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__NOTOC__
 
__NOTOC__
 
= Welcome to MWiki =
 
= Welcome to MWiki =
== Theorem of the month ==
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== Theorems of the month ==
=== Green's theorem ===
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=== Leibniz' differentiation rule ===
  
Given neighbourhood relations <math>B \subseteq {D}^{2}</math> for some <math>h</math>-domain <math>D \subseteq {}^{(\omega)}\mathbb{R}^{2}</math>, infinitesimal <math>h = |dBx|= |dBy| = |\curvearrowright B \gamma(t) - \gamma(t)| = \mathcal{O}({\hat{\omega}}^{m})</math>, sufficiently large <math>m \in \mathbb{N}^{*}, (x, y) \in D, {D}^{-} := \{(x, y) \in D : (x + h, y + h) \in D\}</math>, and a simply closed path <math>\gamma: [a, b[\rightarrow \partial D</math> followed anticlockwise, choosing <math>\curvearrowright B \gamma(t) = \gamma(\curvearrowright A t)</math> for <math>t \in [a, b[, A \subseteq {[a, b]}^{2}</math>, the following equation holds for sufficiently <math>\alpha</math>-continuous functions <math>u, v: D \rightarrow \mathbb{R}</math> with not necessarily continuous partial derivatives <math>\partial Bu/\partial Bx, \partial Bu/\partial By, \partial Bv/\partial Bx</math> and <math>\partial Bv/\partial By</math>:<div style="text-align:center;"><math>\int\limits_{\gamma }{(u\,dBx+v\,dBy)}=\int\limits_{(x,y)\in {{D}^{-}}}{\left( \frac{\partial Bv}{\partial Bx}-\frac{\partial Bu}{\partial By} \right)dB(x,y)}.</math></div>
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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 B x := {(s, {x}_{2}, ..., {x}_{n})}^{T}</math>, and <math>s \in {}^{(\omega)}\mathbb{K} \setminus \{{x}_{1}\}</math>, choosing <math>\curvearrowright D a(x) = a(\curvearrowright B x)</math> and <math>\curvearrowright D b(x) = b(\curvearrowright B 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}Dt} \right)={\uparrow}_{a(x)}^{b(x)}{\tfrac{{\downarrow} f(x,t)}{{\downarrow} {{x}_{1}}}{\downarrow}Dt}+\tfrac{{\downarrow} b(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright Bx,b(x))-\tfrac{{\downarrow} a(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright Bx,a(x)).</math></div>
  
 
==== Proof: ====
 
==== Proof: ====
Wlog the case <math>D := \{(x, y) : r \le x \le s, f(x) \le y \le g(x)\}, r, s \in {}^{(\omega)}\mathbb{R}, f, g : \partial D \rightarrow {}^{(\omega)}\mathbb{R}</math> is proved, since the proof is analogous for each case rotated by <math>\iota</math>, and every <math>h</math>-domain is a union of such sets. It is simply shown that<div style="text-align:center;"><math>\int\limits_{\gamma }{u\,dBx}=-\int\limits_{(x,y)\in {{D}^{-}}}{\frac{\partial Bu}{\partial By}dB(x,y)}</math></div>since the other relation is given analogously. Since the regions of <math>\gamma</math> where <math>dBx = 0</math> do not contribute to the integral, for negligibly small <math>t := h(u(s, g(s)) - u(r, g(r)))</math>, it holds that<div style="text-align:center;"><math>-\int\limits_{\gamma }{u\,dBx}-t=\int\limits_{r}^{s}{u(x,g(x))dBx}-\int\limits_{r}^{s}{u(x,f(x))dBx}=\int\limits_{r}^{s}{\int\limits_{f(x)}^{g(x)}{\frac{\partial Bu}{\partial By}}dBydBx}=\int\limits_{(x,y)\in {{D}^{-}}}{\frac{\partial Bu}{\partial By}dB(x,y)}.\square</math></div>
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<div style="text-align:center;"><math>\begin{aligned}\tfrac{{\downarrow} }{{\downarrow} {{x}_{1}}}\left( {\uparrow}_{a(x)}^{b(x)}{f(x,t){\downarrow}Dt} \right) &={\left( {\uparrow}_{a(\curvearrowright Bx)}^{b(\curvearrowright Bx)}{f(\curvearrowright Bx,t){\downarrow}Dt}-{\uparrow}_{a(x)}^{b(x)}{f(x,t){\downarrow}Dt} \right)}/{{\downarrow} {{x}_{1}}}\; \\ &={\left( {\uparrow}_{a(x)}^{b(x)}{(f(\curvearrowright Bx,t)-f(x,t)){\downarrow}Dt}+{\uparrow}_{b(x)}^{b(\curvearrowright Bx)}{f(\curvearrowright Bx,t){\downarrow}Dt}-{\uparrow}_{a(x)}^{a(\curvearrowright Bx)}{f(\curvearrowright Bx,t){\downarrow}Dt} \right)}/{{\downarrow} {{x}_{1}}}\; \\ &={\uparrow}_{a(x)}^{b(x)}{\tfrac{{\downarrow} f(x,t)}{{\downarrow} {{x}_{1}}}{\downarrow}Dt}+\tfrac{{\downarrow} b(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright Bx,b(x))-\tfrac{{\downarrow} a(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright Bx,a(x)).\square\end{aligned}</math></div>
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=== Beal's theorem ===
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Equation <math>a^m + b^n = c^k</math> where <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>
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==== Proof: ====
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Proof: Real points between <math>r, s \in {}^{\omega}\mathbb{Q}</math> do not avoid that every nontrivial equation for <math>c^k > 1</math> is given by <math>(a^{m-r} + ib^{n-s})(a^r - ib^s) =c^k +i(a^rb^{n-s} - a^{m-r}b^s)</math> where all relations <math>a^{m-\hat{r}} = b^{n-\hat{s}}</math> then imply <math>p \mid</math> gcd<math>(a, b, c)</math> as well as the claim for <math>p \in {}^{\omega} \mathbb{P}.\square</math>
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=== Conclusion: ===
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The preceding theorem enables an infinite descent because of gcd<math>(a, b, c) > 1</math> such 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>
  
