\[\left\{ \begin{array}{c} a_1x+b_1y+c_1z=d_1 \\ a_2x+b_2y+c_2z=d_2 \\ a_3x+b_3y+c_3z=d_3 \end{array} \right. \]\[J(\theta) = \frac{1}{2m}\sum_{i = 0} ^m(y^i - h_\theta (x^i))^2 \]\[\frac{\partial J(\theta)}{\partial\theta_j}=-\frac1m\sum_{i=0}^m(y^i-h_\theta(x^i))x^i_j \]\[\begin{aligned} \frac{\partial J(\theta)}{\partial\theta_j} & = -\frac1m\sum_{i=0}^m(y^i-h_\theta(x^i)) \frac{\partial}{\partial\theta_j}(y^i-h_\theta(x^i)) \\ & = -\frac1m\sum_{i=0}^m(y^i-h_\theta(x^i)) \frac{\partial}{\partial\theta_j}(\sum_{j=0}^n\theta_jx_j^i-y^i) \\ & = -\frac1m\sum_{i=0}^m(y^i-h_\theta(x^i))x^i_j \end{aligned} \]\[\begin{gathered} \operatorname{arg\,max}_a f(a) = \operatorname*{arg\,max}_b f(b) \\ \operatorname{arg\,min}_c f(c) = \operatorname*{arg\,min}_d f(d) \end{gathered} \]\[\begin{aligned} \int_a^b x^2 \mathrm{d} x \end{aligned} \]\[\begin{aligned} \int\limits_a^b x^2 \mathrm{d} x \end{aligned} \]\[\begin{aligned} J(\mathbf{w})&=\frac{1}{2m}\sum_{i=1}^m(f(\mathbf{x_i})-y_i)^2\\ &=\frac{1}{2m}\sum_{i=1}^m [f(\mathbf{x_i})]^2-2f(\mathbf{x_i)}y_i+y_i^2 \end{aligned} \]\[\begin{aligned} J(\mathbf{θ})=-\frac{1}{m}∑_{i=1}^{m}y_ilogh_θ(x_i)+(1−y_i)log(1−h_θ(x_i)) \end{aligned} \]\[\begin{aligned} \left.\begin{aligned} B'&=-\partial \times E,\\ %加&指定对齐位置 E'&=\partial \times B - 4\pi j, \end{aligned} \right\} %加右} \qquad \text{Maxwell's equations} \end{aligned} \]\[\begin{aligned} \sigma_1 &= x + y &\quad \sigma_2 &= \frac{x}{y} \\ \sigma_1' &= \frac{\partial x + y}{\partial x} & \sigma_2' &= \frac{\partial \frac{x}{y}}{\partial x} \end{aligned} \]\[\begin{aligned} a_n&=\frac{1}{\pi}\int\limits_{-\pi}^{\pi}f(x)\cos nx\,\mathrm{d}x\\ &=\frac{1}{\pi}\int\limits_{-\pi}^{\pi}x^2\cos nx\,\mathrm{d}x\\[6pt] \end{aligned} \]\[x=a_0 + \frac{1^2}{a_ 1+\frac{2^2}{a_2+\frac{3^2}{a_3+ \frac{4^2}{a_4+...}}}} \] 标签：begin,Markdown,frac,公式,end,测试,theta,partial,aligned From： https://www.cnblogs.com/Euouae/p/16726606.html