ex-4: revised and typo-fixed

ex-4/main.c

@@ -41,6 +41,9 @@ int main(int argc, char **argv)
   size_t n = 50;
   double p_max = 10;
   size_t go = 0;
+
+  // Get eventual CLI arguments.
+  //
   int res = parser(&N, &n, &p_max, argc, argv, &go);
   if (go == 0 && res == 1)
     {

ex-4/plot.py

@@ -15,7 +15,7 @@ edges = np.linspace(0, bins*step, bins+1)
 plt.rcParams['font.size'] = 17
 
 plt.hist(edges[:-1], edges, weights=counts,
-         color='#eacf00', edgecolor='#3d3d3c')
+         histtype='stepfilled', color='#e3c5ca', edgecolor='#92182b')
 plt.title('Vertical component distribution', loc='right')
 plt.xlabel(r'$p_h$')
 plt.ylabel(r'$\left<|p_v|\right>$')

notes/sections/4.md

@@ -3,14 +3,14 @@
 ## Kinematic dip PDF derivation
 
 Consider a great number of non-interacting particles, each of which with a
-random momentum $\vec{p}$ with module between 0 and $p_{\text{max}}$ randomly
+random momentum $\vec{P}$ with module between 0 and $P_{\text{max}}$ randomly
 angled with respect to a coordinate system {$\hat{x}$, $\hat{y}$, $\hat{z}$}.
 Once the polar angle $\theta$ is defined, the momentum vertical and horizontal
-components of a particle, which will be referred as $\vec{p_v}$ and $\vec{p_h}$
+components of a particle, which will be referred as $\vec{P_v}$ and $\vec{P_h}$
 respectively, are the ones shown in @fig:components.  
 If $\theta$ is evenly distributed on the sphere and the same holds for the
-module $p$, which distribution will the average value of $|p_v|$ follow as a
+module $P$, which distribution will the average value of $|P_v|$ follow as a
-function of $p_h$?
+function of $P_h$?
 
 \begin{figure}
 \hypertarget{fig:components}{%
@@ -30,9 +30,9 @@ function of $p_h$?
   \draw [thick, dashed, cyclamen]   (5,7.2)   -- (3.8,6);
   \draw [ultra thick, ->, pink]     (5,2)     -- (5,7.2);
   \draw [ultra thick, ->, pink]     (5,2)     -- (3.8,0.8);
-  \node at (4.8,1.1) {$\vec{p_h}$};
+  \node at (4.8,1.1) {$\vec{P_h}$};
-  \node at (5.5,6.6) {$\vec{p_v}$};
+  \node at (5.5,6.6) {$\vec{P_v}$};
-  \node at (3.3,5.5) {$\vec{p}$};
+  \node at (3.3,5.5) {$\vec{P}$};
   % Angle
   \draw [thick, cyclamen] (4.4,4) to [out=80,in=210] (5,5);
   \node at (4.7,4.2) {$\theta$};
@@ -41,11 +41,11 @@ function of $p_h$?
 }
 \end{figure}
 
-Since the aim is to compute $\langle |p_v| \rangle (p_h)$, the conditional
+Since the aim is to compute $\langle |P_v| \rangle (P_h)$, the conditional
-distribution probability of $p_v$ given a fixed value of $p_h = x$ must first
+distribution probability of $P_v$ given a fixed value of $P_h = x$ must first
-be determined. It can be computed as the ratio between the probablity of
+be determined. It can be computed as the ratio between the probability of
-getting a fixed value of $P_v$ given $x$ over the total probability of $x$:
+getting a fixed value of $P_v$ given $x$ over the total probability of
-
+getting that $x$:
 $$
   f (P_v | P_h = x) = \frac{f_{P_h , P_v} (x, P_v)}
   {\int_{\{ P_v \}} d P_v f_{P_h , P_v} (x, P_v)}
@@ -53,35 +53,31 @@ $$
 $$
 
