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Generating $ \tilde{s}$.

Let us define $ V:=\omega$, $ D:=H \omega$, $ r:=\beta / \alpha$, equation 5.8, can now be rewritten:
$\displaystyle \frac{(V + r D)^T H ( V + r D)}{(V+rD)^2}$ $\displaystyle =$ $\displaystyle \frac{V^T H V + r
D + r D^T H V + r^2 D^T H D}{ V^2 + 2 r V^T D + r^2 D^2}$  
  $\displaystyle =$ $\displaystyle \frac{r^2 D^T H D + 2 r V^T H D+ V^T H V}{ V^2 + 2 r V^T D+
r^2 D^2} = f(r)$  

We will now search for $ r^*$, root of the Equation

  $\displaystyle \frac{
 \partial f(r^*)}{\partial r} =0 \nonumber$    
$\displaystyle \Leftrightarrow$ $\displaystyle \quad ( 2 r D^T H D + 2 V^T H D) (V^2 + 2 r V^T
 D + r^2 D^2)$    
  $\displaystyle \quad - (r^2 D^T H D + 2 r V^T H D + V^T H V)(2 r D^2 + 2 V^T D)
$\displaystyle \Leftrightarrow$ $\displaystyle \quad \bigg[ (D^T H D) (V^T D) - D^2 D^2 \bigg]
 r^2 + \bigg[ (D^T H D) V^2 - D^2 (V^T D) \bigg] r + \bigg[ D^2 V^2
 - (V^T D)^2 \bigg]=0$    

(Using the fact that $ D=HV \Leftrightarrow D^T=V^T H^T=V^T H$.)
We thus obtain a simple equation $ a x^2+b x+c=0$. We find the two roots of this equation and choose the one $ r^*$ which maximize 5.8. $ \tilde{s}$ is thus $ V+ r^* D$.

Frank Vanden Berghen 2004-04-19