Controlling H t-A-B/B 0: Difference between revisions

Latest revision as of 07:09, 30 March 2018

As computed in Effective bounds on H_t - second approach, there is an effective bound

[math]\displaystyle{ |H^{eff} - A^{eff} - B^{eff}| \leq E_1 + E_2 + E_3 }[/math]

where

[math]\displaystyle{ 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{ 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{ \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^{eff}+B^{eff} }[/math] is a good approximation to [math]\displaystyle{ H_t }[/math], [math]\displaystyle{ A+B-C }[/math] is better, and [math]\displaystyle{ A^{eff}+B^{eff}-C^{eff} }[/math] is excellent source source source source source

[math]\displaystyle{ x }[/math]	[math]\displaystyle{ \frac{\|H_t-(A+B)\|}{\|B_0\|} }[/math]	[math]\displaystyle{ \frac{\|H_t-(A^{eff}+B^{eff})\|}{\|B_0^{eff}\|} }[/math]	[math]\displaystyle{ \frac{\|H_t-(A+B-C)\|}{\|B_0\|} }[/math]	[math]\displaystyle{ \frac{\|H_t-(A^{eff}+B^{eff}-C^{eff})\|}{\|B_0^{eff}\|} }[/math]
160	0.174873661533	0.1675083979955609185	0.06993270565802375041	0.00887362155217
320	0.278624615745	0.2776948344513698276	0.006716674125965016299	0.000708716878236
480	0.167598495339	0.1675667240356922231	0.005332893070605698501	0.000327585584191
640	0.165084846603	0.1635077306008453928	0.003363431256036816251	0.000523969818792
800	0.201954876756	0.2045038601879677257	0.1548144749150572349	0.002644344570
960	0.103387669714	0.1031837988358064657	0.03009229958121352990	0.000819848578351
1120	0.0767779295558	0.07541968034203085865	0.004507664238680722472	0.000978838228690
1280	0.132886551163	0.1339118061014743863	0.002283591962997851167	0.000679785836479
1440	0.0802159981813	0.07958929988050262854	0.01553727684468691873	0.000655447626435
1600	0.0777462698681	0.07700542235140914608	0.001778051951547709718	0.000439823567873
1760	0.0950946156489	0.09568042045936396570	0.02763769444052338578	0.000231103881282
1920	0.0629013452776	0.06385275621986742745	0.002108779890256530964	0.000849398936325
2080	0.0949328843573	0.09421231232885752514	0.02746770886040058927	0.000505410233739
2240	0.0591497767926	0.05888587520703223358	0.001567020041379128455	0.000107206342271
2400	0.0785798163298	0.07899341548208345822	0.01801417530687959747	0.000229146425813
2560	0.0621868667021	0.06283843631123482445	0.001359561117436848149	0.000492123116208
2720	0.0585282736442	0.05966972584730198272	0.008503327577240081269	0.000656180976718
2880	0.0787554869341	0.07980560515423855917	0.001089253262122934826	0.000878298302262
3040	0.0462460274843	0.04636072344121703969	0.003004181560093288747	0.0000470113907733
3200	0.0963053589535	0.09664223832561922043	0.02931455383125538672	0.000354582706466

A closer look at the "spike" in error near [math]\displaystyle{ x=800 \approx 256 \pi \approx 804 }[/math]:

[math]\displaystyle{ x }[/math]	[math]\displaystyle{ \frac{\|H_t-(A+B-C)\|}{\|B_0\|} }[/math]
622.035345	0.003667321
631.460123	0.004268055
640.884901	0.003284407
650.309679	0.004453589
659.734457	0.003872174
669.159235	0.005048162
678.584013	0.005009254
688.008791	0.007418686
697.433569	0.007464541
706.858347	0.010692337
716.283125	0.012938629
725.707903	0.017830524
735.132681	0.022428596
744.557459	0.030907876
753.982237	0.040060298
763.407015	0.053652069
772.831793	0.071092824
782.256571	0.094081856
791.681349	0.123108726
801.106127	0.159299234
810.530905	0.002870724

In practice [math]\displaystyle{ E_1/B^{eff}_0 }[/math] is smaller than [math]\displaystyle{ E_2/B^{eff}_0 }[/math], which is mostly dominated by the first term in the sum which is close to [math]\displaystyle{ \frac{t^2}{16 x} \log^2 \frac{x}{4\pi} }[/math]:

[math]\displaystyle{ x }[/math]	[math]\displaystyle{ E_1 / B^{eff}_0 }[/math]	[math]\displaystyle{ E_2 / B^{eff}_0 }[/math]	[math]\displaystyle{ \frac{t^2}{16x} \log^2 \frac{x}{4\pi} }[/math]
10^3	[math]\displaystyle{ 1.389 \times 10^{-3} }[/math]	[math]\displaystyle{ 2.341 \times 10^{-3} }[/math]	[math]\displaystyle{ 1.915 \times 10^{-4} }[/math]
10^4	[math]\displaystyle{ 1.438 \times 10^{-4} }[/math]	[math]\displaystyle{ 3.156 \times 10^{-4} }[/math]	[math]\displaystyle{ 4.461 \times 10^{-5} }[/math]
10^5	[math]\displaystyle{ 1.118 \times 10^{-5} }[/math]	[math]\displaystyle{ 3.574 \times 10^{-5} }[/math]	[math]\displaystyle{ 8.067 \times 10^{-6} }[/math]
10^6	[math]\displaystyle{ 7.328 \times 10^{-7} }[/math]	[math]\displaystyle{ 3.850 \times 10^{-6} }[/math]	[math]\displaystyle{ 1.273 \times 10^{-6} }[/math]
10^7	[math]\displaystyle{ 4.414 \times 10^{-8} }[/math]	[math]\displaystyle{ 4.197 \times 10^{-7} }[/math]	[math]\displaystyle{ 1.846 \times 10^{-7} }[/math]

Estimation of [math]\displaystyle{ E_1,E_2 }[/math]

First let us obtain bounds for [math]\displaystyle{ |E_1/B^{eff}_0|, |E_2/B^{eff}_0| }[/math], assuming for instance that [math]\displaystyle{ x \geq 100 }[/math], that only depend on [math]\displaystyle{ N }[/math] and not on [math]\displaystyle{ x }[/math]. For fixed [math]\displaystyle{ N }[/math], one has [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]

In particular [math]\displaystyle{ x_N/2 \leq T \leq x_{N+1}/2 }[/math].

We begin with [math]\displaystyle{ E_2 }[/math]. We have

[math]\displaystyle{ |E_2/B^{eff}_0| = \frac{1}{T-3.33} \epsilon'(\frac{1+y+ix}{2}) (2.1) }[/math]

where

[math]\displaystyle{ \epsilon'(s) = \frac{1}{2} \sum_{n=1}^N \frac{1}{n^{\mathrm{Re}(s) + \frac{t}{2} \mathrm{Re} \alpha_1(s) - \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]

and

[math]\displaystyle{ \alpha_1(s) := \frac{1}{2s} + \frac{1}{s-1} + \frac{1}{2} \log \frac{s}{2\pi}. }[/math]

We have

[math]\displaystyle{ \alpha'_1(s) = \frac{-1}{2s^2} - \frac{1}{(s-1)^2} + \frac{1}{2s} }[/math]

and thus for [math]\displaystyle{ s }[/math] between [math]\displaystyle{ \frac{1+y+ix}{2} }[/math] and [math]\displaystyle{ s^+_N := \frac{1+y+i(x_N+x_{N+1})/2}{2} }[/math] one has

[math]\displaystyle{ \alpha'_1(s) = O_{\leq}( \frac{2}{x_N^2} + \frac{4}{x_N^2} + \frac{1}{x_N} ) = O_{\leq}( \frac{1}{x_N - 6} ). }[/math]

(here we use Lemma 1.1 of Effective bounds on H_t - second approach.) Thus we have

[math]\displaystyle{ \alpha_1(s) = \alpha_1(s^+_N) + O_{\leq}( \kappa ) }[/math]

where

[math]\displaystyle{ \kappa := \frac{x_{N+1}-x_N}{4 (x_N-6)} }[/math]

(asymptotically this is [math]\displaystyle{ \sim 1/N }[/math]). Thus

[math]\displaystyle{ |\alpha_1(s) - \log n|^2 = |\alpha_1(s^+_N) - \log n|^2 + O_{\leq}( 2 \kappa |\alpha_1(s^+_N) - \log n| + \kappa^2 ) }[/math]

and we conclude that

[math]\displaystyle{ \epsilon'(\frac{1+y+ix}{2}) \leq \frac{1}{2} \sum_{n=1}^N \frac{1}{n^{\frac{1+y}{2} + \frac{t}{2} \mathrm{Re} \alpha_1(s^+_N) - \frac{t}{2} \kappa - \frac{t}{4} \log n}} \exp( \frac{1}{2(T - 3.33)} c^+_n ) c^+_n }[/math]

where

[math]\displaystyle{ c^+_n := \frac{t^2}{4} (|\alpha_1(s^+_N) - \log n|^2 + 2 \kappa |\alpha_1(s^+_N) - \log n| + \kappa^2) + \frac{1}{3} + t. }[/math]

When combined with (2.1), this gives a uniform upper bound on [math]\displaystyle{ |E_2/B^{eff}_0| }[/math] for a fixed value of [math]\displaystyle{ N }[/math].

