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= Welcome to MWiki =
 
= Welcome to MWiki =
== Theorem of the month ==
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== Theorems of the month ==
The intex method solves every solvable LP in <math>\mathcal{O}({\vartheta}^{3})</math>.
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=== Prime number theorem ===
  
== Proof and algorithm ==
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For <math>\pi(x) := |\{p \in {\mathbb{P}_{\le x}} : x \in {}^{\omega}{\mathbb{R}}\}|</math>, it holds that <math>\pi(\omega) = \widetilde{{_e}\omega}\omega + \mathcal{O}({_\epsilon}\omega\;{\omega}^{\tilde{2}})</math>.
First, normalise and scale <math>{b}^{T}y - {c}^{T}x \le 0, Ax \le b</math> as well as <math>{A}^{T}y \ge c</math>. Let the <em>height</em> <math>h</math> have the initial value <math>h_0 := s |\min \; \{b_1, ..., b_m, -d_1, ..., -d_n\}|</math> for the <em>elongation factor</em> <math>s \in \, ]1, 2]</math>.</br>
 
The LP min <math>\{h \in [0, h_0] : x \in {}^{\omega}\mathbb{R}_{\ge 0}^{n}, y \in {}^{\omega}\mathbb{R}_{\ge 0}^{m},{b}^{T}y - {c}^{T}x \le h, Ax - b \le (h, ..., h)^T \in {}^{\omega}\mathbb{R}_{\ge 0}^{m}, c - {A}^{T}y \le (h, ..., h)^T \in {}^{\omega}\mathbb{R}_{\ge 0}^{n}\}</math> has <math>k</math> constraints and the feasible starting point <math>(x_0, y_0, h_0/s)^{T} \in {}^{\omega}\mathbb{R}_{\ge 0}^{m+n+1}</math>, e.g. <math>(0, 0, h_0/s)^{T}</math>.</br>
 
It identifies the mutually dual LPs max <math>\{{c}^{T}x : c \in {}^{\omega}\mathbb{R}^{n}, x \in {P}_{\ge 0}\}</math> and min <math>\{{b}^{T}y : y \in {}^{\omega}\mathbb{R}_{\ge 0}^{m}, {A}^{T}y \ge c\}</math>.
 
  
Let the point <math>p := (x, y, h)^T</math> approximate the subpolytope’s centre of gravity <math>P^*</math> as <math>p_k^* := (\min p_k + \max p_k)/2</math> until <math>{|| \Delta p ||}_{1}</math> is sufficiently small. Here <math>x</math> takes precedence over <math>y</math>. Then extrapolate <math>p</math> via <math>{p}^{*}</math> into <math>\partial P^*</math> as <math>u</math>. Put <math>p := p^* + (u - p^*)/s</math> to shun <math>\partial P^*</math>. Hereon approximate <math>p</math> more deeply again as centre of gravity. After optionally solving all LPs min<math>{}_{k} {h}_{k}</math> by bisection methods for <math>{h}_{k} \in {}^{\omega}\mathbb{R}_{\ge 0}</math> in <math>\mathcal{O}({\vartheta}^{2})</math> each time, <math>v \in {}^{\omega}\mathbb{R}^{k}</math> may be determined such that <math>v_k := \Delta{p}_{k} \Delta{h}_{k}/r</math> and <math>r :=</math> min<math>{}_{k} \Delta{h}_{k}</math>. Simplified let <math>|\Delta{p}_{1}| = ... = |\Delta{p}_{m+n}|</math>.
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==== Proof: ====
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From intervals of fix length <math>y \in {}^{\omega}{\mathbb{R}_{>0}}, \check{y}</math> set-2-tuples of prime numbers are formed such that the first interval has the unchanged representative prime number density and the second interval is empty, then the interval with the second most prime number density is followed by the second least one etc. The Stirling formula suggests the prime gap <math>n = {\epsilon}^{\sigma} = \mathcal{O}({_\epsilon}(n!))</math>.
  
