latex · markdown | 如何编写矩阵、大公式

发布时间 2023-04-12 09:41:08作者: MoonOut

1

\left[\begin{array}{c}
	a & b \\
	c & d
\end{array}\right]

效果:

\[\left[\begin{array}{c} a & b \\ c & d \end{array}\right] \]

2

\min_{T_t^{set}}J=\lim_{N\rarr\infin}E\bigg\{\sum_{t=0}^{N-1}\Delta t\cdot P_t(T_t^{set},T_t^{sup})\bigg\}
\\
s.t.\left\{\begin{array}{c}
	[\boldsymbol  T^{out}, T^{sup}]_{t+1}
	=
	f([\boldsymbol  T^{out}, T^{sup}]_{t}, \boldsymbol H_{t}^{ite}, T_{t}^{set} )
	\\
	T^{out}_{min} \le T^{out} \le T^{out}_{max} \\
	0 \le H_{t}^{ite} \le 1
\end{array}\right.
\tag 4

效果:

\[\min_{T_t^{set}}J=\lim_{N\rightarrow\infty}E\bigg\{\sum_{t=0}^{N-1}\Delta t\cdot P_t(T_t^{set},T_t^{sup})\bigg\} \]

\[s.t.\left\{\begin{array}{c} [\boldsymbol T^{out}, T^{sup}]_{t+1} = f([\boldsymbol T^{out}, T^{sup}]_{t}, \boldsymbol H_{t}^{ite}, T_{t}^{set} ) \\ T^{out}_{min} \le T^{out} \le T^{out}_{max} \\ 0 \le H_{t}^{ite} \le 1 \end{array}\right. \tag 4 \]