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Differential Equations Solutions 38

# Differential Equations Solutions 38 - p T p = h 2 CHALLENGE...

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48 Chapter 9. Solutions: Numerical Methods for Unconstrained Optimization CHALLENGE 9.3. x = [1;2]; for i=1:5, g = [4*(x(1) - 5)^3 - x(2); 4*(x(2) + 1)^3 - x(1)]; H = [12*(x(1) - 5)^2, -1; -1, 12*(x(2) + 1)^2]; p = -H \ g; x = x + p; end CHALLENGE 9.4. The Lagrangian function is L ( p , λ ) = f ( x ) + p T g + 1 2 p T Hp + λ 2 ( p T p h 2 ) . Setting the partial derivatives to zero yields g + Hp + λ p = 0 , 1 2 ( p T p h 2 ) = 0 . Thus the first equation is equivalent to ( H + λ I ) p = g , or Hp = g , as in the Levenberg-Marquardt algorithm. The parameter λ is chosen so that
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Unformatted text preview: p T p = h 2 . CHALLENGE 9.5. If f is quadratic, then H ( k ) s ( k ) = g ( x ( k +1) ) − g ( x ( k ) ) , where H is the Hessian matrix of f . Close to x ( k +1) , a quadratic model is a close Ft to any function, so we demand this property to hold for our approximation to H . CHALLENGE 9.6. The formula for Broyden’s good method is B ( k +1) = B ( k ) − ( B ( k ) s ( k ) − y ( k ) ) s ( k ) T s ( k ) T s ( k ) ....
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