Controlling A+B/B 0 - Revision history

Teorth: /* Derivative bounds */

2018-04-15T20:52:29Z

Derivative bounds

← Older revision		Revision as of 13:52, 15 April 2018
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	Also, as established above, we have		Also, as established above, we have
	:<math> \mathrm{Re}(s_B) \geq \frac{1+y}{2} + \frac{t}{2} (\frac{3y-1}{x_{N+1}^2} + \log N)</math>		:<math> \mathrm{Re}(s_B) \geq \frac{1+y}{2} + \frac{t}{2} (\frac{3y-1}{x_{N+1}^2} + \log N)</math>
	(for <math>y \geq 1/3</math>) and a similar argument gives		(for <math>y \geq 1/3 + 4y(1+y)/3x^2</math>) and a similar argument gives
	:<math> \mathrm{Re}(s_A) \geq \frac{1-y}{2} + \frac{t}{2} (\frac{2-3y}{x_{N}^2} + \log N).</math>		:<math> \mathrm{Re}(s_A) \geq \frac{1-y}{2} + \frac{t}{2} (\frac{2-3y}{x_{N}^2} + \log N).</math>

Teorth: /* Analysing the effective model */

2018-04-15T20:45:09Z

Analysing the effective model

← Older revision		Revision as of 13:45, 15 April 2018
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	:<math> \sigma = \frac{1+y}{2} + \frac{t}{2} \mathrm{Re}(\frac{1}{1+y+ix} + \frac{2}{-1+y+ix} + \frac{1}{2} \log \frac{1+y+ix}{4\pi})</math>		:<math> \sigma = \frac{1+y}{2} + \frac{t}{2} \mathrm{Re}(\frac{1}{1+y+ix} + \frac{2}{-1+y+ix} + \frac{1}{2} \log \frac{1+y+ix}{4\pi})</math>
	:<math> = \frac{1+y}{2} + \frac{t}{2} (\frac{1+y}{(1+y)^2+x^2} + \frac{-2+2y}{(-1+y)^2+x^2} + \frac{1}{2} \log \frac{\|1+y+ix\|}{4\pi})</math>		:<math> = \frac{1+y}{2} + \frac{t}{2} (\frac{1+y}{(1+y)^2+x^2} + \frac{-2+2y}{(-1+y)^2+x^2} + \frac{1}{2} \log \frac{\|1+y+ix\|}{4\pi})</math>
	:<math> \geq \frac{1+y}{2} + \frac{t}{2} (\frac{1+y}{(-1+y)^2+x^2} + \frac{-2+2y}{(-1+y)^2+x^2} + \frac{1}{2} \log \frac{x}{4\pi})</math>		:<math> \geq \frac{1+y}{2} + \frac{t}{2} (\frac{4 y(1+y)}{((-1+y)^2+x^2)((1+y)^2+x^2)} - \frac{1+3y}{(-1+y)^2+x^2} + \frac{1}{2} \log \frac{x}{4\pi})</math>
	:<math> \geq \frac{1+y}{2} + \frac{t}{2} (\frac{3y-1}{(-1+y)^2+x^2} + \log N)</math>		:<math> \geq \frac{1+y}{2} + \frac{t}{2} (-\frac{1-3y-\frac{4y(1+y)}{x^2}}{(-1+y)^2+x^2} + \log N)</math>
	:<math> \geq \frac{1+y}{2} + \frac{t}{2} \log N</math>		:<math> \geq \frac{1+y}{2} + \frac{t}{2} \log N</math>
	assuming that <math>y \geq 1/3</math>. Hence we can actually just use the same value of <math>\sigma</math> as in the toy case.		assuming that <math>y \geq \frac{1}{3} + \frac{4y(1+y)}{3x^2}</math>. Hence we can actually just use the same value of <math>\sigma</math> as in the toy case.