 where $f_{P_h , P_v}$ is the joint pdf of the two variables $P_v$ and $P_h$ and
-the integral runs over all the possible values of $P_v$ given $P_h$.
+the integral $I$ runs over all the possible values of $P_v$ given a certain
-
+$P_h$.  
-This joint pdf can simly be computed from the joint pdf of $\theta$ and $p$ with
+$f_{P_h , P_v}$ can simply be computed from the joint pdf of $\theta$ and $P$
-a change of variables. For the pdf of $\theta$ $f_{\theta} (\theta)$, the same
+with a change of variables. For the pdf of $\theta$ $f_{\theta} (\theta)$, the
-considerations done in @sec:3 lead to:
+same considerations done in @sec:3 lead to:
-
 $$
   f_{\theta} (\theta) = \frac{1}{2} \sin{\theta} \chi_{[0, \pi]} (\theta)
 $$
 
-whereas, being $p$ evenly distributed:
+whereas, being $P$ evenly distributed:
-
 $$
-  f_p (p) = \chi_{[0, p_{\text{max}}]} (p)
+  f_P (P) = \chi_{[0, P_{\text{max}}]} (P)
 $$
 
 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
+is $1/N$ between $a$ and $b$ (where $N$ is the normalization term) and 0
 elsewhere. Being a couple of independent variables, their joint pdf is simply
 given by the product of their pdfs:
-
 $$
-  f_{\theta , p} (\theta, p) = f_{\theta} (\theta) f_p (p)
+  f_{\theta , P} (\theta, P) = f_{\theta} (\theta) f_P (P)
   = \frac{1}{2} \sin{\theta} \chi_{[0, \pi]} (\theta)
-    \chi_{[0, p_{\text{max}}]} (p)
+    \chi_{[0, P_{\text{max}}]} (P)
 $$
 
 Given the new variables:
-
 $$
   \begin{cases}
     P_v = P \cos(\theta) \\
@@ -89,12 +85,11 @@ $$
   \end{cases}
 $$
 
-with $\theta \in [0, \pi]$, the previous ones are written as:
+with $\theta \in [0, \pi]$, the previous ones can be written as:
-
 $$
   \begin{cases}
     P = \sqrt{P_v^2 + P_h^2} \\
-    \theta = \text{atan}^{\star} ( P_h/P_v ) :=
+    \theta = \text{atan}_2 ( P_h/P_v ) :=
       \begin{cases}
         \text{atan} ( P_h/P_v )       &\incase P_v > 0 \\
         \text{atan} ( P_h/P_v ) + \pi &\incase P_v < 0
@@ -103,24 +98,21 @@ $$
 $$
 
 which can be shown having Jacobian:
-
 $$
   J = \frac{1}{\sqrt{P_v^2 + P_h^2}}
 $$
 
 Hence:
-
 $$  
   f_{P_h , P_v} (P_h, P_v) =
-    \frac{1}{2} \sin[ \text{atan}^{\star} ( P_h/P_v )]
+    \frac{1}{2} \sin[ \text{atan}_2 ( P_h/P_v )]
-    \chi_{[0, \pi]} (\text{atan}^{\star} ( P_h/P_v )) \cdot \\
+    \chi_{[0, \pi]} (\text{atan}_2 ( P_h/P_v )) \cdot \\
     \frac{\chi_{[0, p_{\text{max}}]} \left( \sqrt{P_v^2 + P_h^2} \right)}
     {\sqrt{P_v^2 + P_h^2}}
 $$
 
 from which, the integral $I$ can now be computed. The edges of the integral
 are fixed by the fact that the total momentum can not exceed $P_{\text{max}}$:
-
 $$
   I = \int
   \limits_{- \sqrt{P_{\text{max}}^2 - P_h}}^{\sqrt{P_{\text{max}}^2 - P_h}}
@@ -128,39 +120,32 @@ $$
 $$
 
 after a bit of maths, using the identity:
-
 $$
-  \sin[ \text{atan}^{\star} ( P_h/P_v )] = \frac{P_h}{\sqrt{P_h^2 + P_v^2}}
+  \sin[ \text{atan}_2 ( P_h/P_v )] = \frac{P_h}{\sqrt{P_h^2 + P_v^2}}
 $$
 
