latex 公式如何放在底部代码:\end{multicols}%开始通栏显示\rule{\textwidth}{0.5pt}\begin{eqnarray}\begin{array}{l} \left\| {{{\tilde y}_{k + 1}}(t)} \right\| \le \left\| {I - C{B_f}{k_4}} \right\|\cdot\left\| {{{\tilde y}_k}(t)} \right

来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/18 21:47:16
latex公式如何放在底部代码:\end{multicols}%开始通栏显示\rule{\textwidth}{0.5pt}\begin{eqnarray}\begin{array}{l}\left\

latex 公式如何放在底部代码:\end{multicols}%开始通栏显示\rule{\textwidth}{0.5pt}\begin{eqnarray}\begin{array}{l} \left\| {{{\tilde y}_{k + 1}}(t)} \right\| \le \left\| {I - C{B_f}{k_4}} \right\|\cdot\left\| {{{\tilde y}_k}(t)} \right
latex 公式如何放在底部

代码:
\end{multicols}%开始通栏显示
\rule{\textwidth}{0.5pt}
\begin{eqnarray}
\begin{array}{l}
 \left\| {{{\tilde y}_{k + 1}}(t)} \right\| \le \left\| {I - C{B_f}{k_4}} \right\|\cdot\left\| {{{\tilde y}_k}(t)} \right\| + \:\left\| {CL + C{B_f}{k_5}} \right\|\cdot\int_0^t {\left\| {{{\tilde y}_k}(\tau )} \right\|} {\rm{d}}\tau  + \left[ {\left\| {CA + C{B_f}{k_1}} \right\| + \left\| {C{k_x}} \right\|} \right]\int_0^t {\left\| {{{\hat x}_{k + 1}}(\tau ) - {{\hat x}_k}(\tau )} \right\|} {\rm{d}}\tau  \\
 \;\;\;\;\;\;\;\;\;\;\;\; \le \left\| {I - C{B_f}{k_4}} \right\|\cdot\left\| {{{\tilde y}_k}(t)} \right\| + \:\left\| {CL + C{B_f}{k_5}} \right\|\cdot\int_0^t {\left\| {{{\tilde y}_k}(\tau )} \right\|} {\rm{d}}\tau \; + \left[ {\left\| {A + {B_f}{k_1}} \right\| + \left\| {{k_x}} \right\|} \right]\int_0^t {\left( {\left\| {{{\tilde y}_{k + 1}}(\tau )} \right\|{\rm{ + }}\left\| {{{\tilde y}_k}(\tau )} \right\|} \right)} {\rm{d}}\tau  \\
 \;\;\;\;\;\;\;\;\;\;\;\; = {\Xi _1}\left\| {{{\tilde y}_k}(t)} \right\| + {\Xi _2}\int_0^t {\left\| {{{\tilde y}_k}(\tau )} \right\|} {\rm{d}}\tau \; + {\Xi _3}\int_0^t {\left\| {{{\tilde y}_{k + 1}}(\tau )} \right\|} {\rm{d}}\tau  + {\Xi _3}\int_0^t {\left\| {{{\tilde y}_k}(\tau )} \right\|} {\rm{d}}\tau  \\
 \end{array}
\end{eqnarray}
\rule{\textwidth}{0.5pt}
\begin{multicols}{2}%开始双栏显示




我想让这个长公式放在这一页的底部,

latex 公式如何放在底部代码:\end{multicols}%开始通栏显示\rule{\textwidth}{0.5pt}\begin{eqnarray}\begin{array}{l} \left\| {{{\tilde y}_{k + 1}}(t)} \right\| \le \left\| {I - C{B_f}{k_4}} \right\|\cdot\left\| {{{\tilde y}_k}(t)} \right
把公式放在figure 环境中,用[b!] 控制显示在底部.