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如何让方程数组看起来更加整齐紧凑

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如何让方程数组看起来更加整齐紧凑

我正在编写一个包含两个方程数组的长列表的文档。目前，这些数组看起来非常繁琐和拥挤，我不确定什么才是让它看起来更整洁的最佳方法，而不必将它们分成几部分。

这是我的代码：

    \documentclass[12pt,twoside]{article}
    \usepackage{amsmath}
    \usepackage[makeroom]{cancel}
    \usepackage{amssymb}
    \usepackage{amsthm}
    \usepackage{amsfonts}
    \usepackage{graphicx}
    \usepackage{soul}
    \usepackage{calc} 

            \begin{document}
            \begin{eqnarray*}
    & & \frac{ \displaystyle g(v)}{\displaystyle \beta^{'}(v)} \cdot (v - b - \epsilon (b - \rho )) - G(v) \cdot (1 + \epsilon ) = 0. \\
    & \Longrightarrow & \frac{ \displaystyle g(v)}{\displaystyle \beta^{'}(v)} (v - \beta(v) - \epsilon (\beta(v) - \rho )) - G(v)(1 + \epsilon ) = 0, \\
    & \Longrightarrow & \frac{ \displaystyle g(v)}{\displaystyle \beta^{'}(v)} (v - \beta(v) - \epsilon (\beta(v) - \rho )) = G(v)(1 + \epsilon ), \\ 
    & \Longrightarrow & g(v) (v - \beta(v) - \epsilon (\beta(v) - \rho )) = G(v) \cdot \beta^{'}(v) (1 + \epsilon ), \\ 
    & \Longrightarrow & g(v) v - g(v) \beta(v) - g(v) \epsilon (\beta(v) - \rho ) = G(v) \beta^{'}(v) (1 + \epsilon ), \\
    & \Longrightarrow & g(v) v - g(v) \beta(v)(1 + \epsilon ) + g(v) \epsilon \rho = G(v) \beta^{'}(v) \cdot (1 + \epsilon ), \\
    & \Longrightarrow & g(v)v + g(v) \epsilon \rho = G(v) \beta^{'}(v) \cdot (1 + \epsilon ) + g(v) \beta(v)(1 + \epsilon ), \\
    & \Longrightarrow &  [ G(v) \beta^{'}(v) + g(v) \beta(v) ] (1 + \epsilon ) = g(v)\cdot v + g(v) \epsilon \rho, \\
    & \Longrightarrow & \frac{d}{dv}[G(v) \beta(v)] = \frac{1}{1 + \epsilon} \Big [ g(v) v + g(v) \epsilon \rho \Big ], \\
    & \Longrightarrow & \int_{\bar{v}(r)}^{v} \frac{d}{dw}[G(w) \beta(w)] = \frac{1}{1 + \epsilon} \int_{\bar{v}(r)}^{v} \Big [ g(w)w + g(w)\cdot \epsilon \rho \Big ], \\
    & \Longrightarrow & G(v) \beta(v) = \frac{1}{1 + \epsilon} \int_{\bar{v}(r)}^{v} \Big [ g(w) (w + \epsilon \cdot \rho ) \Big ], \\
    \end{eqnarray*}

     Integrating by parts, we have:

