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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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=== Leibniz' differentiation rule ===
  
== Proof and algorithm ==
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For <math>f: {}^{(\omega)}\mathbb{K}^{\grave{n}} \rightarrow {}^{(\omega)}\mathbb{K}, a, b: {}^{(\omega)}\mathbb{K}^{n} \rightarrow {}^{(\omega)}\mathbb{K}, \curvearrowright B x := {(s, {x}_{2}, ..., {x}_{n})}^{T}</math>, and <math>s \in {}^{(\omega)}\mathbb{K} \setminus \{{x}_{1}\}</math>, choosing <math>\curvearrowright D a(x) = a(\curvearrowright B x)</math> and <math>\curvearrowright D b(x) = b(\curvearrowright B x)</math> it holds that<div style="text-align:center;"><math>\tfrac{{\downarrow} }{{\downarrow} {{x}_{1}}}\left( {\uparrow}_{a(x)}^{b(x)}{f(x,t){\downarrow}Dt} \right)={\uparrow}_{a(x)}^{b(x)}{\tfrac{{\downarrow} f(x,t)}{{\downarrow} {{x}_{1}}}{\downarrow}Dt}+\tfrac{{\downarrow} b(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright Bx,b(x))-\tfrac{{\downarrow} a(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright Bx,a(x)).</math></div>
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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<div style="text-align:center;"><math>\begin{aligned}\tfrac{{\downarrow} }{{\downarrow} {{x}_{1}}}\left( {\uparrow}_{a(x)}^{b(x)}{f(x,t){\downarrow}Dt} \right) &={\left( {\uparrow}_{a(\curvearrowright Bx)}^{b(\curvearrowright Bx)}{f(\curvearrowright Bx,t){\downarrow}Dt}-{\uparrow}_{a(x)}^{b(x)}{f(x,t){\downarrow}Dt} \right)}/{{\downarrow} {{x}_{1}}}\; \\ &={\left( {\uparrow}_{a(x)}^{b(x)}{(f(\curvearrowright Bx,t)-f(x,t)){\downarrow}Dt}+{\uparrow}_{b(x)}^{b(\curvearrowright Bx)}{f(\curvearrowright Bx,t){\downarrow}Dt}-{\uparrow}_{a(x)}^{a(\curvearrowright Bx)}{f(\curvearrowright Bx,t){\downarrow}Dt} \right)}/{{\downarrow} {{x}_{1}}}\; \\ &={\uparrow}_{a(x)}^{b(x)}{\tfrac{{\downarrow} f(x,t)}{{\downarrow} {{x}_{1}}}{\downarrow}Dt}+\tfrac{{\downarrow} b(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright Bx,b(x))-\tfrac{{\downarrow} a(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright Bx,a(x)).\square\end{aligned}</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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=== Beal'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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Equation <math>a^m + b^n = c^k</math> where <math>a, b, c \in \mathbb{N}^{*}</math> and <math>k, m, n \in \mathbb{N}_{\ge 3}</math> implies gcd<math>(a, b, c) > 1.</math>
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==== Proof: ====
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Proof: Real points between <math>r, s \in {}^{\omega}\mathbb{Q}</math> do not avoid that every nontrivial equation for <math>c^k > 1</math> is given by <math>(a^{m-r} + ib^{n-s})(a^r - ib^s) =c^k +i(a^rb^{n-s} - a^{m-r}b^s)</math> where all relations <math>a^{m-\hat{r}} = b^{n-\hat{s}}</math> then imply <math>p \mid</math> gcd<math>(a, b, c)</math> as well as the claim for <math>p \in {}^{\omega} \mathbb{P}.\square</math>
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=== Conclusion: ===
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The preceding theorem enables an infinite descent because of gcd<math>(a, b, c) > 1</math> such that no <math>n \in {}^{\omega}\mathbb{N}_{\ge 3}</math> satisfies <math>a^n + b^n = c^n</math> for arbitrary <math>a, b, c \in {}^{\omega}\mathbb{N}^{*}.\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 15:58, 1 March 2023