In a similar vein, we have

[math]\displaystyle{ |E_2/B^{eff}_0| = \frac{1}{T-3.33} \lambda \epsilon'(\frac{1-y+ix}{2}) (2.2) }[/math]

where [math]\displaystyle{ \lambda }[/math] is the quantity defined in this page. In that page the upper bound

[math]\displaystyle{ \lambda \leq e^\delta N^{-y} }[/math]

was established, 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+i x_{N+1}|}{4\pi}. }[/math]

Also, by repeating previous arguments (with [math]\displaystyle{ y }[/math] replaced by [math]\displaystyle{ -y }[/math]) we have

[math]\displaystyle{ \epsilon'(\frac{1-y+ix}{2}) \leq \frac{1}{2} \sum_{n=1}^N \frac{1}{n^{\frac{1-y}{2} + \frac{t}{2} \mathrm{Re} \alpha_1(s^-_N) - \frac{t}{2} \kappa - \frac{t}{4} \log n}} \exp( \frac{1}{2(T - 3.33)} c^-_n ) c^-_n }[/math]

where

[math]\displaystyle{ c^-_n := \frac{t^2}{4} (|\alpha_1(s^-_N) - \log n|^2 + 2 \kappa |\alpha_1(s^-_N) - \log n| + \kappa^2) + \frac{1}{3} + t }[/math]

and

[math]\displaystyle{ s^-_N := \frac{1-y+i(x_N + x_{N+1})/2}{2}. }[/math]

Tables of upper bounds for [math]\displaystyle{ |E_1|/|B_0^{eff}| }[/math], [math]\displaystyle{ |E_2|/|B_0^{eff}| }[/math] can be found here.

Estimation for large [math]\displaystyle{ x }[/math]

We can crudely bound [math]\displaystyle{ \epsilon'(\frac{1+y+ix}{2}), \epsilon'(\frac{1-y+ix}{2}) }[/math] as follows. In Controlling A+B/B_0 it is shown that

[math]\displaystyle{ \mathrm{Re} \alpha_1(\frac{1+y+ix}{2}) \geq \log N }[/math]

for [math]\displaystyle{ y \geq 1/3 }[/math], and so

[math]\displaystyle{ \epsilon'(\frac{1+y+ix}{2}) \leq \exp( \frac{1}{2(T-3.33)} (\frac{t^2}{4} |\alpha_1(\frac{1+y+ix}{2})|^2 + \frac{1}{3} + t)) (\frac{t^2}{4} |\alpha_1(\frac{1+y+ix}{2})|^2 + \frac{1}{3} + t) \sum_{n=1}^N \frac{1}{n^{\frac{1+y}{2} + \frac{t}{4} \log \frac{N^2}{n}}} }[/math]

and thus in the language of Estimating a sum

[math]\displaystyle{ \epsilon'(\frac{1+y+ix}{2}) \leq \exp( \frac{1}{2(T-3.33)} (\frac{t^2}{4} |\alpha_1(\frac{1+y+ix}{2})|^2 + \frac{1}{3} + t)) (\frac{t^2}{4} |\alpha_1(\frac{1+y+ix}{2})|^2 + \frac{1}{3} + t) S_{\frac{1+y}{2}, t}(N). }[/math]

Thus for instance if [math]\displaystyle{ y=t=0.4 }[/math] and [math]\displaystyle{ N \geq 2000 }[/math] then

[math]\displaystyle{ \epsilon'(\frac{1+y+ix}{2}) \leq \exp( \frac{1}{2(T-3.33)} (\frac{1}{25} |\alpha_1(\frac{0.7+ix}{2})|^2 + \frac{11}{15})) (\frac{1}{25} |\alpha_1(\frac{0.7+ix}{2})|^2 + \frac{11}{15}) \times 1.706. }[/math]

Similarly we have

[math]\displaystyle{ \epsilon'(\frac{1-y+ix}{2}) \leq \exp( \frac{1}{2(T-3.33)} (\frac{1}{25} |\alpha_1(\frac{0.3+ix}{2})|^2 + \frac{11}{15})) (\frac{1}{25} |\alpha_1(\frac{0.3+ix}{2})|^2 + \frac{11}{15}) \times 3.469. }[/math]

Thus

[math]\displaystyle{ E_2 / |B^{eff}_0| \leq \frac{1.706}{T-3.33} \exp( \frac{1}{2(T-3.33)} (\frac{1}{25} |\alpha_1(\frac{0.7+ix}{2})|^2 + \frac{11}{15})) (\frac{1}{25} |\alpha_1(\frac{0.7+ix}{2})|^2 + \frac{11}{15}) }[/math]

and

[math]\displaystyle{ E_1 / |B^{eff}_0| \leq \frac{3.469}{T-3.33} e^{\delta} N^{-0.4} \exp( \frac{1}{2(T-3.33)} (\frac{1}{25} |\alpha_1(\frac{0.3+ix}{2})|^2 + \frac{11}{15})) (\frac{1}{25} |\alpha_1(\frac{0.3+ix}{2})|^2 + \frac{11}{15}). }[/math]

If [math]\displaystyle{ N \geq 2000 }[/math], then [math]\displaystyle{ T \geq 2\pi N^2 - \frac{\pi t}{8} \geq 2 \times 10^7 }[/math] and [math]\displaystyle{ x \geq 4 \times 10^7 }[/math]. One can bound

[math]\displaystyle{ |\alpha_1(\frac{0.7+ix}{2})| \leq \frac{3}{x} + \frac{1}{2} \log \frac{\sqrt{1+x^2}}{4\pi} + \frac{\pi}{4} }[/math]

and hence

[math]\displaystyle{ \frac{1}{T-3.33} (\frac{1}{25} |\alpha_1(\frac{0.7+ix}{2})|^2 + \frac{11}{15}) \leq \frac{1}{T-3.33} (\frac{1}{25} (\frac{3}{x} + \frac{1}{2} \log \frac{\sqrt{1+x^2}}{4\pi} + \frac{\pi}{4})^2 + \frac{11}{15}). }[/math]

this is decreasing in [math]\displaystyle{ x }[/math] and bounded by [math]\displaystyle{ 1.7 \times 10^{-7} }[/math]. From this we conclude that [math]\displaystyle{ E_2 / |B^{eff}_0| \leq 2.9 \times 10^{-7} }[/math]. One can also calculate that [math]\displaystyle{ \delta \leq 1.6 \times 10^{-7} }[/math] and that [math]\displaystyle{ E_1 / |B^{eff}_0| \leq 2.8 \times 10^{-8} }[/math].

Estimation of [math]\displaystyle{ E_3 }[/math]

Here we assume that [math]\displaystyle{ T_0 \geq 100 }[/math], which implies also [math]\displaystyle{ T'_0 \geq 100 }[/math].

We first bound [math]\displaystyle{ w }[/math] by a Gaussian type quantity.

We have

[math]\displaystyle{ 1 + \frac{\sigma^2}{(T'_0)^2} \leq \exp( \frac{\sigma^2}{(T'_0)^2}) }[/math]

and

[math]\displaystyle{ 1 + \frac{(1-\sigma)^2}{(T'_0)^2} \leq \exp( \frac{(1-\sigma)^2}{(T'_0)^2}) }[/math]

and thus

[math]\displaystyle{ ( 1 + \frac{\sigma^2}{(T'_0)^2} )^{1/2} (1 + \frac{(1-\sigma)^2}{(T'_0)^2})^{1/2} \leq \exp( \frac{1}{2} \frac{\sigma^2}{(T'_0)^2} + \frac{1}{2} \frac{(1-\sigma)^2}{(T'_0)^2} ) }[/math]

[math]\displaystyle{ = \exp( \frac{(\sigma-1/2)^2}{(T'_0)^2} + \frac{1}{4 (T'_0)^2} ). }[/math]

Next, from calculus one can verify the bounds

[math]\displaystyle{ \log(1+x^2) \leq 1.479 \sqrt{x} }[/math]

and

[math]\displaystyle{ x - \mathrm{arctan}(x) \leq 0.230 x^2 }[/math]

for any [math]\displaystyle{ x \geq 0 }[/math], and hence

[math]\displaystyle{ \frac{(\sigma-1)_+}{4} \log (1 + \frac{\sigma^2}{(T'_0)^2}) \leq \frac{1}{4} 1.479 \frac{\sigma(\sigma-1)}{T'_0} 1_{\sigma \geq 1} }[/math]

[math]\displaystyle{ \leq 0.37 \frac{(\sigma-1/2)^2}{T'_0} 1_{\sigma \geq 1} }[/math]

and

[math]\displaystyle{ (\frac{T'_0}{2} \arctan \frac{\sigma}{T'_0} - \frac{\sigma}{2}) 1_{\sigma \lt 0} \leq \frac{T'_0}{2} 1_{\sigma\lt 0} 0.230 (\frac{|\sigma|}{T'_0})^2 }[/math]

[math]\displaystyle{ \leq 0.115 \frac{(\sigma-1/2)^2}{T'_0} 1_{\sigma \lt 0}. }[/math]

We conclude that

[math]\displaystyle{ w(\sigma) \leq \exp( \frac{(\sigma-1/2)^2}{(T'_0)^2} + \frac{1}{4 (T'_0)^2} + 0.37 \frac{(\sigma-1/2)^2}{T'_0} + \frac{1}{12(T'_0 - 0.33)}) }[/math]

[math]\displaystyle{ \leq \exp( 0.37 \frac{(\sigma-1/2)^2}{T'_0-2.71} + \frac{1}{12(T'_0 - 3.33)}). }[/math]

Now we work on [math]\displaystyle{ v }[/math]. Observe that if [math]\displaystyle{ k \leq \frac{T'_0}{2.42 \pi} = \frac{a_0^2}{1.21} }[/math] then

[math]\displaystyle{ (1.1)^{k+2} \frac{\Gamma(\frac{k+2}{2})}{a_0^{k+2}} = \frac{1.21 k}{2 a_0^2} \frac{\Gamma(\frac{k}{2})}{a_0^k} \leq \frac{1}{2} (1.1)^{k} \frac{\Gamma(\frac{k}{2})}{a_0^{k}}, }[/math]

and hence

[math]\displaystyle{ \sum_{2 \leq k \leq \frac{T'_0}{2.24 \pi}; k\ \mathrm{even}} (1.1)^{k} \frac{\Gamma(\frac{k}{2})}{a_0^{k}} \leq 2 (1.1)^2 \frac{\Gamma(\frac{2}{2})}{a_0^2} = \frac{2.42 \sqrt{\pi}}{a_0^2} }[/math]

and similarly

[math]\displaystyle{ \sum_{3 \leq k \leq \frac{T'_0}{2.42 \pi}; k\ \mathrm{odd}} (1.1)^{k} \frac{\Gamma(\frac{k}{2})}{a_0^{k}} \leq 2 (1.1)^3 \frac{\Gamma(\frac{3}{2})}{a_0^2} = \frac{1.331}{a_0^3} }[/math]

and hence

[math]\displaystyle{ \sum_{1 \leq k \leq \frac{T'_0}{2.42 \pi}} (1.1)^{k} \frac{\Gamma(\frac{k}{2})}{a_0^{k}} \leq \frac{1.1 \sqrt{\pi}}{a_0} + \frac{2.42}{a_0^2} + \frac{1.331 \sqrt{\pi}}{a_0^3} }[/math]

[math]\displaystyle{ \leq \frac{1.1 \sqrt{\pi}}{a_0 - 1.25}; }[/math]

also

[math]\displaystyle{ (0.400 \frac{9^\sigma}{a_0} + 0.346 \frac{2^{3\sigma/2}}{a_0^2})1_{\sigma \geq 0} \leq 0.400 \times 9^\sigma (\frac{1}{a_0} + 0.865 \frac{1}{a_0^2}) }[/math]

[math]\displaystyle{ \leq 0.4 \frac{9^\sigma}{a_0 - 0.865} }[/math]

and hence (bounding [math]\displaystyle{ (0.9)^{\lceil -\sigma \rceil} \leq \frac{1}{1.1} }[/math])

[math]\displaystyle{ v(\sigma) \leq 1 + 0.400 \frac{9^\sigma}{a_0-0.865} + \frac{\sqrt{\pi}}{a_0-1.25} + \sum_{\frac{T'_0}{2.42 \pi} \lt k \leq 4-\sigma} \frac{(1.1)^{k-1}}{a_0^k} \Gamma(k/2). }[/math]

We conclude (using Fubini's theorem) that

[math]\displaystyle{ \int_{-\infty}^\infty vwf(\sigma)\ d\sigma \leq \exp(\frac{1}{12(T'_0 - 3.33)}) ( (1 + \frac{\sqrt{\pi}}{a_0-1.25}) \int_{-\infty}^\infty f(\sigma) \exp( 0.37 \frac{(\sigma-1/2)^2}{T'_0-2.71} )\ d\sigma }[/math]

[math]\displaystyle{ + \frac{0.4}{a_0-0.865} \int_{-\infty}^\infty 9^\sigma f(\sigma) \exp( 0.37 \frac{(\sigma-1/2)^2}{T'_0-2.71} )\ d\sigma }[/math]

[math]\displaystyle{ + \sum_{k \gt \frac{T'_0}{2.42\pi}} \frac{(1.1)^{k-1}}{a_0^k} \Gamma(k/2) \int_{-\infty}^{4-k} f(\sigma) \exp( 0.37 \frac{(\sigma-1/2)^2}{T'_0-2.71} )\ d\sigma.) }[/math]

Now we estimate the integrals appearing in the right-hand side. By symmetry we have

[math]\displaystyle{ \int_{-\infty}^\infty f(\sigma) \exp( 0.37 \frac{(\sigma-1/2)^2}{T'_0-2.71} )\ d\sigma = \frac{1}{\sqrt{\pi t}} \int_{-\infty}^\infty \exp( - \frac{(\sigma - (1+y)/2)^2}{t} + 0.37 \frac{(\sigma-1/2)^2}{T'_0-2.71} )\ d\sigma. }[/math]

[math]\displaystyle{ = \frac{1}{\sqrt{\pi t}} \int_{-\infty}^\infty \exp( - \frac{\sigma^2}{t} + 0.37 \frac{(\sigma-y/2)^2}{T'_0-2.71} )\ d\sigma. }[/math]

Using the Gaussian identity

[math]\displaystyle{ \int_{-\infty}^\infty \exp( - (a\sigma^2 + b \sigma + c) )\ d\sigma = \sqrt{\pi} a^{-1/2} \exp( - c + \frac{b^2}{4a} ), }[/math]

valid for any [math]\displaystyle{ a,b,c }[/math] with [math]\displaystyle{ a }[/math] positive, we can write the above expression as

[math]\displaystyle{ (1 - \frac{0.37 t}{T'_0 - 2.71})^{-1/2} \exp( \frac{0.37 y^2}{4 (T'_0 - 2.71 - 0.37 t)} ). }[/math]

Similarly, since [math]\displaystyle{ 9^\sigma }[/math] is larger for [math]\displaystyle{ \sigma \geq 1/2 }[/math] than for [math]\displaystyle{ \sigma \lt 1/2 }[/math], we have

[math]\displaystyle{ \int_{-\infty}^\infty 9^\sigma f(\sigma) \exp( 0.37 \frac{(\sigma-1/2)^2}{T'_0-2.71} )\ d\sigma \leq \frac{1}{\sqrt{\pi t}} \int_{-\infty}^\infty 9^\sigma \exp( - \frac{(\sigma - (1+y)/2)^2}{t} + 0.37 \frac{(\sigma-1/2)^2}{T'_0-2.71} )\ d\sigma. }[/math]

[math]\displaystyle{ = \frac{3^{1+y}}{\sqrt{\pi t}} \int_{-\infty}^\infty \exp( - \frac{\sigma^2}{t} + 0.37 \frac{(\sigma-y/2)^2}{T'_0-2.71} + \sigma \log 9)\ d\sigma. }[/math]

[math]\displaystyle{ = 3^{1+y} (1 - \frac{0.37 t}{T'_0 - 2.71})^{-1/2} \exp( 0.37 \frac{y^2}{4 (T'_0-2.71)} + \frac{b^2}{4 (\frac{1}{t} - \frac{0.37}{T'_0-2.71})} ) }[/math]

where

[math]\displaystyle{ b := - \log 9 + 0.37 \frac{y}{T'_0 - 2.71}. }[/math]

