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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