pi}} \rfloor = \lfloor \sqrt{\frac{x}{4\pi}} \rfloor. }[/math]

There is also the refinement [math]\displaystyle{ A+B-C }[/math], where

[math]\displaystyle{ C:= \frac{1}{8} \exp(-\frac{t\pi^2}{64}) \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \frac{e^{-i\pi s} \Gamma(1-s)}{2\pi i} (2\pi i N)^{s-1} \Psi( \frac{s}{2\pi i N}-N ) }[/math]
[math]\displaystyle{ \Psi(\alpha) := 2\pi \frac{\cos \pi(\frac{1}{2} \alpha^2 - \alpha - \frac{\pi}{8})}{\cos(\pi \alpha)} \exp( \frac{i \pi}{2} \alpha^2 - \frac{5 \pi i}{8}). }[/math]

The first approximation was modified slightly to [math]\displaystyle{ A'+B' }[/math], where

[math]\displaystyle{ A' := \frac{2}{8} \pi^{-s/2} \sqrt{2\pi} \exp( (\frac{s+4}{2}-\frac{1}{2}) \log \frac{s+4}{2} - \frac{s+4}{2}) \sum_{n=1}^N \frac{\exp(\frac{t}{16} \log^2 \frac{s+4}{2\pi n^2})}{n^s} }[/math]
[math]\displaystyle{ B' := \frac{2}{8} \pi^{-(1-s)/2} \sqrt{2\pi} \exp( (\frac{5-s}{2}-\frac{1}{2}) \log \frac{5-s}{2} - \frac{5-s}{2}) \sum_{n=1}^N \frac{\exp(\frac{t}{16} \log^2 \frac{5-s}{2\pi n^2})}{n^{1-s}} }[/math]
[math]\displaystyle{ B'_0 := \frac{2}{8} \pi^{-(1-s)/2} \sqrt{2\pi} \exp( (\frac{5-s}{2}-\frac{1}{2}) \log \frac{5-s}{2} - \frac{5-s}{2}) \exp( \frac{t}{16} \log^2 \frac{5-s}{2\pi} ) }[/math]
[math]\displaystyle{ s := \frac{1-y+ix}{2} }[/math]
[math]\displaystyle{ N := \lfloor \sqrt{\frac{\mathrm{Im} s}{2\pi}} \rfloor = \lfloor \sqrt{\frac{x}{4\pi}} \rfloor. }[/math]

In Effective bounds on H_t - second approach, a more refined approximation [math]\displaystyle{ A^{eff} + B^{eff} }[/math] was introduced:

[math]\displaystyle{ A^{eff} := \frac{1}{8} \exp( \frac{t}{4} \alpha_1(\frac{1-y+ix}{2})^2 ) H_{0,1}(\frac{1-y+ix}{2}) \sum_{n=1}^N \frac{1}{n^{\frac{1-y+ix}{2} + \frac{t \alpha_1(\frac{1-y+ix}{2})}{2} - \frac{t}{4} \log n}} }[/math]
[math]\displaystyle{ B^{eff} := \frac{1}{8} \exp( \frac{t}{4} \overline{\alpha_1(\frac{1+y+ix}{2})}^2 ) \overline{H_{0,1}(\frac{1+y+ix}{2})} \sum_{n=1}^N \frac{1}{n^{\frac{1+y-ix}{2} + \frac{t \overline{\alpha_1(\frac{1+y+ix}{2})}}{2} - \frac{t}{4} \log n}} }[/math]
[math]\displaystyle{ B^{eff}_0 := \frac{1}{8} \exp( \frac{t}{4} \overline{\alpha_1(\frac{1+y+ix}{2})}^2 ) \overline{H_{0,1}(\frac{1+y+ix}{2})} }[/math]
[math]\displaystyle{ H_{0,1}(s) := \frac{s (s-1)}{2} \pi^{-s/2} \sqrt{2\pi} \exp( (\frac{s}{2} - \frac{1}{2}) \log \frac{s}{2} - \frac{s}{2} ) }[/math]
[math]\displaystyle{ \alpha_1(s) := \frac{1}{2s} + \frac{1}{s-1} + \frac{1}{2} \log \frac{s}{2\pi} }[/math]
[math]\displaystyle{ N := \lfloor \sqrt{ \frac{T'}{2\pi}} \rfloor }[/math]
[math]\displaystyle{ T' := \frac{x}{2} + \frac{\pi t}{8}. }[/math]