	Next we control <math>\|\lambda\|</math>. Note that we can increase <math>\|\lambda\|</math> (thus multiplying <math>\sum_{n=1}^N \frac{a_n}{n^s}</math> by a quantity greater than 1) without affecting (2.1), so we just need upper bounds on <math>\|\lambda\|</math>. We may factor		Next we control <math>\|\lambda\|</math>. Note that we can increase <math>\|\lambda\|</math> (thus multiplying <math>\sum_{n=1}^N \frac{a_n}{n^s}</math> by a quantity greater than 1) without affecting (2.1), so we just need upper bounds on <math>\|\lambda\|</math>. We may factor

Teorth at 15:10, 30 March 2018

2018-03-30T15:10:27Z

← Older revision		Revision as of 08:10, 30 March 2018
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	:<math>\|\frac{d}{dx} \frac{A^{eff}+B^{eff}}{B^{eff}_0}\| \leq (1 + \frac{t}{2(x_N-6)}) \sum_{n=1}^N \frac{b_n}{n^{\frac{1+y}{2} + \frac{t}{2} (\frac{3y-1}{x_{N+1}^2} + \log N)}} \frac{\log n}{2} + e^\delta N^{-y} \sum_{n=1}^N \frac{b_n}{n^{\frac{1-y}{2} + \frac{t}{2} (\frac{2-3y}{x_{N}^2} + \log N)}} (\frac{t}{4(x_N-6)} \log n + \frac{1}{4} \log \frac{\|1-y+ix_{N+1}\| \|1+y-ix_{N+1}\|}{16\pi^2 n^2} + \frac{3t+1}{4 x_N(x_N-6)})</math>		:<math>\|\frac{d}{dx} \frac{A^{eff}+B^{eff}}{B^{eff}_0}\| \leq (1 + \frac{t}{2(x_N-6)}) \sum_{n=1}^N \frac{b_n}{n^{\frac{1+y}{2} + \frac{t}{2} (\frac{3y-1}{x_{N+1}^2} + \log N)}} \frac{\log n}{2} + e^\delta N^{-y} \sum_{n=1}^N \frac{b_n}{n^{\frac{1-y}{2} + \frac{t}{2} (\frac{2-3y}{x_{N}^2} + \log N)}} (\frac{t}{4(x_N-6)} \log n + \frac{1}{4} \log \frac{\|1-y+ix_{N+1}\| \|1+y-ix_{N+1}\|}{16\pi^2 n^2} + \frac{3t+1}{4 x_N(x_N-6)})</math>
	where <math>b_n = \exp( \frac{t}{4} \log^2 n)</math>, <math>x_N = 4\pi N^2 - \frac{\pi t}{4}</math>, and <math>\delta</math> was defined in (1) above.		where <math>b_n = \exp( \frac{t}{4} \log^2 n)</math>, <math>x_N = 4\pi N^2 - \frac{\pi t}{4}</math>, and <math>\delta</math> was defined in (1) above.

			[[Category:Polymath15]]