    \begin{eqnarray*}
    \beta(v) & = & \frac{1}{1 + \epsilon} \int_{\bar{v}(r)}^{v} \Bigg [ \frac{g(w)w}{G(v)} + \frac{g(w) \epsilon \rho}{G(v)} \Bigg ], \\
    & = & \frac{1}{1 + \epsilon} \Bigg [ \int_{\bar{v}(r)}^{v} \frac{g(w)w}{G(v)} + \int_{\bar{v}(r)}^{v} \frac{g(w)\epsilon \rho}{G(v)} \Bigg ], \\ 
    & = & \frac{1}{1 + \epsilon} \int_{\bar{v}(r)}^{v} \Bigg [ \frac{g(w)w}{G(v)} + \frac{g(w) \epsilon \rho}{G(v)} \Bigg ], \\
    & = & \frac{1}{1 + \epsilon} \Bigg [ \frac{G(w)w}{G(v)} \Bigg |_{\bar{v}(r)}^{v} -   \int_{\bar{v}(r)}^{v} \frac{G(w)}{G(v)} + \epsilon \rho \Bigg ( \frac{G(w)}{G(v)} \Bigg |_{\bar{v}(r)}^{v} \Bigg ) \Bigg ], \\
    & = & \frac{1}{1 + \epsilon} \Bigg [ \frac{1}{G(v)} \Bigg ( G(w)w \Bigg |_{\bar{v}(r)}^{v} + \epsilon \rho G(w) \Bigg |_{\bar{v}(r)}^{v} \Bigg ) - \int_{\bar{v}(r)}^{v} \frac{G(w)}{G(v)} \Bigg ], \\
    & = & \frac{1}{1 + \epsilon} \Bigg [ \frac{1}{G(v)} \Bigg ( G(v)v - \cancel{G(\bar{v}(r))} \bar{v}(r) + \epsilon \rho (G(v) - \cancel{G(\bar{v}(r)})  \Bigg ) - \int_{\bar{v}(r)}^{v} \frac{G(w)}{G(v)} \Bigg ], \\
    & = & \frac{1}{1 + \epsilon} \Bigg [ \frac{1}{\cancel{G(v)}} \Bigg ( \cancel{G(v)}v + \epsilon \rho \cancel{G(v)}  \Bigg ) - \int_{\bar{v}(r)}^{v} \frac{G(w)}{G(v)} \Bigg ] \\
    & = & \frac{1}{1 + \epsilon} \Bigg [ v + \epsilon \rho  - \int_{\bar{v}(r)}^{v} \frac{G(w)}{G(v)} \Bigg ], \\
\end{eqnarray*}
            \end{document}

您可以在下面看到方程式的第一部分：

在此处输入图片描述

答案1

我认为如果你用文字来解释必要的步骤，你可以展示更少的方程式。第一次积分的步骤可以如下呈现。

\documentclass[12pt,twoside]{article}
\usepackage{amsmath}
\usepackage{amssymb}
\begin{document}
We start from the fact that
\begin{equation*} 
 \frac{ \displaystyle g(v)}{\displaystyle \beta^{'}(v)} (v - b - \epsilon (b - \rho )) - G(v) (1 + \epsilon ) = 0. 
\end{equation*}
Since $b=\beta(v)$, a straightforward rearrangement shows that
\begin{equation*}
G(v) \beta'(v) + g(v) \beta(v) = g(v) \, \frac{v + \epsilon \rho}{1 + \epsilon}.
\end{equation*}
The left-hand side is now the derivative of a product, so that integrating yields
\begin{equation*} 
G(v)\beta(v) = \int g(v) \, \frac{v + \epsilon \rho}{1 + \epsilon} \, \mathrm{d} v     
\end{equation*}
\end{document}

方程式

答案2

我不确定这是否看起来更好，但这里有一个使用 align* 的版本。

\documentclass[12pt,twoside]{article}
    \usepackage{amsmath}
    \usepackage[makeroom]{cancel}
    \usepackage{amssymb}
    \usepackage{amsthm}
    \usepackage{amsfonts}
    \usepackage{graphicx}
    \usepackage{soul}
    \usepackage{calc}
    \usepackage{showframe}

