% 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
\[
x((m+k)T)-x(mT) = k\left[x((m+k)T)-x((m+k-1)T)\right],
\]
allows larger envelope steps.
Here $x((m+k-1)T)$ is the unknown variable to be solved
-by Newton iteration\index{Newton iteration},
+by Newton iteration\index{iterative method!Newton iteration},
and $x((m+k)T)$ is dependent on $x((m+k-1)T)$
in each iteration.
% Forward-Euler may be used to generate the initial guess.
the computation of sensitivity matrix.
It can be easily derived that,
%for one signal period with $M$ time steps,
-if the DAE is integrated with backward-Euler rule,
+if the DAE is integrated with backward Euler rule,
the sensitivity is
\[
J = \frac{\ud x_M}{\ud x_0}
