Controlling H t-A-B/B 0
As computed in Effective bounds on H_t - second approach, there is an effective bound
- [math]\displaystyle{ |H^{eff} - A^{eff} - B^{eff}| \leq E_1 + E_2 + E_3 }[/math]
where
- [math]\displaystyle{ 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}. }[/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.
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) F_{\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_1 / |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_2 / |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]
...
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^{\math{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 \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.