\begin{document}
\begin{align*}
 &\frac{ \displaystyle g(v)}{\displaystyle \beta^{'}(v)} \cdot (v - b - \epsilon (b - \rho )) - G(v) \cdot (1 + \epsilon ) = 0. \\
 &\Longrightarrow\quad \frac{ \displaystyle g(v)}{\displaystyle \beta^{'}(v)} (v - \beta(v) - \epsilon (\beta(v) - \rho )) - G(v)(1 + \epsilon ) = 0, \\
 &\Longrightarrow\quad \frac{ \displaystyle g(v)}{\displaystyle \beta^{'}(v)} (v - \beta(v) - \epsilon (\beta(v) - \rho )) = G(v)(1 + \epsilon ), \\ 
 &\Longrightarrow\quad g(v) (v - \beta(v) - \epsilon (\beta(v) - \rho )) = G(v) \cdot \beta^{'}(v) (1 + \epsilon ), \\ 
 &\Longrightarrow\quad g(v) v - g(v) \beta(v) - g(v) \epsilon (\beta(v) - \rho ) = G(v) \beta^{'}(v) (1 + \epsilon ), \\
 &\Longrightarrow\quad g(v) v - g(v) \beta(v)(1 + \epsilon ) + g(v) \epsilon \rho = G(v) \beta^{'}(v) \cdot (1 + \epsilon ), \\
 &\Longrightarrow\quad g(v)v + g(v) \epsilon \rho = G(v) \beta^{'}(v) \cdot (1 + \epsilon ) + g(v) \beta(v)(1 + \epsilon ), \\
 &\Longrightarrow\quad  [ G(v) \beta^{'}(v) + g(v) \beta(v) ] (1 + \epsilon ) = g(v)\cdot v + g(v) \epsilon \rho, \\
 &\Longrightarrow\quad \frac{d}{dv}[G(v) \beta(v)] = \frac{1}{1 + \epsilon} \Big [ g(v) v + g(v) \epsilon \rho \Big ], \\
 &\Longrightarrow\quad \int_{\bar{v}(r)}^{v} \frac{d}{dw}[G(w) \beta(w)] = \frac{1}{1 + \epsilon} \int_{\bar{v}(r)}^{v} 
  \Big [ g(w)w + g(w)\cdot \epsilon \rho \Big ], \\
 &\Longrightarrow\quad G(v) \beta(v) = \frac{1}{1 + \epsilon} \int_{\bar{v}(r)}^{v} \Big [ g(w) (w + \epsilon \cdot \rho ) \Big ],
\end{align*}

By integrating by parts we have
\begin{align*}
\beta(v) &= \frac{1}{1 + \epsilon} \int_{\bar{v}(r)}^{v} \Bigg [ \frac{g(w)w}{G(v)} + \frac{g(w) \epsilon \rho}{G(v)} \Bigg ], \\
&= \frac{1}{1 + \epsilon} \Bigg [ \int_{\bar{v}(r)}^{v} \frac{g(w)w}{G(v)} + \int_{\bar{v}(r)}^{v} \frac{g(w)\epsilon \rho}{G(v)} \Bigg ], \\ 
&= \frac{1}{1 + \epsilon} \int_{\bar{v}(r)}^{v} \Bigg [ \frac{g(w)w}{G(v)} + \frac{g(w) \epsilon \rho}{G(v)} \Bigg ], \\
&= \frac{1}{1 + \epsilon} \Bigg [ \frac{G(w)w}{G(v)} \Bigg |_{\bar{v}(r)}^{v} -   \int_{\bar{v}(r)}^{v} \frac{G(w)}{G(v)} + \epsilon \rho \Bigg ( \frac{G(w)}{G(v)} \Bigg |_{\bar{v}(r)}^{v} \Bigg ) \Bigg ], \\
&= \frac{1}{1 + \epsilon} \Bigg [ \frac{1}{G(v)} \Bigg ( G(w)w \Bigg |_{\bar{v}(r)}^{v} + \epsilon \rho G(w) \Bigg |_{\bar{v}(r)}^{v} \Bigg ) - \int_{\bar{v}(r)}^{v} \frac{G(w)}{G(v)} \Bigg ], \\
&= \frac{1}{1 + \epsilon} \Bigg [ \frac{1}{G(v)} \Bigg (G(v)v - \cancel{G(\bar{v}(r))} \bar{v}(r) + \epsilon \rho (G(v) - \cancel{G(\bar{v}(r)})  \Bigg ) \\
&\phantom{\quad = \frac{1}{1 + \epsilon}} - \int_{\bar{v}(r)}^{v} \frac{G(w)}{G(v)} \Bigg ], \\
&= \frac{1}{1 + \epsilon} \Bigg [ \frac{1}{\cancel{G(v)}} \Bigg ( \cancel{G(v)}v + \epsilon \rho \cancel{G(v)}  \Bigg ) - \int_{\bar{v}(r)}^{v} \frac{G(w)}{G(v)} \Bigg ] \\
&= \frac{1}{1 + \epsilon} \Bigg [ v + \epsilon \rho  - \int_{\bar{v}(r)}^{v} \frac{G(w)}{G(v)} \Bigg ], 
\end{align*}
\end{document}

