Difference between revisions of "Main Page"

From MWiki
Jump to: navigation, search
(Representation theorems for integrals and derivatives)
(70 intermediate revisions by the same user not shown)
Line 1: Line 1:
 
__NOTOC__
 
__NOTOC__
 
= Welcome to MWiki =
 
= Welcome to MWiki =
−
== Theorem of the month ==
+
== Theorems of the month ==
−
=== Fermat's Last Theorem ===
+
=== Definition ===
  
−
For all <math>p \in {}^{\omega }{\mathbb{P}_{\ge 3}}</math> and <math>x, y, z \in {}^{\omega }{\mathbb{N}^{*}}</math>, always <math>x^p + y^p \ne z^p</math> holds and thus for all <math>m \in {}^{\omega }{\mathbb{N}_{\ge 3}}</math> instead of <math>p</math>.
+
Let <math>f_n^*(z) = f(\eta_nz)</math> <em>sisters</em> of the Taylor series <math>f(z) \in \mathcal{O}(\mathbb{D})</math> centred on 0 on the domain <math>\mathbb{D} \subseteq {}^{\omega}\mathbb{C}</math> where <math>m, n \in {}^{\omega}\mathbb{N}^{*}</math> and <math>\eta_n^m := \underline{1}^{2^{\lceil m/n \rceil}}</math>. Then let <math>\delta_n^*f = \tilde{2}(f - f_n^*)</math> the <em>halved sister distances</em> of <math>f.</math> For <math>\mu_n^m := m!n!/(m + n)!</math>, <math>\mu</math> and <math>\eta</math> form an calculus, which can be resolved on the level of Taylor series and allows an easy and finite closed representation of integrals and derivatives.<math>\triangle</math>
  
−
==== Proof: ====
+
=== Representation theorem for integrals ===
−
Because of [[w:Fermat's little theorem|<span class="wikipedia">Fermat's little theorem</span>]], rewritten, <math>f_{akp}(n) := (2n + a - kp)^p - n^p - (n + a)^p \ne 0</math> is to show for <math>a, k, n \in {}^{\omega }{\mathbb{N}^{*}}</math> where <math>kp &lt; n</math>.
+
 
−
<div class="toccolours mw-collapsible mw-collapsed" style="width:100%; overflow:auto;">
+
The Taylor series (see below) <math>f(z) \in \mathcal{O}(\mathbb{D})</math> centred on 0 on <math>\mathbb{D} \subseteq {}^{\omega}\mathbb{C}</math> gives for <math>\grave{m}, n \in {}^{\omega}\mathbb{N}^*</math><div style="text-align:center;"><math>{\uparrow}_0^z...{\uparrow}_0^{\zeta_2}{f(\zeta_1){\downarrow}\zeta_1\;...\;{\downarrow}\zeta_n} = \widetilde{n!} f(z\mu_n) z^n.\square</math></div>
−
<div style="font-weight:bold;line-height:1.6;">Proof details</div>
 
−
<div class="mw-collapsible-content">From <math>x := n, y:= n + a</math> and <math>z := 2n + a + d</math> where <math>d \in {}^{\omega }{\mathbb{N}^{*}}</math>, it follows due to <math>z^p \equiv y, y^p \equiv y</math> and <math>z^p \equiv z</math> first <math>d \equiv 0 \mod p</math>, then <math>d = \pm kp</math>. Since <math>x + y = 2n + a &gt; z</math> is required, <math>f_{akp}(n)</math> is chosen properly.</div></div>
 
  
−
[[w:Mathematical induction|<span class="wikipedia">Induction</span>]] for <math>n</math> implies the claim due to the case <math>m = 4</math><ref name="Ribenboim">[[w:Paulo Ribenboim|<span class="wikipedia">Ribenboim, Paulo</span>]]: ''Thirteen Lectures on Fermat's Last Theorem'' : 1979; Springer; New York; ISBN 9780387904320, p. 35 - 38.</ref> and <math>y &gt; x &gt; p</math><ref name="loccit">loc. cit., p. 226.</ref>:
+
=== Representation theorem for derivatives ===
  
−
'''Induction basis''' <math>(n \le p): f_{akp}(n) \ne 0</math> for all <math>a, k</math> and <math>p</math>. Let <math>r \in {}^{\omega }{\mathbb{N}_{&lt; p}}</math>.
+
For <math>{}^{\widetilde{\nu}}\dot{\mathbb{C}} \subset \mathbb{D} \subseteq {}^{\omega}\mathbb{C},</math> the Taylor series<div style="text-align:center;"><math>f(z):=f(0) + {\LARGE{\textbf{+}}}_{m=1}^{\omega }{\widetilde{m!}\,{{f}^{(m)}}(0){z^m}},</math></div><math>\varepsilon := \tilde{2}^j\tilde{r}, j \in {}^{\omega}\mathbb{Z}, n = \epsilon^{\sigma} \in {}^{\omega}\mathbb{N}^{*}, u :=\epsilon^{\tilde{n} \hat{\underline{\pi}}}</math> and <math>f</math>'s radius of convergence <math>r \in {}^{\nu}{\mathbb{R}}_{&gt;0}</math> imply<div style="text-align:center;"><math>{{f}^{(n)}}(0)=2^{jn}\acute{n}!{\LARGE{\textbf{+}}}_{k=1}^{n}{\delta_n^* f(\tilde{2}^j u^k)}.</math></div>
  
−
'''Induction step''' <math>\,(n = q + r \; \rightarrow \; n^{*} = n + p):</math> Let <math>f_{akp}(n^{*}) \ge 0</math>, but <math>f_{akp}(n) &lt; 0</math>, since <math>f_{akp}(n)</math> is [[w:Monotonic function|<span class="wikipedia">strictly monotonically increasing</span>]] and otherwise nothing to prove.
+
==== Proof: ====
−
<div class="toccolours mw-collapsible mw-collapsed" style="width:100%; overflow:auto;">
+
Taylor's theorem<ref name="Remmert">[[w:Reinhold Remmert|<span class="wikipedia">Remmert, Reinhold</span>]]: ''Funktionentheorie 1'' : 3., verb. Aufl.; 1992; Springer; Berlin; ISBN 9783540552338, S. 165 f.</ref> and the properties of the roots of unity.<math>\square</math>
−
<div style="font-weight:bold;line-height:1.6;">Proof details</div>
 
−
<div class="mw-collapsible-content">The strict monotonicity follows from (continuously) differentiating by <math>n</math> such that <math>f_{akp}(n)' = p(2(2n + a - kp)^{p - 1} - n^{p - 1} - (n + a)^{p - 1}) &gt; 0</math>.</div></div>
 
  
−
It holds <math>f_{akp}(n^{*}) = (\int_0^{n^{*}}{f_{akp}(v)}dv)' \ne 0</math>, since <math>(n^{*})^{p + 1} + (n^{*} + a)^{p + 1}</math> does not divide <math>((n^{*})^p + (n^{*} + a)^p)^2</math> after separating the positive factor as [[w:Polynomial long division|<span class="wikipedia">polynomial division</span>]] shows.<math>\square</math>
+
== Reference ==
−
<div class="toccolours mw-collapsible mw-collapsed" style="width:100%; overflow:auto;">
+
<references />
−
<div style="font-weight:bold;line-height:1.6;">Proof details</div>
 
−
<div class="mw-collapsible-content"><math>\int_0^{n^{*}}{f_{akp}(v)}dv = ((2n^{*} + a - kp)^{p + 1} / 2 - (n^{*})^{p + 1} - (n^{*} + a)^{p + 1})/(p + 1) + t = ((2n^{*} + a - kp)^{(p + 1)/2} \pm \sqrt{2(n^{*})^{p + 1} + 2(n^{*} + a)^{p + 1}})^2/(2p + 2) + t</math> for <math>t \in {}^{\omega}{\mathbb{Q}}</math> where the third binomial formula <math>r^2 - s^2 = (r \pm s)^2 := (r + s)(r - s)</math> was used. After separating the negligible factor <math>\hat{2}((2n^{*} + a - kp)^{(p + 1)/2} + \sqrt{2(n^{*})^{p + 1} + 2(n^{*} + a)^{p + 1}})/(p + 1)</math>, the derivative is just <math>(\hat{2}(2n^{*} + a - kp)^{(p - 1)/2} - \hat{2}((n^{*})^p + (n^{*} + a)^p)/\sqrt{2(n^{*})^{p + 1} + 2(n^{*} + a)^{p + 1}})</math>. After squaring the terms, the polynomial division gives <math>(n^{*})^{p - 1} + (n^{*} + a)^{p - 1} + a^2(n^{*})^{p - 1}(n^{*} + a)^{p - 1}/((n^{*})^{p + 1} + (n^{*} + a)^{p + 1})</math> as recalculating by multiplication confirms.</div></div>
 
  
 
== Recommended reading ==
 
== Recommended reading ==
  
 
[https://en.calameo.com/books/003777977258f7b4aa332 Nonstandard Mathematics]
 
[https://en.calameo.com/books/003777977258f7b4aa332 Nonstandard Mathematics]
−
 
−
== References ==
 
−
<references />
 
  
 
[[de:Hauptseite]]
 
[[de:Hauptseite]]

Revision as of 23:05, 31 March 2024

Welcome to MWiki

Theorems of the month

Definition

Let [math]\displaystyle{ f_n^*(z) = f(\eta_nz) }[/math] sisters of the Taylor series [math]\displaystyle{ f(z) \in \mathcal{O}(\mathbb{D}) }[/math] centred on 0 on the domain [math]\displaystyle{ \mathbb{D} \subseteq {}^{\omega}\mathbb{C} }[/math] where [math]\displaystyle{ m, n \in {}^{\omega}\mathbb{N}^{*} }[/math] and [math]\displaystyle{ \eta_n^m := \underline{1}^{2^{\lceil m/n \rceil}} }[/math]. Then let [math]\displaystyle{ \delta_n^*f = \tilde{2}(f - f_n^*) }[/math] the halved sister distances of [math]\displaystyle{ f. }[/math] For [math]\displaystyle{ \mu_n^m := m!n!/(m + n)! }[/math], [math]\displaystyle{ \mu }[/math] and [math]\displaystyle{ \eta }[/math] form an calculus, which can be resolved on the level of Taylor series and allows an easy and finite closed representation of integrals and derivatives.[math]\displaystyle{ \triangle }[/math]

Representation theorem for integrals

The Taylor series (see below) [math]\displaystyle{ f(z) \in \mathcal{O}(\mathbb{D}) }[/math] centred on 0 on [math]\displaystyle{ \mathbb{D} \subseteq {}^{\omega}\mathbb{C} }[/math] gives for [math]\displaystyle{ \grave{m}, n \in {}^{\omega}\mathbb{N}^* }[/math]

[math]\displaystyle{ {\uparrow}_0^z...{\uparrow}_0^{\zeta_2}{f(\zeta_1){\downarrow}\zeta_1\;...\;{\downarrow}\zeta_n} = \widetilde{n!} f(z\mu_n) z^n.\square }[/math]

Representation theorem for derivatives

For [math]\displaystyle{ {}^{\widetilde{\nu}}\dot{\mathbb{C}} \subset \mathbb{D} \subseteq {}^{\omega}\mathbb{C}, }[/math] the Taylor series

[math]\displaystyle{ f(z):=f(0) + {\LARGE{\textbf{+}}}_{m=1}^{\omega }{\widetilde{m!}\,{{f}^{(m)}}(0){z^m}}, }[/math]

[math]\displaystyle{ \varepsilon := \tilde{2}^j\tilde{r}, j \in {}^{\omega}\mathbb{Z}, n = \epsilon^{\sigma} \in {}^{\omega}\mathbb{N}^{*}, u :=\epsilon^{\tilde{n} \hat{\underline{\pi}}} }[/math] and [math]\displaystyle{ f }[/math]'s radius of convergence [math]\displaystyle{ r \in {}^{\nu}{\mathbb{R}}_{>0} }[/math] imply

[math]\displaystyle{ {{f}^{(n)}}(0)=2^{jn}\acute{n}!{\LARGE{\textbf{+}}}_{k=1}^{n}{\delta_n^* f(\tilde{2}^j u^k)}. }[/math]

Proof:

Taylor's theorem[1] and the properties of the roots of unity.[math]\displaystyle{ \square }[/math]

Reference

  1. ↑ Remmert, Reinhold: Funktionentheorie 1 : 3., verb. Aufl.; 1992; Springer; Berlin; ISBN 9783540552338, S. 165 f.

Recommended reading

Nonstandard Mathematics

Retrieved from "https://en.mwiki.de/index.php?title=Main_Page&oldid=298"