midterm: slight correction

Claudio_Maggioni_midterm/Claudio_Maggioni_midterm.md

@@ -11,19 +11,22 @@ header-includes:
 - \usepackage[ruled,vlined]{algorithm2e}
 - \usepackage{float}
 - \floatplacement{figure}{H}
+- \hypersetup{colorlinks=true,linkcolor=blue}
 
 ---
 \maketitle
 
 # Acknowledgements on group work
 
-- Gianmarco De Vita suggested me the use of MATLAB's equation solver for parts
+- **Gianmarco De Vita** suggested me the use of MATLAB's equation solver for parts
   of `dogleg.m`'s implementation.
 - I have discussed my solutions for exercise 1.2 and exercise 3 with several
   people, namely:
-  - Gianmarco De Vita
-  - Tommaso Rodolfo Masera
-  - Andrea Brites Marto
+  - **Gianmarco De Vita**
+  - **Tommaso Rodolfo Masera**
+  - **Andrea Brites Marto**
+- [This song](https://www.youtube.com/watch?v=gOPqXG0f7rE) for the divine
+  inspiration that made me write the proof for Exercise 1.2.
 
 # Exercise 1
 
@@ -199,7 +202,7 @@ to be a global minimizer.
 
 ### (b) Write down the quadratic model around a current iterate xk and explain the meaning of each term.
 
-$$m(p) = f + g^T p + \frac12 p^T B p \;\; \text{ s.t. } \|p\| < \Delta$$
+$$m(p) = f + g^T p + \frac12 p^T B p \;\; \text{ s.t. } \|p\| \leq \Delta$$
 
 Here's an explaination of the meaning of each term:
 
@@ -557,7 +560,9 @@ $$\hat{h}'(\alpha) = g^T (p^B - p^U) + \frac12 (p^U)^T B (p^B - p^U) + \frac12
 = (p^B - p^U)^T (g + \frac12 \cdot 2 \cdot B p^U) + \alpha (p^B - p^U)^T B
 (p^B - p^U) \leq $$$$
 \leq (p^B - p^U)^T(g + B p^U + B (p^B - p^U)) = $$$$
-=(p^B - p^U)^T(g+Bp^B) = 0$$
+=(p^B - p^U)^T(g+Bp^B) = $$$$
+=(p^B - p^U)^T(g+B \cdot (-1) \cdot B^{-1} g) = $$$$
+=(p^B - p^U)^T(g - g) = 0$$
 
 and we therefore obtain $\hat{h}(\alpha) \leq 0$, thus finding that the
 $m(\tilde{p})$ is indeed a decreasing function of $\tau$ or $\alpha = \tau - 1$