Differential Equations Solutions 39

Differential Equations Solutions 39 - 49 To verify the...

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Unformatted text preview: 49 To verify the secant condition, compute B(k+1) s(k) = B(k) s(k) − (B(k) s(k) − y(k) ) s(k) T s(k) s(k) T s(k) = B(k) s(k) − (B(k) s(k) − y(k) ) = y(k) , as desired. If vT s(k) = 0, then B(k+1) v = B(k) v − (B(k) s(k) − y(k) ) s(k) T v s(k) T s(k) = B(k) v, as desired. CHALLENGE 9.7. B(k+1) s(k) = B(k) s(k) − = B(k) s(k) − B(k) s(k) s(k) T B(k) s(k) + y(k) y(k) T (k) s y(k) T s(k) + y(k) y(k) T s(k) y(k) T s(k) s(k) T B(k) s(k) B(k) s(k) Bss (k ) (k ) (k ) T s(k) T B(k) s(k) s(k) T B(k) s(k) y(k) T s(k) = B(k) s(k) − B(k) s(k) + y(k) (k) T (k) (k ) (k ) y s s(k) T B s (k ) (k ) (k ) (k ) (k ) =B s −B s +y = y(k) . CHALLENGE 9.8. Dropping superscripts for brevity, and taking advantage of symmetry of H, we obtain 1 (x + αp)T H(x + αp) − (x + αp)T b 2 1 1 = xT Hx − xT b + αpT Hx + α2 pT Hp − αpT b. 2 2 Differentiating with respect to α we obtain f (x(0) + αp(0) ) = pT Hx + αpT Hp − pT b = 0, so α= pT b − pT Hx pT r =T , T Hp p p Hp where r = b − Hx. If we differentiate a second time, we find that the second derivative of f with respect to α is pT Hp > 0 (when p = 0), so we have found a minimizer. ...
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