Oops

5 files changed, 35 insertions, 35 deletions

diff --git a/mne/n1.lyx b/mne/n1.lyx
index bc4abd2..fb28c61 100644
--- a/mne/n1.lyx
+++ b/mne/n1.lyx
@@ -182,7 +182,7 @@ Como
 teorema
 \series default
 , si 
-\begin_inset Formula $D:=[a,b]\times\mathbb{R}$
+\begin_inset Formula $D\coloneqq [a,b]\times\mathbb{R}$
 \end_inset
 
 , 
@@ -301,7 +301,7 @@ Si el polinomio interpolador de
 \end_inset
 
 , llamamos 
-\begin_inset Formula $f[x_{0},\dots,x_{n}]:=a_{n}$
+\begin_inset Formula $f[x_{0},\dots,x_{n}]\coloneqq a_{n}$
 \end_inset
 
 .
diff --git a/mne/n2.lyx b/mne/n2.lyx
index 44b1b3a..8e04b88 100644
--- a/mne/n2.lyx
+++ b/mne/n2.lyx
@@ -106,7 +106,7 @@ paso
 \end_inset
 
  con 
-\begin_inset Formula $t_{i}:=a+hi$
+\begin_inset Formula $t_{i}\coloneqq a+hi$
 \end_inset
 
 , aunque esto se suele calcular como 
@@ -134,11 +134,11 @@ El
 método de Euler
 \series default
  viene dado por 
-\begin_inset Formula $\omega_{0}:=x_{0}$
+\begin_inset Formula $\omega_{0}\coloneqq x_{0}$
 \end_inset
 
  y 
-\begin_inset Formula $\omega_{i+1}:=\omega_{i}+hf(t_{i},\omega_{i})$
+\begin_inset Formula $\omega_{i+1}\coloneqq \omega_{i}+hf(t_{i},\omega_{i})$
 \end_inset
 
 .
@@ -204,7 +204,7 @@ Teorema de convergencia del método de Euler:
 \end_inset
 
 , 
-\begin_inset Formula $h:=\frac{b-a}{n}$
+\begin_inset Formula $h\coloneqq \frac{b-a}{n}$
 \end_inset
 
 , 
@@ -241,7 +241,7 @@ con
  para dicho problema con redondeo, dado por
 \begin_inset Formula 
 \[
-\left\{ \begin{aligned}\omega_{0} & \mid =x_{0}+\delta_{0},\\
+\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.
@@ -258,7 +258,7 @@ con cada
 \end_inset
 
 , y 
-\begin_inset Formula $x_{i}:=x(t_{i})$
+\begin_inset Formula $x_{i}\coloneqq x(t_{i})$
 \end_inset
 
  para cada 
@@ -451,7 +451,7 @@ Como
 
 \end_deeper
 \begin_layout Enumerate
-\begin_inset Formula $y_{i}:=2\xi_{2i}-\omega_{i}$
+\begin_inset Formula $y_{i}\coloneqq 2\xi_{2i}-\omega_{i}$
 \end_inset
 
  es un método de paso fijo 
@@ -517,11 +517,11 @@ El método de Euler es el método de Taylor de orden 1.
 
 \begin_layout Standard
 Dado un método de paso fijo de la forma 
-\begin_inset Formula $\omega_{0}:=\alpha$
+\begin_inset Formula $\omega_{0}\coloneqq \alpha$
 \end_inset
 
 , 
-\begin_inset Formula $\omega_{i+1}:=\omega_{i}+hØ(t_{i},\omega_{i})$
+\begin_inset Formula $\omega_{i+1}\coloneqq \omega_{i}+hØ(t_{i},\omega_{i})$
 \end_inset
 
 , llamamos 
diff --git a/mne/n3.lyx b/mne/n3.lyx
index 2c12115..507713a 100644
--- a/mne/n3.lyx
+++ b/mne/n3.lyx
@@ -222,7 +222,7 @@ teorema
 \end_inset
 
 , sean 
-\begin_inset Formula $h:=h_{i}$
+\begin_inset Formula $h\coloneqq h_{i}$
 \end_inset
 
  y 
@@ -338,7 +338,7 @@ Dar un paso con
 
 \begin_layout Enumerate
 Obtener el error 
-\begin_inset Formula $E:=\frac{2^{k}}{2^{k}-1}\Vert Y-\omega_{i+1}\Vert\approx\Vert z_{i}(t_{i+1})-\omega_{i+1}\Vert$
+\begin_inset Formula $E\coloneqq \frac{2^{k}}{2^{k}-1}\Vert Y-\omega_{i+1}\Vert\approx\Vert z_{i}(t_{i+1})-\omega_{i+1}\Vert$
 \end_inset
 
