Selberg sieve variational problem - Revision history

WikiSysop at 03:30, 24 October 2014

2014-10-24T03:30:22Z

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	~~Test~~

WikiSysop at 03:29, 24 October 2014

2014-10-24T03:29:45Z

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			Test

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2014-05-20T00:43:43Z

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	\| [http://terrytao.wordpress.com/2014/01/17/polymath8b-vi-a-low-dimensional-variational-problem/#comments 1.9400]		\| [http://terrytao.wordpress.com/2014/01/17/polymath8b-vi-a-low-dimensional-variational-problem/#comments 1.9400]
	\| [http://terrytao.wordpress.com/~~2013~~/12/20/polymath8b-iv-~~enlarging~~-the-~~sieve~~-~~support-more-efficient~~-~~numerics~~-~~and~~-~~explicit~~-~~asymptotics~~/#comment-~~262156~~ 1.~~87459~~]		\| [http://terrytao.wordpress.com/2014/04/14/polymath8b-x-writing-the-paper-and-chasing-down-loose-ends/#comment-326155 1.91726]
	\| [http://terrytao.wordpress.com/2014/01/17/polymath8b-vi-a-low-dimensional-variational-problem/#comment-268443 1.992]		\| [http://terrytao.wordpress.com/2014/01/17/polymath8b-vi-a-low-dimensional-variational-problem/#comment-268443 1.992]
	\| [http://terrytao.wordpress.com/2014/02/09/polymath8b-viii-time-to-start-writing-up-the-results/#comment-270725 2.0012]		\| [http://terrytao.wordpress.com/2014/02/09/polymath8b-viii-time-to-start-writing-up-the-results/#comment-270725 2.0012]

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2014-05-18T15:52:24Z

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	The crude upper bound of <math>k</math> for any of the <math>M_k</math> type quantities comes from the parity problem obstruction that each separate event "<math>n+h_i</math> prime" can occur with probability at most 1/2.		The crude upper bound of <math>k</math> for any of the <math>M_k</math> type quantities comes from the parity problem obstruction that each separate event "<math>n+h_i</math> prime" can occur with probability at most 1/2.

	Here are some [notes on polytope decomposition] for the k=3 case.		Here are some [[notes on polytope decomposition]] for the k=3 case.

	{\| class="wikitable" border=1		{\| class="wikitable" border=1

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	The crude upper bound of <math>k</math> for any of the <math>M_k</math> type quantities comes from the parity problem obstruction that each separate event "<math>n+h_i</math> prime" can occur with probability at most 1/2.		The crude upper bound of <math>k</math> for any of the <math>M_k</math> type quantities comes from the parity problem obstruction that each separate event "<math>n+h_i</math> prime" can occur with probability at most 1/2.

			Here are some [notes on polytope decomposition] for the k=3 case.

	{\| class="wikitable" border=1		{\| class="wikitable" border=1

Teorth: /* Asymptotic analysis */

2014-04-22T18:37:21Z

Asymptotic analysis

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	and so the constraint <math>1-k\mu > \tau</math> becomes		and so the constraint <math>1-k\mu > \tau</math> becomes

	:<math>\alpha + \log \beta ~~- 1~~ \~~geq~~ \~~gamma~~.</math>		:<math>\alpha + \log \beta + \gamma \leq 1.</math>

	With <math>r =1 + \frac{u\gamma}{\log k}</math> for <math>0 \leq u \leq 1</math>, one has		With <math>r =1 + \frac{u\gamma}{\log k}</math> for <math>0 \leq u \leq 1</math>, one has

	:<math> a = \frac{\alpha+\log \beta-1+u \gamma +o(1)}{\beta}</math>		:<math> a = \frac{1-\alpha-\log \beta+u \gamma +o(1)}{\beta}</math>
	:<math> Z_1 \leq \log a + \frac{1}{4\beta a \log^2 a}</math>		:<math> Z_1 \leq \log a + \frac{1}{4\beta a \log^2 a}</math>

