Difference between revisions of "Math Diagnostic"

Revision as of 20:44, 14 April 2016

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${\displaystyle \pi ={\frac {3}{4}}{\sqrt {3}}+24\int _{0}^{1/4}{{\sqrt {x-x^{2}}}dx}}$

${\displaystyle {\text{cost}}={\text{base}}\times 2^{{\text{level}}-1}}$

${\displaystyle \operatorname {erfc} (x)={\frac {2}{\sqrt {\pi }}}\int _{x}^{\infty }e^{-t^{2}}\,dt={\frac {e^{-x^{2}}}{x{\sqrt {\pi }}}}\sum _{n=0}^{\infty }(-1)^{n}{\frac {(2n)!}{n!(2x)^{2n}}}}$

${\displaystyle {\begin{bmatrix}0&\cdots &0\\\vdots &\ddots &\vdots \\0&\cdots &0\end{bmatrix}}}$

${\displaystyle \left.{\frac {A}{B}}\right\}\to X}$

${\displaystyle {\color {Blue}x^{2}}+{\color {YellowOrange}2x}-{\color {OliveGreen}1}}$

${\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}}$

${\displaystyle \int _{a}^{x}\!\!\!\int _{a}^{s}f(y)\,dy\,ds=\int _{a}^{x}f(y)(x-y)\,dy}$

${\displaystyle |{\bar {z}}|=|z|,|({\bar {z}})^{n}|=|z|^{n},\arg(z^{n})=n\arg(z)}$

${\displaystyle \lim _{z\rightarrow z_{0}}f(z)=f(z_{0})}$

${\displaystyle \phi _{n}(\kappa )={\frac {1}{4\pi ^{2}\kappa ^{2}}}\int _{0}^{\infty }{\frac {\sin(\kappa R)}{\kappa R}}{\frac {\partial }{\partial R}}\left[R^{2}{\frac {\partial D_{n}(R)}{\partial R}}\right]\,dR}$

${\displaystyle {}_{p}F_{q}(a_{1},\dots ,a_{p};c_{1},\dots ,c_{q};z)=\sum _{n=0}^{\infty }{\frac {(a_{1})_{n}\cdots (a_{p})_{n}}{(c_{1})_{n}\cdots (c_{q})_{n}}}{\frac {z^{n}}{n!}}}$

${\displaystyle \varpi \varrho \varsigma \varphi }$

${\displaystyle \left\uparrow {\frac {a}{b}}\right\downarrow \quad \left\Uparrow {\frac {a}{b}}\right\Downarrow \quad \left\updownarrow {\frac {a}{b}}\right\Updownarrow }$

• Making this assumption gives ${\displaystyle 1/\tau (\omega )=1/\tau _{0}+\kappa \omega ^{2}/2\approx 1/\tau _{0}\lambda _{0}\omega ^{2}/\omega _{\lambda }\Rightarrow \omega _{\lambda }=2\lambda _{0}/\kappa ,}$ so that:${\displaystyle \lambda (\omega ,T)={\frac {\lambda _{0}}{1+(\kappa \omega /2\lambda _{0})^{2}+(T/T_{\lambda })^{2}}}.}$