Controlling H t-A-B/B 0 - Revision history

Teorth at 15:09, 30 March 2018

2018-03-30T15:09:38Z

← Older revision		Revision as of 08:09, 30 March 2018
Line 600:		Line 600:

	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].		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]]

Teorth at 07:14, 9 March 2018

2018-03-09T07:14:10Z

← Older revision		Revision as of 00:14, 9 March 2018
Line 383:		Line 383:
	:<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>		:<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]]		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) F_{\frac{1+y}{2}, t}(N).</math>		:<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		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>		:<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>

Teorth: /* Estimation of E_3 */

2018-03-08T04:48:55Z

Estimation of E_3

← Older revision		Revision as of 21:48, 7 March 2018
Line 533:		Line 533:
	:<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}{(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>		:<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 \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		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))-		:<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>		\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>= \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>		:<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		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

Teorth: /* Estimation of E_1,E_2 */

2018-03-08T04:44:57Z

Estimation of E_1,E_2

← Older revision		Revision as of 21:44, 7 March 2018
Line 366:		Line 366:
	:<math> \lambda \leq e^\delta N^{-y}</math>		:<math> \lambda \leq e^\delta N^{-y}</math>
	was established, where		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>		:<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		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>		:<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>
Line 375:		Line 375:

	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].		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		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
Line 387:		Line 389:
	:<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>		:<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		Thus
	:<math>~~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>		:<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		and
	:<math>~~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>		:<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> ===		=== Estimation of <math>E_3</math> ===

Teorth: /* Estimation of E_1,E_2 */

2018-03-07T16:28:53Z

Estimation of E_1,E_2

← Older revision		Revision as of 09:28, 7 March 2018
Line 372:		Line 372:
	:<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>		:<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		and
	:<math> s^-_N := \frac{1-y+i(x_N + x_{N+1})}{2}.</math>		:<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].		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].

Teorth: /* Estimation of E_3 */

2018-03-06T18:05:40Z

Estimation of E_3

← Older revision		Revision as of 11:05, 6 March 2018
Line 518:		Line 518:
	:<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>		:<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		and
	:<math> 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>		:<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>.		where <math>s := \frac{1-y+ix}{2}</math>.

Teorth: /* Estimation of E_1,E_2 */

2018-03-06T18:04:42Z

Estimation of E_1,E_2

← Older revision		Revision as of 11:04, 6 March 2018
Line 385:		Line 385:
	:<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>		:<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		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>		:<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		Thus
	:<math>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>		:<math>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>

