Fitting multiline equation into margins

Question

I have this equation. It is quite wide, and I've had to split things over two lines. However, the part to the left of the = is quite long, and it's causing me some issues.

I've posted a MWE here. This produces the following that I could just about live with (it is 7.6 pt too wide apparently):

The problem is that when I have the same equation in my full document, it adds extra space after the = and is now definitely not OK:

I presume this is a consequence of the longer equation number. Can anyone suggest a way to get my equation fitted into the margins? I have thought of breaking the line after the = sign, but I'd like the = sign on the lower line to line up and that's the one that is spilling off to the right. I have also thought of breaking before the =, and aligning the = signs at the left hand end, which would be the best solution, but I honestly can't work out how to do it. I suspect I need some combination of gather/split or something, but it's doing my head in!

\documentclass[12pt, a4paper, oneside, fleqn]{report}

% Page geometry etc
%----------------------------
\usepackage{setspace}                % allow different line spacing
\renewcommand{\topfraction}{0.85}
\renewcommand{\textfraction}{0.1}
\usepackage[top=2.5cm, left=3.5cm, bottom=2.5cm, right=2.5cm, includehead]{geometry}
\geometry{headheight=28pt, headsep=18pt}

% Maths stuff
%------------
\usepackage{amsmath}

\begin{document}

\onehalfspacing
\mathindent=\parindent

The equation below is too wide for the page margins. The question is how best to format it so that it looks sensible but doesn't spill off the right hand margin?
\begin{alignat}{2}
{\textstyle \frac{1}{2}} \textrm{Cov}(\Delta_{i,p1}-\Delta_{i,p2},\Delta_{j,p1}-\Delta_{j,p2})
& = && {\textstyle \frac{1}{2}}\textrm{Cov}(\Delta_{i,p1},\Delta_{j,p1})
+ {\textstyle \frac{1}{2}}\textrm{Cov}(\Delta_{i,p2},\Delta_{j,p2})  - \notag \\
&   && {\textstyle \frac{1}{2}}\textrm{Cov}(\Delta_{i,p1},\Delta_{j,p2})
- {\textstyle \frac{1}{2}}\textrm{Cov}(\Delta_{i,p2},\Delta_{j,p1}) \\
& = && \textrm{Cov}(\Delta_{i,p1},\Delta_{j,p2})|_{p1=p2}
- \textrm{Cov}(\Delta_{i,p1},\Delta_{j,p2})|_{p1 \ne p2}
\end{alignat}


\end{document}

UPDATE: I've just had a go with gather, but now my equation numbers are way off to the right, even in my test document:

\begin{gather}
{\textstyle \frac{1}{2}} \textrm{Cov}(\Delta_{i,p1}-\Delta_{i,p2},\Delta_{j,p1}-\Delta_{j,p2}) \notag \\
\begin{alignat}{2}
& = {} && {\textstyle \frac{1}{2}}\textrm{Cov}(\Delta_{i,p1},\Delta_{j,p1})
+ {\textstyle \frac{1}{2}}\textrm{Cov}(\Delta_{i,p2},\Delta_{j,p2})  - \notag \\
&   && {\textstyle \frac{1}{2}}\textrm{Cov}(\Delta_{i,p1},\Delta_{j,p2})
- {\textstyle \frac{1}{2}}\textrm{Cov}(\Delta_{i,p2},\Delta_{j,p1}) \\
& = {} && \textrm{Cov}(\Delta_{i,p1},\Delta_{j,p2})|_{p1=p2}
- \textrm{Cov}(\Delta_{i,p1},\Delta_{j,p2})|_{p1 \ne p2}
\end{alignat}
\end{gather}

Answer 1

The answer by @JF Meier can be edited easily to correct the alignment of the '=' signs by using

\begin{align}
{\textstyle \frac{1}{2}} \textrm{Cov}(\Delta_{i,p1}-\Delta_{i,p2},&\Delta_{j,p1}-\Delta_{j,p2}) \notag \\
 = & {\textstyle \frac{1}{2}}\textrm{Cov}(\Delta_{i,p1},\Delta_{j,p1})
+ {\textstyle \frac{1}{2}}\textrm{Cov}(\Delta_{i,p2},\Delta_{j,p2})  - \notag \\
&    {\textstyle \frac{1}{2}}\textrm{Cov}(\Delta_{i,p1},\Delta_{j,p2})
- {\textstyle \frac{1}{2}}\textrm{Cov}(\Delta_{i,p2},\Delta_{j,p1}) \\
= & \textrm{Cov}(\Delta_{i,p1},\Delta_{j,p2})|_{p1=p2}
- \textrm{Cov}(\Delta_{i,p1},\Delta_{j,p2})|_{p1 \ne p2}
\end{align}

Answer 2

I'd go with align, the easiest way is masking the width of the common left hand side, rather than fixing an arbitrary alignment point.

