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First fundamental theorem of exact differential and integral calculus for line integrals: The function <math>F(z)=\int\limits_{\gamma }{f(\zeta )dB\zeta }</math> where <math>\gamma: [d, x[C \rightarrow A \subseteq {}^{(\omega)}\mathbb{K}, C \subseteq \mathbb{R}, f: A \rightarrow {}^{(\omega)}\mathbb{K}, d \in [a, b[C</math>, and choosing <math>\curvearrowright B \gamma(x) = \gamma(\curvearrowright D x)</math> is exactly <math>B</math>-differentiable, and for all <math>x \in [a, b[C</math> and <math>z = \gamma(x)</math>
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__NOTOC__
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= Welcome to MWiki =
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== Theorems of the month ==
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=== Cauchy's integral 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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<center><math>F' \curvearrowright B(z) = f(z).</math></center>
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=== Fundamental theorem of algebra ===
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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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Proof: <math>dB(F(z))=\int\limits_{t\in [d,x]C}{f(\gamma (t)){{{{\gamma }'}}_{\curvearrowright }}D(t)dDt}-\int\limits_{t\in [d,x[C}{f(\gamma (t)){{{{\gamma }'}}_{\curvearrowright }}D(t)dDt}=\int\limits_{x}{f(\gamma (t))\frac{\gamma (\curvearrowright Dt)-\gamma (t)}{\curvearrowright Dt-t}dDt}=f(\gamma (x)){{{\gamma }'}_{\curvearrowright }}D(x)dDx=\,f(\gamma (x))(\curvearrowright B\gamma (x)-\gamma (x))=f(z)dBz.\square</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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Second fundamental theorem of exact differential and integral calculus for line integrals: According to the conditions from above, we have with <math>\gamma: [a, b[C \rightarrow {}^{(\omega)}\mathbb{K}</math> that
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== Recommended reading ==
  
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[https://en.calameo.com/books/003777977258f7b4aa332 Nonstandard Mathematics]
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<center><math>F(\gamma (b))-F(\gamma (a))=\int\limits_{\gamma }{{{{{F}'}}_{\curvearrowright }}B(\zeta )dB\zeta }.</math></center>
 
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Proof: <math>F(\gamma (b))-F(\gamma (a))=\sum\limits_{t\in [a,b[C}{F(\curvearrowright B\,\gamma (t))}-F(\gamma (t))=\sum\limits_{t\in [a,b[C}{{{{{F}'}}_{\curvearrowright }}B(\gamma (t))(\curvearrowright B\,\gamma (t)-\gamma (t))}=\int\limits_{t\in [a,b[C}{{{{{F}'}}_{\curvearrowright }}B(\gamma (t)){{{{\gamma }'}}_{\curvearrowright }}D(t)dDt}=\int\limits_{\gamma }{{{{{F}'}}_{\curvearrowright }}B(\zeta )dB\zeta }.\square</math>
 
  
 
[[de:Hauptseite]]
 
[[de:Hauptseite]]

Revision as of 18:06, 16 February 2020

Welcome to MWiki

Theorems of the month

Cauchy's integral theorem

Given the neighbourhood relations [math]\displaystyle{ B \subseteq {A}^{2} }[/math] and [math]\displaystyle{ D \subseteq [a, b] }[/math] for some simply connected [math]\displaystyle{ h }[/math]-set [math]\displaystyle{ A \subseteq {}^{\omega}\mathbb{C} }[/math], infinitesimal [math]\displaystyle{ h }[/math], a holomorphic function [math]\displaystyle{ f: A \rightarrow {}^{\omega}\mathbb{C} }[/math] and a closed path [math]\displaystyle{ \gamma: [a, b[\rightarrow \partial A }[/math], choosing [math]\displaystyle{ \curvearrowright B \gamma(t) = \gamma(\curvearrowright D t) }[/math] for [math]\displaystyle{ t \in [a, b[ }[/math], we have that

[math]\displaystyle{ \int\limits_{\gamma }{f(z)dBz}=0. }[/math]

Proof: By the Cauchy-Riemann partial differential equations and Green's theorem, with [math]\displaystyle{ x := \text{Re} \, z, y := \text{Im} \, z, u := \text{Re} \, f, v := \text{Im} \, f }[/math] and [math]\displaystyle{ {A}^{-} := \{z \in A : z + h + ih \in A\} }[/math], we have that

[math]\displaystyle{ \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]

Fundamental theorem of algebra

For every non-constant polynomial [math]\displaystyle{ p \in {}^{(\omega)}\mathbb{C} }[/math], there exists some [math]\displaystyle{ z \in {}^{(\omega)}\mathbb{C} }[/math] such that [math]\displaystyle{ p(z) = 0 }[/math].

Indirect proof: By performing an affine substitution of variables, we can reduce to the case [math]\displaystyle{ 1/p(0) \ne \mathcal{O}(\text{d0}) }[/math]. Suppose that [math]\displaystyle{ p(z) \ne 0 }[/math] for all [math]\displaystyle{ z \in {}^{(\omega)}\mathbb{C} }[/math].

Since [math]\displaystyle{ f(z) := 1/p(z) }[/math] is holomorphic, we have that [math]\displaystyle{ f(1/\text{d0}) = \mathcal{O}(\text{d0}) }[/math]. By the mean value inequality [math]\displaystyle{ |f(0)| \le {|f|}_{\gamma} }[/math] for [math]\displaystyle{ \gamma = \partial\mathbb{B}_{r}(0) }[/math] and arbitrary [math]\displaystyle{ r \in {}^{(\omega)}\mathbb{R}_{>0} }[/math], and hence [math]\displaystyle{ f(0) = \mathcal{O}(\text{d0}) }[/math], which is a contradiction.[math]\displaystyle{ \square }[/math]

Recommended reading

Nonstandard Mathematics

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