Long brackets in flalign*

Question

I'm trying to type a nice looking long sum of functions:

I managed to do it with combination of flalign* and \mspace:

\begin{flalign*}
f(a+h)-f(a)=&f(a_1+h_1,\dots\mspace{25mu}\dots,a_n+h_n)-f(a_1,\dots\mspace{120mu}\dots,a_n)=\\
=&f(a_1+h_1,a_2+h_2,\dots\mspace{1mu}\dots,a_n)-f(a_1,a_2,\dots\mspace{95mu}\dots,a_n)+\\
+&f(a_1+h_1,a_2+h_2,a_3,\dots,a_n)-f(a_1+h_1,a_2,\dots\mspace{55mu}\dots,a_n)+\\
&\dots\\
+&f(a_1+h_1,\dots\mspace{24mu}\dots,a_n+h_n)-f(a_1+h1,\dots,a_{n-1}+h_{n-1},a_n)\\
\end{flalign*}

I'm pretty happy with what I got, but here's my question: is it possible to make right brackets place themselves automatically in the right place? \phantom could help, if not for a_n+h_n instead of a_n bit. And is it possible to use another & pointer (say, &2), to mark the line of "-"s?

On the side note: how come this whole "reputation" does not allow me to make new tags? Aren't they suppose to, you know, HELP people?

Answer 1

You need to put the & before the + or = not after otherwise you do not get the correct spacing, note your image has operators without infix spacing at both ends, I think you just want more alignment points.

\documentclass{article}

\usepackage{amsmath}

\begin{document}

\begin{flalign*}
f(a+h)-f(a)&=f(a_1+h_1,\dots&\dots,a_n+h_n)&-f(a_1,\dots&\dots,a_n)={}\\
&=f(a_1+h_1,a_2+h_2,\dots&\dots,a_n)&-f(a_1,a_2,\dots&\dots,a_n)+{}\\
&+f(a_1+h_1,a_2+h_2,a_3,&\dots,a_n)&-f(a_1+h_1,a_2,\dots &\dots,a_n)+{}\\
&\dots\\
&+f(a_1+h_1,\dots&\dots,a_n+h_n)&-f(a_1+h1,&\dots,a_{n-1}+h_{n-1},a_n)\phantom{{}+{}}\\
\end{flalign*}
`



\end{document}

Answer 2

Notes:

• I used the scrartcl class only because your original equation was to wide for the standard article class.
• mathtools provides the \math?lap family and \vdotswithin.
• I removed one of \dots if they are followed without space.
• \hphantom intends the equations so that the second part is better aligned.
• I used the alignat* environment because for one equation the flalign* environment doesn't do much. (There is of course no flalignat environment.)
• I also removed = and + at the end of rows because I don't think that they make any sense there.

Code

\documentclass{scrartcl}
\usepackage{amsmath,mathtools}
\begin{document}
\begin{alignat*}{4}
f(a+h)-f(a) & = f(a_1 + h_1, \dots                             & \mathllap{\dots, a_n + h_n)} & {} - f(a_1, \dots          & \dots, a_n)   \\
            & = \hphantom{{}+{}} f(a_1 + h_1, a_2 + h_2,       &           \dots, a_n)        & {} - f(a_1, a_2, \mathrlap{\dots}     & \dots, a_n)   \\
            & \hphantom{{}={}} + f(a_1 + h_1, a_2 + h_2, a_3,  &        {} \dots, a_n)        & {} - f(a_1+h_1, \mathrlap{a_2, \dots} & \dots,a_n) \\
            & \hphantom{{}={}} \vdotswithin{+} \\
            & \hphantom{{}={}} + f(a_1+h_1,\dots               & \mathllap{\dots, a_n + h_n)} & {} - f(a_1+h_1,{} & \dots, a_{n-1}+h_{n-1},a_n)\\
\end{alignat*}
\end{document}

Output

Answer 3

I don't think that this is a particularly nice way to express your thought.

However, this is one way that requires only to guess the widest vector, which can be done at a first stage. This is surely easier than guessing many spacings.

\documentclass{article}
\usepackage{amsmath}

\newlength{\providewdlen}
\newcommand{\providewd}[2]{
  \settowidth{\providewdlen}{$\displaystyle#1$}
  \makebox[\providewdlen][s]{$\displaystyle#2$}
}
\newcommand{\alignplus}{\mathrel{\providewd{=}{\hfill+}}}

\begin{document}
\begin{align*}
f(a+h)-f(a)
 &= 
    f(\providewd{a_1+h_1,a_2+h_2,\dots\dots,a_n}{a_1+h_1,\dots\hfill\dots,a_n+h_n})
   -f(\providewd{a_1+h1,\dots,a_{n-1}+h_{n-1},a_n}{a_1,\dots\hfill\dots,a_n})\\
 &=
    f(\providewd{a_1+h_1,a_2+h_2,a_3,\dots,a_n}{a_1+h_1,a_2,\dots\hfill\dots,a_n})
   -f(\providewd{a_1+h1,\dots,a_{n-1}+h_{n-1},a_n}{a_1,a_2,\dots\hfill\dots,a_n})\\
 &\alignplus
   f(a_1+h_1,a_2+h_2,a_3,\dots,a_n)
  -f(\providewd{a_1+h1,\dots,a_{n-1}+h_{n-1},a_n}{a_1+h_1,a_2,\dots\hfill\dots,a_n})\\
 & \dots\\
 &\alignplus
   f(\providewd{a_1+h_1,a_2+h_2,a_3,\dots,a_n}{a_1+h_1,\dots\hfill\dots,a_n+h_n})
  -f(a_1+h_1,\dots,a_{n-1}+h_{n-1},a_n)
\end{align*}


\end{document}

Notice that the symbols should not to be repeated at the end of the lines and that & goes before the alignment point. There should not be a trailing \\ in alignment environments.

Since the + is less wide than the =, we need to take care of that too. Note also that flalign* does nothing more than align* in this case; the differences can be seen when there is more than one alignment point.

My feeling is that a simpler approach could be clearer:

\begin{align*}
f(a+h)-f(a)
 &=f(a_1+h_1,\dots,a_n+h_n)-f(a_1,\dots,a_n)\\
 &=
   f(a_1+h_1,a_2,\dots,a_n)-f(a_1,a_2,\dots,a_n)\\
 &\alignplus
   f(a_1+h_1,a_2+h_2,a_3,\dots,a_n)-f(a_1+h_1,a_2,\dots,a_n)\\
 & \dots\\
 &\alignplus
   f(a_1+h_1,\dots,a_n+h_n)-f(a_1+h_1,\dots,a_{n-1}+h_{n-1},a_n)
\end{align*}