There is a refinement [math]\displaystyle{ A^{eff}+B^{eff}-C^{eff} }[/math], where

[math]\displaystyle{ C^{eff} := \frac{1}{8} \exp( \frac{t\pi^2}{64}) \frac{s'(s'-1)}{2} (-1)^N ( \pi^{-s'/2} \Gamma(s'/2) a^{-\sigma} C_0(p) U + \pi^{-(1-s')/2} \Gamma((1-s')/2) a^{-(1-\sigma)} \overline{C_0(p)} \overline{U}) }[/math]
[math]\displaystyle{ s' := \frac{1-y}{2} + iT' = \frac{1-y+ix}{2} + \frac{\pi i t}{8} }[/math]
[math]\displaystyle{ a := \sqrt{\frac{T'}{2\pi}} }[/math]
[math]\displaystyle{ p := 1 - 2(a-N) }[/math]
[math]\displaystyle{ \sigma := \mathrm{Re} s' = \frac{1-y}{2} }[/math]
[math]\displaystyle{ U := \exp( -i (\frac{T'}{2} \log \frac{T'}{2\pi} - \frac{T'}{2} - \frac{\pi}{8} )) }[/math]
[math]\displaystyle{ C_0(p) := \frac{ \exp( \pi i (p^2/2 + 3/8) )- i \sqrt{2} \cos(\pi p/2)}{2 \cos(\pi p)}. }[/math]

One can also replace [math]\displaystyle{ C^{eff} }[/math] by the very slightly different quantity

[math]\displaystyle{ \tilde C^{eff} :=\frac{2 e^{-\pi i y/8}}{8} \exp( \frac{t\pi^2}{64}) (-1)^N \mathrm{Re}( H_{0,1}(iT') C_0(p) U e^{\pi i/8} ). }[/math]

Finally, a simplified approximation is [math]\displaystyle{ A^{toy} + B^{toy} }[/math], where

[math]\displaystyle{ A^{toy} := B^{toy}_0 \exp(i ((\frac{x}{2} + \frac{\pi t}{8}) \log \frac{x}{4\pi} - \frac{x}{2} - \frac{\pi}{4} )) N^{-y} \sum_{n=1}^N \frac{1}{n^{\frac{1-y+ix}{2} + \frac{t}{4} \log \frac{N^2}{n} + \pi i t/8}} }[/math]
[math]\displaystyle{ B^{toy} := B^{toy}_0 \sum_{n=1}^N \frac{1}{n^{\frac{1+y-ix}{2} + \frac{t}{4} \log \frac{N^2}{n} - \pi i t/8}} }[/math]
[math]\displaystyle{ B^{toy}_0 := \frac{\sqrt{2}}{4} \pi^2 N^{\frac{7+y}{2}} \exp( i (-\frac{x}{4} \log \frac{x}{4\pi} + \frac{x}{4} + \frac{9-y}{8} \pi) + \frac{t}{16} (\log \frac{x}{4\pi} - \frac{\pi i}{2})^2 ) e^{-\pi x/8} }[/math]
[math]\displaystyle{ N := \lfloor \sqrt{\frac{x}{4\pi}} \rfloor. }[/math]

Here is a table comparing the size of the various main terms:

[math]\displaystyle{ x }[/math] [math]\displaystyle{ B_0 }[/math] [math]\displaystyle{ B'_0 }[/math] [math]\displaystyle{ B^{eff}_0 }[/math] [math]\displaystyle{ B^{toy}_0 }[/math]
[math]\displaystyle{ 10^3 }[/math] [math]\displaystyle{ (3.4405 + 3.5443 i) \times 10^{-167} }[/math] [math]\displaystyle{ (3.4204 + 3.5383 i) \times 10^{-167} }[/math] [math]\displaystyle{ (3.4426 + 3.5411 i) \times 10^{-167} }[/math] [math]\displaystyle{ (2.3040 + 2.3606 i) \times 10^{-167} }[/math]
[math]\displaystyle{ 10^4 }[/math] [math]\displaystyle{ (-1.1843 - 7.7882 i) \times 10^{-1700} }[/math] [math]\displaystyle{ (-1.1180 - 7.7888 i) \times 10^{-1700} }[/math] [math]\displaystyle{ (-1.1185 - 7.7879 i) \times 10^{-1700} }[/math] [math]\displaystyle{ (-1.1155 - 7.5753 i) \times 10^{-1700} }[/math]
[math]\displaystyle{ 10^5 }[/math] [math]\displaystyle{ (-7.6133 + 2.5065 i) * 10^{-17047} }[/math] [math]\displaystyle{ (-7.6134 + 2.5060 i) * 10^{-17047} }[/math] [math]\displaystyle{ (-7.6134 + 2.5059 i) * 10^{-17047} }[/math] [math]\displaystyle{ (-7.5483 + 2.4848 i) * 10^{-17047} }[/math]
[math]\displaystyle{ 10^6 }[/math] [math]\displaystyle{ (-3.1615 - 7.7093 i) * 10^{-170537} }[/math] [math]\displaystyle{ (-3.1676 - 7.7063 i) * 10^{-170537} }[/math] [math]\displaystyle{ (-3.1646 - 7.7079 i) * 10^{-170537} }[/math] [math]\displaystyle{ (-3.1590 - 7.6898 i) * 10^{-170537} }[/math]
[math]\displaystyle{ 10^7 }[/math] [math]\displaystyle{ (2.1676 - 9.6330 i) * 10^{-1705458} }[/math] [math]\displaystyle{ (2.1711 - 9.6236 i) * 10^{-1705458} }[/math] [math]\displaystyle{ (2.1571 - 9.6329 i) * 10^{-1705458} }[/math] [math]\displaystyle{ (2.2566 - 9.6000 i) * 10^{-1705458} }[/math]

Here some typical values of [math]\displaystyle{ B/B_0 }[/math] (note that [math]\displaystyle{ B/B_0 }[/math] and [math]\displaystyle{ B'/B'_0 }[/math] are identical):

[math]\displaystyle{ x }[/math] [math]\displaystyle{ B/B_0 }[/math] [math]\displaystyle{ B'/B'_0 }[/math] [math]\displaystyle{ B^{eff}/B^{eff}_0 }[/math] [math]\displaystyle{ B^{toy}/B^{toy}_0 }[/math]
[math]\displaystyle{ 10^3 }[/math] [math]\displaystyle{ 0.7722 + 0.6102 i }[/math] [math]\displaystyle{ 0.7722 + 0.6102 i }[/math] [math]\displaystyle{ 0.7733 + 0.6101 i }[/math] [math]\displaystyle{ 0.7626 + 0.6192 i }[/math]
[math]\displaystyle{ 10^4 }[/math] [math]\displaystyle{ 0.7434 - 0.0126 i }[/math] [math]\displaystyle{ 0.7434 - 0.0126 i }[/math] [math]\displaystyle{ 0.7434 - 0.0126 i }[/math] [math]\displaystyle{ 0.7434 - 0.0124 i }[/math]
[math]\displaystyle{ 10^5 }[/math] [math]\displaystyle{ 1.1218 - 0.3211 i }[/math] [math]\displaystyle{ 1.1218 - 0.3211 i }[/math] [math]\displaystyle{ 1.1218 - 0.3211 i }[/math] [math]\displaystyle{ 1.1219 - 0.3213 i }[/math]
[math]\displaystyle{ 10^6 }[/math] [math]\displaystyle{ 1.3956 - 0.5682 i }[/math] [math]\displaystyle{ 1.3956 - 0.5682 i }[/math] [math]\displaystyle{ 1.3955 - 0.5682 i }[/math] [math]\displaystyle{ 1.3956 - 0.5683 i }[/math]
[math]\displaystyle{ 10^7 }[/math] [math]\displaystyle{ 1.6400 + 0.0198 i }[/math] [math]\displaystyle{ 1.6400 + 0.0198 i }[/math] [math]\displaystyle{ 1.6401 + 0.0198 i }[/math] [math]\displaystyle{ 1.6400 - 0.0198 i }[/math]

