我正在努力尝试对齐一组长度差异很大的方程式。所有方程式都应与公式和变量声明之间的逗号对齐。
尽管使用了，公式 3 仍然太长。如果我在和split之间再换行，就会出现错误\left(\right)"Extra }, or forgotten right"。
拆分完成后，我想将这种多行公式的第一行左对齐，接下来的几行左对齐并在前面略微跳过，最后一行右对齐到逗号。有可能这样做吗？或者还有其他建议，可以在一个环境中以漂亮的方式一起呈现许多这样的长公式吗？
我找到了将一个长公式排列在多行中或将多个公式围绕逗号或其他内容排列在一起或将一个公式左对齐（如我所述）（mathtools' \MoveEqLeft）的解决方案，但我不知道如何同时完成所有这些操作。
\documentclass[]{scrreprt}
\usepackage{amsmath}
\begin{document}
\begin{align}
\sum_{k=K_1+1}^{K_{21}-1} s_{ik} \tilde{h}_{k+1}
\leq h_i
\leq \sum_{k=K_1+1}^{K_{21}} s_{ik} \tilde{h}_k
+ \sum_{k=K_1+1}^{K_{21}-1} \bar{s}_{ik} \tilde{h}_{k+1}
& , i \in \mathcal{I}_{21}
\label{eq_1} \\
%
\begin{split}
\sum_{l=K_{21}+1}^{K_{22}} \alpha_{lk} (s_{il} + \bar{s}_{il}) \left( 1 + \varepsilon \right)
+ w_i + \sum_{j=I_1+1}^{I_2} \alpha_{ji} (s_{jk} + \bar{s}_{jk}) w_j
& \\
\geq (1 + \varepsilon)(1-s_{ik}-\bar{s}_{ik})
& , i \in \mathcal{I}_{22}, k \in \mathcal{K}_{22}
\end{split}
\label{eq_2} \\
%
\begin{split}
\left( 1- \sum_{j=I_1+1}^{I_2}\alpha_{ji}t_{jk}\right)
\cdot \left( w_i - \frac{1}{m+2} - \varepsilon -|w_i - \frac{1}{m+2} - \varepsilon|\right) \cdot \ldots
& \\
\ldots \cdot \left( h_i + \sum_{l=K_1+1}^{K_2} \alpha_{lk} \tilde{h}_l - H_{12} - \varepsilon -
| h_i + \sum_{l=K_1+1}^{K_2} \alpha_{lk} \tilde{h}_l - H_{12} - \varepsilon | \right)
= 0
& , i \in \mathcal{I}_{22}, k \in\mathcal{K}_{21}
\end{split}
\label{eq_3} \\
%
H_{21} = H_{12}
&
\label{eq_4}
\end{align}
\end{document}
补充：
到目前为止，我使用的是旧的 eqnarray 环境。使用它，它看起来几乎应该是这样的：
\documentclass{scrreprt}
\begin{document}
\begin{eqnarray}
\sum_{k=K_1+1}^{K_{21}-1} s_{ik} \tilde{h}_{k+1}
\leq h_i
\leq \sum_{k=K_1+1}^{K_{21}} s_{ik} \tilde{h}_k + \sum_{k=K_1+1}^{K_{21}-1} \bar{s}_{ik} \tilde{h}_{k+1}
& , {} & i \in \mathcal{I}_{21}
\label{eq:sf_height_fits_levelheight_i21}\\
%
\left( 1- \sum_{j=I_1+1}^{I_2}\alpha_{ji}t_{jk}\right)
\times \left( w_i - \frac{1}{m+2} - \varepsilon -|w_i - \frac{1}{m+2} - \varepsilon|\right) \qquad \qquad \quad
& & \nonumber \\
\qquad \times \left( h_i + \sum_{l=K_1+1}^{K_2} \alpha_{lk} \tilde{h}_l - H_{12} - \varepsilon - | h_i + \sum_{l=K_1+1}^{K_2} \alpha_{lk} \tilde{h}_l - H_{12} - \varepsilon |\right)
\geq 0
%
& , & i \in \mathcal{I}_{22}, k \in\mathcal{K}_{21} \qquad \quad
\label{eq:sf_I22_not_in_new_I21} \\
H_{21} = H_{12}
& &
\label{eq:sf_def_H21}
\end{eqnarray}
\end{document}
唯一的问题：公式 3 太宽，公式数字超出了文本区域（并且 eqnarray 环境不是第一选择）。
答案1
这是一个aligned在环境中使用环境的解决方案align。它加载\mathtools包（及其\smashoperator宏）以压缩求和符号周围的空间。第三个等式增加了一个换行符，现在跨越了三行。
\documentclass{scrreprt}
\usepackage{mathtools}
