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(Centre Method)
(Cauchy's integral theorem and Fundamental theorem of algebra)
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= Welcome to MWiki =
 
= Welcome to MWiki =
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== Theorem of the month ==
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== Theorems of the month ==
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The centre method solves every solvable LP in <math>\mathcal{O}(\omega{\vartheta}^{2})</math>.
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=== Cauchy's integral theorem ===
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Given the neighbourhood relations <math>B \subseteq {A}^{2}</math> and <math>D \subseteq [a, b]</math> for some simply connected <math>h</math>-set <math>A \subseteq {}^{\omega}\mathbb{C}</math>, infinitesimal <math>h</math>, a holomorphic function <math>f: A \rightarrow {}^{\omega}\mathbb{C}</math> and a closed path <math>\gamma: [a, b[\rightarrow \partial A</math>, choosing <math>\curvearrowright B \gamma(t) = \gamma(\curvearrowright D t)</math> for <math>t \in [a, b[</math>, it holds that
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<div style="text-align:center;"><math>\int\limits_{\gamma }{f(z)dBz}=0.</math></div>
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'''Proof:''' By the Cauchy-Riemann partial differential equations and Green's theorem, with <math>x := \text{Re} \, z, y := \text{Im} \, z, u := \text{Re} \, f, v := \text{Im} \, f</math> and <math>{A}^{-} := \{z \in A : z + h + ih \in A\}</math>, it holds that
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<div style="text-align:center;"><math>\int\limits_{\gamma }{f(z)dBz}=\int\limits_{\gamma }{\left( u+iv \right)\left( dBx+idBy \right)}=\int\limits_{z\in {{A}^{-}}}{\left( i\left( \frac{\partial Bu}{\partial Bx}-\frac{\partial Bv}{\partial By} \right)-\left( \frac{\partial Bv}{\partial Bx}+\frac{\partial Bu}{\partial By} \right) \right)dB(x,y)}=0.\square</math></div>
  
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== Proof and algorithm ==
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=== Fundamental theorem of algebra ===
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Let <math>z := \grave{m} + n</math> and <math>d \in [0, 1]</math> the density of <math>A</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 <math>P_r := \{(x, y)^T \in {}^{\omega}\mathbb{R}_{\ge 0}^{z} : {b}^{T}y - {c}^{T}x \le r \in [0, \check{r}], Ax - b \le \underline{r}_m, c - {A}^{T}y \le \underline{r}_n\}</math> have the radius <math>\check{r} := s|\min \; \{b_1, ..., b_m, -c_1, ..., -c_n\}|</math> and the scaling factor <math>s \in [1, 2]</math>. It follows <math>\underline{0}_{z} \in \partial P_{\check{r}}</math>. By the strong duality theorem, the LP min <math>\{ r \in [0, \check{r}] : (x, y)^T \in P_r\}</math> solves the 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>.
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For every non-constant polynomial <math>p \in {}^{(\omega)}\mathbb{C}</math>, there exists some <math>z \in {}^{(\omega)}\mathbb{C}</math> such that <math>p(z) = 0</math>.
  
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'''Indirect proof:''' By performing an affine substitution of variables, reduce to the case <math>1/p(0) \ne \mathcal{O}(\text{d0})</math>. Suppose that <math>p(z) \ne 0</math> for all <math>z \in {}^{(\omega)}\mathbb{C}</math>.
  
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Its solution is the geometric centre <math>g</math> of the polytope <math>P_0</math>. For <math>p_k^* := (\text{min}\,p_k + \text{max}\,p_k)/2</math> and <math>k = 1, ..., \grave{z}</math> approximate <math>g</math> by <math>p_0 := (x_0, y_0, r_0)^T</math> until <math>||\Delta p||_1</math> is sufficiently small. The solution <math>t^o(x^o, y^o, r^o)^T</math> of the two-dimensional LP min <math>\{ r \in [0, \check{r}] : t \in {}^{\omega}\mathbb{R}_{&gt; 0}, t(x_0, y_0)^T \in P_r\}</math> approximates <math>g</math> better and achieves <math>r \le \check{r}/\sqrt{\grave{z}}</math>. Repeat this for <math>t^o(x^o, y^o)^T</math> until <math>g \in P_0</math> is computed in <math>\mathcal{O}({}_z\check{r} {}_e\check{r}dmn)</math> if it exists. Numbers of length <math>\mathcal{O}({\omega})</math> can only be processed in <math>\mathcal{O}(\vartheta)</math> as is generally known.
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Since <math>f(z) := 1/p(z)</math> is holomorphic, it holds that <math>f(1/\text{d0}) = \mathcal{O}(\text{d0})</math>. By the mean value inequality <math>|f(0)| \le {|f|}_{\gamma}</math> for <math>\gamma = \partial\mathbb{B}_{r}(0)</math> and arbitrary <math>r \in {}^{(\omega)}\mathbb{R}_{&gt;0}</math>, and hence <math>f(0) = \mathcal{O}(\text{d0})</math>, which is a contradiction.<math>\square</math>
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Solving all two-dimensional LPs <math>\text{min}_k r_k</math> by bisection methods for <math>r_k \in {}^{\omega}\mathbb{R}_{\ge 0}</math> and <math>k = 1, ..., z</math> in <math>\mathcal{O}({\vartheta}^2)</math> each time determines <math>q \in {}^{\omega}\mathbb{R}^k</math> where <math>q_k := \Delta p_k \Delta r_k/r</math> and <math>r := \text{min}_k \Delta r_k</math>. Let simplified <math>|\Delta p_1| = … = |\Delta p_{z}|</math>. Here min <math>r_z</math> for <math>p^* := p + wq</math> and <math>w \in {}^{\omega}\mathbb{R}_{\ge 0}</math> would be also to solve. If <math>\text{min}_k \Delta r_k r = 0</math> follows, stop, otherwise repeat until min <math>r = 0</math> or min <math>r &gt; 0</math> is sure. If necessary, constraints are temporarily relaxed by the same small modulus.<math>\square</math>
 