Here some typical values of [math]\displaystyle{ A/B_0 }[/math], which seems to be about an order of magnitude smaller than [math]\displaystyle{ B/B_0 }[/math] in many cases:

[math]\displaystyle{ x }[/math] [math]\displaystyle{ A/B_0 }[/math] [math]\displaystyle{ A'/B'_0 }[/math] [math]\displaystyle{ A^{eff}/B^{eff}_0 }[/math] [math]\displaystyle{ A^{toy}/B^{toy}_0 }[/math]
[math]\displaystyle{ 10^3 }[/math] [math]\displaystyle{ -0.3856 - 0.0997 i }[/math] [math]\displaystyle{ -0.3857 - 0.0953 i }[/math] [math]\displaystyle{ -0.3854 - 0.1002 i }[/math] [math]\displaystyle{ -0.4036 - 0.0968 i }[/math]
[math]\displaystyle{ 10^4 }[/math] [math]\displaystyle{ -0.2199 - 0.0034 i }[/math] [math]\displaystyle{ -0.2199 - 0.0036 i }[/math] [math]\displaystyle{ -0.2199 - 0.0033 i }[/math] [math]\displaystyle{ -0.2208 - 0.0033 i }[/math]
[math]\displaystyle{ 10^5 }[/math] [math]\displaystyle{ 0.1543 + 0.1660 i }[/math] [math]\displaystyle{ 0.1543 + 0.1660 i }[/math] [math]\displaystyle{ 0.1543 + 0.1660 i }[/math] [math]\displaystyle{ 0.1544 + 0.1663 i }[/math]
[math]\displaystyle{ 10^6 }[/math] [math]\displaystyle{ -0.1013 - 0.1887 i }[/math] [math]\displaystyle{ -0.1010 - 0.1889 i }[/math] [math]\displaystyle{ -0.1011 - 0.1890 i }[/math] [math]\displaystyle{ -0.1012 - 0.1888 i }[/math]
[math]\displaystyle{ 10^7 }[/math] [math]\displaystyle{ -0.1018 + 0.1135 i }[/math] [math]\displaystyle{ -0.1022 + 0.1133 i }[/math] [math]\displaystyle{ -0.1025 + 0.1128 i }[/math] [math]\displaystyle{ -0.0986 + 0.1163 i }[/math]

Here some typical values of [math]\displaystyle{ C/B_0 }[/math], which is significantly smaller than either [math]\displaystyle{ A/B_0 }[/math] or [math]\displaystyle{ B/B_0 }[/math]:


[math]\displaystyle{ x }[/math] [math]\displaystyle{ C/B_0 }[/math] [math]\displaystyle{ C^{eff}/B^{eff}_0 }[/math]
[math]\displaystyle{ 10^3 }[/math] [math]\displaystyle{ -0.1183 + 0.0697i }[/math] [math]\displaystyle{ -0.0581 + 0.0823 i }[/math]
[math]\displaystyle{ 10^4 }[/math] [math]\displaystyle{ -0.0001 - 0.0184 i }[/math] [math]\displaystyle{ -0.0001 - 0.0172 i }[/math]
[math]\displaystyle{ 10^5 }[/math] [math]\displaystyle{ -0.0033 - 0.0005i }[/math] [math]\displaystyle{ -0.0031 - 0.0005i }[/math]
[math]\displaystyle{ 10^6 }[/math] [math]\displaystyle{ -0.0001 - 0.0006 i }[/math] [math]\displaystyle{ -0.0001 - 0.0006 i }[/math]
[math]\displaystyle{ 10^7 }[/math] [math]\displaystyle{ -0.0000 - 0.0001 i }[/math] [math]\displaystyle{ -0.0000 - 0.0001 i }[/math]

Some values of [math]\displaystyle{ H_t }[/math] and its approximations at small values of [math]\displaystyle{ x }[/math] source source:

[math]\displaystyle{ x }[/math] [math]\displaystyle{ H_t }[/math] [math]\displaystyle{ A+B }[/math] [math]\displaystyle{ A'+B' }[/math] [math]\displaystyle{ A^{eff}+B^{eff} }[/math] [math]\displaystyle{ A^{toy}+B^{toy} }[/math] [math]\displaystyle{ A+B-C }[/math] [math]\displaystyle{ A^{eff}+B^{eff}-C^{eff} }[/math]
[math]\displaystyle{ 10 }[/math] [math]\displaystyle{ (3.442 - 0.168 i) \times 10^{-2} }[/math] 0 0 0 N/A N/A [math]\displaystyle{ (3.501 - 0.316 i) \times 10^{-2} }[/math]
[math]\displaystyle{ 30 }[/math] [math]\displaystyle{ (-1.000 - 0.071 i) \times 10^{-4} }[/math] [math]\displaystyle{ (-0.650 - 0.188 i) \times 10^{-4} }[/math] [math]\displaystyle{ (-0.211 - 0.192 i) \times 10^{-4} }[/math] [math]\displaystyle{ (-0.670 - 0.114 i) \times 10^{-4} }[/math] [math]\displaystyle{ (-0.136 + 0.021 i) \times 10^{-4} }[/math] [math]\displaystyle{ (-1.227 - 0.058 i) \times 10^{-4} }[/math] [math]\displaystyle{ (-1.032 - 0.066 i) \times 10^{-4} }[/math]
[math]\displaystyle{ 100 }[/math] [math]\displaystyle{ (6.702 + 3.134 i) \times 10^{-16} }[/math] [math]\displaystyle{ (2.890 + 3.667 i) \times 10^{-16} }[/math] [math]\displaystyle{ (2.338 + 3.742 i) \times 10^{-16} }[/math] [math]\displaystyle{ (2.955 + 3.650 i) \times 10^{-16} }[/math] [math]\displaystyle{ (0.959 + 0.871 i) \times 10^{-16} }[/math] [math]\displaystyle{ (6.158 + 12.226 i) \times 10^{-16} }[/math] [math]\displaystyle{ (6.763 + 3.074 i) \times 10^{-16} }[/math]
[math]\displaystyle{ 300 }[/math] [math]\displaystyle{ (-4.016 - 1.401 i) \times 10^{-49} }[/math] [math]\displaystyle{ (-5.808 - 1.140 i) \times 10^{-49} }[/math] [math]\displaystyle{ (-5.586 - 1.228 i) \times 10^{-49} }[/math] [math]\displaystyle{ (-5.824 - 1.129 i) \times 10^{-49} }[/math] [math]\displaystyle{ (-2.677 - 0.327 i) \times 10^{-49} }[/math] [math]\displaystyle{ (-3.346 + 6.818 i) \times 10^{-49} }[/math] [math]\displaystyle{ (-4.032 - 1.408 i) \times 10^{-49} }[/math]
[math]\displaystyle{ 1000 }[/math] [math]\displaystyle{ (0.015 + 3.051 i) \times 10^{-167} }[/math] [math]\displaystyle{ (-0.479 + 3.126 i) \times 10^{-167} }[/math] [math]\displaystyle{ (-0.516 + 3.135 i) \times 10^{-167} }[/math] [math]\displaystyle{ (-0.474 + 3.124 i) \times 10^{-167} }[/math] [math]\displaystyle{ (-0.406 + 2.051 i) \times 10^{-167} }[/math] [math]\displaystyle{ (0.175 + 3.306 i) \times 10^{-167} }[/math] [math]\displaystyle{ (0.017 + 3.047 i) \times 10^{-167} }[/math]
[math]\displaystyle{ 3000 }[/math] [math]\displaystyle{ (-1.144+ 1.5702 i) 10^{-507} }[/math] [math]\displaystyle{ (-1.039+ 1.5534 i) 10^{-507} }[/math] [math]\displaystyle{ (-1.039+ 1.5552 i) 10^{-507} }[/math] [math]\displaystyle{ (-1.038+ 1.5535 i) 10^{-507} }[/math] [math]\displaystyle{ (-0.925+ 1.3933 i) 10^{-507} }[/math] [math]\displaystyle{ (-1.155+ 1.5686 i) 10^{-507} }[/math] [math]\displaystyle{ (-1.144+ 1.5701 i) 10^{-507} }[/math]
[math]\displaystyle{ 10000 }[/math] [math]\displaystyle{ (-0.558 - 4.088 i) \times 10^{-1700} }[/math] [math]\displaystyle{ (-0.692 - 4.067 i) \times 10^{-1700} }[/math] [math]\displaystyle{ (-0.687 - 4.067 i) \times 10^{-1700} }[/math] [math]\displaystyle{ (-0.692 - 4.066 i) \times 10^{-1700} }[/math] [math]\displaystyle{ (-0.673 - 3.948 i) \times 10^{-1700} }[/math] [math]\displaystyle{ (-0.548 - 4.089 i) \times 10^{-1700} }[/math] [math]\displaystyle{ (-0.558 - 4.088 i) \times 10^{-1700} }[/math]
[math]\displaystyle{ 30000 }[/math] [math]\displaystyle{ (3.160 - 6.737) \times 10^{-5110} }[/math] [math]\displaystyle{ (3.065 - 6.722) \times 10^{-5100} }[/math] [math]\displaystyle{ (3.066 - 6.722) \times 10^{-5100} }[/math] [math]\displaystyle{ (3.065 - 6.722) \times 10^{-5100} }[/math] [math]\displaystyle{ (2.853 - 6.286) \times 10^{-5100} }[/math] [math]\displaystyle{ (3.170 - 6.733) \times 10^{-5100} }[/math] [math]\displaystyle{ (3.160 - 6.737) \times 10^{-5100} }[/math]