Teorth: /* Derivative bounds */

2018-03-08T21:56:38Z

Derivative bounds

← Older revision		Revision as of 14:56, 8 March 2018
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	:<math> = O_{\leq}( \frac{1}{2} + \frac{t}{4(x_N-6)} ).</math>		:<math> = O_{\leq}( \frac{1}{2} + \frac{t}{4(x_N-6)} ).</math>
	Similarly		Similarly
	:<math> \frac{d}{dx} s_A = -\frac{i}{2} + O_{\leq}( \frac{t}{4(x_N-6)} ).</math>		:<math> \frac{d}{dx} s_A = \frac{i}{2} + O_{\leq}( \frac{t}{4(x_N-6)} ).</math>
	Finally, writing <math>s := \frac{1+y-ix}{2}</math>, we have		Finally, writing <math>s := \frac{1-y+ix}{2}</math>, we have
	:<math> \log \lambda = \frac{t}{4} (\alpha_1(s)^2 - \alpha_1(1-s)^2) + (-\frac{s}{2} \log \pi + (\frac{s}{2}-\frac{1}{2}) \log \frac{s}{2} - \frac{s}{2}) - (-\frac{1-s}{2} \log \pi + (\frac{1-s}{2}-\frac{1}{2}) \log \frac{1-s}{2} - \frac{1-s}{2})</math>		:<math> \log \lambda = \frac{t}{4} (\alpha_1(s)^2 - \alpha_1(1-s)^2) + (-\frac{s}{2} \log \pi + (\frac{s}{2}-\frac{1}{2}) \log \frac{s}{2} - \frac{s}{2}) - (-\frac{1-s}{2} \log \pi + (\frac{1-s}{2}-\frac{1}{2}) \log \frac{1-s}{2} - \frac{1-s}{2})</math>
	and hence		and hence
	:<math> \frac{i}{2} \log n + \frac{d}{dx} \log \lambda = \frac{i}{2} \log n -\frac{i}{2} \frac{d}{ds} \log \lambda</math>		:<math> -\frac{i}{2} \log n + \frac{d}{dx} \log \lambda = -\frac{i}{2} \log n +\frac{i}{2} \frac{d}{ds} \log \lambda</math>
	:<math> = -\frac{it}{4} (\alpha_1(s) \alpha'_1(s) + \alpha_1(1-s) \alpha'_1(1-s)) + \frac{-i}{4} \log \frac{s}{2\pi n} + \frac{i}{4s} + \frac{-i}{4} \log \frac{1-s}{2\pi n} + \frac{i}{4(1-s)} </math>		:<math> = \frac{it}{4} (\alpha_1(s) \alpha'_1(s) + \alpha_1(1-s) \alpha'_1(1-s)) + \frac{i}{4} \log \frac{s}{2\pi n} - \frac{i}{4s} + \frac{i}{4} \log \frac{1-s}{2\pi n} - \frac{i}{4(1-s)} </math>
	:<math> = O_{\leq}( \frac{t}{4(x_N-6)} (\|\alpha_1(s)\| + \|\alpha_1(1-s)\| + \frac{1}{4} \log \frac{\|1-y+ix\| \|1+y-ix\|}{16\pi^2 n^2} + \frac{1}{4\|s\| \|1-s\|} );</math>		:<math> = O_{\leq}( \frac{t}{4(x_N-6)} (\|\alpha_1(s)\| + \|\alpha_1(1-s)\| + \frac{1}{4} \log \frac{\|1-y+ix\| \|1+y-ix\|}{16\pi^2 n^2} + \frac{1}{4\|s\| \|1-s\|} );</math>
	bounding		bounding

Teorth: /* Derivative bounds */

2018-03-07T21:52:21Z

Derivative bounds

← Older revision		Revision as of 14:52, 7 March 2018
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	:<math>\frac{d}{dx} \frac{A^{eff}+B^{eff}}{B^{eff}_0} = \sum_{n=1}^N \frac{-b_n \log n}{n^{s_B}} \frac{d}{dx}(s_B) + \lambda \sum_{n=1}^N \frac{b_n}{n^{s_A}} (-\log n \frac{d}{dx}(s_A) + \frac{d}{dx} \log \lambda)</math>		:<math>\frac{d}{dx} \frac{A^{eff}+B^{eff}}{B^{eff}_0} = \sum_{n=1}^N \frac{-b_n \log n}{n^{s_B}} \frac{d}{dx}(s_B) + \lambda \sum_{n=1}^N \frac{b_n}{n^{s_A}} (-\log n \frac{d}{dx}(s_A) + \frac{d}{dx} \log \lambda)</math>
	and hence		and hence
	:<math>\|\frac{d}{dx} \frac{A^{eff}+B^{eff}}{B^{eff}_0}\| \leq \sum_{n=1}^N \frac{b_n \log n}{n^{\mathrm{Re}(s_B)}} \|\frac{d}{dx}(s_B)\| + \|\lambda\| \sum_{n=1}^N \frac{b_n}{n^{\mathrm{Re}(s_A)}} \|\log n \frac{d}{dx}(s_A) + \frac{d}{dx} \log \lambda\|.</math>		:<math>\|\frac{d}{dx} \frac{A^{eff}+B^{eff}}{B^{eff}_0}\| \leq \sum_{n=1}^N \frac{b_n \log n}{n^{\mathrm{Re}(s_B)}} \|\frac{d}{dx}(s_B)\| + \|\lambda\| \sum_{n=1}^N \frac{b_n}{n^{\mathrm{Re}(s_A)}} \|-\log n \frac{d}{dx}(s_A) + \frac{d}{dx} \log \lambda\|.</math>

