Use math also for the 'x1' and 'x2' of linspace

chapters/arrays.tex

@@ -317,13 +317,13 @@ \subsection{Specialized Array Constructor Functions}\doublelabel{specialized-arr
 \lstinline[mathescape=true]!fill(s, $n$)! = \lstinline[mathescape=true]!{s, s, $\ldots$, s}!
 
 The function needs two or more arguments; that is \lstinline!fill(s)! is not legal.\\ \hline
-\lstinline[mathescape=true]!linspace(x1, x2, $n$)!
+\lstinline[mathescape=true]!linspace($x_{1}$, $x_{2}$, $n$)!
 &
-Returns a \lstinline!Real! vector with $n$ equally spaced elements, such that
-\lstinline[mathescape=true]!v = linspace(x1, x2, $n$)! results in
-\lstinline[mathescape=true]!v[$i$] = x1 + (x2-x1)*($i$-1)/($n$-1)! for $1 \leq i \leq n$.
-It is required that $n \geq 2$. The arguments \lstinline!x1! and \lstinline!x2! shall
-be numeric scalar expressions.\\ \hline
+Returns a \lstinline!Real! vector with $n$ equally spaced elements, such that \lstinline[mathescape=true]!v = linspace($x_{1}$, $x_{2}$, $n$)! results in
+\begin{equation*}
+\text{\lstinline[mathescape=true]!v[$i$]!} = x_{1} + (x_{2} - x_{1}) \frac{i - 1}{n - 1} \quad \text{for $1 \leq i \leq n$}
+\end{equation*}
+It is required that $n \geq 2$.  The arguments $x_{1}$ and $x_{2}$ shall be numeric scalar expressions.\\ \hline
 \end{longtable}
 
 \subsection{Reduction Functions and Operators}\doublelabel{reduction-functions-and-operators}