Controlling |A+B|/|B_0|

See Controlling A+B/B_0.

Mesh evaluations of [math]\displaystyle{ A^{eff}+B^{eff}/B^{eff}_0 }[/math] in the ranges

Here is a table of analytic lower bounds for [math]\displaystyle{ A^{eff}+B^{eff}/B^{eff}_0 }[/math] for [math]\displaystyle{ 3 \leq N \leq 2000 }[/math].

Controlling |H_t-A-B|/|B_0|

See Controlling H_t-A-B/B_0.

Here is a table on bounds on error terms [math]\displaystyle{ E_1/B^{eff}_0, E_2/B^{eff}_0, E_3^*/B^{eff}_0 }[/math] for N=3 to 2000. Here is a table with some sharpened estimates from the PDF writeup.

Here is a graph depicting [math]\displaystyle{ |H_t-A^{eff}-B^{eff}/B_0^{eff}| }[/math] and [math]\displaystyle{ E_1+E_2+E_3^*/|B_0^{eff}| }[/math] for [math]\displaystyle{ x \leq 1600 }[/math].

Small values of x

Tables of [math]\displaystyle{ H_t(x+iy) }[/math] for small values of [math]\displaystyle{ x }[/math]:


Here are some snapshots of [math]\displaystyle{ H_t/B^{eff}_0 }[/math].

In this range we will need Bounding the derivative of H_t or Bounding the derivative of H_t - second approach or Bounding the derivative of H_t - third approach.

Tables of [math]\displaystyle{ H'_t(x+iy) }[/math]:

Here is a table of [math]\displaystyle{ x }[/math], pari/gp prec, [math]\displaystyle{ H_{t}, H^{'}_{t}, |H_{t}|, |H^{'}_{t}|, \frac{|H_{t}|}{|B_{0}^{eff}|}, \frac{|H^{'}_{t}|}{|B_{0}^{eff}|} }[/math] for x=0 to x=30 with step size 0.01.


Here is a plot of [math]\displaystyle{ H_t/B_0 }[/math] for a rectangle [math]\displaystyle{ \{x+iy: 0 \leq x \leq 300; 0.4 \leq y \leq 0.45\} }[/math]. Here is an adaptive mesh plot; here is a closeup near the origin.

Here is a script for verifying the absence of zeroes of [math]\displaystyle{ H_t }[/math] in a rectangle. It can eliminate zeros in the rectangle [math]\displaystyle{ \{0 \leq x \leq 1000, 0.4 \leq y \leq 0.45\} }[/math] when t = 0.4.


Large negative values of [math]\displaystyle{ t }[/math]

See also Second attempt at computing H_t(x) for negative t.

We heuristically compute [math]\displaystyle{ H_t(x) }[/math] in the regime where [math]\displaystyle{ x }[/math] is large and [math]\displaystyle{ t }[/math] is large and negative with [math]\displaystyle{ |t|/x \asymp 1 }[/math]. We shall only be interested in the zeroes and so we discard any multiplicative factor which is non-zero: we write [math]\displaystyle{ X \sim Y }[/math] if X is equal (or approximately equal) to Y times something that is explicit and non-zero.