	Recall from above that <math>\alpha'_1(s) = O_{\leq}( \frac{1}{x_N-6} )</math>. Thus we have		Recall from above that <math>\alpha'_1(s) = O_{\leq}( \frac{1}{x_N-6} )</math>. Thus we have

Teorth: /* Derivative bounds */

2018-03-07T16:46:59Z

Derivative bounds

Teorth: /* Large x case */

2018-03-07T16:30:59Z

Large x case

← Older revision		Revision as of 09:30, 7 March 2018
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	:<math>\|\sum_{n=1}^N \frac{b_n}{n^s}\| - \|\sum_{n=1}^N \frac{a_n}{n^s}\| \geq 2 - \sum_{n=1}^N \frac{\|b_n\|}{n^\sigma} - \sum_{n=1}^N \frac{\|a_n\|}{n^\sigma}</math>		:<math>\|\sum_{n=1}^N \frac{b_n}{n^s}\| - \|\sum_{n=1}^N \frac{a_n}{n^s}\| \geq 2 - \sum_{n=1}^N \frac{\|b_n\|}{n^\sigma} - \sum_{n=1}^N \frac{\|a_n\|}{n^\sigma}</math>
	:<math> \geq 2 - \sum_{n=1}^N \frac{1}{n^{0.7 + 0.1 \log \frac{N^2}{n}}} - e^\delta \exp( \frac{t y \log N}{2(x_N-6)} ) N^{-0.4} \sum_{n=1}^N \frac{1}{n^{0.3 + 0.1 \log \frac{N^2}{n}}} </math>		:<math> \geq 2 - \sum_{n=1}^N \frac{1}{n^{0.7 + 0.1 \log \frac{N^2}{n}}} - e^\delta \exp( \frac{t y \log N}{2(x_N-6)} ) N^{-0.4} \sum_{n=1}^N \frac{1}{n^{0.3 + 0.1 \log \frac{N^2}{n}}} </math>
	:<math> \geq 2 - 1.706 - e^\delta \exp( \frac{0.8 \log N}{4\pi N^2 - 6.315} ) N^{-0.4} \times 3.469</math>		:<math> \geq 2 - 1.706 - e^\delta \exp( \frac{0.08 \log N}{4\pi N^2 - 6.315} ) N^{-0.4} \times 3.469</math>
	using the estimates from [[Estimating a sum]]. We can bound		using the estimates from [[Estimating a sum]]. We can bound
	:<math>\delta := \frac{0.2 \pi}{4\pi N^2 - 6.315 - \frac{14.8}{\pi}} + \frac{5.6}{(4\pi N^2 - 0.315)^2} \log \|\frac{1.4}{4\pi}+i (N+1)^2\| </math>		:<math>\delta := \frac{0.2 \pi}{4\pi N^2 - 6.315 - \frac{14.8}{\pi}} + \frac{5.6}{(4\pi N^2 - 0.315)^2} \log \|\frac{1.4}{4\pi}+i (N+1)^2\| </math>
	which for <math>N \geq 2000</math> may be bounded by <math>1.26 \times 10^{-8}</math>. One may similarly bound <math>\frac{0.8 \log N}{4\pi N^2 - 6.315} </math> by <math>1.21 \times 10^{-7}</math>, we obtain the lower bound		which for <math>N \geq 2000</math> may be bounded by <math>1.26 \times 10^{-8}</math>. One may similarly bound <math>\frac{0.08 \log N}{4\pi N^2 - 6.315} </math> by <math>1.21 \times 10^{-8}</math>, so we obtain the lower bound
	:<math>\|\frac{A^{eff}+B^{eff}\|}{\|B^{eff}_0\|} \geq \|\sum_{n=1}^N \frac{b_n}{n^s}\| - \|\sum_{n=1}^N \frac{a_n}{n^s}\| \geq 2 - 1.706 - 0.166 = 0.128.</math>		:<math>\|\frac{A^{eff}+B^{eff}\|}{\|B^{eff}_0\|} \geq \|\sum_{n=1}^N \frac{b_n}{n^s}\| - \|\sum_{n=1}^N \frac{a_n}{n^s}\| \geq 2 - 1.706 - 0.166 = 0.128.</math>