If [math]\displaystyle{ T'_0 \geq 100 }[/math] and [math]\displaystyle{ y \leq 1/2 }[/math] then [math]\displaystyle{ |b| \leq \log 9 }[/math], thus the above integral is at most

[math]\displaystyle{ = 3^{1+y} (1 - \frac{0.37 t}{T'_0 - 2.71})^{-1/2} \exp( 0.37 \frac{y^2}{4 (T'_0-2.71)} + \frac{t \log^2 9}{4 (1 - \frac{0.37 t}{T'_0-2.71})} ). }[/math]

Now we consider the integral

[math]\displaystyle{ \int_{-\infty}^{4-k} f(\sigma) \exp( 0.37 \frac{(\sigma-1/2)^2}{T'_0-2.71} )\ d\sigma. }[/math]

If we assume that [math]\displaystyle{ T_0 \geq 100 }[/math], then [math]\displaystyle{ 4-k \leq 4 - \frac{100}{2.42 \pi} \leq -9 }[/math] is negative, so this expression is at most

[math]\displaystyle{ \leq \frac{1}{\sqrt{\pi t}} \int_{-\infty}^{4-k} \exp( - \frac{(\sigma - (1-y)/2)^2}{t} + 0.37 \frac{(\sigma-1/2)^2}{T'_0-2.71} )\ d\sigma }[/math]

[math]\displaystyle{ \leq \frac{1}{\sqrt{\pi t}} \int_{-\infty}^{4-k} \exp( - \sigma^2 (\frac{1}{t} - \frac{0.37}{T'_0-2.71}) )\ d\sigma. }[/math]

With [math]\displaystyle{ t \leq 0.4 }[/math] and [math]\displaystyle{ T'_0 \geq 100 }[/math], one can verify numerically that

[math]\displaystyle{ \frac{1}{t} - \frac{0.37}{T'_0-2.71} \geq 2 + \frac{1}{2} \log t }[/math]

and so (since [math]\displaystyle{ \sigma^2 \geq 1 }[/math]) one can bound the above by

[math]\displaystyle{ \leq \frac{1}{\sqrt{\pi}} \int_{-\infty}^{4-k} \exp( - 2 \sigma^2 )\ d\sigma }[/math]

[math]\displaystyle{ \leq \frac{1}{\sqrt{\pi}} \exp( - 2 (k - 4)^2 ) \frac{1}{4 (k - 4)} }[/math]

and so the contribution to [math]\displaystyle{ \int_{-\infty}^\infty vwf(\sigma)\ d\sigma) }[/math] is at most

[math]\displaystyle{ \frac{1}{4 (\frac{T'_0}{2.42\pi} - 4) \sqrt{\pi}} \sum_{k \gt \frac{T'_0}{2.42\pi}} c_k }[/math]

where

[math]\displaystyle{ c_k := \frac{(1.1)^{k-1}}{a_0^k} \Gamma(k/2) \exp( - 2(k-4)^2 ). }[/math]

Observe that

[math]\displaystyle{ c_{k+2}/c_k = \frac{(1.1)^2}{a_0^2} \frac{k}{2} \exp( - 4 (k+5) ) }[/math]

and this can be shown to be less than [math]\displaystyle{ 1/2 }[/math] if [math]\displaystyle{ T_0 \geq 100 }[/math], and [math]\displaystyle{ k \gt \frac{T'_0}{2.42 \pi} }[/math]. Thus

[math]\displaystyle{ \sum_{k \gt \frac{T'_0}{2.42\pi}} c_k \leq 4 \sup_{\frac{T'_0}{2.42\pi} \lt k \leq \frac{T'_0}{2.42\pi}+2} a_k }[/math]

[math]\displaystyle{ \leq 4 (\frac{1.1}{a_0})^{\frac{T'_0}{2.42\pi}} \Gamma( \frac{T'_0}{4.84\pi}+1 ) \exp( - 4 (\frac{T'_0}{2.42\pi}-4)^2 ). }[/math]

Putting all this together, we obtain

[math]\displaystyle{ \int_{-\infty}^\infty vwf(\sigma)\ d\sigma \leq \exp(\frac{1}{12(T'_0 - 3.33)}) (1 - \frac{0.37 t}{T'_0 - 2.71})^{-1/2} \exp( \frac{0.37 y^2}{4 (T'_0 - 2.71 - 0.37 t)} ) \times }[/math]

[math]\displaystyle{ (1 + \frac{\sqrt{\pi}}{a_0-1.25} + \frac{0.4}{a_0-0.85} 3^{1+y} \exp( \frac{t \log^2 9}{4 (1 - \frac{0.37}{T'_0-2.71})} ) ) }[/math]

[math]\displaystyle{ + \varepsilon }[/math]

where [math]\displaystyle{ \varepsilon }[/math] is the exponentially small quantity

[math]\displaystyle{ \varepsilon := \exp(\frac{1}{12(T'_0 - 3.33)}) \frac{1}{(\frac{T'_0}{2.42\pi} - 4) \sqrt{\pi}} (\frac{1.1}{a_0})^{\frac{T'_0}{2.42\pi}} \Gamma( \frac{T'_0}{4.84\pi}+1 ) \exp( - 4 (\frac{T'_0}{2.42\pi}-4)^2 ) }[/math]

which looks fearsome but is extremely negligible in practice. For instance, one can check that

[math]\displaystyle{ \varepsilon \leq \frac{10^{-10}}{a_0^2} \leq 0.4 (\frac{1}{a_0-0.85} - \frac{1}{a_0-1.25}) }[/math]

whenever [math]\displaystyle{ T_0 \geq 100 }[/math], and hence

[math]\displaystyle{ \int_{-\infty}^\infty vwf(\sigma)\ d\sigma \leq (1 - \frac{0.37 t}{T'_0 - 2.71})^{-1/2} \exp( \frac{1}{12(T'_0 - 3.33)} + \frac{0.37 y^2}{4 (T'_0 - 2.71 - 0.37 t)} ) \times }[/math]

[math]\displaystyle{ (1 + \frac{\sqrt{\pi}}{a_0-1.25} + \frac{0.4}{a_0-1.25} 3^{1+y} \exp( \frac{t \log^2 9}{4 (1 - \frac{0.37}{T'_0-2.71})} ) ). }[/math]

To clean this up, we write

[math]\displaystyle{ 1 - \frac{0.37 t}{T'_0 - 2.71} = \exp( O_{\leq}( \frac{0.37 t}{T'_0 - 2.71 - 0.37 t} ) }[/math]

and note that [math]\displaystyle{ T'_0 - 2.71 - 0.37t \geq T'_0 - 3.33 }[/math] to obtain

[math]\displaystyle{ \int_{-\infty}^\infty vwf(\sigma)\ d\sigma \leq \exp( \frac{6 \times 0.37t + 1 + 3 \times 0.37 y^2}{12(T'_0 - 3.33)}) \times }[/math]

[math]\displaystyle{ (1 + \frac{1}{a_0-1.25} (\sqrt{\pi} + 1.2 \times 3^y \exp( \frac{t \log^2 9}{4 (1 - \frac{0.37 t}{T'_0-2.71})} ) ). }[/math]

We bound [math]\displaystyle{ (6 \times 0.37t + 1 + 3 \times 0.37 y^2)/12 \leq 0.181 }[/math] and [math]\displaystyle{ 1.2 \times 3^y \exp( \frac{t \log^2 9}{4(1 - \frac{0.37 t}{T'_0-2.71}}) \leq 5.15 }[/math] for [math]\displaystyle{ y \leq 1/2 }[/math], thus

[math]\displaystyle{ \int_{-\infty}^\infty vwf(\sigma)\ d\sigma \leq \exp( \frac{0.181}{T'_0 - 3.33}) (1 + \frac{5.15}{a_0-1.25}). }[/math]

We conclude that

[math]\displaystyle{ E_3 \leq E_3^* }[/math]

where

[math]\displaystyle{ E_3^* := \frac{1}{8} \sqrt{\pi} \exp( - \frac{t\pi^2}{64}) (T'_0)^{3/2} e^{-\pi T_0/4} \exp( \frac{0.181}{T'_0 - 3.33}) (1 + \frac{5.15}{a_0-1.25}). }[/math]

The main term here is

[math]\displaystyle{ E_3^{main} := (T'_0)^{3/2} e^{-\pi T_0/4}; }[/math]

in particular, we can factor the ratio [math]\displaystyle{ E_3^* / |B^{eff}_0| }[/math] as the product of

[math]\displaystyle{ E_3^* / E_3^{main} = \frac{1}{8} \sqrt{\pi} \exp( - \frac{t\pi^2}{64}) \exp( \frac{0.181}{T'_0 - 3.33}) (1 + \frac{5.15}{a_0-1.25}) }[/math]

and

[math]\displaystyle{ E_3^{main} / |B^{eff}_0| = \frac{16}{\sqrt{2\pi} |s(s-1)|} \pi^{\mathrm{Re} (1-s)/2} \exp( - \frac{\pi T_0}{4} + \frac{3}{2} \log T'_0 - \frac{t}{4} \mathrm{Re}(\alpha_1(1-s)^2) - \mathrm{Re}( (\frac{1-s}{2} - \frac{1}{2}) \log \frac{1-s}{2} + \frac{1-s}{2}) ) ) }[/math]

where [math]\displaystyle{ s := \frac{1-y+ix}{2} }[/math].

