8

I am currently trying to get the display

enter image description here

By doing this code, i was able to produce

\begin{gather}
   1 \uparrow 1 = {}^1 1 = 1 \nonumber \\
   2 \uparrow \uparrow 2 = {}^2 2 = 4 \nonumber \\
   3 \uparrow \uparrow \uparrow 3 = {}^{{}^{{}^3}} {}^3 3 = 3 \uparrow \uparrow 3 \uparrow \uparrow 3 = \underbrace{3^{3^{3^{3^{3^{3^{.^{.^{.^{3}}}}}}}}}}_{3^{3^3} \text{ threes}}\tag{14}\label{eq.14}
\end{gather}

enter image description here

Of course, this is still not the correct display, so i then tried to use an array environment and use negative \vspace, i got close, but not quite the perfect display. Here is the code for that

\[\begin{array}{c}
   \vspace{0.2cm}
   1 \uparrow 1 = {}^1 1 = 1 \nonumber \\
   \vspace{-0.3cm}
   2 \uparrow \uparrow 2 = {}^2 2 = 4 \nonumber \\
   3 \uparrow \uparrow \uparrow 3 = {}^{{}^{{}^3}}{}^3 3 = 3 \uparrow \uparrow 3 \uparrow \uparrow 3 = \underbrace{3^{3^{3^{3^{3^{3^{.^{.^{.^{3}}}}}}}}}}_{3^{3^3} \text{ threes}} \vspace{1cm}\tag{28}\label{eq.28}
\end{array}\]

enter image description here

I probably could tinker around with that, but I am pretty sure that a better method exists, so if anyone can help it would be very appreciated.

PS: If a better method exists to only get the equation tag on the last line instead using \nonumber, please let me know

Edit: One of the comments suggested to use \smash, so i tried and the code gave

\begin{gather}
  1 \uparrow 1 = {}^1 1 = 1 \nonumber \\
  2 \uparrow \uparrow 2 = {}^2 2 = 4 \nonumber \\
  \smash{3 \uparrow \uparrow \uparrow 3 = {}^{{}^{{}^3}} {}^3 3 = 3 \uparrow \uparrow 3 \uparrow \uparrow 3 = \underbrace{3^{3^{3^{3^{3^{3^{.^{.^{.^{3}}}}}}}}}}_{3^{3^3} \text{ threes}}}\tag{14}\label{eq.14}
\end{gather}

enter image description here

So using \smash seem to have only worked to remove the vertical spacing, but it has pushed the line to the left...

10
  • 3
    use \smash{.....} around the power tower and it will hide its height completely Feb 14, 2021 at 14:38
  • You can use \begin{equation} \begin{gathered}[b] …… \end{gathered} \end{equation}.
    – Bernard
    Feb 14, 2021 at 14:54
  • @DavidCarlisle using \smash does help to remove the vertical space, but the problem is that the centering seems kind of off to the left...
    – needle
    Feb 14, 2021 at 15:57
  • @Bernard I tried to use \begin{equation}\being{gathered}[b] ... \end{gathered}\end{equation} and it doesn't seem to work..
    – needle
    Feb 14, 2021 at 15:58
  • you have not provided an example that anone can run to see any centering (the use of \nonumber` with array in your second fragment looks very odd by the way). It always makes a question clearer if it contains an example that shows the problem, not just fragments that can not easily be tested. Feb 14, 2021 at 15:59

2 Answers 2

6

You wrote,

If a better method exists to only get the equation tag on the last line ..., please let me know.

I think you need a combination of an equation environment, an aligned[b] environment, and a \smash[t]{...} directive.

enter image description here

\documentclass{article}
\usepackage{amsmath} 

\begin{document}
\begin{equation} \label{eq.14}
\begin{aligned}[b] % 'b' means "place equation number on bottom line"
   1 \uparrow 1                  &= {}^1 1 = 1  \\
   2 \uparrow\uparrow 2          &= {}^2 2 = 4  \\
   3 \uparrow\uparrow \uparrow 3 &= 
     \smash[t]{ {}^{{}^{{}^3\!}} {}^3 3 
   =   3 \uparrow\uparrow 3 \uparrow\uparrow 3 
   =   \underbrace{3^{3^{3^{3^{3^{3^{.^{.^{.^{3}}}}}}}}}}_{3^{3^3}\text{ threes}}
     }
\end{aligned}
\end{equation}
\end{document}

Addendum: Here's an additional solution, which (a) employs an alignedat environment to perform alignment on both the first and second instance of =, (b) changes the math status of \uparrow, \uparrow\uparrow, and \uparrow\uparrow\uparrow to math-bin (the default math status is math-rel), and (c) implements @egreg's suggestion to provide a bit more distance between the rows and to enlarge the contents of the _{...} component of the \underbrace directive in row 3.

enter image description here

\documentclass{article}
\usepackage{mathtools} 

\begin{document}
\begin{equation} \label{eq.14}
\addtolength\jot{4pt}
\begin{alignedat}[b]{2} % 'b' means "place equation number on bottom line"
   1 \mathbin{\uparrow} 1                  &= {}^1 1 &&= 1  \\
   2 \mathbin{\uparrow\uparrow} 2          &= {}^2 2 &&= 4  \\
   3 \mathbin{\uparrow\uparrow\uparrow} 3 &= 
     \smash[t]{ {}^{{}^{{}^3\!}} {}^3 3} 
   &&=   \smash[t]{3 \mathbin{\uparrow\uparrow} 3 \mathbin{\uparrow\uparrow} 3 
     =   \underbrace{3^{3^{3^{3^{3^{3^{.^{.^{.^{3}}}}}}}}}}_{\mathclap{\textstyle 3^{3^3}\text{ threes}}}
     }
\end{alignedat}
\end{equation}
\end{document}
0
4

The main idea is the same as Mico's about smashing the top of the big object, but with some improvements on spacing and size of characters.

\documentclass{article}
\usepackage{amsmath}

\ExplSyntaxOn
\NewDocumentCommand{\hyop}{m}
 {
  \mathbin{ \prg_replicate:nn { #1 } { \uparrow } }
 }
\ExplSyntaxOff
\newcommand{\preexp}[2]{\vphantom{#2}^{#1}\kern-\scriptspace#2}

\begin{document}
\begin{equation} \label{eq.14}
\begin{aligned}[b] % 'b' means "place equation number on bottom line"
   1 \hyop{1} 1 &= \preexp{1}{1} = 1  \\
   2 \hyop{2} 2 &= \preexp{2}{2} = 4  \vphantom{\preexp{\preexp{3}{3}}{3}}\\
   3 \hyop{3} 3 &= \preexp{\preexp{3}{3}}{3}
   = 3 \hyop{2} ( 3 \hyop{2} 3 )
   = \smash[t]{
       \underbrace{
         3^{3^{3^{3^{3^{3^{{\cdot}^{{\cdot}^{{\cdot}^{3}}}}}}}}}
       }_{\text{$\displaystyle 3^{3^3}$ threes}}
     }
\end{aligned}
\end{equation}

\end{document}

enter image description here

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