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				<section>
					<h2>reveal.js Math Plugin</h2>
					<p>A thin wrapper for MathJax</p>
				</section>

				<section>
					<h3>The Lorenz Equations</h3>

					\[\begin{aligned}
					\dot{x} &amp; = \sigma(y-x) \\
					\dot{y} &amp; = \rho x - y - xz \\
					\dot{z} &amp; = -\beta z + xy
					\end{aligned} \]
				</section>

				<section>
					<h3>The Cauchy-Schwarz Inequality</h3>

					<script type="math/tex; mode=display">
						\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)
					</script>
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				<section>
					<h3>A Cross Product Formula</h3>

					\[\mathbf{V}_1 \times \mathbf{V}_2 =  \begin{vmatrix}
					\mathbf{i} &amp; \mathbf{j} &amp; \mathbf{k} \\
					\frac{\partial X}{\partial u} &amp;  \frac{\partial Y}{\partial u} &amp; 0 \\
					\frac{\partial X}{\partial v} &amp;  \frac{\partial Y}{\partial v} &amp; 0
					\end{vmatrix}  \]
				</section>

				<section>
					<h3>The probability of getting \(k\) heads when flipping \(n\) coins is</h3>

					\[P(E)   = {n \choose k} p^k (1-p)^{ n-k} \]
				</section>

				<section>
					<h3>An Identity of Ramanujan</h3>

					\[ \frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} =
					1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}}
					{1+\frac{e^{-8\pi}} {1+\ldots} } } } \]
				</section>

				<section>
					<h3>A Rogers-Ramanujan Identity</h3>

					\[  1 +  \frac{q^2}{(1-q)}+\frac{q^6}{(1-q)(1-q^2)}+\cdots =
					\prod_{j=0}^{\infty}\frac{1}{(1-q^{5j+2})(1-q^{5j+3})}\]
				</section>

				<section>
					<h3>Maxwell&#8217;s Equations</h3>

					\[  \begin{aligned}
					\nabla \times \vec{\mathbf{B}} -\, \frac1c\, \frac{\partial\vec{\mathbf{E}}}{\partial t} &amp; = \frac{4\pi}{c}\vec{\mathbf{j}} \\   \nabla \cdot \vec{\mathbf{E}} &amp; = 4 \pi \rho \\
					\nabla \times \vec{\mathbf{E}}\, +\, \frac1c\, \frac{\partial\vec{\mathbf{B}}}{\partial t} &amp; = \vec{\mathbf{0}} \\
					\nabla \cdot \vec{\mathbf{B}} &amp; = 0 \end{aligned}
					\]
				</section>

				<section>
					<section>
						<h3>The Lorenz Equations</h3>

						<div class="fragment">
							\[\begin{aligned}
							\dot{x} &amp; = \sigma(y-x) \\
							\dot{y} &amp; = \rho x - y - xz \\
							\dot{z} &amp; = -\beta z + xy
							\end{aligned} \]
						</div>
					</section>

					<section>
						<h3>The Cauchy-Schwarz Inequality</h3>

						<div class="fragment">
							\[ \left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right) \]
						</div>
					</section>

					<section>
						<h3>A Cross Product Formula</h3>

						<div class="fragment">
							\[\mathbf{V}_1 \times \mathbf{V}_2 =  \begin{vmatrix}
							\mathbf{i} &amp; \mathbf{j} &amp; \mathbf{k} \\
							\frac{\partial X}{\partial u} &amp;  \frac{\partial Y}{\partial u} &amp; 0 \\
							\frac{\partial X}{\partial v} &amp;  \frac{\partial Y}{\partial v} &amp; 0
							\end{vmatrix}  \]
						</div>
					</section>

					<section>
						<h3>The probability of getting \(k\) heads when flipping \(n\) coins is</h3>

						<div class="fragment">
							\[P(E)   = {n \choose k} p^k (1-p)^{ n-k} \]
						</div>
					</section>

					<section>
						<h3>An Identity of Ramanujan</h3>

						<div class="fragment">
							\[ \frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} =
							1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}}
							{1+\frac{e^{-8\pi}} {1+\ldots} } } } \]
						</div>
					</section>

					<section>
						<h3>A Rogers-Ramanujan Identity</h3>

						<div class="fragment">
							\[  1 +  \frac{q^2}{(1-q)}+\frac{q^6}{(1-q)(1-q^2)}+\cdots =
							\prod_{j=0}^{\infty}\frac{1}{(1-q^{5j+2})(1-q^{5j+3})}\]
						</div>
					</section>

					<section>
						<h3>Maxwell&#8217;s Equations</h3>

						<div class="fragment">
							\[  \begin{aligned}
							\nabla \times \vec{\mathbf{B}} -\, \frac1c\, \frac{\partial\vec{\mathbf{E}}}{\partial t} &amp; = \frac{4\pi}{c}\vec{\mathbf{j}} \\   \nabla \cdot \vec{\mathbf{E}} &amp; = 4 \pi \rho \\
							\nabla \times \vec{\mathbf{E}}\, +\, \frac1c\, \frac{\partial\vec{\mathbf{B}}}{\partial t} &amp; = \vec{\mathbf{0}} \\
							\nabla \cdot \vec{\mathbf{B}} &amp; = 0 \end{aligned}
							\]
						</div>
					</section>
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