The first ratio [math]\displaystyle{ E_3^* / E_3^{main} }[/math] decreases monotonically to [math]\displaystyle{ \frac{1}{8} \sqrt{\pi} \exp( - \frac{t\pi^2}{64}) }[/math], which equals [math]\displaystyle{ 0.2038\dots }[/math] when [math]\displaystyle{ t=0.4 }[/math]. We claim the second ratio decreases for [math]\displaystyle{ x \geq 100 }[/math]. To see this, we compute the log-derivative [math]\displaystyle{ \frac{d}{dx} \log E_3^{main} / |B^{eff}_0| }[/math] as

[math]\displaystyle{ \mathrm{Re}( -\frac{i}{2s} - \frac{i}{2(s-1)} - \frac{\pi}{8} + \frac{3}{4 T'_0} + \frac{i t}{4} \alpha_1(1-s) \alpha'_1(1-s) + \frac{i}{4} \log \frac{1-s}{2} - \frac{i}{4 (1-s)} ) }[/math]

[math]\displaystyle{ = \frac{-x}{(1-y)^2 + x^2} + \frac{-x}{(1+y)^2+x^2} - \frac{\pi}{8} + \frac{3}{2 (x + \pi t/4)} - \frac{t}{4} \mathrm{Im}(\alpha_1(1-s) \alpha'_1(1-s)) + \frac{1}{4} (\frac{\pi}{2} - \mathrm{arctan} \frac{1+y}{x}) + \frac{x/2}{(1+y)^2+x^2} }[/math]

[math]\displaystyle{ = \frac{-x}{(1-y)^2 + x^2} + \frac{-x/2}{(1+y)^2+x^2} + \frac{3}{2 (x + \pi t/4)} - \frac{t}{4} \mathrm{Im}(\alpha_1(1-s) \alpha'_1(1-s)) - \frac{1}{4} \mathrm{arctan} \frac{1+y}{x}. }[/math]

For [math]\displaystyle{ x \geq 100 }[/math], we have [math]\displaystyle{ \mathrm{arctan} \frac{1+y}{x} \geq \frac{1+y}{2x} }[/math] (say), and we bound [math]\displaystyle{ \frac{3}{2(x+\pi t/4)} }[/math] by [math]\displaystyle{ \frac{1}{x} + \frac{1}{2x} }[/math] to obtain an upper bound of

[math]\displaystyle{ \leq (\frac{1}{x} - \frac{x}{(1-y)^2 + x^2}) + (\frac{1}{2x} - \frac{x/2}{(1+y)^2+x^2}) - \frac{t}{4} \mathrm{Im}(\alpha_1(1-s) \alpha'_1(1-s))- \frac{1+y}{8x} }[/math]

[math]\displaystyle{ = \frac{(1-y)^2}{x ((1-y)^2 + x^2)} + \frac{(1+y)^2}{2x ((1+y)^2 + x^2)} - \frac{t}{4} \mathrm{Im}(\alpha_1(1-s) \alpha'_1(1-s)) }[/math]

[math]\displaystyle{ \leq \frac{(1-y)^2 + (1+y)^2/2}{x^3} - \frac{t}{4} \mathrm{Im}(\alpha_1(1-s) \alpha'_1(1-s)) - \frac{1+y}{8x}. }[/math]

For [math]\displaystyle{ x \geq 100 }[/math], we have [math]\displaystyle{ \frac{1+y}{8x} \gt \frac{(1-y)^2 + (1+y)^2/2}{x^3} }[/math], so to establish decrease it suffices to show that

[math]\displaystyle{ \mathrm{Im}(\alpha_1(1-s) \alpha'_1(1-s)) \gt 0. }[/math]

We have

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

[math]\displaystyle{ = O_{\leq}( \frac{1}{x} + \frac{2}{x} ) + \frac{1}{2} \log \frac{|1+y+ix|}{4\pi} + i O_{\leq}( \frac{\pi}{2} ) }[/math]

and

[math]\displaystyle{ \alpha'_1(1-s) := -\frac{1}{2(1-s)^2} - \frac{1}{(-s)^2} + \frac{1}{2(1-s)} }[/math]

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

and hence

[math]\displaystyle{ \mathrm{Im}(\alpha_1(1-s) \alpha'_1(1-s)) = \frac{1}{2} \log \frac{|1+y+ix|}{4\pi} (\frac{x}{(1+y)^2+x^2} + O_{\leq}(\frac{6}{x^2}) ) + O_{\leq}( \frac{\pi}{2} (\frac{(1+y)^2}{(1+y)^2+x^2} + \frac{6}{x^2})) + O_{\leq}( \frac{3}{x} (\frac{6}{x^2} + \frac{1}{x}) ) }[/math]

[math]\displaystyle{ \geq \frac{1}{2(x-6)} \log \frac{x}{4\pi} - \frac{\pi}{2} \frac{6+(1+y)^2}{x^2} - \frac{3}{x(x-6)} }[/math]

[math]\displaystyle{ \geq \frac{1}{2(x-6)} (\log \frac{x}{4\pi} - \frac{\pi(6+(1.5)^2) + 3}{x}) }[/math]

[math]\displaystyle{ \geq \frac{1}{2(x-6)} (\log \frac{x}{4\pi} - \frac{28.92}{x}) }[/math]

and this is positive for [math]\displaystyle{ x \geq 100 }[/math].

[math]\displaystyle{ x }[/math]	[math]\displaystyle{ \|C^{eff}\|/\|B^{eff}_0\| }[/math]	[math]\displaystyle{ E_3^*/\|B^{eff}_0\| }[/math]	[math]\displaystyle{ E_1/\|B^{eff}_0\| }[/math]	[math]\displaystyle{ E_2/\|B^{eff}_0\| }[/math]
[math]\displaystyle{ 10^3 }[/math]	[math]\displaystyle{ 0.1008 }[/math]	[math]\displaystyle{ 0.2238 }[/math]	[math]\displaystyle{ 0.0014 }[/math]	[math]\displaystyle{ 0.0024 }[/math]
[math]\displaystyle{ 10^4 }[/math]	[math]\displaystyle{ 0.0172 }[/math]	[math]\displaystyle{ 0.0377 }[/math]	[math]\displaystyle{ 0.0001 }[/math]	[math]\displaystyle{ 0.0003 }[/math]
[math]\displaystyle{ 10^5 }[/math]	[math]\displaystyle{ 0.0031 }[/math]	[math]\displaystyle{ 0.0061 }[/math]	[math]\displaystyle{ 0.0000 }[/math]	[math]\displaystyle{ 0.0000 }[/math]
[math]\displaystyle{ 10^6 }[/math]	[math]\displaystyle{ 0.0006 }[/math]	[math]\displaystyle{ 0.0008 }[/math]	[math]\displaystyle{ 0.0000 }[/math]	[math]\displaystyle{ 0.0000 }[/math]
[math]\displaystyle{ 10^7 }[/math]	[math]\displaystyle{ 0.0001 }[/math]	[math]\displaystyle{ 0.0001 }[/math]	[math]\displaystyle{ 0.0000 }[/math]	[math]\displaystyle{ 0.0000 }[/math]

Graphical comparisons between [math]\displaystyle{ |C^{eff}|/|B^{eff}_0| }[/math] and [math]\displaystyle{ |E_3^*|/|B^{eff}_0| }[/math] can be found here, here, here, here, and here. Tables of upper bounds for [math]\displaystyle{ |E_3|/|B_0^{eff}| }[/math] can be found here.

Controlling H t-A-B/B 0: Difference between revisions

Latest revision as of 07:09, 30 March 2018

Estimation of [math]\displaystyle{ E_1,E_2 }[/math]

Estimation for large [math]\displaystyle{ x }[/math]

Estimation of [math]\displaystyle{ E_3 }[/math]

Navigation menu

Page actions

Page actions

Personal tools

Navigation

Search

Tools

@@ Line 1: / Line 1: @@
 As computed in [[Effective bounds on H_t - second approach]], there is an effective bound
-:<math>|H_{eff} - A^{eff} - B^{eff}| \leq E_1 + E_2 + E_3</math>
+:<math>|H^{eff} - A^{eff} - B^{eff}| \leq E_1 + E_2 + E_3</math>
 where
-:<math>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> 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> 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> 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>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> \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))
@@ Line 94: / Line 94: @@
 |}
-<math>A+B-C</math> is a good approximation to <math>H_t</math> [https://terrytao.wordpress.com/2018/02/12/polymath15-third-thread-computing-and-approximating-h_t/#comment-492695 source] [https://terrytao.wordpress.com/2018/02/24/polymath15-fourth-thread-closing-in-on-the-test-problem/#comment-493282 source] [https://terrytao.wordpress.com/2018/02/24/polymath15-fourth-thread-closing-in-on-the-test-problem/#comment-493319 source]
+<math>A^{eff}+B^{eff}</math> is a good approximation to <math>H_t</math>, <math>A+B-C</math> is better, and <math>A^{eff}+B^{eff}-C^{eff}</math> is excellent [https://terrytao.wordpress.com/2018/02/12/polymath15-third-thread-computing-and-approximating-h_t/#comment-492695 source] [https://terrytao.wordpress.com/2018/02/24/polymath15-fourth-thread-closing-in-on-the-test-problem/#comment-493282 source] [https://terrytao.wordpress.com/2018/02/24/polymath15-fourth-thread-closing-in-on-the-test-problem/#comment-493319 source] [https://terrytao.wordpress.com/2018/02/24/polymath15-fourth-thread-closing-in-on-the-test-problem/#comment-493313 source]
+[https://terrytao.wordpress.com/2018/02/24/polymath15-fourth-thread-closing-in-on-the-test-problem/#comment-493365 source]
 {| border=1
 |-
 ! style="text-align:left;"| <math>x</math>
+! <math>\frac{|H_t-(A+B)|}{|B_0|}</math>
+! <math>\frac{|H_t-(A^{eff}+B^{eff})|}{|B_0^{eff}|}</math>
 ! <math>\frac{|H_t-(A+B-C)|}{|B_0|}</math>
 ! <math>\frac{|H_t-(A^{eff}+B^{eff}-C^{eff})|}{|B_0^{eff}|}</math>
 |-
 |160
+|0.174873661533
+|0.1675083979955609185
 |0.06993270565802375041
-|0.009155667752
+|0.00887362155217
 |-
 |320
+| 0.278624615745
+|0.2776948344513698276
 |0.006716674125965016299
-|0.0005529962481
+|0.000708716878236
 |-
 |480
+|0.167598495339
+|0.1675667240356922231
 |0.005332893070605698501
-|0.0004966282128
+|0.000327585584191
 |-
 |640
+|0.165084846603
+|0.1635077306008453928
 |0.003363431256036816251
-|0.0004482768972
+|0.000523969818792
 |-
 |800
+|0.201954876756
+|0.2045038601879677257
 |0.1548144749150572349
 |0.002644344570
 |-
 |960
+|0.103387669714
+|0.1031837988358064657
 |0.03009229958121352990
-|0.001270168744
+|0.000819848578351
 |-
 |1120
+|0.0767779295558
+|0.07541968034203085865
 |0.004507664238680722472
-|0.0009957229500
+|0.000978838228690
 |-
 |1280
+|0.132886551163
+|0.1339118061014743863
 |0.002283591962997851167
-|0.0007024411378
+|0.000679785836479
 |-
 |1440
+| 0.0802159981813
+|0.07958929988050262854
 |0.01553727684468691873
-|0.0007000473085
+|0.000655447626435
 |-
 |1600
+|0.0777462698681
+|0.07700542235140914608
 |0.001778051951547709718
-|0.0004882487218
+|0.000439823567873
 |-
 |1760
+|0.0950946156489
+|0.09568042045936396570
 |0.02763769444052338578
-|0.0002518910919
+|0.000231103881282
 |-
 |1920
+|0.0629013452776
+|0.06385275621986742745
 |0.002108779890256530964
-|0.0008378989413
+|0.000849398936325
 |-
 |2080
+|0.0949328843573
+|0.09421231232885752514
 |0.02746770886040058927
-|0.0004924765754
+|0.000505410233739
 |-
 |2240
+| 0.0591497767926
+|0.05888587520703223358
 |0.001567020041379128455
-|0.0001171320991
+|0.000107206342271
 |-
 |2400
+|0.0785798163298
+|0.07899341548208345822
 |0.01801417530687959747
-|0.0002443802551
+|0.000229146425813
 |-
 |2560
+|0.0621868667021
+|0.06283843631123482445
 |0.001359561117436848149
-|0.0004569058755
+|0.000492123116208
 |-
 |2720
+|0.0585282736442
+|0.05966972584730198272
 |0.008503327577240081269
-|0.0006355966221
+|0.000656180976718
 |-
 |2880
+|0.0787554869341
+|0.07980560515423855917
 |0.001089253262122934826
-|0.0008864917365
+|0.000878298302262
 |-
 |3040
+|0.0462460274843
+|0.04636072344121703969
 |0.003004181560093288747
-|0.00004326840265
+|0.0000470113907733
 |-
 |3200
+|0.0963053589535
+|0.09664223832561922043
 |0.02931455383125538672
-|0.0003598521453
+|0.000354582706466
 |}
@@ Line 292: / Line 334: @@
 === Estimation of <math>E_1,E_2</math> ===
-...
+First let us obtain bounds for <math>|E_1/B^{eff}_0|, |E_2/B^{eff}_0|</math>, assuming for instance that <math>x \geq 100</math>, that only depend on <math>N</math> and not on <math>x</math>.  For fixed <math>N</math>, one has <math>x_N \leq x < x_{N+1}</math> where
+:<math>x_N := 4 \pi N^2 - \frac{\pi t}{4}.</math>
+In particular <math>x_N/2 \leq T \leq x_{N+1}/2</math>.
+We begin with <math>E_2</math>.  We have
+:<math>|E_2/B^{eff}_0| = \frac{1}{T-3.33}  \epsilon'(\frac{1+y+ix}{2}) (2.1)</math>
+where
+:<math> \epsilon'(s) = \frac{1}{2} \sum_{n=1}^N \frac{1}{n^{\mathrm{Re}(s) + \frac{t}{2} \mathrm{Re} \alpha_1(s) - \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>
+and
+:<math> \alpha_1(s) := \frac{1}{2s} + \frac{1}{s-1} + \frac{1}{2} \log \frac{s}{2\pi}.</math>
+We have
+:<math> \alpha'_1(s) = \frac{-1}{2s^2} - \frac{1}{(s-1)^2} + \frac{1}{2s}</math>
+and thus for <math>s</math> between <math>\frac{1+y+ix}{2}</math> and <math>s^+_N := \frac{1+y+i(x_N+x_{N+1})/2}{2}</math> one has
+:<math> \alpha'_1(s) = O_{\leq}( \frac{2}{x_N^2} + \frac{4}{x_N^2} + \frac{1}{x_N} ) = O_{\leq}( \frac{1}{x_N - 6} ).</math>
+(here we use Lemma 1.1 of [[Effective bounds on H_t - second approach]].) Thus we have
+:<math> \alpha_1(s) = \alpha_1(s^+_N) + O_{\leq}( \kappa )</math>
+where
+:<math> \kappa := \frac{x_{N+1}-x_N}{4 (x_N-6)} </math>
+(asymptotically this is <math>\sim 1/N</math>).  Thus
+:<math> |\alpha_1(s) - \log n|^2 = |\alpha_1(s^+_N) - \log n|^2 + O_{\leq}( 2 \kappa |\alpha_1(s^+_N) - \log n| + \kappa^2 )</math>
+and we conclude that
+:<math> \epsilon'(\frac{1+y+ix}{2}) \leq \frac{1}{2} \sum_{n=1}^N \frac{1}{n^{\frac{1+y}{2} + \frac{t}{2} \mathrm{Re} \alpha_1(s^+_N) - \frac{t}{2} \kappa - \frac{t}{4} \log n}} \exp( \frac{1}{2(T - 3.33)} c^+_n ) c^+_n </math>
+where
+:<math> c^+_n := \frac{t^2}{4} (|\alpha_1(s^+_N) - \log n|^2 + 2 \kappa |\alpha_1(s^+_N) - \log n| + \kappa^2) + \frac{1}{3} + t.</math>
+When combined with (2.1), this gives a uniform upper bound on <math>|E_2/B^{eff}_0|</math> for a fixed value of <math>N</math>.
+In a similar vein, we have
+:<math>|E_2/B^{eff}_0| = \frac{1}{T-3.33} \lambda \epsilon'(\frac{1-y+ix}{2}) (2.2)</math>
+where <math>\lambda</math> is the quantity defined in [http://michaelnielsen.org/polymath1/index.php?title=Controlling_A%2BB/B_0 this page].  In that page the upper bound
+:<math> \lambda \leq e^\delta N^{-y}</math>
+was established, where
+:<math> \delta := \frac{\pi y}{2(x_N-6 - \frac{14+2y}{\pi})} + \frac{2y(7+y)}{x_N^2} \log \frac{|1+y+i x_{N+1}|}{4\pi}.</math>
+Also, by repeating previous arguments (with <math>y</math> replaced by <math>-y</math>) we have
+:<math> \epsilon'(\frac{1-y+ix}{2}) \leq \frac{1}{2} \sum_{n=1}^N \frac{1}{n^{\frac{1-y}{2} + \frac{t}{2} \mathrm{Re} \alpha_1(s^-_N) - \frac{t}{2} \kappa - \frac{t}{4} \log n}} \exp( \frac{1}{2(T - 3.33)} c^-_n ) c^-_n </math>
+where
+:<math> c^-_n := \frac{t^2}{4} (|\alpha_1(s^-_N) - \log n|^2 + 2 \kappa |\alpha_1(s^-_N) - \log n| + \kappa^2) + \frac{1}{3} + t</math>
+and
+:<math> s^-_N := \frac{1-y+i(x_N + x_{N+1})/2}{2}.</math>
+Tables of upper bounds for <math>|E_1|/|B_0^{eff}|</math>, <math>|E_2|/|B_0^{eff}|</math> can be found [https://github.com/km-git-acc/dbn_upper_bound/tree/master/dbn_upper_bound/python/research here].
+==== Estimation for large <math>x</math> ====
+We can crudely bound <math>\epsilon'(\frac{1+y+ix}{2}), \epsilon'(\frac{1-y+ix}{2})</math> as follows. In [[Controlling A+B/B_0]] it is shown that
+:<math>\mathrm{Re} \alpha_1(\frac{1+y+ix}{2}) \geq \log N</math>
+for <math>y \geq 1/3</math>, and so
+:<math>\epsilon'(\frac{1+y+ix}{2}) \leq \exp( \frac{1}{2(T-3.33)} (\frac{t^2}{4} |\alpha_1(\frac{1+y+ix}{2})|^2 + \frac{1}{3} + t)) (\frac{t^2}{4} |\alpha_1(\frac{1+y+ix}{2})|^2 + \frac{1}{3} + t) \sum_{n=1}^N \frac{1}{n^{\frac{1+y}{2} + \frac{t}{4} \log \frac{N^2}{n}}}</math>
+and thus in the language of [[Estimating a sum]]
+:<math>\epsilon'(\frac{1+y+ix}{2}) \leq \exp( \frac{1}{2(T-3.33)} (\frac{t^2}{4} |\alpha_1(\frac{1+y+ix}{2})|^2 + \frac{1}{3} + t)) (\frac{t^2}{4} |\alpha_1(\frac{1+y+ix}{2})|^2 + \frac{1}{3} + t) S_{\frac{1+y}{2}, t}(N).</math>
+Thus for instance if <math>y=t=0.4</math> and <math>N \geq 2000</math> then
+:<math>\epsilon'(\frac{1+y+ix}{2}) \leq \exp( \frac{1}{2(T-3.33)} (\frac{1}{25} |\alpha_1(\frac{0.7+ix}{2})|^2 + \frac{11}{15})) (\frac{1}{25} |\alpha_1(\frac{0.7+ix}{2})|^2 + \frac{11}{15}) \times 1.706.</math>
+Similarly we have
+:<math>\epsilon'(\frac{1-y+ix}{2}) \leq \exp( \frac{1}{2(T-3.33)} (\frac{1}{25} |\alpha_1(\frac{0.3+ix}{2})|^2 + \frac{11}{15})) (\frac{1}{25} |\alpha_1(\frac{0.3+ix}{2})|^2 + \frac{11}{15}) \times 3.469.</math>
+Thus
+:<math>E_2 / |B^{eff}_0| \leq \frac{1.706}{T-3.33} \exp( \frac{1}{2(T-3.33)} (\frac{1}{25} |\alpha_1(\frac{0.7+ix}{2})|^2 + \frac{11}{15})) (\frac{1}{25} |\alpha_1(\frac{0.7+ix}{2})|^2 + \frac{11}{15})</math>
+and
+:<math>E_1 / |B^{eff}_0| \leq \frac{3.469}{T-3.33} e^{\delta} N^{-0.4} \exp( \frac{1}{2(T-3.33)} (\frac{1}{25} |\alpha_1(\frac{0.3+ix}{2})|^2 + \frac{11}{15})) (\frac{1}{25} |\alpha_1(\frac{0.3+ix}{2})|^2 + \frac{11}{15}).</math>
+If <math>N \geq 2000</math>, then <math>T \geq 2\pi N^2 - \frac{\pi t}{8} \geq 2 \times 10^7</math> and <math>x \geq 4 \times 10^7</math>.  One can bound
+:<math>|\alpha_1(\frac{0.7+ix}{2})| \leq \frac{3}{x} + \frac{1}{2} \log \frac{\sqrt{1+x^2}}{4\pi} + \frac{\pi}{4}</math>
+and hence
+:<math> \frac{1}{T-3.33} (\frac{1}{25} |\alpha_1(\frac{0.7+ix}{2})|^2 + \frac{11}{15})
+\leq \frac{1}{T-3.33} (\frac{1}{25} (\frac{3}{x} + \frac{1}{2} \log \frac{\sqrt{1+x^2}}{4\pi} + \frac{\pi}{4})^2 + \frac{11}{15}).
+</math>
+this is decreasing in <math>x</math> and bounded by <math>1.7 \times 10^{-7}</math>.  From this we conclude that <math>E_2 / |B^{eff}_0| \leq 2.9 \times 10^{-7}</math>.  One can also calculate that <math>\delta \leq 1.6 \times 10^{-7}</math> and that <math>E_1 / |B^{eff}_0| \leq 2.8 \times 10^{-8}</math>.
 === Estimation of <math>E_3</math> ===
@@ Line 322: / Line 429: @@
 :<math> \leq \exp(  0.37 \frac{(\sigma-1/2)^2}{T'_0-2.71} + \frac{1}{12(T'_0 - 3.33)}).</math>
-Now we work on <math>\nu</math>.  Observe that if <math>k \leq \frac{T'_0}{2.42 \pi} = \frac{a_0^2}{1.21}</math> then
+Now we work on <math>v</math>.  Observe that if <math>k \leq \frac{T'_0}{2.42 \pi} = \frac{a_0^2}{1.21}</math> then
 :<math> (1.1)^{k+2} \frac{\Gamma(\frac{k+2}{2})}{a_0^{k+2}} = \frac{1.21 k}{2 a_0^2} \frac{\Gamma(\frac{k}{2})}{a_0^k} \leq \frac{1}{2} (1.1)^{k} \frac{\Gamma(\frac{k}{2})}{a_0^{k}},</math>
 and hence
@@ Line 405: / Line 512: @@
 :<math>
 (1 + \frac{1}{a_0-1.25} (\sqrt{\pi} + 1.2 \times 3^y \exp( \frac{t \log^2 9}{4 (1 - \frac{0.37 t}{T'_0-2.71})} ) ).</math>
