midterm: slight correction

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Claudio Maggioni 2021-05-14 14:55:38 +02:00
parent 27da90d89b
commit 3a8c08701e
2 changed files with 11 additions and 6 deletions

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@ -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 $\alpha = \tau - 1$