User:Ripe

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Wikipedia:Babel

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de-2Dieser Benutzer hat fortgeschrittene Deutschkenntnisse.

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la-1Hic usor simplici latinitate contribuere potest.

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$dx^{2}={\frac {\int _{-\infty }^{\infty }{x^{2}f(x)f^{*}(x)dx}}{\int _{-\infty }^{\infty }{f(x)f^{*}(x)dx}}}$

$ds^{2}={\frac {\int _{-\infty }^{\infty }{s^{2}f(s)f^{*}(s)dx}}{\int _{-\infty }^{\infty }{f(s)f^{*}(s)ds}}}$

$dx^{2}\cdot ds^{2}={\frac {\int _{-\infty }^{\infty }{x^{2}f(x)f^{*}(x)dx}}{\int _{-\infty }^{\infty }{f(x)f^{*}(x)dx}}}\cdot {\frac {\int _{-\infty }^{\infty }{s^{2}f(s)f^{*}(s)dx}}{\int _{-\infty }^{\infty }{f(s)f^{*}(s)ds}}}$

$dx\cdot ds\geq {\frac {1}{4\pi }}$

$dx^{2}ds^{2}\geq {\frac {|\int _{-\infty }^{\infty }dxxf^{*}f'+xff'^{*}|^{2}}{16\pi (\int _{-infty}^{\infty }ff^{*}dx)^{2}}}$

integrate by parts with u=x and dv = d/dx:

$dx^{2}ds^{2}\geq {\frac {|\int _{-\infty }^{\infty }dxxf^{*}f'+xff'^{*}|^{2}}{16\pi (\int _{-\infty }^{\infty }ff^{*}dx)^{2}}}$

${\frac {|\int _{-\infty }^{\infty }dxff^{*}|^{2}}{16\pi (\int _{-\infty }^{\infty }dxff^{*})^{2}}}\sim {\frac {1}{16\pi }}\leq dx^{2}ds^{2}$

${\frac {1}{4\pi }}\leq dxds$

$\int _{0}^{\infty }\Delta p\geq {\frac {\hbar }{2}}$

Gabor function: $G(t)\propto e^{[{\frac {-t^{2}}{2a^{2}}}+i(kt+\theta )]}$