\documentclass[12pt, a4paper, oneside, fleqn]{report}

% Page geometry etc
%----------------------------
\usepackage[top=2.5cm, left=3.5cm, bottom=2.5cm, right=2.5cm, includehead]{geometry}
\geometry{headheight=28pt, headsep=18pt}

% Maths stuff
%------------
\usepackage{mathtools}

\DeclareMathOperator{\Cov}{Cov}

\setlength{\mathindent}{\parindent}

\begin{document}
The equation below is too wide for the page margins. The question is how best 
to format it so that it looks sensible but doesn't spill off the right hand margin?
\begin{align}
{\textstyle \frac{1}{2}} \textrm{Cov}(\Delta_{i,p1}-\Delta_{i,p2},&\Delta_{j,p1}-\Delta_{j,p2}) \notag \\
 = & {\textstyle \frac{1}{2}}\textrm{Cov}(\Delta_{i,p1},\Delta_{j,p1})
+ {\textstyle \frac{1}{2}}\textrm{Cov}(\Delta_{i,p2},\Delta_{j,p2})  - \notag \\
&    {\textstyle \frac{1}{2}}\textrm{Cov}(\Delta_{i,p1},\Delta_{j,p2})
- {\textstyle \frac{1}{2}}\textrm{Cov}(\Delta_{i,p2},\Delta_{j,p1}) \\
= & \textrm{Cov}(\Delta_{i,p1},\Delta_{j,p2})|_{p1=p2}
- \textrm{Cov}(\Delta_{i,p1},\Delta_{j,p2})|_{p1 \ne p2}
\end{align}
Above is Chaplin's code, below it's mine.
\begin{align}
\mathmakebox[3\mathindent][l]{
  \tfrac{1}{2} \Cov(\Delta_{i,p1}-\Delta_{i,p2},\Delta_{j,p1}-\Delta_{j,p2})
}\notag \\
 ={}& \tfrac{1}{2}\Cov(\Delta_{i,p1},\Delta_{j,p1}) +
      \tfrac{1}{2}\Cov(\Delta_{i,p2},\Delta_{j,p2}) - {} \notag \\
    & \tfrac{1}{2}\Cov(\Delta_{i,p1},\Delta_{j,p2}) -
      \tfrac{1}{2}\Cov(\Delta_{i,p2},\Delta_{j,p1}) \\[\jot]
 ={}& \Cov(\Delta_{i,p1},\Delta_{j,p2})|_{p1=p2} -
      \Cov(\Delta_{i,p1},\Delta_{j,p2})|_{p1 \ne p2}
\end{align}


\end{document}

Note the vertical spacing between the rows containing the numbers, that helps in distinguishing the two block.

You can also easily change the apparent width of the left hand side, here set to three times the math indent. If you don't want to load mathtools just for that, the same appearance is obtained by

\hspace{3\mathindent}
\lefteqn{
  \hspace{-3\mathindent}
  \tfrac{1}{2} \Cov(\Delta_{i,p1}-\Delta_{i,p2},\Delta_{j,p1}-\Delta_{j,p2})
}\notag \\

Answer 3

I'm with JF Meier on this, but here is how I'd to it. Note the minus belong at the start of the line in displayed math, not the end.

\documentclass{report}
\usepackage{amsmath,mathtools}
\DeclareMathOperator\Cov{Cov}
\begin{document}
\begin{align}
  \MoveEqLeft[3] 
  \tfrac{1}{2}
  \Cov(\Delta_{i,p1}-\Delta_{i,p2},\Delta_{j,p1}-\Delta_{j,p2}) 
  \notag \\
  = {} &
  \tfrac{1}{2}\Cov(\Delta_{i,p1},\Delta_{j,p1}) +
  \tfrac{1}{2}\Cov(\Delta_{i,p2},\Delta_{j,p2}) \notag
  \\
  & - \tfrac{1}{2}\Cov(\Delta_{i,p1},\Delta_{j,p2}) -
  \tfrac{1}{2}\Cov(\Delta_{i,p2},\Delta_{j,p1})
  \\
  = {} & \Cov(\Delta_{i,p1},\Delta_{j,p2})|_{p1=p2} -
  \Cov(\Delta_{i,p1},\Delta_{j,p2})|_{p1 \ne p2}
\end{align}
\end{document}

---

Edit, as to the pull back, I think the image below explains it. In the top image the vertical line illustrate where the alignment is, notice how the two subsequent lines are indented compared to where the alignment is.

What \MoveEqLeft does is the same as in the upper image, but instead of indenting line 2 and 3, we do a negative indent (we pull it back) on the first line, giving the impression that line 2 and 3 are indented. (mathematicians are lazy ;-)

Answer 4

Just to add my own gather solution in case it helps anyone in the future - which works as long as I didn't need the equation numbers on the aligned section

\begin{gather}
\tfrac{1}{2} \Cov(\Delta_{i,p1}-\Delta_{i,p2},\Delta_{j,p1}-\Delta_{j,p2}) \\
\begin{alignedat}{2}
& = {} && \tfrac{1}{2}\Cov(\Delta_{i,p1},\Delta_{j,p1})
+ \tfrac{1}{2}\Cov(\Delta_{i,p2},\Delta_{j,p2})  - {} \\
&   && \tfrac{1}{2}\Cov(\Delta_{i,p1},\Delta_{j,p2})
- \tfrac{1}{2}\Cov(\Delta_{i,p2},\Delta_{j,p1}) \notag \\
& = {} && \Cov(\Delta_{i,p1},\Delta_{j,p2})|_{p1=p2}
- \Cov(\Delta_{i,p1},\Delta_{j,p2})|_{p1 \ne p2} 
\end{alignedat}
\end{gather}

As it happens I'd made a note to remove the extra equation number anyway, but I'd prefer it on the bottom line than the top

Answer 5

I would suggest something like this:

\begin{align}
{\textstyle \frac{1}{2}} \textrm{Cov}(\Delta_{i,p1}-\Delta_{i,p2},&\Delta_{j,p1}-\Delta_{j,p2}) \notag \\
& =  {\textstyle \frac{1}{2}}\textrm{Cov}(\Delta_{i,p1},\Delta_{j,p1})
+ {\textstyle \frac{1}{2}}\textrm{Cov}(\Delta_{i,p2},\Delta_{j,p2})  - \notag \\
&    {\textstyle \frac{1}{2}}\textrm{Cov}(\Delta_{i,p1},\Delta_{j,p2})
- {\textstyle \frac{1}{2}}\textrm{Cov}(\Delta_{i,p2},\Delta_{j,p1}) \\
& =  \textrm{Cov}(\Delta_{i,p1},\Delta_{j,p2})|_{p1=p2}
- \textrm{Cov}(\Delta_{i,p1},\Delta_{j,p2})|_{p1 \ne p2}
\end{align}

Answer 6

Another suggestion: put the first left-hand side on a single line of its own, and starting slightly left aligned to the first = sign. See the question Multiline equation with LHS alone on first line? for how to achieve this.

Answer 7

I would like to add my proposal of using the IEEEeqnarray environment from the IEEEtrantools package. It provides the IEEEeqnarraymulticol command that has a similar syntax to the tabular multicolumn command.

\documentclass{report}

\usepackage{mathtools}
\usepackage{IEEEtrantools}

\DeclareMathOperator\Cov{Cov}

\begin{document}
dalief's version:
\begin{align}
  \MoveEqLeft[3]
  \tfrac{1}{2}
  \Cov(\Delta_{i,p1}-\Delta_{i,p2},\Delta_{j,p1}-\Delta_{j,p2})
  \notag \\
  = {} &
  \tfrac{1}{2}\Cov(\Delta_{i,p1},\Delta_{j,p1}) +
  \tfrac{1}{2}\Cov(\Delta_{i,p2},\Delta_{j,p2}) \notag
  \\
  & - \tfrac{1}{2}\Cov(\Delta_{i,p1},\Delta_{j,p2}) -
  \tfrac{1}{2}\Cov(\Delta_{i,p2},\Delta_{j,p1})
  \\
  = {} & \Cov(\Delta_{i,p1},\Delta_{j,p2})|_{p1=p2} -
  \Cov(\Delta_{i,p1},\Delta_{j,p2})|_{p1 \ne p2}
\end{align}

Using my proposal:
\begin{IEEEeqnarray}{rCl}
  \IEEEeqnarraymulticol{3}{l}{
  \tfrac{1}{2}\Cov(\Delta_{i,p1}-\Delta_{i,p2},\Delta_{j,p1}-\Delta_{j,p2})}
  \nonumber \\ \qquad
  & = &
        \tfrac{1}{2}\Cov(\Delta_{i,p1},\Delta_{j,p1}) +
        \tfrac{1}{2}\Cov(\Delta_{i,p2},\Delta_{j,p2}) \notag
  \\
  &  &
        \negmedspace {} - \tfrac{1}{2}\Cov(\Delta_{i,p1},\Delta_{j,p2}) -
        \tfrac{1}{2}\Cov(\Delta_{i,p2},\Delta_{j,p1})
  \\
  & = &
        \Cov(\Delta_{i,p1},\Delta_{j,p2})|_{p1=p2} -
        \Cov(\Delta_{i,p1},\Delta_{j,p2})|_{p1 \ne p2}
\end{IEEEeqnarray}
\end{document}