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= Welcome to MWiki =
 
= Welcome to MWiki =
 
== Theorems of the month ==
 
== Theorems of the month ==
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=== Cauchy's integral theorem===
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=== Definition ===
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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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Let <math>f_n^*(z) = f(\eta_nz)</math> <em>sisters</em> of the Taylor series <math>f(z) \in \mathcal{O}(\mathbb{D})</math> centred on 0 on the domain <math>\mathbb{D} \subseteq {}^{\omega}\mathbb{C}</math> where <math>m, n \in {}^{\omega}\mathbb{N}^{*}</math> and <math>\eta_n^m := \underline{1}^{2^{\lceil m/n \rceil}}</math>. Then let <math>\delta_n^*f = \tilde{2}(f - f_n^*)</math> the <em>halved sister distances</em> of <math>f.</math> For <math>\mu_n^m := m!n!/(m + n)!</math>, <math>\mu</math> and <math>\eta</math> form an calculus, which can be resolved on the level of Taylor series and allows an easy and finite closed representation of integrals and derivatives.<math>\triangle</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, 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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=== Representation theorem for integrals ===
  
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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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The Taylor series (see below) <math>f(z) \in \mathcal{O}(\mathbb{D})</math> centred on 0 on <math>\mathbb{D} \subseteq {}^{\omega}\mathbb{C}</math> gives for <math>\grave{m}, n \in {}^{\omega}\mathbb{N}^*</math><div style="text-align:center;"><math>{\uparrow}_0^z...{\uparrow}_0^{\zeta_2}{f(\zeta_1){\downarrow}\zeta_1\;...\;{\downarrow}\zeta_n} = \widetilde{n!} f(z\mu_n) z^n.\square</math></div>
  
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== Recommended readings ==
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=== Representation theorem for derivatives ===
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[http://www.epubli.de/shop/buch/Relil-Boris-Haase-9783844208726/11049 Relil - Religion und Lebensweg]
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For <math>{}^{\widetilde{\nu}}\dot{\mathbb{C}} \subset \mathbb{D} \subseteq {}^{\omega}\mathbb{C},</math> the Taylor series<div style="text-align:center;"><math>f(z):=f(0) + {\LARGE{\textbf{+}}}_{m=1}^{\omega }{\widetilde{m!}\,{{f}^{(m)}}(0){z^m}},</math></div><math>\varepsilon := \tilde{2}^j\tilde{r}, j \in {}^{\omega}\mathbb{Z}, n = \epsilon^{\sigma} \in {}^{\omega}\mathbb{N}^{*}, u :=\epsilon^{\tilde{n} \hat{\underline{\pi}}}</math> and <math>f</math>'s radius of convergence <math>r \in {}^{\nu}{\mathbb{R}}_{&gt;0}</math> imply<div style="text-align:center;"><math>{{f}^{(n)}}(0)=2^{jn}\acute{n}!{\LARGE{\textbf{+}}}_{k=1}^{n}{\delta_n^* f(\tilde{2}^j u^k)}.</math></div>
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==== Proof: ====
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Taylor's theorem<ref name="Remmert">[[w:Reinhold Remmert|<span class="wikipedia">Remmert, Reinhold</span>]]: ''Funktionentheorie 1'' : 3., verb. Aufl.; 1992; Springer; Berlin; ISBN 9783540552338, S. 165 f.</ref> and the properties of the roots of unity.<math>\square</math>
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== Reference ==
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<references />
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== Recommended reading ==
  
 
[https://en.calameo.com/books/003777977258f7b4aa332 Nonstandard Mathematics]
 
[https://en.calameo.com/books/003777977258f7b4aa332 Nonstandard Mathematics]
  
 
[[de:Hauptseite]]
 
[[de:Hauptseite]]

Latest revision as of 23:05, 31 March 2024

Welcome to MWiki

Theorems of the month

Definition

Let [math]\displaystyle{ f_n^*(z) = f(\eta_nz) }[/math] sisters of the Taylor series [math]\displaystyle{ f(z) \in \mathcal{O}(\mathbb{D}) }[/math] centred on 0 on the domain [math]\displaystyle{ \mathbb{D} \subseteq {}^{\omega}\mathbb{C} }[/math] where [math]\displaystyle{ m, n \in {}^{\omega}\mathbb{N}^{*} }[/math] and [math]\displaystyle{ \eta_n^m := \underline{1}^{2^{\lceil m/n \rceil}} }[/math]. Then let [math]\displaystyle{ \delta_n^*f = \tilde{2}(f - f_n^*) }[/math] the halved sister distances of [math]\displaystyle{ f. }[/math] For [math]\displaystyle{ \mu_n^m := m!n!/(m + n)! }[/math], [math]\displaystyle{ \mu }[/math] and [math]\displaystyle{ \eta }[/math] form an calculus, which can be resolved on the level of Taylor series and allows an easy and finite closed representation of integrals and derivatives.[math]\displaystyle{ \triangle }[/math]

Representation theorem for integrals

The Taylor series (see below) [math]\displaystyle{ f(z) \in \mathcal{O}(\mathbb{D}) }[/math] centred on 0 on [math]\displaystyle{ \mathbb{D} \subseteq {}^{\omega}\mathbb{C} }[/math] gives for [math]\displaystyle{ \grave{m}, n \in {}^{\omega}\mathbb{N}^* }[/math]

[math]\displaystyle{ {\uparrow}_0^z...{\uparrow}_0^{\zeta_2}{f(\zeta_1){\downarrow}\zeta_1\;...\;{\downarrow}\zeta_n} = \widetilde{n!} f(z\mu_n) z^n.\square }[/math]

Representation theorem for derivatives

For [math]\displaystyle{ {}^{\widetilde{\nu}}\dot{\mathbb{C}} \subset \mathbb{D} \subseteq {}^{\omega}\mathbb{C}, }[/math] the Taylor series

[math]\displaystyle{ f(z):=f(0) + {\LARGE{\textbf{+}}}_{m=1}^{\omega }{\widetilde{m!}\,{{f}^{(m)}}(0){z^m}}, }[/math]

[math]\displaystyle{ \varepsilon := \tilde{2}^j\tilde{r}, j \in {}^{\omega}\mathbb{Z}, n = \epsilon^{\sigma} \in {}^{\omega}\mathbb{N}^{*}, u :=\epsilon^{\tilde{n} \hat{\underline{\pi}}} }[/math] and [math]\displaystyle{ f }[/math]'s radius of convergence [math]\displaystyle{ r \in {}^{\nu}{\mathbb{R}}_{>0} }[/math] imply

[math]\displaystyle{ {{f}^{(n)}}(0)=2^{jn}\acute{n}!{\LARGE{\textbf{+}}}_{k=1}^{n}{\delta_n^* f(\tilde{2}^j u^k)}. }[/math]

Proof:

Taylor's theorem[1] and the properties of the roots of unity.[math]\displaystyle{ \square }[/math]

Reference

  1. ↑ Remmert, Reinhold: Funktionentheorie 1 : 3., verb. Aufl.; 1992; Springer; Berlin; ISBN 9783540552338, S. 165 f.

Recommended reading

Nonstandard Mathematics

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