1

Is there a way to make only a part of an equation larger? I'm working with the code below, but I feel like the s-angle-n and a-angle-n are far too small and would really like to make them both bigger without affecting the rest of the equation.

\DeclareRobustCommand{\lcroof}[1]{
  \hbox{\vtop{\vbox{%
      \hrule\kern 1pt\hbox{%
        $\scriptstyle #1$%
        \kern 1pt}}\kern1pt}%
    \vrule\kern1pt}}
\DeclareRobustCommand{\angle}[1]{
  _{\lcroof{#1}}}


\begin{align*}
s\angle{n}&=\dfrac{(1+r)^n-1}{r}\\
a\angle{n}&=\dfrac{1-\dfrac{1}{(1+r)^n}}{r}
\end{align*}

Thanks for any help.

2
  • 1
    How much bigger would be better than "far too small"?
    – Werner
    Commented Oct 2, 2013 at 20:47
  • They seem to have the proper size.
    – egreg
    Commented Oct 2, 2013 at 21:55

2 Answers 2

1

How about put them in text mode and then use \Huge, \Large, etc.

\begin{align*}
\mbox{\Huge{$s\angle{n}$}}  &=  \dfrac{(1+r)^n -1}{r}  \\
\mbox{\Huge{$a\angle{n}$}}  &=  \dfrac{1 - \dfrac{1}{(1+r)^n}}{r}
\end{align*}
0

Probably not the nicest solution, but you could use: \usepackage{graphicx}, and then have:

\begin{align*}
    \scalebox{1.5}{$s\angle{n}$}  &=  \dfrac{(1+r)^n -1}{r}  \\
    \scalebox{1.5}{$a\angle{n}$}  &=  \dfrac{1 - \dfrac{1}{(1+r)^n}}{r}
\end{align*}

Where you can change the 1.5 to suit your taste.

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