7

I would like to know what is the latex command to include a double lower index using sigma notation (i.e: second index lies underneath the first index)?

I tried typing:

\displaystyle\sum_{m,n=-\infty, \; (m,n)\neq (0,0)}^{\infty}
\left[
  \frac{1}{(z-2 m\omega_{1} - 2 n\omega_{2})^{2}}
 -\frac{1}{   4(m\omega_{1} +   n\omega_{2})^{2}}
\right] 

I would appreciate the assistance. Thanks

3
  • 2
    Welcome to TeX.SX! Could you please add some context? What's \dsy, for instance?
    – egreg
    Feb 10, 2014 at 14:39
  • Hi Greg. \dsy means displaystyle for the sigma sign
    – user12015
    Feb 10, 2014 at 14:42
  • 2
    You don't need \displaystyle; if your formula is inline, use \sum\limits (or simply \sum, which is better).
    – egreg
    Feb 10, 2014 at 14:46

2 Answers 2

6

The \substack command from the amsmath package can do what you are after:

\documentclass{article}

\usepackage{amsmath}

\begin{document}
\begin{equation*}
  \sum_{\substack{m,n=-\infty,\\(m,n)\neq (0,0)}}^{\infty}\left[\frac{1}{(z-2 m\omega_{1}-2 n\omega_{2})^{2}}-\frac{1}{4(m\omega_{1}+n\omega_{2})^{2}}\right]
\end{equation*}
\end{document}

output1

Or without the comma, since the next condition is on a new line:

\documentclass{article}

\usepackage{amsmath}

\begin{document}
\begin{equation*}
  \sum_{\substack{m,n=-\infty\\(m,n)\neq (0,0)}}^{\infty}\left[\frac{1}{(z-2 m\omega_{1}-2 n\omega_{2})^{2}}-\frac{1}{4(m\omega_{1}+n\omega_{2})^{2}}\right]
\end{equation*}
\end{document}

output2

1
  • Thank you very much for all your comments and i appreciate the assistance!
    – user12015
    Feb 10, 2014 at 15:01
1

Try using substack:

\usepackage{amsmath}

...

\begin{equation}
  \prod_{\substack{
    1 \le i \le n \\
    1 \le j \le m} }
  M_{i,j}
\end{equation}

This example is taken from the Latex Wikibook.

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