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value: for <math display=inline>x \ge 0</math><math display=inline> \begin{aligned} 1-\Phi\left(x\right) & = \left(\frac{0.39894228040143268}{x+2...

\begin{aligned} 4x + y &= 6x \\ 6x + 3y &= 6y\end{aligned} \right. </math> {{spaces|4}} that is {{spaces|4}} <math> \left\{ \begin{aligned} -2x +...

({{EquationNote|Eq.2}}) has these forms: <math display="block">\begin{aligned}\widehat f(\xi)\cdot e^{i 2 \pi \xi x} &= A e^{i \theta} \cdot e^{i 2 \pi...

three events, this proceeds as follows:<math display="block"> \begin{aligned}P\left( A\cup B\cup C\right) =&P\left( \left( A\cup B\right) \cup C\right)...

layer for each vector individually. Schematically, we have:<math>\begin{aligned} \text{given input vectors } & h_0, h_1, \dots\\ \text{combine them into...

representations (Formula 06.34.10.0003) }}</ref> <math display="block">\begin{aligned} \Gamma (z) &= \cfrac{e^{-1}}{ 2 + 0 - z + 1\cfrac{z-1}{ ...

<math display="block">\begin{aligned} rV_{e} &= w + q(V_{u}-V_{e}) \\ rV_{u} &= z + \theta m(\theta) (V_{e}-V_{u}) \end{aligned}</math>Let <math>w</math>...

colour=#0073CF|background colour=#F5FFFA|cellpadding=6 |equation={{NumBlk| |<math> \begin{aligned} A_0 &= \frac{1}{P}\int_P s(x)\,dx \\ A_n &= \frac{2}{P}\int_P s(x) \cos...

solutions for <math>\lambda_{i}</math> are:<math display="block">\begin{aligned} \lambda_{1} &= {-\left(r-q-{1\over{2}}\sigma^{2} \right ) +...

<math>\theta</math> the previous equations simplify to: <math display="block">\begin{aligned} \vec r &= r(\theta) \hat e_r\\ \frac {d\vec r} {d\theta} &= \frac {dr}...

transformation]] <math display=block> T : \left\{\begin{aligned} K^n &\to K^m \\ x &\mapsto \mathbf{M}x \end{aligned}\right. </math> has a particularly simple...

−2}} enabling further expansion by integration by parts. :<math>\begin{aligned} \zeta(s)= & 1+\frac{1}{s-1}-\frac{s}{2 !}[\zeta(s+1)-1] \\ - & \frac{s(s+1)}{3...

differentiation with respect to <math>\epsilon</math>: :<math> \begin{aligned} \lim_{\epsilon \to 0} P(Y \in U \mid x_0-\epsilon < X < x_0+\epsilon)...

display="block">\begin{aligned} \underset{x,y}{\text{maximize}} \quad& f(x,y) \\ \text{subject to}\quad& g(x,y) = 0. \end{aligned}</math> (Sometimes an...

display="block">\left. \begin{aligned} A^2 (1-\omega^2)^2 &= \cos^2\varphi \\ (2 \zeta \omega A)^2 &= \sin^2\varphi \end{aligned} \right\} \Rightarrow A^2[(1...

{aligned} \frac{\partial J}{\partial K} &= (1-\tau)f'(k) - r \\ \frac{\partial J}{\partial L} &= (1-\tau)\left[f(k) - f'(k)k - w\right] \end{aligned}</math>Therefore...

display="inline">\hat\theta-\theta</math>, we have<math display="block">\begin{aligned} \operatorname{MSE}(\hat{\theta}) &= \mathbb{E}[(\hat\theta-\theta)^2]...

uniformly and independently. Then we have <math display="block">\begin{aligned} E[\tau_A^2] &= E\left[\left(1-\frac{4\sum_i l_i}{n(n-1)}\right)^2\right]...

University|access-date=22 March 2013}}</ref> <math display="block">\begin{aligned} &\frac{dx}{dy} = x\quad \Rightarrow \quad \frac{dx}{x} = dy \\[10mu]...

Fourier transform is the [[sinc function]]<math display="block">\begin{aligned} G_1(t, x) &= \frac{1}{2\pi} \int_\R \frac{\sin(|\omega| t)}{|\omega|}...