Perhaps this as an answer to your first question
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.
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}