diff --git a/451-Signal_Image/Cours/chap4.tex b/451-Signal_Image/Cours/chap4.tex
index afc9362..f5ff58a 100644
--- a/451-Signal_Image/Cours/chap4.tex
+++ b/451-Signal_Image/Cours/chap4.tex
@@ -160,7 +160,10 @@ On remarque que $DA = I_n$.
 \subparagraph{Méthode 2} Pour $A^TMA>0$.
 
 \[
-J_{MC} = \underbracket{(D(Y-m_B)-\Theta)^TA^TMA(D(Y-m_B)-\theta)}_ {J_1(Y,\theta)} + \underbracket{(Y-m_B)^T(M-D^TA^TMAD)(Y-m_B)}_{J_2(Y)}
+  \begin{aligned}
+    J_{MC} &= \underbracket{(D(Y-m_B)-\Theta)^TA^TMA(D(Y-m_B)-\theta)}_ {J_1(Y,\theta)}\\
+    &+ \underbracket{(Y-m_B)^T(M-D^TA^TMAD)(Y-m_B)}_{J_2(Y)}
+\end{aligned}
 \]
 Alors $\nabla J_{MC} = 0  \implies J_1 = 0 \implies D(Y-m_B) = \hat{\theta}_{MC}$
 
@@ -376,8 +379,8 @@ On considère un cout uniforme.
 \begin{defin}
   En prenant:
   \begin{align*}
-    E_{\theta|Y}[C(\hat{\theta},\theta)] &= \int_{\R^m}(1-\Pi_{\Delta}(\tilde{\theta}))f_{\theta|Y=y}(\theta)d\theta
-    &= 1 - \int_{\hat{\theta}-\Delta/2}^{{\hat{\theta}+\Delta/2}}f_{\theta|Y=y}(\theta)d\theta
+    E_{\theta|Y}[C(\hat{\theta},\theta)] &= \int_{\R^m}(1-\Pi_{\Delta}(\tilde{\theta}))f_{\theta|Y=y}(\theta)d\theta \\
+    &= 1 - \int_{\hat{\theta}-\Delta/2}^{{\hat{\theta}+\Delta/2}}f_{\theta|Y=y}(\theta)d\theta \\
       &\simeq 1- \Delta^nf_{\theta|Y=y}(\hat{\theta})
   \end{align*}
   Soit \[