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