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Generating $ \hat{u}$ and $ \tilde{u} $ from $ \hat{s}$ and $ \tilde{s}$

Having generated $ \tilde{s}$ and $ \hat{s}$ in the ways that have been described, the algorithm sets $ s$ to a linear combination of these vectors, but the choice is not restricted to $ \pm \hat{s}$ or $ \pm \tilde{s}$ as suggested in the introduction of this chapter(unless $ \tilde{s}$ and $ \hat{s}$ are nearly or exactly parallel). Instead, the vectors $ \hat{u}$ and $ \tilde{u} $ of unit length are found in the span of $ \tilde{s}$ and $ \hat{s}$ that satisfy the condition $ \hat{u}^T \tilde{u} = 0$ and $ \hat{u}^T H
\tilde{u} = 0$. The final $ s$ will be a combination of $ \hat{u}$ and $ \tilde{u} $.
If we set:

$\displaystyle \begin{cases}G = \tilde{s} & \\  V = \hat{s} &
\end{cases} $

We have

\begin{displaymath}\begin{cases}\tilde{u}= \cos(\theta) G+ \sin(\theta)
 V \\  \...
...es}\quad \quad \text{ we have directly }\hat{u}^T
 \tilde{u}= 0\end{displaymath} (5.8)

We will now find $ \theta$ such that $ \hat{u}^T H
\tilde{u} = 0$:

  $\displaystyle \hat{u}^T H \tilde{u} = 0$    
$\displaystyle \Leftrightarrow$ $\displaystyle (-\sin(\theta) G+ \cos(\theta) V)^T H
 (\cos(\theta) G+ \sin(\theta) V) =0$    
$\displaystyle \Leftrightarrow$ $\displaystyle (\cos^2(\theta)-\sin^2(\theta)) V^T H G + (G^T H
 G - V^T H V) \sin(\theta) \cos(\theta) =0$    

Using

$\displaystyle \begin{cases}\sin(2 \theta) = 2\sin(\theta)
\cos(\theta) \\  \co...
...in^2(\theta) \\
tg (\theta) = \frac{\sin(\theta)}{\cos(\theta)} \end{cases} $

We obtain

  $\displaystyle (V^T H G) \cos(2 \theta) + \frac{G^T H G-
 V^T H V}{2} \sin(2 \theta)=0$    
$\displaystyle \Leftrightarrow$ $\displaystyle \theta
 = \frac{1}{2} arctg (\frac{2 V^T H G}{V^T H V - G^T H G})$ (5.9)

Using the value of $ \theta$ from Equation 5.10 in Equation 5.9 give the required $ \hat{u}$ and $ \tilde{u} $.
next up previous contents
Next: Generating the final from Up: The secondary Trust-Region subproblem Previous: Generating .   Contents
Frank Vanden Berghen 2004-04-19