1

I'd like to have something like this

\documentclass[a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage[a4paper,top=3cm,bottom=3cm,left=1.5cm,right=1.5cm]{geometry}
\usepackage{mathtools}
\usepackage{stix}

\newcommand{\tonde}[1]{\bigl(#1\bigr)} \newcommand{\Quadre}[1]{\left[#1\right]} \newcommand{\normadue}[1]{\left|#1\right|}

\makeatletter \newcommand{\pushright}[1]{\ifmeasuring@#1\else\omit\hfill$\displaystyle#1$\fi\ignorespaces} \newcommand{\pushleft}[1]{\ifmeasuring@#1\else\omit$\displaystyle#1$\hfill\fi\ignorespaces} \makeatother

\begin{document}

\Large

\section{Introduction}

\begin{equation} \begin{split} L\tonde{f\circ\phi}&=\int_I\normadue{\nabla\tonde{f\circ\phi}\tonde{t}}dt=\int_I\normadue{df_{\phi(t)}\Quadre{\nabla\phi\tonde{t}}}dt\&=\int_I\normadue{\nabla\phi\tonde{t}}dt=L\tonde{\phi}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\square\notag \end{split} \end{equation}

\end{document}

But I used \qquad 7 times. Is there I way to simplify my code using a hfill (or other else) such that the \square is in the same place of the tag of the equation? I tried with this solution but I dislike the result. Could you help me? Thank you so much

notag square

egreg
  • 1,121,712
Puck
  • 1,208

1 Answers1

1

Do you mean something like \qedhere?

\documentclass[a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage[a4paper,top=3cm,bottom=3cm,left=1.5cm,right=1.5cm]{geometry}
\usepackage{mathtools}
\usepackage{amsthm}
\usepackage{stix}

%\newcommand{\tonde}[1]{\bigl(#1\bigr)} %\newcommand{\Quadre}[1]{\left[#1\right]} %\newcommand{\normadue}[1]{\left|#1\right|} \DeclarePairedDelimiter{\tonde}{(}{)} \DeclarePairedDelimiter{\Quadre}{[}{]} \DeclarePairedDelimiter{\normadue}{|}{|}

%\makeatletter %\newcommand{\pushright}[1]{\ifmeasuring@#1\else\omit\hfill$\displaystyle#1$\fi\ignorespaces} %\newcommand{\pushleft}[1]{\ifmeasuring@#1\else\omit$\displaystyle#1$\hfill\fi\ignorespaces} %\makeatother

\begin{document}

\section{Introduction}

\begin{proof}[First part of the proof] \begin{equation} \begin{split} L\tonde{f\circ\phi} &=\int_I\normadue{\nabla\tonde{f\circ\phi}\tonde{t}},dt =\int_I\normadue{df_{\phi(t)}\Quadre{\nabla\phi\tonde{t}}}dt\ &=\int_I\normadue{\nabla\phi\tonde{t}}dt=L\tonde{\phi}\qedhere \end{split} \end{equation} \end{proof}

\end{document}

enter image description here

You get the same output with

\documentclass[a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage[a4paper,top=3cm,bottom=3cm,left=1.5cm,right=1.5cm]{geometry}
\usepackage{mathtools}
\usepackage{amsthm}
\usepackage{stix}

%\newcommand{\tonde}[1]{\bigl(#1\bigr)} %\newcommand{\Quadre}[1]{\left[#1\right]} %\newcommand{\normadue}[1]{\left|#1\right|} \DeclarePairedDelimiter{\tonde}{(}{)} \DeclarePairedDelimiter{\Quadre}{[}{]} \DeclarePairedDelimiter{\normadue}{|}{|}

%\makeatletter %\newcommand{\pushright}[1]{\ifmeasuring@#1\else\omit\hfill$\displaystyle#1$\fi\ignorespaces} %\newcommand{\pushleft}[1]{\ifmeasuring@#1\else\omit$\displaystyle#1$\hfill\fi\ignorespaces} %\makeatother

\begin{document}

\section{Introduction}

\begin{align} L\tonde{f\circ\phi} &=\int_I\normadue{\nabla\tonde{f\circ\phi}\tonde{t}},dt =\int_I\normadue{df_{\phi(t)}\Quadre{\nabla\phi\tonde{t}}}dt\ &=\int_I\normadue{\nabla\phi\tonde{t}}dt=L\tonde{\phi}\tag{\qedsymbol} \end{align*}

\end{document}

but I would ask myself what would be the meaning of ending a proof without ever having started it.

I fixed just the first space before “dt”. You want to fix all those you have in your document.

By the way, you can spot the differences between my version of \normadue and yours, where the delimiters are far too big.

egreg
  • 1,121,712
  • Thank you. Unfortunately Overleaf gives me 2 errors (I copied both of your solutions), if you want I can load the Overleaf window. And about delimiters, I prefer a lot that they are a little bit bigger, it's just my style choice. – Puck Jun 08 '22 at 14:21
  • @Puck Did you add \usepackage{amsthm}? – egreg Jun 08 '22 at 14:29
  • Ok now it is correct. Thank you so much – Puck Jun 08 '22 at 14:43