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Sesión 7: Otros símbolos

Aprenderás a crear nuevos símbolos matemáticos usando código fuente LaTeX.


Galería de ecuaciones

En los siguientes ejemplos se muestra el código primero y después el resultado que usted verá en el documento.

Estas ecuaciones se incluyen para que pueda tener una mejor idea de qué cosas se pueden hacer usando LaTeX.

Si $P$, $Q$ y $R$ son tres puntos entonces, 
$\overrightarrow{PQ} + \overrightarrow{QR} = \overrightarrow{PR}$.

 \begin{minipage}{10cm} Si $P$, $Q$ y $R$ son tres puntos entonces,  $\overrightarrow{PQ} + \overrightarrow{QR} = \overrightarrow{PR}$.  \end{minipage}

\begin{equation*}
   \det(\mathbf{AB}) = \det(\mathbf{A})\cdot\det(\mathbf{B})
\end{equation*}

     \begin{minipage}{10cm} \begin{equation*} \det(\mathbf{AB}) = \det(\mathbf{A})\cdot\det(\mathbf{B}) \end{equation*} \end{minipage}

\begin{equation*}
\delta_{ij} = \left\{
\begin{array}{ll}
1 & \mbox{ si } i = j\\
0 & \mbox{ si } i \neq j
\end{array}
\right.
\end{equation*}

     \begin{equation*} \delta_{ij} = \left\{ \begin{array}{ll} 1 & \mbox{ si } i = j\\ 0 & \mbox{ si } i \neq j \end{array} \right. \end{equation*}

Por definición, el coeficiente de correlación de Pearson es: 
%
\begin{equation*}
r = \displaystyle\frac{S_{xy}}{\sqrt{S_{xx}\,S_{yy}}}
\end{equation*}
%

     \begin{minipage}{10cm} Por definici\'on, el coeficiente de correlaci\'on de Pearson es:  % \begin{equation*} r = \displaystyle\frac{S_{xy}}{\sqrt{S_{xx}\,S_{yy}}} \end{equation*} % \end{minipage}

Si consideramos los vectores:
\begin{equation*}
\vec{x} = \left(
	\begin{array}{c}
	x_1 - \bar{x}\\
	x_2 - \bar{x}\\
	\vdots\\
	x_n - \bar{x}
	\end{array}
\right)
\qquad\mbox{ y }\qquad
\vec{y} = \left(
	\begin{array}{c}
	y_1 - \bar{y}\\
	y_2 - \bar{y}\\
	\vdots\\
	y_n - \bar{y}
	\end{array}
\right)
\end{equation*}

     \begin{minipage}{10cm} Si consideramos los vectores: \begin{equation*} \vec{x} = \left( \begin{array}{c} x_1 - \bar{x}\\ x_2 - \bar{x}\\ \vdots\\ x_n - \bar{x} \end{array} \right) \qquad\mbox{ y }\qquad \vec{y} = \left( \begin{array}{c} y_1 - \bar{y}\\ y_2 - \bar{y}\\ \vdots\\ y_n - \bar{y} \end{array} \right) \end{equation*} \end{minipage}

\begin{equation*}
\cos\theta = \displaystyle
	  \frac{\vec{x}\cdot\vec{y}}{\|\vec{x}\|\cdot \|\vec{y}\|} 
	= \frac{\sum\limits_{i=1}^{n}{(x_i - \bar{x})(y_i - \bar{y})}}
	{\sqrt{\sum\limits_{i=1}^{n}{(x_i - \bar{x})^2}}
	\sqrt{\sum\limits_{i=1}^{n}{(y_i - \bar{y})^2}}} = r
\end{equation*}

     \begin{minipage}{10cm} \begin{equation*} \cos\theta = \displaystyle\frac{\vec{x}\cdot \vec{y}}{\|\vec{x}\|\cdot \|\vec{y}\|}  	= \frac{\sum\limits_{i=1}^{n}{(x_i - \bar{x})(y_i - \bar{y})}}{ 	\sqrt{\sum\limits_{i=1}^{n}{(x_i - \bar{x})^2}} 	\sqrt{\sum\limits_{i=1}^{n}{(y_i - \bar{y})^2}}} = r \end{equation*} \end{minipage}

\begin{equation*}
\frac{d^2y}{dx^2} + p(x)\cdot\frac{dy}{dx} + q(x)\cdot y = 0
\end{equation*}

     \begin{minipage}{10cm} \begin{equation*} \frac{d^2y}{dx^2} + p(x)\cdot\frac{dy}{dx} + q(x)\cdot y = 0 \end{equation*} \end{minipage}

