diff options
Diffstat (limited to 'fvv1')
| -rw-r--r-- | fvv1/n1.lyx | 32 | ||||
| -rw-r--r-- | fvv1/n2.lyx | 16 | ||||
| -rw-r--r-- | fvv1/n3.lyx | 18 | ||||
| -rw-r--r-- | fvv1/n4.lyx | 4 | 
4 files changed, 35 insertions, 35 deletions
| diff --git a/fvv1/n1.lyx b/fvv1/n1.lyx index e3422b2..8eede13 100644 --- a/fvv1/n1.lyx +++ b/fvv1/n1.lyx @@ -137,7 +137,7 @@ espacio normado  distancia asociada a la norma  \series default   a  -\begin_inset Formula $d(x,y):=\Vert x-y\Vert$ +\begin_inset Formula $d(x,y)\coloneqq \Vert x-y\Vert$  \end_inset  . @@ -150,7 +150,7 @@ Ejemplos de normas en  \end_inset   son las dadas por  -\begin_inset Formula $\Vert(x_{1},\dots,x_{n})\Vert_{p}:=\sqrt[p]{\sum_{i=1}^{n}|x_{i}|^{p}}$ +\begin_inset Formula $\Vert(x_{1},\dots,x_{n})\Vert_{p}\coloneqq \sqrt[p]{\sum_{i=1}^{n}|x_{i}|^{p}}$  \end_inset   y @@ -158,16 +158,16 @@ Ejemplos de normas en  \end_inset -\begin_inset Formula $\Vert(x_{1},\dots,x_{n})\Vert_{\infty}:=\max\{|x_{i}|\}_{i=1}^{n}$ +\begin_inset Formula $\Vert(x_{1},\dots,x_{n})\Vert_{\infty}\coloneqq \max\{|x_{i}|\}_{i=1}^{n}$  \end_inset  .   Además,  -\begin_inset Formula $V:={\cal C}[a,b]:=\{f\mid [a,b]\rightarrow\mathbb{R}\text{ continua}\}$ +\begin_inset Formula $V\coloneqq {\cal C}[a,b]\coloneqq \{f\mid [a,b]\rightarrow\mathbb{R}\text{ continua}\}$  \end_inset   con  -\begin_inset Formula $\Vert f\Vert_{\infty}:=\sup\{|f(x)|\}_{x\in[a,b]}$ +\begin_inset Formula $\Vert f\Vert_{\infty}\coloneqq \sup\{|f(x)|\}_{x\in[a,b]}$  \end_inset   es un espacio normado. @@ -274,7 +274,7 @@ Definimos la norma de una aplicación  \end_inset   como  -\begin_inset Formula $\Vert L\Vert:=\Vert L\Vert_{\Vert\cdot\Vert}^{\Vert\cdot\Vert'}:=\sup\{\Vert L(x)\Vert'\}_{x\in E,\Vert x\Vert\leq1}$ +\begin_inset Formula $\Vert L\Vert\coloneqq \Vert L\Vert_{\Vert\cdot\Vert}^{\Vert\cdot\Vert'}\coloneqq \sup\{\Vert L(x)\Vert'\}_{x\in E,\Vert x\Vert\leq1}$  \end_inset  , y tenemos como  @@ -377,7 +377,7 @@ Veamos primero que  \end_inset  , tomando  -\begin_inset Formula $\delta:=\frac{\varepsilon}{\Vert L\Vert+1}$ +\begin_inset Formula $\delta\coloneqq \frac{\varepsilon}{\Vert L\Vert+1}$  \end_inset   entonces  @@ -423,11 +423,11 @@ Dos normas  Demostración:  \series default   Sean  -\begin_inset Formula $L:=id_{E}:(E,\Vert\cdot\Vert)\rightarrow(E,\Vert\cdot\Vert')$ +\begin_inset Formula $L\coloneqq id_{E}:(E,\Vert\cdot\Vert)\rightarrow(E,\Vert\cdot\Vert')$  \end_inset   y  -\begin_inset Formula $L':=L^{-1}$ +\begin_inset Formula $L'\coloneqq L^{-1}$  \end_inset  , entonces  @@ -472,7 +472,7 @@ Si  \end_inset  , luego  -\begin_inset Formula $\Vert x\Vert'=\Vert L(x)\Vert'\leq\Vert L\Vert\Vert x\Vert\overset{\beta:=\Vert L\Vert}{=}\beta\Vert