11

I am typing a series that involves higher order time derivatives, and I have two main problems.

First, the subscript of the derived variable seems to move further away from it with the more characters on top, is there a way to fix that?

Second, what is the best way to typeset an nth order derivative without it being too big as in \overset{(n)}{V}, or too high as in \overset{{}^{(n)}}{V}?

Here is an example that shows all the mentioned problems:

\documentclass[border=10pt]{standalone}

\usepackage{mathtools}
\begin{document}
 $\displaystyle
 \dot{V}_t = \dot{V}_{t-1} + \ddot{V}_{t-1} dt + \frac{\dddot{V}_{t-1}dt^2}{2} +
 \frac{\overset{(4)}{V}_{t-1}dt^3}{6} + \dots + \frac{\overset{{}^{(n+1)}}{V}_{t-1}dt^n}{n!}
 $
\end{document}

example

EDIT:

\overset{\text{\tiny $(n)$}}{V} works for reducing the size of the overset, but the subscript distance is still a problem

EDIT 2:

The subscript distance can be reduced by adding negative spaces as in \frac{\overset{{}^{(n+1)}}{V}_{\!\!\!t-1}dt^n}{n!}. I'm not sure if it's the cleanest way to do it, but it works

10

There's probably no single or unique "correct" way to typeset this expression, but

  • using \scriptscriptstyle for the overset terms (4) and (n+1),

  • using \mathclap to "smash" the width of the term (n+1), and

  • inserting a thinspace, \,, in front of the dt terms

may be what you need:

enter image description here

\documentclass{article}
\usepackage{mathtools} % for \mathclap macro
\begin{document}
\[
 \dot{V}_t = \dot{V}_{t-1} + 
 \ddot{V}_{t-1} \,dt + 
 \frac{\dddot{V}_{t-1}\,dt^2}{2} +
 \frac{\overset{\scriptscriptstyle(4)}{V}_{t-1}\,dt^3}{6} + \dots + 
 \frac{\overset{\mathclap{\scriptscriptstyle(n+1)}}{V}_{t-1}\,dt^n}{n!}
\]
\end{document} 
0
6

I would just put them as ordinary superscripts. For the third derivative it seems to be necessary to backspace the subscript a bit:

Sample output

\documentclass{article}

\usepackage{mathtools}

\begin{document}

\begin{equation*}
  \dot{V}_t = \dot V_{t-1} + \ddot V_{t-1} dt +
  \frac{\dddot V_{\!t-1}dt^2}{2} + \frac{V^{(4)}_{t-1}dt^3}{6} +
  \dots + \frac{V^{(n+1)}_{t-1}dt^n}{n!}
\end{equation*}

\end{document}

If, as in your case, this notation is used for something else than I would recommend using a differential operator instead:

Second sample

\documentclass{article}

\usepackage{mathtools}

\begin{document}

\begin{equation*}
  \dot{V}_t = \dot V_{t-1} + \ddot V_{t-1} dt +
  \frac{(D^3_t V)_{t-1}dt^2}{2} + \frac{(D^4_tV)_{t-1}dt^3}{6} +
  \dots + \frac{(D^{n+1}_tV)_{t-1}dt^n}{n!}
\end{equation*}

\end{document}
4
  • 1
    In the notation I'm using, ordinary subscripts are reserved for the positional derivatives $\frac{dy}{dx}$. I use oversets for time derivatives – Tymric Jun 24 '14 at 11:08
  • Especially for the 3rd derivative, you could need, something nicer, like ![enter image description here][1] $\overset{\therefore}{V}$ [1]: i.stack.imgur.com/zF9DT.png – cis Jun 24 '14 at 11:20
  • 3
    @cis That is not a standard notation. – Andrew Swann Jun 24 '14 at 11:36
  • I know, but you and I are not standard-human-beeings, so we could think in categories, like meaningful and nice too ;) PS: I saw this notatation in some scripts from professors, so: why not. – cis Jun 24 '14 at 12:56
3

Just to complete the answers of Mico and Andrew:

In Wikipedia, the dot notation for derivatives higher than the third (i.e., \ddot{V}   third dot derivative V) is \overset{n}{\dot{V}}

enter image description here

(that is, the n-th dot derivative)

Example:

enter image description here

(code: \overset{5}{\dot{V}} \equiv \frac{d^5V}{dt^5})

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