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(Counter-directional theorem)
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= Welcome to MWiki =
 
= Welcome to MWiki =
 
== Theorem of the month ==
 
== Theorem of the month ==
=== Finite representation for odd <math>\zeta</math>-arguments ===
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=== Counter-directional theorem ===
  
Using the digamma function <math>\psi</math>, it holds for <math>n \in {}^{\omega}2\mathbb{N}^{*}</math>, small <math>\varepsilon \in ]0, 1]</math> and <math>{{d}_{\varepsilon k n}}:={{\varepsilon}^{{\hat{n}}}}{e}^{\hat{n}2k\pi i}</math> that<div style="text-align:center;"><math>\zeta(\grave{n}) = \underset{\varepsilon \to 0}{\mathop{\lim }}\,\widehat{-\varepsilon n}\sum\limits_{k=1}^{n}{\left( \gamma +\psi ({{d}_{\varepsilon k n}}) \right)}+\mathcal{O}(\varepsilon )</math></div>and<div style="text-align:center;"><math>\zeta(\grave{n}) = \underset{\varepsilon \to 0}{\mathop{\lim }}\,\widehat{2\varepsilon n}\sum\limits_{k=1}^{n}{\left( \psi ({{d}_{\varepsilon k n}}{{i}^{\hat{n}2}})-\psi ({{d}_{\varepsilon k n}}) \right)}+\mathcal{O}({{\varepsilon }^{2}}).</math></div>
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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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<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>
  
 
==== Proof: ====
 
==== Proof: ====
The claim results easily via the geometric series from <div style="text-align:center;"><math>\psi (z)+\gamma +\hat{z}=\sum\limits_{m=1}^{\omega }{\left( \hat{m}-\widehat{m+z} \right)}=-\sum\limits_{m=1}^{\omega }{\zeta(\grave{m}){{(-z)}^{m}}}=z\sum\limits_{m=1}^{\omega }{\hat{m}\widehat{m+z}}.\square</math></div>
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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 04:04, 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