x\Vert$ +\begin_inset Formula $\Vert x\Vert'=\Vert L(x)\Vert'\leq\Vert L\Vert\Vert x\Vert\overset{\beta\coloneqq \Vert L\Vert}{=}\beta\Vert x\Vert$  \end_inset  . @@ -644,7 +644,7 @@ Demostración:  \end_inset   y tomando  -\begin_inset Formula $\delta:=\varepsilon$ +\begin_inset Formula $\delta\coloneqq \varepsilon$  \end_inset  , si  @@ -706,7 +706,7 @@ teorema  , que es continua por ser composición de dos funciones continuas (la identidad   es continua por la otra cota y la demostración del teorema anterior), entonces -\begin_inset Formula $S:=\{x\in\mathbb{R}^{n}\mid \Vert x\Vert_{1}=1\}$ +\begin_inset Formula $S\coloneqq \{x\in\mathbb{R}^{n}\mid \Vert x\Vert_{1}=1\}$  \end_inset   es cerrado dentro del compacto  @@ -719,7 +719,7 @@ teorema   es compacto y alcanza su máximo y su mínimo.   Sea ahora  -\begin_inset Formula $\mu:=\min\{\Vert x\Vert\}_{x\in S}>0$ +\begin_inset Formula $\mu\coloneqq \min\{\Vert x\Vert\}_{x\in S}>0$  \end_inset   (pues  @@ -1338,7 +1338,7 @@ 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  : @@ -1403,7 +1403,7 @@ 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  : @@ -1449,7 +1449,7 @@ 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  . 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  diff --git a/fvv1/n3.lyx b/fvv1/n3.lyx index 91f5019..b6138cf 100644 --- a/fvv1/n3.lyx +++ b/fvv1/n3.lyx @@ -343,12 +343,12 @@ Demostración:  \end_inset  , y consideramos  -\begin_inset Formula $\Delta_{t,s}:=f(t,s)-f(t,y_{0})-f(x_{0},s)+f(x_{0},y_{0})$ +\begin_inset Formula $\Delta_{t,s}\coloneqq f(t,s)-f(t,y_{0})-f(x_{0},s)+f(x_{0},y_{0})$  \end_inset  .   Si ahora llamamos  -\begin_inset Formula $F_{\overline{s}}(\overline{t}):=f(\overline{t},\overline{s})-f(\overline{t},y_{0})$ +\begin_inset Formula $F_{\overline{s}}(\overline{t})\coloneqq f(\overline{t},\overline{s})-f(\overline{t},y_{0})$  \end_inset  , vemos que  @@ -360,12 +360,12 @@ Demostración:  \end_inset   y que entonces  -\begin_inset Formula $\Delta_{t,s}=F_{s}(t)-F_{s}(x_{0})=F'_{\overline{s}}(\xi_{t,s})(t-x_{0})=\left(\frac{\partial f}{\partial x}(\xi_{t,s},s)-\frac{\partial f}{\partial x}(\xi_{t,s},y_{0})\right)(t-x_{0})\overset{\Phi(\overline{s}):=\frac{\partial f}{\partial x}(\xi_{t,s},\overline{s})}{=}(\Phi(s)-\Phi(y_{0}))(t-x_{0})\overset{\Phi\text{ derivable por hipótesis}}{=}\Phi'(\eta_{t,s})(s-y_{0})(t-x_{0})=\frac{\partial^{2}f}{\partial x\partial y}(\xi_{t,s},\eta_{t,s})(s-y_{0})(t-x_{0})$ +\begin_inset Formula $\Delta_{t,s}=F_{s}(t)-F_{s}(x_{0})=F'_{\overline{s}}(\xi_{t,s})(t-x_{0})=\left(\frac{\partial f}{\partial x}(\xi_{t,s},s)-\frac{\partial f}{\partial x}(\xi_{t,s},y_{0})\right)(t-x_{0})\overset{\Phi(\overline{s})\coloneqq \frac{\partial f}{\partial x}(\xi_{t,s},\overline{s})}{=}(\Phi(s)-\Phi(y_{0}))(t-x_{0})\overset{\Phi\text{ derivable por hipótesis}}{=}\Phi'(\eta_{t,s})(s-y_{0})(t-x_{0})=\frac{\partial^{2}f}{\partial x\partial y}(\xi_{t,s},\eta_{t,s})(s-y_{0})(t-x_{0})$  \end_inset  .   