10

how I can make a little the following equation smaller?

\begin{equation}
\dfrac{1}{\psi_{m}}=\dfrac{1}{\sum_{k=0}^{n-1}c_{k}\exp(\frac{-2\pi imk}{n})}}=\dfrac{1}{\langle V_{m},c \rangle+i\langle U_{m},c\rangle}
=\dfrac{\langle V_{m},c \rangle-i\langle U_{m},c\rangle}{\langle V_{m},c \rangle^{2}+\langle U_{m},c\rangle^{2}} 
\end{equation}

I tried to split it by useing multline but it doesn't give a number to this term.

3
  • multline does give equation number. Please extend your sniplet to a real example, as the code sits here it cannot even be compiled because of the two blank lines.
    – daleif
    Commented Nov 13, 2019 at 16:37
  • tex.stackexchange.com/questions/4139/… Commented Nov 13, 2019 at 16:57
  • @HumaidAlkalbani Changing font size to accommodate a long equation should be used only as last resort. And even then it shouldn't be used :-)
    – campa
    Commented Nov 13, 2019 at 16:59

3 Answers 3

15

with use of the nccmath package:

\documentclass{article}
\usepackage{nccmath}

\begin{document}
\begin{equation}\medmath{  % <--- reduce equation size for about 20 %
\frac{1}{\psi_{m}}
    =\frac{1}{\sum\limits_{k=0}^{n-1}c_{k}\exp\Bigl(\mfrac{-2\pi imk}{n}\Bigr)}
    =\frac{1}{\langle V_{m},c \rangle+i\langle U_{m},c\rangle}
    =\frac{\langle V_{m},c \rangle-i\langle U_{m},c\rangle}{\langle V_{m},c \rangle^{2}+\langle U_{m},c\rangle^{2}}
                        }
\end{equation}
or without \verb+\limits+ at \verb+\sum+, i.e. sumation limits are in inline mode:
\begin{equation}\medmath{
\frac{1}{\psi_{m}}
    =\frac{1}{\sum_{k=0}^{n-1}c_{k}\exp\Bigl(\mfrac{-2\pi imk}{n}\Bigr)}
    =\frac{1}{\langle V_{m},c \rangle+i\langle U_{m},c\rangle}
    =\frac{\langle V_{m},c \rangle-i\langle U_{m},c\rangle}{\langle V_{m},c \rangle^{2}+\langle U_{m},c\rangle^{2}}
                        }
\end{equation}
\end{document}

enter image description here

Edit: added is a case where summation limits consider @Mico comment below.

2
  • 4
    +1 for \medmath. Please don't use \sum\limits, though.
    – Mico
    Commented Nov 13, 2019 at 17:16
  • 2
    @Mico, thank you very much! Using \sum\limits make fraction a bit shorter and it is only a little bit higher than argument of \exp. To me not look so bad :-). Of course, OP can freely remove \limits. I will add this option asap.
    – Zarko
    Commented Nov 13, 2019 at 18:11
11

I suggest that you employ a split environment inside the equation environment, split the math material across three lines, and use the = symbols as the alignment points.

For extra legibility, I would also replace \exp(\frac{-2\pi imk}{n}) with \exp(-2\pi imk/n), i.e., use inline-fraction notation in the denominator term of the first fraction expression.

enter image description here

\documentclass{article}
\usepackage{amsmath} % for "split" environment
\begin{document}
\begin{equation}
\begin{split}
\frac{1}{\psi_{m}}
&=\frac{1}{\sum_{k=0}^{n-1}c_{k}\exp(-2\pi imk/n)}\\
&=\frac{1}{\langle V_{m},c\rangle + i\langle U_{m},c\rangle}\\[1ex]
&=\frac{\langle V_{m},c\rangle    - i\langle U_{m},c\rangle}%
       {\langle V_{m},c\rangle^{2}+  \langle U_{m},c\rangle^{2}}
\end{split}
\end{equation}
\end{document}
2

I'm not sure what dfrac is, so I replaced it by frac and simply used an eqnarray environment.

\documentclass{article}

\begin{document}
\begin{eqnarray}
\frac{1}{\psi_{m}} &=&\frac{1}{\displaystyle\sum_{k=0}^{n-1}c_{k}
\exp\left(\frac{-2\pi imk}{n}\right)} \nonumber \\
&=& \frac{1}{\langle V_{m},c \rangle+i\langle U_{m},c\rangle} \nonumber \\[7pt]
&=& \frac{\langle V_{m},c \rangle-i\langle U_{m},c\rangle
         }{\langle V_{m},c \rangle^{2}+\langle U_{m},c\rangle^{2}} 
\end{eqnarray}

\end{document}

enter image description here

2
  • 3
    \dfrac is similar to {\displaystyle\frac ...}. Also, why are you still using eqnarray? You should have a read of eqnarray vs align.
    – Werner
    Commented Nov 13, 2019 at 21:27
  • I did not know that about eqnarray; I learned to use it some 30 years ago and never looked back. Sorry - and thanks for the info.
    – André
    Commented Nov 13, 2019 at 21:30

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