<jupyter_start><jupyter_text>Beautiful Mathematics Typesetting[Tex](https://en.wikipedia.org/wiki/TeX)[LaTex](https://www.latex-project.org/)[Motivating Examples](http://jupyter-notebook.readthedocs.io/en/latest/examples/Notebook/Typesetting%20Equations.html)  The Lorenz Equations<jupyter_code>\begin{align}
\dot{x} & = \sigma(y-x) \\
\dot{y} & = \rho x - y - xz \\
\dot{z} & = -\beta z + xy
\end{align}<jupyter_output><empty_output><jupyter_text>\begin{align}\dot{x} & = \sigma(y-x) \\\dot{y} & = \rho x - y - xz \\\dot{z} & = -\beta z + xy\end{align}  The Cauchy-Schwarz Inequality<jupyter_code>\begin{equation*}
\left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)
\end{equation*}<jupyter_output><empty_output><jupyter_text>\begin{equation*}\left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)\end{equation*}  Cross Product Formula<jupyter_code>\begin{equation*}
\mathbf{V}_1 \times \mathbf{V}_2 =  \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial X}{\partial u} &  \frac{\partial Y}{\partial u} & 0 \\
\frac{\partial X}{\partial v} &  \frac{\partial Y}{\partial v} & 0
\end{vmatrix}
\end{equation*}<jupyter_output><empty_output><jupyter_text>\begin{equation*}\mathbf{V}_1 \times \mathbf{V}_2 =  \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\\frac{\partial X}{\partial u} &  \frac{\partial Y}{\partial u} & 0 \\\frac{\partial X}{\partial v} &  \frac{\partial Y}{\partial v} & 0\end{vmatrix}\end{equation*}  Probability of getting (k) heads when flipping (n) coins<jupyter_code>\begin{equation*}
P(E)   = {n \choose k} p^k (1-p)^{ n-k}
\end{equation*}<jupyter_output><empty_output><jupyter_text>\begin{equation*}P(E)   = {n \choose k} p^k (1-p)^{ n-k}\end{equation*}  Identity of Ramanujan[Srinivasa Ramanujan](https://en.wikipedia.org/wiki/Srinivasa_Ramanujan)Self-taught, no formal training in mathematics, made contributions to:- mathematical analysis- number theory- infinite series- continued fractions<jupyter_code>\begin{equation*}
\frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} =
1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}}
{1+\frac{e^{-8\pi}} {1+\ldots} } } }
\end{equation*}<jupyter_output><empty_output><jupyter_text>\begin{equation*}\frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} =1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}}{1+\frac{e^{-8\pi}} {1+\ldots} } } }\end{equation*}  Maxwell’s Equations<jupyter_code>\begin{align}
\nabla \times \vec{\mathbf{B}} -\, \frac1c\, \frac{\partial\vec{\mathbf{E}}}{\partial t} & = \frac{4\pi}{c}\vec{\mathbf{j}} \\   \nabla \cdot \vec{\mathbf{E}} & = 4 \pi \rho \\
\nabla \times \vec{\mathbf{E}}\, +\, \frac1c\, \frac{\partial\vec{\mathbf{B}}}{\partial t} & = \vec{\mathbf{0}} \\
\nabla \cdot \vec{\mathbf{B}} & = 0
\end{align}<jupyter_output><empty_output>
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