-We bound <math>(6 \times 0.37t + 1 + 3 \times 0.37 y^2)/12 \leq 0.181</math> and <math>1.2 \times 3^y \exp( \frac{t \log^2 9}{4(1 - \frac{0.37 t}{T'_0-2.71}) \leq 5.15</math> for <math>y \leq 1/2</math>, thus
+We bound <math>(6 \times 0.37t + 1 + 3 \times 0.37 y^2)/12 \leq 0.181</math> and <math>1.2 \times 3^y \exp( \frac{t \log^2 9}{4(1 - \frac{0.37 t}{T'_0-2.71}}) \leq 5.15</math> for <math>y \leq 1/2</math>, thus
 :<math>\int_{-\infty}^\infty vwf(\sigma)\ d\sigma \leq \exp( \frac{0.181}{T'_0 - 3.33}) (1 + \frac{5.15}{a_0-1.25}).</math>
 We conclude that
-:<math>E_3 \leq \frac{1}{8} \sqrt{\pi} \exp( - \frac{t\pi^2}{64}) (T'_0)^{3/2} e^{-\pi T_0/4} \exp( \frac{0.181}{T'_0 - 3.33}) (1 + \frac{5.15}{a_0-1.25}).</math>
+:<math>E_3 \leq E_3^*</math>
+where
+:<math>E_3^* := \frac{1}{8} \sqrt{\pi} \exp( - \frac{t\pi^2}{64}) (T'_0)^{3/2} e^{-\pi T_0/4} \exp( \frac{0.181}{T'_0 - 3.33}) (1 + \frac{5.15}{a_0-1.25}).</math>
+The main term here is
+:<math>E_3^{main} := (T'_0)^{3/2} e^{-\pi T_0/4};</math>
+in particular, we can factor the ratio <math>E_3^* / |B^{eff}_0|</math> as the product of
+:<math> E_3^* / E_3^{main} = \frac{1}{8} \sqrt{\pi} \exp( - \frac{t\pi^2}{64}) \exp( \frac{0.181}{T'_0 - 3.33}) (1 + \frac{5.15}{a_0-1.25})</math>
+and
+:<math> E_3^{main} / |B^{eff}_0| = \frac{16}{\sqrt{2\pi} |s(s-1)|} \pi^{\mathrm{Re} (1-s)/2} \exp( - \frac{\pi T_0}{4} + \frac{3}{2} \log T'_0 - \frac{t}{4} \mathrm{Re}(\alpha_1(1-s)^2) - \mathrm{Re}( (\frac{1-s}{2} - \frac{1}{2}) \log \frac{1-s}{2} + \frac{1-s}{2}) ) )</math>
+where <math>s := \frac{1-y+ix}{2}</math>.
+The first ratio <math>E_3^* / E_3^{main} </math> decreases monotonically to <math>\frac{1}{8} \sqrt{\pi} \exp( - \frac{t\pi^2}{64})</math>, which equals <math>0.2038\dots</math> when <math>t=0.4</math>.  We claim the second ratio decreases for <math>x \geq 100</math>.  To see this, we compute the log-derivative <math> \frac{d}{dx} \log E_3^{main} / |B^{eff}_0|</math> as
+:<math>\mathrm{Re}( -\frac{i}{2s} - \frac{i}{2(s-1)} - \frac{\pi}{8} + \frac{3}{4 T'_0} + \frac{i t}{4} \alpha_1(1-s) \alpha'_1(1-s) + \frac{i}{4} \log \frac{1-s}{2} - \frac{i}{4 (1-s)} )</math>
+:<math>= \frac{-x}{(1-y)^2 + x^2} + \frac{-x}{(1+y)^2+x^2} - \frac{\pi}{8} + \frac{3}{2 (x + \pi t/4)} - \frac{t}{4} \mathrm{Im}(\alpha_1(1-s) \alpha'_1(1-s)) + \frac{1}{4} (\frac{\pi}{2} - \mathrm{arctan} \frac{1+y}{x}) + \frac{x/2}{(1+y)^2+x^2}</math>
+:<math>= \frac{-x}{(1-y)^2 + x^2} + \frac{-x/2}{(1+y)^2+x^2} + \frac{3}{2 (x + \pi t/4)} - \frac{t}{4} \mathrm{Im}(\alpha_1(1-s) \alpha'_1(1-s)) - \frac{1}{4} \mathrm{arctan} \frac{1+y}{x}.</math>
+For <math>x \geq 100</math>, we have <math>\mathrm{arctan} \frac{1+y}{x} \geq \frac{1+y}{2x}</math> (say), and we bound <math>\frac{3}{2(x+\pi t/4)}</math> by <math>\frac{1}{x} + \frac{1}{2x}</math> to obtain an upper bound of
+:<math>\leq (\frac{1}{x} - \frac{x}{(1-y)^2 + x^2}) + (\frac{1}{2x} - \frac{x/2}{(1+y)^2+x^2}) - \frac{t}{4} \mathrm{Im}(\alpha_1(1-s) \alpha'_1(1-s))-
+ \frac{1+y}{8x}</math>
+:<math>= \frac{(1-y)^2}{x ((1-y)^2 + x^2)} + \frac{(1+y)^2}{2x ((1+y)^2 + x^2)} - \frac{t}{4} \mathrm{Im}(\alpha_1(1-s) \alpha'_1(1-s))</math>
+:<math> \leq \frac{(1-y)^2 + (1+y)^2/2}{x^3} -  \frac{t}{4} \mathrm{Im}(\alpha_1(1-s) \alpha'_1(1-s)) - \frac{1+y}{8x}.</math>
+For <math>x \geq 100</math>, we have <math>\frac{1+y}{8x} > \frac{(1-y)^2 + (1+y)^2/2}{x^3} </math>, so to establish decrease it suffices to show that
+:<math>\mathrm{Im}(\alpha_1(1-s) \alpha'_1(1-s)) > 0.</math>
+We have
+:<math>\alpha_1(1-s) := \frac{1}{2(1-s)} + \frac{1}{-s} + \frac{1}{2} \log \frac{1-s}{2\pi} </math>
+:<math> = O_{\leq}( \frac{1}{x} + \frac{2}{x} ) + \frac{1}{2} \log \frac{|1+y+ix|}{4\pi} + i O_{\leq}( \frac{\pi}{2} )</math>
+and
+:<math>\alpha'_1(1-s) := -\frac{1}{2(1-s)^2} - \frac{1}{(-s)^2} + \frac{1}{2(1-s)}</math>
+:<math>= O_{\leq}( \frac{2}{x^2} + \frac{4}{x^2} ) + \frac{1+y+ix}{(1+y)^2+x^2}</math>
+and hence
+:<math>\mathrm{Im}(\alpha_1(1-s) \alpha'_1(1-s)) = \frac{1}{2} \log \frac{|1+y+ix|}{4\pi} (\frac{x}{(1+y)^2+x^2} + O_{\leq}(\frac{6}{x^2}) )
++ O_{\leq}( \frac{\pi}{2} (\frac{(1+y)^2}{(1+y)^2+x^2} + \frac{6}{x^2})) + O_{\leq}( \frac{3}{x} (\frac{6}{x^2} + \frac{1}{x}) )</math>
+:<math> \geq \frac{1}{2(x-6)} \log \frac{x}{4\pi} - \frac{\pi}{2} \frac{6+(1+y)^2}{x^2} - \frac{3}{x(x-6)} </math>
+:<math> \geq \frac{1}{2(x-6)} (\log \frac{x}{4\pi} - \frac{\pi(6+(1.5)^2) + 3}{x})</math>
+:<math> \geq \frac{1}{2(x-6)} (\log \frac{x}{4\pi} - \frac{28.92}{x}) </math>
+and this is positive for <math>x \geq 100</math>.
+{| border=1
+|-
+! style="text-align:left;"| <math>x</math>
+! <math>|C^{eff}|/|B^{eff}_0|</math>
+! <math>E_3^*/|B^{eff}_0|</math>
+! <math>E_1/|B^{eff}_0|</math>
+! <math>E_2/|B^{eff}_0|</math>
+|-
+| <math>10^3</math>
+| <math>0.1008</math>
+| <math>0.2238</math>
+| <math>0.0014</math>
+| <math>0.0024</math>
+|-
+| <math>10^4</math>
+| <math>0.0172</math>
+| <math>0.0377</math>
+| <math>0.0001</math>
+| <math>0.0003</math>
+|-
+| <math>10^5</math>
+| <math>0.0031</math>
+| <math>0.0061</math>
+| <math>0.0000</math>
+| <math>0.0000</math>
+|-
+| <math>10^6</math>
+| <math>0.0006</math>
+| <math>0.0008</math>
+| <math>0.0000</math>
+| <math>0.0000</math>
+|-
+| <math>10^7</math>
+| <math>0.0001</math>
+| <math>0.0001</math>
+| <math>0.0000</math>
+| <math>0.0000</math>
+|}
+Graphical comparisons between <math>|C^{eff}|/|B^{eff}_0|</math> and <math>|E_3^*|/|B^{eff}_0|</math> can be found [https://ibb.co/d7zDv7 here], [https://ibb.co/efY6F7 here], [https://ibb.co/eYdZ2n here], [https://ibb.co/fjSu2n here], and [https://ibb.co/gDsf8S here].  Tables of upper bounds for <math>|E_3|/|B_0^{eff}|</math> can be found [https://github.com/km-git-acc/dbn_upper_bound/tree/master/dbn_upper_bound/python/research here].
+[[Category:Polymath15]]