由于方程式超过两行导致方括号大小不合适 - 如何使它们的大小相等？

Question

考虑以下：

\documentclass[12pt,a4paper,oneside]{book}
\usepackage{amsfonts, graphicx, verbatim, mathtools,amssymb, amsthm, mathrsfs,amsmath}
\begin{document}
\begin{align*}
M_3&=\frac{(\mu(k)+\rho(k)+\xi(k))\left(\sigma_1^2(k)+2\int_M D_1^2(k,y)\nu(dy)\right)}{2M_1}S_k^{*^2}\\[1ex]    &+\frac{(\mu(k)+\alpha(k)+\theta(k)+\gamma(k))\left(\sigma_2^2(k)+2\int_M D_2^2(k,y)\nu(dy)\right)}{2M_2}I_k^{*^2}\\[1ex]
&+\frac{2\mu(k)+\rho(k)+\xi(k)+\alpha(k)+\theta(k)+\gamma(k)}{2\beta(k)(1-\rho(k))(1-\xi(k))}\left[\sigma_2^2(k)\right.\\
&\left.+2\int_M (D_2(k,y)-\log(1+D_2(k,y))\nu(dy)) \right]I_k^*
\end{align*}
\end{document}

我如何才能使\left[第 3 行中的“大”方括号与\right]第 4 行中的大括号相匹配？

Answer 1

为了保留 的三个加法分量的分子项的成对内容M_3，我建议您使用第 3 行分子包\splitdfrac的宏。mathtools

另一个问题：我认为第 1 行和第 2 行中的术语S_k^{*^2}和I_k^{*^2}是错误的，因为指数2太小。我相信{S_k^*}^2和{I_k^*}^2更好，因为指数将在数学模式中呈现，而不是在数学模式中2呈现。\scriptstyle\scriptscriptstyle

请注意，任何地方都没有\leftand的实例\right。这是经过深思熟虑的。

\documentclass[12pt,a4paper,oneside]{book}
\usepackage{%amsfonts, %% amsfonts is loaded automatically by amssymb
      graphicx, verbatim,
      mathtools, % for '\splitdfrac' macro    
      amssymb, amsthm, mathrsfs,
      %amsmath %% amsmath is loaded automatically by mathtools
      }
\begin{document}
\begin{align*}
M_3
&=\frac{\bigl[\mu(k)+\rho(k)+\xi(k)\bigr]
        \bigl[\sigma_1^2(k)+2\int_M D_1^2(k,y)\nu(dy)\bigr]}{%
        2M_1} \, {S_k^*}^2 \\[1ex]
&\quad+
  \frac{\bigl[\mu(k)+\alpha(k)+\theta(k)+\gamma(k)\bigr]
        \bigl[\sigma_2^2(k)+2\int_M D_2^2(k,y)\nu(dy)\bigr]}{%
        2M_2} \, {I_k^*}^2 \\[1ex]
&\quad+
  \frac{\splitdfrac{\bigl[2\mu(k)+\rho(k)+\xi(k)
            +\alpha(k)+\theta(k)+\gamma(k)\bigr]}
        {\cdot\bigl[\sigma_2^2(k) +2\int_M
            \bigl(D_2(k,y)-\log(1+D_2(k,y))\nu(dy) \bigr) \bigr]}}%
     {2\beta(k)(1-\rho(k))(1-\xi(k))} \, I_k^* \\
\end{align*}
\end{document}

Answer 2

Mico这个建议很好，但如果您仍然想按原样使用编码，请尝试以下操作：

使用left和right

\documentclass[12pt,a4paper,oneside]{book}
\usepackage{amsfonts, graphicx, verbatim, mathtools,amssymb, amsthm, mathrsfs,amsmath}
\begin{document}
\begin{align*}
M_3&=\frac{(\mu(k)+\rho(k)+\xi(k))\left(\sigma_1^2(k)+2\int_M D_1^2(k,y)\nu(dy)\right)}{2M_1}S_k^{*^2}\\[1ex]
&\quad +\frac{(\mu(k)+\alpha(k)+\theta(k)+\gamma(k))\left(\sigma_2^2(k)+2\int_M D_2^2(k,y)\nu(dy)\right)}{2M_2}I_k^{*^2}\\[1ex]
&\quad+\frac{2\mu(k)+\rho(k)+\xi(k)+\alpha(k)+\theta(k)+\gamma(k)}{2\beta(k)(1-\rho(k))(1-\xi(k))}\left[\vphantom{+2\int_M (D_2(k,y)-\log(1+D_2(k,y))\nu(dy))}\sigma_2^2(k)\right.\\
&\quad\left.+2\int_M (D_2(k,y)-\log(1+D_2(k,y))\nu(dy)) \right]I_k^*
\end{align*}

使用biggl和biggr

\begin{align*}
M_3&=\frac{(\mu(k)+\rho(k)+\xi(k))\left(\sigma_1^2(k)+2\int_M D_1^2(k,y)\nu(dy)\right)}{2M_1}S_k^{*^2}\\[1ex]
&\quad +\frac{(\mu(k)+\alpha(k)+\theta(k)+\gamma(k))\left(\sigma_2^2(k)+2\int_M D_2^2(k,y)\nu(dy)\right)}{2M_2}I_k^{*^2}\\[1ex]
&\quad+\frac{2\mu(k)+\rho(k)+\xi(k)+\alpha(k)+\theta(k)+\gamma(k)}{2\beta(k)(1-\rho(k))(1-\xi(k))}\biggl[\sigma_2^2(k)\biggr.\\
&\quad\biggl.+2\int_M (D_2(k,y)-\log(1+D_2(k,y))\nu(dy)) \biggr]I_k^*
\end{align*}