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${\begin{aligned}X_{k_{1},k_{2}}&=\sum _{n_{1}=0}^{N_{1}-1}\left(\sum _{n_{2}=0}^{N_{2}-1}x_{n_{1},n_{2}}\cos \left[{\frac {\pi }{N_{2}}}\left(n_{2}+{\frac {1}{2}}\right)k_{2}\right]\right)\cos \left[{\frac {\pi }{N_{1}}}\left(n_{1}+{\frac {1}{2}}\right)k_{1}\right]\\&=\sum _{n_{1}=0}^{N_{1}-1}\sum _{n_{2}=0}^{N_{2}-1}x_{n_{1},n_{2}}\cos \left[{\frac {\pi }{N_{1}}}\left(n_{1}+{\frac {1}{2}}\right)k_{1}\right]\cos \left[{\frac {\pi }{N_{2}}}\left(n_{2}+{\frac {1}{2}}\right)k_{2}\right].\end{aligned}}$

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{\displaystyle {\begin{aligned}X_{k_{1},k_{2}}&...

${\begin{array}{lcl}{\tilde {x}}(n_{1},n_{2},n_{3})=x(2n_{1},2n_{2},2n_{3})\\{\tilde {x}}(n_{1},n_{2},N-n_{3}-1)=x(2n_{1},2n_{2},2n_{3}+1)\\{\tilde {x}}(n_{1},N-n_{2}-1,n_{3})=x(2n_{1},2n_{2}+1,2n_{3})\\{\tilde {x}}(n_{1},N-n_{2}-1,N-n_{3}-1)=x(2n_{1},2n_{2}+1,2n_{3}+1)\\{\tilde {x}}(N-n_{1}-1,n_{2},n_{3})=x(2n_{1}+1,2n_{2},2n_{3})\\{\tilde {x}}(N-n_{1}-1,n_{2},N-n_{3}-1)=x(2n_{1}+1,2n_{2},2n_{3}+1)\\{\tilde {x}}(N-n_{1}-1,N-n_{2}-1,n_{3})=x(2n_{1}+1,2n_{2}+1,2n_{3})\\{\tilde {x}}(N-n_{1}-1,N-n_{2}-1,N-n_{3}-1)=x(2n_{1}+1,2n_{2}+1,2n_{3}+1)\\\end{array}}$

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{\displaystyle {\begin{array}{lcl}{\tilde {x}}(...

${\displaystyle {\begin{aligned}\|f\|_{\mathcal {H}}^{2}&=\left\langle \sum _{i=1}^{n}\gamma ((x_{i},t_{i}),\cdot )c_{i}^{t_{i}},\sum _{j=1}^{n}\gamma ((x_{j},t_{j}),\cdot )c_{j}^{t_{j}}\right\rangle _{\mathcal {H}}\\&=\sum _{i,j=1}^{n}c_{i}^{t_{i}}c_{j}^{t_{j}}\gamma ((x_{i},t_{i}),(x_{j},t_{j}))\\&=\sum _{i,j=1}^{n}\sum _{s,t=1}^{T}c_{i}^{t}c_{j}^{s}k(x_{i},x_{j})A_{s,t}\\&=\sum _{i,j=1}^{n}k(x_{i},x_{j})\langle c_{i},Ac_{j}\rangle _{\mathbb {R} ^{T}}\\&=\sum _{i,j=1}^{n}k(x_{i},x_{j})\langle c_{i},AA^{\dagger }Ac_{j}\rangle _{\mathbb {R} ^{T}}\\&=\sum _{i,j=1}^{n}k(x_{i},x_{j})\langle Ac_{i},A^{\dagger }Ac_{j}\rangle _{\mathbb {R} ^{T}}\\&=\sum _{i,j=1}^{n}\sum _{s,t=1}^{T}(Ac_{i})^{t}(Ac_{j})^{s}k(x_{i},x_{j})A_{s,t}^{\dagger }\\&=\sum _{s,t=1}^{T}A_{s,t}^{\dagger }\langle \sum _{i=1}^{n}k(x_{i},\cdot )(Ac_{i})^{t},\sum _{j=1}^{n}k(x_{j},\cdot )(Ac_{j})^{s}\rangle _{{\mathcal {H}}_{k}}\\&=\sum _{s,t=1}^{T}A_{s,t}^{\dagger }\langle f_{t},f_{s}\rangle _{{\mathcal {H}}_{k}}\end{aligned$