Una serie de potencias tiene la forma:
\begin{equation*}
y = \sum_{i=0}^{\infty}{c_i\,x^{i}}
\end{equation*}

     \begin{minipage}{10cm} Una serie de potencias tiene la forma: \begin{equation*} y = \sum_{i=0}^{\infty}{c_i\,x^{i}} \end{equation*} \end{minipage}

\begin{eqnarray*}
f(k+1) &=& f(k) + f(k-1)\\
a_{k+1} &=& a_{k} + a_{k-1}\\
a_1\cdot r^{k+1} &=& a_1\cdot r^{k} + a_1\cdot r^{k-1}\\
r^{k+1} &=& r^{k} + r^{k-1}\\
r^2 &=& r + 1
\end{eqnarray*}

     \begin{minipage}{10cm} \begin{eqnarray*} f(k+1) &=& f(k) + f(k-1)\\ a_{k+1} &=& a_{k} + a_{k-1}\\ a_1\cdot r^{k+1} &=& a_1\cdot r^{k} + a_1\cdot r^{k-1}\\ r^{k+1} &=& r^{k} + r^{k-1}\\ r^2 &=& r + 1 \end{eqnarray*} \end{minipage}

\begin{equation*}
y_n = \left[c + \sum_{j=0}^{n-1}{\frac{b_j}{a^{j+1}}}\right]\,a^{n}
\end{equation*}

     \begin{minipage}{10cm} \begin{equation*} y_n = \left[c + \sum_{j=0}^{n-1}{\frac{b_j}{a^{j+1}}}\right]\,a^{n} \end{equation*} \end{minipage}

\begin{equation*}
\int\limits_{a}^{b}\!
	\sqrt{\left(\frac{dx_1}{dt}\right)^2 + \cdots + 
	\left(\frac{dx_n}{dt}\right)^2}\,dt
\end{equation*}

     \begin{minipage}{10cm} \begin{equation*} \displaystyle \int\limits_{a}^{b}\! 	\sqrt{\left(\frac{dx_1}{dt}\right)^2 + \cdots +  	\left(\frac{dx_n}{dt}\right)^2}\,dt \end{equation*} \end{minipage}

\begin{equation*}
\nabla f = \mathrm{grad} (f) = 
	\left(\frac{df}{dx}, \frac{df}{dy}, \frac{df}{dz}\right)
\end{equation*}

     \begin{minipage}{10cm} \begin{equation*} \nabla f = \mathrm{grad} (f) = \left(\frac{df}{dx}, \frac{df}{dy}, \frac{df}{dz}\right) \end{equation*} \end{minipage}

\begin{equation*}
\mathrm{div}(f) = \frac{\partial f}{\partial x} + 
	\frac{\partial f}{\partial y} + \frac{\partial f}{\partial z}
\end{equation*}

     \begin{minipage}{10cm} \begin{equation*} \mathrm{div} (f) = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}  	+ \frac{\partial f}{\partial z} \end{equation*} \end{minipage}

\begin{equation*}
\mathrm{rot} (F) = \nabla \times F = \left\vert \begin{array}{ccc}
	\hat{\imath} & \hat{\jmath} & \hat{k} \\\displaystyle
	\frac{\partial}{\partial x} & \displaystyle\frac{\partial}{\partial y} 
		& \displaystyle\frac{\partial}{\partial z}\\
	f_1 & f_2 & f_3
	\end{array}
	\right\vert
\end{equation*}

     \begin{minipage}{10cm} \begin{equation*} \mathrm{rot} (F) = \nabla \times F = \left\vert \begin{array}{ccc} 	\hat{\imath} & \hat{\jmath} & \hat{k} \\\displaystyle 	\frac{\partial}{\partial x} & \displaystyle\frac{\partial}{\partial y}  	& \displaystyle\frac{\partial}{\partial z}\\ 	f_1 & f_2 & f_3 	\end{array} 	\right\vert \end{equation*} \end{minipage}