Here min <math>{h}_{m+n+1}</math> may be solved for <math>p^* := p + tv</math> where <math>t \in {}^{\omega}\mathbb{R}_{\ge 0}</math> and <math>{v}_{m+n+1} = 0</math>. If min<math>{}_{k} {h}_{k} r = 0</math> follows, end, otherwise start over until min <math>h = 0</math> or min <math>h &gt; 0</math> is certain. If necessary, relax the constraints temporarily by the same small modulus.</br>
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For induction basis <math>n = 2</math> resp. 3, the hypothesis states the first interval to contain <math>x_n/{_\epsilon}x_n</math> primes for <math>n \in {}^{\omega}{\mathbb{N}_{\ge2}}</math> and <math>x_4 \in [2, 4[</math>. Stepping from <math>x_n</math> to <math>x_n^2</math> finds <math>\pi(x_n^2) = \pi(x_n) \check{x}_n</math> primes only from <math>\pi(x_n) = x_n/{_\epsilon}x_n</math>. The average prime gap is <math>{_\epsilon}x_n</math>, the maximal one <math>{_\epsilon}x_n^2</math> and the maximal <math>x_n^2</math> to <math>x_n</math> behaves like <math>\omega</math> to <math>{\omega}^{\tilde{2}}.\square</math>
Since almost every iteration step in <math>\mathcal{O}({\omega\vartheta}^{2})</math> halves <math>h</math> at least, the strong duality theorem yields the result.<math>\square</math>
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=== Gelfond-Schneider theorem ===
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It holds <math>a^b \notin {}_{\omega}^{\omega} \mathbb{A}_\mathbb{C}</math> where <math>a, c \in {}^{\omega} \mathbb{A}_\mathbb{C} \setminus \mathbb{B}</math> and infinitesimal <math>\varepsilon, b \in {}^{\omega}\mathbb{A}_\mathbb{C} \setminus {}_{\omega}^{\omega}\mathbb{R}</math>.
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==== Proof: ====
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The minimal polynomials <math>p</math> (and <math>q</math>) of <math>c^r</math> resp. <math>c^{r\pm\varepsilon} = a^b</math> for maximal <math>r \in {}_{\omega}^{\omega}\mathbb{R}_{>0}</math> and <math>f = p\;(q)</math> lead to the contradiction <math>{}^1f(c^{r(\pm\varepsilon)}) \ne 0 = (f(c^r) - f(c^{r\pm\varepsilon})) / (c^r - c^{r\pm\varepsilon}) = {}^1f(c^{r(\pm\varepsilon)}).\square</math>
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== MWiki has moved ==
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The new URL is: [https://en.hwiki.de/maths.html HWiki]
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== Recommended reading ==
  
== Recommended readings ==
 
 
[https://en.calameo.com/books/003777977258f7b4aa332 Nonstandard Mathematics]
 
[https://en.calameo.com/books/003777977258f7b4aa332 Nonstandard Mathematics]
  
 
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[[de:Hauptseite]]

Latest revision as of 18:01, 31 July 2024

Welcome to MWiki

Theorems of the month

Prime number theorem

For [math]\displaystyle{ \pi(x) := |\{p \in {\mathbb{P}_{\le x}} : x \in {}^{\omega}{\mathbb{R}}\}| }[/math], it holds that [math]\displaystyle{ \pi(\omega) = \widetilde{{_e}\omega}\omega + \mathcal{O}({_\epsilon}\omega\;{\omega}^{\tilde{2}}) }[/math].

Proof:

From intervals of fix length [math]\displaystyle{ y \in {}^{\omega}{\mathbb{R}_{\gt 0}}, \check{y} }[/math] set-2-tuples of prime numbers are formed such that the first interval has the unchanged representative prime number density and the second interval is empty, then the interval with the second most prime number density is followed by the second least one etc. The Stirling formula suggests the prime gap [math]\displaystyle{ n = {\epsilon}^{\sigma} = \mathcal{O}({_\epsilon}(n!)) }[/math].

For induction basis [math]\displaystyle{ n = 2 }[/math] resp. 3, the hypothesis states the first interval to contain [math]\displaystyle{ x_n/{_\epsilon}x_n }[/math] primes for [math]\displaystyle{ n \in {}^{\omega}{\mathbb{N}_{\ge2}} }[/math] and [math]\displaystyle{ x_4 \in [2, 4[ }[/math]. Stepping from [math]\displaystyle{ x_n }[/math] to [math]\displaystyle{ x_n^2 }[/math] finds [math]\displaystyle{ \pi(x_n^2) = \pi(x_n) \check{x}_n }[/math] primes only from [math]\displaystyle{ \pi(x_n) = x_n/{_\epsilon}x_n }[/math]. The average prime gap is [math]\displaystyle{ {_\epsilon}x_n }[/math], the maximal one [math]\displaystyle{ {_\epsilon}x_n^2 }[/math] and the maximal [math]\displaystyle{ x_n^2 }[/math] to [math]\displaystyle{ x_n }[/math] behaves like [math]\displaystyle{ \omega }[/math] to [math]\displaystyle{ {\omega}^{\tilde{2}}.\square }[/math]

Gelfond-Schneider theorem

It holds [math]\displaystyle{ a^b \notin {}_{\omega}^{\omega} \mathbb{A}_\mathbb{C} }[/math] where [math]\displaystyle{ a, c \in {}^{\omega} \mathbb{A}_\mathbb{C} \setminus \mathbb{B} }[/math] and infinitesimal [math]\displaystyle{ \varepsilon, b \in {}^{\omega}\mathbb{A}_\mathbb{C} \setminus {}_{\omega}^{\omega}\mathbb{R} }[/math].

Proof:

The minimal polynomials [math]\displaystyle{ p }[/math] (and [math]\displaystyle{ q }[/math]) of [math]\displaystyle{ c^r }[/math] resp. [math]\displaystyle{ c^{r\pm\varepsilon} = a^b }[/math] for maximal [math]\displaystyle{ r \in {}_{\omega}^{\omega}\mathbb{R}_{\gt 0} }[/math] and [math]\displaystyle{ f = p\;(q) }[/math] lead to the contradiction [math]\displaystyle{ {}^1f(c^{r(\pm\varepsilon)}) \ne 0 = (f(c^r) - f(c^{r\pm\varepsilon})) / (c^r - c^{r\pm\varepsilon}) = {}^1f(c^{r(\pm\varepsilon)}).\square }[/math]

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Recommended reading

Nonstandard Mathematics