From equation (35) of the writeup we have

[math]\displaystyle{ H_t(x) = \int_{\bf R} \frac{1}{8} \xi(\frac{1+ix}{2} + i |t|^{1/2} v) \frac{1}{\pi} e^{-v^2}\ dv \quad (3.1) }[/math]
[math]\displaystyle{ \sim \int_{\bf R} \xi(\frac{1+ix}{2} + i |t|^{1/2} v) e^{-v^2}\ dv. \quad (3.2) }[/math]

To cancel off an exponential decay factor in the [math]\displaystyle{ \xi }[/math] function, it is convenient to shift the v variable by [math]\displaystyle{ \pi |t|^{1/2}/8 }[/math], thus

[math]\displaystyle{ H_t(x) \sim \int_{\bf R} \xi(\frac{1+ix}{2} + i |t|^{1/2} v - \pi i |t|/8) e^{-(v - \pi |t|^{1/2}/8)^2}\ dv \quad (3.3) }[/math]
[math]\displaystyle{ \sim \int_{\bf R} \xi(\frac{1+i\tilde x}{2} + i |t|^{1/2} v) e^{-v^2 + \pi |t|^{1/2} v / 4}\ dv \quad (3.4) }[/math]

where

[math]\displaystyle{ \tilde x := x - \pi |t|/4 = x + \frac{\pi t}{4}. \quad (3.5) }[/math]

Now from the definition of [math]\displaystyle{ \xi }[/math] and the Stirling approximation we have

[math]\displaystyle{ \xi(s) \sim M_0(s) \zeta(s)\quad (3.6) }[/math]

where [math]\displaystyle{ M_0 }[/math] is defined in (6) of the writeup. Thus

[math]\displaystyle{ H_t(x) \sim \int_{\bf R} M_0(\frac{1+i\tilde x}{2} + i |t|^{1/2} v) \zeta(\frac{1+i\tilde x}{2} + i |t|^{1/2} v) e^{-v^2 + \pi |t|^{1/2} v / 4}\ dv.\quad (3.7) }[/math]

By Taylor expansion we have

[math]\displaystyle{ M_0(\frac{1+i\tilde x}{2} + i |t|^{1/2} v) \sim M_0(\frac{1+i\tilde x}{2}) \exp( \alpha( \frac{1+i\tilde x}{2} ) i |t|^{1/2} v + \alpha'(\frac{1+i \tilde x}{2}) \frac{-|t| v^2}{2} )\quad (3.8) }[/math]
[math]\displaystyle{ \sim \exp( \alpha( \frac{1+i\tilde x}{2} ) i |t|^{1/2} v + \alpha'(\frac{1+i \tilde x}{2}) \frac{-|t| v^2}{2} )\quad (3.9) }[/math]

where [math]\displaystyle{ \alpha }[/math] is defined in equation (8) of the writeup. We have the approximations

[math]\displaystyle{ \alpha(\frac{1+i\tilde x}{2} ) \approx \frac{1}{2} \log \frac{\tilde x}{4\pi} + \frac{i\pi}{4} \quad (3.10) }[/math]

and

[math]\displaystyle{ \alpha'(\frac{1+i\tilde x}{2} ) \approx \frac{-i}{\tilde x} \quad (3.11) }[/math]

and hence

[math]\displaystyle{ H_t(x) \sim \int_{\bf R} \exp( \frac{i |t|^{1/2} v}{2} \log \frac{\tilde x}{4\pi} - \pi |t|^{1/2} v/4 + i |t| v^2 / 2\tilde x) \zeta(\frac{1+i\tilde x}{2} + i |t|^{1/2} v) e^{-v^2 + \pi |t|^{1/2} v / 4}\ dv.\quad (3.12) }[/math]

The two factors of [math]\displaystyle{ \exp( \pi |t|^{1/2} v/4 ) }[/math] cancel. If we now write