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Multi-task learning - Wikipedia

$x_{n_{1},n_{2},n_{3}}=\sum _{k_{1}=0}^{N_{1}-1}\sum _{k_{2}=0}^{N_{2}-1}\sum _{k_{3}=0}^{N_{3}-1}X_{k_{1},k_{2},k_{3}}\cos \left[{\frac {\pi }{N_{1}}}\left(n_{1}+{\frac {1}{2}}\right)k_{1}\right]\cos \left[{\frac {\pi }{N_{2}}}\left(n_{2}+{\frac {1}{2}}\right)k_{2}\right]\cos \left[{\frac {\pi }{N_{3}}}\left(n_{3}+{\frac {1}{2}}\right)k_{3}\right],\quad {\text{for }}n_{i}=0,1,2,\dots ,N_{i}-1.$

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{\displaystyle x_{n_{1},n_{2},n_{3}}=\sum _{k_{...

$X_{k_{1},k_{2},k_{3}}=\sum _{n_{1}=0}^{N_{1}-1}\sum _{n_{2}=0}^{N_{2}-1}\sum _{n_{3}=0}^{N_{3}-1}x_{n_{1},n_{2},n_{3}}\cos \left[{\frac {\pi }{N_{1}}}\left(n_{1}+{\frac {1}{2}}\right)k_{1}\right]\cos \left[{\frac {\pi }{N_{2}}}\left(n_{2}+{\frac {1}{2}}\right)k_{2}\right]\cos \left[{\frac {\pi }{N_{3}}}\left(n_{3}+{\frac {1}{2}}\right)k_{3}\right],\quad {\text{for }}k_{i}=0,1,2,\dots ,N_{i}-1.$

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{\displaystyle X_{k_{1},k_{2},k_{3}}=\sum _{n_{...

${\begin{aligned}\sigma &={\sqrt {{\frac {1}{N}}\sum _{i=1}^{N}(x_{i}-{\overline {x}})^{2}}}\\&={\sqrt {{\frac {1}{4}}\sum _{i=1}^{4}(x_{i}-{\overline {x}})^{2}}}&(N=4)\\&={\sqrt {{\frac {1}{4}}\sum _{i=1}^{4}(x_{i}-7)^{2}}}&({\overline {x}}=7)\\&={\sqrt {{\frac {1}{4}}\left[(x_{1}-7)^{2}+(x_{2}-7)^{2}+(x_{3}-7)^{2}+(x_{4}-7)^{2}\right]}}\\&={\sqrt {{\frac {1}{4}}\left[(5-7)^{2}+(6-7)^{2}+(8-7)^{2}+(9-7)^{2}\right]}}\\&={\sqrt {{\frac {1}{4}}\left((-2)^{2}+(-1)^{2}+1^{2}+2^{2}\right)}}\\&={\sqrt {{\frac {1}{4}}\left(4+1+1+4\right)}}\\&={\sqrt {\frac {10}{4}}}\\&\approx 1.58114\,.\end{aligned}}$

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標準偏差 - 维基百科，自由的百科全书

${\begin{aligned}&\operatorname {E} \left[X^{p}\right]=\sigma ^{p}\cdot (-i{\sqrt {2}}\operatorname {sgn} \mu )^{p}\;U\left({-{\frac {1}{2}}p},\,{\frac {1}{2}},\,-{\frac {1}{2}}(\mu /\sigma )^{2}\right),\\&\operatorname {E} \left[|X|^{p}\right]=\sigma ^{p}\cdot 2^{\frac {p}{2}}{\frac {\Gamma \left({\frac {1+p}{2}}\right)}{\sqrt {\pi }}}\;_{1}F_{1}\left({-{\frac {1}{2}}p},\,{\frac {1}{2}},\,-{\frac {1}{2}}(\mu /\sigma )^{2}\right).\\\end{aligned}}$

