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- :<math>H_t(x) = \int_{\bf R} \frac{1}{8} \xi(\frac{1+ix}{2} + i |t|^{1/2} v) \frac{1}{\sqrt{\pi}} e^{-v^2}\ dv \quad (1.1)</math> in the regime when <math>x</math> is large and <math>t</math> is large and negative. ...22 KB (3,943 words) - 08:56, 10 January 2019
- ...\gg 1 </math>. If we have <math>s = \sigma+iT</math> for some large <math>T</math> and bounded <math> \sigma \gg 1 </math>, this gives ...mma(s) \approx \sqrt{2\pi} T^{\sigma -1/2} e^{-\pi T/2} \exp(i (T \log T - T + \pi \sigma/2 - \pi/4)). (1.7)</math> ...19 KB (3,626 words) - 07:07, 30 March 2018
- For any <math>t \geq 0</math>, <math>s, b \in {\bf C}</math> with <math>\mathrm{Im} s > \ma :<math>I_t(s,b) := \int_C \exp( s(1 + u - e^u) + \frac{t}{16} ((u+b)^2 - b^2) )\ du</math> ...20 KB (3,730 words) - 07:07, 30 March 2018
- :<math>\xi_t(s) = \int_{-\infty}^\infty \xi(s+\sqrt{t} u) \frac{1}{\sqrt{\pi}} e^{-u^2}\ du.</math> Now suppose that <math>s = \sigma+iT</math> for some <math>T \geq T_0 \geq 10</math> (say). From equations (1.1), (3.1) of [A2011] (see ...24 KB (3,915 words) - 07:07, 30 March 2018
- :<math> 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})| :<math> 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})| ...28 KB (3,817 words) - 07:09, 30 March 2018
- ...n from [[Effective bounds on H_t - second approach]]. We also assume <math>T \geq 10</math> (so <math>x \geq 20</math>). :<math> \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> ...4 KB (809 words) - 07:08, 30 March 2018
- ...{2} \sum_{n=1}^\infty 2\pi^2 n^4 I_{t,\theta}(z-9i, \pi n^2) - 3\pi n^2 I_{t,\theta}(z-5i, \pi n^2) + ...\overline{I_{t,\theta}(\overline{z}-9i, \pi n^2)} - 3\pi n^2 \overline{I_{t,\theta}(\overline{z}-5i, \pi n^2)} </math> ...12 KB (2,253 words) - 07:10, 30 March 2018
- ..._t(x+iy) = \frac{1}{8} \int_{-\infty}^\infty \xi( \frac{1-y+ix}{2} + \sqrt{t} v) \frac{1}{\sqrt{\pi}} e^{-v^2}\ dv</math> which on making the change of variables <math>s = \frac{1-y+ix}{2} + \sqrt{t} v</math> is ...7 KB (1,245 words) - 07:10, 30 March 2018
Page text matches
- :<math>\xi_t(s) = \int_{-\infty}^\infty \xi(s+\sqrt{t} u) \frac{1}{\sqrt{\pi}} e^{-u^2}\ du.</math> Now suppose that <math>s = \sigma+iT</math> for some <math>T \geq T_0 \geq 10</math> (say). From equations (1.1), (3.1) of [A2011] (see ...24 KB (3,915 words) - 07:07, 30 March 2018
- :<math> 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})| :<math> 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})| ...28 KB (3,817 words) - 07:09, 30 March 2018
- For any <math>t \geq 0</math>, <math>s, b \in {\bf C}</math> with <math>\mathrm{Im} s > \ma :<math>I_t(s,b) := \int_C \exp( s(1 + u - e^u) + \frac{t}{16} ((u+b)^2 - b^2) )\ du</math> ...20 KB (3,730 words) - 07:07, 30 March 2018
- ...to be chosen later (in practice we take c close to <math>1/\log k</math>, T a small multiple of c, and <math>\tau</math> a small multiple of c/k. ...\to {\mathbf R}</math>, not identically zero, with <math>m_2 := \int_0^T g(t)^2\ dt</math>. We have the probabilistic interpretations ...30 KB (4,511 words) - 19:30, 23 October 2014
- ..._t(x+iy) = \frac{1}{8} \int_{-\infty}^\infty \xi( \frac{1-y+ix}{2} + \sqrt{t} v) \frac{1}{\sqrt{\pi}} e^{-v^2}\ dv</math> which on making the change of variables <math>s = \frac{1-y+ix}{2} + \sqrt{t} v</math> is ...7 KB (1,245 words) - 07:10, 30 March 2018
- ...\gg 1 </math>. If we have <math>s = \sigma+iT</math> for some large <math>T</math> and bounded <math> \sigma \gg 1 </math>, this gives ...mma(s) \approx \sqrt{2\pi} T^{\sigma -1/2} e^{-\pi T/2} \exp(i (T \log T - T + \pi \sigma/2 - \pi/4)). (1.7)</math> ...19 KB (3,626 words) - 07:07, 30 March 2018
- :'''Test problem''' For <math>t=y=0.4</math>, can one prove that <math>H_t(x+iy) \neq 0</math> for all <mat ...}{8} \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \sum_{n=1}^N \frac{\exp(\frac{t}{16} \log^2 \frac{s+4}{2\pi n^2})}{n^s}</math> ...22 KB (3,658 words) - 22:26, 1 January 2019
- ...n from [[Effective bounds on H_t - second approach]]. We also assume <math>T \geq 10</math> (so <math>x \geq 20</math>). :<math> \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> ...4 KB (809 words) - 07:08, 30 March 2018
- ...{2} \sum_{n=1}^\infty 2\pi^2 n^4 I_{t,\theta}(z-9i, \pi n^2) - 3\pi n^2 I_{t,\theta}(z-5i, \pi n^2) + ...\overline{I_{t,\theta}(\overline{z}-9i, \pi n^2)} - 3\pi n^2 \overline{I_{t,\theta}(\overline{z}-5i, \pi n^2)} </math> ...12 KB (2,253 words) - 07:10, 30 March 2018
- For each real number <math>t</math>, define the entire function <math>H_t: {\mathbf C} \to {\mathbf C}</ :<math>\displaystyle H_t(z) = \frac{1}{2} \int_0^\infty e^{t\log^2 x} \Phi(\log x) e^{iz \log x}\ \frac{dx}{x}.</math> ...12 KB (1,840 words) - 12:37, 22 April 2020
- For any <math>\sigma, t>0</math> and natural number <math>N</math>, introduce the sum :<math>S_{\sigma,t}(N) := \sum_{n=1}^N \frac{1}{n^{\sigma + \frac{t}{4} \log\frac{N^2}{n}}}.</math> ...11 KB (2,025 words) - 06:17, 6 July 2018
- :<math>H_t(x) = \int_{\bf R} \frac{1}{8} \xi(\frac{1+ix}{2} + i |t|^{1/2} v) \frac{1}{\sqrt{\pi}} e^{-v^2}\ dv \quad (1.1)</math> in the regime when <math>x</math> is large and <math>t</math> is large and negative. ...22 KB (3,943 words) - 08:56, 10 January 2019
- ...<math>t</math>. By implicitly differentiating the equation <math>H_t(z_j(t)) = 0</math>, we see that ...rtial_z H_t(z_j(t))} = \frac{\partial_{zz} H_t(z_j(t))}{\partial_z H_t(z_j(t))}.</math> ...12 KB (2,472 words) - 08:39, 5 April 2018
- ...nal bounded connected domain D and let u(x,t) (the heat at point x at time t) satisfy the heat equation with Neumann boundary conditions. We then conjec : For sufficiently large t > 0, u(x,t) achieves its maximum on the boundary of D ...61 KB (9,824 words) - 16:22, 21 March 2018
- ! style="text-align:left;"| range of <math>T=x/2</math> .../math>, and <math>s := \frac{1+y+ix}{2} + \frac{t}{2} \log N + \frac{\pi i t}{8}</math>. For the effective approximation one can write ...21 KB (3,501 words) - 12:52, 15 April 2018
- ...t{t}, \sqrt{t}, 1</math> isosceles triangle, and the distances <math>\sqrt{t}</math> are the distances that arise in the triangular lattice. The rings | [https://arxiv.org/abs/1804.02385 T] ...51 KB (7,013 words) - 00:34, 9 February 2021
- The table below lists various regions of the <math>(t,y,x)</math> parameter space where <math>H_t(x+iy)</math> is known to be non :<math> N := \lfloor \sqrt{\frac{x}{4\pi} + \frac{t}{16}} \rfloor</math> ...10 KB (1,300 words) - 16:44, 10 May 2018
- ...0,0). So if <math>|S|>9</math> S doesn’t contain any of these. Also, S can’t contain all of <math>(0,1,3), (0,3,1), (2,1,1)</math>. Similar for <math>(3 ...math>\Delta_5</math> with exactly one of a,b,c=0 has 12 elements and doesn’t contain any equilateral triangles. ...26 KB (4,003 words) - 20:25, 29 September 2015
- ...d you must decide whether there is some subset of N_1, …, N_k that sums to T. The problem is to beat the “trivial algorithm”, which I shall describe pre ...ents, until you find a disjoint pair of subsets that sums to either T or S-T. The running time of this procedure should be comparable to (k choose k/4) ...10 KB (1,641 words) - 07:20, 27 March 2020
- ...ath> on the right, thus creating a new coloring <math>\sigma \circ c \circ T^{-1}: {\bf C} \to \{1,2,3,4\}</math> of the complex plane with the same pro ...S_4 \times E(2)</math>, the coloring <math> \sigma \circ {\mathbf c} \circ T^{-1}</math> has the same law as <math>{\mathbf c}</math>. This gives the c ...53 KB (8,763 words) - 03:58, 7 May 2019