答案3

这是使用 AMSalign*环境对 John Kormylo 的解决方案的略微变化。align*环境以“成对”的方式组织对齐，因此我用\Longrightarrowa 将 和方程分开&&，然后在等号周围进行第二次对齐。对于第一组方程，这会产生（我认为第二组方程与 John 的方程非常相似）：

在此处输入图片描述

\Bigg ....\Bigg我还用更正确的s替换了您的 ，并将第二个方程中的\Biggl ... \Biggr降级为，因为我认为这看起来更好。\Bigg(...\Bigg)\bigg(...\bigg)

代码如下：

\documentclass[12pt,twoside]{article}
\usepackage{amsmath}
\usepackage[makeroom]{cancel}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{amsfonts}
\usepackage{graphicx}
\usepackage{soul}
\usepackage{calc}

\begin{document}
\begin{align*}
    &&\frac{ \displaystyle g(v)}{\displaystyle \beta^{'}(v)} \cdot (v - b - \epsilon (b - \rho )) - G(v) \cdot (1 + \epsilon ) &= 0. \\
    \Longrightarrow && \frac{ \displaystyle g(v)}{\displaystyle \beta^{'}(v)} (v - \beta(v) - \epsilon (\beta(v) - \rho )) - G(v)(1 + \epsilon ) &= 0, \\
    \Longrightarrow && \frac{ \displaystyle g(v)}{\displaystyle \beta^{'}(v)} (v - \beta(v) - \epsilon (\beta(v) - \rho )) &= G(v)(1 + \epsilon ), \\
    \Longrightarrow && g(v) (v - \beta(v) - \epsilon (\beta(v) - \rho )) &= G(v) \cdot \beta^{'}(v) (1 + \epsilon ), \\
    \Longrightarrow && g(v) v - g(v) \beta(v) - g(v) \epsilon (\beta(v) - \rho ) &= G(v) \beta^{'}(v) (1 + \epsilon ), \\
    \Longrightarrow && g(v) v - g(v) \beta(v)(1 + \epsilon ) + g(v) \epsilon \rho &= G(v) \beta^{'}(v) \cdot (1 + \epsilon ), \\
    \Longrightarrow && g(v)v + g(v) \epsilon \rho &= G(v) \beta^{'}(v) \cdot (1 + \epsilon ) + g(v) \beta(v)(1 + \epsilon ), \\
    \Longrightarrow &&  [ G(v) \beta^{'}(v) + g(v) \beta(v) ] (1 + \epsilon ) &= g(v)\cdot v + g(v) \epsilon \rho, \\
    \Longrightarrow && \frac{d}{dv}[G(v) \beta(v)] &= \frac{1}{1 + \epsilon} \Bigl[ g(v) v + g(v) \epsilon \rho \Bigr], \\
    \Longrightarrow && \int_{\bar{v}(r)}^{v} \frac{d}{dw}[G(w) \beta(w)] &= \frac{1}{1 + \epsilon} \int_{\bar{v}(r)}^{v} \Bigl[ g(w)w + g(w)\cdot \epsilon \rho \Bigr], \\
    \Longrightarrow && G(v) \beta(v) &= \frac{1}{1 + \epsilon} \int_{\bar{v}(r)}^{v} \Bigl[ g(w) (w + \epsilon \cdot \rho ) \Bigr], \\
\end{align*}

