# Align: Text and centered formulas

I got an align-environment starting off with a sentence. After this introductory sentence I want to have the equations centered in the page, each equal-sign under each other. Following MEW:

\documentclass{article}

\usepackage{amsmath,bm}
\usepackage{mathtools}

\begin{document}

\begin{align} \label{eq:gen_alpha}
\begin{aligned}
\text{Given } \mathbf{d}_{n},\mathbf{v}_{n} \text{ and } \mathbf{a}_{n} \text{. Find } \mathbf{d}_{n+1},\mathbf{v}_{n+1}, \mathbf{a}_{n+1},\mathbf{d}_{n+\alpha_{f}},\mathbf{v}_{n+\alpha_{f}}, \mathbf{a}_{n+\alpha_{m}} \text {and } \bm{\Phi}_{n+1} \text{ so that:} \\
\mathbf{R}^{u}\left(\mathbf{d}_{n+\alpha_{f}},\mathbf{v}_{n+\alpha_{f}},\mathbf{a}_{n+\alpha_{m}},\mathbf{\Phi}_{n+1}\right) &= 0, \\
\mathbf{R}^{c}\left(\mathbf{d}_{n+\alpha_{f}},\mathbf{\Phi}_{n+1}\right) &= 0, \\
\mathbf{d}_{n+\alpha_{f}} &= \mathbf{d}_{n}+\alpha_{f}\left(\mathbf{d}_{n+1}-\mathbf{d}_{n}\right), \\
\mathbf{v}_{n+\alpha_{f}} &= \mathbf{v}_{n}+\alpha_{f}\left(\mathbf{v}_{n+1}-\mathbf{v}_{n}\right), \\
\mathbf{a}_{n+\alpha_{m}} &= \mathbf{a}_{n}+\alpha_{m}\left(\mathbf{a}_{n+1}-\mathbf{a}_{n}\right), \\
\mathbf{v}_{n+1} &= \mathbf{v}_{n}+\Delta t\left(\left(1-\gamma\right)\mathbf{a}_{n}+\gamma\mathbf{a}_{n+1}\right), \\
\mathbf{d}_{n+\alpha f} &= \mathbf{d}_{n}+\alpha_{f}\left(\mathbf{d}_{n+1}-\mathbf{d}_{n}\right).
\end{aligned}
\end{align}

\end{document}


The text alignment is great, only all the equations should be centered.

Thanks for help :)

• Welcome. Simply remove the first sentence from the align environment. Or enclose it within \intertext{sentence}. – Johannes_B Oct 31 '17 at 12:35

Remove the sentence from the aligned definition. Likewise, with only one numbered element, use equation instead of align.

\documentclass{article}

\usepackage{amsmath,bm}
\usepackage{mathtools}

\begin{document}

Given $\mathbf{d}_{n}$, $\mathbf{v}_{n}$ and $\mathbf{a}_{n}$. Find $\mathbf{d}_{n+1}$, $\mathbf{v}_{n+1}$, $\mathbf{a}_{n+1}$, $\mathbf{d}_{n+\alpha_{f}}$, $\mathbf{v}_{n+\alpha_{f}}$, $\mathbf{a}_{n+\alpha_{m}}$ and $\bm{\Phi}_{n+1}$ so that:
\label{eq:gen_alpha} \begin{aligned} \mathbf{R}^{u}\left(\mathbf{d}_{n+\alpha_{f}},\mathbf{v}_{n+\alpha_{f}},\mathbf{a}_{n+\alpha_{m}},\mathbf{\Phi}_{n+1}\right) &= 0, \\ \mathbf{R}^{c}\left(\mathbf{d}_{n+\alpha_{f}},\mathbf{\Phi}_{n+1}\right) &= 0, \\ \mathbf{d}_{n+\alpha_{f}} &= \mathbf{d}_{n}+\alpha_{f}\left(\mathbf{d}_{n+1}-\mathbf{d}_{n}\right), \\ \mathbf{v}_{n+\alpha_{f}} &= \mathbf{v}_{n}+\alpha_{f}\left(\mathbf{v}_{n+1}-\mathbf{v}_{n}\right), \\ \mathbf{a}_{n+\alpha_{m}} &= \mathbf{a}_{n}+\alpha_{m}\left(\mathbf{a}_{n+1}-\mathbf{a}_{n}\right), \\ \mathbf{v}_{n+1} &= \mathbf{v}_{n}+\Delta t\left(\left(1-\gamma\right)\mathbf{a}_{n}+\gamma\mathbf{a}_{n+1}\right), \\ \mathbf{d}_{n+\alpha f} &= \mathbf{d}_{n}+\alpha_{f}\left(\mathbf{d}_{n+1}-\mathbf{d}_{n}\right). \end{aligned}

\end{document}


This might be what you are looking for.

