Euler

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diff --git a/mne/n.lyx b/mne/n.lyx
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--- a/mne/n.lyx
+++ b/mne/n.lyx
@@ -152,5 +152,19 @@ filename "n1.lyx"
 
 \end_layout
 
+\begin_layout Chapter
+Métodos de paso fijo
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "n2.lyx"
+
+\end_inset
+
+
+\end_layout
+
 \end_body
 \end_document
diff --git a/mne/n2.lyx b/mne/n2.lyx
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+++ b/mne/n2.lyx
@@ -0,0 +1,469 @@
+#LyX 2.3 created this file. For more info see http://www.lyx.org/
+\lyxformat 544
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
+\use_default_options true
+\maintain_unincluded_children false
+\language spanish
+\language_package default
+\inputencoding auto
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+\use_package amsmath 1
+\use_package amssymb 1
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+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_minted 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
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+\html_math_output 0
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+\html_be_strict false
+\end_header
+
+\begin_body
+
+\begin_layout Standard
+Dado un problema de valores iniciales
+\begin_inset Formula 
+\[
+\left\{ \begin{aligned}\dot{x} & =f(t,x),\\
+x(a) & =x_{0},
+\end{aligned}
+\right.
+\]
+
+\end_inset
+
+los 
+\series bold
+métodos de paso fijo
+\series default
+ toman un 
+\series bold
+paso
+\series default
+ 
+\begin_inset Formula $h>0$
+\end_inset
+
+ y particionan 
+\begin_inset Formula $[a,b]$
+\end_inset
+
+ con 
+\begin_inset Formula $t_{i}:=a+hi$
+\end_inset
+
+, aunque esto se suele calcular como 
+\begin_inset Formula $t_{0}=a$
+\end_inset
+
+ y 
+\begin_inset Formula $t_{i}=t_{i-1}+h$
+\end_inset
+
+ para 
+\begin_inset Formula $i\geq1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Método de Euler
+\end_layout
+
+\begin_layout Standard
+El 
+\series bold
+método de Euler
+\series default
+ viene dado por 
+\begin_inset Formula $\omega_{0}:=x_{0}$
+\end_inset
+
+ y 
+\begin_inset Formula $\omega_{i+1}:=\omega_{i}+hf(t_{i},\omega_{i})$
+\end_inset
+
+.
+ Para obtener el valor para un 
+\begin_inset Formula $t\in(t_{i-1},t_{i})$
+\end_inset
+
+, podemos interpolar con el propio método, lo que si se hace desde 
+\begin_inset Formula $t_{i-1}$
+\end_inset
+
+ equivale a una interpolación lineal o una interpolación de Newton en los
+ puntos 
+\begin_inset Formula $(t_{i-1},\omega_{i-1})$
+\end_inset
+
+ y 
+\begin_inset Formula $(t_{i},\omega_{i})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Teorema de convergencia del método de Euler:
+\series default
+ Sean 
+\begin_inset Formula $a,b\in\mathbb{R}$
+\end_inset
+
+ con 
+\begin_inset Formula $a<b$
+\end_inset
+
+, 
+\begin_inset Formula $D\subseteq\mathbb{R}^{2}$
+\end_inset
+
+ un abierto conexo, 
+\begin_inset Formula $x_{0}\in\mathbb{R}$
+\end_inset
+
+ con 
+\begin_inset Formula $(a,x_{0})\in D$
+\end_inset
+
+, 
+\begin_inset Formula $f:D\to\mathbb{R}$
+\end_inset
+
+ lipschitziana en 
+\begin_inset Formula $D$
+\end_inset
+
+ en la segunda variable con constante de lipschitzianidad 
+\begin_inset Formula $K\geq0$
+\end_inset
+
+, 
+\begin_inset Formula $n\in\mathbb{N}^{*}$
+\end_inset
+
+, 
+\begin_inset Formula $h:=\frac{b-a}{n}$
+\end_inset
+
+, 
+\begin_inset Formula $x:[a,b]\to\mathbb{R}$
+\end_inset
+
+ una solución de 
+\begin_inset Formula 
+\[
+\left\{ \begin{aligned}\dot{x} & =f(t,x),\\
+x(a) & =x_{0}
+\end{aligned}
+\right.