Welcome to MWiki

Theorems of the month

Leibniz' differentiation rule

For [math]\displaystyle{ f: {}^{(\omega)}\mathbb{K}^{\grave{n}} \rightarrow {}^{(\omega)}\mathbb{K}, a, b: {}^{(\omega)}\mathbb{K}^{n} \rightarrow {}^{(\omega)}\mathbb{K}, \curvearrowright B x := {(s, {x}_{2}, ..., {x}_{n})}^{T} }[/math], and [math]\displaystyle{ s \in {}^{(\omega)}\mathbb{K} \setminus \{{x}_{1}\} }[/math], choosing [math]\displaystyle{ \curvearrowright D a(x) = a(\curvearrowright B x) }[/math] and [math]\displaystyle{ \curvearrowright D b(x) = b(\curvearrowright B x) }[/math] it holds that

[math]\displaystyle{ \tfrac{{\downarrow} }{{\downarrow} {{x}_{1}}}\left( {\uparrow}_{a(x)}^{b(x)}{f(x,t){\downarrow}Dt} \right)={\uparrow}_{a(x)}^{b(x)}{\tfrac{{\downarrow} f(x,t)}{{\downarrow} {{x}_{1}}}{\downarrow}Dt}+\tfrac{{\downarrow} b(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright Bx,b(x))-\tfrac{{\downarrow} a(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright Bx,a(x)). }[/math]

Proof:

[math]\displaystyle{ \begin{aligned}\tfrac{{\downarrow} }{{\downarrow} {{x}_{1}}}\left( {\uparrow}_{a(x)}^{b(x)}{f(x,t){\downarrow}Dt} \right) &={\left( {\uparrow}_{a(\curvearrowright Bx)}^{b(\curvearrowright Bx)}{f(\curvearrowright Bx,t){\downarrow}Dt}-{\uparrow}_{a(x)}^{b(x)}{f(x,t){\downarrow}Dt} \right)}/{{\downarrow} {{x}_{1}}}\; \\ &={\left( {\uparrow}_{a(x)}^{b(x)}{(f(\curvearrowright Bx,t)-f(x,t)){\downarrow}Dt}+{\uparrow}_{b(x)}^{b(\curvearrowright Bx)}{f(\curvearrowright Bx,t){\downarrow}Dt}-{\uparrow}_{a(x)}^{a(\curvearrowright Bx)}{f(\curvearrowright Bx,t){\downarrow}Dt} \right)}/{{\downarrow} {{x}_{1}}}\; \\ &={\uparrow}_{a(x)}^{b(x)}{\tfrac{{\downarrow} f(x,t)}{{\downarrow} {{x}_{1}}}{\downarrow}Dt}+\tfrac{{\downarrow} b(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright Bx,b(x))-\tfrac{{\downarrow} a(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright Bx,a(x)).\square\end{aligned} }[/math]

Beal's theorem

Equation [math]\displaystyle{ a^m + b^n = c^k }[/math] where [math]\displaystyle{ a, b, c \in \mathbb{N}^{*} }[/math] and [math]\displaystyle{ k, m, n \in \mathbb{N}_{\ge 3} }[/math] implies gcd[math]\displaystyle{ (a, b, c) \gt 1. }[/math]

Proof:

Proof: Real points between [math]\displaystyle{ r, s \in {}^{\omega}\mathbb{Q} }[/math] do not avoid that every nontrivial equation for [math]\displaystyle{ c^k \gt 1 }[/math] is given by [math]\displaystyle{ (a^{m-r} + ib^{n-s})(a^r - ib^s) =c^k +i(a^rb^{n-s} - a^{m-r}b^s) }[/math] where all relations [math]\displaystyle{ a^{m-\hat{r}} = b^{n-\hat{s}} }[/math] then imply [math]\displaystyle{ p \mid }[/math] gcd[math]\displaystyle{ (a, b, c) }[/math] as well as the claim for [math]\displaystyle{ p \in {}^{\omega} \mathbb{P}.\square }[/math]

Conclusion:

The preceding theorem enables an infinite descent because of gcd[math]\displaystyle{ (a, b, c) \gt 1 }[/math] such that no [math]\displaystyle{ n \in {}^{\omega}\mathbb{N}_{\ge 3} }[/math] satisfies [math]\displaystyle{ a^n + b^n = c^n }[/math] for arbitrary [math]\displaystyle{ a, b, c \in {}^{\omega}\mathbb{N}^{*}.\square }[/math]

Recommended reading

Nonstandard Mathematics