Teorth: /* Estimation of E_1,E_2 */

2018-03-05T00:48:04Z

Estimation of E_1,E_2

← Older revision		Revision as of 17:48, 4 March 2018
Line 379:		Line 379:
	:<math>\mathrm{Re} \alpha_1(\frac{1+y+ix}{2}) \geq \log N</math>		:<math>\mathrm{Re} \alpha_1(\frac{1+y+ix}{2}) \geq \log N</math>
	for <math>y \geq 1/3</math>, and so		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>		:<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>
	~~The summand is decreasing for~~ <math>1 \~~leq n \leq N</math>~~ (~~one can write it for instance as <math>~~\frac{\exp( \frac{t}{4}(~~\log N~~ - ~~\log n~~)^2}~~{\exp~~(\frac{t}{4} \~~log^2 N) n^{~~\frac{1+y}{2}~~}}</math>~~) ~~and hence by the integral test one may bound~~		and thus in the language of [[Estimating a sum]]
	~~:<math> \sum_{n=1}~~^N \frac{1}{~~n^{\frac{1+y}{2~~} + \frac{t}{4} \~~log~~ \frac{~~N^2~~}{n}~~}} \leq 1~~ + ~~\int_1^N~~ \frac{1}{a^{\frac{1+y}{2} ~~+ \frac{~~t}~~{4} \log \frac{~~N~~^2}{a}}\ da~~.</math>		:<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) F_{\frac{1+y}{2}, t}(N).</math>
	~~Making the change of variables~~ <math>a = N ~~e^{-u}~~</math>~~, the right-hand side becomes~~		Thus for instance if <math>y=t=0.4</math> and <math>N \geq 2000</math> then
	:<math>~~1 + e^{-~~\~~frac{t}{4} \log^2 N} N^{~~\frac{1-y}{2}} \~~int_0^{\log N}~~ \exp( \frac{t}{~~4} u^~~2 - \frac{1-y}{2} u)\ ~~du.</math>~~		:<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>
	~~Completing the square and making the change of variables <math>v := \sqrt{t/4}~~ (~~u -~~ \frac{~~1-y~~}{t})~~</math>, this is~~		Similarly we have
	~~:<math>1~~ + ~~e^{-~~\frac{t}{4} ~~\log^2 N} N^{~~\frac{1-y}{2}} \~~exp~~( -\frac{~~(1-y)^2~~}{4t} )		:<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>
	\~~int_{-(1-y)/\sqrt~~{~~4t}~~}^{V} ~~\exp( v^2~~)\ dv</math>		Thus
	~~where~~		:<math>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>
	:<math>~~V :=~~ \~~sqrt{t/4} \log N -~~ (~~1-y)/~~\~~sqrt~~{~~4t}</math>.~~		and
	~~Applying Lemma~~ 1~~.6 of [[Effective bounds on H_t~~ - ~~second approach]], we may bound this by~~		:<math>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>
	~~:<math>1~~ + e^{-\frac{t}{~~4} \log^~~2 N} ~~N^{~~\frac{1~~-y}~~{2}} \~~exp~~( -\frac{~~(1-y)^2~~}{4t} ) ~~(\theta + e^{V~~^2~~} (~~\frac{1}{2V} + \frac{1}{~~V^3~~}~~) )</math>~~
	~~where~~
	~~:<math>~~ \~~theta :=~~ \~~int_~~{~~-(1-y~~)/\~~sqrt~~{4t}~~} e^~~{~~v^2~~}\ dv.</math>
	~~Since~~
	:<math>e^{~~V^2~~} ~~= e^{~~\frac{t}{4} \~~log^2 N} N^{-~~\frac{1-y}{2}~~} \exp~~( \frac{(1~~-y)^2~~}{4t}~~)</math>~~
	~~we may rewrite this as~~
	~~:<math>1 +~~ \frac{1}{2V} + \frac{1}{~~V^3~~} ~~+ e^{-~~\frac{t}{4} \~~log^2 N} N^{~~\frac{~~1-y~~}{2}~~} \exp( -~~\frac{~~(1-y)^2~~}{4t} ) ~~\theta.~~</math>
	~~Thus we have~~
	:<math>E_2/\|B^{eff}_0}\| = \~~epsilon'(~~\frac{~~1+y+ix~~}{2}~~) \leq e_2~~^*</math>
	~~where~~
	~~:<math> e_2^* :=~~ \~~frac{1~~}{T-3.33} \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) ( 1 + \frac{1}{2V} + \frac{1}{V^3} + e^{-\frac{t~~}{4} ~~\log^2 N} N^{\frac{1-y}{2}} \exp( -\frac{(1-y)^2}{4t} ) \theta~~ ).</math>


	...		...