-and the fact that both the characteristic functions are automatically satisfied
+and the fact that both the characteristic functions play no role within the
-within the integral limits, the following result arises:
+integral limits, the following result arises:
-
 $$
   I = 2 \, \text{atan} \left( \sqrt{\frac{P_{\text{max}}^2}{x^2} - 1} \right)
 $$
 
 from which:
-
 $$
   f (P_v | P_h = x) = \frac{x}{P_v^2 + x^2} \cdot
   \frac{1}{2 \, \text{atan}
            \left( \sqrt{\frac{P_{\text{max}}^2}{x^2} - 1} \right)}
 $$
 
-Finally, putting all the pieces together, the average value of $|P_v|$ can now
+Finally, putting all the pieces together, the average value of $|P_v|$ can be
-be computed:
+computed:
-
+$$
-\begin{align*}
+  \langle |P_v| \rangle = \int
-  \langle |P_v| \rangle &= \int
   \limits_{- \sqrt{P_{\text{max}}^2 - P_h}}^{\sqrt{P_{\text{max}}^2 - P_h}}
-  f (P_v | P_h = x)
+  f (P_v | P_h = x) = [ \dots ]
-  \\
+  = x \, \frac{\ln \left( \frac{P_{\text{max}}}{x} \right)}
-  &= [ \dots ]
-  \\
-  &= x \, \frac{\ln \left( \frac{P_{\text{max}}}{x} \right)}
   {\text{atan} \left( \sqrt{ \frac{P^2_{\text{max}}}{x^2} - 1} \right)}
-\end{align*}
+$$ {#eq:dip}
 
 Namely:
 
@@ -171,80 +156,71 @@ Namely:
 ## Monte Carlo simulation
 
 The same distribution should be found by generating and binning points in a
-proper way.  
+proper way. A number of $N = 50'000$ points were generated as a couple of values
-A number of $N = 50'000$ points were generated as a couple of values ($p$,
+($P$, $\theta$), with $P$ evenly distributed between 0 and $P_{\text{max}}$ and
-$\theta$), with $p$ evenly distrinuted between 0 and $p_{\text{max}}$ and
 $\theta$ given by the same procedure described in @sec:3, namely:
-
 $$
   \theta = \text{acos}(1 - 2x)
 $$
 
-with $x$ uniformely distributed between 0 and 1.  
+with $x$ uniformly distributed between 0 and 1.  
-The data binning turned out to be a bit challenging. Once $p$ was sorted and
+The data binning turned out to be a bit challenging. Once $P$ was sampled and
-$p_h$ was computed, the bin in which the latter goes in must be found. If $n$ is
+$P_h$ was computed, the bin containing the latter's value must be found. If $n$
-the number of bins in which the range [0, $p_{\text{max}}$] is willing to be
+is the number of bins in which the range $[0, P_{\text{max}}]$ is divided into,
-divided into, then the width $w$ of each bin is given by:
+then the width $w$ of each bin is given by:
-
 $$
-  w = \frac{p_{\text{max}}}{n}
+  w = \frac{P_{\text{max}}}{n}
 $$
 
-and the $j^{th}$ bin in which $p_h$ goes in is:
+and the $i^{th}$ bin in which $P_h$ goes in is:
-
 $$
-  j = \text{floor} \left( \frac{p_h}{w} \right)
+  i = \text{floor} \left( \frac{P_h}{w} \right)
 $$
 
-where 'floor' is the function which gives the upper integer lesser than its
+where 'floor' is the function which gives the bigger integer smaller than its
 argument and the bins are counted starting from zero.  
-Then, a vector in which the $j^{\text{th}}$ entry contains the sum $S_j$ of all
+Then, a vector in which the $j^{\text{th}}$ entry contains both the sum $S_j$
-the $|p_v|$s relative to each $p_h$ fallen into the $j^{\text{th}}$ bin and the
+of all the $|P_v|$s relative to each $P_h$ fallen into the $j^{\text{th}}$ bin
-number num$_j$ of entries in the $j^{\text{th}}$ bin was reiteratively updated.
+itself and the number num$_j$ of the bin entries was reiteratively updated. At
-At the end, the average value of $|p_v|_j$ was computed as $S_j /
+the end, the average value of $|P_v|_j$ was computed as $S_j / \text{num}_j$.  
-\text{num}_j$.  
+For the sake of clarity, for each sampled couple the procedure is the
-For the sake of clarity, for each sorted couple, it works like this:
+following. At first $S_j = 0 \wedge \text{num}_j = 0 \, \forall \, j$, then:
 