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இயல்நிலைப் பரவல் - தமிழ் விக்கிப்பீடியா

${\begin{aligned}{\Bigl (}1+{\frac {1}{x^{2}}}{\Bigr )}(1-\Phi (x))&={\Bigl (}1+{\frac {1}{x^{2}}}{\Bigr )}\int _{x}^{\infty }\varphi (u)\,du\\&=\int _{x}^{\infty }{\Bigl (}1+{\frac {1}{x^{2}}}{\Bigr )}\varphi (u)\,du\\&>\int _{x}^{\infty }{\Bigl (}1+{\frac {1}{u^{2}}}{\Bigr )}\varphi (u)\,du=-{\biggl .}{\frac {\varphi (u)}{u}}{\biggr |}_{x}^{\infty }={\frac {\varphi (x)}{x}}.\end{aligned}}$

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Distribución normal - Wikipedia, la enciclopedi...

${\begin{aligned}{\Bigl (}1+{\frac {1}{x^{2}}}{\Bigr )}(1-\Phi (x))&={\Bigl (}1+{\frac {1}{x^{2}}}{\Bigr )}\int _{x}^{\infty }\varphi (o)\,du\\&=\int _{x}^{\infty }{\Bigl (}1+{\frac {1}{x^{2}}}{\Bigr )}\varphi (o)\,du\\&>\int _{x}^{\infty }{\Bigl (}1+{\frac {1}{o^{2}}}{\Bigr )}\varphi (o)\,du=-{\biggl .}{\frac {\varphi (o)}{o}}{\biggr |}_{x}^{\infty }={\frac {\varphi (x)}{x}}.\end{aligned}}$

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Distribución normal - Wikipedia

$\sigma \ ={\sqrt {\left({\frac {1}{N}}\sum _{i=1}^{N}{{x_{i}}^{2}}\right)-{\overline {x}}^{2}}}={\sqrt {\left({\frac {1}{N}}\sum _{i=1}^{N}{{x_{i}}^{2}}\right)-\left({\frac {1}{N}}\sum _{i=1}^{N}{x_{i}}\right)^{2}}}={\frac {1}{N}}{\sqrt {N\left(\sum _{i=1}^{N}{{x_{i}}^{2}}\right)-\left(\sum _{i=1}^{N}{x_{i}}\right)^{2}}}$

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Sisihan piawai - Wikipedia Bahasa Melayu, ensik...

${\begin{aligned}\sigma \ &={\sqrt {\left({\frac {1}{N}}\sum _{i=1}^{N}{{x_{i}}^{2}}\right)-{\overline {x}}^{2}}}={\sqrt {\left({\frac {1}{N}}\sum _{i=1}^{N}{{x_{i}}^{2}}\right)-\left({\frac {1}{N}}\sum _{i=1}^{N}{x_{i}}\right)^{2}}}\\\\&={\frac {1}{N}}{\sqrt {N\left(\sum _{i=1}^{N}{{x_{i}}^{2}}\right)-\left(\sum _{i=1}^{N}{x_{i}}\right)^{2}}}\end{aligned}}$

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Стандардно отстапување — Википедија

$X(k_{1},k_{2},k_{3})=\sum _{n_{1}=1}^{{\tfrac {N}{2}}-1}\sum _{n_{2}=1}^{{\tfrac {N}{2}}-1}\sum _{n_{1}=1}^{{\tfrac {N}{2}}-1}{\tilde {x}}_{ijl}(n_{1},n_{2},n_{3})\cos(\varphi (2k_{1}+i)\cos(\varphi (2k_{2}+j)\cos(\varphi (2k_{3}+l))$

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{\displaystyle X(k_{1},k_{2},k_{3})=\sum _{n_{1...

${\begin{aligned}a_{1,1}x_{1}+a_{1,2}x_{2}+&\cdots +a_{1,n}x_{n}=b_{1}\\&\ \ \vdots \\a_{m,1}x_{1}+a_{m,2}x_{2}+&\cdots +a_{m,n}x_{n}=b_{m}\end{aligned}}$

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{\displaystyle {\begin{aligned}a_{1,1}x_{1}+a_{...

