3

I am trying to make the first 3 lines of this equation flush left. How can I do so? Also, is there an easier way to enter the parentheses and have the size I want? If I don't use the \left( and \right commands, the parenthesis are too small around my summation symbols.

\begin{align*}
\frac{A_{j+1,n-1}\left( t\right) }{
A_{j,n-1}\left( t\right) }\leq \frac{j}{j-1}\Longleftrightarrow\MoveEqLeft[30]\\
A_{j+1,n-1}\left( t\right) \leq \left( \frac{j}{j-1}\right) A_{j,n-1}\left(
t\right) \Longleftrightarrow \MoveEqLeft[30]\\
\sum\limits_{i=0}^{j}\binom{n-1}{i}F\left( t\right) ^{n-1-i}\left(
1-F\left( t\right) \right) ^{i}+\sum\limits_{i=j+1}^{n-1}\binom{n-1}{i}%
F\left( t\right) ^{n-1-i}\left( 1-F\left( t\right) \right) ^{i}\frac{j+1}{1+i%
} \MoveEqLeft[30]\\
\leq \left( \frac{j}{j-1}\right) \left( \sum\limits_{i=0}^{j-1}\binom{n-1%
}{i}F\left( t\right) ^{n-1-i}\left( 1-F\left( t\right) \right)
^{i}+\sum\limits_{i=j}^{n-1}\binom{n-1}{i}F\left( t\right) ^{n-1-i}\left(
1-F\left( t\right) \right) ^{i}\frac{j}{1+i}\right) \MoveEqLeft[30]\\
=\sum\limits_{i=0}^{j-1}\binom{n-1}{i}\left( \frac{j}{j-1}\right) F\left(
t\right) ^{n-1-i}\left( 1-F\left( t\right) \right)
^{i}+\sum\limits_{i=j}^{n-1}\binom{n-1}{i}F\left( t\right) ^{n-1-i}\left(
1-F\left( t\right) \right) ^{i}\frac{j\left( \frac{j}{j-1}\right) } {1+i}\MoveEqLeft[30]
 \end{align*}

This is what the equation looks like currently

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  • insert an & before each line. align takes input strings in pairs -- the first set flush right (as it would be before a sign of relation), and the second, flush left. (it wouldn't be a bad idea to look at the documentation: texdoc amsmath if you're using a system based on tex live.) Commented Jun 6, 2016 at 15:10
  • 1
    don't do \left( t\right) it makes unwanted horizontal space just use (t) also you don't need \sum\limits just \sum will use the limits layout in display math. Commented Jun 6, 2016 at 15:19

1 Answer 1

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Here are two versions where the leading & force left alignment; since lines are long, I also provided a version with split lines.

Note that \left( t\right) and \left( 1-F\left( t\right) \right) just waste space. I also removed useless parentheses around fractions. The \limits declaration is not needed in displays.

\documentclass{article}
\usepackage[margin=2cm]{geometry}
\usepackage{amsmath}

\begin{document}

Version with long lines
\begin{align*}
& \frac{A_{j+1,n-1}(t)}{A_{j,n-1}(t)}\leq \frac{j}{j-1}\Longleftrightarrow
\\
& A_{j+1,n-1}(t) \leq \frac{j}{j-1} A_{j,n-1}(t) \Longleftrightarrow
\\
&\! \sum_{i=0}^{j} \binom{n-1}{i} F(t)^{n-1-i}(1-F(t))^{i}
  +
  \sum_{i=j+1}^{n-1} \binom{n-1}{i} F(t)^{n-1-i}(1-F(t))^{i}\frac{j+1}{1+i}
\\
&\qquad \leq 
  \frac{j}{j-1} \biggl(\,
    \sum_{i=0}^{j-1}\binom{n-1}{i}F(t)^{n-1-i}(1-F(t))^{i}
    +
    \sum_{i=j}^{n-1}\binom{n-1}{i}F(t)^{n-1-i}(1-F(t))^{i}\frac{j}{1+i}
  \biggr)
\\
&\qquad = 
  \sum_{i=0}^{j-1}\binom{n-1}{i}\frac{j}{j-1}F(t)^{n-1-i}(1-F(t))^{i}
  +
  \sum_{i=j}^{n-1}\binom{n-1}{i}F(t)^{n-1-i}(1-F(t))^{i}\frac{j\frac{j}{j-1}}{1+i}
\end{align*}
and a version with short lines
\begin{align*}
& \frac{A_{j+1,n-1}(t)}{A_{j,n-1}(t)}\leq \frac{j}{j-1}\Longleftrightarrow
\\[2ex]
& A_{j+1,n-1}(t) \leq \frac{j}{j-1} A_{j,n-1}(t) \Longleftrightarrow
\\[2ex]
&\! \sum_{i=0}^{j} \binom{n-1}{i} F(t)^{n-1-i}(1-F(t))^{i}
\\&\qquad\qquad\qquad+
  \sum_{i=j+1}^{n-1} \binom{n-1}{i} F(t)^{n-1-i}(1-F(t))^{i}\frac{j+1}{1+i}
\\[2ex]
&\qquad \leq 
  \frac{j}{j-1} \biggl(\,
    \sum_{i=0}^{j-1}\binom{n-1}{i}F(t)^{n-1-i}(1-F(t))^{i}
\\&\qquad\qquad\qquad+
    \sum_{i=j}^{n-1}\binom{n-1}{i}F(t)^{n-1-i}(1-F(t))^{i}\frac{j}{1+i}
  \biggr)
\\[2ex]
&\qquad = 
  \sum_{i=0}^{j-1}\binom{n-1}{i}\frac{j}{j-1}F(t)^{n-1-i}(1-F(t))^{i}
\\&\qquad\qquad\qquad+
  \sum_{i=j}^{n-1}\binom{n-1}{i}F(t)^{n-1-i}(1-F(t))^{i}\frac{j\frac{j}{j-1}}{1+i}
\end{align*}

\end{document}

enter image description here

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  • 1
    +1. You may also want to mention that all \limits directives may be omitted as the material is in displaymath mode.
    – Mico
    Commented Jun 6, 2016 at 15:33

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