Better looking equations

Question

I am trying to find a way to make the equations that appear in this text a bit more better looking...

\documentclass{article}
\usepackage{amsmath, amssymb}
\usepackage{relsize}

\begin{document}

\begin{equation}
    N_\text{c}(i) = \dfrac{N_\text{o}(i)}{1 - \mathlarger{‎‎\sum}_{j=i_0}^{i-1}N_\text{o}(j)/N_\text{b} - N_\text{o}(i)/2N_\text{b}}\label{eq:Bollinger}
\end{equation}

\begin{equation}
    N_\text{c}(i) = -N_\text{b}\ln{\left(1-\dfrac{N_\text{o}(i)/N_\text{b}}{1-\mathlarger{\sum}_{j=i_0}^{i-1}N_\text{o}(j)/N_\text{b}}\right)}
\end{equation}

\begin{equation}
    N_\text{c}(i) = -N_\text{b}\dfrac{\ln{\left(1-\dfrac{N_\text{o}(i)/N_\text{b}}{1-\mathlarger{\sum}_{j=i_0}^{i-1}N_\text{o}(j)/N_\text{b}}\right)}}{1-\sigma\tanh\left(\sigma\mathlarger{\sum}_{j=i_0}^{i-1}N_\text{c}(j)/N_\text{b}\right)}
\end{equation}

\end{document}

Any idea on how to make them look more beautiful?

Answer 1

Avoid having blank lines before display mat, or consecutive display math (TeX can not really handle either in a sane way) and keep control over the delimiters by avoiding \left\right I guessed you want to keep the \limits setting of the summation so I kept the normal summation (which has better vertical alignment) and made a larger but still fixed bracket size to cope with the large numerator.

\documentclass{article}
\usepackage{amsmath, amssymb}
\makeatletter
\def\Biggg#1{{\hbox{$\left#1\vbox to21\p@{}\right.\n@space$}}}
\def\Bigggl{\mathopen\Biggg}
\def\Bigggm{\mathrel\Biggg}
\def\Bigggr{\mathclose\Biggg}
\makeatother

\begin{document}

align (or gather if no alignement)
\begin{align}
    N_{\mathrm{c}}(i) &= \frac{N_\text{o}(i)}{1 - ‎‎\sum\limits_{j=i_0}^{i-1}N_\text{o}(j)/N_{\mathrm{b}} - N_\text{o}(i)/2N_{\mathrm{b}}}\label{eq:Bollinger}
\\[\jot]
    N_{\mathrm{c}}(i) &= -N_{\mathrm{b}}\ln{\Biggl(1-\frac{N_\text{o}(i)/N_{\mathrm{b}}}{1-\sum\limits_{j=i_0}^{i-1}N_\text{o}(j)/N_{\mathrm{b}}}\Biggr)}
\\[\jot]
    N_{\mathrm{c}}(i) &= -N_{\mathrm{b}}\frac{\ln{\Bigggl(1-\dfrac{N_\text{o}(i)/N_{\mathrm{b}}}{1-\sum\limits_{j=i_0}^{i-1}N_\text{o}(j)/N_{\mathrm{b}}}\Bigggr)}}{1-\sigma\tanh\left(\sigma\sum\limits_{j=i_0}^{i-1}N_{\mathrm{c}}(j)/N_{\mathrm{b}}\right)}
\end{align}

\end{document}

Answer 2

Instead of \mathlarger{\sum}, just make sure the numerators and denominators are in display-style math. To avoid excessively large parentheses, use \sum\nolimits. Use \Biggl( and \Biggr) for the larger parentheses in rows 2 and 3, and use \Bigl( and \Bigr) for the parentheses in the denominator of row 3. Finally, I would use \mathrm rather than \text to render the items c, i and o, to get the math-specific spacing.

\documentclass{article}
\usepackage{amsmath, amssymb}
\newcommand\ddfrac[2]{\dfrac{\displaystyle #1}{\displaystyle #2}}

\begin{document}

\begin{align}
N_{\mathrm{c}}(i) 
&= \ddfrac{N_{\mathrm{o}}(i)}{1 - \sum\nolimits_{j=i^{}_0}^{i-1}N_{\mathrm{o}}(j)/N_{\mathrm{b}} - N_{\mathrm{o}}(i)/2N_{\mathrm{b}}}\label{eq:Bollinger}\\[2ex]
N_{\mathrm{c}}(i) 
&= -N_{\mathrm{b}}\ln\Biggl(1-\ddfrac{N_{\mathrm{o}}(i)/N_{\mathrm{b}}}{1-\sum\nolimits_{j=i^{}_0}^{i-1}N_{\mathrm{o}}(j)/N_{\mathrm{b}}}\Biggr) \\[2ex]
N_{\mathrm{c}}(i) 
&= -N_{\mathrm{b}}\,\ddfrac{\ln\Biggl(1-\ddfrac{N_{\mathrm{o}}(i)/N_{\mathrm{b}}}{1-\sum\nolimits_{j=i^{}_0}^{i-1}N_{\mathrm{o}}(j)/N_{\mathrm{b}}}\Biggr)}{1-\sigma\tanh\Bigl(\sigma\sum\nolimits_{j=i^{}_0}^{i-1}N_{\mathrm{c}}(j)/N_{\mathrm{b}}\Bigr)}
\end{align}

\end{document}

Answer 3

If you must keep the equation as it is and just remove the excess space (nasty thing to do but...) you can use fixit from this solution:

\fixit[<mathstyle>]{<left-delim>}{<content>}{<right-delim>}

The command definition is:

\usepackage{amsmath}
\newcommand\fixit[4][\displaystyle]{
  \setbox0=\hbox{$#1#3$}
  \setbox2=\hbox{$\vcenter{\copy0}$}
  \raisebox{\dimexpr\ht0-\ht2}{$#1\left#2\copy2\right#4$}
}

MWE for your case:

\documentclass{article}
\usepackage{amsmath, amssymb}
\usepackage{relsize}
\newcommand\fixit[4][\displaystyle]{
  \setbox0=\hbox{$#1#3$}
  \setbox2=\hbox{$\vcenter{\copy0}$}
  \raisebox{\dimexpr\ht0-\ht2}{$#1\left#2\copy2\right#4$}
}
\begin{document}

\begin{equation}
    N_\text{c}(i) = -N_\text{b}\ln{\fixit{(}{1-\dfrac{N_\text{o}(i)/N_\text{b}}{1-\mathlarger{\sum}_{j=i_0}^{i-1}N_\text{o}(j)/N_\text{b}}}{)}}
\end{equation}

\end{document}

The result:

Answer 4

One way to reduce the white space (I do not think that the following is really more beautiful) is to make your \mathlarger small again.

The white space (AFAIK) results from the computations of \left etc. which stupidly scales up the parenthesis. See e.g. "(" or "\left(" parentheses?

\documentclass{article}
\usepackage{amsmath, amssymb}
\usepackage{relsize}

\begin{document}

\begin{equation}
    N_\text{c}(i) = \frac{N_\text{o}(i)}{1 - ‎‎\sum_{j=i_0}^{i-1}N_\text{o}(j)/N_\text{b} - N_\text{o}(i)/2N_\text{b}}\label{eq:Bollinger}
\end{equation}

\begin{equation}
    N_\text{c}(i) = -N_\text{b}\ln{\Biggl(1-\frac{N_\text{o}(i)/N_\text{b}}{1-\sum_{j=i_0}^{i-1}N_\text{o}(j)/N_\text{b}}\Biggr)}
\end{equation}

\begin{equation}
    N_\text{c}(i) = -N_\text{b}\frac{\ln{\Biggl(1-\dfrac{N_\text{o}(i)/N_\text{b}}{1-\sum_{j=i_0}^{i-1}N_\text{o}(j)/N_\text{b}}\Biggr)}}{1-\sigma\tanh\left(\sigma\sum_{j=i_0}^{i-1}N_\text{c}(j)/N_\text{b}\right)}
\end{equation}

\end{document}