Teorth: /* Derivative bounds */

2018-03-06T17:46:07Z

Derivative bounds

← Older revision		Revision as of 10:46, 6 March 2018
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	Also, as established above, we have		Also, as established above, we have
	:<math> \mathrm{Re}(s_B) \geq \frac{1+y}{2} + \frac{t}{2} (\frac{3y-1}{x_{N+1}^2} + \log N)</math>		:<math> \mathrm{Re}(s_B) \geq \frac{1+y}{2} + \frac{t}{2} (\frac{3y-1}{x_{N+1}^2} + \log N)</math>
	and a similar argument gives		(for <math>y \geq 1/3</math>) and a similar argument gives
	:<math> \mathrm{Re}(s_A) \geq \frac{1-y}{2} + \frac{t}{2} (\frac{2-3y}{x_{N+1}^2} + \log N).</math>		:<math> \mathrm{Re}(s_A) \geq \frac{1-y}{2} + \frac{t}{2} (\frac{2-3y}{x_{N}^2} + \log N).</math>

	We conclude that		We conclude that
	:<math>\|\frac{d}{dx} \frac{A^{eff}+B^{eff}}{B^{eff}_0}\| \leq (1 + \frac{t}{2(x_N-6)}) \sum_{n=1}^N \frac{b_n}{n^{\frac{1+y}{2} + \frac{t}{2} (\frac{3y-1}{x_{N+1}^2} + \log N)}} \frac{\log n}{2} + e^\delta N^{-y} \sum_{n=1}^N \frac{b_n}{n^{\frac{1-y}{2} + \frac{t}{2} (\frac{2-3y}{x_{N+1}^2} + \log N)}} (\frac{t}{4(x_N-6)} \log n + \frac{1}{4} \log \frac{\|1-y+ix_{N+1}\| \|1+y-ix_{N+1}\|}{4\pi n^2} + \frac{3t+1}{4 x_N(x_N-6)})</math>		:<math>\|\frac{d}{dx} \frac{A^{eff}+B^{eff}}{B^{eff}_0}\| \leq (1 + \frac{t}{2(x_N-6)}) \sum_{n=1}^N \frac{b_n}{n^{\frac{1+y}{2} + \frac{t}{2} (\frac{3y-1}{x_{N+1}^2} + \log N)}} \frac{\log n}{2} + e^\delta N^{-y} \sum_{n=1}^N \frac{b_n}{n^{\frac{1-y}{2} + \frac{t}{2} (\frac{2-3y}{x_{N}^2} + \log N)}} (\frac{t}{4(x_N-6)} \log n + \frac{1}{4} \log \frac{\|1-y+ix_{N+1}\| \|1+y-ix_{N+1}\|}{4\pi n^2} + \frac{3t+1}{4 x_N(x_N-6)})</math>
	where <math>b_n = \exp( \frac{t}{4} \log^2 n)</math>, <math>x_N = 4\pi N^2 - \frac{\pi t}{4}</math>, and <math>\delta</math> was defined in (1) above.		where <math>b_n = \exp( \frac{t}{4} \log^2 n)</math>, <math>x_N = 4\pi N^2 - \frac{\pi t}{4}</math>, and <math>\delta</math> was defined in (1) above.

