🧮 Math

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Table of Contents
  1. References
$$E = mc^2$$
\[E = mc^2\] 
$$\text{EMA}_t = 0.3 \cdot x_t + 0.7 \cdot \text{EMA}_{t-1}$$
\[\text{EMA}_t = 0.3 \cdot x_t + 0.7 \cdot \text{EMA}_{t-1}\] 
$$\displaystyle \frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} = 1+\frac{-2\pi} {1+\frac{-4\pi} {1+\frac{-6\pi} {1+\frac{-8\pi} {1+\cdots} } } }$$
\[\displaystyle \frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}} {1+\frac{e^{-8\pi}} {1+\cdots} } } }\] 
$$\displaystyle \left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)$$
\[\displaystyle \left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)\] 
$$\displaystyle {1 +  \frac{q^2}{(1-q)}+\frac{q^6}{(1-q)(1-q^2)}+\cdots }= \prod_{j=0}^{\infty}\frac{1}{(1-q^{5j+2})(1-q^{5j+3})}, \quad\quad \text{for }\lvert q\rvert<1.$$
\[\displaystyle {1 + \frac{q^2}{(1-q)}+\frac{q^6}{(1-q)(1-q^2)}+\cdots }= \prod_{j=0}^{\infty}\frac{1}{(1-q^{5j+2})(1-q^{5j+3})}, \quad\quad \text{for }\lvert q\rvert<1.\]

References