Oops

1 files changed, 55 insertions, 55 deletions

diff --git a/fuvr1/n2.lyx b/fuvr1/n2.lyx
index 6312a4f..b046ed1 100644
--- a/fuvr1/n2.lyx
+++ b/fuvr1/n2.lyx
@@ -139,7 +139,7 @@ sucesión
 \end_inset
 
 , con elementos 
-\begin_inset Formula $a_{n}:=\phi(n)$
+\begin_inset Formula $a_{n}\coloneqq \phi(n)$
 \end_inset
 
 .
@@ -369,7 +369,7 @@ intervalo cerrado
 \end_inset
 
  al conjunto 
-\begin_inset Formula $[a,b]:=\{x\in\mathbb{R}\mid a\leq x\leq b\}$
+\begin_inset Formula $[a,b]\coloneqq \{x\in\mathbb{R}\mid a\leq x\leq b\}$
 \end_inset
 
 , 
@@ -377,7 +377,7 @@ intervalo cerrado
 intervalo abierto
 \series default
  a 
-\begin_inset Formula $(a,b):=\{x\in\mathbb{R}\mid a<x<b\}$
+\begin_inset Formula $(a,b)\coloneqq \{x\in\mathbb{R}\mid a<x<b\}$
 \end_inset
 
  e 
@@ -385,11 +385,11 @@ intervalo abierto
 intervalos semiabiertos
 \series default
  por la derecha e izquierda, respectivamente, a 
-\begin_inset Formula $[a,b):=\{x\in\mathbb{R}\mid a\leq x<b\}$
+\begin_inset Formula $[a,b)\coloneqq \{x\in\mathbb{R}\mid a\leq x<b\}$
 \end_inset
 
  y 
-\begin_inset Formula $(a,b]:=\{x\in\mathbb{R}\mid a<x\leq b\}$
+\begin_inset Formula $(a,b]\coloneqq \{x\in\mathbb{R}\mid a<x\leq b\}$
 \end_inset
 
 .
@@ -415,7 +415,7 @@ bola cerrada
 \end_inset
 
  al conjunto 
-\begin_inset Formula $B[x_{0},r]:=\{x\in K\mid |x-x_{0}|\leq r\}$
+\begin_inset Formula $B[x_{0},r]\coloneqq \{x\in K\mid |x-x_{0}|\leq r\}$
 \end_inset
 
 , y 
@@ -423,7 +423,7 @@ bola cerrada
 bola abierta
 \series default
  a 
-\begin_inset Formula $B(x_{0},r):=\{x\in K\mid |x-x_{0}|<r\}$
+\begin_inset Formula $B(x_{0},r)\coloneqq \{x\in K\mid |x-x_{0}|<r\}$
 \end_inset
 
 .
@@ -475,7 +475,7 @@ Demostración:
 
 .
  Sea 
-\begin_inset Formula $n_{0}:=\max\{n_{1},n_{2}\}$
+\begin_inset Formula $n_{0}\coloneqq \max\{n_{1},n_{2}\}$
 \end_inset
 
 , entonces 
@@ -538,7 +538,7 @@ Demostración:
 
 .
  Llamando 
-\begin_inset Formula $M:=\max\{|a_{1}|,\dots,|a_{n_{0}}|,1+|a|\}$
+\begin_inset Formula $M\coloneqq \max\{|a_{1}|,\dots,|a_{n_{0}}|,1+|a|\}$
 \end_inset
 
 , se tiene que 
@@ -682,7 +682,7 @@ Pero entonces, fijado
 \end_inset
 
 , si 
-\begin_inset Formula $n>n_{0}:=\max\{n_{1},n_{2}\}$
+\begin_inset Formula $n>n_{0}\coloneqq \max\{n_{1},n_{2}\}$
 \end_inset
 
 , entonces
@@ -728,7 +728,7 @@ Si tomamos
 \end_inset
 
  tal que 
-\begin_inset Formula $\alpha:=\frac{|b|}{2}<|b_{n}|$
+\begin_inset Formula $\alpha\coloneqq \frac{|b|}{2}<|b_{n}|$
 \end_inset
 
  para 
@@ -770,7 +770,7 @@ Ahora, fijado
 
 .
  Ahora, si 
-\begin_inset Formula $n>n_{0}:=\max\{n_{1},n_{2},n_{\text{3}}\}$
+\begin_inset Formula $n>n_{0}\coloneqq \max\{n_{1},n_{2},n_{\text{3}}\}$
 \end_inset
 
 , entonces
@@ -828,11 +828,11 @@ status open
 
 \begin_layout Plain Layout
 Sean 
-\begin_inset Formula $a:=\lim_{n}a_{n}$
+\begin_inset Formula $a\coloneqq \lim_{n}a_{n}$
 \end_inset
 
  y 
-\begin_inset Formula $b:=\lim_{n}b_{n}$
+\begin_inset Formula $b\coloneqq \lim_{n}b_{n}$
 \end_inset
 
