What is the correct way to typeset higher order time derivatives in LaTeX?
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)
, andinserting a thinspace,
\,
, in front of thedt
terms
may be what you need:
\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}
I would just put them as ordinary superscripts. For the third derivative it seems to be necessary to backspace the subscript a bit:
\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:
\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}
Just to complete the answers of Mico and Andrew:
In Wikipedia, the dot notation for derivatives higher than the third (i.e., \ddot{V}
) is \overset{n}{\dot{V}}
(that is, the n-th dot derivative)
Example:
(code: \overset{5}{\dot{V}} \equiv \frac{d^5V}{dt^5}
)