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Borishaase (talk | contribs) (Cauchy's integral theorem, fundamental theorem of algebra and Newton’s method) |
Borishaase (talk | contribs) (Leibniz' differentiation rule and Beal's theorem) |
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= Welcome to MWiki = | = Welcome to MWiki = | ||
== Theorems of the month == | == Theorems of the month == | ||
− | === | + | === Leibniz' differentiation rule === |
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− | === | + | For <math>f: {}^{(\omega)}\mathbb{K}^{\grave{n}} \rightarrow {}^{(\omega)}\mathbb{K}, a, b: {}^{(\omega)}\mathbb{K}^{n} \rightarrow {}^{(\omega)}\mathbb{K}, \curvearrowright x := {(s, {x}_{2}, ..., {x}_{n})}^{T}</math> and <math>s \in {}^{(\omega)}\mathbb{K} \setminus \{{x}_{1}\}</math>, choosing <math>\curvearrowright a(x) = a(\curvearrowright x)</math> and <math>\curvearrowright b(x) = b(\curvearrowright x)</math>, it holds that<div style="text-align:center;"><math>\tfrac{{\downarrow} }{{\downarrow} {{x}_{1}}}\left( {\uparrow}_{a(x)}^{b(x)}{f(x,t){\downarrow}t} \right)={\uparrow}_{a(x)}^{b(x)}{\tfrac{{\downarrow} f(x,t)}{{\downarrow} {{x}_{1}}}{\downarrow}t}+\tfrac{{\downarrow} b(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright x,b(x))-\tfrac{{\downarrow} a(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright x,a(x)).</math></div> |
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− | + | ==== Proof: ==== | |
+ | <div style="text-align:center;"><math>\begin{aligned}\tfrac{{\downarrow} }{{\downarrow} {{x}_{1}}}\left( {\uparrow}_{a(x)}^{b(x)}{f(x,t){\downarrow}t} \right) &={\left( {\uparrow}_{a(\curvearrowright x)}^{b(\curvearrowright x)}{f(\curvearrowright x,t){\downarrow}t}-{\uparrow}_{a(x)}^{b(x)}{f(x,t){\downarrow}t} \right)}/{{\downarrow} {{x}_{1}}}\; \\ &={\left( {\uparrow}_{a(x)}^{b(x)}{(f(\curvearrowright x,t)-f(x,t)){\downarrow}t}+{\uparrow}_{b(x)}^{b(\curvearrowright x)}{f(\curvearrowright x,t){\downarrow}t}-{\uparrow}_{a(x)}^{a(\curvearrowright x)}{f(\curvearrowright x,t){\downarrow}t} \right)}/{{\downarrow} {{x}_{1}}}\; \\ &={\uparrow}_{a(x)}^{b(x)}{\tfrac{{\downarrow} f(x,t)}{{\downarrow} {{x}_{1}}}{\downarrow}t}+\tfrac{{\downarrow} b(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright x,b(x))-\tfrac{{\downarrow} a(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright x,a(x)).\square\end{aligned}</math></div> | ||
− | + | === Beal's theorem === | |
+ | Equation <math>a^m + b^n = c^k</math> for <math>a, b, c \in \mathbb{N}^{*}</math> and <math>k, m, n \in \mathbb{N}_{\ge 3}</math> implies gcd<math>(a, b, c) > 1.</math> | ||
− | === | + | ==== Proof: ==== |
− | + | From <math>b^n = (c^{k-r} – a^m)(c^r + 1) = c^k – a^m + c^{k-r} – a^mc^r</math>, it follows that <math>a^m =c^{k-\hat{r}}</math> and <math>r \in {}^{\omega}\mathbb{Q}_{>0}</math> by the Gelfond-Schneider theorem, which proves the claim with gcd<math>(a,c)>1</math> after exponentiation.<math>\square</math> | |
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+ | === Conclusion: === | ||
+ | The Fermat-Catalan conjecture can be proven analogously and an infinite descent implies because of gcd<math>(a, b, c) > 1</math> that no <math>n \in {}^{\omega}\mathbb{N}_{\ge 3}</math> satisfies <math>a^n + b^n = c^n</math> for arbitrary <math>a, b, c \in {}^{\omega}\mathbb{N}^{*}.\square</math> | ||
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+ | == Recommended reading == | ||
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[https://en.calameo.com/books/003777977258f7b4aa332 Nonstandard Mathematics] | [https://en.calameo.com/books/003777977258f7b4aa332 Nonstandard Mathematics] | ||
[[de:Hauptseite]] | [[de:Hauptseite]] |
Revision as of 18:39, 29 February 2024
Welcome to MWiki
Theorems of the month
Leibniz' differentiation rule
For [math]\displaystyle{ f: {}^{(\omega)}\mathbb{K}^{\grave{n}} \rightarrow {}^{(\omega)}\mathbb{K}, a, b: {}^{(\omega)}\mathbb{K}^{n} \rightarrow {}^{(\omega)}\mathbb{K}, \curvearrowright x := {(s, {x}_{2}, ..., {x}_{n})}^{T} }[/math] and [math]\displaystyle{ s \in {}^{(\omega)}\mathbb{K} \setminus \{{x}_{1}\} }[/math], choosing [math]\displaystyle{ \curvearrowright a(x) = a(\curvearrowright x) }[/math] and [math]\displaystyle{ \curvearrowright b(x) = b(\curvearrowright x) }[/math], it holds that
Proof:
Beal's theorem
Equation [math]\displaystyle{ a^m + b^n = c^k }[/math] for [math]\displaystyle{ a, b, c \in \mathbb{N}^{*} }[/math] and [math]\displaystyle{ k, m, n \in \mathbb{N}_{\ge 3} }[/math] implies gcd[math]\displaystyle{ (a, b, c) \gt 1. }[/math]
Proof:
From [math]\displaystyle{ b^n = (c^{k-r} – a^m)(c^r + 1) = c^k – a^m + c^{k-r} – a^mc^r }[/math], it follows that [math]\displaystyle{ a^m =c^{k-\hat{r}} }[/math] and [math]\displaystyle{ r \in {}^{\omega}\mathbb{Q}_{\gt 0} }[/math] by the Gelfond-Schneider theorem, which proves the claim with gcd[math]\displaystyle{ (a,c)\gt 1 }[/math] after exponentiation.[math]\displaystyle{ \square }[/math]
Conclusion:
The Fermat-Catalan conjecture can be proven analogously and an infinite descent implies because of gcd[math]\displaystyle{ (a, b, c) \gt 1 }[/math] that no [math]\displaystyle{ n \in {}^{\omega}\mathbb{N}_{\ge 3} }[/math] satisfies [math]\displaystyle{ a^n + b^n = c^n }[/math] for arbitrary [math]\displaystyle{ a, b, c \in {}^{\omega}\mathbb{N}^{*}.\square }[/math]