Bounding the derivative of H t: Difference between revisions
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We can bound | We can bound | ||
:<math> \frac{1}{4 (T - 3.08)} ( |\sqrt{t} u + \frac{t}{2} \alpha_n(s)|^2 + \frac{2}{3} ) \leq \frac{1}{2(T-3.08)} ( tu^2 + \frac{t^2}{4} |\alpha_n(s)|^2 + \frac{1}{3} )</math> | :<math> \frac{1}{4 (T - 3.08)} ( |\sqrt{t} u + \frac{t}{2} \alpha_n(s)|^2 + \frac{2}{3} ) \leq \frac{1}{2(T-3.08)} ( tu^2 + \frac{t^2}{4} |\alpha_n(s)|^2 + \frac{1}{3} )</math> | ||
and also write | |||
:<math>\exp( \frac{t}{4} \mathrm{Re}(\alpha_n(s)^2) ) |H_{0,n}(s)| = \exp( \frac{t}{4} \mathrm{Re}(\alpha_1(s)^2) ) |H_{0,1}(s)| \frac{b_n}{n^{\mathrm{Re} s_A}}</math> | |||
:<math>= 8 |\lambda| |B^{eff}_0| \frac{b_n^2}{n^{\sigma + \frac{t}{2} \mathrm{Re} \alpha_1(s)}}</math> | |||
to obtain | to obtain | ||
:<math> |F'_{t,n}(s)| \leq \ | :<math> \frac{1}{8} |F'_{t,n}(s)| \leq |\lambda| |B^{eff}_0| \frac{b_n}{n^{\mathrm{Re} s_A}} \frac{\exp( \frac{1}{2(T-3.08)} (\frac{t^2}{4} |\alpha_n(s)|^2 + \frac{1}{3}) ) }{\sqrt{\pi}} \int_{-\infty}^\infty (|\alpha_n(s)| + |\frac{2u}{\sqrt{t}}|) \exp( -(1-\frac{t}{2(T-3.08)}) u^2 )\ du.</math> | ||
Noting that | Noting that | ||
:<math>\int_{-\infty}^\infty e^{-a u^2}\ du = a^{-1/2} \sqrt{\pi}</math> | :<math>\int_{-\infty}^\infty e^{-a u^2}\ du = a^{-1/2} \sqrt{\pi}</math> | ||
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for any <math>a>0</math>, | for any <math>a>0</math>, | ||
we thus have | we thus have | ||
:<math> |F'_{t,n}(s)| \leq \ | :<math> \frac{1}{8} |F'_{t,n}(s)| \leq |\lambda| |B^{eff}_0| \frac{b_n}{n^{\mathrm{Re} s_A}} \exp( \frac{1}{2(T-3.08)} (\frac{t^2}{4} |\alpha_n(s)|^2 + \frac{1}{3}) ) ( |\alpha_n(s)| (1-\frac{t}{2(T-3.08)})^{-1/2} + \frac{2}{\sqrt{\pi t}} (1-\frac{t}{2(T-3.08)})^{-1}).</math> | ||
We can bound <math>|\alpha_n(s)| \leq |\alpha_1(s)|</math> and <math>T \geq T_N</math> and conclude that | We can bound <math>|\alpha_n(s)| \leq |\alpha_1(s)|</math> and <math>T \geq T_N</math> and conclude that | ||
:<math>\sum_{n=1}^N |F'_{t,n}(s)| \leq | | :<math>\frac{1}{8} \sum_{n=1}^N |F'_{t,n}(s)| \leq |\lambda| |B^{eff}_0|\exp( \frac{1}{2(T_N-3.08)} (\frac{t^2}{4} |\alpha_1(s)|^2 + \frac{1}{3}) ) ( |\alpha_1(s)| (1-\frac{t}{2(T_N-3.08)})^{-1/2} + \frac{2}{\sqrt{\pi t}} (1-\frac{t}{2(T_N-3.08)})^{-1}) F_{{\mathrm{Re} s_A - \frac{t}{2} \log N, t}(N).</math> | ||
Similarly | |||
:<math>\frac{1}{8} \sum_{n=1}^N |F'_{t,n}(1-s)| \leq |B^{eff}_0|\exp( \frac{1}{2(T_N-3.08)} (\frac{t^2}{4} |\alpha_1(1-s)|^2 + \frac{1}{3}) ) ( |\alpha_1(1-s)| (1-\frac{t}{2(T_N-3.08)})^{-1/2} + \frac{2}{\sqrt{\pi t}} (1-\frac{t}{2(T_N-3.08)})^{-1}) F_{{\mathrm{Re} s_B - \frac{t}{2} \log N, t}(N).</math> |
Revision as of 22:55, 8 March 2018
We continue using the notation from Effective bounds on H_t - second approach. We also assume [math]\displaystyle{ T \geq 10 }[/math] (so [math]\displaystyle{ x \geq 20 }[/math]).
