4

I'd like the \times symbol to be under the exponent in the following example, how can I achieve this ?

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{mathtools}
\usepackage{amsmath}
\usepackage{amsthm}

\begin{document}

\begin{equation}
    \left(\frac{k+1}{k^\frac{k}{k+1}}\right)^{k+1} \times \frac{1}{(2k+2)(2k+1)}
\end{equation}

\end{document}

Output :

enter image description here

I tried to use \smashoperator but failed to achieve the desired result.

5 Answers 5

7

I personally would not do that but since you are asking...

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{mathtools}
\usepackage{amsmath}
\usepackage{amsthm}

\begin{document}

\begin{equation}
    \left(\frac{k+1}{k^\frac{k}{k+1}}\right)^{\mathrlap{k+1}} \times \frac{1}{(2k+2)(2k+1)}
\end{equation}

\end{document}

enter image description here

2
  • Why wouldn't you do that ?
    – Sileo
    Commented Apr 22, 2020 at 18:14
  • 12
    @Rphad This is only a personal opinion which you may ignore. I personally would not add the \times here and would put the bracketed expression in the numerator of the fraction that follows. If you put the k+1 on top of the \times it looks a bit like a math operator like a sum with some limits and I personally would try to avoid this possible confusion.
    – user194703
    Commented Apr 22, 2020 at 18:18
10

I could understand a desire to avoid the hole, but your readers would have a hard time in interpreting the expression if the exponent is completely above \times.

I'd possibly use (3) or (4), but would prefer (5), removing the \times altogether, or better yet (6).

Notice the \! in the exponent to avoid the other hole.

\documentclass{article}
\usepackage{amsmath}

\begin{document}

\begin{equation}
\left(\frac{k+1}{k^\frac{k}{k+1}}\right)^{\!k+1}\times\frac{1}{(2k+2)(2k+1)}
\end{equation}

\begin{equation}
\left(\frac{k+1}{k^\frac{k}{k+1}}\right)^{\!k+1}
\mspace{-\medmuskip}
\times \frac{1}{(2k+2)(2k+1)}
\end{equation}

\begin{equation}
\left(\frac{k+1}{k^\frac{k}{k+1}}\right)^{\!k+1}
\mspace{-1.5\medmuskip}
\times \frac{1}{(2k+2)(2k+1)}
\end{equation}

\begin{equation}
\left(\frac{k+1}{k^\frac{k}{k+1}}\right)^{\!k+1}
\mspace{-2\medmuskip}
\times \frac{1}{(2k+2)(2k+1)}
\end{equation}

\begin{equation}
\left(\frac{k+1}{k^\frac{k}{k+1}}\right)^{\!k+1}
\frac{1}{(2k+2)(2k+1)}
\end{equation}

\begin{equation}
\frac{1}{(2k+2)(2k+1)}
\left(\frac{k+1}{k^\frac{k}{k+1}}\right)^{\!k+1}
\end{equation}

\end{document}

enter image description here

0
4

I'm not sure if this is what you mean, but you can do this with the \rlap command.

enter image description here

Here is the code:

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{mathtools}
\usepackage{amsmath}
\usepackage{amsthm}

\begin{document}

\begin{equation}
    \left(\frac{k+1}{k^\frac{k}{k+1}}\right)^{\rlap{$\scriptstyle{k+1}$}} \times \frac{1}{(2k+2)(2k+1)}
\end{equation}

\end{document}
2

Since there seems now to be some additional interest in typesetting this equation, let me spell out my above comment.

\documentclass{article}
\usepackage{mathtools}

\begin{document}

\begin{equation}
     \frac{(k+1)^{k+1}}{k^k}\frac{1}{(2k+2)(2k+1)}
\end{equation}
or
\begin{equation}
     \frac{1}{(2k+2)(2k+1)}\frac{(k+1)^{k+1}}{k^k}
\end{equation}
or
\begin{equation}
     \frac{\dfrac{(k+1)^{k+1}}{k^k}}{(2k+2)(2k+1)}
\end{equation}
\end{document}

enter image description here

I think, though, that all these are off-topic as they do not answer the original question.

1

A solution with \mathrlap from `mathtools. Adding some mayj kening, one can have the × symbol under the + in the exponent.

Comments aside: loading inputenc is not necessary nowadays, since LaTeX expects utf8 by default, and amsmath is already loaded by mathtools.

\documentclass{article}
\usepackage{mathtools}
\usepackage{amsthm}

\begin{document}

\begin{equation}
    \left(\frac{k+1}{k^{\frac{k}{k+1}}}\right)^{\mkern-6mu \mathrlap{k+1}}\mkern 1.5mu\times \frac{1}{(2k+2)(2k+1)}
\end{equation}

\end{document} 

enter image description here

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .