Markdown公式测试

$$\{\begin{array}{c}{a}_{1}x+b_{1}y+c_{1}z=d_1\\ a_{2}x+b_{2}y+c_{2}z=d_2\\ a_{3}x+b_{3}y+c_{3}z=d_3\end{array}$$

$$J(\theta )=\frac{1}{2m}\sum _{i=0}^m(y^i-h_{\theta }(x^i){)}^2$$

$$\frac{\partial J(\theta )}{\partial {\theta }_j}=-\frac{1}{m}\sum _{i=0}^m(y^i-h_{\theta }(x^i))x_j^i$$

$$\begin{array}{rl}\frac{\partial J(\theta )}{\partial {\theta }_j}& =-\frac{1}{m}\sum _{i=0}^m(y^i-h_{\theta }(x^i))\frac{\partial }{\partial {\theta }_j}(y^i-h_{\theta }(x^i))\\ & =-\frac{1}{m}\sum _{i=0}^m(y^i-h_{\theta }(x^i))\frac{\partial }{\partial {\theta }_j}(\sum _{j=0}^n{\theta }_j{x}_j^i-y^i)\\ & =-\frac{1}{m}\sum _{i=0}^m(y^i-h_{\theta }(x^i))x_j^i\end{array}$$

$$\begin{array}{c}{\arg \max }_{a}f(a)=\underset{b}{\arg \max }f(b)\\ {\arg \min }_{c}f(c)=\underset{d}{\arg \min }f(d)\end{array}$$

$$\begin{array}{r}{\int }_a^b{x}^2\mathrm{d}x\end{array}$$

$$\begin{array}{r}\underset{a}{\overset{b}{\int }}{x}^2\mathrm{d}x\end{array}$$

$$\begin{array}{rl}J(\mathbf{w})& =\frac{1}{2m}\sum _{i=1}^m(f({\mathbf{x}}_{\mathbf{i}})-y_i{)}^2\\ & =\frac{1}{2m}\sum _{i=1}^m[f({\mathbf{x}}_{\mathbf{i}}){]}^2-2f({\mathbf{x}}_{\mathbf{i}}\mathbf{)}{y}_i+y_i^2\end{array}$$

$$\begin{array}{r}J(\mathbf{\theta })=-\frac{1}{m}{\sum }_{i=1}^m{y}_{i}log{h}_{\theta }(x_i)+(1-y_i)log(1-h_{\theta }(x_i))\end{array}$$

$$\begin{array}{r}\begin{array}{rl}{B}^{\prime }& =-\partial \times E,\\ E^{\prime }& =\partial \times B-4\pi j,\end{array}\}\text{Maxwell's equations}\end{array}$$

$$\begin{array}{rlrl}{\sigma }_1& =x+y& {\sigma }_2& =\frac{x}{y}\\ {\sigma }_1^{\prime }& =\frac{\partial x+y}{\partial x}& {\sigma }_2^{\prime }& =\frac{\partial \frac{x}{y}}{\partial x}\end{array}$$

$$\begin{array}{rl}{a}_n& =\frac{1}{\pi }\underset{-\pi }{\overset{\pi }{\int }}f(x)\cos nx\mathrm{d}x\\ & =\frac{1}{\pi }\underset{-\pi }{\overset{\pi }{\int }}{x}^2\cos nx\mathrm{d}x\end{array}$$

$$x=a_0+\frac{1^2}{a_1+\frac{2^2}{a_2+\frac{3^2}{a_3+\frac{4^2}{a_4+...}}}}$$