 , y supongamos por reducción al absurdo que 
@@ -841,7 +841,7 @@ Sean
 
 .
  Tomando 
-\begin_inset Formula $\varepsilon:=\frac{a-b}{4}$
+\begin_inset Formula $\varepsilon\coloneqq \frac{a-b}{4}$
 \end_inset
 
 , debería existir 
@@ -1036,7 +1036,7 @@ Demostración:
 \end_inset
 
  es creciente y acotada superiormente, existe 
-\begin_inset Formula $\alpha:=\sup\{a_{n}\}_{n\in\mathbb{N}}$
+\begin_inset Formula $\alpha\coloneqq \sup\{a_{n}\}_{n\in\mathbb{N}}$
 \end_inset
 
 .
@@ -1087,7 +1087,7 @@ A continuación definimos el número
 \end_layout
 
 \begin_layout Enumerate
-\begin_inset Formula $e:=\lim_{n}a_{n}=\lim_{n}b_{n}$
+\begin_inset Formula $e\coloneqq \lim_{n}a_{n}=\lim_{n}b_{n}$
 \end_inset
 
 .
@@ -1374,7 +1374,7 @@ principio de encaje de Cantor
 Demostración:
 \series default
  Sea 
-\begin_inset Formula $I_{n}:=[a_{n},b_{n}]$
+\begin_inset Formula $I_{n}\coloneqq [a_{n},b_{n}]$
 \end_inset
 
 .
@@ -1405,7 +1405,7 @@ Demostración:
 
  converge.
  Si 
-\begin_inset Formula $a:=\lim_{n}a_{n}$
+\begin_inset Formula $a\coloneqq \lim_{n}a_{n}$
 \end_inset
 
  entonces 
@@ -1468,11 +1468,11 @@ subsucesión
 
 .
  Si 
-\begin_inset Formula $(a_{n})_{n\in\mathbb{N}}:=(\phi(n))_{n\in\mathbb{N}}$
+\begin_inset Formula $(a_{n})_{n\in\mathbb{N}}\coloneqq (\phi(n))_{n\in\mathbb{N}}$
 \end_inset
 
 , entonces 
-\begin_inset Formula $(a_{n_{k}})_{k\in\mathbb{N}}:=(\phi\circ\tau(k))_{k\in\mathbb{N}}$
+\begin_inset Formula $(a_{n_{k}})_{k\in\mathbb{N}}\coloneqq (\phi\circ\tau(k))_{k\in\mathbb{N}}$
 \end_inset
 
 .
@@ -1558,11 +1558,11 @@ Demostración:
 
 .
  Sea entonces 
-\begin_inset Formula $I_{0}:=[c_{0},d_{0}]$
+\begin_inset Formula $I_{0}\coloneqq [c_{0},d_{0}]$
 \end_inset
 
  y 
-\begin_inset Formula $m_{0}:=\frac{c_{0}+d_{0}}{2}$
+\begin_inset Formula $m_{0}\coloneqq \frac{c_{0}+d_{0}}{2}$
 \end_inset
 
 .
@@ -1576,7 +1576,7 @@ Demostración:
 
  es infinito.
  Llamamos a este 
-\begin_inset Formula $I_{1}:=[c_{1},d_{1}]$
+\begin_inset Formula $I_{1}\coloneqq [c_{1},d_{1}]$
 \end_inset
 
  y tomamos 
@@ -1593,7 +1593,7 @@ Demostración:
 \end_inset
 
  por 
-\begin_inset Formula $m_{1}:=\frac{c_{1}+d_{1}}{2}$
+\begin_inset Formula $m_{1}\coloneqq \frac{c_{1}+d_{1}}{2}$
 \end_inset
 
  y obtenemos, del mismo modo que antes, 
@@ -1782,7 +1782,7 @@ status open
 \end_inset
 
 Sea 
-\begin_inset Formula $a:=\lim_{n}a_{n}$
+\begin_inset Formula $a\coloneqq \lim_{n}a_{n}$
 \end_inset
 
 .
@@ -1832,7 +1832,7 @@ Primero probamos que una sucesión de Cauchy es acotada: Dado
 \end_inset
 
 y si llamamos 
-\begin_inset Formula $M:=\max\{|a_{1}|,\dots,|a_{n_{0}}|,1+|a_{n_{0}}|\}$
+\begin_inset Formula $M\coloneqq \max\{|a_{1}|,\dots,|a_{n_{0}}|,1+|a_{n_{0}}|\}$
 \end_inset
 
  entonces 
@@ -1923,7 +1923,7 @@ Para
 \end_inset
 
 , definimos 
-\begin_inset Formula $a^{n}:=a\cdots a$
+\begin_inset Formula $a^{n}\coloneqq a\cdots a$
 \end_inset
 
