Revision as of 08:15, 4 March 2024

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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

[math]\displaystyle{ \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]

Proof:

[math]\displaystyle{ \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]

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:

For [math]\displaystyle{ b^n = (c^{kq}-a^{mr})\left(\tilde{c}^{k\acute{q}} + \tilde{a}^{m\acute{r}}\right) = c^k - a^m + c^{kq} \tilde{a}^{m\acute{r}} - \tilde{c}^{k\acute{q}} a^{mr} }[/math], the function [math]\displaystyle{ f(q,r) := c^{k(\hat{q}-1)} - a^{m(\hat{r}-1)} = 0 }[/math] is continuous in [math]\displaystyle{ q, r \in {}^{\omega} \mathbb{R}_{\gt 0} }[/math] and [math]\displaystyle{ (q_0, r_0) = \left(\check{1}, \check{1}\right) }[/math] solves the equation. Every further solution in fractions yields after exponentiation gcd[math]\displaystyle{ (a, c) \gt 1 }[/math] and thus proves the claim.[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]

@@ Line 2: / Line 2: @@
 = Welcome to MWiki =
 == Theorems of the month ==
-=== First fundamental theorem of exact differential and integral calculus for <abbr title="line integral">LI</abbr>s ===
+=== Leibniz' differentiation rule ===
-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>
+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>
-<div style="text-align:center;"><math>F^{\prime}(z) = f(z).</math></div>
+==== 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>
-==== Proof ====
+=== Beal's theorem ===
-<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>
+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>
-=== Second fundamental theorem of exact differential and integral calculus for <abbr title="line integral">LI</abbr>s ===
+==== Proof: ====
-Conditions above imply with <math>\gamma: G \rightarrow {}^{(\omega)}\mathbb{K}</math> that
+For <math>b^n = (c^{kq}-a^{mr})\left(\tilde{c}^{k\acute{q}} + \tilde{a}^{m\acute{r}}\right) = c^k - a^m + c^{kq} \tilde{a}^{m\acute{r}} - \tilde{c}^{k\acute{q}} a^{mr}</math>, the function <math>f(q,r) := c^{k(\hat{q}-1)} - a^{m(\hat{r}-1)} = 0</math> is continuous in <math>q, r \in {}^{\omega} \mathbb{R}_{>0}</math> and <math>(q_0, r_0) = \left(\check{1}, \check{1}\right)</math> solves the equation. Every further solution in fractions yields after exponentiation gcd<math>(a, c) > 1</math> and thus proves the claim.<math>\square</math>
+=== 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>
-<div style="text-align:center;"><math>F(\gamma (b))-F(\gamma (a))={\uparrow}_{\gamma }{{F^{\prime}}(\zeta ){\downarrow}\zeta }.</math></div>
+== Recommended reading ==
-==== Proof ====
-<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>
-=== Approximation theorem ===
-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 ==
 [https://en.calameo.com/books/003777977258f7b4aa332 Nonstandard Mathematics]
 [[de:Hauptseite]]

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Revision as of 08:15, 4 March 2024

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Theorems of the month

Leibniz' differentiation rule

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Beal's theorem

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Conclusion:

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