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The Cauchy-Schwarz Inequality

$\left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)$

Maxwell’s Equations

$\begin{aligned} \nabla \times \vec{\mathbf{B}} -\, \frac1c\, \frac{\partial\vec{\mathbf{E}}}{\partial t} & = \frac{4\pi}{c}\vec{\mathbf{j}} \\ \nabla \cdot \vec{\mathbf{E}} & = 4 \pi \rho \\ \nabla \times \vec{\mathbf{E}}\, +\, \frac1c\, \frac{\partial\vec{\mathbf{B}}}{\partial t} & = \vec{\mathbf{0}} \\ \nabla \cdot \vec{\mathbf{B}} & = 0 \end{aligned}$

An Identity of Ramanujan

$\frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}} {1+\frac{e^{-8\pi}} {1+\ldots} } } }$

Curl of a Vector Field


Definition of Christoffel Symbols


Standard Deviation

Solving the quadratic equation. Suppose `a x^2+b x+c=0` and `a!=0`. We first divide by `\a` to get `x^2+b/a x+c/a=0`. Then we complete the square and obtain `x^2+b/a x+(b/(2a))^2-(b/(2a))^2+c/a=0`. The first three terms factor to give `(x+b/(2a))^2=(b^2)/(4a^2)-c/a`. Now we take square roots on both sides and get `x+b/(2a)=+-sqrt((b^2)/(4a^2)-c/a)`. Finally we move the `b/(2a)` to the right and simplify to get the two solutions: `x_(1,2)=(-b+-sqrt(b^2-4a c))/(2a)`
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