     Integrating by parts, we have:

\begin{align*}
    \beta(v) & = \frac{1}{1 + \epsilon} \int_{\bar{v}(r)}^{v} \Biggl[ \frac{g(w)w}{G(v)} + \frac{g(w) \epsilon \rho}{G(v)} \Biggr], \\
    & = \frac{1}{1 + \epsilon} \Biggl[ \int_{\bar{v}(r)}^{v} \frac{g(w)w}{G(v)} + \int_{\bar{v}(r)}^{v} \frac{g(w)\epsilon \rho}{G(v)} \Biggr], \\
    & = \frac{1}{1 + \epsilon} \int_{\bar{v}(r)}^{v} \Biggl[ \frac{g(w)w}{G(v)} + \frac{g(w) \epsilon \rho}{G(v)} \Biggr], \\
    & = \frac{1}{1 + \epsilon} \Biggl[ \frac{G(w)w}{G(v)} \Bigg |_{\bar{v}(r)}^{v} -   \int_{\bar{v}(r)}^{v} \frac{G(w)}{G(v)} + \epsilon \rho \biggl( \frac{G(w)}{G(v)} \Bigg |_{\bar{v}(r)}^{v} \biggr)\Biggr], \\
    & = \frac{1}{1 + \epsilon} \Biggl[ \frac{1}{G(v)} \biggl( G(w)w \Bigg |_{\bar{v}(r)}^{v} + \epsilon \rho G(w) \Bigg |_{\bar{v}(r)}^{v} \biggr)- \int_{\bar{v}(r)}^{v} \frac{G(w)}{G(v)} \Biggr], \\
    & = \frac{1}{1 + \epsilon} \Biggl[ \frac{1}{G(v)} \biggl( G(v)v - \cancel{G(\bar{v}(r))} \bar{v}(r) + \epsilon \rho (G(v) - \cancel{G(\bar{v}(r)})  \biggr)- \int_{\bar{v}(r)}^{v} \frac{G(w)}{G(v)} \Biggr], \\
    & = \frac{1}{1 + \epsilon} \Biggl[ \frac{1}{\cancel{G(v)}} \biggl( \cancel{G(v)}v + \epsilon \rho \cancel{G(v)}  \biggr)- \int_{\bar{v}(r)}^{v} \frac{G(w)}{G(v)} \Biggr] \\
    & = \frac{1}{1 + \epsilon} \Biggl[ v + \epsilon \rho  - \int_{\bar{v}(r)}^{v} \frac{G(w)}{G(v)} \Biggr], \\
\end{align*}
\end{document}

答案4

我建议array在第一个系列和align*第二个系列中使用。我更改了分隔符的大小：大多数分隔符都太大了，至少对我来说是这样的：

\documentclass[a4paper]{article}
 \usepackage[utf8]{inputenc}
 \usepackage[showframe]{geometry}
\usepackage{mathtools, cancel, array, amssymb}