  (
@@ -1936,7 +1936,7 @@ Para
 \end_inset
 
  definiendo 
-\begin_inset Formula $a^{0}:=1$
+\begin_inset Formula $a^{0}\coloneqq 1$
 \end_inset
 
  y 
@@ -1949,7 +1949,7 @@ Para
 
 .
  Con exponentes racionales, se define 
-\begin_inset Formula $a^{\frac{m}{n}}:=\sqrt[n]{a^{m}}$
+\begin_inset Formula $a^{\frac{m}{n}}\coloneqq \sqrt[n]{a^{m}}$
 \end_inset
 
 , y podemos probar fácilmente que si 
@@ -2099,7 +2099,7 @@ Demostración:
 \end_inset
 
  a partir de cierto elemento, y entonces 
-\begin_inset Formula $a^{r_{n}}\leq a^{K}:=M$
+\begin_inset Formula $a^{r_{n}}\leq a^{K}\coloneqq M$
 \end_inset
 
  si 
@@ -2107,7 +2107,7 @@ Demostración:
 \end_inset
 
  o 
-\begin_inset Formula $a^{r_{n}}<a^{0}=1:=M$
+\begin_inset Formula $a^{r_{n}}<a^{0}=1\coloneqq M$
 \end_inset
 
 .
@@ -2156,7 +2156,7 @@ Demostración:
 
 .
  Sea ahora 
-\begin_inset Formula $y:=\lim_{n}a^{r_{n}}$
+\begin_inset Formula $y\coloneqq \lim_{n}a^{r_{n}}$
 \end_inset
 
  y 
@@ -2518,7 +2518,7 @@ status open
 
 \begin_layout Plain Layout
 Sea 
-\begin_inset Formula $x:=\lim_{n}x_{n}$
+\begin_inset Formula $x\coloneqq \lim_{n}x_{n}$
 \end_inset
 
 .
@@ -2703,7 +2703,7 @@ status open
 
 \begin_layout Plain Layout
 Tomamos 
-\begin_inset Formula $b:=\frac{1}{a}>1$
+\begin_inset Formula $b\coloneqq \frac{1}{a}>1$
 \end_inset
 
  y aplicamos el apartado anterior.
@@ -2744,12 +2744,12 @@ Demostración:
 \end_inset
 
  y sea 
-\begin_inset Formula $A:=\{z\in\mathbb{R}\mid a^{z}\leq x\}$
+\begin_inset Formula $A\coloneqq \{z\in\mathbb{R}\mid a^{z}\leq x\}$
 \end_inset
 
 , que sabemos acotado superiormente.
  Sea entonces 
-\begin_inset Formula $y:=\sup A$
+\begin_inset Formula $y\coloneqq \sup A$
 \end_inset
 
  y 
@@ -2812,11 +2812,11 @@ Demostración:
 \end_inset
 
  y sea 
-\begin_inset Formula $a^{\prime}:=\frac{1}{a}>1$
+\begin_inset Formula $a^{\prime}\coloneqq \frac{1}{a}>1$
 \end_inset
 
  y 
-\begin_inset Formula $x^{\prime}:=\frac{1}{x}$
+\begin_inset Formula $x^{\prime}\coloneqq \frac{1}{x}$
 \end_inset
 
 .
@@ -3089,7 +3089,7 @@ status open
 
 \begin_layout Plain Layout
 Sea 
-\begin_inset Formula $x:=\lim_{n}x_{n}>0$
+\begin_inset Formula $x\coloneqq \lim_{n}x_{n}>0$
 \end_inset
 
  y queremos demostrar que 
@@ -3123,7 +3123,7 @@ Sea
 
 .
  Sea 
-\begin_inset Formula $\beta_{n}:=\log_{a}c_{n}$
+\begin_inset Formula $\beta_{n}\coloneqq \log_{a}c_{n}$
 \end_inset
 
  y supongamos que 
@@ -3180,7 +3180,7 @@ Sea
 .
  Podemos suponer que todos son positivos o negativos.
  Pero entonces, para el primer caso, 
-\begin_inset Formula $c_{n_{k}}=a^{\beta_{n_{k}}}>a^{\varepsilon}:=M>a^{0}=1$
+\begin_inset Formula $c_{n_{k}}=a^{\beta_{n_{k}}}>a^{\varepsilon}\coloneqq M>a^{0}=1$
 \end_inset
 