Since
- [math]\displaystyle{ H_t(z) = \frac{1}{8} \xi_t( s) }[/math]
with [math]\displaystyle{ s := \frac{1+iz}{2} }[/math], we have
- [math]\displaystyle{ \frac{d}{dz} H_t(z) = \frac{i}{16} \frac{d}{ds} \xi_t(s). }[/math]
Next, we have
- [math]\displaystyle{ \xi_t(s) = \sum_{n=1}^N F_{t,n}(s) + F_{t,n}(1-s) + G_{t,N}(s) + G_{t,N}(1-s) }[/math]
(using the convention [math]\displaystyle{ F(\bar{s}) = \bar{F(s)} }[/math] for [math]\displaystyle{ s }[/math] in the lower half-plane). Thus (assuming that we are not at a discontinuity for $latex N$) we have
- [math]\displaystyle{ \frac{d}{ds} \xi_t(s) = \sum_{n=1}^N F'_{t,n}(s) - F'_{t,n}(1-s) + G'_{t,N}(s) - G'_{t,N}(1-s). }[/math]
Now we have for any [math]\displaystyle{ \alpha_n }[/math] that
- [math]\displaystyle{ F_{t,n}(s) = \exp( - \frac{t}{4} \alpha_n^2 ) \int_{-\infty}^\infty \exp( - \sqrt{t} \alpha_n u) F_{0,n}( s + \sqrt{t} u + \frac{t}{2} \alpha_n ) \frac{1}{\sqrt{\pi}} e^{-u^2}\ du, }[/math]
hence in differentiation under the integral sign (justifiable for instance using the Cauchy integral formula and Fubini's theorem)
- [math]\displaystyle{ F'_{t,n}(s) = \exp( - \frac{t}{4} \alpha_n^2 ) \int_{-\infty}^\infty \exp( - \sqrt{t} \alpha_n u) \frac{\partial}{\partial s} F_{0,n}( s + \sqrt{t} u + \frac{t}{2} \alpha_n ) \frac{1}{\sqrt{\pi}} e^{-u^2}\ du. }[/math]
This identity is true for any [math]\displaystyle{ \alpha_n }[/math]; we now set [math]\displaystyle{ \alpha_n = \alpha_n(s) }[/math] as in the above wiki page. One can replace [math]\displaystyle{ \frac{\partial}{\partial s} }[/math] on the RHS by [math]\displaystyle{ \frac{1}{\sqrt{t}} \frac{\partial}{\partial u} }[/math] and integrate by parts to conclude that
- [math]\displaystyle{ F'_{t,n}(s) = \exp( - \frac{t}{4} \alpha_n(s)^2 ) \int_{-\infty}^\infty \exp( - \sqrt{t} \alpha_n(s) u) F_{0,n}( s + \sqrt{t} u + \frac{t}{2} \alpha_n ) \frac{1}{\sqrt{\pi}} (\alpha_n(s) + \frac{2u}{\sqrt{t}}) e^{-u^2}\ du. }[/math]
We have
- [math]\displaystyle{ F_{0,n}( s + \sqrt{t} u + \frac{t}{2} \alpha_n(s)) = H_{0,n}(s) \exp( (\sqrt{t} u + \frac{t}{2} \alpha_n(s)) \alpha_n(s) + O_{\leq}( \frac{1}{4 (T - 3.08)} ( |\sqrt{t} u + \frac{t}{2} \alpha_n(s)|^2 + \frac{2}{3} ) ) ) }[/math]
and hence
