6

enter image description here

While writing down the above expression on Overleaf, I encountered the problem of drawing the dashed line in between the Left Hand Side and the Right Hand Side expressions. Do help me in writing it exactly that way.

(I had written it in my own unique way, but without the dashed line, and would be intrigued to know if such a function existed.)

My code & Output:

\documentclass{article}
\usepackage{amsmath}

\begin{document}

ARCTAN TERMS:


\[\boxed{\sqrt{2}\zeta_1+1 \to \infty \implies \tan^{-1}(\sqrt{2}\zeta_1+1)\rightarrow \dfrac{\pi}{2}}\text{ and } \boxed{\sqrt{2}\zeta_1+1 \rightarrow -\infty \implies \tan^{-1}(\sqrt{2}\zeta_1+1)\to \dfrac{-\pi}{2}}\]

\[\boxed{\sqrt{2}\zeta_1-1 \to \infty \implies \tan^{-1}(\sqrt{2}\zeta_1-1)\rightarrow \dfrac{\pi}{2}}\text{ and } \boxed{\sqrt{2}\zeta_1-1 \rightarrow -\infty \implies \tan^{-1}(\sqrt{2}\zeta_1-1)\to \dfrac{-\pi}{2}}\]

Hence, \[\text{ Arctan terms evaluates to } \frac{\sqrt{2}}{4}\left(\frac{\pi}{2}+\frac{\pi}{2}-\left(\frac{-\pi}{2}-\frac{\pi}{2}\right)\right)=\frac{\pi}{\sqrt2}\]

\end{document}

Output:

enter image description here

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18
  • 3
    Please make your code compilable (if possible), or at least complete it with \documentclass{...}, the required \usepackage's, \begin{document}, and \end{document}. That may seem tedious to you, but think of the extra work it represents for the users willing to give you a hand. Help them help you: remove that one hurdle between you and a solution to your problem. Commented Feb 10 at 10:53
  • 1
    Please add a class and move the package into the preamble. Commented Feb 10 at 10:59
  • Please read Why is \[ … \] preferable to $$? Commented Feb 10 at 11:00
  • 1
    yes exactly and \usepackage{amsmath} will generate errors in the place you show it, so you will need to move that. Commented Feb 10 at 11:10
  • 1
    @DhairyaKumar Exactly. And this needs to be fixed before asking questions about equations. Commented Feb 10 at 11:14

2 Answers 2

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5

I will address the part of the question that concerns the dashed vertical line. To draw a dashed vertical line between two columns, the simplest approach is to use a regular table/array layout with a dashed vertical rule. For example, you can use the arydshln package: in array/tabular it interprets the symbol : in the preamble as a dashed vertical separator. Here is one way to implement it:

\documentclass{article}
\usepackage{amsmath}
\usepackage{arydshln}

\begin{document}

\paragraph{ARCTAN TERMS:}

\[
\begin{array}{@{}c: c@{}}
\boxed{\sqrt2\,\zeta_1+1 \to \infty \implies \tan^{-1}(\sqrt2\,\zeta_1+1)\to \dfrac{\pi}{2}}
&
\boxed{\sqrt2\,\zeta_1+1 \to -\infty \implies \tan^{-1}(\sqrt2\,\zeta_1+1)\to \dfrac{-\pi}{2}}
\\[1.2ex]
\boxed{\sqrt2\,\zeta_1-1 \to \infty \implies \tan^{-1}(\sqrt2\,\zeta_1-1)\to \dfrac{\pi}{2}}
&
\boxed{\sqrt2\,\zeta_1-1 \to -\infty \implies \tan^{-1}(\sqrt2\,\zeta_1-1)\to \dfrac{-\pi}{2}}
\end{array}
\]    

\end{document}

enter image description here

Edit.
Here is the same thing, but without the boxes:

\documentclass{article}
\usepackage{amsmath}
\usepackage{arydshln}

\begin{document}

\paragraph{ARCTAN TERMS:}

\[
\begin{array}{@{}c: c@{}}
\sqrt2\,\zeta_1+1 \to \infty \implies \tan^{-1}(\sqrt2\,\zeta_1+1)\to \dfrac{\pi}{2}
&
\sqrt2\,\zeta_1+1 \to -\infty \implies \tan^{-1}(\sqrt2\,\zeta_1+1)\to \dfrac{-\pi}{2}
\\[1.2ex]
\sqrt2\,\zeta_1-1 \to \infty \implies \tan^{-1}(\sqrt2\,\zeta_1-1)\to \dfrac{\pi}{2}
&
\sqrt2\,\zeta_1-1 \to -\infty \implies \tan^{-1}(\sqrt2\,\zeta_1-1)\to \dfrac{-\pi}{2}
\end{array}
\]

\end{document}

enter image description here

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3
  • I tried excecuting it without the \boxed{} but it won't excecute for whatsoever no reason. Commented Feb 10 at 14:04
  • @DhairyaKumar I updated my answer by adding code where the formulas are shown without boxes. Commented Feb 10 at 14:44
  • Brilliant. Thanks. Commented Feb 10 at 14:48
4

You can use nicematrix

\documentclass{article}
\usepackage{amsmath}
\usepackage{nicematrix}
\usepackage{tikz}

\NiceMatrixOptions{
  custom-line={
    letter=I,
    tikz=dashed,
  },
}

\begin{document}

\[
\begin{NiceArray}{l I l}
\sqrt{2}\zeta_1+1 \to \infty
\implies
\arctan(\sqrt{2}\zeta_1+1) \to \dfrac{\pi}{2} &
\sqrt{2}\zeta_1+1 \to -\infty
\implies
\arctan(\sqrt{2}\zeta_1+1) \to -\dfrac{\pi}{2} \\[4ex]
\sqrt{2}\zeta_1-1 \to \infty
\implies
\arctan(\sqrt{2}\zeta_1-1)\to \dfrac{\pi}{2} &
\sqrt{2}\zeta_1-1 \to -\infty
\implies
\arctan(\sqrt{2}\zeta_1-1)\to -\dfrac{\pi}{2}
\end{NiceArray}
\]

\end{document}

output

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