${\overline {X}}^{(1)}={\frac {1}{n_{1}}}\sum _{i=1}^{n_{1}}X_{i}^{(1)},\quad {\overline {X}}^{(2)}={\frac {1}{n_{2}}}\sum _{i=1}^{n_{2}}X_{i}^{(2)}$

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Normalverteilung – Wikipedia

$Y_{n_{1}-1,n_{2}-1}:={\frac {\sigma _{2}^{2}(n_{2}-1)\sum \limits _{i=1}^{n_{1}}(X_{i}^{(1)}-{\overline {X}}^{(1)})^{2}}{\sigma _{1}^{2}(n_{1}-1)\sum \limits _{j=1}^{n_{2}}(X_{i}^{(2)}-{\overline {X}}^{(2)})^{2}}}$

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Normalverteilung – Wikipedia

${\sqrt {{\frac {1}{N}}\sum _{i=1}^{N}(x_{i}-{\overline {x}})^{2}}}={\sqrt {{\frac {1}{N}}\left(\sum _{i=1}^{N}x_{i}^{2}\right)-{\overline {x}}^{2}}}={\sqrt {{\frac {1}{N}}\sum _{i=1}^{N}x_{i}^{2}-\left({\frac {1}{N}}\sum _{i=1}^{N}x_{i}\right)^{2}}}.$

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நியமவிலகல் - தமிழ் விக்கிப்பீடியா

$\sigma {\bigl (}X{\bigr )}={\sqrt {{\frac {1}{6}}\cdot {\bigl (}1-3,5{\bigr )}^{2}+{\frac {1}{6}}\cdot {\bigl (}2-3,5{\bigr )}^{2}+{\frac {1}{6}}\cdot {\bigl (}3-3,5{\bigr )}^{2}+{\frac {1}{6}}\cdot {\bigl (}4-3,5{\bigr )}^{2}+{\frac {1}{6}}\cdot {\bigl (}5-3,5{\bigr )}^{2}+{\frac {1}{6}}\cdot {\bigl (}6-3,5{\bigr )}^{2}}}\approx 1,71$

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Стандартне відхилення — Вікіпедія

${\sqrt {{\frac {1}{N}}\sum _{i=1}^{N}\left(x_{i}-{\bar {x}}\right)^{2}}}={\sqrt {{\frac {1}{N}}\left(\sum _{i=1}^{N}x_{i}^{2}\right)-{\bar {x}}^{2}}}={\sqrt {\left({\frac {1}{N}}\sum _{i=1}^{N}x_{i}^{2}\right)-\left({\frac {1}{N}}\sum _{i=1}^{N}x_{i}\right)^{2}}},$

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Tipico desviacion - Chavacano de Zamboanga Wiki...

${\sqrt {{\frac {1}{N}}\sum _{i=1}^{N}(x_{i}-{\overline {x}})^{2}}}={\sqrt {{\frac {1}{N}}\left(\sum _{i=1}^{N}x_{i}^{2}\right)-{\overline {x}}^{2}}}={\sqrt {\left({\frac {1}{N}}\sum _{i=1}^{N}x_{i}^{2}\right)-\left({\frac {1}{N}}\sum _{i=1}^{N}x_{i}\right)^{2}}}.$

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Τυπική απόκλιση - Βικιπαίδεια

$\sigma ={\sqrt {\sigma ^{2}}}={\sqrt {{\frac {1}{n-1}}\sum _{i=1}^{n}\left(x_{i}-{\bar {x}}\right)^{2}}},\quad \sigma ={\sqrt {{\frac {1}{n-1}}\sum _{i=1}^{n}\left(x_{i}-{\bar {x}}\right)^{2}}},\quad {\bar {x}}={\frac {1}{n}}\sum _{i=1}^{n}x_{i}={\frac {1}{n}}(x_{1}+\cdots +x_{n}),$

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Стандартно отклонение – Уикипедия

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