2

I'm sorry if this has been asked. I'm new to this and to Latex. I have this piece of code:

\[
  e^{-\xi_1 Y_1^N} = \exp\left(-\frac{\xi_1}{N}\ \sum_{n\geq1}{}V_n^1
  e^{-s_n}\1_{\{ s_n \leq T_1\}}\right)
\]

And when compiling the line of fraction gets placed at the height of the middle of the sigma symbol of the \sum. I would like to place it at the height of the Base of the sigma symbol.

Thank you in advance!

5
  • 2
    How is the \1 macro defined?
    – Mico
    Commented Feb 4, 2016 at 21:39
  • 2
    What are you trying to achieve by shifting the fraction term down so much?
    – Mico
    Commented Feb 4, 2016 at 21:42
  • 1
    Please always post a complete document, but the fraction position is not affected by the summation, if that were not there it would be placed at the same position relative to the V, why do you want to lower it to subscript position? Commented Feb 4, 2016 at 21:50
  • @Mico \1 is defined as \def\1{\mathbbm{1}} Looking at your answer I realize that in fact what I "needed" was to shift up the Summation, rather than shifting sown the fraction, and it is only to fit what I do when handwriting in paper, thought it would be more tidy... If it gets similar to the first way; I'd rather keep it like it is now, too. Thank you for your help!
    – Max
    Commented Feb 4, 2016 at 22:22
  • 1
    @Max - Rather than raise the summation symbol and its associated lower limit, you may want to place the summation's limit to the side of the summation symbol. This may be achieved by inserting \nolimits immediately after \sum. I've provided an addendum to my answer to show the resulting look.
    – Mico
    Commented Feb 5, 2016 at 7:28

1 Answer 1

4

The first equation below features the frational term shifted down so that the fraction bar is at the same height as the lower horizontal stroke of the summation symbol. The second equation features a more standard look.

Speaking for myself, I'd go with the look of the second equation.

enter image description here

\documentclass{article}
\usepackage{amsmath,amsfonts}
\DeclareMathOperator{\1}{\mathbf{1}} % educated guess...
\newcommand\e{\mathrm{e}}
\begin{document}

fractional term shifted down:
\[ 
\e^{-\xi_1^{} Y_1^N} 
= \exp\biggl(- \lower1.6ex\hbox{$\dfrac{\xi_1}{N}$} 
     \sum_{n\geq1}{}V_n^1 
     \e^{-s_n} \1_{\{ s_n \leq T_1\}}\biggr) 
\]

\bigskip
normal look:
\[ 
\e^{-\xi_1^{} Y_1^N} 
= \exp\biggl(- \frac{\xi_1}{N} 
     \sum_{n\geq1}{}V_n^1 
     \e^{-s_n} \1_{\{ s_n \leq T_1\}}\biggr) 
\]

\end{document}

Addendum: Judging by the comment the OP left, it may be that the following look -- placing the limit of the summation to the side rather than below the summation symbol -- is really what is supposed to be achieved.

enter image description here

\documentclass{article}
\usepackage{amsmath}
\DeclareMathOperator{\1}{\mathbf{1}} 
\newcommand\e{\mathrm{e}}
\begin{document}
\[ 
\e^{-\xi_1^{} Y_1^N} 
= \exp\Bigl(- \frac{\xi_1}{N} 
     \sum\nolimits_{n\geq1}V_n^1 
     \e^{-s_n} \1_{\{ s_n \leq T_1\}}\Bigr) 
\]
\end{document}
2
  • 2
    You're using understatement. ;-) In my opinion the first way lacks any meaning.
    – egreg
    Commented Feb 4, 2016 at 22:08
  • After six upvotes, one un-upvote. I'd find it helpful to know what caused the un-upvote.
    – Mico
    Commented Feb 5, 2016 at 17:20

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