diff --git a/notes/sections/4.md b/notes/sections/4.md
index 2852479..5c39187 100644
--- a/notes/sections/4.md
+++ b/notes/sections/4.md
@@ -24,7 +24,6 @@ $|P_v|$ of the particles with a given $P_h$?
   \node at (8.5,0.9) {$y$};
   \node at (5,8.4)   {$z$};
   % Momentum
-  \definecolor{cyclamen}{RGB}{146, 24, 43}
   \draw [ultra thick, ->, cyclamen] (5,2)     -- (3.8,6);
   \draw [thick, dashed, cyclamen]   (3.8,0.8) -- (3.8,6);
   \draw [thick, dashed, cyclamen]   (5,7.2)   -- (3.8,6);
@@ -52,7 +51,6 @@ $$
   {\int_{\{ P_v \}} d P_v f_{P_h , P_v} (x, P_v)}
   = \frac{f_{P_h , P_v} (x, P_v)}{I}
 $$
-
 where $f_{P_h , P_v}$ is the joint PDF of the two variables $P_v$ and $P_h$ and
 the integral $I$ runs over all the possible values of $P_v$ given a certain
 $P_h$.  
@@ -62,12 +60,10 @@ same considerations done in @sec:3 lead to:
 $$
   f_{\theta} (\theta) = \frac{1}{2} \sin{\theta} \chi_{[0, \pi]} (\theta)
 $$
-
 whereas, being $P$ uniform:
 $$
   f_P (P) = \chi_{[0, P_{\text{max}}]} (P)
 $$
-
 where $\chi_{[a, b]} (y)$ is the normalized characteristic function which value
 is $1/N$ between $a$ and $b$ (where $N$ is the normalization term) and 0
 elsewhere. Since $P,\theta$ are independent variables, their joint PDF is
@@ -77,9 +73,8 @@ $$
   = \frac{1}{2} \sin{\theta} \chi_{[0, \pi]} (\theta)
     \chi_{[0, P_{\text{max}}]} (P)
 $$
-and they are related to the vertical and horizontal components
-by a standard polar coordinate transformation:
-
+and they are related to the vertical and horizontal components by a standard
+polar coordinate transformation:
 \begin{align*}
   \begin{cases}
     P_v = P \cos(\theta) \\
@@ -91,12 +86,12 @@ by a standard polar coordinate transformation:
     \theta = \text{atan2}(P_h, P_v)
   \end{cases}
 \end{align*}
-
 where:
 
 - $\theta \in [0, \pi]$,
 
 - and atan2 is defined by:
+
 $$
 \begin{cases}
   \arctan(P_h/P_v) &\incase P_v > 0 \\
@@ -104,7 +99,6 @@ $$
   \arctan(P_h/P_v) + \pi &\incase P_v < 0
 \end{cases}
 $$
-
 The Jacobian of the inverse transformation is easily found to be:
 $$
   |J^{-1}| = \frac{1}{\sqrt{P_v^2 + P_h^2}}
@@ -117,9 +111,8 @@ $$
     \frac{\chi_{[0, p_{\text{max}}]} \left(\sqrt{P_v^2 + P_h^2} \right)}
     {\sqrt{P_v^2 + P_h^2}}
 $$
-
-The integral $I$ can now be computed. Note that the domain
-is implicit in the characteristic function:
+The integral $I$ can now be computed. Note that the domain is implicit in the
+characteristic functions:
 $$
   I(x) = \int_{-\infty}^{+\infty} dP_v \, f_{P_h , P_v} (x, P_v)
        = \int \limits_{- \sqrt{P_{\text{max}}^2 - P_h}}
@@ -143,7 +136,6 @@ $$
     \chi_{[0, p_{\text{max}}]} \left(\sqrt{P_v^2 + P_h^2}\right)}{2 \, \arctan
            \left( \sqrt{\frac{P_{\text{max}}^2}{x^2} - 1} \right)}
 $$
-
 Finally, putting all the pieces together, the average value of $|P_v|$ can be
 computed:
 $$
@@ -152,7 +144,6 @@ $$
   = \frac{x \ln \left( \frac{P_{\text{max}}}{x} \right)}
   {\arctan \left( \sqrt{ \frac{P^2_{\text{max}}}{x^2} - 1} \right)}
 $$ {#eq:dip}
-
 The result is plotted in the figure below:
 
 ![Plot of the expected dependence of $\langle |P_v| \rangle$ with
@@ -163,30 +154,28 @@ The result is plotted in the figure below:
 
 This dependence should be found by running a Monte Carlo simulation and
 computing a binned average of the vertical momentum. A number of $N = 50000$
-particles were generated as pair of values ($P$, $\theta$), with $P$
-uniformly distributed between 0 and $P_{\text{max}}$ and $\theta$ given by the
-same procedure described in @sec:3, namely:
+particles was generated as pairs of values ($P$, $\theta$), with $P$ uniformly
+distributed between 0 and $P_{\text{max}}$ and $\theta$ given by the same
+procedure described in @sec:3, namely:
 $$
   \theta = \arccos(1 - 2x)
 $$
-
 where $x$ is uniformly distributed between 0 and 1.  
 The binning turned out to be quite a challenge: once a $P$ is sampled and
 $P_h$ computed, the bin containing the latter has to be found. If
 the range $[0, P_{\text{max}}]$ is divided into $n$ equal bins
-of the width
+of width:
 $$
   w = \frac{P_{\text{max}}}{n}
 $$
-then (counting from zero) $P_h$ goes into the $i$-th bin where
+then (counting from zero) $P_h$ goes into the $i$-th bin, where:
 $$
   i = \left\lfloor \frac{P_h}{w} \right\rfloor
 $$
-
 Then, the sum $S_j$ of all the $|P_v|$ values relative to the $P_h$ of the
-$j$-th bin itself and number num$_j$ of the bin counts are stored in an array
-and iteratively updated.  Once every bin has been updated, the average value of
-$|P_v|_j$ is computed as $S_j / \text{num}_j$.
+$j$-th bin and the number num$_j$ of the bin counts are stored in an array
+and iteratively updated. Once every point has been sampled, the average value
+of $|P_v|_j$ is computed as $S_j / \text{num}_j$.
 
 For the sake of clarity, for each sampled couple the procedure is the
 following. At first $S_j = 0 \wedge \text{num}_j = 0 \, \forall \, j$, then:
@@ -217,14 +206,12 @@ The following results were obtained:
 
 The $\chi^2$ and $p$-value show a very good agreement.
 In order to compare $P^{\text{oss}}_{\text{max}}$ with the expected value
-$P_{\text{max}} = 10$, the following compatibility $t$-test was applied:
-
+$P_{\text{max}} = 10$, the usual compatibility $t$-test was applied:
 $$
   p = 1 - \text{erf}\left(\frac{t}{\sqrt{2}}\right)\ \with
   t = \frac{|P^{\text{oss}}_{\text{max}} - P_{\text{max}}|}
            {\Delta P^{\text{oss}}_{\text{max}}}
 $$
-
 where $\Delta P^{\text{oss}}_{\text{max}}$ is the $P^{\text{oss}}_{\text{max}}$
 uncertainty. At 95% confidence level, the values are compatible if $p > 0.05$.
 In this case: