# Break down long summation with fraction

How can I break down this equation and make it looks nice in two column format:

\begin{align}
G=-\frac{\sum_{k=0}^{N}A_{nk}\left(2\frac{(2\frac{q_{sm}}{q^{mex}}-1)^{k+1}
(k+1)}{q^{kex}(2\frac{q_{sm}}{q^{mex}}-1)}-\frac{2k(1-\frac{q_{sm}}
{q^{mex}})}{(q^{kex}(2\frac{q_{sm}}{q^{mex}}-1)^{1-k})}+\frac{2kq_{sm}}
{(q^{kex})^2(\frac{q_{sm}}{q^{mex}}-1)^{1-k}}+\frac{4kq_{sm}(1-\frac{q_{sm}}
{q^{mex}})(1-k)}{(q^{kex})^2(2\frac{q_{sm}}{q^{mex}}-1)^{1-k}(2\frac{q_{sm}}
{q^{mex}}-1)}\right)}{nF}
\end{align}

• You are probably more likely to get help if you provide a full minimal example, that others can copy and run immediately without having to add anything. Also an idea to add a picture of what it looks like right now (I, for example, is currently not at a pc so cannot test, but I can still make suggestions based on code or images). Since two column formats are very restrictive, best solution is often to rewrite the equation. Feb 27, 2017 at 19:06
• the answer depends crucially on information that you have not shown, such as the column width and font size, that is why it is always best to provide a complete small document that shows the problem. Also are kex, mex , sm products or multi-letter identifiers? Feb 27, 2017 at 19:24

Something like this? (Note that I've enlarged some of the round parentheses, replaced the outer-most round parentheses with curly braces, and rendered the sub- and superscript terms in math-roman.)

\documentclass[twocolumn]{article}
\usepackage{amsmath}
\begin{document}
\hrule % just to illustrate width of textblock

\begin{align*}
G&=-\frac{1}{nF}\sum_{k=0}^{N}A_{nk}\Biggl\{
2\frac{\bigl(2\frac{q^{}_{\mathrm{sm}}}{q^{\mathrm{mex}}}-1\bigr)^{k+1}(k+1)}{
q^{\mathrm{kex}}\bigl(2\frac{q^{}_{\mathrm{sm}}}{q^{\mathrm{mex}}}-1\bigr)}\\
q^{\mathrm{kex}}\bigl(2\frac{q^{}_{\mathrm{sm}}}{q^{\mathrm{mex}}}-1\bigr)^{1-k}}\\
{\bigl(q^{\mathrm{kex}}\bigr)^2
\bigl(\frac{q^{}_{\mathrm{sm}}}{q^{\mathrm{mex}}}-1\bigr)^{1-k}}\\
{q^{\mathrm{mex}}}\bigr)(1-k)}
{\bigl(q^{\mathrm{kex}}\bigr)^2
\bigl(2\frac{q^{}_{\mathrm{sm}}}{q^{\mathrm{mex}}}-1\bigr)^{1-k}
\bigl(2\frac{q^{}_{\mathrm{sm}}}
{q^{\mathrm{mex}}}-1\bigr)}\Biggr\}
\end{align*}

\end{document}


Addendum: You should consider making the substitution $Q=q_{\mathrm{sm}}/q^{\mathrm{mex}}$, which would greatly simplify (and clarify) the appearance of the four-line formula. (You're obviously free to choose a different symbol in lieu of Q.) Also, it looks like one take the common factor 2 and place it outside the summation.

\documentclass[twocolumn]{article}
\usepackage{amsmath}
\begin{document}
\hrule % just to illustrate width of textblock

\bigskip\noindent
Set $Q=q_{\mathrm{sm}}/q^{\mathrm{mex}}$. Then
\begin{align*}
G&=-\frac{2}{nF}\sum_{k=0}^{N}A_{nk}\biggl\{
\frac{(2Q-1)^{k+1}(k+1)}{q^{\mathrm{kex}}(2Q-1)}\\
{(q^{\mathrm{kex}})^2 (Q-1)^{1-k}}\\
{(q^{\mathrm{kex}})^2(2Q-1)^{1-k}(2Q-1)}\biggr\}
\end{align*}

\end{document}


This works, but I'd also get rid of some of the nested \fracs and replace them with a / b.

\documentclass{article}

\usepackage{amsmath}

\begin{document}

\begin{multline}
G=-\frac{1}{nF}\sum_{k=0}^{N}A_{nk}\Biggl(2\frac{(2\frac{q_{sm}}{q^{mex}}-1)^{k+1}
(k+1)}{q^{kex}(2\frac{q_{sm}}{q^{mex}}-1)}-\frac{2k(1-\frac{q_{sm}}
{q^{mex}})}{(q^{kex}(2\frac{q_{sm}}{q^{mex}}-1)^{1-k})}\\+\frac{2kq_{sm}}
{(q^{kex})^2(\frac{q_{sm}}{q^{mex}}-1)^{1-k}}+\frac{4kq_{sm}(1-\frac{q_{sm}}
{q^{mex}})(1-k)}{(q^{kex})^2(2\frac{q_{sm}}{q^{mex}}-1)^{1-k}(2\frac{q_{sm}}
{q^{mex}}-1)}\biggr)
\end{multline}

\end{document}

• Does this solution take into account the condition that the document is in two-column format?
– Mico
Feb 27, 2017 at 19:20