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I have two equations. One is big and the other is small. I want to align them and want these two equations to have only one equation number. Another question is : Are there ways to make these equations look more appealing?

enter image description here

\documentclass{article}

\usepackage{amsmath,mathrsfs,xcolor,mathtools}

\begin{document}

\begin{equation}\label{eq:lmlt}
  \begin{gathered}
t_0 = 0, \quad s_0 = t_0+t_1, \quad t_{n+1} = s_n + \dfrac{M(1+M(s_n-t_n))
(s_n-t_n)^2}{2(1-M_0t_n)^5},\\
s_{n+m} = t_{n+m-1}+ \dfrac{1}{1-M_0t_{n+1}}\big[\dfrac{M(t_{n+1}-s_n)^2}{2} +
\dfrac{13L(s_n-t_n)^4}{108} +\dfrac{{\color{red}N}M(s_n-t_n)^4}{9(1-M_0t_n)}\\
+\dfrac{M^3(s_n-t_n)^4}{3(1-M_0t_n)^2}\big]
  \end{gathered}
\end{equation}

\end{document}
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1 Answer 1

up vote 5 down vote accepted

Perhaps this as an answer to your first question

enter image description here

which uses the aligned environment- note that using \big[ doesn't give the correct size of []; we'll fix that in the next solution below.

\subsection*{Original}
\begin{equation}\label{eq:lmlt}
    \begin{gathered}
        t_0 = 0, \quad s_0 = t_0+t_1\\
        \begin{aligned}
            t_{n+1} & = s_n + \dfrac{M(1+M(s_n-t_n)) (s_n-t_n)^2}{2(1-M_0t_n)^5},                                              \\
            s_{n+m} & = t_{n+m-1}+ \dfrac{1}{1-M_0t_{n+1}}\big[\dfrac{M(t_{n+1}-s_n)^2}{2} + \dfrac{13L(s_n-t_n)^4}{108}       \\
                    & \phantom{=} +\dfrac{{\color{red}N}M(s_n-t_n)^4}{9(1-M_0t_n)} +\dfrac{M^3(s_n-t_n)^4}{3(1-M_0t_n)^2}\big] 
        \end{aligned}
    \end{gathered}
\end{equation}

There are lots of different ways to present these equations- so this will be quite subjective. Here's one alternative- it's not drastically different, it just splits up the initial and subsequent iterations which can be a little easier to read.

enter image description here

Note that this solution uses \left[ ... \right. and \left. ... \right] to get the correct sizing of your [ ]; you should implement this in whichever solution you use.

\subsection*{Alternative}
The intial values of $t$ and $s$ are defined by
\begin{equation*}
    t_0  = 0,   \qquad s_0  = t_0+t_1 
\end{equation*}
with subsequent iterations following the formulas
\begin{align*}
    t_{n+1} & = s_n + \dfrac{M(1+M(s_n-t_n)) (s_n-t_n)^2}{2(1-M_0t_n)^5},                                                    \\
    s_{n+m} & = t_{n+m-1}+ \dfrac{1}{1-M_0t_{n+1}}\left[\dfrac{M(t_{n+1}-s_n)^2}{2} + \dfrac{13L(s_n-t_n)^4}{108}\right.     \\
            & \phantom{=}+\left.\dfrac{{\color{red}N}M(s_n-t_n)^4}{9(1-M_0t_n)}+\dfrac{M^3(s_n-t_n)^4}{3(1-M_0t_n)^2}\right] 
\end{align*}

Here's the complete MWE- note that the mathtools package loads amsmath so there's no need to load amsmath if you load mathtools

\documentclass{article}

\usepackage{xcolor,mathtools}

\begin{document}

\subsection*{Original}
\begin{equation}\label{eq:lmlt}
    \begin{gathered}
        t_0 = 0, \quad s_0 = t_0+t_1\\
        \begin{aligned}
            t_{n+1} & = s_n + \dfrac{M(1+M(s_n-t_n)) (s_n-t_n)^2}{2(1-M_0t_n)^5},                                              \\
            s_{n+m} & = t_{n+m-1}+ \dfrac{1}{1-M_0t_{n+1}}\big[\dfrac{M(t_{n+1}-s_n)^2}{2} + \dfrac{13L(s_n-t_n)^4}{108}       \\
                    & \phantom{=} +\dfrac{{\color{red}N}M(s_n-t_n)^4}{9(1-M_0t_n)} +\dfrac{M^3(s_n-t_n)^4}{3(1-M_0t_n)^2}\big] 
        \end{aligned}
    \end{gathered}
\end{equation}


\subsection*{Alternative}
The intial values of $t$ and $s$ are defined by
\begin{equation*}
    t_0  = 0,   \qquad s_0  = t_0+t_1 
\end{equation*}
with subsequent iterations following the formulas
\begin{align*}
    t_{n+1} & = s_n + \dfrac{M(1+M(s_n-t_n)) (s_n-t_n)^2}{2(1-M_0t_n)^5},                                                    \\
    s_{n+m} & = t_{n+m-1}+ \dfrac{1}{1-M_0t_{n+1}}\left[\dfrac{M(t_{n+1}-s_n)^2}{2} + \dfrac{13L(s_n-t_n)^4}{108}\right.     \\
            & \phantom{=}+\left.\dfrac{{\color{red}N}M(s_n-t_n)^4}{9(1-M_0t_n)}+\dfrac{M^3(s_n-t_n)^4}{3(1-M_0t_n)^2}\right] 
\end{align*}
\end{document}
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