X-Git-Url: https://bilbo.iut-bm.univ-fcomte.fr/and/gitweb/book_gpu.git/blobdiff_plain/17bff40b83bcdcc39769f9e59c70ffae1c525b72..0d39f3bfb1736ae41805f75a779e0bb01f4f5139:/BookGPU/Chapters/chapter16/ef.tex?ds=inline

diff --git a/BookGPU/Chapters/chapter16/ef.tex b/BookGPU/Chapters/chapter16/ef.tex
index 400b467..1cc747b 100644
--- a/BookGPU/Chapters/chapter16/ef.tex
+++ b/BookGPU/Chapters/chapter16/ef.tex
@@ -52,7 +52,7 @@ significant savings in simulation time.
 % 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.,
 \[
    x((m+k)T)
@@ -61,7 +61,7 @@ i.e.,
 \]
 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],