midterm: done all except 2.1{a,d,f}

Claudio_Maggioni_midterm/Claudio_Maggioni_midterm.md

@@ -376,7 +376,7 @@ We first show that the lemma holds for $\tau \in [0,1]$. Since
 
 $$\|\tilde{p}(\tau)\| = \|\tau p^U\| = \tau \|p^U\| \text{ for } \tau \in [0,1]$$
 
-Then the norm of the step $\tilde{p}$ clearly increases monotonically as $\tau$
+Then the norm of the step $\tilde{p}$ clearly increases as $\tau$
 increases. For the second criterion, we compute the quadratic model for a
 generic $\tau \in [0,1]$:
 
@@ -388,17 +388,18 @@ term in function
 of $\tau^2$ is negative and thus the model for an increasing $\tau \in [0,1]$
 decreases monotonically (to be precise quadratically).
 
-Now we show that the monotonicity claims hold also for $\tau \in [1,2]$. We
+Now we show that the two claims on gradients hold also for $\tau \in [1,2]$. We
-define a function $h(\alpha)$ (where $\alpha = \tau - 1$) with same monotonicity
+define a function $h(\alpha)$ (where $\alpha = \tau - 1$) with same gradient
-as $\|\tilde{p}(\tau)\|$ and we show that this function monotonically increases:
+"sign"
+as $\|\tilde{p}(\tau)\|$ and we show that this function increases:
 
 $$h(\alpha) = \frac12 \|\tilde{p}(1 - \alpha)\|^2 = \frac12 \|p^U + \alpha(p^B -
 p^U)\|^2 = \frac12 \|p^U\|^2 + \frac12 \alpha^2 \|p^B - p^U\|^2 + \alpha (p^U)^T
 (p^B - p^U)$$
 
 We now take the derivative of $h(\alpha)$ and we show it is always positive,
-i.e. that $h(\alpha)$ has always positive gradient and thus that is it
+i.e. that $h(\alpha)$ has always positive gradient and thus that it is
-monotonically increasing w.r.t. $\alpha$:
+increasing w.r.t. $\alpha$:
 
 $$h'(\alpha) = \alpha \|p^B - p^U\|^2 + (p^U)^T (p^B - p^U) \geq (p^U)^T (p^B - p^U)
 = \frac{g^Tg}{g^TBg}g^T\left(- \frac{g^Tg}{g^TBg}g + B^{-1}g\right) =$$$$= \|g\|^2
@@ -445,7 +446,50 @@ Which, since $B$ is symmetric, in turn is equivalent to writing:
 
 $$g^T g \leq (g^TBg) (g^T B^{-1} g)$$
 
-which is what we needed to show to prove that the first monotonicity constraint
+which is what we needed to show to prove that the first gradient constraint
 holds for $\tau \in [1,2]$.
 
-**TBD**
+For the second constraint, we adopt a similar strategy as for before and we
+define a new function $\hat{h}(\alpha) = m(\tilde{p}(1 + \alpha))$, thus
+plugging the Dogleg step in the quadratic model:
+
+$$\hat{h}(\alpha) = m(\tilde{p}(1+\alpha)) = f + g^T (p^U + \alpha (p^B - p^U)) +
+\frac12 (p^U + \alpha (p^B - p^U))^T B (p^U + \alpha (p^B - p^U)) = $$$$ =
+f + g^T p^U + \alpha g^T (p^B - p^U) + (p^U)^T B p^U + \frac12 \alpha (p^U)^T B
+(p^B - p^U) + \frac12 \alpha (p^B - p^U)^T B p^U + \frac12 \alpha^2
+(p^B - p^U)^T B (p^B - p^U)$$
+
+We now derive $\hat{h}(\alpha)$:
+
+$$\hat{h}'(\alpha) = g^T (p^B - p^U) + \frac12 (p^U)^T B (p^B - p^U) + \frac12
+(p^B - p^U)^T B p^U + \alpha (p^B - p^U)^T B (p^B - p^U) = $$$$
+= (p^B - p^U)^T g + \frac12 ((p^U)^T B (p^B - p^U))^T +
+\frac12 (p^B - p^U)^T B p^U + \alpha (p^B - p^U)^T B (p^B - p^U) = $$$$
+= (p^B - p^U)^T g + \frac12 (p^B - p^U) B^T (p^U)^T +
+\frac12 (p^B - p^U)^T B p^U + \alpha (p^B - p^U)^T B (p^B - p^U) = $$$$
+= (p^B - p^U)^T (g + \frac12 \cdot 2 \cdot B p^U) + \alpha (p^B - p^U)^T B
+(p^B - p^U) \leq $$$$
+\leq (p^B - p^U)^T(g + B p^U + B (p^B - p^U)) = $$$$
+=(p^B - p^U)^T(g+Bp^B) = 0$$
+
+and we therefore obtain $\hat{h}(\alpha) \leq 0$, thus finding that the
+$m(\tilde{p})$ is indeed a decreasing function of $\tau$ or $\alpha = \tau - 1$
+also for $\tau \in [1,2]$, thus completing the proof for the lemma.
+
+<!--
+$$\hat{h}'(\alpha) = g^T (p^B - p^U) + \frac12 (p^U)^T B (p^B - p^U) + \frac12
+(p^B - p^U)^T B p^U + \alpha (p^B - p^U)^T B (p^B - p^U) = $$$$
+
+= (p^B - p^U)^T g + \frac12 ((p^U)^T B (p^B - p^U))^T +
+\frac12 (p^B - p^U)^T B p^U + \alpha (p^B - p^U)^T B (p^B - p^U) = $$$$
+
+= (p^B - p^U)^T g + \frac12 (p^B - p^U) B^T (p^U)^T +
+\frac12 (p^B - p^U)^T B p^U + \alpha (p^B - p^U)^T B (p^B - p^U) = $$$$
+
+= (p^B - p^U)^T (g + \frac12 \cdot 2 \cdot B p^U) + \alpha (p^B - p^U)^T B
+(p^B - p^U) \leq $$$$
+
+\leq (p^B - p^U)^T(g + B p^U + B (p^B - p^U)) = $$$$
+
+=(p^B - p^U)T(g+Bp^B) = 0$$
+-->

Claudio_Maggioni_midterm/dogleg.m

@@ -1,3 +1,5 @@
+% Discussed with: Gianmarco De Vita (MATLAB solver for determining \tau)
+
 function pk = dogleg(B, g, deltak) 
     pnewton = - (B \ g);