4

I have an equation which has a fraction to the power of another fraction. The fraction in the power is really small. I tried using \cfrac to change the size of that fraction but then the constant and exponential at the start of the equation became too small compared to the rest of the equation.

enter image description here

What is the best way to make this equation look good? Here is a MWE.

\documentclass{article}

\usepackage{amsmath}
\usepackage{geometry}

\begin{document} 

\begin{equation}
\rho(T) = \rho_{0}\exp\left[\left(\frac{T_{0}}{T}\right)^{\frac{1}{d+1}}\right]
\end{equation}

\end{document}
9

A larger size for the exponent can be used, e.g. via \tfrac. Or the fraction expression can be written with a slash:

\documentclass{article}

\usepackage{amsmath}
\usepackage{geometry}

\begin{document}

\begin{equation}
\rho(T) =
\rho_{0}\exp\left[\left(\frac{T_{0}}{T}\right)^{\tfrac{1}{d+1}}\right]
\end{equation}
\begin{equation}
\rho(T) =
\rho_{0}\exp\left[\left(\frac{T_{0}}{T}\right)^{1/(d+1)}\right]
\end{equation}

\end{document}

Result

| improve this answer | |
  • The use of \tfrac does the trick. Many thanks! – Shawn P Apr 8 '16 at 19:44
3

There is no single "best" way to make an equation look "good".

For the equation at hand, I'd like to suggest you use "inline" or "slash" fractional notation. That way, T will be rendered at "textstyle" size and symbols in the subscripts and the superscripts will be rendered at "scriptstyle" size.

enter image description here

\documentclass{article}
\begin{document} 
\[
\rho(T) = \rho_{0} \exp\Bigl[(T_{0}/T)^{1/(d+1)}\Bigr]
\]
\end{document}
| improve this answer | |
3

Have your choice!

\documentclass{article}

\usepackage{amsmath}
\usepackage{geometry}
\usepackage{nicefrac}

 \begin{document}

\begin{align*}
    \rho(T) & = \rho_{0}\exp\left[\left(\frac{T_{0}}{T}\right)^{\tfrac{1}{d+1}}\right]\\
    \rho(T) & = \rho_{0}\exp\left[\left(\frac{T_{0}}{T}\right)^{\frac{1}{d+1}}\right]\\
    \rho(T) & = \rho_{0}\exp\left[\left(\frac{T_{0}}{T}\right)^{\nicefrac{1}{(d+1)}}\right]\\
    \rho(T) & = \rho_{0}\exp\left[\left(\frac{T_{0}}{T}\right)^{\nicefrac[\scriptstyle]{1}{(d+1)}}\right]
\end{align*}

\end{document} 

enter image description here

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.