El teorema de Green establece:
\begin{equation*}
\int\limits_{C}\!P\,dx + Q\,dy = \int\!\!\int\limits_{A}\!\left(
	\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}
	\right)\,dy\,dx
\end{equation*}

     \begin{minipage}{10cm} El teorema de Green establece: \begin{equation*} \int\limits_{C}\!P\,dx + Q\,dy = \int\!\!\int\limits_{A}\!\left( 	\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} 	\right)\,dy\,dx \end{equation*} \end{minipage}

Entonces, el área de la superficie parametrizada es:
\begin{equation*}
\int\!\!\int\limits_{S}\!d\sigma = \int\!\!\int\limits_{R}\left\|
	\frac{\partial X}{\partial t}\times\frac{\partial X}{\partial u}
	\right\|\,dt\,du
\end{equation*}

     \begin{minipage}{10cm} Entonces, el \'area de la superficie parametrizada es: \begin{equation*} \int\!\!\int\limits_{S}\!d\sigma = \int\!\!\int\limits_{R}\left\| 	\frac{\partial X}{\partial t}\times\frac{\partial X}{\partial u} 	\right\|\,dt\,du \end{equation*} \end{minipage}

Teorema de Gauss:
\begin{equation*}
\int\limits_{S}\!\mathbf{u}\cdot\mathbf{n}\,dS 
	= \int\limits_{V}\!\nabla\mathbf{u}\,dV
\end{equation*}

     \begin{minipage}{10cm} Teorema de Gauss: \begin{equation*} \int\limits_{S}\!\mathbf{u}\cdot\mathbf{n}\,dS = \int\limits_{V}\!\nabla\mathbf{u}\,dV \end{equation*} \end{minipage}

\begin{equation*}
P\{N(t) \geq k\} = P\left\{S_k \leq t \right\} = 1 - 
	\sum_{j=0}^{k-1}{e^{-\lambda t}\frac{(\lambda t)^j}{j!}}
\end{equation*}

     \begin{minipage}{10cm} \begin{equation*} P\{N(t) \geq k\} = P\left\{S_k \leq t \right\} = 1 -  	\sum_{j=0}^{k-1}{e^{-\lambda t}\frac{(\lambda t)^j}{j!}} \end{equation*} \end{minipage}

\begin{equation*}
P\{\gamma_t > x\} = e^{-\lambda (t + x)} + 
	\sum_{n=1}^{\infty}{\int\limits_{0}^{t}{\!e^{-\lambda (t + x - y)}
	\lambda^n\frac{y^{n-1}}{(n-1)!}e^{-\lambda y}dy}}
\end{equation*}

     \begin{minipage}{10cm} \begin{equation*} P\{\gamma_t > x\} = e^{-\lambda (t + x)} +  	\sum_{n=1}^{\infty}{\int\limits_{0}^{t}{\!e^{-\lambda (t + x - y)} 	\lambda^n\frac{y^{n-1}}{(n-1)!}e^{-\lambda y}dy}} \end{equation*} \end{minipage}

\begin{equation*}
P\{\mbox{un arribo en }(t + \Delta t)\} 
	= \sum_{i=1}^{2}{P\left\{
	\begin{array}{c}
	\mbox{un arribo del tipo $i$ sin arribo}\\
	\mbox{del otro tipo en }t + \Delta t
	\end{array}
	\right\}}
\end{equation*}

     \begin{minipage}{10cm} \begin{equation*} P\{\mbox{un arribo en }(t + \Delta t)\}  	= \sum_{i=1}^{2}{P\left\{ 	\begin{array}{c} 	\mbox{un arribo del tipo $i$ sin arribo}\\ 	\mbox{del otro tipo en }t + \Delta t 	\end{array} 	\right\}} \end{equation*} \end{minipage}

\begin{equation*}
\sum\limits_{\nu,\mu=1}^{2}\alpha_{\nu\mu}\xi_{\nu}\xi^{\mu} 
	+ 2\,\sum\limits_{\nu=1}^{2}\beta_{\nu}\xi^{\nu} = \alpha
\end{equation*}
donde $\alpha_{\nu\mu}$, $\beta_{\nu}$ y $\alpha$ son constantes.

