3

Consider the following:

\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}

enter image description here

How to i get the \left[ -- the "big" square bracket in line 3 -- to match the \right] big bracket in line 4?

3
  • don't use \left\right use \bigl[ ... \bigr] or whatever size you need (this will be a duplicate) Commented Oct 27, 2022 at 13:10
  • @DavidCarlisle This gives me an error..
    – Math
    Commented Oct 27, 2022 at 13:15
  • then you did something wrong. You probably left in \right. or some such Commented Oct 27, 2022 at 13:22

2 Answers 2

3

To preserve the pairwise contents of the numerator terms of the three additive components of M_3, I would like to suggest you employ the \splitdfrac macro of the mathtools package for the numerator on line 3.

A separate issue: I think that the terms S_k^{*^2} and I_k^{*^2} in lines 1 and 2 are wrong, as the exponent 2 is too small. I believe {S_k^*}^2 and {I_k^*}^2 is better, as the exponent 2 will be rendered in \scriptstyle math mode instead of in \scriptscriptstyle math mode.

enter image description here

Note that there are no instances of \left and \right anywhere. That's quite deliberate.

\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}
3
  • This is nice but don't you think it is a little confusing compared to what i tried? I'm no expert(you are!) but that's my two cents :) Also, unrelated to this question(actually somewhat it is), I had asked David if he could "fix" the mistakes in my thesis but unfortunately he couldn't do it! Maybe can you? I will pay for your time :)
    – Math
    Commented Oct 27, 2022 at 16:22
  • @Math - I guess it really depends on what you're looking to emphasize to the reader. My solution brings out the overall structure -- e.g., three additive terms, the numerators of the additive terms consists of a product of two components -- quite forcefully, at the cost of making the reader slow down a bit to take in what the numerator of the third term is all about. Put differently, readers may have to slow down a bit, but I don't think that the choice of notation will confuse them.
    – Mico
    Commented Oct 27, 2022 at 16:32
  • Fair enough, I'll stick with your solution. So to my other question, would you be able to edit?
    – Math
    Commented Oct 27, 2022 at 16:42
1

Mico suggestion is the good one, but if you still want to use your coding as is, then try with the below:

Using left and 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*}

Using biggl and 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*}
1
  • You don't need \biggr. on line 3 or \biggl. on line 4. Furthermore, placing \biggl[\sigma_2^2(k) at the end of line 3 would appear to be inferior to placing it at the start of line 4..
    – Mico
    Commented Oct 27, 2022 at 16:30

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