% One simple way to estimate $x((m+k)T)$ is to use the information at
% $mT$ and $(m-1)T$, which leads to the forward-Euler method as
To estimate $x((m+k)T)$,
-a forward Euler\index{forward Euler} style jumping relies only on $x(mT)$ and $x((m-1)T)$,
+a forward Euler\index{Euler!forward Euler} style jumping relies only on $x(mT)$ and $x((m-1)T)$,
i.e.,
\[
x((m+k)T)
\]
However, this approach is inefficient due to its restriction on
envelope step $k$, since larger $k$ usually causes instability.
-Instead, backward Euler\index{backward Euler} jumping,
+Instead, backward Euler\index{Euler!backward Euler} jumping,
%and the equation becomes
\[
x((m+k)T)-x(mT) = k\left[x((m+k)T)-x((m+k-1)T)\right],
