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(Fundamental theorems of calculus and approximation theorem)
(Cauchy's integral theorem, fundamental theorem of algebra and Newton’s method)
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= Welcome to MWiki =
 
= Welcome to MWiki =
 
== Theorems of the month ==
 
== Theorems of the month ==
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=== First fundamental theorem of exact differential and integral calculus for <abbr title="line integral">LI</abbr>s ===
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=== Cauchy's integral theorem ===
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The function <math>F(z)={\uparrow}_{\gamma }{f(\zeta ){\downarrow}\zeta }</math> where <math>\gamma: [d, x[ \, \cap \, C \rightarrow A \subseteq {}^{(\omega)}\mathbb{K}, C \subseteq \mathbb{R}, f: A \rightarrow {}^{(\omega)}\mathbb{K}, d \in G = [a, b[ \, \cap \, C</math>, and choosing <math>{}^\curvearrowright \gamma(x) = \gamma({}^\curvearrowright x)</math> is exactly differentiable, and for all <math>x \in G</math> and <math>z = \gamma(x)</math>
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Given the <abbr title="neighbourhood relation">NR</abbr>s <math>B \subseteq {D}^{2}</math> and <math>A \subseteq [a, b]</math> for some <math>h</math>-domain <math>D \subseteq {}^{\omega}\mathbb{C}</math>, infinitesimal <math>h</math>, <math>f \in \mathcal{O}(D)</math> and a <abbr title="closed path">CP</abbr> <math>\gamma: [a, b[\rightarrow \partial D</math>, choosing <math>{}^\curvearrowright \gamma(t) = \gamma({}^\curvearrowright t)</math> for <math>t \in [a, b[</math> gives
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<div style="text-align:center;"><math>{\uparrow}_{\gamma }{f(z){\downarrow}z}=0.</math></div>
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'''Proof:''' By the Cauchy-Riemann 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>{D}^{-} := \{z \in D : z + h + \underline{h} \in D\}</math>, it holds that
 +
<div style="text-align:center;"><math>{\uparrow}_{\gamma }{f(z){\downarrow}z}={\uparrow}_{\gamma }{\left( u+\underline{v} \right)\left( {\downarrow}x+{\downarrow}\underline{y} \right)}={\uparrow}_{z\in {{D}^{-}}}{\left( \left( \tfrac{{\downarrow} \underline{u}}{{\downarrow} x}-\tfrac{{\downarrow} \underline{v}}{{\downarrow} y} \right)-\left( \tfrac{{\downarrow} v}{{\downarrow} x}+\tfrac{{\downarrow} u}{{\downarrow} y} \right) \right){\downarrow}(x,y)}=0.\square</math></div>
  
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=== Fundamental theorem of algebra ===
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Every non-constant polynomial <math>p \in {}^{(\omega)}\mathbb{C}</math> has at least one complex root.
  
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<div style="text-align:center;"><math>F^{\prime}(z) = f(z).</math></div>
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'''Indirect proof:''' By performing an affine substitution of variables, reduce to the case <math>\widetilde{p(0)} \ne \mathcal{O}(\iota)</math>. Suppose that <math>p(z) \ne 0</math> for all <math>z \in {}^{(\omega)}\mathbb{C}</math>.
  
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==== Proof ====
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Since <math>f(z) := \widetilde{p(z)}</math> is holomorphic, it holds that <math>f(\tilde{\iota}) = \mathcal{O}(\iota)</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}(\iota)</math>, which is a contradiction (and hence exactly <math>z(m) = m</math> holds).<math>\square</math>
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<math>{\downarrow}F(z)</math> <math>={\uparrow}_{t\in [d,x] \cap C}{f(\gamma (t)){{\gamma }^{\prime}}(t){\downarrow}t}-{\uparrow}_{t\in [d,x[ \, \cap \, C}{f(\gamma (t)){{\gamma }^{\prime}}(t){\downarrow}t}</math> <math>={\uparrow}_{x}{f(\gamma (t))\tfrac{\gamma ({}^\curvearrowright t)-\gamma (t)}{{}^\curvearrowright t-t}{\downarrow}t}</math> <math>=f(\gamma (x)){{\gamma}^{\prime}}(x){\downarrow}x=</math> <math>\,f(\gamma (x))({}^\curvearrowright\gamma (x)-\gamma (x))</math> <math>=f(z){\downarrow}z.\square</math>
 
  
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=== Second fundamental theorem of exact differential and integral calculus for <abbr title="line integral">LI</abbr>s ===
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=== Newton’s method ===
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Conditions above imply with <math>\gamma: G \rightarrow {}^{(\omega)}\mathbb{K}</math> that
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Demanding above <math>f(\curvearrowright z)=f(z)+f^\prime(z){\downarrow}z=0</math> implies <math>z_{\grave{n}} := z_n-{f^\prime(z_n)}^{-1}f(z_n)</math> if <math>{f^\prime(z_n)}^{-1}</math> is invertible resulting in quadratic convergence close to a zero.<math>\square</math>
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−
 
 
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<div style="text-align:center;"><math>F(\gamma (b))-F(\gamma (a))={\uparrow}_{\gamma }{{F^{\prime}}(\zeta ){\downarrow}\zeta }.</math></div>
 
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==== Proof ====
 
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<math>F(\gamma (b))-F(\gamma (a))</math> <math>={+}_{t\in G}{F({}^\curvearrowright\,\gamma (t))}-F(\gamma (t))</math> <math>={+}_{t\in G}{{{F}^{\prime}}(\gamma (t))({}^\curvearrowright\,\gamma (t)-\gamma (t))}</math> <math>={\uparrow}_{t\in G}{{F^{\prime}}(\gamma (t)){{\gamma }^{\prime}}(t){\downarrow}t}</math> <math>={\uparrow}_{\gamma }{{F_{{}^\curvearrowright }^{\prime}}(\zeta ){\downarrow}\zeta }.\square</math>
 
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=== Approximation theorem ===
 
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The derivatives <math>f^{(s)}(x) \in {}^{\omega}\mathbb{R}</math> for <math>x \in {}^{\omega}\mathbb{R}</math> allow computing the interpolating function <math>g(x) := {+}_{r=0}^{\acute{m}}{\chi_{]x_r, x_{\grave{r}}[}(x)((x_{\grave{r}}-x)p_r(x)+(x-x_r)p_{\grave{r}}(x))/(x_{\grave{r}}-x_r)}+{+}_{r=0}^m{\chi_{\{x_r\}}(x)p_r(x)}</math> for <math>m, n \in {}^{\nu}\mathbb{N}</math> and <math>p_r(x) := {+}_{s=0}^n{f^{(s)}(x_r){(x-x_r)}^s/s!}</math> in <math>\mathcal{O}(\sigma mn)</math> where <math>f^{(s)}(x_r) = g^{(s)}(x_r)</math> holds for every <math>x_r \in {}^{\omega}\mathbb{R}</math>. Replace in the complex case <math>{}^{\omega}\mathbb{R}</math> by <math>{}^{\omega}\mathbb{C}</math> and put <math>x = \gamma(t) \in {}^{\omega}\mathbb{C}</math> for the path <math>\gamma(t)</math> where <math>t \in {}^{\omega}\mathbb{R}.\square</math>
 
  
 
== Recommended readings ==
 
== Recommended readings ==

Revision as of 17:11, 31 January 2024

Welcome to MWiki

Theorems of the month

Cauchy's integral theorem

Given the NRs [math]\displaystyle{ B \subseteq {D}^{2} }[/math] and [math]\displaystyle{ A \subseteq [a, b] }[/math] for some [math]\displaystyle{ h }[/math]-domain [math]\displaystyle{ D \subseteq {}^{\omega}\mathbb{C} }[/math], infinitesimal [math]\displaystyle{ h }[/math], [math]\displaystyle{ f \in \mathcal{O}(D) }[/math] and a CP [math]\displaystyle{ \gamma: [a, b[\rightarrow \partial D }[/math], choosing [math]\displaystyle{ {}^\curvearrowright \gamma(t) = \gamma({}^\curvearrowright t) }[/math] for [math]\displaystyle{ t \in [a, b[ }[/math] gives

[math]\displaystyle{ {\uparrow}_{\gamma }{f(z){\downarrow}z}=0. }[/math]

Proof: By the Cauchy-Riemann 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{ {D}^{-} := \{z \in D : z + h + \underline{h} \in D\} }[/math], it holds that

[math]\displaystyle{ {\uparrow}_{\gamma }{f(z){\downarrow}z}={\uparrow}_{\gamma }{\left( u+\underline{v} \right)\left( {\downarrow}x+{\downarrow}\underline{y} \right)}={\uparrow}_{z\in {{D}^{-}}}{\left( \left( \tfrac{{\downarrow} \underline{u}}{{\downarrow} x}-\tfrac{{\downarrow} \underline{v}}{{\downarrow} y} \right)-\left( \tfrac{{\downarrow} v}{{\downarrow} x}+\tfrac{{\downarrow} u}{{\downarrow} y} \right) \right){\downarrow}(x,y)}=0.\square }[/math]

Fundamental theorem of algebra

Every non-constant polynomial [math]\displaystyle{ p \in {}^{(\omega)}\mathbb{C} }[/math] has at least one complex root.

Indirect proof: By performing an affine substitution of variables, reduce to the case [math]\displaystyle{ \widetilde{p(0)} \ne \mathcal{O}(\iota) }[/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) := \widetilde{p(z)} }[/math] is holomorphic, it holds that [math]\displaystyle{ f(\tilde{\iota}) = \mathcal{O}(\iota) }[/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}(\iota) }[/math], which is a contradiction (and hence exactly [math]\displaystyle{ z(m) = m }[/math] holds).[math]\displaystyle{ \square }[/math]

Newton’s method

Demanding above [math]\displaystyle{ f(\curvearrowright z)=f(z)+f^\prime(z){\downarrow}z=0 }[/math] implies [math]\displaystyle{ z_{\grave{n}} := z_n-{f^\prime(z_n)}^{-1}f(z_n) }[/math] if [math]\displaystyle{ {f^\prime(z_n)}^{-1} }[/math] is invertible resulting in quadratic convergence close to a zero.[math]\displaystyle{ \square }[/math]

Recommended readings

Nonstandard Mathematics

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