\DeclarePairedDelimiter\abs\lvert\rvert
\begin{document}
\setcounter{chapter}{1} % just for this example
\begin{align}
&\smashoperator[r]{\sum_{k=K_1+1}^{K_{21}-1}}
s_{ik} \tilde{h}_{k+1}
\leq h_i \leq
\smashoperator{\sum_{k=K_1+1}^{K_{21}}}
s_{ik} \tilde{h}_k +
\smashoperator{\sum_{k=K_1+1}^{K_{21}-1}}
\bar{s}_{ik} \tilde{h}_{k+1},
\quad i\in\mathcal{I}_{21} \label{eq_1} \\[2ex]
%
&\smashoperator[r]{\sum_{l=K_{21}+1}^{K_{22}}}
\begin{aligned}[t]
&\alpha_{lk} (s_{il} + \bar{s}_{il}) (1+\varepsilon)
+ w_i +
\smashoperator{\sum_{j=I_1+1}^{I_2}}
\alpha_{ji} (s_{jk} + \bar{s}_{jk}) w_j\\
&\geq (1 + \varepsilon)(1-s_{ik}-\bar{s}_{ik})
,\quad i \in \mathcal{I}_{22}, k \in \mathcal{K}_{22}
\end{aligned}
\label{eq_2} \\[2ex]
%
&\begin{aligned}
&\biggl( 1- \smashoperator{\sum_{j=I_1+1}^{I_2}}
\alpha_{ji}t_{jk}\biggr)
\times \biggl( w_i - \frac{1}{m+2} - \varepsilon -
\abs[\Big]{w_i - \frac{1}{m+2} - \varepsilon}
\biggr) \times \dotsb \\
&\quad\dotsb\times\biggl( h_i +
\smashoperator{\sum_{l=K_1+1}^{K_2}}
\alpha_{lk} \tilde{h}_l - H_{12} - \varepsilon -
\abs[\Big]{ h_i +
\smashoperator{\sum_{l=K_1+1}^{K_2}}
\alpha_{lk} \tilde{h}_l - H_{12} - \varepsilon }
\biggr)
= 0,\\
&\qquad\quad i \in \mathcal{I}_{22}, k \in\mathcal{K}_{21}
\end{aligned} \label{eq_3} \\[2ex]
%
&H_{21} = H_{12}
\label{eq_4}
\end{align}
\end{document}
答案2
与来自align和multlined环境的组合mathtools。我也不会将其用于\cdot乘法（括号内的项是相乘的，这是不言而喻的）：
\documentclass{scrreprt}
\usepackage{mathtools}
\DeclarePairedDelimiter\abs{\lvert}{\rvert}
\setcounter{chapter}{1} % just for this example
\begin{document}
\begin{align}
& \sum_{k=K_1+1}^{K_{21}-1} s_{ik} \tilde{h}_{k+1}
\leq h_i \leq
\sum_{k=K_1+1}^{K_{21}} s_{ik} \tilde{h}_k
+ \sum_{k=K_1+1}^{K_{21}-1} \bar{s}_{ik} \tilde{h}_{k+1},
\qquad i \in \mathcal{I}_{21}
\label{eq_1} \\[1em]
%
& \begin{multlined}[0.7\linewidth]
\sum_{l=K_{21}+1}^{K_{22}} \alpha_{lk}(s_{il} + \bar{s}_{il})
(1 + \varepsilon) + w_i \\
+ \sum_{j=I_1+1}^{I_2} \alpha_{ji}(s_{jk} + \bar{s}_{jk}) w_j
\geq (1 + \varepsilon)(1-s_{ik}-\bar{s}_{ik}),
\qquad i \in \mathcal{I}_{22}, k \in \mathcal{K}_{22}
\end{multlined}
\label{eq_2} \\[1em]
%
& \begin{multlined}[0.7\linewidth]
\biggl( 1- \sum_{j=I_1+1}^{I_2}\alpha_{ji}t_{jk}\biggr)
\biggl( w_i - \frac{1}{m+2} - \varepsilon -
\abs*{w_i - \frac{1}{m+2} - \varepsilon}\biggr) \dotsm \\
\dotsm \biggl( h_i + \sum_{l=K_1+1}^{K_2} \alpha_{lk} \tilde{h}_l
- H_{12} - \varepsilon - \abs[\bigg]{h_i + \sum_{l=K_1+1}^{K_2} \alpha_{lk}
\tilde{h}_l - H_{12} - \varepsilon }\biggr) = 0, \\
i \in \mathcal{I}_{22}, k \in\mathcal{K}_{21}
