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= Welcome to MWiki =
 
= Welcome to MWiki =
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== Theorems of the month ==
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== Theorem of the month ==
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=== Cauchy's integral theorem===
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=== Counter-directional 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>, we have 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>, we have 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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=== Fundamental theorem of algebra ===
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If the path <math>\gamma: [a, b[ \, \cap \, C \rightarrow V</math> with <math>C \subseteq \mathbb{R}</math> passes the edges of every <math>n</math>-cube of side length d0 in the <math>n</math>-volume <math>V \subseteq {}^{(\omega)}\mathbb{R}^{n}</math> with <math>n \in \mathbb{N}_{\ge 2}</math> exactly once, where the opposite edges in all two-dimensional faces of every <math>n</math>-cube are traversed in reverse direction, but uniformly, then, for <math>D \subseteq \mathbb{R}^{2}, B \subseteq {V}^{2}, f = ({f}_{1}, ..., {f}_{n}): V \rightarrow {}^{(\omega)}\mathbb{R}^{n}, \gamma(t) = x, \gamma(\curvearrowright D t) = \curvearrowright B x</math> and <math>{V}_{\curvearrowright } := \{\curvearrowright B x \in V: x \in V, \curvearrowright B x \ne \curvearrowleft B x\}</math>, it holds that
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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, we can 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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Since <math>f(z) := 1/p(z)</math> is holomorphic, we have 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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<div style="text-align:center;"><math>\int\limits_{t \in [a,b[ \, \cap \, C}{f(\gamma (t)){{{{\gamma }'}}_{\curvearrowright }}(t)dDt}=\int\limits_{\begin{smallmatrix} (x,\curvearrowright B\,x) \\ \in V\times {{V}_{\curvearrowright}} \end{smallmatrix}}{f(x)dBx}=\int\limits_{\begin{smallmatrix} t \in [a,b[ \, \cap \, C, \\ \gamma | {\partial{}^{\acute{n}}} V \end{smallmatrix}}{f(\gamma (t)){{{{\gamma }'}}_{\curvearrowright }}(t)dDt}.</math></div>
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==== Proof: ====
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If two arbitrary squares are considered with common edge of length d0 included in one plane, then only the edges of <math>V\times{V}_{\curvearrowright}</math> are not passed in both directions for the same function value. They all, and thus the path to be passed, are exactly contained in <math>{\partial}^{\acute{n}}V.\square</math>
  
 
== Recommended reading ==
 
== Recommended reading ==

Revision as of 05:04, 1 August 2020

Welcome to MWiki

Theorem of the month

Counter-directional theorem

If the path [math]\displaystyle{ \gamma: [a, b[ \, \cap \, C \rightarrow V }[/math] with [math]\displaystyle{ C \subseteq \mathbb{R} }[/math] passes the edges of every [math]\displaystyle{ n }[/math]-cube of side length d0 in the [math]\displaystyle{ n }[/math]-volume [math]\displaystyle{ V \subseteq {}^{(\omega)}\mathbb{R}^{n} }[/math] with [math]\displaystyle{ n \in \mathbb{N}_{\ge 2} }[/math] exactly once, where the opposite edges in all two-dimensional faces of every [math]\displaystyle{ n }[/math]-cube are traversed in reverse direction, but uniformly, then, for [math]\displaystyle{ D \subseteq \mathbb{R}^{2}, B \subseteq {V}^{2}, f = ({f}_{1}, ..., {f}_{n}): V \rightarrow {}^{(\omega)}\mathbb{R}^{n}, \gamma(t) = x, \gamma(\curvearrowright D t) = \curvearrowright B x }[/math] and [math]\displaystyle{ {V}_{\curvearrowright } := \{\curvearrowright B x \in V: x \in V, \curvearrowright B x \ne \curvearrowleft B x\} }[/math], it holds that


[math]\displaystyle{ \int\limits_{t \in [a,b[ \, \cap \, C}{f(\gamma (t)){{{{\gamma }'}}_{\curvearrowright }}(t)dDt}=\int\limits_{\begin{smallmatrix} (x,\curvearrowright B\,x) \\ \in V\times {{V}_{\curvearrowright}} \end{smallmatrix}}{f(x)dBx}=\int\limits_{\begin{smallmatrix} t \in [a,b[ \, \cap \, C, \\ \gamma | {\partial{}^{\acute{n}}} V \end{smallmatrix}}{f(\gamma (t)){{{{\gamma }'}}_{\curvearrowright }}(t)dDt}. }[/math]

Proof:

If two arbitrary squares are considered with common edge of length d0 included in one plane, then only the edges of [math]\displaystyle{ V\times{V}_{\curvearrowright} }[/math] are not passed in both directions for the same function value. They all, and thus the path to be passed, are exactly contained in [math]\displaystyle{ {\partial}^{\acute{n}}V.\square }[/math]

Recommended reading

Nonstandard Mathematics

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