Teorth: /* Estimation of E_1,E_2 */

2018-03-04T19:23:42Z

Estimation of E_1,E_2

← Older revision		Revision as of 12:23, 4 March 2018
Line 380:		Line 380:
	for <math>y \geq 1/3</math>, and so		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>		:<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>
	The summand is decreasing for <math>1 \leq n \leq N</math> (one can write it for instance as <math>\frac{\exp( \frac{t}{4}(\log N - \log n)^2}{\exp(\frac{t}{4} \log^2 N) n^{\frac{1+y}{2}}}) and hence by the integral test one may bound		The summand is decreasing for <math>1 \leq n \leq N</math> (one can write it for instance as <math>\frac{\exp( \frac{t}{4}(\log N - \log n)^2}{\exp(\frac{t}{4} \log^2 N) n^{\frac{1+y}{2}}}</math>) and hence by the integral test one may bound
	:<math> \sum_{n=1}^N \frac{1}{n^{\frac{1+y}{2} + \frac{t}{4} \log \frac{N^2}{n}}} \leq 1 + \int_1^N \frac{1}{a^{\frac{1+y}{2} + \frac{t}{4} \log \frac{N^2}{a}}\ da.</math>		:<math> \sum_{n=1}^N \frac{1}{n^{\frac{1+y}{2} + \frac{t}{4} \log \frac{N^2}{n}}} \leq 1 + \int_1^N \frac{1}{a^{\frac{1+y}{2} + \frac{t}{4} \log \frac{N^2}{a}}\ da.</math>
	Making the change of variables <math>a = N e^{-u}</math>, the right-hand side becomes		Making the change of variables <math>a = N e^{-u}</math>, the right-hand side becomes

Teorth: /* Estimation of E_1,E_2 */

2018-03-04T19:22:31Z

Estimation of E_1,E_2

← Older revision		Revision as of 12:22, 4 March 2018
Line 375:		Line 375:

	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].		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].

			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>
			The summand is decreasing for <math>1 \leq n \leq N</math> (one can write it for instance as <math>\frac{\exp( \frac{t}{4}(\log N - \log n)^2}{\exp(\frac{t}{4} \log^2 N) n^{\frac{1+y}{2}}}) and hence by the integral test one may bound
			:<math> \sum_{n=1}^N \frac{1}{n^{\frac{1+y}{2} + \frac{t}{4} \log \frac{N^2}{n}}} \leq 1 + \int_1^N \frac{1}{a^{\frac{1+y}{2} + \frac{t}{4} \log \frac{N^2}{a}}\ da.</math>
			Making the change of variables <math>a = N e^{-u}</math>, the right-hand side becomes
			:<math>1 + e^{-\frac{t}{4} \log^2 N} N^{\frac{1-y}{2}} \int_0^{\log N} \exp( \frac{t}{4} u^2 - \frac{1-y}{2} u)\ du.</math>
			Completing the square and making the change of variables <math>v := \sqrt{t/4} (u - \frac{1-y}{t})</math>, this is
			:<math>1 + e^{-\frac{t}{4} \log^2 N} N^{\frac{1-y}{2}} \exp( -\frac{(1-y)^2}{4t} )
			\int_{-(1-y)/\sqrt{4t}}^{V} \exp( v^2)\ dv</math>
			where
			:<math>V := \sqrt{t/4} \log N - (1-y)/\sqrt{4t}</math>.
			Applying Lemma 1.6 of [[Effective bounds on H_t - second approach]], we may bound this by
			:<math>1 + e^{-\frac{t}{4} \log^2 N} N^{\frac{1-y}{2}} \exp( -\frac{(1-y)^2}{4t} ) (\theta + e^{V^2} (\frac{1}{2V} + \frac{1}{V^3}) )</math>
			where
			:<math> \theta := \int_{-(1-y)/\sqrt{4t}} e^{v^2}\ dv.</math>
			Since
			:<math>e^{V^2} = e^{\frac{t}{4} \log^2 N} N^{-\frac{1-y}{2}} \exp( \frac{(1-y)^2}{4t})</math>
			we may rewrite this as
			:<math>1 + \frac{1}{2V} + \frac{1}{V^3} + e^{-\frac{t}{4} \log^2 N} N^{\frac{1-y}{2}} \exp( -\frac{(1-y)^2}{4t} ) \theta.</math>
			Thus we have
			:<math>E_2/\|B^{eff}_0}\| = \epsilon'(\frac{1+y+ix}{2}) \leq e_2^*</math>
			where
			:<math> e_2^* := \frac{1}{T-3.33} \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) ( 1 + \frac{1}{2V} + \frac{1}{V^3} + e^{-\frac{t}{4} \log^2 N} N^{\frac{1-y}{2}} \exp( -\frac{(1-y)^2}{4t} ) \theta ).</math>


	...		...