\documentclass{article}

\usepackage{amsmath,bm}
\usepackage{mathtools}

\begin{document}

\begin{align*} \label{eq:gen_alpha}
\intertext{Given  $\mathbf{d}_{n},\mathbf{v}_{n}$  and  $\mathbf{a}_{n}$. Find  $\mathbf{d}_{n+1},\mathbf{v}_{n+1}, \mathbf{a}_{n+1},\mathbf{d}_{n+\alpha_{f}},\mathbf{v}_{n+\alpha_{f}}, \mathbf{a}_{n+\alpha_{m}}$ and  $\bm{\Phi}_{n+1}$  so that:}
\mathbf{R}^{u}\left(\mathbf{d}_{n+\alpha_{f}},\mathbf{v}_{n+\alpha_{f}},\mathbf{a}_{n+\alpha_{m}},\mathbf{\Phi}_{n+1}\right) &= 0, \\
\mathbf{R}^{c}\left(\mathbf{d}_{n+\alpha_{f}},\mathbf{\Phi}_{n+1}\right) &= 0, \\
\mathbf{d}_{n+\alpha_{f}} &= \mathbf{d}_{n}+\alpha_{f}\left(\mathbf{d}_{n+1}-\mathbf{d}_{n}\right), \\
\mathbf{v}_{n+\alpha_{f}} &= \mathbf{v}_{n}+\alpha_{f}\left(\mathbf{v}_{n+1}-\mathbf{v}_{n}\right), \\
\mathbf{a}_{n+\alpha_{m}} &= \mathbf{a}_{n}+\alpha_{m}\left(\mathbf{a}_{n+1}-\mathbf{a}_{n}\right), \\
\mathbf{v}_{n+1} &= \mathbf{v}_{n}+\Delta t\left(\left(1-\gamma\right)\mathbf{a}_{n}+\gamma\mathbf{a}_{n+1}\right), \\
\mathbf{d}_{n+\alpha f} &= \mathbf{d}_{n}+\alpha_{f}\left(\mathbf{d}_{n+1}-\mathbf{d}_{n}\right).
\end{align*}

\end{document}


I'd not align the equality signs of the first two relations, because they're of a different kind and the display would be very unbalanced.

\documentclass{article}

\usepackage{amsmath,bm}
\usepackage{mathtools}

\newcommand{\bv}[1]{%
\ifcat\noexpand#1\relax
\bm{#1}%
\else
\mathbf{#1}%
\fi
}

\begin{document}

Given $\bv{d}_{n}$, $\bv{v}_{n}$ and $\bv{a}_{n}$, find $\bv{d}_{n+1}$,
$\bv{v}_{n+1}$, $\bv{a}_{n+1}$, $\bv{d}_{n+\alpha_{f}}$, $\bv{v}_{n+\alpha_{f}}$,
$\bv{a}_{n+\alpha_{m}}$ and $\bv{\Phi}_{n+1}$ so that:
\label{eq:gen_alpha} \begin{aligned} &\bv{R}^{u}(\bv{d}_{n+\alpha_{f}},\bv{v}_{n+\alpha_{f}},\bv{a}_{n+\alpha_{m}},\bv{\Phi}_{n+1})=0,\\ &\bv{R}^{c}(\bv{d}_{n+\alpha_{f}},\bv{\Phi}_{n+1}) = 0, \\ &\begin{aligned} \bv{d}_{n+\alpha_{f}} &= \bv{d}_{n}+\alpha_{f}(\bv{d}_{n+1}-\bv{d}_{n}), \\ \bv{v}_{n+\alpha_{f}} &= \bv{v}_{n}+\alpha_{f}(\bv{v}_{n+1}-\bv{v}_{n}), \\ \bv{a}_{n+\alpha_{m}} &= \bv{a}_{n}+\alpha_{m}(\bv{a}_{n+1}-\bv{a}_{n}), \\ \bv{v}_{n+1} &= \bv{v}_{n}+\Delta t((1-\gamma)\bv{a}_{n}+\gamma\bv{a}_{n+1}), \\ \bv{d}_{n+\alpha f} &= \bv{d}_{n}+\alpha_{f}(\bv{d}_{n+1}-\bv{d}_{n}). \end{aligned} \end{aligned}

\end{document}


Find a better name for \bv (the command expects the argument to be a single Latin letter or a single command for a Greek letter). I also removed all \left and \right that serve no purpose, other than introducing unwanted spaces.

Alternatively, center the first two and leave some vertical space after them.

\documentclass{article}

\usepackage{amsmath,bm}
\usepackage{mathtools}

\newcommand{\bv}[1]{%
\ifcat\noexpand#1\relax
\bm{#1}%
\else
\mathbf{#1}%
\fi
}

\begin{document}

Given $\bv{d}_{n}$, $\bv{v}_{n}$ and $\bv{a}_{n}$, find $\bv{d}_{n+1}$,
$\bv{v}_{n+1}$, $\bv{a}_{n+1}$, $\bv{d}_{n+\alpha_{f}}$, $\bv{v}_{n+\alpha_{f}}$,
$\bv{a}_{n+\alpha_{m}}$ and $\bv{\Phi}_{n+1}$ so that:
\label{eq:gen_alpha} \begin{gathered} \bv{R}^{u}(\bv{d}_{n+\alpha_{f}},\bv{v}_{n+\alpha_{f}},\bv{a}_{n+\alpha_{m}},\bv{\Phi}_{n+1})=0,\\ \bv{R}^{c}(\bv{d}_{n+\alpha_{f}},\bv{\Phi}_{n+1}) = 0, \\[1ex] \begin{aligned} \bv{d}_{n+\alpha_{f}} &= \bv{d}_{n}+\alpha_{f}(\bv{d}_{n+1}-\bv{d}_{n}), \\ \bv{v}_{n+\alpha_{f}} &= \bv{v}_{n}+\alpha_{f}(\bv{v}_{n+1}-\bv{v}_{n}), \\ \bv{a}_{n+\alpha_{m}} &= \bv{a}_{n}+\alpha_{m}(\bv{a}_{n+1}-\bv{a}_{n}), \\ \bv{v}_{n+1} &= \bv{v}_{n}+\Delta t((1-\gamma)\bv{a}_{n}+\gamma\bv{a}_{n+1}), \\ \bv{d}_{n+\alpha f} &= \bv{d}_{n}+\alpha_{f}(\bv{d}_{n+1}-\bv{d}_{n}). \end{aligned} \end{gathered}

\end{document}