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authorJuan Marin Noguera <juan@mnpi.eu>2022-12-04 22:49:17 +0100
committerJuan Marin Noguera <juan@mnpi.eu>2022-12-04 22:49:17 +0100
commitc34b47089a133e58032fe4ea52f61efacaf5f548 (patch)
tree4242772e26a9e7b6f7e02b1d1e00dfbe68981345 /fvv1/n2.lyx
parent214b20d1614b09cd5c18e111df0f0d392af2e721 (diff)
Oops
Diffstat (limited to 'fvv1/n2.lyx')
-rw-r--r--fvv1/n2.lyx16
1 files changed, 8 insertions, 8 deletions
diff --git a/fvv1/n2.lyx b/fvv1/n2.lyx
index 1b761e2..f9a9447 100644
--- a/fvv1/n2.lyx
+++ b/fvv1/n2.lyx
@@ -542,7 +542,7 @@ suma telescópica
se supone lo suficientemente pequeño.
Ahora llamamos
-\begin_inset Formula $\varphi_{i}(t):=f(a+h_{1}\vec{e}_{1}+\dots+h_{i-1}\vec{e}_{i-1}+t\vec{e}_{i})$
+\begin_inset Formula $\varphi_{i}(t)\coloneqq f(a+h_{1}\vec{e}_{1}+\dots+h_{i-1}\vec{e}_{i-1}+t\vec{e}_{i})$
\end_inset
, con lo que
@@ -550,7 +550,7 @@ suma telescópica
\end_inset
, y
-\begin_inset Formula $\Delta_{i}:=\varphi_{i}(h_{i})-\varphi_{i}(0)=f(a+h_{1}\vec{e}_{1}+\dots+h_{i}\vec{e}_{i})-f(a+h_{1}\vec{e}_{1}+\dots+h_{i-1}\vec{e}_{i-1})=\varphi'_{i}(\xi_{i})h_{i}$
+\begin_inset Formula $\Delta_{i}\coloneqq \varphi_{i}(h_{i})-\varphi_{i}(0)=f(a+h_{1}\vec{e}_{1}+\dots+h_{i}\vec{e}_{i})-f(a+h_{1}\vec{e}_{1}+\dots+h_{i-1}\vec{e}_{i-1})=\varphi'_{i}(\xi_{i})h_{i}$
\end_inset
para algún
@@ -662,11 +662,11 @@ regla de la cadena
Demostración:
\series default
Sean
-\begin_inset Formula $L:=df(a):\mathbb{R}^{m}\rightarrow\mathbb{R}^{n}$
+\begin_inset Formula $L\coloneqq df(a):\mathbb{R}^{m}\rightarrow\mathbb{R}^{n}$
\end_inset
y
-\begin_inset Formula $S:=dg(f(a)):\mathbb{R}^{n}\rightarrow\mathbb{R}^{k}$
+\begin_inset Formula $S\coloneqq dg(f(a)):\mathbb{R}^{n}\rightarrow\mathbb{R}^{k}$
\end_inset
, tenemos que
@@ -686,7 +686,7 @@ y queremos ver que
\end_inset
Si llamamos
-\begin_inset Formula $\eta:=f(a+h)-f(a)$
+\begin_inset Formula $\eta\coloneqq f(a+h)-f(a)$
\end_inset
, que tiende a 0 por la continuidad de
@@ -897,17 +897,17 @@ to por abiertos de
\end_inset
y
-\begin_inset Formula $\{B_{i}\}_{i=1}^{k}\mid =\{B(x_{i},\frac{\delta_{x_{i}}}{2})\}_{i=1}^{k}$
+\begin_inset Formula $\{B_{i}\}_{i=1}^{k}\coloneqq \{B(x_{i},\frac{\delta_{x_{i}}}{2})\}_{i=1}^{k}$
\end_inset
un subrecubrimiento finito del que suponemos que no podemos quitar ninguna
bola.
Ahora llamamos
-\begin_inset Formula $x_{0}:=a$
+\begin_inset Formula $x_{0}\coloneqq a$
\end_inset
y
-\begin_inset Formula $x_{k+1}:=b$
+\begin_inset Formula $x_{k+1}\coloneqq b$
\end_inset
y suponemos