 .
@@ -3189,7 +3189,7 @@ Sea
 \end_inset
 
  y por tanto 
-\begin_inset Formula $c_{n_{k}}=a^{\beta_{n_{k}}}<a^{-\varepsilon}:=M<a^{0}=1$
+\begin_inset Formula $c_{n_{k}}=a^{\beta_{n_{k}}}<a^{-\varepsilon}\coloneqq M<a^{0}=1$
 \end_inset
 
 .
@@ -3693,7 +3693,7 @@ Demostración:
 \end_inset
 
 , tomamos 
-\begin_inset Formula $z_{n}:=\frac{n^{b}}{c^{n}}$
+\begin_inset Formula $z_{n}\coloneqq \frac{n^{b}}{c^{n}}$
 \end_inset
 
  y entonces 
@@ -3823,7 +3823,7 @@ Supongamos
 
 .
  Entonces, si 
-\begin_inset Formula $y_{n}:=\frac{1}{x_{n}}$
+\begin_inset Formula $y_{n}\coloneqq \frac{1}{x_{n}}$
 \end_inset
 
 , 
@@ -3883,7 +3883,7 @@ status open
 
 \begin_layout Plain Layout
 Sea 
-\begin_inset Formula $y_{n}:=e^{x_{n}}-1$
+\begin_inset Formula $y_{n}\coloneqq e^{x_{n}}-1$
 \end_inset
 
 , entonces 
@@ -4083,7 +4083,7 @@ status open
 Demostración:
 \series default
  Sea 
-\begin_inset Formula $L:=\lim_{n}\frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}}$
+\begin_inset Formula $L\coloneqq \lim_{n}\frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}}$
 \end_inset
 
 .
@@ -4318,12 +4318,12 @@ Si
 \end_inset
 
 con 
-\begin_inset Formula $M:=a_{1}\cdots a_{n_{0}}$
+\begin_inset Formula $M\coloneqq a_{1}\cdots a_{n_{0}}$
 \end_inset
 
 .
  Si 
-\begin_inset Formula $\alpha_{n}:=\varepsilon\frac{n-n_{0}}{n}$
+\begin_inset Formula $\alpha_{n}\coloneqq \varepsilon\frac{n-n_{0}}{n}$
 \end_inset
 
 , 
@@ -4420,7 +4420,7 @@ Se tiene que
 
 .
  Sea entonces 
-\begin_inset Formula $A_{n}:=\frac{a_{n}}{a_{n-1}}$
+\begin_inset Formula $A_{n}\coloneqq \frac{a_{n}}{a_{n-1}}$
 \end_inset
 
 , para 
@@ -4620,7 +4620,7 @@ Demostración:
 \end_inset
 
 , 
-\begin_inset Formula $S_{n}:=\lambda A_{n}+\mu B_{n}=\sum_{k=1}^{n}\lambda a_{n}+\sum_{k=1}^{n}\mu b_{n}=\sum_{k=1}^{n}(\lambda a_{n}+\mu b_{n})$
+\begin_inset Formula $S_{n}\coloneqq \lambda A_{n}+\mu B_{n}=\sum_{k=1}^{n}\lambda a_{n}+\sum_{k=1}^{n}\mu b_{n}=\sum_{k=1}^{n}(\lambda a_{n}+\mu b_{n})$
 \end_inset
 
 .
@@ -4701,7 +4701,7 @@ Dadas
 \end_inset
 
  y existe 
-\begin_inset Formula $l:=\lim_{n}\frac{a_{n}}{b_{n}}\in\mathbb{R}\cup\{+\infty\}$
+\begin_inset Formula $l\coloneqq \lim_{n}\frac{a_{n}}{b_{n}}\in\mathbb{R}\cup\{+\infty\}$
 \end_inset
 
 :
@@ -4856,7 +4856,7 @@ Criterio de la raíz:
 \end_inset
 
  y 
-\begin_inset Formula $a:=\lim_{n}\sqrt[n]{a_{n}}\in\mathbb{R}$
+\begin_inset Formula $a\coloneqq \lim_{n}\sqrt[n]{a_{n}}\in\mathbb{R}$
 \end_inset
 
 :
@@ -4963,7 +4963,7 @@ Criterio del cociente:
 \end_inset
 
  y 
-\begin_inset Formula $a:=\lim_{n}\frac{a_{n+1}}{a_{n}}\in\mathbb{R}$
+\begin_inset Formula $a\coloneqq \lim_{n}\frac{a_{n+1}}{a_{n}}\in\mathbb{R}$
 \end_inset
 
 .
@@ -5209,7 +5209,7 @@ Demostración:
 
  es decreciente.
  Definimos la sucesión de intervalos cerrados acotados y encajados 
-\begin_inset Formula $I_{n}:=[S_{2n},S_{2n+1}]$
+\begin_inset Formula $I_{n}\coloneqq [S_{2n},S_{2n+1}]$
 \end_inset
 
 .