The values of the upper and lower limits of a summation are usually printed below and above sigma. But I find TeX something very unusual. How to correct this?

Here is the MWE:

<\mu_{m}> = \frac{\sum_{-J}^{J}{-g\mu_{\beta}M_{J}B.\exp{\left(\frac{-g\mu_{\beta}M_{J}B}{kT}\right)}}}{\sum_{-J}^{J}{\exp{\left(\frac{-g\mu_{\beta}M_{J}B}{kT}\right)}}}

The output:
enter image description here

  • 3
    Don't use <>. Use \langle \rangle instead.
    – Sigur
    Jul 19 '14 at 16:49
  • 4
    It's because you have the sums in a fraction. Use \sum\limits_{-J}^{J} to change the limit position. Jul 19 '14 at 16:49
  • @Sigur \langle \rangle, makes it soooo thin Jul 19 '14 at 16:54
  • 4
    OK, but < is a relation symbol and \langle is a delimiter. They are different in some sense and the space around them are different.
    – Sigur
    Jul 19 '14 at 16:56

LaTeX by default uses scriptstyle for sums inside fractions. You can change the sums to displaystyle like this:

\langle \mu_{m} \rangle = 
\frac{{\displaystyle \sum_{-J}^{J}}{-g\mu_{\beta}M_{J}B.\exp{\left(\frac{-g\mu_{\beta}M_{J}B}{kT}\right)}}}
{{\displaystyle \sum_{-J}^{J}}{\exp{\left(\frac{-g\mu_{\beta}M_{J}B}{kT}\right)}}}

PLEASE, use \langle \rangle instead of < >. Here's the output: enter image description here

  • 5
    \displaystyle is not recommended here. See @Paul's comment above. The ideal would be \sum\limits.
    – Sigur
    Jul 19 '14 at 16:57
  • 2
    Also, \left and \right add certain unwanted space, better \Bigl( and \Bigr). And one more, if you want to use the . as a multiplication symbol (binary operator), wether you define a command, or use \mathbin{.}.
    – Manuel
    Jul 19 '14 at 17:02
  • @Sigur Err... ok. Displaystyle also makes the sigma bigger, but maybe it looks better this way. Jul 19 '14 at 17:02
  • 3
    @JairoBochi I too would use \sum\limits precisely in order to avoid big summation signs; you might show both alternatives. The period is never used for multiplication in math, as far as I know. Either \cdot or nothing. In this case nothing is surely better.
    – egreg
    Jul 19 '14 at 20:04
  • 1
    I think the best solution is to use the nccmath package, and \medopsum\limits_{-J}^{J}` to have medium-sized sum operator.
    – Bernard
    Jul 19 '14 at 21:03

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