对齐环境多行方程式 - 从左边稍微开始等式和不等式以避免页面截断

Question

我正在使用align环境编写相当大的多行方程。以下 latex 源代码生成以下屏幕截图

\begin{align}
        \Bigl\|\sum_{i=1}^{k-1}\phi(s_t^i,a_t^i)[\bigl(\overline{V}^{k}_{t+1} - \mathbb{P}_t\overline{V}^{k}_{t+1}\bigr) (s_t^i, a_t^i)]\Bigr\|_{(\Sigma_t^k)^{-1}} &\leq 2H\sqrt{d}\Bigl[\frac{1}{2}\log(k+1) + \log(1/\delta)  \notag\\ & \quad + \log \frac{3k[2H\sqrt{dk} + c(\sqrt{\beta_k(\delta)} + 2H\sqrt{d})\sqrt{Hd\log(d/\delta)}]}{H\sqrt{d}} \Bigr]^{1/2} \notag \\
        & \quad + 2\sqrt{2}H\sqrt{d} \notag \\
        & = 2H\sqrt{d}\Bigl[\log \frac{3k\sqrt{k+1} [2H\sqrt{dk} + c(\sqrt{\beta_k(\delta)} + 2H\sqrt{d})\sqrt{Hd\log(d/\delta)}]}{H\sqrt{d}\delta}\Bigr]^{1/2} \notag\\
        & \quad + 2\sqrt{2}H\sqrt{d} \notag \\
        & \leq 2H\sqrt{d}\Bigl[\log \frac{3k\sqrt{k+1}[2H\sqrt{dk} + 2H\sqrt{d}c(\sqrt{\beta_k(\delta)} + 1)\sqrt{Hd\log(d/\delta)}]}{H\sqrt{d}\delta}\Bigr]^{1/2}  \notag \\
        & \quad + 2\sqrt{2}H\sqrt{d} \notag\\
        & = 2H\sqrt{d}\Bigl[\log \frac{3k\sqrt{k+1}[2\sqrt{k}  + 2c(\sqrt{\beta_k(\delta)} + 1)\sqrt{Hd\log(d/\delta)}]}{\delta}\Bigr]^{1/2} \notag \\
        & \quad + 2\sqrt{2}H\sqrt{d} \notag\\
        & \leq c_1 H\sqrt{d}\sqrt{\log(Hdk)/\delta} \notag\\
        & = \sqrt{\beta_k(\delta)}
\end{align}

请注意，在第一个等式中，末尾缺少幂 1/2 幂。第二个不等式也是如此。是否可以将不等式和等式从左边稍微往后拉一点？

Answer 1

为了解决主要问题，我建议您（a）&在第一行的开头插入一个新的对齐点，以及（b）\\在第一个实例之前插入一个新的换行符指令\leq。

我还建议您align用嵌套的equation和aligned环境替换环境。这将让您摆脱所有\notag指令。

\documentclass{article}
\usepackage{mathtools,amssymb}
\DeclarePairedDelimiter{\norm}{\lVert}{\rVert} % "\norm" macro
\begin{document}
\begin{equation}
\begin{aligned}[b]
&\norm[\bigg]{\sum_{i=1}^{k-1}\phi(s_t^i,a_t^i)
 \bigl[ \bigl(\, \overline{V}^{k}_{t+1} 
 - \mathbb{P}_t \overline{V}^{k}_{t+1}\bigr) 
 (s_t^i, a_t^i)\bigr] }_{{(\Sigma_t^k)}^{-1}} \\
&\leq 2H\sqrt{d}\biggl[\frac{1}{2}\log(k+1)+\log(1/\delta)\\ 
&\quad + \log\frac{3k\bigl[2H\sqrt{dk} + c\bigl(\! 
 \sqrt{\beta_k(\delta)} + 2H\sqrt{d}\,\bigr)
 \sqrt{Hd\log(d/\delta)}\,\bigr]}{H\sqrt{d}} \biggr]^{1/2}\\
&\quad + 2\sqrt{2}H\sqrt{d} \\
&= 2H\sqrt{d}\biggl[\log \frac{3k\sqrt{k+1}\,\bigl[
 2H\sqrt{dk} + c\bigl(\!\sqrt{\beta_k(\delta)} 
 + 2H\sqrt{d}\,\bigr)\sqrt{Hd\log(d/\delta)}\,\bigr]}{
 H\sqrt{d}\delta}\biggr]^{1/2} \\
&\quad + 2\sqrt{2}H\sqrt{d} \\
&\leq 2H\sqrt{d}\biggl[\log \frac{3k\sqrt{k+1}\,\bigl[ 
 2H\sqrt{dk} + 2H\sqrt{d}c\bigl(\!\sqrt{\beta_k(\delta)} 
 + 1\bigr)\sqrt{Hd\log(d/\delta)}\,\bigr]}{
 H\sqrt{d}\delta}\biggr]^{1/2} \\
&\quad + 2\sqrt{2}H\sqrt{d} \\
&= 2H\sqrt{d}\biggl[\log \frac{3k\sqrt{k+1}\,\bigl[2\sqrt{k} 
 + 2c\bigl(\!\sqrt{\beta_k(\delta)} + 1\bigr)
 \sqrt{Hd\log(d/\delta)}\,\bigr]}{\delta}\biggr]^{1/2} \\
&\quad + 2\sqrt{2}H\sqrt{d} \\
&\leq c_1 H\sqrt{d}\sqrt{\log(Hdk)/\delta} \\
&= \sqrt{\beta_k(\delta)}
\end{aligned}
\end{equation}
\end{document}