\end{multlined}
\label{eq_3} \\[1em]
%
& H_{21} = H_{12}
\label{eq_4}
\end{align}
\end{document}
答案3
您可以尝试使用aligned而不是split。这就是您想要的吗？
\documentclass[]{scrreprt}
\usepackage{amsmath}
\begin{document}
\begin{align}
\begin{aligned}
\sum_{k=K_1+1}^{K_{21}-1} s_{ik} \tilde{h}_{k+1}
\leq h_i
\leq \sum_{k=K_1+1}^{K_{21}} s_{ik} \tilde{h}_k
+ \sum_{k=K_1+1}^{K_{21}-1} \bar{s}_{ik} \tilde{h}_{k+1}
& , i \in \mathcal{I}_{21}
\end{aligned}
\label{eq_1} \\
%
\begin{aligned}
\sum_{l=K_{21}+1}^{K_{22}} \alpha_{lk} (s_{il} + \bar{s}_{il}) \left( 1 + \varepsilon \right)
+ w_i + \sum_{j=I_1+1}^{I_2} \alpha_{ji} (s_{jk} + \bar{s}_{jk}) w_j
& \\
\geq (1 + \varepsilon)(1-s_{ik}-\bar{s}_{ik})
& , i \in \mathcal{I}_{22}, k \in \mathcal{K}_{22}
\end{aligned}
\label{eq_2} \\
%
\begin{aligned}
\left( 1- \sum_{j=I_1+1}^{I_2}\alpha_{ji}t_{jk}\right)
\cdot \left( w_i - \frac{1}{m+2} - \varepsilon -|w_i - \frac{1}{m+2} - \varepsilon|\right) \cdot \ldots
& \\
\ldots \cdot \left( h_i + \sum_{l=K_1+1}^{K_2} \alpha_{lk} \tilde{h}_l - H_{12} - \varepsilon -
| h_i + \sum_{l=K_1+1}^{K_2} \alpha_{lk} \tilde{h}_l - H_{12} - \varepsilon | \right)
= 0
& , i \in \mathcal{I}_{22}, k \in\mathcal{K}_{21}
\end{aligned}
\label{eq_3} \\
%
H_{21} = H_{12}
&
\label{eq_4}
\end{align}
\end{document}
答案4
也许这更符合您的要求。我能够将最后一个等式拆分到括号中间（使用 $\left.$ 和 $\right.$），并且我选择使用水平间距来实现缩进。
\documentclass[]{scrreprt}
\usepackage{amsmath}
\begin{document}
\begin{align}
& \sum_{k=K_1+1}^{K_{21}-1} s_{ik} \tilde{h}_{k+1}
\leq h_i
\leq \sum_{k=K_1+1}^{K_{21}} s_{ik} \tilde{h}_k
+ \sum_{k=K_1+1}^{K_{21}-1} \bar{s}_{ik} \tilde{h}_{k+1}
& , & i \in \mathcal{I}_{21}
\label{eq_1} \\
%
\begin{split}
& \sum_{l=K_{21}+1}^{K_{22}} \alpha_{lk} (s_{il} + \bar{s}_{il}) \left( 1 + \varepsilon \right)
+ w_i + \sum_{j=I_1+1}^{I_2} \alpha_{ji} (s_{jk} + \bar{s}_{jk}) w_j
\\
& \kern 2cm \geq (1 + \varepsilon)(1-s_{ik}-\bar{s}_{ik})
\end{split}
& , & i \in \mathcal{I}_{22}, k \in \mathcal{K}_{22}
\label{eq_2} \\
%
\begin{split}
& \left( 1- \sum_{j=I_1+1}^{I_2}\alpha_{ji}t_{jk}\right)
\cdot \left( w_i - \frac{1}{m+2} - \varepsilon -|w_i - \frac{1}{m+2} - \varepsilon|\right) \cdot \ldots
\\
& \kern 2cm \ldots \cdot \left( h_i + \sum_{l=K_1+1}^{K_2} \alpha_{lk} \tilde{h}_l - H_{12} - \varepsilon -
| h_i + \right.
\\
& \kern 3.85cm \left. \sum_{l=K_1+1}^{K_2} \alpha_{lk} \tilde{h}_l - H_{12} - \varepsilon | \right)
= 0
\end{split}
& , & i \in \mathcal{I}_{22}, k \in\mathcal{K}_{21}
\label{eq_3} \\
%
& H_{21} = H_{12}
\label{eq_4}
\end{align}
\end{document}