[math]\displaystyle{ N := \sqrt{\frac{\tilde x}{4\pi}}\quad (3.13) }[/math]

and

[math]\displaystyle{ u := |t|/N^2 = 4\pi |t|/\tilde x,\quad (3.14) }[/math]

we conclude that

[math]\displaystyle{ H_t(x) \sim \int_{\bf R} \exp( i |t|^{1/2} v \log N + i u v^2 / 8 \pi) \zeta(\frac{1+i\tilde x}{2} + i |t|^{1/2} v) e^{-v^2}\ dv.\quad (3.15) }[/math]

If we formally write [math]\displaystyle{ \zeta(s) = \sum_n \frac{1}{n^s} }[/math] (ignoring convergence issues) we obtain

[math]\displaystyle{ H_t(x) \sim \sum_n \int_{\bf R} \exp( i |t|^{1/2} v \log N + i u v^2 / 8 \pi) n^{-\frac{1+i\tilde x}{2} - i |t|^{1/2} v} e^{-v^2}\ dv\quad (3.16) }[/math]
[math]\displaystyle{ \sim \sum_n \int_{\bf R} \exp( - i |t|^{1/2} v \log \frac{n}{N} + i u v^2 / 8 \pi -\frac{1+i\tilde x}{2} \log \frac{n}{N} ) e^{-v^2}\ dv\quad (3.17) }[/math]

We can compute the [math]\displaystyle{ v }[/math] integral to obtain

[math]\displaystyle{ H_t(x) \sim \sum_n \exp( - \frac{|t| \log^2 \frac{n}{N}}{4 (1 - iu / 8 \pi)} -\frac{1+i\tilde x}{2} \log \frac{n}{N}).\quad (3.18) }[/math]

Using the Taylor approximation

[math]\displaystyle{ \log \frac{n}{N} \approx \frac{n-N}{N} - \frac{(n-N)^2}{2N^2} \quad (3.19) }[/math]

and dropping some small terms, we obtain

[math]\displaystyle{ H_t(x) \sim \sum_n \exp( - \frac{|t| (n-N)^2}{4 N^2 (1 - iu/8\pi)} -\frac{i\tilde x}{2} \frac{n-N}{N} + \frac{i \tilde x (n-N)^2}{4N^2} ).\quad (3.20) }[/math]

Writing [math]\displaystyle{ \tilde x = 4\pi N^2 }[/math] and [math]\displaystyle{ |t| = u N^2 }[/math], this becomes

[math]\displaystyle{ H_t(x) \sim \sum_n \exp( - \frac{2\pi u (n-N)^2}{8\pi - iu} -2 \pi i N(n-N) + \pi i (n-N)^2 ).\quad (3.21) }[/math]

Writing

[math]\displaystyle{ N(n-N) = \frac{1}{2} n^2 - \frac{1}{2} (n-N)^2 - \frac{1}{2} N^2 \quad (3.22) }[/math]

we thus have

[math]\displaystyle{ H_t(x) \sim \sum_n \exp( - \frac{2 \pi u (n-N)^2}{8 \pi - iu} - \pi i n^2 + 2 \pi i (n-N)^2 )\quad (3.23) }[/math]
[math]\displaystyle{ \sim \sum_n \exp( \frac{16 \pi^2 i (n-N)^2}{8 \pi - iu} ) e^{\pi i n}\quad (3.24) }[/math]
[math]\displaystyle{ \sim \theta_{01}( \frac{16 \pi N}{8\pi - iu}, \frac{16 \pi}{8\pi - iu} )\quad (3.25) }[/math]

where [math]\displaystyle{ \theta_{01} }[/math] is the theta function defined in this Wikipedia page. Using the Jacobi identity we then have

[math]\displaystyle{ H_t(x) \sim \theta_{10}(N, \frac{iu - 8\pi}{16 \pi} )\quad (3.26) }[/math]
[math]\displaystyle{ \sim \theta( N + \frac{1}{2} \frac{iu - 8\pi}{16 \pi}, \frac{iu - 8\pi}{16 \pi})\quad (3.27) }[/math]
[math]\displaystyle{ \sim \sum_n \exp( - \pi i n(n+1) / 2 ) e^{2\pi i (n+1/2) N} e^{-u n(n+1)/16}.\quad (3.28) }[/math]

As a sanity check, one can verify that the RHS is real-valued, just as [math]\displaystyle{ H_t(x) }[/math] is (by the functional equation).