-  - the couple $[p; \theta]$ is generated;
+  - the couple $(P, \theta)$ is generated,
-  - $p_h$ and $p_v$ are computed;
+  - $P_h$ and $P_v$ are computed,
-  - the $j^{\text{th}}$ bin in which $p_h$ goes in is computed;
+  - the $j^{\text{th}}$ bin containing $P_h$ is found,
-  - num$_j$ is increased by 1;
+  - num$_j$ is increased by 1,
-  - $S_j$ (which is zero at the beginning of everything) is increased by a factor
+  - $S_j$ is increased by $|P_v|$.
-    $|p_v|$.
 
-At the end, $\langle |p_h| \rangle_j$ = $\langle |p_h| \rangle (p_h)$ where:
+For $P_{\text{max}} = 10$ and $n = 50$, the following result was obtained:
 
-$$
+![Sampled points histogram.](images/dip.pdf)
-  p_h = j \cdot w + \frac{w}{2} = w \left( 1 + \frac{1}{2} \right)
-$$
 
-For $p_{\text{max}} = 10$, the following result was obtained:
+In order to check whether the expected distribution (@eq:dip) properly matches
-
+the produced histogram, a chi-squared minimization was applied. Being a simple
-![Histogram of the obtained distribution.](images/dip.pdf)
-
-In order to check wheter the expected distribution properly metches the
-produced histogram, a chi-squared minimization was applied. Being a simple
 one-parameter fit, the $\chi^2$ was computed without a suitable GSL function
-and the error of the so obtained estimation of $p_{\text{max}}$ was given as
+and the error of the so obtained estimation of $P_{\text{max}}$ was given as
-the inverese of the $\chi^3$ second derivative in its minimum, according to the
+the inverse of the $\chi^2$ second derivative in its minimum, according to the
-Cramér-Rao bound.  
+Cramér-Rao bound.
+
 The following results were obtained:
 
 $$
-  p^{\text{oss}}_{\text{max}} = 10.005 \pm 0.018 \with \chi^2 = 0.071
+  P^{\text{oss}}_{\text{max}} = 10.005 \pm 0.018 \with \chi_r^2 = 0.071
 $$
 
-In order to compare $p^{\text{oss}}_{\text{max}}$ with the expected value
+where $\chi_r^2$ is the reduced $\chi^2$, proving a good convergence.  
-$p_{\text{max}} = 10$, the following compatibility t-test was applied:
+In order to compare $P^{\text{oss}}_{\text{max}}$ with the expected value
+$P_{\text{max}} = 10$, the following compatibility t-test was applied:
 
 $$
   p = 1 - \text{erf}\left(\frac{t}{\sqrt{2}}\right)\ \with
-  t = \frac{|p^{\text{oss}}_{\text{max}} - p_{\text{max}}|}
+  t = \frac{|P^{\text{oss}}_{\text{max}} - P_{\text{max}}|}
-           {\Delta p_{\text{max}}}
+           {\Delta P^{\text{oss}}_{\text{max}}}
 $$
 
-where $\Delta p_{\text{max}}$ is the absolute error of $p_{\text{max}}$. At 95%
+where $\Delta P^{\text{oss}}_{\text{max}}$ is the $P^{\text{oss}}_{\text{max}}$
-confidence level, the values are compatible if $p > 0.05$.
+uncertainty. At 95% confidence level, the values are compatible if $p > 0.05$.
 In this case:
 
   - t = 0.295
@@ -254,4 +230,5 @@ which allows to assert that the sampled points actually follow the predicted
 distribution. In @fig:fit, the fit function superimposed on the histogram is
 shown.
 
-![Fitted data.](images/fit.pdf){#fig:fit}
+![Fitted sampled data. $P^{\text{oss}}_{\text{max}} = 10.005
+  \pm 0.018$, $\chi_r^2 = 0.071$.](images/fit.pdf){#fig:fit}