\begin{document}

\[\renewcommand\arraystretch{1.5} \begin{array}{c@{\implies{}} > {\displaystyle}l}
  \multicolumn{2}{l}{ \frac{ \displaystyle g(v)}{\displaystyle \beta^{'}(v)} · (v - b - ϵ(b - ρ)) - G(v) · (1 + ϵ) = 0. } \\
  \makebox[2em]{} & \frac{ \displaystyle g(v)}{\displaystyle \beta^{'}(v)} (v - β(v) - ϵ(β(v) - ρ)) - G(v)(1 + ϵ) = 0, \\
  & \frac{ \displaystyle g(v)}{\displaystyle \beta^{'}(v)} (v - β(v) - ϵ(β(v) - ρ)) = G(v)(1 + ϵ), \\
  & g(v) (v - β(v) - ϵ(β(v) - ρ)) = G(v) · \beta^{'}(v) (1 + ϵ), \\
  & g(v) v - g(v) β(v) - g(v) ϵ(β(v) - ρ) = G(v) \beta^{'}(v) (1 + ϵ), \\
  & g(v) v - g(v) β(v)(1 + ϵ) + g(v) ϵρ= G(v) \beta^{'}(v) · (1 + ϵ), \\
  & g(v)v + g(v) ϵρ= G(v) \beta^{'}(v) · (1 + ϵ) + g(v) β(v)(1 + ϵ), \\
  & [ G(v) \beta^{'}(v) + g(v) β(v) ] (1 + ϵ) = g(v) · v + g(v) ϵ\rho, \\[0.5ex]
    & \frac{d}{dv}[G(v) β(v)] = \frac{1}{1 + ϵ} \Big [ g(v) v + g(v) ϵρ\Big ], \\[2ex]
      & ∫_{\bar{v}(r)}^{v} \frac{d}{dw}[G(w) β(w)] = \frac{1}{1 + ϵ} ∫_{\bar{v}(r)}^{v} \Big [ g(w)w + g(w) · ϵρ\Big ], \\[2ex]
        & G(v) β(v) = \frac{1}{1 + ϵ} ∫_{\bar{v}(r)}^{v} \Big [ g(w) (w + ϵ · ρ) \Big ],
        \end{array} \]
        Integrating by parts, we have
        \begin{align*}
          β(v) & = \frac{1}{1 + ϵ} ∫_{\bar{v}(r)}^{v} \biggl[ \frac{g(w)w}{G(v)} + \frac{g(w) ϵρ}{G(v)} \biggr]
          = \frac{1}{1 + ϵ} \biggl[ ∫_{\bar{v}(r)}^{v} \frac{g(w)w}{G(v)} + ∫_{\bar{v}(r)}^{v} \frac{g(w)ϵρ}{G(v)} \biggr ], \\
                   & = \frac{1}{1 + ϵ} ∫_{\bar{v}(r)}^{v} \biggl [ \frac{g(w)w}{G(v)} + \frac{g(w) ϵρ}{G(v)} \biggr], \\
                   & = \frac{1}{1 + ϵ} \biggl[ \frac{G(w)w}{G(v)} \biggr\rvert_{\bar{v}(r)}^{v} - ∫_{\bar{v}(r)}^{v} \frac{G(w)}{G(v)} + ϵρ\biggr( \frac{G(w)}{G(v)} \biggr\rvert_{\bar{v}(r)}^{v} \biggr ) \biggr ], \\
                   & = \frac{1}{1 + ϵ} \biggl [ \frac{1}{G(v)} \biggl ( G(w)w \biggr \rvert_{\bar{v}(r)}^{v} + ϵρG(w) \biggr \rvert_{\bar{v}(r)}^{v} \biggr ) - ∫_{\bar{v}(r)}^{v} \frac{G(w)}{G(v)} \biggr ], \\
                   & = \frac{1}{1 + ϵ} \biggl [ \frac{1}{G(v)} \Bigl ( G(v)v - \cancel{G(\bar{v}(r))} \bar{v}(r) + ϵρ(G(v) - \cancel{G(\bar{v}(r)}) \Bigr ) - ∫_{\bar{v}(r)}^{v} \frac{G(w)}{G(v)} \biggr ], \\
                   & = \frac{1}{1 + ϵ} \biggl [ \frac{1}{\cancel{G(v)}} \Bigl ( \cancel{G(v)}v + ϵρ\cancel{G(v)} \Bigr ) - ∫_{\bar{v}(r)}^{v} \frac{G(w)}{G(v)} \biggr ] \\
                   & = \frac{1}{1 + ϵ} \biggl [ v + ϵρ- ∫_{\bar{v}(r)}^{v} \frac{G(w)}{G(v)} \biggr ], \\
        \end{align*}

\end{document}

在此处输入图片描述

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