From 5026da324a478098b7b5e1566d5ebe07cc69b5c4 Mon Sep 17 00:00:00 2001 From: Claudio Maggioni <maggicl@usi.ch> Date: Mon, 10 May 2021 17:30:38 +0200 Subject: [PATCH] midterm: done all except 2.1{a,d,f} --- .../Claudio_Maggioni_midterm.md | 60 +++++++++++++++--- .../Claudio_Maggioni_midterm.pdf | Bin 618131 -> 619869 bytes Claudio_Maggioni_midterm/dogleg.m | 2 + 3 files changed, 54 insertions(+), 8 deletions(-) diff --git a/Claudio_Maggioni_midterm/Claudio_Maggioni_midterm.md b/Claudio_Maggioni_midterm/Claudio_Maggioni_midterm.md index d8fb9e8..62eb807 100644 --- a/Claudio_Maggioni_midterm/Claudio_Maggioni_midterm.md +++ b/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 -define a function $h(\alpha)$ (where $\alpha = \tau - 1$) with same monotonicity -as $\|\tilde{p}(\tau)\|$ and we show that this function monotonically increases: +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 gradient +"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 -monotonically increasing w.r.t. $\alpha$: +i.e. that $h(\alpha)$ has always positive gradient and thus that it is +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 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a/Claudio_Maggioni_midterm/dogleg.m +++ b/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);