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	:<math>X = \frac{1}{\gamma} + o(1)</math>		:<math>X = \frac{1}{\gamma} + o(1)</math>
	:<math> V = \int_0^1\frac{dt}{(1+t)^2 (2+t)} + o(1) = 1 - \log(2) + o(1)</math>		:<math> V = \int_0^1\frac{dt}{(1+t)^2 (2+t)} + o(1) = 1 - \log(2) + o(1)</math>
	:<math> U = \int_0^1 (u\gamma +\alpha+\log \beta ~~- 2~~)^2\ du + \beta + o(1)</math>		:<math> U = \int_0^1 (u\gamma -\alpha-\log \beta)^2\ du + \beta + o(1)</math>
	:<math> = \frac{(\gamma+\alpha+\log\beta -2)^3 - (\alpha+\log\beta-2)^3}{3}+ \beta + o(1)</math>		:<math> = \frac{(\gamma-\alpha-\log\beta)^3 + (\alpha+\log\beta)^3}{3}+ \beta + o(1)</math>

	and hence		and hence

	:<math>\Delta \leq \frac{ \int_0^1 (\log a(u) + \frac{1}{4\beta a(u) \log^2 a(u)})\ du + \frac{\log(\gamma)}{\gamma-1} + \frac{1-\log(2)}{3} ((\gamma+\alpha+\log\beta -2)^3 - (\alpha+\log\beta-2)^3+3\beta) }{1 - \frac{\beta}{(\alpha+\log \beta-1+\gamma)^2} }+ o(1)</math>		:<math>\Delta \leq \frac{ \int_0^1 (\log a(u) + \frac{1}{4\beta a(u) \log^2 a(u)})\ du + \frac{\log(\gamma)}{\gamma-1} + \frac{1-\log(2)}{3} ((\gamma-\alpha-\log\beta)^3 + (\alpha+\log\beta)^3+3\beta) }{1 - \frac{\beta}{(1-\alpha-\log \beta-\gamma)^2} }+ o(1)</math>

	assuming that <math>\alpha+\log \beta~~- 1>~~ \gamma</math> and		assuming that <math>\alpha+\log \beta + \gamma< 1</math> and
	:<math> 1 - \frac{\beta}{(\alpha+\log \beta-1+\gamma)^2} > 0,</math>		:<math> 1 - \frac{\beta}{(1-\alpha-\log \beta-\gamma)^2} > 0,</math>
	and where <math>a(u):=\frac{\alpha+\log \beta-1+u \gamma}{\beta}</math>.		and where <math>a(u):=\frac{1-\alpha-\log \beta+u \gamma}{\beta}</math>.

	In particular, by setting <math>\alpha,\beta,\gamma</math> as absolute constants obeying these constraints, we have <math>\Delta \leq O(1)</math>, and so		In particular, by setting <math>\alpha,\beta,\gamma</math> as absolute constants obeying these constraints, we have <math>\Delta \leq O(1)</math>, and so

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2014-04-22T17:34:07Z

Asymptotic analysis

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	:<math> M_k = \log k+O(1).</math>		:<math> M_k = \log k+O(1).</math>

	If we set <math>c = \tau := 1/\log k</math>, with <math>T</math> a small multiple of c, then <math>\mu \approx \frac{1}{k}(1 - \frac{A}{\log k})</math> for a large absolute constant A, and <math>\sigma^2</math> is a small multiple of <math>\frac{1}{k \log^2 k}</math>. This makes the denominator comparable to 1; one can check that all the terms in the numerator are O(1), finally giving the bound

	~~:<math>\Delta_k = O(1)</math>~~

	~~and thus we have the lower bound~~

	~~:<math>M_k \geq\frac{k}{k-1} \log k - O(1) = \log k - O(1)</math>.~~

	== More general variational problems ==		== More general variational problems ==

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2014-04-14T16:55:23Z

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2014-03-31T18:10:43Z

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2014-03-09T17:25:50Z

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