 .
@@ -374,11 +374,11 @@ Para ajustar el paso:
 
 \begin_layout Enumerate
 Calcular 
-\begin_inset Formula $q:=\left(\frac{\varepsilon h}{2E}\right)^{1/k}$
+\begin_inset Formula $q\coloneqq \left(\frac{\varepsilon h}{2E}\right)^{1/k}$
 \end_inset
 
  y 
-\begin_inset Formula $q':=\min\{4,\max\{0.1,q\}\}$
+\begin_inset Formula $q'\coloneqq \min\{4,\max\{0.1,q\}\}$
 \end_inset
 
 , y hacer 
diff --git a/mne/n4.lyx b/mne/n4.lyx
index 9aed707..c480167 100644
--- a/mne/n4.lyx
+++ b/mne/n4.lyx
@@ -321,7 +321,7 @@ Consideremos un método multipaso de paso fijo que, para un problema en un
 \end_inset
 
  dada por 
-\begin_inset Formula $\tau(h):=\max_{i=0}^{n_{h}}\Vert\tau_{i}(h)\Vert$
+\begin_inset Formula $\tau(h)\coloneqq \max_{i=0}^{n_{h}}\Vert\tau_{i}(h)\Vert$
 \end_inset
 
 , el método es 
@@ -385,7 +385,7 @@ estable
 \end_inset
 
 , sea 
-\begin_inset Formula $(t_{i},\omega_{i})_{i=0}^{n}:=(t_{hi},\omega_{hi})_{i=0}^{n_{h}}$
+\begin_inset Formula $(t_{i},\omega_{i})_{i=0}^{n}\coloneqq (t_{hi},\omega_{hi})_{i=0}^{n_{h}}$
 \end_inset
 
 , si se puede generar una solución 
@@ -393,7 +393,7 @@ estable
 \end_inset
 
  con 
-\begin_inset Formula $\tilde{\omega}_{i}:=\omega_{i}$
+\begin_inset Formula $\tilde{\omega}_{i}\coloneqq \omega_{i}$
 \end_inset
 
  para 
@@ -468,7 +468,7 @@ Demostración:
 \end_inset
 
  los coeficientes del método y 
-\begin_inset Formula $\varepsilon_{i}:=h\tau_{i}(h)$
+\begin_inset Formula $\varepsilon_{i}\coloneqq h\tau_{i}(h)$
 \end_inset
 
 , como 
@@ -561,11 +561,11 @@ teorema
 \end_inset
 
  dado por 
-\begin_inset Formula $\omega_{0}:=x(t_{0})$
+\begin_inset Formula $\omega_{0}\coloneqq x(t_{0})$
 \end_inset
 
  y 
-\begin_inset Formula $\omega_{i+1}:=\omega_{i}+hØ(t_{i},\omega_{i},h)$
+\begin_inset Formula $\omega_{i+1}\coloneqq \omega_{i}+hØ(t_{i},\omega_{i},h)$
 \end_inset
 
  con 
@@ -594,11 +594,11 @@ Fijado
 \end_inset
 
  dados por 
-\begin_inset Formula $\omega_{i+1}:=\omega_{i}+hØ(t_{i},\omega_{i},h)$
+\begin_inset Formula $\omega_{i+1}\coloneqq \omega_{i}+hØ(t_{i},\omega_{i},h)$
 \end_inset
 
  y 
-\begin_inset Formula $\tilde{\omega}_{i+1}:=\tilde{\omega}_{i}+hØ(t_{i},\tilde{\omega}_{i},h)+\varepsilon_{i+1}$
+\begin_inset Formula $\tilde{\omega}_{i+1}\coloneqq \tilde{\omega}_{i}+hØ(t_{i},\tilde{\omega}_{i},h)+\varepsilon_{i+1}$
 \end_inset
 
  para ciertos 
@@ -641,7 +641,7 @@ Con esto, como
 \end_inset
 
 , llamando 
-\begin_inset Formula $M:=(1+hL)^{n}$
+\begin_inset Formula $M\coloneqq (1+hL)^{n}$
 \end_inset
 