Teorth: /* Derivative bounds */

2018-03-06T17:44:08Z

Derivative bounds

← Older revision		Revision as of 10:44, 6 March 2018
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	We conclude that		We conclude that
	:<math>\|\frac{d}{dx} \frac{A^{eff}+B^{eff}}{B^{eff}_0}\| \leq (1 + \frac{t}{2(x_N-6)}) \sum_{n=1}^N \frac{b_n}{n^{\frac{1+y}{2} + \frac{t}{2} (\frac{3y-1}{x_{N+1}^2} + \log N)}} \frac{\log n}{2} + e^\delta N^{-y} \sum_{n=1}^N \frac{b_n}{n^{\frac{1-y}{2} + \frac{t}{2} (\frac{2-3y}{x_{N+1}^2} + \log N)}} (\frac{t}{4(x_N-6)} \log n + \frac{1}{4} \log \frac{\|1-y+ix_{N+1}\| \|1+y-ix_{N+1}\|}{4\pi n^2} + \frac{3t+1}{4 x_N(x_N-6)})</math>		:<math>\|\frac{d}{dx} \frac{A^{eff}+B^{eff}}{B^{eff}_0}\| \leq (1 + \frac{t}{2(x_N-6)}) \sum_{n=1}^N \frac{b_n}{n^{\frac{1+y}{2} + \frac{t}{2} (\frac{3y-1}{x_{N+1}^2} + \log N)}} \frac{\log n}{2} + e^\delta N^{-y} \sum_{n=1}^N \frac{b_n}{n^{\frac{1-y}{2} + \frac{t}{2} (\frac{2-3y}{x_{N+1}^2} + \log N)}} (\frac{t}{4(x_N-6)} \log n + \frac{1}{4} \log \frac{\|1-y+ix_{N+1}\| \|1+y-ix_{N+1}\|}{4\pi n^2} + \frac{3t+1}{4 x_N(x_N-6)})</math>
	where <math>b_n = \exp( \frac{t}{4} \log^2 n)</math>, <math>x_N = 4\pi N^2 + \frac{\pi t}{4}</math>, and <math>\delta</math> was defined in (1) above.		where <math>b_n = \exp( \frac{t}{4} \log^2 n)</math>, <math>x_N = 4\pi N^2 - \frac{\pi t}{4}</math>, and <math>\delta</math> was defined in (1) above.

Teorth: /* Derivative bounds */

2018-03-06T17:43:01Z

Derivative bounds

← Older revision		Revision as of 10:43, 6 March 2018
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	We conclude that		We conclude that
	:<math>\|\frac{d}{dx} \frac{A^{eff}+B^{eff}}{B^{eff}_0}\| \leq (1 + \frac{t}{2(x_N-6)}) \sum_{n=1}^N \frac{b_n}{n^{\frac{1+y}{2} + \frac{t}{2} (\frac{3y-1}{x_{N+1}^2} + \log N)}} \frac{\log n}{2} + e^\delta N^{-y} \sum_{n=1}^N \frac{b_n}{n^{\frac{1-y}{2} + \frac{t}{2} (\frac{2-3y}{x_{N+1}^2} + \log N)}} (\frac{t}{4(x_N-6)} \log n + \frac{1}{4} \log \frac{\|1-y+ix_{N+1}\| \|1+y-ix_{N+1}\|}{4\pi n^2} + \frac{3t+1}{4 x_N(x_N-6)})</math>		:<math>\|\frac{d}{dx} \frac{A^{eff}+B^{eff}}{B^{eff}_0}\| \leq (1 + \frac{t}{2(x_N-6)}) \sum_{n=1}^N \frac{b_n}{n^{\frac{1+y}{2} + \frac{t}{2} (\frac{3y-1}{x_{N+1}^2} + \log N)}} \frac{\log n}{2} + e^\delta N^{-y} \sum_{n=1}^N \frac{b_n}{n^{\frac{1-y}{2} + \frac{t}{2} (\frac{2-3y}{x_{N+1}^2} + \log N)}} (\frac{t}{4(x_N-6)} \log n + \frac{1}{4} \log \frac{\|1-y+ix_{N+1}\| \|1+y-ix_{N+1}\|}{4\pi n^2} + \frac{3t+1}{4 x_N(x_N-6)})</math>
	where <math>\delta</math> was defined in (1) above.		where <math>b_n = \exp( \frac{t}{4} \log^2 n)</math>, <math>x_N = 4\pi N^2 + \frac{\pi t}{4}</math>, and <math>\delta</math> was defined in (1) above.