Permutando los papeles de las dos coordenadas (definiendo  -\begin_inset Formula $\sigma_{\overline{t}}(\overline{s}):=f(\overline{t},\overline{s})-f(x,\overline{s})$ +\begin_inset Formula $\sigma_{\overline{t}}(\overline{s})\coloneqq f(\overline{t},\overline{s})-f(x,\overline{s})$  \end_inset  ) obtenemos que  @@ -448,7 +448,7 @@ teorema  Demostración:  \series default   Sea  -\begin_inset Formula $R(h):=f(a+h)-f(a)-df(a)(h)-\frac{1}{2}d^{2}f(a)(h,h)$ +\begin_inset Formula $R(h)\coloneqq f(a+h)-f(a)-df(a)(h)-\frac{1}{2}d^{2}f(a)(h,h)$  \end_inset  , y hemos de ver que  @@ -631,7 +631,7 @@ Si  \end_inset  -espacio vectorial con  -\begin_inset Formula $k:=\dim_{K}(V)<+\infty$ +\begin_inset Formula $k\coloneqq \dim_{K}(V)<+\infty$  \end_inset   y  @@ -751,7 +751,7 @@ Podemos suponer que alcanza un máximo.  \end_inset   definimos  -\begin_inset Formula $\varphi_{i}(t):=f(a_{1},\dots,a_{i-1},t,a_{i+1},\dots,a_{m})$ +\begin_inset Formula $\varphi_{i}(t)\coloneqq f(a_{1},\dots,a_{i-1},t,a_{i+1},\dots,a_{m})$  \end_inset  , fijado  @@ -848,7 +848,7 @@ suponiendo  \end_inset   dada por  -\begin_inset Formula $\Phi(u):=d^{2}f(a)(u,u)=\sum_{i,j}\frac{\partial f}{\partial x_{i}\partial x_{j}}(a)u_{i}u_{j}$ +\begin_inset Formula $\Phi(u)\coloneqq d^{2}f(a)(u,u)=\sum_{i,j}\frac{\partial f}{\partial x_{i}\partial x_{j}}(a)u_{i}u_{j}$  \end_inset   es continua, luego  @@ -914,7 +914,7 @@ Como  \end_inset   y definimos  -\begin_inset Formula $\varphi(t):=a+tu$ +\begin_inset Formula $\varphi(t)\coloneqq a+tu$  \end_inset   como la función  diff --git a/fvv1/n4.lyx b/fvv1/n4.lyx index f95baae..2edff4c 100644 --- a/fvv1/n4.lyx +++ b/fvv1/n4.lyx @@ -341,7 +341,7 @@ gradiente  \end_inset   al vector  -\begin_inset Formula $\nabla f(a):=\left(\frac{\partial f}{\partial x_{1}},\dots,\frac{\partial f}{\partial x_{n}}\right)\in\mathbb{R}^{n}$ +\begin_inset Formula $\nabla f(a)\coloneqq \left(\frac{\partial f}{\partial x_{1}},\dots,\frac{\partial f}{\partial x_{n}}\right)\in\mathbb{R}^{n}$  \end_inset  , la matriz de la diferencial de  @@ -495,7 +495,7 @@ teorema de los multiplicadores de Lagrange  \end_inset  , entonces  -\begin_inset Formula $\nabla f(a)\in\text{span}(\nabla g_{1}(a),\dots,\nabla g_{k}(a)):=<\nabla g_{1}(a),\dots,\nabla g_{k}(a)>$ +\begin_inset Formula $\nabla f(a)\in\text{span}(\nabla g_{1}(a),\dots,\nabla g_{k}(a))\coloneqq <\nabla g_{1}(a),\dots,\nabla g_{k}(a)>$  \end_inset   (el espacio generado por los vectores). | 