+\]
+
+\end_inset
+
+con 
+\begin_inset Formula $\ddot{x}(D)$
+\end_inset
+
+ acotada por un 
+\begin_inset Formula $C\geq0$
+\end_inset
+
+, 
+\begin_inset Formula $(t_{i},\omega_{i})_{i=0}^{n}$
+\end_inset
+
+ los puntos dados por el método de Euler con paso 
+\begin_inset Formula $h$
+\end_inset
+
+ para dicho problema con redondeo, dado por
+\begin_inset Formula 
+\[
+\left\{ \begin{aligned}\omega_{0} & :=x_{0}+\delta_{0},\\
+\omega_{i+1} & :=\omega_{i}+hf(t_{i},\omega_{i})+\delta_{i+1},
+\end{aligned}
+\right.
+\]
+
+\end_inset
+
+con cada 
+\begin_inset Formula $|\delta_{i}|<\delta$
+\end_inset
+
+ para un cierto 
+\begin_inset Formula $\delta\geq0$
+\end_inset
+
+, y 
+\begin_inset Formula $x_{i}:=x(t_{i})$
+\end_inset
+
+ para cada 
+\begin_inset Formula $i$
+\end_inset
+
+, entonces
+\begin_inset Formula 
+\[
+\max_{0\leq i\leq n}|x_{i}-\omega_{i}|\leq e^{(b-a)K}\delta+\left(\frac{e^{(b-a)K}-1}{K}\right)\left(\frac{1}{2}Ch+\frac{\delta}{h}\right).
+\]
+
+\end_inset
+
+En particular, sin redondeo el método de Euler tiene una precisión de 
+\begin_inset Formula $O(h)$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Por Taylor, 
+\begin_inset Formula 
+\[
+x_{i+1}=x(t_{i}+h)=x(t_{i})+h\dot{x}(t_{i})+\frac{1}{2}h^{2}\ddot{x}(\xi_{i})=x_{i}+hf(t_{i},x_{i})+\frac{1}{2}h^{2}\ddot{x}(\xi_{i})
+\]
+
+\end_inset
+
+ para algún 
+\begin_inset Formula $\xi_{i}\in[t_{i},t_{i+1}]$
+\end_inset
+
+.
+ Queremos ver que, para 
+\begin_inset Formula $i\in\{0,\dots,n\}$
+\end_inset
+
+,
+\begin_inset Formula 
+\[
+|x_{i}-\omega_{i}|\leq(1+hK)^{i}\delta+\sum_{j=0}^{i-1}(1+hK)^{j}\left(\frac{1}{2}Ch^{2}+\delta\right).
+\]
+
+\end_inset
+
+Para 
+\begin_inset Formula $i=0$
+\end_inset
+
+ esto es obvio, y supuesto esto probado para un 
+\begin_inset Formula $i\in\{0,\dots,n-1\}$
+\end_inset
+
+,
+\begin_inset Formula 
+\begin{align*}
+|y_{i+1}-\omega_{i+1}| & =\left|(y_{i}-\omega_{i})+h(f(t_{i},y_{i})-f(t_{i},\omega_{i}))+\frac{1}{2}h^{2}\ddot{y}(\xi_{i})-\delta_{i+1}\right|\\
+ & \leq|y_{i}-\omega_{i}|+h|f(t_{i},y_{i})-f(t_{i},\omega_{i})|+\frac{1}{2}h^{2}|\ddot{y}(\xi_{i})|+|\delta_{i+1}|\\
+ & \leq(1+hK)|y_{i}-\omega_{i}|+\frac{1}{2}Ch^{2}+\delta\\
+ & \leq(1+hK)\left((1+hK)^{i}\delta+\sum_{j=0}^{i-1}(1+hK)^{j}\left(\frac{1}{2}Ch^{2}+\delta\right)\right)+\frac{1}{2}Ch^{2}+\delta\\
+ & =(1+hK)^{i+1}\delta+\sum_{j=0}^{i}(1+hK)^{j}\left(\frac{1}{2}Ch^{2}+\delta\right).
