Vertical alignment of align* in enumerate

Question

Related (but not the same):
Align number from enumerate with equation
Displaying an equation in a list

Earlier today I asked a question about aligning equations in an enumerate environment and got an answer that seemed to work for the example I gave, but it turns out I wasn't really specific enough for my problem.

I'm using align* environments, often as the only part of a list item. (If you're wondering why I'm not using more words to introduce the align*, these are mostly homework solutions where most of the work is algebra, etc.)

Here's my code and what it makes (above the middle line):

\begin{enumerate}
  \item \begin{align*}
    x^2 + y^2 &= x^2 + (iy)^2 \\
              &= (x + iy) (x - iy)
  \end{align*}
  \item \begin{align*}
    \frac{1}{n^2-4} &= \frac14 \frac{4}{(n-2)(n+2)} \\
                    &= \frac14 \left( \frac{1}{n-2}-\frac{1}{n+2} \right)
  \end{align*}
\end{enumerate}

I'm trying to make the baseline of the first line of the align* environment the same as the number's baseline—I want the first line of the align* to have the same vertical position (centered horizontally in the same way) relative to its number as in the second half of my picture. (Side question: why is there so much space before the align* starts?)

Answer 1

I had been using Philippe Goutet's solution posted here, but have fairly recently found a simpler solution of using the aligned environment with the optional [t] alignment:

  \item $\begin{aligned}[t]
    x^2 + y^2 &= x^2 + (iy)^2 \\
              &= (x + iy) (x - iy)
  \end{aligned}$

which yields:

Notes:

• The showframe package was used to show the page margins.

Code:

\documentclass{article}
\usepackage{showframe}
\usepackage{amsmath}

\begin{document}
\begin{enumerate}
  \item $\begin{aligned}[t]
    x^2 + y^2 &= x^2 + (iy)^2 \\
              &= (x + iy) (x - iy)
  \end{aligned}$
  \item $\begin{aligned}[t]
    \frac{1}{n^2-4} &= \frac14 \frac{4}{(n-2)(n+2)} \\
                    &= \frac14 \left( \frac{1}{n-2}-\frac{1}{n+2} \right)
  \end{aligned}$
\end{enumerate}
\end{document}

---

If you want them horizontally centered as is the default with the align environment you could add an \hfill on either side:

  \item \hfill$\begin{aligned}[t]
    x^2 + y^2 &= x^2 + (iy)^2 \\
              &= (x + iy) (x - iy)
  \end{aligned}$\hfill\null

Code:

\documentclass{article}
\usepackage{showframe}
\usepackage{amsmath}

\begin{document}
\begin{enumerate}
  \item \hfill$\begin{aligned}[t]
    x^2 + y^2 &= x^2 + (iy)^2 \\
              &= (x + iy) (x - iy)
  \end{aligned}$\hfill\null
  \item \hfill$\begin{aligned}[t]
    \frac{1}{n^2-4} &= \frac14 \frac{4}{(n-2)(n+2)} \\
                    &= \frac14 \left( \frac{1}{n-2}-\frac{1}{n+2} \right)
  \end{aligned}$\hfill\null
\end{enumerate}
\end{document}

Answer 2

There is a way to automatically measure the space between the beginning of the \item and the first line of the {align*} with the pdftex primitive \pdfsavepos. The savepos option of the zref package allows a rather nice interface for this feature.

The code works like this: you place a \mi macro where you want the material to be and you place a \md (you can change these names if you don't like them) where the material is. For example,

  \item \mi\begin{align*}\md
      x^2 + y^2 &= x^2 + (iy)^2 \\
              &= (x + iy) (x - iy)
  \end{align*}

After a few compilation (up to 4 may be needed), you will obtain perfect alignment of the equation and the item number.

Here's the full code, showing how the alignment works with a few equations and a table. I've also put a variant in which I redefined {align*} so that \mi and \md are not necessary anymore (making it easier to type).

\documentclass[12pt]{article}

\usepackage{amsmath}
\usepackage[savepos]{zref}

\makeatletter
\newcounter{autoenumvspace}
\setcounter{autoenumvspace}{0}
\newcommand{\mi}{\markitemstart}
\newcommand{\markitemstart}{%
  \addtocounter{autoenumvspace}{1}%
  \@ifundefined{autoenumvspacevalue\romannumeral\value{autoenumvspace}}%
    {\edef\autoenumvspace@value{0pt}}%
    {\edef\autoenumvspace@value{\csname autoenumvspacevalue\romannumeral\value{autoenumvspace}\endcsname}%
  }
  \ifnum\zposy{auto@enum@\number\value{autoenumvspace}@bottom}=%
        \zposy{auto@enum@\number\value{autoenumvspace}@top}
     \@ifundefined{autoenumvspacevalue\romannumeral\value{autoenumvspace}}%
     {}%
     {\immediate\write\@mainaux{\gdef\expandafter\noexpand\csname autoenumvspacevalue\romannumeral\value{autoenumvspace}\endcsname{\csname autoenumvspacevalue\romannumeral\value{autoenumvspace}\endcsname}}}%
  \else
     \ifdim\dimexpr\zposy{auto@enum@\number\value{autoenumvspace}@bottom}sp-\zposy{auto@enum@\number\value{autoenumvspace}@top}sp\relax=\autoenumvspace@value
        \immediate\write\@mainaux{\gdef\expandafter\noexpand\csname autoenumvspacevalue\romannumeral\value{autoenumvspace}\endcsname{\the\dimexpr\zposy{auto@enum@\number\value{autoenumvspace}@bottom}sp-\zposy{auto@enum@\number\value{autoenumvspace}@top}sp\relax}}%
     \else
        \edef\autoenumvspace@value{0pt}%
        \immediate\write\@mainaux{\gdef\expandafter\noexpand\csname autoenumvspacevalue\romannumeral\value{autoenumvspace}\endcsname{\the\dimexpr\zposy{auto@enum@\number\value{autoenumvspace}@bottom}sp-\zposy{auto@enum@\number\value{autoenumvspace}@top}sp\relax}}%
     \fi
  \fi
  \vspace*{-\autoenumvspace@value}%
  \leavevmode
  \zsavepos{auto@enum@\number\value{autoenumvspace}@top}%
  \vspace*{\autoenumvspace@value}%
}
\newcommand{\md}{\markdisplaystart}
\newcommand{\markdisplaystart}{%
  \zsavepos{auto@enum@\number\value{autoenumvspace}@bottom}%
}
\makeatother

\begin{document}

\section{Manual version}

\begin{enumerate}\belowdisplayskip=0pt \abovedisplayskip=0pt % optional
  \item \mi\begin{align*}\md
      x^2 + y^2 &= x^2 + (iy)^2 \\
              &= (x + iy) (x - iy)
  \end{align*}
  \item \mi\begin{align*}\md
     \frac{1}{n^2-4} &= \frac14 \frac{4}{(n-2)(n+2)} \\
                    &= \frac14 \left( \frac{1}{n-2}-\frac{1}{n+2} \right)
  \end{align*}
  \begin{enumerate}
      \item \mi\[\md x^n + y^n = z^n\]
      \item \mi\[\md \sum_{n=1}^{+\infty}{\frac{1}{n^2}} = \frac{\pi^2}{6}\]
      \item \mi
\begin{center}\begin{tabular}{|c|c|c|}
\hline
\md text & text & text \\
\hline
text & text & text \\
\hline
\end{tabular}\end{center}
  \end{enumerate}
\end{enumerate}

\section{Automatic version}

\begin{enumerate}\belowdisplayskip=0pt \abovedisplayskip=0pt % optional
\makeatletter
\renewenvironment{align*}{%
  \mi\start@align\@ne\st@rredtrue\m@ne\md
}{%
  \math@cr \black@\totwidth@
  \egroup
  \ifingather@
    \restorealignstate@
    \egroup
    \nonumber
    \ifnum0=`{\fi\iffalse}\fi
  \else
    $$%
  \fi
  \ignorespacesafterend
}
\expandafter\def\expandafter\[\expandafter{\expandafter\mi\[\md}
\makeatother
  \item \begin{align*}
      x^2 + y^2 &= x^2 + (iy)^2 \\
              &= (x + iy) (x - iy)
  \end{align*}
  \item \begin{align*}
     \frac{1}{n^2-4} &= \frac14 \frac{4}{(n-2)(n+2)} \\
                    &= \frac14 \left( \frac{1}{n-2}-\frac{1}{n+2} \right)
  \end{align*}
  \begin{enumerate}
      \item \[ x^n + y^n = z^n\]
      \item \[ \sum_{n=1}^{+\infty}{\frac{1}{n^2}} = \frac{\pi^2}{6}\]
      \item \mi
\begin{center}\begin{tabular}{|c|c|c|}
\hline
\md text & text & text \\
\hline
text & text & text \\
\hline
\end{tabular}\end{center}
  \end{enumerate}
\end{enumerate}


\end{document}

Answer 3

You can put the numbers as equation tags into the first line of the align* environment if you use the leqno option: \tag*{1.}. If you want to get automatic numbering, the following code can help you:

\documentclass[12pt,leqno]{article}
\usepackage{amsmath}
\newcounter{exercisenumber}
\setcounter{exercisenumber}{0}
\newcommand\leftnum{%
    \addtocounter{exercisenumber}{1}%
    \tag*{\phantom{99.}\llap{\arabic{exercisenumber}.}}%
    }
\begin{document}

\begin{align*}\leftnum
  x^2 + y^2 &= x^2 - (iy)^2 \\
            &= (x + iy) (x - iy)
\end{align*}
\begin{align*}\leftnum
  \frac{1}{n^2-4} &= \frac14 \frac{4}{(n-2)(n+2)} \\
                  &= \frac14 \left( \frac{1}{n-2}-\frac{1}{n+2} \right)
\end{align*}

\end{document}

Concerning your additional question in the comment: I don't know exactly know what output you're aiming at, but you can try this:

\newcounter{exercisenumber}
\newcounter{subnumber}
\setcounter{exercisenumber}{0}
\newcommand\leftnum{%
    \addtocounter{exercisenumber}{1}%
    \tag*{\phantom{99.}\llap{\arabic{exercisenumber}.}}%
    }
\newenvironment{subnumbering}{%
    \addtocounter{exercisenumber}{1}%
    \setcounter{subnumber}{0}
    \renewcommand\leftnum{%
        \addtocounter{subnumber}{1}%
        \tag*{\phantom{99(a)}\llap{%
              \ifnum\value{subnumber}=1 \arabic{exercisenumber}\fi
              (\alph{subnumber})}}%
        }
    }{}

Answer 4

Here is a solution using xparse. I define new listalign and listgather environments, which behave as align and gather, have starred versions, but add no vertical skip at the beginning. They're intended to be used in list environments, if there is no text after an \item command:

\documentclass{article}
\usepackage{enumitem}
\usepackage{mathtools}
\usepackage{xparse}

\makeatletter
\DeclareDocumentEnvironment{listgather}{s}{%
\setlength{\abovedisplayskip}{0pt}\leavevmode\vspace*{-\baselineskip}\IfBooleanTF{#1}{\gather\st@rredtrue}{\gather}}%
{\endgather}
\DeclareDocumentEnvironment{listalign}{s}{%
\setlength{\abovedisplayskip}{0pt}\leavevmode\vspace*{-\baselineskip}\IfBooleanTF{#1}{\start@align\@ne\st@rredtrue\m@ne}{\align}}%
{\endalign}
\makeatother

\ExplSyntaxOn
\cs_new:cpn {listgather*} {\listgather*}
\cs_new_eq:cN {endlistgather*}\endlistgather
\cs_new:cpn {listalign*} {\listalign*}
\cs_new_eq:cN {endlistalign*}\endlistalign
\ExplSyntaxOff

\begin{document}

Some text. Some text.Some text. Some text.
\begin{enumerate}[label=\arabic*.]
    \item \begin{listgather}a + b = c + d\end{listgather}
    \item \begin{listgather*}c + d = m + n\end{listgather*}
  \item \begin{listalign*} x^2 + y^2 &= x^2 + (iy)^2 \\
              &= (x + iy) (x - iy)
  \end{listalign*}
  \item \begin{listalign} \frac{1}{n^2-4} &= \frac14 \frac{4}{(n-2)(n+2)} \\
                    &= \frac14 \left( \frac{1}{n-2}-\frac{1}{n+2} \right)
  \end{listalign}
\end{enumerate}

\end{document} 

Answer 5

A little variant version of @Herbert.

\documentclass[12pt]{article}
\usepackage{amsmath}
\begin{document}

\begin{enumerate}
   \item  \parbox[t]{\textwidth}{
   \vspace{-2.2em}\begin{align*}
        x^2 + y^2 &= x^2 + (iy)^2 \\
                  &= (x + iy) (x - iy)
   \end{align*}}
   \item \parbox[t]{\textwidth}{
   \vspace{-2.2em}\begin{align*}
        \frac{1}{n^2-4} &= \frac14 \frac{4}{(n-2)(n+2)} \\
                        &= \frac14 \left( \frac{1}{n-2}-\frac{1}{n+2} \right)
   \end{align*}}
\end{enumerate}

\end{document}

Answer 6

\documentclass[12pt]{article}
\usepackage{amsmath}
\begin{document}

\begin{enumerate}\belowdisplayskip=0pt \abovedisplayskip=0pt 
  \item \parbox{\linewidth}{\begin{align*}
    x^2 + y^2 &= x^2 + (iy)^2 \\
              &= (x + iy) (x - iy)
  \end{align*}}
  \item 
    \parbox[t]{\linewidth}{\begin{align*}
    \frac{1}{n^2-4} &= \frac14 \frac{4}{(n-2)(n+2)} \\
                    &= \frac14 \left( \frac{1}{n-2}-\frac{1}{n+2} \right)
  \end{align*}}
\end{enumerate}

\end{document}

Answer 7

You can do that with the package witharrows. This package provides an environment {DispWithArrows*} very similar to {align*} (the main difference is that by default, the elements are not composed in displaystyle but this point may be changed with \WithArrowsOptions).

By default, {DispWithArrows*} has the behaviour required in this question (there is an option standard-behaviour-with-items to go back to the behaviour of {align}).

Of course, the main goal of witharrows was the ability to add arrows. That's why I have added an arrow as example.

\documentclass{article}

\usepackage{witharrows}
\WithArrowsOptions{displaystyle}

\begin{document}

\begin{enumerate}
  \item \begin{DispWithArrows*}
    x^2 + y^2 &= x^2 - (iy)^2 \Arrow{factorization} \\
              &= (x + iy) (x - iy)
  \end{DispWithArrows*}
  \item \begin{DispWithArrows*}
    \frac{1}{n^2-4} &= \frac14 \frac{4}{(n-2)(n+2)} \\
                    &= \frac14 \left( \frac{1}{n-2}-\frac{1}{n+2} \right)
  \end{DispWithArrows*}
\end{enumerate}

\end{document}