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= Welcome to MWiki =
 
= Welcome to MWiki =
 
== Theorem of the month ==
 
== Theorem of the month ==
=== Leibniz' differentiation rule ===
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=== Finite representation for odd <math>\zeta</math>-arguments ===
  
For <math>f: {}^{(\omega)}\mathbb{K}^{n+1} \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>\frac{\partial }{\partial {{x}_{1}}}\left( \int\limits_{a(x)}^{b(x)}{f(x,t)dDt} \right)=\int\limits_{a(x)}^{b(x)}{\frac{\partial f(x,t)}{\partial {{x}_{1}}}dDt}+\frac{\partial b(x)}{\partial {{x}_{1}}}f(\curvearrowright Bx,b(x))-\frac{\partial a(x)}{\partial {{x}_{1}}}f(\curvearrowright Bx,a(x)).</math></div>
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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>
  
 
==== Proof: ====
 
==== Proof: ====
<div style="text-align:center;"><math>\begin{aligned}\frac{\partial }{\partial {{x}_{1}}}\left( \int\limits_{a(x)}^{b(x)}{f(x,t)dDt} \right) &amp;={\left( \int\limits_{a(\curvearrowright Bx)}^{b(\curvearrowright Bx)}{f(\curvearrowright Bx,t)dDt}-\int\limits_{a(x)}^{b(x)}{f(x,t)dDt} \right)}/{\partial {{x}_{1}}}\;={\left( \int\limits_{a(x)}^{b(x)}{(f(\curvearrowright Bx,t)-f(x,t))dDt}+\int\limits_{b(x)}^{b(\curvearrowright Bx)}{f(\curvearrowright Bx,t)dDt}-\int\limits_{a(x)}^{a(\curvearrowright Bx)}{f(\curvearrowright Bx,t)dDt} \right)}/{\partial {{x}_{1}}}\; \\ &amp;=\int\limits_{a(x)}^{b(x)}{\frac{\partial f(x,t)}{\partial {{x}_{1}}}dDt}+\frac{\partial b(x)}{\partial {{x}_{1}}}f(\curvearrowright Bx,b(x))-\frac{\partial a(x)}{\partial {{x}_{1}}}f(\curvearrowright Bx,a(x)).\square\end{aligned}</math></div>
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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>
  
 
== Recommended reading ==
 
== Recommended reading ==

Revision as of 17:59, 3 March 2020

Welcome to MWiki

Theorem of the month

Finite representation for odd [math]\displaystyle{ \zeta }[/math]-arguments

Using the digamma function [math]\displaystyle{ \psi }[/math], it holds for [math]\displaystyle{ n \in {}^{\omega}2\mathbb{N}^{*} }[/math], small [math]\displaystyle{ \varepsilon \in ]0, 1] }[/math] and [math]\displaystyle{ {{d}_{\varepsilon k n}}:={{\varepsilon}^{{\hat{n}}}}{e}^{\hat{n}2k\pi i} }[/math] that

[math]\displaystyle{ \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]

and

[math]\displaystyle{ \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]

Proof:

The claim results easily via the geometric series from

[math]\displaystyle{ \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]

Recommended reading

Nonstandard Mathematics