[math][/math]
\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}
\forall \epsilon \in \mathbb{R}_{+} \Bigl[ \exists \delta \in \mathbb{R}_{+} \bigl[ \left| x - d \right| < \delta \Rightarrow \left| f(x) - f(d) \right| < \epsilon \bigr] \Bigr]
\begin{aligned} \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{aligned}
[math] and [ /math]
[math]\begin{align} \sum_{i=1}^n r^2 = r\ \frac{1-r^{n}}{1-r} \end{align}[/math]
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