  
 
== Recommended readings ==
 
== Recommended readings ==

Revision as of 20:55, 31 January 2022

Welcome to MWiki

Theorems of the month

Cauchy's integral theorem

Given the neighbourhood relations [math]\displaystyle{ B \subseteq {A}^{2} }[/math] and [math]\displaystyle{ D \subseteq [a, b] }[/math] for some simply connected [math]\displaystyle{ h }[/math]-set [math]\displaystyle{ A \subseteq {}^{\omega}\mathbb{C} }[/math], infinitesimal [math]\displaystyle{ h }[/math], a holomorphic function [math]\displaystyle{ f: A \rightarrow {}^{\omega}\mathbb{C} }[/math] and a closed path [math]\displaystyle{ \gamma: [a, b[\rightarrow \partial A }[/math], choosing [math]\displaystyle{ \curvearrowright B \gamma(t) = \gamma(\curvearrowright D t) }[/math] for [math]\displaystyle{ t \in [a, b[ }[/math], it holds that

[math]\displaystyle{ \int\limits_{\gamma }{f(z)dBz}=0. }[/math]

Proof: By the Cauchy-Riemann partial differential equations and Green's theorem, with [math]\displaystyle{ x := \text{Re} \, z, y := \text{Im} \, z, u := \text{Re} \, f, v := \text{Im} \, f }[/math] and [math]\displaystyle{ {A}^{-} := \{z \in A : z + h + ih \in A\} }[/math], it holds that

[math]\displaystyle{ \int\limits_{\gamma }{f(z)dBz}=\int\limits_{\gamma }{\left( u+iv \right)\left( dBx+idBy \right)}=\int\limits_{z\in {{A}^{-}}}{\left( i\left( \frac{\partial Bu}{\partial Bx}-\frac{\partial Bv}{\partial By} \right)-\left( \frac{\partial Bv}{\partial Bx}+\frac{\partial Bu}{\partial By} \right) \right)dB(x,y)}=0.\square }[/math]

Fundamental theorem of algebra

For every non-constant polynomial [math]\displaystyle{ p \in {}^{(\omega)}\mathbb{C} }[/math], there exists some [math]\displaystyle{ z \in {}^{(\omega)}\mathbb{C} }[/math] such that [math]\displaystyle{ p(z) = 0 }[/math].

Indirect proof: By performing an affine substitution of variables, reduce to the case [math]\displaystyle{ 1/p(0) \ne \mathcal{O}(\text{d0}) }[/math]. Suppose that [math]\displaystyle{ p(z) \ne 0 }[/math] for all [math]\displaystyle{ z \in {}^{(\omega)}\mathbb{C} }[/math].

Since [math]\displaystyle{ f(z) := 1/p(z) }[/math] is holomorphic, it holds that [math]\displaystyle{ f(1/\text{d0}) = \mathcal{O}(\text{d0}) }[/math]. By the mean value inequality [math]\displaystyle{ |f(0)| \le {|f|}_{\gamma} }[/math] for [math]\displaystyle{ \gamma = \partial\mathbb{B}_{r}(0) }[/math] and arbitrary [math]\displaystyle{ r \in {}^{(\omega)}\mathbb{R}_{>0} }[/math], and hence [math]\displaystyle{ f(0) = \mathcal{O}(\text{d0}) }[/math], which is a contradiction.[math]\displaystyle{ \square }[/math]

Recommended readings

Nonstandard Mathematics

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