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= Welcome to MWiki =
 
= Welcome to MWiki =
 
== Theorem of the month ==
 
== Theorem of the month ==
The intex method solves every solvable LP in <math>\mathcal{O}({\vartheta}^{3})</math>.
 
  
== Proof and algorithm ==
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=== Universal multistep theorem ===
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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For <math>n \in {}^{\nu}\mathbb{N}_{\le p}, k, m, p \in {}^{\nu}\mathbb{N}^{*}, d_{\curvearrowright B} x \in\, ]0, 1[, x \in [a, b] \subseteq {}^{\omega}\mathbb{R}, y : [a, b] \rightarrow {}^{\omega}\mathbb{R}^q, f : [a, b] \times {}^{\omega}\mathbb{R}^{q \times n} \rightarrow {}^{\omega}\mathbb{R}^q, g_k(\curvearrowright B x) := g_{\acute{k}}(x)</math> and <math>g_0(a) = f((\curvearrowleft B)a, y_0, ... , y_{\acute{n}})</math>, the Taylor series of the initial value problem <math>y^\prime(x) = f(x, y((\curvearrowright B)^0 x), ... , y((\curvearrowright B)^{\acute{n}} x))</math> of order <math>n</math> implies <div style="text-align:center;"><math>y(\curvearrowright B x) = y(x) - d_{\curvearrowright B}x\sum\limits_{k=1}^{p}{i^{2k} g_{p-k}((\curvearrowright B) x)\sum\limits_{m=k}^{p}{\widehat{m!}\binom{\acute{m}}{\acute{k}}}} + \mathcal{O}((d_{\curvearrowright B} x)^{\grave{p}}).\square</math></div>
  
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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=== Goldbach’s theorem ===
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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Every even whole number greater than 2 is the sum of two primes.
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==== Proof: ====
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Induction over all prime gaps until the maximally possible one each time.<math>\square</math>
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=== Foundation theorem ===
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Only the postulation of the axiom of foundation that every nonempty subset <math>X \subseteq Y</math> contains an element <math>x_0</math> such that <math>X</math> und <math>x_0</math> are disjoint guarantees cycle freedom.
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==== Proof: ====
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Set <math>X := \{x_m : x_0 := \{\emptyset\}, x_{\omega} := \{x_1\}</math> and <math>x_{\acute{n}} := \{x_n\}</math> for <math>m \in {}^{\omega}\mathbb{N}</math> and <math>n \in {}^{\omega}\mathbb{N}_{\ge 2}\}</math> .<math>\square</math>
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== Recommended reading ==
  
== Recommended readings ==
 
 
[https://en.calameo.com/books/003777977258f7b4aa332 Nonstandard Mathematics]
 
[https://en.calameo.com/books/003777977258f7b4aa332 Nonstandard Mathematics]
  
 
[[de:Hauptseite]]
 
[[de:Hauptseite]]

Revision as of 01:05, 1 June 2021

Welcome to MWiki

Theorem of the month

Universal multistep theorem

For [math]\displaystyle{ n \in {}^{\nu}\mathbb{N}_{\le p}, k, m, p \in {}^{\nu}\mathbb{N}^{*}, d_{\curvearrowright B} x \in\, ]0, 1[, x \in [a, b] \subseteq {}^{\omega}\mathbb{R}, y : [a, b] \rightarrow {}^{\omega}\mathbb{R}^q, f : [a, b] \times {}^{\omega}\mathbb{R}^{q \times n} \rightarrow {}^{\omega}\mathbb{R}^q, g_k(\curvearrowright B x) := g_{\acute{k}}(x) }[/math] and [math]\displaystyle{ g_0(a) = f((\curvearrowleft B)a, y_0, ... , y_{\acute{n}}) }[/math], the Taylor series of the initial value problem [math]\displaystyle{ y^\prime(x) = f(x, y((\curvearrowright B)^0 x), ... , y((\curvearrowright B)^{\acute{n}} x)) }[/math] of order [math]\displaystyle{ n }[/math] implies

[math]\displaystyle{ y(\curvearrowright B x) = y(x) - d_{\curvearrowright B}x\sum\limits_{k=1}^{p}{i^{2k} g_{p-k}((\curvearrowright B) x)\sum\limits_{m=k}^{p}{\widehat{m!}\binom{\acute{m}}{\acute{k}}}} + \mathcal{O}((d_{\curvearrowright B} x)^{\grave{p}}).\square }[/math]

Goldbach’s theorem

Every even whole number greater than 2 is the sum of two primes.

Proof:

Induction over all prime gaps until the maximally possible one each time.[math]\displaystyle{ \square }[/math]

Foundation theorem

Only the postulation of the axiom of foundation that every nonempty subset [math]\displaystyle{ X \subseteq Y }[/math] contains an element [math]\displaystyle{ x_0 }[/math] such that [math]\displaystyle{ X }[/math] und [math]\displaystyle{ x_0 }[/math] are disjoint guarantees cycle freedom.

Proof:

Set [math]\displaystyle{ X := \{x_m : x_0 := \{\emptyset\}, x_{\omega} := \{x_1\} }[/math] and [math]\displaystyle{ x_{\acute{n}} := \{x_n\} }[/math] for [math]\displaystyle{ m \in {}^{\omega}\mathbb{N} }[/math] and [math]\displaystyle{ n \in {}^{\omega}\mathbb{N}_{\ge 2}\} }[/math] .[math]\displaystyle{ \square }[/math]

Recommended reading

Nonstandard Mathematics