- [math]\displaystyle{ |F'_{t,n}(s)| \leq \exp( \frac{t}{4} \mathrm{Re}(\alpha_n(s)^2) ) |H_{0,n}(s)| \int_{-\infty}^\infty \exp( \frac{1}{4 (T - 3.08)} ( |\sqrt{t} u + \frac{t}{2} \alpha_n(s)|^2 + \frac{2}{3} ) ) ) \frac{1}{\sqrt{\pi}} (|\alpha_n(s)| + |\frac{2u}{\sqrt{t}}|) e^{-u^2}\ du. }[/math]
We can bound
- [math]\displaystyle{ \frac{1}{4 (T - 3.08)} ( |\sqrt{t} u + \frac{t}{2} \alpha_n(s)|^2 + \frac{2}{3} ) \leq \frac{1}{2(T-3.08)} ( tu^2 + \frac{t^2}{4} |\alpha_n(s)|^2 + \frac{1}{3} ) }[/math]
and also write
- [math]\displaystyle{ \exp( \frac{t}{4} \mathrm{Re}(\alpha_n(s)^2) ) |H_{0,n}(s)| = \exp( \frac{t}{4} \mathrm{Re}(\alpha_1(s)^2) ) |H_{0,1}(s)| \frac{b_n}{n^{\mathrm{Re} s_A}} }[/math]
- [math]\displaystyle{ = 8 |\lambda| |B^{eff}_0| \frac{b_n^2}{n^{\sigma + \frac{t}{2} \mathrm{Re} \alpha_1(s)}} }[/math]
to obtain
- [math]\displaystyle{ \frac{1}{8} |F'_{t,n}(s)| \leq |\lambda| |B^{eff}_0| \frac{b_n}{n^{\mathrm{Re} s_A}} \frac{\exp( \frac{1}{2(T-3.08)} (\frac{t^2}{4} |\alpha_n(s)|^2 + \frac{1}{3}) ) }{\sqrt{\pi}} \int_{-\infty}^\infty (|\alpha_n(s)| + |\frac{2u}{\sqrt{t}}|) \exp( -(1-\frac{t}{2(T-3.08)}) u^2 )\ du. }[/math]
Noting that
- [math]\displaystyle{ \int_{-\infty}^\infty e^{-a u^2}\ du = a^{-1/2} \sqrt{\pi} }[/math]
and
- [math]\displaystyle{ \int_{-\infty}^\infty e^{-a u^2} |u|\ du = a^{-1} }[/math]
for any [math]\displaystyle{ a\gt 0 }[/math], we thus have
- [math]\displaystyle{ \frac{1}{8} |F'_{t,n}(s)| \leq |\lambda| |B^{eff}_0| \frac{b_n}{n^{\mathrm{Re} s_A}} \exp( \frac{1}{2(T-3.08)} (\frac{t^2}{4} |\alpha_n(s)|^2 + \frac{1}{3}) ) ( |\alpha_n(s)| (1-\frac{t}{2(T-3.08)})^{-1/2} + \frac{2}{\sqrt{\pi t}} (1-\frac{t}{2(T-3.08)})^{-1}). }[/math]
We can bound [math]\displaystyle{ |\alpha_n(s)| \leq |\alpha_1(s)| }[/math] and [math]\displaystyle{ T \geq T_N }[/math] and conclude that
- [math]\displaystyle{ \frac{1}{8} \sum_{n=1}^N |F'_{t,n}(s)| \leq |\lambda| |B^{eff}_0|\exp( \frac{1}{2(T_N-3.08)} (\frac{t^2}{4} |\alpha_1(s)|^2 + \frac{1}{3}) ) ( |\alpha_1(s)| (1-\frac{t}{2(T_N-3.08)})^{-1/2} + \frac{2}{\sqrt{\pi t}} (1-\frac{t}{2(T_N-3.08)})^{-1}) F_{{\mathrm{Re} s_A - \frac{t}{2} \log N, t}(N). }[/math]
Similarly
- [math]\displaystyle{ \frac{1}{8} \sum_{n=1}^N |F'_{t,n}(1-s)| \leq |B^{eff}_0|\exp( \frac{1}{2(T_N-3.08)} (\frac{t^2}{4} |\alpha_1(1-s)|^2 + \frac{1}{3}) ) ( |\alpha_1(1-s)| (1-\frac{t}{2(T_N-3.08)})^{-1/2} + \frac{2}{\sqrt{\pi t}} (1-\frac{t}{2(T_N-3.08)})^{-1}) F_{{\mathrm{Re} s_B - \frac{t}{2} \log N, t}(N). }[/math]