← Older revision		Revision as of 09:46, 7 March 2018
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	:<math> \frac{i}{2} \log n + \frac{d}{dx} \log \lambda = \frac{i}{2} \log n -\frac{i}{2} \frac{d}{ds} \log \lambda</math>		:<math> \frac{i}{2} \log n + \frac{d}{dx} \log \lambda = \frac{i}{2} \log n -\frac{i}{2} \frac{d}{ds} \log \lambda</math>
	:<math> = -\frac{it}{4} (\alpha_1(s) \alpha'_1(s) + \alpha_1(1-s) \alpha'_1(1-s)) + \frac{-i}{4} \log \frac{s}{2\pi n} + \frac{i}{4s} + \frac{-i}{4} \log \frac{1-s}{2\pi n} + \frac{i}{4(1-s)} </math>		:<math> = -\frac{it}{4} (\alpha_1(s) \alpha'_1(s) + \alpha_1(1-s) \alpha'_1(1-s)) + \frac{-i}{4} \log \frac{s}{2\pi n} + \frac{i}{4s} + \frac{-i}{4} \log \frac{1-s}{2\pi n} + \frac{i}{4(1-s)} </math>
	:<math> = O_{\leq}( \frac{t}{4(x_N-6)} (\|\alpha_1(s)\| + \|\alpha_1(1-s)\| + \frac{1}{4} \log \frac{\|1-y+ix\| \|1+y-ix\|}{4\pi n^2} + \frac{1}{4\|s\| \|1-s\|} );</math>		:<math> = O_{\leq}( \frac{t}{4(x_N-6)} (\|\alpha_1(s)\| + \|\alpha_1(1-s)\| + \frac{1}{4} \log \frac{\|1-y+ix\| \|1+y-ix\|}{16\pi^2 n^2} + \frac{1}{4\|s\| \|1-s\|} );</math>
	bounding		bounding
	:<math>\|\alpha_1(s)\| \leq \frac{1}{2} \log \frac{\|1-y+ix\|}{4\pi} + \frac{3}{2x_N} </math>		:<math>\|\alpha_1(s)\| \leq \frac{1}{2} \log \frac{\|1-y+ix\|}{4\pi} + \frac{3}{2x_N} </math>
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	:<math>\|\alpha_1(1-s)\| \leq \frac{1}{2} \log \frac{\|1+y+ix\|}{4\pi} + \frac{3}{2x_N} </math>		:<math>\|\alpha_1(1-s)\| \leq \frac{1}{2} \log \frac{\|1+y+ix\|}{4\pi} + \frac{3}{2x_N} </math>
	we thus have		we thus have
	:<math>\|\frac{i}{2} \log n + \frac{d}{dx} \log \lambda\| \leq \frac{1}{4} (1 + \frac{t}{2(x_N-6)}) \log \frac{\|1-y+ix\| \|1+y-ix\|}{4\pi n^2} + \frac{3t}{4 x_N(x_N-6)} + \frac{1}{4 x_N^2}</math>		:<math>\|\frac{i}{2} \log n + \frac{d}{dx} \log \lambda\| \leq \frac{1}{4} (1 + \frac{t}{2(x_N-6)}) \log \frac{\|1-y+ix\| \|1+y-ix\|}{16\pi^2 n^2} + \frac{3t}{4 x_N(x_N-6)} + \frac{1}{4 x_N^2}</math>
	:<math> \leq \frac{1}{4} (1 + \frac{t}{2(x_N-6)}) \log \frac{\|1-y+ix_{N+1}\| \|1+y-ix_{N+1}\|}{4\pi} + \frac{3t+1}{4 x_N(x_N-6)}.</math>		:<math> \leq \frac{1}{4} (1 + \frac{t}{2(x_N-6)}) \log \frac{\|1-y+ix_{N+1}\| \|1+y-ix_{N+1}\|}{16\pi^2} + \frac{3t+1}{4 x_N(x_N-6)}.</math>
	Also, as established above, we have		Also, as established above, we have
	:<math> \mathrm{Re}(s_B) \geq \frac{1+y}{2} + \frac{t}{2} (\frac{3y-1}{x_{N+1}^2} + \log N)</math>		:<math> \mathrm{Re}(s_B) \geq \frac{1+y}{2} + \frac{t}{2} (\frac{3y-1}{x_{N+1}^2} + \log N)</math>
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	We conclude that		We conclude that
	:<math>\|\frac{d}{dx} \frac{A^{eff}+B^{eff}}{B^{eff}_0}\| \leq (1 + \frac{t}{2(x_N-6)}) \sum_{n=1}^N \frac{b_n}{n^{\frac{1+y}{2} + \frac{t}{2} (\frac{3y-1}{x_{N+1}^2} + \log N)}} \frac{\log n}{2} + e^\delta N^{-y} \sum_{n=1}^N \frac{b_n}{n^{\frac{1-y}{2} + \frac{t}{2} (\frac{2-3y}{x_{N}^2} + \log N)}} (\frac{t}{4(x_N-6)} \log n + \frac{1}{4} \log \frac{\|1-y+ix_{N+1}\| \|1+y-ix_{N+1}\|}{4\pi n^2} + \frac{3t+1}{4 x_N(x_N-6)})</math>		:<math>\|\frac{d}{dx} \frac{A^{eff}+B^{eff}}{B^{eff}_0}\| \leq (1 + \frac{t}{2(x_N-6)}) \sum_{n=1}^N \frac{b_n}{n^{\frac{1+y}{2} + \frac{t}{2} (\frac{3y-1}{x_{N+1}^2} + \log N)}} \frac{\log n}{2} + e^\delta N^{-y} \sum_{n=1}^N \frac{b_n}{n^{\frac{1-y}{2} + \frac{t}{2} (\frac{2-3y}{x_{N}^2} + \log N)}} (\frac{t}{4(x_N-6)} \log n + \frac{1}{4} \log \frac{\|1-y+ix_{N+1}\| \|1+y-ix_{N+1}\|}{16\pi^2 n^2} + \frac{3t+1}{4 x_N(x_N-6)})</math>
	where <math>b_n = \exp( \frac{t}{4} \log^2 n)</math>, <math>x_N = 4\pi N^2 - \frac{\pi t}{4}</math>, and <math>\delta</math> was defined in (1) above.		where <math>b_n = \exp( \frac{t}{4} \log^2 n)</math>, <math>x_N = 4\pi N^2 - \frac{\pi t}{4}</math>, and <math>\delta</math> was defined in (1) above.