User:Ripe From Wikipedia, the free encyclopedia edit count | edit summary usage More information Wikipedia:Babel ...Wikipedia:BabelenThis user is a native speaker of the English language.de-2Dieser Benutzer hat fortgeschrittene Deutschkenntnisse.zh-2該用戶能以一般的中文進行交流。该用户能以一般的中文进行交流。la-1Hic usor simplici latinitate contribuere potest.Search user languagesClose mathml scratchpad d x 2 = ∫ − ∞ ∞ x 2 f ( x ) f ∗ ( x ) d x ∫ − ∞ ∞ f ( x ) f ∗ ( x ) d x {\displaystyle dx^{2}={\frac {\int _{-\infty }^{\infty }{x^{2}f(x)f^{*}(x)dx}}{\int _{-\infty }^{\infty }{f(x)f^{*}(x)dx}}}} d s 2 = ∫ − ∞ ∞ s 2 f ( s ) f ∗ ( s ) d x ∫ − ∞ ∞ f ( s ) f ∗ ( s ) d s {\displaystyle ds^{2}={\frac {\int _{-\infty }^{\infty }{s^{2}f(s)f^{*}(s)dx}}{\int _{-\infty }^{\infty }{f(s)f^{*}(s)ds}}}} d x 2 ⋅ d s 2 = ∫ − ∞ ∞ x 2 f ( x ) f ∗ ( x ) d x ∫ − ∞ ∞ f ( x ) f ∗ ( x ) d x ⋅ ∫ − ∞ ∞ s 2 f ( s ) f ∗ ( s ) d x ∫ − ∞ ∞ f ( s ) f ∗ ( s ) d s {\displaystyle 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}}}} d x ⋅ d s ≥ 1 4 π {\displaystyle dx\cdot ds\geq {\frac {1}{4\pi }}} d x 2 d s 2 ≥ | ∫ − ∞ ∞ d x x f ∗ f ′ + x f f ′ ∗ | 2 16 π ( ∫ − i n f t y ∞ f f ∗ d x ) 2 {\displaystyle 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: d x 2 d s 2 ≥ | ∫ − ∞ ∞ d x x f ∗ f ′ + x f f ′ ∗ | 2 16 π ( ∫ − ∞ ∞ f f ∗ d x ) 2 {\displaystyle dx^{2}ds^{2}\geq {\frac {|\int _{-\infty }^{\infty }dxxf^{*}f'+xff'^{*}|^{2}}{16\pi (\int _{-\infty }^{\infty }ff^{*}dx)^{2}}}} | ∫ − ∞ ∞ d x f f ∗ | 2 16 π ( ∫ − ∞ ∞ d x f f ∗ ) 2 ∼ 1 16 π ≤ d x 2 d s 2 {\displaystyle {\frac {|\int _{-\infty }^{\infty }dxff^{*}|^{2}}{16\pi (\int _{-\infty }^{\infty }dxff^{*})^{2}}}\sim {\frac {1}{16\pi }}\leq dx^{2}ds^{2}} 1 4 π ≤ d x d s {\displaystyle {\frac {1}{4\pi }}\leq dxds} ∫ 0 ∞ Δ p ≥ ℏ 2 {\displaystyle \int _{0}^{\infty }\Delta p\geq {\frac {\hbar }{2}}} Gabor function: G ( t ) ∝ e [ − t 2 2 a 2 + i ( k t + θ ) ] {\displaystyle G(t)\propto e^{[{\frac {-t^{2}}{2a^{2}}}+i(kt+\theta )]}} Related Articles