+\end{align*}
+
+\end_inset
+
+Ahora bien, 
+\begin_inset Formula $\sum_{j=0}^{i}(1+hK)^{j}=\frac{(1+hK)^{i+1}-1}{(1+hK)-1}=\frac{(1+hK)^{i+1}-1}{hK}$
+\end_inset
+
+, y para 
+\begin_inset Formula $t\in\mathbb{R}$
+\end_inset
+
+, existe 
+\begin_inset Formula $\xi$
+\end_inset
+
+ con 
+\begin_inset Formula $e^{t}=1+t+\frac{t^{2}}{2}e^{\xi}\geq1+t$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $(1+t)^{n}\leq e^{nt}$
+\end_inset
+
+, luego en particular 
+\begin_inset Formula $(1+hK)^{i+1}\leq(1+hK)^{n}\leq e^{hKn}=e^{(b-a)K}$
+\end_inset
+
+ y
+\begin_inset Formula 
+\[
+|y_{i}-\omega_{i}|\leq e^{(b-a)K}+\frac{e^{(b-a)K}-1}{hK}\left(\frac{1}{2}Ch^{2}+\delta\right).
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $x:[a,b]\to\mathbb{R}$
+\end_inset
+
+ es de clase 
+\begin_inset Formula ${\cal C}^{3}$
+\end_inset
+
+, 
+\begin_inset Formula $\frac{\partial f}{\partial x}$
+\end_inset
+
+ y 
+\begin_inset Formula $\frac{\partial^{2}f}{\partial x^{2}}$
+\end_inset
+
+ son continuas y 
+\begin_inset Formula $(t_{i},\omega_{i})_{i=0}^{n}$
+\end_inset
+
+ son los puntos dados por el método de Euler en 
+\begin_inset Formula $[a,b]$
+\end_inset
+
+ con paso 
+\begin_inset Formula $h$
+\end_inset
+
+, 
+\begin_inset Formula $x_{i}-\omega_{i}=hD(t_{i})+O(h^{2})$
+\end_inset
+
+, donde 
+\begin_inset Formula $D$
+\end_inset
+
+ es la solución del problema
+\begin_inset Formula 
+\[
+\left\{ \begin{aligned}\dot{D}(t) & =\frac{\partial f}{\partial x}(t,x(t))D(t)+\frac{1}{2}\ddot{x}(t),\\
+D(t_{0}) & =0.
+\end{aligned}
+\right.
+\]
+
+\end_inset
+
+De aquí, si 
+\begin_inset Formula $(t_{i},\xi_{i})_{i=0}^{2n}$
+\end_inset
+
+ son los puntos dados por el método de Euler en 
+\begin_inset Formula $[a,b]$
+\end_inset
+
+ con paso 
+\begin_inset Formula $\frac{h}{2}$
+\end_inset
+
+,
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $|x_{i}-\xi_{2i}|=(\xi_{2i}-\omega_{i})+O(h^{2})$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Como 
+\begin_inset Formula $x_{i}-\omega_{i}=hD(t)+O(h^{2})$
+\end_inset
+
+ y 
+\begin_inset Formula $x_{i}-\xi_{2i}=\frac{h}{2}D(t)+O(h^{2})$
+\end_inset
+
+, despejando, 
+\begin_inset Formula $x_{i}-2\xi_{i}+\omega_{i}=O(h^{2})$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $y_{i}:=2\xi_{2i}-\omega_{i}$
+\end_inset
+
+ es un método de paso fijo 
+\begin_inset Formula $h$
+\end_inset
+
+ de orden 
+\begin_inset Formula $O(h^{2})$
+\end_inset
+
+.
+\end_layout
+
+\end_body
+\end_document