     \begin{minipage}{10cm} \begin{equation*} \sum\limits_{\nu,\mu=1}^{2}\alpha_{\nu\mu}\xi_{\nu}\xi^{\mu}  	+ 2\,\sum\limits_{\nu=1}^{2}\beta_{\nu}\xi^{\nu} = \alpha \end{equation*} donde $\alpha_{\nu\mu}$, $\beta_{\nu}$ y $\alpha$ son constantes. \end{minipage}

\begin{equation*}
\left.\lim\limits_{\Delta x\rightarrow0}{\left(
	\frac{\Delta y}{\Delta x}\right)}
	\right\vert_{x=2} = 3\cdot(2)^2 = 12
\end{equation*}

     \begin{minipage}{10cm} \begin{equation*} \left.\lim\limits_{\Delta x\rightarrow0}{	\left(\dsf{\Delta y}{\Delta x}\right)}       \right\vert_{x=2} = 3\cdot(2)^2 = 12 \end{equation*} \end{minipage}

\begin{equation*}
m = \left.\frac{dy}{dx}\right\vert_{x=3,y=2.4} 
	= -\frac{9\,(3)}{25\,(2.4)} = -\frac{27}{60} = -\frac{9}{20}
\end{equation*}

     \begin{minipage}{10cm} \begin{equation*} m = \left.\frac{dy}{dx}\right\vert_{x=3,y=2.4}  	= -\frac{9\,(3)}{25\,(2.4)} = -\frac{27}{60} = -\frac{9}{20} \end{equation*} \end{minipage}

\begin{equation*}
\sum_{j=1}^{n}\left(\sum_{i=1}^{m}a_{ij}\right)\,x_j \geq 
	\sum_{i=1}^{m}b_i
\end{equation*}

     \begin{minipage}{10cm} \begin{equation*} \sum_{j=1}^{n}\left(\sum_{i=1}^{m}a_{ij}\right)\,x_j \geq  	\sum_{i=1}^{m}b_i \end{equation*} \end{minipage}

Determinar: $\vec{x}$ para maximizar: 
\begin{equation*}
z(\vec{x})=\vec{c}\cdot\vec{x}
\end{equation*}
sujeto a:
\begin{eqnarray*}
\mathbf{A}\,\vec{x} & \leq & \vec{b}\\
\vec{x} & \geq & \vec{0}
\end{eqnarray*}

     \begin{minipage}{10cm} Determinar: $\vec{x}$ para maximizar:  \begin{equation*} z(\vec{x})=\vec{c}\cdot\vec{x} \end{equation*} sujeto a: \begin{eqnarray*} \mathbf{A}\,\vec{x} & \leq & \vec{b}\\ \vec{x} & \geq & \vec{0} \end{eqnarray*} \end{minipage}

\begin{equation*} % Transformada de Laplace
\mathcal{L}(f(t)) = F(s) = \int\limits_{0}^{\infty}\!f(t)\,e^{-st}\,dt
\end{equation*}

     \begin{minipage}{10cm} \begin{equation*} \mathcal{L}(f(t)) = F(s) = \int\limits_{0}^{\infty}\!f(t)\,e^{-st}\,dt \end{equation*} \end{minipage}

\begin{equation*}
\mathbf{t} = \frac{\dot{x}\mathbf{i} + \dot{y}\mathbf{j}}
                  {\sqrt{\dot{x}^2 + \dot{y}^2}}
\end{equation*}

     \begin{minipage}{10cm} \begin{equation*} \mathbf{t} = \frac{\dot{x}\mathbf{i} + \dot{y}\mathbf{j}}{\sqrt{\dot{x}^2 + \dot{y}^2}} \end{equation*} \end{minipage}

Definimos: $v(t)=\dot{p}$, entonces, 
\begin{equation*}
\ddot{x} = \dot{v}(t) = \frac{dv}{dx}\cdot\frac{dx}{dt} = v\cdot\frac{dv}{dx}
\end{equation*}

     \begin{minipage}{10cm} Definimos: $v(t)=\dot{p}$, entonces,  \begin{equation*} \ddot{x} = \dot{v}(t) = \frac{dv}{dx}\cdot\frac{dx}{dt} = v\cdot\frac{dv}{dx} \end{equation*} \end{minipage}

Estos ejemplos pueden servirle para aprender a crear sus propias fórmulas a partir del código correspondiente: modifíquelo de acuerdo a su necesidad.

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