From 365b2397016f0598fe9206970ed375fb9dcab2d4 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?Gi=C3=B9=20Marcer?= <giuliamarcer@yahoo.it>
Date: Mon, 27 Apr 2020 23:53:23 +0200
Subject: [PATCH] =?UTF-8?q?ex-4:=20=CF=87=C2=B2=20minimization=20results?=
 =?UTF-8?q?=20reported?=
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---
 ex-4/main.c         |  7 +++++--
 notes/sections/4.md | 30 ++++++++++++++++++++++++------
 2 files changed, 29 insertions(+), 8 deletions(-)

diff --git a/ex-4/main.c b/ex-4/main.c
index 8dda0ec..03ecd2b 100644
--- a/ex-4/main.c
+++ b/ex-4/main.c
@@ -160,7 +160,10 @@ int main(int argc, char **argv)
     }
 
   double result = x;
-  printf("p_max: %.7f\n", result);
+  double res_chi = chi2(result, &params);
+  printf("Results:\n");
+  printf("χ² = %.3f\n", res_chi);
+  printf("p_max = %.3f\n", result);
 
   // Compute the second derivative of χ² in its minimum for the result error.
   //   
@@ -189,7 +192,7 @@ int main(int argc, char **argv)
     error = error + A + B;
     };
     error = 1/error;
-  printf("ΔP_max: %.7f\n\n", error);
+  printf("ΔP_max = %.3f\n\n", error);
 
 
   // Free memory.
diff --git a/notes/sections/4.md b/notes/sections/4.md
index 60adde4..55a07db 100644
--- a/notes/sections/4.md
+++ b/notes/sections/4.md
@@ -151,13 +151,16 @@ $$
 Finally, putting all the pieces together, the average value of $|P_v|$ can now
 be computed:
 
-$$
-  \langle |P_v| \rangle = \int
+\begin{align*}
+  \langle |P_v| \rangle &= \int
   \limits_{- \sqrt{P_{\text{max}}^2 - P_h}}^{\sqrt{P_{\text{max}}^2 - P_h}}
-  f (P_v | P_h = x) = [ \dots ] =
-  x \, \frac{\ln \left( \frac{P_{\text{max}}}{x} \right)}
-  {\text{atan} \left( \sqrt{ \frac{P_{\text{max}}}{x^2} - 1} \right)}
-$$
+  f (P_v | P_h = x)
+  \\
+  &= [ \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*}
 
 Namely:
 
@@ -218,3 +221,18 @@ $$
 The following result was obtained:
 
 ![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
+the inverese of the $\chi^3$ second derivative in its minimum, according to the
+Cramér-Rao bound.  
+The following results were obtained:
+
+$$
+  p_{\text{max}} = 10 \pm 0.016 \with \chi^2 = 0.072
+$$
+
+which allows to assert that the sampled points actually follow the predicted
+distribution.