 
== Recommended reading ==
 
== Recommended reading ==

Revision as of 15:58, 1 March 2023

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 B x := {(s, {x}_{2}, ..., {x}_{n})}^{T} }[/math], and [math]\displaystyle{ s \in {}^{(\omega)}\mathbb{K} \setminus \{{x}_{1}\} }[/math], choosing [math]\displaystyle{ \curvearrowright D a(x) = a(\curvearrowright B x) }[/math] and [math]\displaystyle{ \curvearrowright D b(x) = b(\curvearrowright B x) }[/math] it holds that

[math]\displaystyle{ \tfrac{{\downarrow} }{{\downarrow} {{x}_{1}}}\left( {\uparrow}_{a(x)}^{b(x)}{f(x,t){\downarrow}Dt} \right)={\uparrow}_{a(x)}^{b(x)}{\tfrac{{\downarrow} f(x,t)}{{\downarrow} {{x}_{1}}}{\downarrow}Dt}+\tfrac{{\downarrow} b(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright Bx,b(x))-\tfrac{{\downarrow} a(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright Bx,a(x)). }[/math]

Proof:

[math]\displaystyle{ \begin{aligned}\tfrac{{\downarrow} }{{\downarrow} {{x}_{1}}}\left( {\uparrow}_{a(x)}^{b(x)}{f(x,t){\downarrow}Dt} \right) &={\left( {\uparrow}_{a(\curvearrowright Bx)}^{b(\curvearrowright Bx)}{f(\curvearrowright Bx,t){\downarrow}Dt}-{\uparrow}_{a(x)}^{b(x)}{f(x,t){\downarrow}Dt} \right)}/{{\downarrow} {{x}_{1}}}\; \\ &={\left( {\uparrow}_{a(x)}^{b(x)}{(f(\curvearrowright Bx,t)-f(x,t)){\downarrow}Dt}+{\uparrow}_{b(x)}^{b(\curvearrowright Bx)}{f(\curvearrowright Bx,t){\downarrow}Dt}-{\uparrow}_{a(x)}^{a(\curvearrowright Bx)}{f(\curvearrowright Bx,t){\downarrow}Dt} \right)}/{{\downarrow} {{x}_{1}}}\; \\ &={\uparrow}_{a(x)}^{b(x)}{\tfrac{{\downarrow} f(x,t)}{{\downarrow} {{x}_{1}}}{\downarrow}Dt}+\tfrac{{\downarrow} b(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright Bx,b(x))-\tfrac{{\downarrow} a(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright Bx,a(x)).\square\end{aligned} }[/math]

Beal's theorem

Equation [math]\displaystyle{ a^m + b^n = c^k }[/math] where [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:

Proof: Real points between [math]\displaystyle{ r, s \in {}^{\omega}\mathbb{Q} }[/math] do not avoid that every nontrivial equation for [math]\displaystyle{ c^k \gt 1 }[/math] is given by [math]\displaystyle{ (a^{m-r} + ib^{n-s})(a^r - ib^s) =c^k +i(a^rb^{n-s} - a^{m-r}b^s) }[/math] where all relations [math]\displaystyle{ a^{m-\hat{r}} = b^{n-\hat{s}} }[/math] then imply [math]\displaystyle{ p \mid }[/math] gcd[math]\displaystyle{ (a, b, c) }[/math] as well as the claim for [math]\displaystyle{ p \in {}^{\omega} \mathbb{P}.\square }[/math]

Conclusion:

The preceding theorem enables an infinite descent because of gcd[math]\displaystyle{ (a, b, c) \gt 1 }[/math] such 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]

Recommended reading

Nonstandard Mathematics