 , 
@@ -806,7 +806,7 @@ donde los
 polinomio característico
 \series default
  de la ecuación es 
-\begin_inset Formula $P(\lambda):=\lambda^{m}-a_{m-1}\lambda^{m-1}-\dots-a_{1}\lambda-a_{0}$
+\begin_inset Formula $P(\lambda)\coloneqq \lambda^{m}-a_{m-1}\lambda^{m-1}-\dots-a_{1}\lambda-a_{0}$
 \end_inset
 
 .
@@ -840,7 +840,7 @@ Dados un método multipaso de paso fijo
 \end_inset
 
 y 
-\begin_inset Formula $\omega_{i}:=\alpha_{i}$
+\begin_inset Formula $\omega_{i}\coloneqq \alpha_{i}$
 \end_inset
 
  para 
@@ -950,11 +950,11 @@ begin{sloppypar}
 \end_inset
 
 Dados un método implícito 
-\begin_inset Formula $\omega_{i}:=F(t_{i},h,\omega_{i-1},\dots,\omega_{i-m})$
+\begin_inset Formula $\omega_{i}\coloneqq F(t_{i},h,\omega_{i-1},\dots,\omega_{i-m})$
 \end_inset
 
  y uno explícito 
-\begin_inset Formula $\omega_{i}:=G(t_{i},h,\omega_{i},\dots,\omega_{i-m})$
+\begin_inset Formula $\omega_{i}\coloneqq G(t_{i},h,\omega_{i},\dots,\omega_{i-m})$
 \end_inset
 
 , el 
@@ -1072,7 +1072,7 @@ Sean
 \begin_layout Standard
 El método es de paso variable, ajustando el paso como en los métodos de
  paso fijo pero con error 
-\begin_inset Formula $E:=\frac{19}{270}\Vert\beta_{i}-\omega_{i}\Vert$
+\begin_inset Formula $E\coloneqq \frac{19}{270}\Vert\beta_{i}-\omega_{i}\Vert$
 \end_inset
 
 .
diff --git a/mne/n5.lyx b/mne/n5.lyx
index 79f6ad9..b771594 100644
--- a/mne/n5.lyx
+++ b/mne/n5.lyx
@@ -186,7 +186,7 @@ En general, con los métodos de un paso fijo en
 \end_inset
 
 , esto es 
-\begin_inset Formula $Q(x):=\sum_{i=0}^{n}\frac{x^{i}}{i!}$
+\begin_inset Formula $Q(x)\coloneqq \sum_{i=0}^{n}\frac{x^{i}}{i!}$
 \end_inset
 
 .
@@ -362,7 +362,7 @@ El polinomio
 \end_inset
 
  tiene como única raíz 
-\begin_inset Formula $\beta:=\frac{1}{1-h\lambda}$
+\begin_inset Formula $\beta\coloneqq \frac{1}{1-h\lambda}$
 \end_inset
 
 , y si 
@@ -383,7 +383,7 @@ El polinomio
 \end_deeper
 \begin_layout Standard
 Para implementarlo, sea 
-\begin_inset Formula $F(\omega):=\omega-\omega_{i-1}-hf(t_{i},\omega)$
+\begin_inset Formula $F(\omega)\coloneqq \omega-\omega_{i-1}-hf(t_{i},\omega)$
 \end_inset
 
 , se trata de resolver 
@@ -400,7 +400,7 @@ Para implementarlo, sea
 \end_inset
 
 , dada por 
-\begin_inset Formula $\omega_{i}^{0}:=\omega_{i-1}$
+\begin_inset Formula $\omega_{i}^{0}\coloneqq \omega_{i-1}$
 \end_inset
 
  y
@@ -549,11 +549,11 @@ Dado un método a
 \end_inset
 
 llamamos 
-\begin_inset Formula $\rho(z):=z^{m}-a_{m-1}z^{m-1}-\dots-a_{1}z-a_{0}$
+\begin_inset Formula $\rho(z)\coloneqq z^{m}-a_{m-1}z^{m-1}-\dots-a_{1}z-a_{0}$
 \end_inset
 
  y 
-\begin_inset Formula $\sigma(z):=b_{m}z^{m}+\dots+b_{1}z+b_{0}$
+\begin_inset Formula $\sigma(z)\coloneqq b_{m}z^{m}+\dots+b_{1}z+b_{0}$
 \end_inset
 
 .