Sesión 7: Otros símbolos

Considerando que $2\in\mathbb{Z}$ y que $\pi\notin\mathbb{Q}$ 
tenemos que $\{2,\pi\}$...

Recordando que: $\mathbb{Q}\cup\mathbb{Q}' = \mathbb{R}$,...

\begin{equation*}
S_k = \sum\limits_{i=1}^{k}{a_i}  
    = a_1 +   a_2  +  a_3     + \cdots + a_k
\end{equation*}

\begin{equation*}
	\bar{x} = \frac{\sum{x_i}}{n}
\end{equation*}

\begin{equation*}
	\bar{x} = \frac{\sum\limits_{i=1}^{n}{x_i}}{n}
\end{equation*}

\begin{equation*}
	\cos(\vec{u},\vec{v}) = 
     \frac{\vec{u}\cdot\vec{v}}{\|\vec{u}\| \cdot \|\vec{v}\|}
\end{equation*}

\begin{equation*}
	\ln (x\cdot y) = \ln x + \ln y
\end{equation*}

\begin{equation*}
	\log\left(x^{k}\right) = k\cdot\log x
\end{equation*}

\begin{equation*}
\frac{df}{dx} = \lim\limits_{\Delta x\rightarrow 0}
                \frac{f(x + \Delta x) - f(x)}{\Delta x}
\end{equation*}

\begin{equation*}
	\int\!v^n\,{dv} = \displaystyle\frac{v^{n+1}}{n+1} + C
\end{equation*}

\begin{equation*}
	\int\limits_{1}^{\infty}\!\frac{dx}{x}
\end{equation*}

\begin{equation*}
	\int\limits_{-1}^{1}\!\sqrt{\frac{1 + x}{1 - x}}\,dx
\end{equation*}

\begin{equation*}
	\int_{-1}^{1}\!\sqrt{\frac{1 + x}{1 - x}}\,dx
\end{equation*}

Las tres identidades trigonométricas pitagóricas son:
%
\begin{itemize}
\item $\sin^2 \alpha + \cos^2 \alpha = 1$
\item $\sec^2 \alpha = 1 + \tan^2 \alpha$
\item $\csc^2 \alpha = 1 + \cot^2 \alpha$
\end{itemize}
%

\begin{itemize}
\item $\sin(\alpha + \beta) = \sin\alpha\,\cos\beta+\sin\beta\,\cos\alpha$
\item $\cos(\alpha + \beta) = \cos\alpha\,\cos\beta-\sin\alpha\,\sin\beta$
\item $\tan(\alpha + \beta) = \displaystyle
    \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\,\tan\beta}$
\end{itemize}

De acuerdo a la ley de senos, tenemos:
\begin{equation*}
\displaystyle
	\frac{a}{\sin\alpha} = \frac{b}{\sin\beta}  = \frac{c}{\sin\gamma}
\end{equation*}

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

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

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

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

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

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

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

\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*}

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*}

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*}

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

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

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

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

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*}